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HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Extension 2 => Topic started by: jakesilove on January 28, 2016, 07:04:05 pm

Title: 4U Maths Question Thread
Post by: jakesilove on January 28, 2016, 07:04:05 pm
HSC 4U MATHS Q&A THREAD

To go straight to posts from 2018, click here.

What is this thread for?
If you have general questions about the HSC 4U Maths course or how to improve in certain areas, this is the place to ask! 👌


Who can/will answer questions?
Everyone is welcome to contribute; even if you're unsure of yourself, providing different perspectives is incredibly valuable.

Please don't be dissuaded by the fact that you haven't finished Year 12, or didn't score as highly as others, or your advice contradicts something else you've seen on this thread, or whatever; none of this disqualifies you from helping others. And if you're worried you do have some sort of misconception, put it out there and someone else can clarify and modify your understanding! 

There'll be a whole bunch of other high-scoring students with their own wealths of wisdom to share with you. So you may even get multiple answers from different people offering their insights - very cool.


To ask a question or make a post, you will first need an ATAR Notes account. You probably already have one, but if you don't, it takes about four seconds to sign up - and completely free!

OTHER 4U MATHS RESOURCES

Original post.
Before you can ask a question, you'll have to make an ATAR Notes account here. Once you've done that, a little 'reply' button will come up when you're viewing threads, and you'll be able to post whatever you want! :)

Hey everyone!

A lot of you will have met me at the HSC Head Start lectures, where I lectured in 2U and 3U Maths, Physics and Chemistry.
My role on these forums is to help you. The HSC syllabus is tricky, nuanced and pretty damn huge. To help you out, I thought it would be a great idea to have a forum where you can just post questions, and myself or other forum members can post answers!

This is a community, so we want you to feel like you can post any type of 4U Mathematics question, no matter how "basic" you might think it is. Remember, IF YOU'RE HAVING TROUBLE WITH A TOPIC, THERE ARE THOUSANDS OF OTHERS HAVING THE SAME ISSUE. The best way to learn Maths is by looking through practice questions, and their associated answers. I honestly think a forum like this, and a place where I could always go to have difficult questions answered would have helped me in my HSC year.

Remember that Extension II Maths is a bloody difficult course. There will be lots of answers to the same questions, and I'll try give you the best or easiest to remember ones.

I got an ATAR of 99.80, and a mark of 93 in the 4U Mathematics course. There are similar forums for a bunch of other subjects, so make sure to take a look at them as well!
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: Happy Physics Land on February 05, 2016, 08:24:25 pm
Hey Jake:

Hope you are impressed by this first ever 4 unit question of the year. I have little idea how to sketch this graph would you mind giving me a hand on this question? Thank you fam! :)

Essentially the question is asking me to sketch y = ln (1-|x|)
Thank you so much Jake!

(http://i.imgur.com/bxnzaML.jpg)

Best Regards
Happy Physics Land
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: jakesilove on February 05, 2016, 09:04:59 pm
Hey Jake:

Hope you are impressed by this first ever 4 unit question of the year. I have little idea how to sketch this graph would you mind giving me a hand on this question? Thank you fam! :)

Essentially the question is asking me to sketch y = ln (1-|x|)
Thank you so much Jake!

(http://i.imgur.com/bxnzaML.jpg)

Best Regards
Happy Physics Land

Hey HPL!

Glad I got a question from the Extension Two course! I really enjoyed this course, but seriously struggled with it, so would encourage anyone who needs help to post up some difficult questions (like the one you have asked!)

I'm sorry that the graphing is sort of dodgy, the software that I usually used has magically disappeared but I'll try find it again soon! Hope the answer helps!

(http://i.imgur.com/Rf6NaF1.png?1)
(http://i.imgur.com/jGEatyN.png?1)

I'm thinking of making a "Graph Sketching" Extension Two resource, but that's probably still a while off.

Jake :)
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: Happy Physics Land on February 05, 2016, 09:25:12 pm
Hey HPL!

Glad I got a question from the Extension Two course! I really enjoyed this course, but seriously struggled with it, so would encourage anyone who needs help to post up some difficult questions (like the one you have asked!)

I'm sorry that the graphing is sort of dodgy, the software that I usually used has magically disappeared but I'll try find it again soon! Hope the answer helps!

(http://i.imgur.com/Rf6NaF1.png?1)
(http://i.imgur.com/jGEatyN.png?1)

I'm thinking of making a "Graph Sketching" Extension Two resource, but that's probably still a while off.

Jake :)

Hey Jake!

Thank you so much for your help I'm really grateful towards this very elaborative answer. It took me a while to kind of understand the range part and I think it was a really good idea to use domain and range and to check whether the function is even or odd. My initial approach was kind of like trying to start off with the parent function of y = lnx and just translate it and reflect it about the x axis and then take the absolute value of the graph. Well my method didnt quite work out. But yeah I was just ignorant of the first principles of logarithmic functions and this is definitely an effective approach I can adopt in the future. Thank you very much Jake for always helping me in times of need! :D

Best Regards
Happy Physics Land
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: jakesilove on February 05, 2016, 10:45:53 pm
Hey Jake!

Thank you so much for your help I'm really grateful towards this very elaborative answer. It took me a while to kind of understand the range part and I think it was a really good idea to use domain and range and to check whether the function is even or odd. My initial approach was kind of like trying to start off with the parent function of y = lnx and just translate it and reflect it about the x axis and then take the absolute value of the graph. Well my method didnt quite work out. But yeah I was just ignorant of the first principles of logarithmic functions and this is definitely an effective approach I can adopt in the future. Thank you very much Jake for always helping me in times of need! :D

Best Regards
Happy Physics Land

It was a really difficult question mate, so definitely don't kick yourself for not getting it!

My biggest recommendation, for a graph that you don't have any idea how to sketch, is always to PLOT INTELLIGENT POINTS! Here,you could have just tried x=1,x=2, x=3... x=-1, x=-2, x=-3.... etc. and eventually got a good idea of what the graph should be. Whilst this probably won't get you full marks, at least it will get you something!

Still, having a good approach is always better. Start with Domain/Range, think about Even/Odd, and examine important locations (approaching asymptotes etc.).

Would love you to post any other questions you're struggling with! Love you long time HPL:)

Jake
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: Happy Physics Land on February 05, 2016, 10:52:43 pm
It was a really difficult question mate, so definitely don't kick yourself for not getting it!

My biggest recommendation, for a graph that you don't have any idea how to sketch, is always to PLOT INTELLIGENT POINTS! Here,you could have just tried x=1,x=2, x=3... x=-1, x=-2, x=-3.... etc. and eventually got a good idea of what the graph should be. Whilst this probably won't get you full marks, at least it will get you something!

Still, having a good approach is always better. Start with Domain/Range, think about Even/Odd, and examine important locations (approaching asymptotes etc.).

Would love you to post any other questions you're struggling with! Love you long time HPL:)

Jake

Be prepared for me to flood you with 4u questions the days before half yearly exams Jake ;) But thank you so much for your tips Jake I really appreciate it!
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: Neutron on February 06, 2016, 10:35:18 pm
Hey Jake!

I'm currently doing 3 sciences, 4u maths and 2u english and was planning on dropping to 10 units (I would drop biology) however, I'm slightly hesitant as this would mean every single one of my units would count. Right now, I'm behind on biology which is giving me the incentive to drop it so I wont have to invest time catching up. What are your thoughts? Since you did 12 units? Did you feel it was more beneficial? And also, what would you say is the best way to approach 4U maths and do you think it's doable for everyone? I get stumped on quite a few questions and I feel like I'm more a rote learner than an innovative thinker if that makes sense.. Also, what are your thoughts on dot point summaries for Physics and Chemistry? Necessary or nah? My school puts a lot of emphasis on it but I find that it's extremely time consuming and I don't exactly trust my own notes anyway, but I'm scared I'm disadvantaging myself by not writing them! Thanks Jake!
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: Neutron on February 06, 2016, 10:41:59 pm
And also, I didn't do amazingly well in the first term, got around ranked 20th for my subjects.. What do I do? I really want to improve my ranks but have I already stuffed myself up? :(
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: Happy Physics Land on February 06, 2016, 11:44:33 pm
Hey Jake!

I'm currently doing 3 sciences, 4u maths and 2u english and was planning on dropping to 10 units (I would drop biology) however, I'm slightly hesitant as this would mean every single one of my units would count. Right now, I'm behind on biology which is giving me the incentive to drop it so I wont have to invest time catching up. What are your thoughts? Since you did 12 units? Did you feel it was more beneficial? And also, what would you say is the best way to approach 4U maths and do you think it's doable for everyone? I get stumped on quite a few questions and I feel like I'm more a rote learner than an innovative thinker if that makes sense.. Also, what are your thoughts on dot point summaries for Physics and Chemistry? Necessary or nah? My school puts a lot of emphasis on it but I find that it's extremely time consuming and I don't exactly trust my own notes anyway, but I'm scared I'm disadvantaging myself by not writing them! Thanks Jake!

Hey Neutron:

Im Jake (For sure!) Im also doing physics, chem, ext 1 and ext 2 mathematics, same as you. And to be honest, I stuffed up three assessments already in the first term so Im in the same shoe as you. But what I tell myself is that the assessment is only gonna to contribute to 5% of your RAW atar for THIS ONE SUBJECT. Perhaps you and the person that came first were 20% apart from one another, if you multiply that difference by 0.05, then really you are only losing to him by 1 atar in this subject and if you consider again this 1% might go into 0.5% after scaling and then if you consider how you have 6 subjects all together you pretty much is only 0.08% apart from that guy.... So this is what l tell myself (and it is actually true!) and this is what motivates me to try even harder for my half yearlies which would weigh 5% more.

But thats all not-so-important. I would just like to share some of my extension 2 experience with you so that I can perhaps help you out in some ways. I would list them in dot points form so that it would be easier for you to understand.

1. Do LOTS AND LOTS AND LOTS of exercises on each topic, especially for complex numbers. The outrageously strange questions they can give you in complex numbers is just frustrating, so I would have done as much past HSC questions on complex numbers as possible, you can easily find ext 2 trial papers online and do questions from there. If you just can just manage to do 10 questions everyday, imagine how prepared you will be after 239 days (THATS 2390 QUESTIONS DONE!) when HSC comes.

2. Everyone pretty much knows this first tip. But what Im going to tell you now is something people might not do: FOR ALL THE QUESTIONS THAT YOU DO NOT KNOW HOW TO SOLVE, WRITE THE ENTIRE SOLUTION DOWN AND MEMORIZE HOW TO SOLVE IT. Extension 2 is a weird sort of maths, to my perpsective. Back in the 2unit days, I would be able to solve some questions on the spot without having previously encountered it before. But when I see an unfamiliar extension 2 question, it takes me almost double or even triple the time to solve it. So through memorising questions of diverse styles, there is a higher chance you will be familiarised with the questions in your exam

3. MADE A MISTAKE? JOT IT DOWN! Collect all the mistakes you have made during both your practises and your exams, write a specific question that you have made a mistake on if you want to. Look through it at least once a week. For example, in my notes I have "cos(x) - isin(x) doesnt equal to cis(x)!" and " Simplify cis(x) answers wherever necessary"

4. The first three points are just generally how you would study for extension 2. Now let's talk about some skills for complex numbers. One of the most important thing to remember is SIN (X) is an ODD FUNCTION and COS(X) is an EVEN FUNCTION. This means that sin(-x) = -sinx and cos(-x) = cos (x). This will be very useful in complex number proofs involving de'moivre's theorem

5. Another key thing to do is to MEMORISE ALL THE COMPLEX NUMBER PROPERTIES.These are your best friends in proofs,  For example, |z| = |conjugate of z| = sqrt (x^2 + y^2), arg(conjugate of z) = -arg(z), z(conjugate of z) = |z|^2 = |conjugate of z|^2 = x^2 + y^2, z + conjugate of z = 2x, z - conjugate of z = 2iy. These are only a few of the complex properties, you can find the rest either online or in your textbooks.

6. REMEMBER THE SHAPE OF YOUR LOCUS.  This is so important. Of course, you can always work out your locus through algebraic means, but by remembering the general shape of your complex locus, you would not be required to spend as much time. Just to list a few:
- |z| = r is a circle
- |z-w| = r is the equation of a circle with the centre at w=a+ib and radius r.
- |z - z1| = |z - z2| is the equation of the perpendicular bisector of the line AB
- arg (z-z1) = x, where x is a constant, is the equation of a half-ray, starting at the fixed point z1
- arg(z-z1) - arg(z-z2) = x, 0<x<pi, is the equation of an arc of a circle on the chord AB, where A and B represent z1 and z2 respectively


7. MULTIPLYING A COMPLEX NUMBER THROUGH BY CIS(X) TURNS THE GRAPH ANTI-CLOCKWISE BY X DEGREES, and of course, multiplying a complex number through by i turns the graph by 90 degrees anti-clockwise

8. These are just some of the core tips of complex numbers, I can go into the specifics if you are willing to supply some questions? Now let's have a look at some essential tips for the topics of Graphs.
- -f(x) = reflection of graph over the x-axis
- f(-x) = reflection of graph over the y-axis
- f(x) + t = moving the graph up by t units in the y-direction
- f(x) - t = moving the graph down by t units in the y-direction
- f(x+t) = moving the graph t units to the left along x-direction
- f(x-t) = moving the graph t units to the right along x-direction
- |f(x)| = reflect everything below the y-axis over to above the y-axis
- f(|x|) = delete left hand side, reflect everything on the right over y-axis
-|y| = f(x) = delete everything on the bottom, reflect everything above x-axis over to the bottom of x-axis

And of course, if you want to sketch 1/f(x), sqrt (f(x)), [f(x)]^2 and [f(x)]^3, always draw the lines y = 1 and y = -1 and locate where f(x) = 0

In regards to the dot point summaries I would definitely recommend to do them and begin revising them 1 month before your exam and constantly revise them over time. This is something l do and I would strongly recommend you to make revision notes that would suit your own style.

Anyways I hope my tips would help you in extension 2 and best of luck!

Best Regards
Happy Physics Land
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: jakesilove on February 07, 2016, 12:19:23 pm
Hey Jake!

I'm currently doing 3 sciences, 4u maths and 2u english and was planning on dropping to 10 units (I would drop biology) however, I'm slightly hesitant as this would mean every single one of my units would count. Right now, I'm behind on biology which is giving me the incentive to drop it so I wont have to invest time catching up. What are your thoughts? Since you did 12 units? Did you feel it was more beneficial? And also, what would you say is the best way to approach 4U maths and do you think it's doable for everyone? I get stumped on quite a few questions and I feel like I'm more a rote learner than an innovative thinker if that makes sense.. Also, what are your thoughts on dot point summaries for Physics and Chemistry? Necessary or nah? My school puts a lot of emphasis on it but I find that it's extremely time consuming and I don't exactly trust my own notes anyway, but I'm scared I'm disadvantaging myself by not writing them! Thanks Jake!

Hey Neutron!

I'll address your points one at a time.

Doing 10 units as opposed to 12

I feel like, if you do think you're quite behind on Biology, and you don't feel very confident with it, dropping it isn't the worst idea. That being said, if you feel like you can catch up (despite having to do a lot of work to get there) I would definitely recommend that. Having 12 units, rather than 10, is almost always beneficial. Between scaling, difficulty of the actual HSC etc. having a fallback is seriously a good idea. You never know what will happen throughout the year, and having only 10 units is definitely risky. However if you're 100% sure you're going to do worse in Biology that ALL of your other 10 units, dropping it may make sense.

I will say, however, that there is a direct correlation between high number of units and high marks. Usually, the highest mark students have between 12-13 units under their belt.

Approaching Extension 2 Maths

Firstly, massive thank you to HPL for writing out those tips. Seriously helpful stuff, so take note! If you've gotten through Maths so far, I think that it is definitely doable for you. Whilst it is definitely substantially more difficult than the other maths courses, once you've done practice papers enough time you can answer a paper in a "rote learner" sense rather than innovative. Really, by the HSC you'll just be spouting off answers. As to getting stumped on some questions.... If you weren't stumped of Extension 2 Maths questions, you would be some kind of savant. It is an insanely difficult course: Getting a raw mark of 60 is seriously good. I think that if you are still going, and you just spend lots of time practicing practicing practicing (taking into account HPL's tips) you will definitely get to a comfortable level of knowledge.

I never FULLY understood the course in year 12, so don't expect to. You should just be able to do SOME maths and see what happens, even if you can't get an answer out.

Dotpoint notes

I think the best way to make notes is FROM ANOTHER STUDENT'S NOTES. What I mean by that is: buy a set of notes, find a set of notes, and TAKE NOTES FROM THAT in a more succinct way. I do think it is really important to make your own notes. It just means that you have to think about your notes a little bit more, go through the content a few more times. I would really recommend it: even if it is time consuming it is really really worth it.

General "HOW DO I COPE"

It doesn't matter if your rank is not 1st or 2nd at this stage. You have so much time, never mind half-yearlies and trials, to pick up your marks. That being said, you could come 20th overall and still get a fantastic ATAR. Don't worry about ranks, just do the best you can! Seriously, it sounds cliched and stupid but ignore the stupid competition side and just do as well as possible.

Use this forum if you need support! This is a community of students, built to help one another. I hope to see you on here often!

Great question, glad you asked as this will help so many students. For anyone else interested in contributing or asking question, it would seriously help build this community exponentially. Before you can ask or answer a question, you'll have to make an ATAR Notes account here. Once you've done that, a little 'reply' button will come up when you're viewing threads, and you'll be able to post whatever you want! :)

Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: Neutron on February 07, 2016, 02:10:11 pm
Thank you both Happy Physics Land and Jake for your amazingly comprehensive responses! They were definitely helpful but I have a question on what Jake said about the summarising another student's notes. Do you mean I highlight them and study them as though they were my own or do I continue to write my own but take theirs into consideration? Sorry for my confusion!

Neutron
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: jakesilove on February 07, 2016, 03:34:46 pm
Thank you both Happy Physics Land and Jake for your amazingly comprehensive responses! They were definitely helpful but I have a question on what Jake said about the summarising another student's notes. Do you mean I highlight them and study them as though they were my own or do I continue to write my own but take theirs into consideration? Sorry for my confusion!

Neutron

Hey Neutron,

Totally valid question, sorry I didn't make that clear! I would be writing you own notes, using basically only another set of notes that you know are good. What I mean is going through, highlighting important information, and summarising the notes into a set of your own.
For instance, if you had a chapter of the texbook, or a paragraph of someone else's notes, I would highlight the important parts, and summarise all the information into a single dot-point. Still have the lengthier notes on hand, but having a super succinct set of notes is really really useful for last minute study.
So, basically, my answer is do all of the above! You can never have too-few notes!

Feel free to send through example notes, or just anything you'd like us to look over. Looking forward to hearing from you!

Jake
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: xXCandyDXx on February 10, 2016, 09:55:28 pm
Hey Jake! How's it going? Hope everything is good! Anywho, I have a question from polynomials. The question asks me to solve an equation 'given that the roots form an arithmetic series'... But my teacher hasn't really gone through series ( unless it's just saying 'plus or minus k )... Here's the question! Q6!!!!!
 (http://images.tapatalk-cdn.com/16/02/10/872438724fa260c62a8b25f0b073ccec.jpg)
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: jakesilove on February 10, 2016, 10:10:48 pm
Hey Jake! How's it going? Hope everything is good! Anywho, I have a question from polynomials. The question asks me to solve an equation 'given that the roots form an arithmetic series'... But my teacher hasn't really gone through series ( unless it's just saying 'plus or minus k )... Here's the question! Q6!!!!!
 (http://images.tapatalk-cdn.com/16/02/10/872438724fa260c62a8b25f0b073ccec.jpg)

Hey Candy!

Great question! Plus, you've identified the most difficult part: what the "Arithmetic series" bit actually means! Once you've taken a look through this solution, you should be able to answer any similar question. Hope that this helps!

(http://i.imgur.com/Ak4B5bz.png?1)

Absolutely great question! I really encourage everyone to keep posting questions on the forum, so that everyone can benefit from this community.

Jake :)
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: Happy Physics Land on February 11, 2016, 12:12:53 am
Damn Jake, looks like I missed out on some delicious polynomial questions :/
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: jamonwindeyer on February 11, 2016, 07:42:35 pm
Damn Jake, looks like I missed out on some delicious polynomial questions :/

I love that you described a polynomial question as "delicious." Absolutely fantastic  8)
Title: 93 in 4U Maths: Ask me Anything!
Post by: xXCandyDXx on February 11, 2016, 09:27:17 pm
Thank you Jake! I think that's the same method a friend of mine used! But I don't understand it very well... I'll have to keep looking HAHA HOWEVER I have accidentally come across the same solution using a different method and I think this may benefit others too! Here it is! Also, would this be a valid method to use??? (http://images.tapatalk-cdn.com/16/02/11/2f8545333d5348e2fefc5317609e9435.jpg) oh wait I think I understand what you've done in the working out (THANK YOU!) but I don't know if I will know when to use them ... Are there indicators to use those 'sum of roots' equations :s ? I still feel pretty blind...
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: jakesilove on February 11, 2016, 09:32:36 pm
Thank you Jake! I think that's the same method a friend of mine used! But I don't understand it very well... I'll have to keep looking HAHA HOWEVER I have accidentally come across the same solution using a different method and I think this may benefit others too! Here it is! Also, would this be a valid method to use??? (http://images.tapatalk-cdn.com/16/02/11/2f8545333d5348e2fefc5317609e9435.jpg)

Hey hey!!

That's actually a fantastic method! Definitely a valid method, and use whatever you understand more. Remember, there are always heaps of ways to solve a question.

Thanks for posting that answer though! Very beneficial to the community.

Jake
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: xXCandyDXx on February 11, 2016, 09:37:55 pm
Thank you !!
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: jamonwindeyer on February 11, 2016, 11:23:01 pm
Hey xXCandyDXx! Glad Jake could be of help!

In addition, a tip for spotting when to use the "sums of roots" formulae. Pretty much, if a question follows the following template:

Find the roots of ____________________ given that __________________

or even

The roots of ___________________ form an arithmetic series. Find the value of k (some unknown in the polynomial))

Essentially, if you are given a polynomial with something to find, and all you are given is some information about the roots, chances are that these formulae will help you. They are very powerful in surprising ways.

While I'm here, I have an issue with the method you posted above. It's very possible I am misunderstanding, but it seems like the notions of root and factor are being confused. Remember, factors can be multiplied together to get back to the original, roots are just solutions to the polynomial equal to zero.

So, in the first few lines, we multiply the roots together and then form an identity with the original polynomial. This doesn't make sense to me. We could do it with factors, but multiplying the roots of a polynomial together has no relationship with the original, besides the relationships highlighted in the formulae used by Jake's method.

The factor rule makes perfect sense. The first part of the solution however, while the correct answer is obtained, does not make as much sense to me, and I would argue that it is incorrect. I'd love to hear others interpretations on the matter!  ;D
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: jakesilove on February 12, 2016, 09:36:28 am
Hey xXCandyDXx! Glad Jake could be of help!

In addition, a tip for spotting when to use the "sums of roots" formulae. Pretty much, if a question follows the following template:

Find the roots of ____________________ given that __________________

or even

The roots of ___________________ form an arithmetic series. Find the value of k (some unknown in the polynomial))

Essentially, if you are given a polynomial with something to find, and all you are given is some information about the roots, chances are that these formulae will help you. They are very powerful in surprising ways.

While I'm here, I have an issue with the method you posted above. It's very possible I am misunderstanding, but it seems like the notions of root and factor are being confused. Remember, factors can be multiplied together to get back to the original, roots are just solutions to the polynomial equal to zero.

So, in the first few lines, we multiply the roots together and then form an identity with the original polynomial. This doesn't make sense to me. We could do it with factors, but multiplying the roots of a polynomial together has no relationship with the original, besides the relationships highlighted in the formulae used by Jake's method.

The factor rule makes perfect sense. The first part of the solution however, while the correct answer is obtained, does not make as much sense to me, and I would argue that it is incorrect. I'd love to hear others interpretations on the matter!  ;D

Hey Jamon!

Thanks for clarifying when to use the sum and product formulas for Polynomial questions!

Also, I agree that her method is actually incorrect, but as luck would have it gets to the right answer. Here's why

(http://i.imgur.com/WhUlMoD.png?1)


Jake
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: Happy Physics Land on February 12, 2016, 11:15:56 am
Hey Jamon!

Thanks for clarifying when to use the sum and product formulas for Polynomial questions!

Also, I agree that her method is actually incorrect, but as luck would have it gets to the right answer. Here's why

(http://i.imgur.com/WhUlMoD.png?1)


Jake


Jake you are such a god what an elaborative response!
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: Happy Physics Land on February 13, 2016, 10:22:56 pm
Please send help Almighty Jake when you are available!!! This has been screwing me over!!!  :-[

(http://i.imgur.com/aQWttKH.jpg)
(http://i.imgur.com/D9yBL9R.jpg)

Thank you so much Jake! <3 (I dont know why the images are sideways doe...)
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: xXCandyDXx on February 13, 2016, 11:20:39 pm
Sorry for the very late response!!! >~<
Thanks Jamon and Jake for clearing it up ... I guess it is luck that I got the answer this time hehe ... But YEP thank you guys! I understand a lot more about polynomials and how they work .. Kinda HAHA I'm getting there! But i totally get where you're coming from in the sense that I suddenly multiplied the roots and equated them ...( clearly I don't understand ) but yeah I was actually thinking of trying to multiply the factors together which I totally mistook ... for the roots ahh... But yeah !!! Thanks again !!! I won't be making the same mistake in the future for sure
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: xXCandyDXx on February 13, 2016, 11:22:01 pm



Jake you are such a god what an elaborative response!
He really is a god
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: jakesilove on February 14, 2016, 12:18:30 am
Please send help Almighty Jake when you are available!!! This has been screwing me over!!!  :-[

(http://i.imgur.com/aQWttKH.jpg)
(http://i.imgur.com/D9yBL9R.jpg)

Thank you so much Jake! <3 (I dont know why the images are sideways doe...)
CLICK HERE TO SEE THE IMAGES/QUESTION
(http://i.imgur.com/aQWttKH.jpg)
(http://i.imgur.com/D9yBL9R.jpg)

Hey HPL.

Typical you giving me incredibly difficult 4U maths questions at Midnight. Couldn't resist though, so this is the solution I came up with! An extremely, extremely difficult graphing question, and I definitely may have missed some components, so if anyone in the community wants to correct me please do.

(http://i.imgur.com/ISRRvZQ.png?1)
(http://i.imgur.com/u4CLi2b.png?1)
(http://i.imgur.com/dMXmX2Z.png?1)
(http://i.imgur.com/QzAvog5.png?1)
(http://i.imgur.com/iIOXMHX.png?1)

Like I said, if anyone has any corrections, suggestions, or anything like that, please feel free to post below. Extension 2 is an incredibly difficult course, one that even University professors often struggle with.

Still, this is a lot of fun (to me, I know I'm a bit of a nerd) so keep posting, and keep answering questions! We're starting to build a real community, and I love watching it grow.

Yours,

Jake :)
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: jakesilove on February 14, 2016, 12:19:46 am
He really is a god

Aha thank you all, although I wouldn't go so far as to say that. My head still hurts sometimes when I try to do Extension 2 Maths questions!

Still, thanks for participating in the forums so actively! Hope to see you around here soon :)
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: Happy Physics Land on February 14, 2016, 12:52:26 am
Hey HPL.

Typical you giving me incredibly difficult 4U maths questions at Midnight. Couldn't resist though, so this is the solution I came up with! An extremely, extremely difficult graphing question, and I definitely may have missed some components, so if anyone in the community wants to correct me please do.

(http://i.imgur.com/ISRRvZQ.png?1)
(http://i.imgur.com/u4CLi2b.png?1)
(http://i.imgur.com/dMXmX2Z.png?1)
(http://i.imgur.com/QzAvog5.png?1)
(http://i.imgur.com/iIOXMHX.png?1)

Like I said, if anyone has any corrections, suggestions, or anything like that, please feel free to post below. Extension 2 is an incredibly difficult course, one that even University professors often struggle with.

Still, this is a lot of fun (to me, I know I'm a bit of a nerd) so keep posting, and keep answering questions! We're starting to build a real community, and I love watching it grow.

Yours,

Jake :)


I swear you enlighten me Jake, this is actually more than what my teacher will ever explain for me. Omg thank you sooo much Jake this must have taken you an extensive amount of time and I reallly reallly reallly reallly reallllllyyyy appreciate this. So elaborative and it just helped me to understand so much better (the difficulty of this question essentially dropped from a 10/10 to a 5.5/10, which I reckon is a huge achievement). I am just extremely grateful, thank you so much for everything you have done Jake!!!!!!!!  :D :D
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: keltingmeith on February 14, 2016, 12:52:36 am
Like I said, if anyone has any corrections, suggestions, or anything like that, please feel free to post below. Extension 2 is an incredibly difficult course, one that even University professors often struggle with.

Holy crap that was beautiful to watch. o.o I feel I have a name change coming on...

The only thing I have to add is this:
Just as Jake did, if you're ever in doubt - sub in some points, and draw lines between them. From there, it's a really good idea to pick out the "nice" points to pick out.

For example, in the case of the f(1/x) question, x=1 is easy - both functions share that point! Other easy cases include x=+/- infinity (as they go to zero) and x=0 (because that'll go to infinity). In fact, it's worth going through all the functions you know, and thinking about what the easy points are - for example, f(x)=sin(x) and f(x)=e^x. Not just for answering these types of questions, but any type of graphing question - if you ever get stuck on a graphing question, and you know some "easy" points, then you can very quickly take a guess at the graph's shape by picking some points and drawing lines between them!

(a bunch of that was really just repeating - but I felt it needed a stronger emphasis, soz Jake :P)
(also, hi from Victoria, where our maths subjects don't try to kill us~)
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: katherine123 on February 17, 2016, 11:28:12 pm

For the poly ques how do u find the leading coefficient? I only know that there are 2 known roots
and for the trig ques  i dont know how to simplify it after implicitly differentiating

please help thank you
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: jakesilove on February 18, 2016, 10:41:00 am
For the poly ques how do u find the leading coefficient? I only know that there are 2 known roots
and for the trig ques  i dont know how to simplify it after implicitly differentiating

please help thank you

Katherine, these questions were insanely difficult. I wasn't even able to properly get the last one (I'm incredibly tired right now, so I'll blame it on that). Still, as I'll explain that second question should NOT be a multiple choice. If you find yourself spending too much time on a multiple choice, MOVE ON, because it isn't worth the marks. Still, my head hurts and I'm going to go sleep...

(http://i.imgur.com/qndqJJF.png?1)
(http://i.imgur.com/apiRfa3.png?1)
(http://i.imgur.com/aSRBFOq.png?1)

If someone else can come up with a better, proper answer please feel free to contribute!
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: Spencerr on February 18, 2016, 12:15:11 pm
exsiny + eysinx = 0

After differentiating with respect to x and grouping the dy/dx onto one side we get

dy/dx = -(exsiny + eycosx) / (excosy + eysinx)

Going back to the main equation that they give us (this was the tricky part) Rearranging we get.
exsiny = - eysinx

Then subbing that in to:

dy/dx = -(exsiny + eycosx) / (excosy + eysinx)

We get
dy/dx = -(-eysinx + eycosx) / (excosy - exsiny)
         = (eysinx - eycosx) / (excosy - ex siny)

Factorise out the ey from the top and ex from the bottom and you get the answer

This was a pain to type up haha, what program or website do you use jake to post the solutions?
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: brenden on February 18, 2016, 12:19:18 pm
exsiny + eysinx = 0

After differentiating with respect to x and grouping the dy/dx onto one side we get

dy/dx = -(exsiny + eycosx) / (excosy + eysinx)

Going back to the main equation that they give us (this was the tricky part) Rearranging we get.
exsiny = - eysinx

Then subbing that in to:

dy/dx = -(exsiny + eycosx) / (excosy + eysinx)

We get
dy/dx = -(-eysinx + eycosx) / (excosy - exsiny)
         = (eysinx - eycosx) / (excosy - ex siny)

Factorise out the ey from the top and ex from the bottom and you get the answer

This was a pain to type up haha, what program or website do you use jake to post the solutions?
Hey dii!

Try the tex code :)

Code: [Select]
[tex][/tex]
When you type things in between it it looks like this.



It's super easy to learn, here are a list of resources: List of LaTeX Resources

In particular, this one is great for learning: LaTeX typeset in Maths boards


Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: jakesilove on February 18, 2016, 12:20:24 pm
exsiny + eysinx = 0

After differentiating with respect to x and grouping the dy/dx onto one side we get

dy/dx = -(exsiny + eycosx) / (excosy + eysinx)

Going back to the main equation that they give us (this was the tricky part) Rearranging we get.
exsiny = - eysinx

Then subbing that in to:

dy/dx = -(exsiny + eycosx) / (excosy + eysinx)

We get
dy/dx = -(-eysinx + eycosx) / (excosy - exsiny)
         = (eysinx - eycosx) / (excosy - ex siny)

Factorise out the ey from the top and ex from the bottom and you get the answer

This was a pain to type up haha, what program or website do you use jake to post the solutions?

Wow, that's beautiful! I initially did the same thing Diiiiiii, but didn't think to substitute the original equation back in! Absolute genius.

I literally use Word, then screenshot my response and upload it to Imgur so I can upload it to the forum. On Word, there is an "equation" feature which is actually very intuitive. I would highly recommend it; it's super useful!

Or, use to Tex code! I'm not as proficient at that, but it is also quite intuitive.

Again, thanks for the brilliant answer.

Jake
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: RuiAce on February 18, 2016, 01:22:55 pm




Also, to ALL MX2 students I suggest purchasing the calculator CASIO fx-100 AU PLUS for it's ability to do COMPLEX NUMBER OPERATIONS in the exam! This calculator is indeed, board approved!

For the multiple choice question, whilst the method provided by Jake is essentially what a short response requires, multiple choice can be hastened by simply SUBSTITUTING x=1-2i directly in! If what I typed into the calculator is right, A, B and C ALL don't equate to 0.

Multiple choice tricks.

FURTHER EDITING: In fact, there is one last trick. A and B are automatically wrong because the last root has to be a fraction! Not an integer! Non integral (adjective for integer in this case) roots are the only cause the leading to coefficient to deviate from 1.

In the long method, this means that the (ax+b) could be SAFELY written as (x+b)! Then, only constants have to be equated.
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: RuiAce on February 18, 2016, 01:52:16 pm


Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: katherine123 on February 19, 2016, 05:27:21 pm
hi how do u integrate (x+1)/(x^2+x+1)^1/2
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: RuiAce on February 19, 2016, 06:53:04 pm
hi how do u integrate (x+1)/(x^2+x+1)^1/2




Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: RuiAce on February 19, 2016, 07:04:59 pm




Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: Neutron on February 20, 2016, 05:20:47 pm
Hello! Me again :P Heh so I was trying to complete some 4U curve sketching when I came across a few silly problems I realised I haven't fully understood. The question is as follows:

Sketch the graph of y=lxl-lx-4l . Use this graph to solve the inequality lxl-lx-4l>2

So I managed to sketch it by plotting points and I was wondering whether there's an easier method? Like how would you know whether the absolute value graph resembles a backwards Z like this one or one that looks more like \_/ ? And secondly, I tried solving
2=|x|-|x-4| to find the x coordinate but I kept getting the answer as -1 when the real answer is 3.. Could you please show me how to solve these equations ^^'' heh sorry this is kinda dumb

Neutron
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: RuiAce on February 20, 2016, 06:05:07 pm
Hello! Me again :P Heh so I was trying to complete some 4U curve sketching when I came across a few silly problems I realised I haven't fully understood. The question is as follows:

Sketch the graph of y=lxl-lx-4l . Use this graph to solve the inequality lxl-lx-4l>2

So I managed to sketch it by plotting points and I was wondering whether there's an easier method? Like how would you know whether the absolute value graph resembles a backwards Z like this one or one that looks more like \_/ ? And secondly, I tried solving
2=|x|-|x-4| to find the x coordinate but I kept getting the answer as -1 when the real answer is 3.. Could you please show me how to solve these equations ^^'' heh sorry this is kinda dumb

Neutron





Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: katherine123 on February 20, 2016, 06:43:50 pm
how to integrate x^2sinx  thanks :)
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: Happy Physics Land on February 20, 2016, 07:18:15 pm
how to integrate x^2sinx  thanks :)

Hey Katherine:

For this question you would be required to use IBP (integration by parts) which is an integration technique that you learn from extension 2 mathematics. The technique is actually derived from the production rule that we use when we differentiate. Anyways, this question would require double applications of IBP and if you have any further questions or doubts please dont hesitate to ask!

(http://i.imgur.com/HT2vU5X.jpg)
(http://i.imgur.com/T9hj7nn.jpg)

Best Regards
Happy Physics Land
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: she0071 on February 20, 2016, 09:43:38 pm
hi! There's this question I don't really understand:
'A cuboid tank is open at the top and the internal dimensions of its base are xm and 2xm. The height is hm. The volume of the tank is Vm^3 and the volume is fixed. Let S m^2 denote the internal surface area of the tank.'

hence find S in terms of V and x

Thanks a lot in advance! :)
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: Happy Physics Land on February 20, 2016, 10:16:55 pm
hi! There's this question I don't really understand:
'A cuboid tank is open at the top and the internal dimensions of its base are xm and 2xm. The height is hm. The volume of the tank is Vm^3 and the volume is fixed. Let S m^2 denote the internal surface area of the tank.'

hence find S in terms of V and x

Thanks a lot in advance! :)

Hey Ms She:

To deal with these questions, the most immediate things I would have done are to find the surface area in terms of x and h and the volume in terms of x and h. Because the question asks us to find S in terms of V and x, we know that we need to substitute h with an expression in terms of V and x in order to eliminate the h from the surface area equation. anyways, you can check out what l mean through observing my working out below. If you have any further questions dont hesitate to ask me! :)

(http://i.imgur.com/ixaxogq.jpg)

Best Regards
Happy Physics Land
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: jakesilove on February 21, 2016, 12:53:59 pm
Hey Ms She:

To deal with these questions, the most immediate things I would have done are to find the surface area in terms of x and h and the volume in terms of x and h. Because the question asks us to find S in terms of V and x, we know that we need to substitute h with an expression in terms of V and x in order to eliminate the h from the surface area equation. anyways, you can check out what l mean through observing my working out below. If you have any further questions dont hesitate to ask me! :)

(http://i.imgur.com/ixaxogq.jpg)

Best Regards
Happy Physics Land

Gorgeous responses HPL, looks like you're really on top of the content! Keep it up!

Jake
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: she0071 on February 21, 2016, 06:32:34 pm
Hey Ms She:

To deal with these questions, the most immediate things I would have done are to find the surface area in terms of x and h and the volume in terms of x and h. Because the question asks us to find S in terms of V and x, we know that we need to substitute h with an expression in terms of V and x in order to eliminate the h from the surface area equation. anyways, you can check out what l mean through observing my working out below. If you have any further questions dont hesitate to ask me! :)

Best Regards
Happy Physics Land
Thankyou!!
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: Happy Physics Land on February 21, 2016, 06:59:34 pm
Hey guys would you mind giving me a hand with this cambridge question?? Thank you very much in advance!

(http://i.imgur.com/xgKU3XR.jpg)

Questions 7 Please thank you!!! :)))
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: RuiAce on February 21, 2016, 07:15:57 pm



Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: RuiAce on February 21, 2016, 07:23:27 pm


Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: Happy Physics Land on February 21, 2016, 07:27:23 pm
Ahhh ok so I should have subtracted the area of the sector and the area of the triangle and area of semicircle from the entire area of the circle... hmmm thank you rui that was very elaborate!!! :)
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: amandali on February 23, 2016, 01:38:20 pm
Which integer n do you take?    would u use n=0,1,2,3  or n=0,+-1, +-2
so when the angle is 4 theta   then there will be 4 solutions ,  3theta=3 solutions
I have the answer attached below
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: Happy Physics Land on February 23, 2016, 05:14:59 pm
Which integer n do you take?    would u use n=0,1,2,3  or n=0,+-1, +-2
so when the angle is 4 theta   then there will be 4 solutions ,  3theta=3 solutions
I have the answer attached below

Hey Amanda:

Interesting question indeed. Personally I would always use n=0,1,2,3 because that clearly tells me the four solutions to this problem. So for me, I use n=0,1,2,3 and got cos(pi/8), cos(3pi/8), cos(5pi/8) and cos(7pi/8) as my roots. If you want to manipulate these values, you can always convert cos(5pi/8) into -cos(3pi/8) and cos(7pi/8) into -cos(pi/8). This is just my approach to these questions and I pretty much have never used n = 0, +-1 or +-2 before.

Hope my answer helps

Best Regards
Happy Physics Land
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: RuiAce on February 23, 2016, 06:05:57 pm
When I talk about cosine, I usually only take a few integer values and that's enough. This is because the general formula for cosine has a ± in it.

When I talk about sine, however, I like to stick with negatives. This way, I keep my angles within the boundaries of -π<x≤π, which is what a principal argument of a complex number will be.
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: amandali on February 25, 2016, 04:42:07 pm
how do you simplify this?
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: RuiAce on February 25, 2016, 05:20:16 pm
how do you simplify this?
There is no neat simplification to this: this is simplified completely by brute force.



Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: amandali on February 26, 2016, 02:33:17 pm
how to do part 2 (ii)
why does z1 , z2 and z1+z2  have to be collinear
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: RuiAce on February 26, 2016, 07:14:17 pm
how to do part 2 (ii)
why does z1 , z2 and z1+z2  have to be collinear





Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: Kezzdee on February 29, 2016, 09:40:51 pm
Hey Jake,

I'm currently in Year 12 and I'm doing 4U mathematics. I've always done quite well in maths up until around the beginning of year 12, where I've just been struggling with my 4u and 3u maths marks. I'm working quite hard at the moment, although I definitely could be working harder. I usually have a strong grasp on the basic knowledge (component A) parts of maths papers, however I'm never really quite prepared for Component B style questions. In my latest exam, I received 22/25 for the Component A section and a 6/25 for Component B. I'm also having trouble finding a balance between 3u and 4u, as I generally just do 4U maths most days of the week. I was wondering if you had any tips in regards to this kind of issue.

Thanks for your help,
Kezz
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: jakesilove on February 29, 2016, 11:27:55 pm
Hey Jake,

I'm currently in Year 12 and I'm doing 4U mathematics. I've always done quite well in maths up until around the beginning of year 12, where I've just been struggling with my 4u and 3u maths marks. I'm working quite hard at the moment, although I definitely could be working harder. I usually have a strong grasp on the basic knowledge (component A) parts of maths papers, however I'm never really quite prepared for Component B style questions. In my latest exam, I received 22/25 for the Component A section and a 6/25 for Component B. I'm also having trouble finding a balance between 3u and 4u, as I generally just do 4U maths most days of the week. I was wondering if you had any tips in regards to this kind of issue.

Thanks for your help,
Kezz

Hey Kezz!

I think that your concerns are shared with the majority of the 4U cohort (certainly, I felt the same way during my Extension 2 year!).

I'll try to address each of your issues individually, but I just want to preface by saying that close to 60% in a 4U exam is not a bad mark. Even just passing the 4U course should be a massive achievement, because honestly it is a tough bugger. So don't feel disheartened by where you're at: You have the nuts and bolts of the course down, and now it's all about figuring out the second half to maximise your final mark!

Let's talk about balance between 4U and 3U. Firstly, studying for 4U will absolutely help you with 3U, so don't feel like its "wasted" time.
Secondly, it is absolutely hard to balance your subjects when 4U feels like such a drain on your time. The way that I managed to structure it was by forcing myself to stop studying for 4U after a set amount of time/past papers, and making myself do other subjects. The fact is that, even if you studied every day, you wouldn't perfectly understand the course. Once you've acknowledged that fact, it becomes easier to stop yourself studying. Figure out how much time you want to spend on your other subjects, work 4U in (even if you do spend a bit more time on it than others) and stick to that schedule. Don't get carried away trying to derive De Moivre's theorum. To repeat, CHOOSE A 4U MATHS SCHEDULE AND STICK WITH IT!

My only tip for improving your mark for more difficult questions is to just keep doing HSC past papers, over and over. You'll start to see patterns in how to answer certain types of questions, and be able to quickly take the correct route that leads to the answer (rather than following the hundreds of other tempting routes that the Board of Studies have left for you).

I also wrote out a set of notes describing the best methods for approaching really difficult questions. For instance, for Polynomial questions I wrote a detailedish list of "if it looks like this, try this" etc. etc. I do think it helped in my HSC exam, so if you're so inclined give it a go!

Really though, I don't think you're doing badly at all. The 4U course is immense, difficult and just bloody brutal. Keep slugging on, do billions of practice papers, but don't let 4U take over your life. Best of luck!

Jake
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: amandali on March 06, 2016, 04:59:36 pm
for Part (iv)  how do you know p is greater than q
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: jakesilove on March 06, 2016, 05:56:39 pm
for Part (iv)  how do you know p is greater than q

Hey Amandali!

To be completely honest, I'm not sure! I think that probably it doesn't really matter which one is which: Once you get the answer out, just say "therefore p=.... q=....".

I could be wrong; there could be a specific reason that I'm not seeing, and if so I'd love the community to correct me!

Jake
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: Spencerr on March 08, 2016, 03:56:31 pm
Hey I think by drawing out a unit circle and plotting the roots of z to the power of 5 equals 1 will let you see which p or q is bigger.
Once you've plotted it out the pots you can identify the real vales of alpha, alpha squared and so on.
Comparing the sizes of the real values using the diagram will let you determine which one p or q is bigger. Sorry I'm on my phone at the moment but hopefully jake gets my drift and explains better if needed :)
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: amandali on March 09, 2016, 10:07:23 am
the answer is -inverse cos x - root (1-x^2)  but mine is   -1/2inversecosx -1/2 root (1-x^2)
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: jakesilove on March 09, 2016, 10:37:59 am
the answer is -inverse cos x - root (1-x^2)  but mine is   -1/2inversecosx -1/2 root (1-x^2)

Hey!

Have you made sure that you used the double angle rules correctly? It is likely that when you converted from cos^2(theta), you forgot to multiply everything by a half! Go through, and just make sure your substitutions are absolutely right. If you're still unsure, I can post an answer!

Jake
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: jamonwindeyer on March 09, 2016, 10:55:17 am
the answer is -inverse cos x - root (1-x^2)  but mine is   -1/2inversecosx -1/2 root (1-x^2)

Have a crack again using Jake's tip, and check your working against mine if you are still struggling!  ;D

Spoiler



That last line is from using the original substitution, and the Pythagorean Identities! Let me know if it makes sense. Note that I forgot the proper notation in most of those integrals, from the moment I substitute I am integrating with respect to Theta.
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: xXCandyDXx on March 09, 2016, 11:13:28 pm
Hey guys ... Having some trouble starting this question... I don't know how to start it correctly...

(http://images.tapatalk-cdn.com/16/03/09/33f5e34bbf1ab29843be0ead1b96e750.jpg)
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: Happy Physics Land on March 09, 2016, 11:36:04 pm
Hey guys ... Having some trouble starting this question... I don't know how to start it correctly...

(http://images.tapatalk-cdn.com/16/03/09/33f5e34bbf1ab29843be0ead1b96e750.jpg)

Hey Candy!

This is quite a typical question that you might see in your half yearlies or topic tests, perhaps not so much in the HSC. Sorry I cant give you a working solution on paper because of some technical issues and I will just type out my working procedure here.

Step1: apply the distance formula

Let P be the point (x,y)
sqrt((x-3)^2 + y^2) = 4/5 (2-x)

Step 2: square both sides:

(x-3)^2+y^2 = 16/25 (2-x)^2

Step 3: expand and simplify

25((x-3)^2 + y^2) = 16(2-x)^2
25(x^2 - 6x + 9 + y^2) = 16(4 - 4x + x^2)
25x^2 - 150x + 225 + 25y^2 = 64 - 64x + 16x^2
9x^2 - 86x + 161 + 25y^2 = 0
9x^2 - 86x + 25y^2 = -161
(9x^2 - 86x + 1849/9) + 25y^2 = -161 + 1849/9
9(x^2 - 86x/9 + 1849/81) + 25y^2 = 400/9
9(x-43/9)^2 + 25y^2 = 400/9
[81(x-43/9)^2]/400 + 9y^2/16 = 1

Ok I got some pretty dank answer here because I might have made a mistake somewhere in my calculations due to my late night practise. But you get my point, start with distance formula and then square both sides, and you will end up getting an ellipse. Hope my answer helped!

Best Regards
Happy Physics Land
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: xXCandyDXx on March 10, 2016, 12:43:03 am
Hey Happy Physics Land,
Thanks so much! Sorry, I forgot to also post the solution!!! You got it btw! 9x^2 - 86x + 25y^2 + 161 = 0 :D
The wording of the question is confusing to me ... When I saw the fraction I thought of the definition of an ellipse so like the "PS/PN =e" thing .. I also don't really understand why you put 4/5 with (2-X) ... 
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: xXCandyDXx on March 10, 2016, 12:55:58 am
Oh no wait never mind ... I get it LOL thanks again !!!
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: Happy Physics Land on March 10, 2016, 06:55:43 pm
Oh no wait never mind ... I get it LOL thanks again !!!

Haha no worries, thanks for asking that question! Sometimes intimidating questions can actually be solved using the most basic principles, but ofc, as extension 2 students we are often too ignorant towards those basic formulae haha :)
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: amandali on March 11, 2016, 06:09:21 am
I dont how its progressed from last 3rd line to last 2nd line  :/
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: jakesilove on March 11, 2016, 07:20:26 am
I dont how its progressed from last 3rd line to last 2nd line  :/

Hey Amandali!

So, the only change from the third last line to the second last line is, originally, we have



Now, if we take out a factor of two, we have



And so, we have 2 multiplied (n+1) times! Or, in other words,



Which is what we needed in the second last line!

Hope that this helped :)

Jake
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: amandali on March 11, 2016, 11:49:44 am
im lost :/
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: Happy Physics Land on March 11, 2016, 03:06:26 pm
im lost :/

Hey Amanda:

This is quite an interesting approach adopted by whoever came up with the answer. The first four steps were quite self explanatory, and quite smooth running. Now, when we get up to step 5, there isnt a clear indication of what the persona has exactly done to figure out why the expression 2u^2/(u^2+1)^2 equals to u(derivative of 1/u^2+1). But we do understand that step, because if we do the differentiation in step 5, we can get the same expression as in the fourth step. Step 6 is even more ambiguous, they simply just skipped from the integral to another integral without even telling you what has exactly happened! So what happened between step 5 and step 6, which is the step you underlined, is that they used integration by parts. I have attached a photo below to show how the transformation of the integral has occurred. Hope it helps!

(http://i.imgur.com/tNT0bUC.jpg)

If you are still confused dont hesitate to ask! :D

Best Regards
Happy Physics Land
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: katherine123 on March 15, 2016, 05:38:54 pm
how do you normally start with these questions involving i(n)  I(n-1)   
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: RuiAce on March 15, 2016, 07:22:33 pm
how do you normally start with these questions involving i(n)  I(n-1)   







Edit: Found a mistake
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: Happy Physics Land on March 15, 2016, 07:49:12 pm







You are an absolute god Rui
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: katherine123 on March 17, 2016, 09:33:11 pm
How do you do these questions? thanks
Title: Re: 93 in 4U Maths: Ask me Anything!
Post by: RuiAce on March 17, 2016, 09:58:58 pm
How do you do these questions? thanks



Observe that a way to avoid the partial fractions is to decompose ln(x^2-1) into ln(x+1) + ln(x-1) and perform two applications of integration by parts

Title: Re: 4U Maths Question Thread
Post by: Neutron on March 19, 2016, 08:29:32 pm
Hey guys!

Sort of a dumb question but I was doing some questions and one of the questions was:

"If P(x)=x^4-2x^3-x^2+6x-6 has a zero 1-i, find the zeros of P(x) over C"

Now I know that 1+i is also a root (conjugate root theorem) but how do you know there aren't 4 complex roots? Like how come you can assume it's just two complex roots idk I'm tired and confused so sorry if this is very basic!

Neutron
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 20, 2016, 07:48:18 am
Hey guys!

Sort of a dumb question but I was doing some questions and one of the questions was:

"If P(x)=x^4-2x^3-x^2+6x-6 has a zero 1-i, find the zeros of P(x) over C"

Now I know that 1+i is also a root (conjugate root theorem) but how do you know there aren't 4 complex roots? Like how come you can assume it's just two complex roots idk I'm tired and confused so sorry if this is very basic!

Neutron


Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on March 20, 2016, 01:44:52 pm




And I'll add that, in case you aren't sure, we can infer that first quadratic root by multiplying the two complex roots together:

Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 20, 2016, 06:51:53 pm
And I'll add that, in case you aren't sure, we can infer that first quadratic root by multiplying the two complex roots together:




Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on March 20, 2016, 08:36:10 pm



Ahh cool, wasn't sure if that result is taught explicitly or is just derived in practice  ;D
Title: Re: 4U Maths Question Thread
Post by: xXCandyDXx on March 21, 2016, 06:57:25 pm
Hey guys I need some help here.. How do I do this question? I know that q, r and s are coefficients but ... How would you relate them ?.. I don't know, I'm lost and confused thanks a heap! @.@(http://images.tapatalk-cdn.com/16/03/21/23b64e7b1d37b614fc81612701e0a552.jpg)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 21, 2016, 07:50:43 pm
Hey guys I need some help here.. How do I do this question? I know that q, r and s are coefficients but ... How would you relate them ?.. I don't know, I'm lost and confused thanks a heap! @.@(http://images.tapatalk-cdn.com/16/03/21/23b64e7b1d37b614fc81612701e0a552.jpg)




And I threw the pencil here; Somehow I accidentally got a negative stuck in there so unless the question has a typo, I had a mistake in algebra at some point.
Title: Re: 4U Maths Question Thread
Post by: xXCandyDXx on March 21, 2016, 08:15:54 pm





And I threw the pencil here; Somehow I accidentally got a negative stuck in there so unless the question has a typo, I had a mistake in algebra at some point.


Ohhh I got it!!! (http://images.tapatalk-cdn.com/16/03/21/cc2592586b07a55ab58a9c1e8dea7725.jpg) but how about the next part :((((
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 21, 2016, 08:22:23 pm

Ohhh I got it!!! (http://images.tapatalk-cdn.com/16/03/21/cc2592586b07a55ab58a9c1e8dea7725.jpg) but how about the next part :((((

Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 22, 2016, 08:13:07 am

Ohhh I got it!!! (http://images.tapatalk-cdn.com/16/03/21/cc2592586b07a55ab58a9c1e8dea7725.jpg) but how about the next part :((((

Actually, hang on (in regards to the diagram). Shouldn't the product of roots be negative s?
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on March 22, 2016, 12:16:07 pm
Actually, hang on (in regards to the diagram). Shouldn't the product of roots be negative s?

That's funny, there is also a mistake in the 3rd last line. It should be negative beta, so for the poster, the two incorrect lines:




They happened to cancel each other out  ;D
Title: Re: 4U Maths Question Thread
Post by: amandali on March 25, 2016, 12:03:55 am
Is it better if i memorise this or show working out?
Title: Re: 4U Maths Question Thread
Post by: katherine123 on March 25, 2016, 07:47:38 am
Im stucked with this ques 
What i did was changing sec^3(x) --> secx(1+tan^2(x)   
so  1/cosx(1+sin^2(x)/cos^2(x))
Title: Re: 4U Maths Question Thread
Post by: katherine123 on March 25, 2016, 08:23:32 am
I dont get
-1/n[cosxsin^(n-1)x]
since  I(n)= sin^n(x)
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on March 25, 2016, 12:27:03 pm
Is it better if i memorise this or show working out?

Hey amandali!! It isn't in your table of standard integrals, so I think you best show working out in an exam. If it isn't in that table, you aren't expected to know it. However, it is definitely worth remembering as a security measure to check yourself as you go!!  ;D
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on March 25, 2016, 12:51:12 pm
Im stucked with this ques 
What i did was changing sec^3(x) --> secx(1+tan^2(x)   
so  1/cosx(1+sin^2(x)/cos^2(x))

Hey Katherine! That is not the approach I would use, though it might work! I'll show how I would do it with integration by parts:





Through integration by parts:



Now the integration of sec(x) is another job in itself. There is a common trick for figuring out this integral. Transform the integral like so:



Then, use a substitution:



Follow this process to find that:



Taking the formula from above and adding this, and rearranging, will give us the final solution:



Let me know if anything here is unclear!! It is a long and difficult question  ;D
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on March 25, 2016, 01:04:26 pm
I dont get
-1/n[cosxsin^(n-1)x]
since  I(n)= sin^n(x)

Hey Katherine! We have another job for integration by parts here!



By integration by parts:



Try rearranging this to obtain the solution given above:



I hope this helps!!  ;D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 25, 2016, 06:21:19 pm
Is it better if i memorise this or show working out?

I'm sorry, but Jamon is right. You have to use partial fractions or a trigonometric substitution for that one.

Anyway side note @Jamon I'm sorry but please take care of any questions HPL can't! I'm not sure when Jake is back but I'm on a vacation at Melbourne right now with grandparents so I might miss the posts. Thanks! :)
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on March 25, 2016, 06:33:33 pm
Is it better if i memorise this or show working out?

I'm sorry, but Jamon is right. You have to use partial fractions or a trigonometric substitution for that one.

Anyway side note @Jamon I'm sorry but please take care of any questions HPL can't! I'm not sure when Jake is back but I'm on a vacation at Melbourne right now with grandparents so I might miss the posts. Thanks! :)

All good Rui! I'm on it, enjoy Melbourne and cheers for your help!  ;D
Title: Re: 4U Maths Question Thread
Post by: Happy Physics Land on March 25, 2016, 08:18:34 pm
I'm sorry, but Jamon is right. You have to use partial fractions or a trigonometric substitution for that one.

Anyway side note @Jamon I'm sorry but please take care of any questions HPL can't! I'm not sure when Jake is back but I'm on a vacation at Melbourne right now with grandparents so I might miss the posts. Thanks! :)

Jake should be back tonight actually
Title: Re: 4U Maths Question Thread
Post by: jakesilove on March 25, 2016, 08:33:30 pm
Jake should be back tonight actually

HPL is right! I'm back and scrolling the forums, you guys have all done an incredible job keeping up to date on questions. I seriously, seriously appreciate the hard work you've all put in!

But yeah, back from my break and ready to answer some forum questions!

Thank guys; you are absolutely instrumental in maintaining the growth of this community, and we here at Atar Notes couldn't appreciate you more.

Jake
Title: Re: 4U Maths Question Thread
Post by: Happy Physics Land on March 25, 2016, 09:22:50 pm
HPL is right! I'm back and scrolling the forums, you guys have all done an incredible job keeping up to date on questions. I seriously, seriously appreciate the hard work you've all put in!

But yeah, back from my break and ready to answer some forum questions!

Thank guys; you are absolutely instrumental in maintaining the growth of this community, and we here at Atar Notes couldn't appreciate you more.

Jake

Damn Jamon and Rui, we can't keep all the questions to ourselves anymore...
Jake just rocks up and breaks the equilibrium we have been trying to maintain for the past 2 weeks, and according to Le Chatelier's principle .... hehehe
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 25, 2016, 10:35:53 pm
Damn Jamon and Rui, we can't keep all the questions to ourselves anymore...
Jake just rocks up and breaks the equilibrium we have been trying to maintain for the past 2 weeks, and according to Le Chatelier's principle .... hehehe
Oh my goodness.

WRONG. THREAD.
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on March 25, 2016, 11:51:32 pm
Damn Jamon and Rui, we can't keep all the questions to ourselves anymore...
Jake just rocks up and breaks the equilibrium we have been trying to maintain for the past 2 weeks, and according to Le Chatelier's principle .... hehehe

Hey come on man I didn't even do Chemistry! Don't confuse me  ;)
Title: Re: 4U Maths Question Thread
Post by: Happy Physics Land on March 25, 2016, 11:54:34 pm
Hey come on man I didn't even do Chemistry! Don't confuse me  ;)

Yeah I forgot you did legal studies, well in this CASE, I reckon you do have the right to remain silent ... :) :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 25, 2016, 11:55:15 pm
Yeah I forgot you did legal studies, well in this CASE, I reckon you do have the right to remain silent ... :) :)

Ok go home. You're drunk.
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on March 26, 2016, 02:38:03 pm

Yeah I forgot you did legal studies, well in this CASE, I reckon you do have the right to remain silent ... :) :)

Wow HPL, these puns are just unLAWful...
Title: Re: 4U Maths Question Thread
Post by: Happy Physics Land on March 30, 2016, 05:47:23 pm
Hey lads Im a bit stuck with this conic question, would anyone mind lending me a hand? Thank you very much!!! :D  (I hope rui doesnt hit me for asking this question)

The chord AB of the hyperbola xy=c2 subtends a right angle at a point P on the curve. Prove that AB is parallel to the normal at P.

I couldnt really start this one because Im struggle to even draw the diagram.

Title: Re: 4U Maths Question Thread
Post by: jakesilove on March 30, 2016, 06:46:45 pm
Hey lads Im a bit stuck with this conic question, would anyone mind lending me a hand? Thank you very much!!! :D  (I hope rui doesnt hit me for asking this question)

The chord AB of the hyperbola xy=c2 subtends a right angle at a point P on the curve. Prove that AB is parallel to the normal at P.

I couldnt really start this one because Im struggle to even draw the diagram.

Hey! Below is my answer. Hope it makes sense!

(http://i.imgur.com/5ZnpIcN.png?1)

Jake
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 30, 2016, 06:56:42 pm
Hey lads Im a bit stuck with this conic question, would anyone mind lending me a hand? Thank you very much!!! :D  (I hope rui doesnt hit me for asking this question)

The chord AB of the hyperbola xy=c2 subtends a right angle at a point P on the curve. Prove that AB is parallel to the normal at P.

I couldnt really start this one because Im struggle to even draw the diagram.

You're off for this question cause it's still the point of half yearly exams and this is a bit tricky for an average 4U student. If it were HSC study time and you couldn't do this THEN I'd hit you. :P

But yeah, pretty sure I don't need to input anything else. Jake's method is probably what I would've used
Title: Re: 4U Maths Question Thread
Post by: Happy Physics Land on March 31, 2016, 12:14:38 am
Hey! Below is my answer. Hope it makes sense!

(http://i.imgur.com/5ZnpIcN.png?1)

Jake

Thank you so much Jake! I understand it a lot better now! Im just wondering, were the coordinates meant to be A = (ca, c/a), B = (cb, c/b) and P = (cp, c/p)? Because I've always learnt that you let x=ct and y = c/t, where t is the variable. If Im the one getting confused here please correct me Jake!!! Thank you very much! I would have loved to contribute more to the forum but because half-yearlies are on currently I'm a bit tight with time, sorry for dropping the burden on you guys! I shall be back to join you guys soon! :))))))

Best Regards
Happy Physics Land
Title: Re: 4U Maths Question Thread
Post by: jakesilove on March 31, 2016, 12:30:15 am
Thank you so much Jake! I understand it a lot better now! Im just wondering, were the coordinates meant to be A = (ca, c/a), B = (cb, c/b) and P = (cp, c/p)? Because I've always learnt that you let x=ct and y = c/t, where t is the variable. If Im the one getting confused here please correct me Jake!!! Thank you very much! I would have loved to contribute more to the forum but because half-yearlies are on currently I'm a bit tight with time, sorry for dropping the burden on you guys! I shall be back to join you guys soon! :))))))

Best Regards
Happy Physics Land

You're absolutely right about the co-ordinates, I have no idea how I managed to write out the complete mess that I did in the first line. Honestly it's like I had an aneurysm or something; anyway, thanks for picking that up!

Don't even worry, you contribute literally all the time. The community wouldn't be the same without you, and we totally understand that there will be times where you'll be quite busy and unable to do the same amount as usual! Never feel pressured to be on the forums :)

Jake
Title: Re: 4U Maths Question Thread
Post by: katherine123 on April 01, 2016, 05:39:35 pm
Hey Katherine! That is not the approach I would use, though it might work! I'll show how I would do it with integration by parts:





Through integration by parts:



Now the integration of sec(x) is another job in itself. There is a common trick for figuring out this integral. Transform the integral like so:



Then, use a substitution:



Follow this process to find that:



Taking the formula from above and adding this, and rearranging, will give us the final solution:



Let me know if anything here is unclear!! It is a long and difficult question  ;D


is this method the same as :
Let say   I= integral (x*lnx)
let u=x   v=lnx
I=(Integrating u)*(leave v) - integral of (integrating u*differentiate v)
Title: Re: 4U Maths Question Thread
Post by: jakesilove on April 01, 2016, 06:14:14 pm

is this method the same as :
Let say   I= integral (x*lnx)
let u=x   v=lnx
I=(Integrating u)*(leave v) - integral of (integrating u*differentiate v)

Hey Katherine!

Yep, that's the integration by parts method! It's fairly common in 4U, so I would have a good working knowledge of it. Try a bunch of practice questions, including the one Jamon wrote up (which is a pretty difficult one, if I do say so myself!).

Best of luck!

Jake
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 01, 2016, 07:56:51 pm

is this method the same as :
Let say   I= integral (x*lnx)
let u=x   v=lnx
I=(Integrating u)*(leave v) - integral of (integrating u*differentiate v)

Integration by parts is essentially the reverse of the product rule. One way of describing it is indeed, what you demonstrated.

Many questions will require you to demonstrate knowledge of this integration technique. E.g.


Note - The example I gave is a tiny bit harder, but not as hard as that for the integral of sec3(x)
Title: Re: 4U Maths Question Thread
Post by: katherine123 on April 03, 2016, 07:52:47 am
why isnt this B
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 03, 2016, 09:30:10 am
why isnt this B




Note - You may have accidentally thought we were rotating about the y-axis, not the line x=2
This could potentially explain, for you, why the answer is not B


Title: Re: 4U Maths Question Thread
Post by: amandali on April 03, 2016, 02:41:09 pm
how do you do this ques?
i know that the angle will be 90
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 03, 2016, 03:21:13 pm
how do you do this ques?
i know that the angle will be 90


In this context, wlog just implies ignore the plus or minus sign as all we care is that the angle is indeed, 90 degrees.
(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpscakyguio.png)

C, D and E exist for the sole purpose of labelling angles

Notice that all I am using here is the definition of arg(z-z0) where z0 is some complex number a+ib. That is, on the Argand diagram, you draw out the ray, inclined at an angle of theta to the horizontal.

And now comes the tricky part.


(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpskui9ydpz.png)




Title: Re: 4U Maths Question Thread
Post by: Maz on April 03, 2016, 05:01:34 pm
`hey, can someone please help me with this...i have little/to no idea how to do it... :)
find d/dx of:


thankyou so much in advance
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 03, 2016, 05:17:29 pm
`hey, can someone please help me with this...i have little/to no idea how to do it... :)
find d/dx of:


thankyou so much in advance






Title: Re: 4U Maths Question Thread
Post by: Maz on April 03, 2016, 05:21:25 pm
okay...thankyou
do you know how to do it without use using logs or In?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 03, 2016, 05:22:08 pm
okay...thankyou
do you know how to do it without use using logs or In?

Do they teach you hyperbolic functions in VCE?

Title: Re: 4U Maths Question Thread
Post by: Maz on April 03, 2016, 05:26:47 pm
no they don't...is there a way to integrate the function without using any of those things?
there's another question you might be able to use to explain it (i attached it) ...because we haven't done logs at school yet...so i'm thinking they may not ask those type of questions
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 03, 2016, 05:30:36 pm


I would've expanded that one out tbh. What was your method to attempting this one?
Title: Re: 4U Maths Question Thread
Post by: Maz on April 03, 2016, 05:39:46 pm
i just subbed e^x into  (1+√t)^5 into the equation...then subbed 1 in and then just F(e^x) - F(1)...but it's not working :(
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 03, 2016, 05:47:31 pm
i just subbed e^x into  (1+√t)^5 into the equation...then subbed 1 in and then just F(e^x) - F(1)...but it's not working :(
Facepalm
Of course! Just use the second fundamental theorem in the more clever way...

Ok let's see, in that case the answer would crumble down to


Then what does your given answer say?
Title: Re: 4U Maths Question Thread
Post by: Maz on April 03, 2016, 05:51:57 pm
the given answer says:
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 03, 2016, 06:04:41 pm
Since the method isn't wrong at this point I have to presume the answers are because...
http://www.wolframalpha.com/input/?t=crmtb01&i=1%2Fsqrt(1%2Be%5E(1%2Bx%5E2))-1%2Fsqrt(1%2Be)%3D2x%2Fsqrt(1%2Be%5E(1%2Bx%5E2))

Not true for ALL x
Title: Re: 4U Maths Question Thread
Post by: Maz on April 03, 2016, 06:14:38 pm
okay thankyou- so essentially u always just sub the limits in? because a lot of my answers are coming up as wrong that way...maybe they are all wrong...but then it is a published textbook...hmmm
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 03, 2016, 06:30:36 pm
no they don't...is there a way to integrate the function without using any of those things?
there's another question you might be able to use to explain it (i attached it) ...because we haven't done logs at school yet...so i'm thinking they may not ask those type of questions

Wait I'm just thinking, shouldn't the answer for this one have been



 Because I'm inclined to say we missed something important - the chain rule
Title: Re: 4U Maths Question Thread
Post by: Maz on April 03, 2016, 06:40:43 pm
that is the answer!  :) so do u do the chain rule on that
Title: Re: 4U Maths Question Thread
Post by: Maz on April 03, 2016, 06:42:40 pm
Wait I'm just thinking, shouldn't the answer for this one have been



 Because I'm inclined to say we missed something important - the chain rule
but how does the chain rule fit...we don't even need to do any differentiation or integration cos h equestion is saying to differentiate then integrate...?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 03, 2016, 06:43:23 pm










Basically, I must apologise - my lack of experience with the fundamental theorem meant I applied it wrongly! I learnt something new today.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 03, 2016, 06:45:10 pm
but how does the chain rule fit...we don't even need to do any differentiation or integration cos h equestion is saying to differentiate then integrate...?

Actually it said to integrate then differentiate. So the fundamental theorem of calculus allows us to work our way around that.
Title: Re: 4U Maths Question Thread
Post by: Maz on April 03, 2016, 06:48:53 pm
Actually it said to integrate then differentiate. So the fundamental theorem of calculus allows us to work our way around that.
how would you do that?  :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 03, 2016, 06:50:35 pm
how would you do that?  :)

The old fashioned and very VERY tedious way of doing that is the one I posted at the very start with the logs and stuff.

The more recent post is a much faster way to get things done.
Title: Re: 4U Maths Question Thread
Post by: Maz on April 03, 2016, 06:59:18 pm
ohhh...
thankyou so so soooo much  :)
Title: Re: 4U Maths Question Thread
Post by: amandali on April 04, 2016, 12:20:09 pm
not sure why its D
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on April 04, 2016, 12:49:15 pm
not sure why its D

Hey amandali!

Okay, this question requires a little bit of thought, but it is actually not too difficult!

Consider the equation for the roots (just using normal alphabetical characters instead of Greek):



If all of these roots were real, then this equation wouldn't make sense, since the square root of a real is always positive. Thus, we must have a complex root in there somewhere. However, by the complex conjugate root theorem, one complex root for a polynomial automatically means that the conjugate is also a root. Thus, since we require one complex root for the equation to make sense, we must have at least two complex roots by this theorem. Hence, the answer is D.

I hope this helps!  ;D
Title: Re: 4U Maths Question Thread
Post by: amandali on April 04, 2016, 04:24:26 pm
how to solve this?
please help thanks
Title: Re: 4U Maths Question Thread
Post by: katherine123 on April 04, 2016, 04:36:13 pm
im not sure how to determine whether point P lies above or below the x axis
for questions like   arg(z-1) = arg(z+1) + pie/4
Title: Re: 4U Maths Question Thread
Post by: jakesilove on April 04, 2016, 05:01:19 pm
how to solve this?
please help thanks

Hey!

Below is my solution. Generally, this is the method you'll need to use for any similar question so I would recommend memorising it!

(http://i.imgur.com/pPjzWwN.png?1)

Jake
Title: Re: 4U Maths Question Thread
Post by: jakesilove on April 04, 2016, 05:04:55 pm
im not sure how to determine whether point P lies above or below the x axis
for questions like   arg(z-1) = arg(z+1) + pie/4

Hey!

Presumably, you are being asked to sketch something like the graph



So first, you solve and draw the general curve (I believe that this is a circle?). Then, I would choose an easy point (generally, 0+0i) and see whether it falls within or without the given question.

We can see that



Which is clearly greater than Pi/4! Therefore, whichever section the value 0+0i is in, should be filled in!

I hope I've understood your question correctly.

Jake
Title: Re: 4U Maths Question Thread
Post by: Neutron on April 12, 2016, 07:02:22 pm
Hello friends!

I was wondering whether you could help me integrate tan^2 (x) sec^2 (x) dx ? Thank you, I seem to be having trouble ughhh

Neutron
Title: Re: 4U Maths Question Thread
Post by: jakesilove on April 12, 2016, 07:14:44 pm
Hello friends!

I was wondering whether you could help me integrate tan^2 (x) sec^2 (x) dx ? Thank you, I seem to be having trouble ughhh

Neutron

Hey Neutron!

This question looks deceptively difficult, but the answer ends up being quite straight forward! The trick is recognising what you need to use (ie. reverse chain rule), and not falling into the trap of using a more complicated method (ie. Integration by parts). Anyway, my answer is below, hope it helps!

(http://i.imgur.com/SpAgXmp.png?1)

Jake
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 12, 2016, 09:09:32 pm
Alternatively, if the reverse chain rule is a bit too hard to see, just apply the substitution u=tan(x).

At the Ext 2 level though, you do want to practice identifying the reverse chain rule
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on April 12, 2016, 09:39:45 pm
Alternatively, if the reverse chain rule is a bit too hard to see, just apply the substitution u=tan(x).

At the Ext 2 level though, you do want to practice identifying the reverse chain rule

Just to see that alternative method, it would go something like below! (I would probably not have spotted the reverse chain rule, I rely on substitution way too much  ;)):



Title: Re: 4U Maths Question Thread
Post by: Neutron on April 12, 2016, 10:24:33 pm
Hey Neutron!

This question looks deceptively difficult, but the answer ends up being quite straight forward! The trick is recognising what you need to use (ie. reverse chain rule), and not falling into the trap of using a more complicated method (ie. Integration by parts). Anyway, my answer is below, hope it helps!

(http://i.imgur.com/SpAgXmp.png?1)

Jake


Omg I'm honestly feeling so dumb right now, thank you so much guys :') I kept trying to expand the sec^2 and the tan^2 and ugh that was a mess oops
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on April 12, 2016, 10:28:57 pm

Omg I'm honestly feeling so dumb right now, thank you so much guys :') I kept trying to expand the sec^2 and the tan^2 and ugh that was a mess oops

Don't feel dumb Neutron, not being given the substitution makes it hard to see! Interesting approach, you mean pythagorean identities and such? Must have been nasty  ;)

I once got told, Differentiation is a Science, and Integration is an Art. Differentiation has set procedures and rules we follow to get our answers. Integration is a little more intuitive, and in my opinion, much more difficult. Practice makes perfect I think  ;D

Be sure to keep hitting us up with questions!!   :)
Title: Re: 4U Maths Question Thread
Post by: katherine123 on April 22, 2016, 02:20:28 pm
- make  x/[(x-1)^3(x-2)]  into partial fractions

 why is it   A/(x-1) + (BX+C)/(x-1)^2 + D/(x-2)       instead of AX+B/(x-1)^2 + C/(x-2)



- Is the cover up method commonly used and when does it not work?

- Will 4x-3/[x^3(x+1)]  decompose into A/x+ B/x^2 + C/x^3 + D/(x+1)   and why isnt B linear and C a quadratic
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 22, 2016, 07:04:48 pm
- make  x/[(x-1)^3(x-2)]  into partial fractions

 why is it   A/(x-1) + (BX+C)/(x-1)^2 + D/(x-2)       instead of AX+B/(x-1)^2 + C/(x-2)



- Is the cover up method commonly used and when does it not work?

- Will 4x-3/[x^3(x+1)]  decompose into A/x+ B/x^2 + C/x^3 + D/(x+1)   and why isnt B linear and C a quadratic

Look at your first question again. It is confusing. All I gathered was this:

Which decomposes into


As for Heaviside Cover-up, that only works for linear combinations. The instant you throw in a quadratic (or something higher) into the denominator, it won't work as a neater process.

Question two is rightfully so as the idea is the implementation of powers allows ways to abide by the unnecessarily complicated method of introducing a numerator with linear terms in it. To make it seem a bit more obvious however:

(I will ignore the term with x3 as the process is similar)
Title: Re: 4U Maths Question Thread
Post by: katherine123 on April 23, 2016, 09:33:40 pm
1)how to make 1/(x^2+2x-1)   into impartial fractions 
is it not decomposible since the denominator cant be factorised

2) how to make 1/[x(x^2-1)^2] into impartial fraction

so will it be
A/x + B/(x-1) + C/(x-1)^2 + D/(x+1)  + E/(x+1)^2
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 23, 2016, 10:03:05 pm
1)how to make 1/(x^2+2x-1)   into impartial fractions 
is it not decomposible since the denominator cant be factorised

2) how to make 1/[x(x^2-1)^2] into impartial fraction

so will it be
A/x + B/(x-1) + C/(x-1)^2 + D/(x+1)  + E/(x+1)^2




Title: Re: 4U Maths Question Thread
Post by: katherine123 on April 26, 2016, 12:23:12 am
how to integrate  this x^3/ (x^2-x-3)   thanks
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 26, 2016, 05:39:11 pm
how to integrate  this x^3/ (x^2-x-3)   thanks






First one is done by a u-substitution (or just by inspection), whereas the second is done using completing the square.
Title: Re: 4U Maths Question Thread
Post by: Happy Physics Land on April 26, 2016, 07:55:14 pm





First one is done by a u-substitution (or just by inspection), whereas the second is done using completing the square.

That was awesome! When I attempted it I had no idea what to do after long division with that 4x - 3 / x2 - x - 3 thing
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 27, 2016, 09:25:15 pm
That was awesome! When I attempted it I had no idea what to do after long division with that 4x - 3 / x2 - x - 3 thing

Oh to just complete the rule of thumb



Title: Re: 4U Maths Question Thread
Post by: katherine123 on April 28, 2016, 12:03:51 am
How to make  x^3/ [(x-1)(x^2+2)^2] into impartial fractions:

ans: (8/9-1/9x)/(x^2+2)  + (2/3x-4/3)/(x^2+2)^2  +1/9/(x-1)

doesnt the numerator becomes a constant when
you make something like  1/(x+2)^2 to  A/(x+2) + B/(x+2)^2
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 28, 2016, 06:31:11 am
How to make  x^3/ [(x-1)(x^2+2)^2] into impartial fractions:

ans: (8/9-1/9x)/(x^2+2)  + (2/3x-4/3)/(x^2+2)^2  +1/9/(x-1)

doesnt the numerator becomes a constant when
you make something like  1/(x+2)^2 to  A/(x+2) + B/(x+2)^2





Subject to inaccuracy now:
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on April 28, 2016, 07:32:17 am
How to make  x^3/ [(x-1)(x^2+2)^2] into impartial fractions:

ans: (8/9-1/9x)/(x^2+2)  + (2/3x-4/3)/(x^2+2)^2  +1/9/(x-1)

doesnt the numerator becomes a constant when
you make something like  1/(x+2)^2 to  A/(x+2) + B/(x+2)^2

To clarify on Rui's response:

When you need a partial fraction decomposition of something with a fully factorised quadratic term on the bottom, for example:



Then it gets broken down like so:



For an irreducible quadratic factor, the numerator takes the form (Cx + D) ;D

Also remember this technique only works for proper fractions, where the degree of the denominator is larger than the degree of the numerator.

Hope this helps you a tad!  ;D
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on April 28, 2016, 10:54:54 am
I'll also add, on the topic of Partial Fractions, that over half of my lecture on signal transformations this morning was work on Partial fraction decomposition and the Heaviside method. 2nd year university courses teaching very standard 4U content. Another example of why 4 Unit math is an amazing head start for many Math, Science, and Engineering degrees!!  ;D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 28, 2016, 08:34:32 pm
I'll also add, on the topic of Partial Fractions, that over half of my lecture on signal transformations this morning was work on Partial fraction decomposition and the Heaviside method. 2nd year university courses teaching very standard 4U content. Another example of why 4 Unit math is an amazing head start for many Math, Science, and Engineering degrees!!  ;D

You gotta admit though... without Wolfram or something some really messy partial fractions like that above one are tedious :P
Title: Re: 4U Maths Question Thread
Post by: katherine123 on April 28, 2016, 11:30:37 pm
To clarify on Rui's response:

When you need a partial fraction decomposition of something with a fully factorised quadratic term on the bottom, for example:



Then it gets broken down like so:



For an irreducible quadratic factor, the numerator takes the form (Cx + D) ;D

Also remember this technique only works for proper fractions, where the degree of the denominator is larger than the degree of the numerator.

Hope this helps you a tad!  ;D

so for eg.    1/(x^3+1)^2
the partial fractions will be   (Ax^2+Bx+C)/(x^3+1) +   (Dx^2+Ex+F)/(x^3+1)^2
since the denominator has irreducible cubic factor , the numerator will be quadratic
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 29, 2016, 07:27:09 am
so for eg.    1/(x^3+1)^2
the partial fractions will be   (Ax^2+Bx+C)/(x^3+1) +   (Dx^2+Ex+F)/(x^3+1)^2
since the denominator has irreducible cubic factor , the numerator will be quadratic

Bad idea.






Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on April 29, 2016, 11:01:16 am
Bad idea.








In principle Katherine, you are totally correct! But yes, Rui is right, this does simplify everything a fair bit.

That said, if your fraction wasn't a nice sum of cubes, say:



Then it probably becomes easier to do it your way, unless you like working with surds  ;D of course at that point you've also probably gone beyond what you would actually get asked, it just gets too tedious, they craft these questions to make the numbers work nicely... Most of the time  ;) hope this helps!
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on April 29, 2016, 11:04:53 am
You gotta admit though... without Wolfram or something some really messy partial fractions like that above one are tedious :P

I totally 100% agree, though partial fraction stuff is still probably a little tame compared to some of the stuff I've worked with. Engineering is tedious by nature  ;) Wolfram Alpha has definitely lightened my workload numerous times in the last 3 semesters!  ;D
Title: Re: 4U Maths Question Thread
Post by: Happy Physics Land on April 29, 2016, 10:31:26 pm
Wolfram Alpha has definitely lightened my workload numerous times in the last 3 semesters!  ;D

Cheeky Jamon
Title: Re: 4U Maths Question Thread
Post by: aoifera on April 30, 2016, 11:58:04 pm
Hey,
I'm currently doing the volumes of solids topic and the question is:
By taking slices work out the volume of the area enclosed within the circle (x-1)^2 + y^2 = 1 rotated about the y-axis
I'm not getting to the answer and I don't know where the mistake is. I would appreciate some help
Thank you  :)
Fe
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on May 01, 2016, 11:23:52 am
Hey,
I'm currently doing the volumes of solids topic and the question is:
By taking slices work out the volume of the area enclosed within the circle (x-1)^2 + y^2 = 1 rotated about the y-axis
I'm not getting to the answer and I don't know where the mistake is. I would appreciate some help
Thank you  :)
Fe

Hey there aoifera! Welcome to the forums!  ;D

That's a bit of a nasty one you've got there, but let's have a look  :) using geometric formula for the volume of a horn torus, I'm expecting an answer of 2 pi squared, so let's see how I go  :)

The volume of the solid when you rotate the circle about the y-axis is interesting. What we should do is separate the circle into two halves:



This separates the circle into the two halves, left and right. The volume when we rotate the circle around the y axis is just the volume when we rotate the right hand side, minus the volume when we rotate the left hand side. So, we set up our formula like so;



At this point we should use symmetry to simplify slightly (which we can do because this is an even function we are integrating):



Now we can use a substitution here, the easiest is probably:



Or, you can consider the integral geometrically. The area under the curve from 1 to 0 is just a quarter of the area of a unit circle. Whichever method you choose:



I hope this helps!! It might be ever so slightly different to how the slices method usually works, I didn't do Extension 2 in the HSC, I'm just doing the question how I'd do it at uni ;D
Title: Re: 4U Maths Question Thread
Post by: birdwing341 on May 01, 2016, 03:15:19 pm
Hey,
I'm currently doing the volumes of solids topic and the question is:
By taking slices work out the volume of the area enclosed within the circle (x-1)^2 + y^2 = 1 rotated about the y-axis
I'm not getting to the answer and I don't know where the mistake is. I would appreciate some help
Thank you  :)
Fe

To do in terms of volumes is in attachment (just in case jamon's didn't suffice). Like he said though, it's pretty similar :)
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on May 01, 2016, 03:48:21 pm
To do in terms of volumes is in attachment (just in case jamon's didn't suffice). Like he said though, it's pretty similar :)

Oh wow it's actually very similar, the formal notation makes the proof a little nicer for yours though I'd say  ;D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 01, 2016, 04:00:23 pm
The Extension 2 student is indeed required to formally derive (using a method similar to how integration originated from) what exactly the volume of each disk/washer or cylindrical shell. The following two formulas should only be used as an aid to see if you were headed in the right direction:



When tackling volumes problems, more notoriously with shells however in both cases the first criteria is to determine what the radius is. Especially with shells and determining the correct outer/inner radii of the annulus cross section. Height for cylindrical shells is usually very easy.



Obviously, replace x with y or whatever whenever appropriate

A side note on parallel cross sections - these are done similarly, requiring derivations as well, but they typically need to be handled with more care.
Title: Re: 4U Maths Question Thread
Post by: Happy Physics Land on May 01, 2016, 06:50:24 pm
Hey guys!

I'm a bit stuck on which substitution to use with integrating x/(x2 + x)3/2. Would anyone mind helping me out a bit? Thank you so much!!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 01, 2016, 07:25:58 pm
Hey guys!

I'm a bit stuck on which substitution to use with integrating x/(x2 + x)3/2. Would anyone mind helping me out a bit? Thank you so much!!!





Title: Re: 4U Maths Question Thread
Post by: Happy Physics Land on May 02, 2016, 12:52:31 pm







What a disgusting integral .... Thank you so much rui ~!!!!!!! <3
Title: Re: 4U Maths Question Thread
Post by: katherine123 on May 02, 2016, 07:12:48 pm
how to solve
integral of 1/(1+tanx) thanks
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 02, 2016, 08:58:24 pm
how to solve
integral of 1/(1+tanx) thanks



Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on May 02, 2016, 10:26:19 pm
how to solve
integral of 1/(1+tanx) thanks

The substitution method Rui mentioned above, just in case you were tempted, starts like this. Rui's method is definitely easier if you spot it, but this will get you there too (and admittedly what my first instinct was)  ;D





You then perform a partial fraction decomposition on the integrand, and go from there. If you wanted to see more let me know, but yeah, if you do ever spot shortcuts like the one Rui suggested, snap them up  :D

Title: Re: 4U Maths Question Thread
Post by: amandali on May 03, 2016, 01:39:15 am
how to solve
integral of sinx /[ 2+cosx - cos^2(x) ]   and integral of 1/(e^2x + 4e^x + 9)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 03, 2016, 09:30:51 am
how to solve
integral of sinx /[ 2+cosx - cos^2(x) ]   and integral of 1/(e^2x + 4e^x + 9)
Doing these questions by hand today cause I'm not at home and typing LaTeX.

For Q1, whilst normally I advocate reverse chain rule, when it collapses into partial fractions it's easier to start on a substitution.
(http://uploads.tapatalk-cdn.com/20160502/45afb04e16c5985eb247de9d75b3bfb4.jpg)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 03, 2016, 09:47:24 am
The substitution isn't directly obvious for the second one, but the neat thing about exponentials is that you can "introduce" the substitution by multiplying top and bottom by e^x.

The final answer is subject to inaccuracy(http://uploads.tapatalk-cdn.com/20160502/fb43c67499fe5e2c60eb802047daeaeb.jpg)(http://uploads.tapatalk-cdn.com/20160502/9d1eb17d236f7a94c3094a8e75c85d2e.jpg)
Title: Re: 4U Maths Question Thread
Post by: birdwing341 on May 03, 2016, 11:13:48 am
The substitution isn't directly obvious for the second one, but the neat thing about exponentials is that you can "introduce" the substitution by multiplying top and bottom by e^x.

The final answer is subject to inaccuracy(http://uploads.tapatalk-cdn.com/20160502/fb43c67499fe5e2c60eb802047daeaeb.jpg)(http://uploads.tapatalk-cdn.com/20160502/9d1eb17d236f7a94c3094a8e75c85d2e.jpg)

Tis a good answer, but I think you forgot to carry the 1/9 for the second two parts of the integral so the coefficients of the log should be -1/18 and the tan inverse is -2/9(5)^1/2
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on May 03, 2016, 11:19:57 am
Tis a good answer, but I think you forgot to carry the 1/9 for the second two parts of the integral so the coefficients of the log should be -1/18 and the tan inverse is -2/9(5)^1/2

That's correct. That last inverse tangent term should be:



Nasty integral  ;D
Title: Re: 4U Maths Question Thread
Post by: jakesilove on May 03, 2016, 11:48:49 am
That's correct. That last inverse tangent term should be:



Nasty integral  ;D

Really, really nasty. Had to use a program to get the correct answer, and I really can't imagine that you would be asked this in an exam situation (maybe as a final questions?). Ah well, I guess this is why every other State shakes in fear when they hear that we have an Extension 2 Maths course.

(http://i.imgur.com/NCPMA2q.png?1)

Jake
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 03, 2016, 01:01:11 pm
Tis a good answer, but I think you forgot to carry the 1/9 for the second two parts of the integral so the coefficients of the log should be -1/18 and the tan inverse is -2/9(5)^1/2

Like I said, subject to inaccuracy. I paused the question and got back to it so I dropped 1/9 everywhere. Same goes with the completing the square dropoff
Title: Re: 4U Maths Question Thread
Post by: katherine123 on May 03, 2016, 11:01:17 pm
integral of  1/(x^2-9)^3/2

I let x=3sec(B)

and im left with
integral of   1/9*cos(B)*cosec^2(B)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 04, 2016, 12:15:48 am
integral of  1/(x^2-9)^3/2

I let x=3sec(B)

and im left with
integral of   1/9*cos(B)*cosec^2(B)



Title: Re: 4U Maths Question Thread
Post by: katherine123 on May 05, 2016, 12:06:17 am
how to solve
integral of 1/[x*(x^2+1)^1/2] with limits between 1 and 2?

Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 05, 2016, 12:24:36 am
how to solve
integral of 1/[x*(x^2+1)^1/2] with limits between 1 and 2?





Note: In the absence of the boundaries, to avoid the partial fractions it may have been more beneficial to use the trigonometric substitution x=tan(θ). However, dealing with boundaries that have inverse trigonometric functions in them, whilst not necessarily bad, can get messy because you don't know what the final result is.
Title: Re: 4U Maths Question Thread
Post by: aoifera on May 05, 2016, 06:33:15 pm
Hello,
I'm supposed to do this question with both washers and cylinders but I can't get either to work
Find the volume of a torus formed with the circle (x-2)2 + y2 = 1 is rotated about the y axis
Thank you  :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 05, 2016, 10:02:20 pm
Hello,
I'm supposed to do this question with both washers and cylinders but I can't get either to work
Find the volume of a torus formed with the circle (x-2)2 + y2 = 1 is rotated about the y axis
Thank you  :)










Title: Re: 4U Maths Question Thread
Post by: amandali on May 07, 2016, 10:09:58 pm
how do i integrate  x^4*e^(-x)
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on May 08, 2016, 12:03:10 am
how do i integrate  x^4*e^(-x)

Hey Amandali! You are looking at integration by parts for this question, but it is actually quite difficult, as it involves Recurrence Relations!  ;D



So, the working would look a little like this.





The last part of that line should look familiar, it is the same integral we just performed! The only difference is the power has been reduced by one. This might be enough for you to spot the recurrence relation, but just in case, let's do that last integral by integration by parts.



You should start to see a pattern emerging here. The recurrence relation we are seeing is:



We can use this relation all the way to its conclusion (it breaks once we hit n=1), and find the answer. Be careful to carry the multiples all the way through each recurrence!





I hope this helps! Recurrence relations can be a little strange if you haven't seen them before, so let me know if anything is a little unclear!!  ;D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 08, 2016, 05:54:34 am
how do i integrate  x^4*e^(-x)




You may choose now to work backwards, or forwards. Here is how to work forwards.

Title: Re: 4U Maths Question Thread
Post by: katherine123 on May 08, 2016, 02:00:59 pm
how to solve these
1.integral of x*inverse sin^-1(x^2)
2.integral of ln[x+(x^2+1)^1/2]   after proving   d/dx[ln[x+(x^2+a^2)] = 1/[(x^2+a^2)^1/2]
Title: Re: 4U Maths Question Thread
Post by: jakesilove on May 08, 2016, 02:19:16 pm
how to solve these
1.integral of x*inverse sin^-1(x^2)
2.integral of ln[x+(x^2+1)^1/2]   after proving   d/dx[ln[x+(x^2+a^2)] = 1/[(x^2+a^2)^1/2]

Hey Katherine!

Below is my solution to the first question (Will get to the second in a sec). This is just a straightforward application of integration by parts; it's an important method to learn, as its very standard in the extension 2 course.

(http://i.imgur.com/QZF80nl.png?1)

Jake
Title: Re: 4U Maths Question Thread
Post by: jakesilove on May 08, 2016, 02:39:29 pm
how to solve these
1.integral of x*inverse sin^-1(x^2)
2.integral of ln[x+(x^2+1)^1/2]   after proving   d/dx[ln[x+(x^2+a^2)] = 1/[(x^2+a^2)^1/2]

Here is my solution to the second question!

(http://i.imgur.com/lMZUBIW.jpg?1)

Jake
Title: Re: 4U Maths Question Thread
Post by: amandali on May 09, 2016, 11:01:18 pm
1st ques: im asked to find the integral of  (x^2+4)^1/2   and  i used the substitution x=2tanB  which gave me  integral of 4*sec^3(B) 

2nd ques: integral of e^x*sin(4x)cos(2x) with no limits

i changed the above integral to   e^x*[sin(2x)+sin(6x)]  using trig property

answer for ques 2 = e^x/370*[37sin(2x)+5sin6x-74cos(2x)-30cos(6x)] +C
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 10, 2016, 09:18:36 am
1st ques: im asked to find the integral of  (x^2+4)^1/2   and  i used the substitution x=2tanB  which gave me  integral of 4*sec^3(B) 

2nd ques: integral of e^x*sin(4x)cos(2x) with no limits

i changed the above integral to   e^x*[sin(2x)+sin(6x)]  using trig property

answer for ques 2 = e^x/370*[37sin(2x)+5sin6x-74cos(2x)-30cos(6x)] +C







Title: Re: 4U Maths Question Thread
Post by: katherine123 on May 12, 2016, 12:02:22 pm
question 1 :  integral of (x^2+2x+2)^1/2

i used integration by parts which gives me   
[x*(x^2+2x+2)^1/2] - integral of (x^2+x)/(x^2+2x+2)^1/2

im not sure how to integrate this part (x^2+x)/(x^2+2x+2)^1/2


question2 : integral of [x*(x+1)]^1/2
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 12, 2016, 06:48:43 pm
(http://uploads.tapatalk-cdn.com/20160512/72ff1d4641bdbe467bb48e2ed877c8a8.jpg)
I've already done the integral of secant cubed previously
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 12, 2016, 06:51:31 pm
These are really just unnecessarily tedious integrals if you ask me(http://uploads.tapatalk-cdn.com/20160512/feee17ed8ba91420d9485044b2499a93.jpg)
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on May 12, 2016, 07:13:35 pm
These are really just unnecessarily tedious integrals if you ask me(http://uploads.tapatalk-cdn.com/20160512/feee17ed8ba91420d9485044b2499a93.jpg)

(https://cdn.meme.am/instances/500x/68157159.jpg)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 12, 2016, 08:18:58 pm
(https://cdn.meme.am/instances/500x/68157159.jpg)

Are you ok Jamon

HSC integrals - only partial fractions are tedious :P

(Except for I think 2014 when the last question of the paper was a huge surprise)
Title: Re: 4U Maths Question Thread
Post by: jakesilove on May 12, 2016, 08:52:25 pm
Are you ok Jamon

HSC integrals - only partial fractions are tedious :P

(Except for I think 2014 when the last question of the paper was a huge surprise)

As a Class of 2014 graduate, I concur
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 12, 2016, 09:27:52 pm
As a Class of 2014 graduate, I concur

To be fair that one was a huge surprise. Here's the question done in two methods for anybody interested whilst I'm here






And the rest is easy to finish off.

The IBP was ridiculous dodgy as you had to pick the one to be differentiated: d/du ue^u = (u+1)e^u
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on May 13, 2016, 11:43:12 am
As a Class of 2014 graduate, I concur

Though I didn't take MX2, I actually heard about that question. It's interesting because in my opinion I actually think the MX1 paper for 2014 was a little easier than the standard, or maybe I just got lucky with the subjects  ;D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 13, 2016, 01:24:45 pm
Though I didn't take MX2, I actually heard about that question. It's interesting because in my opinion I actually think the MX1 paper for 2014 was a little easier than the standard, or maybe I just got lucky with the subjects  ;D

That's cause it was easier
Title: Re: 4U Maths Question Thread
Post by: amandali on May 15, 2016, 12:17:37 am
(http://uploads.tapatalk-cdn.com/20160514/ea8c6a7d4e9b7cbc39be832d47f38c1e.jpg)


im stuck after making sin^n(x) to sin^n-2(x)sin^2(x)
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on May 15, 2016, 02:04:12 am
(http://uploads.tapatalk-cdn.com/20160514/ea8c6a7d4e9b7cbc39be832d47f38c1e.jpg)


im stuck after making sin^n(x) to sin^n-2(x)sin^2(x)

Hey Amandali! Your approach might work, but I think it will be easier to change that split a little, try splitting into sinx and sin^(n-1)x and doing integration by parts:





Check the sign changes and substitutions in that second term to make sure you follow what I did. That first term will disappear, it evaluates to zero! So, we manipulate the second term:




Let me know if anything here was a little unclear! I skipped a little bit of algebra  ;D hope it helps!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 15, 2016, 09:53:52 am
(http://uploads.tapatalk-cdn.com/20160514/ea8c6a7d4e9b7cbc39be832d47f38c1e.jpg)


im stuck after making sin^n(x) to sin^n-2(x)sin^2(x)





I recommend that for the classic examples such as sinn(x) you simply memorise the natural approach, which is to split it up by a power of one and not two in this case. (Method of course demonstrated by Jamon already.)
Title: Re: 4U Maths Question Thread
Post by: jakesilove on May 15, 2016, 10:16:23 am




I recommend that for the classic examples such as sinn(x) you simply memorise the natural approach, which is to split it up by a power of one and not two in this case. (Method of course demonstrated by Jamon already.)

One day, I'm going to get to some Maths and Physics questions before you or Jamon.... One day!
Title: Re: 4U Maths Question Thread
Post by: amandali on May 17, 2016, 12:31:25 am
(http://uploads.tapatalk-cdn.com/20160516/a6eb482deff5136ea9b679311af571bc.jpg)

 got stuck after using integration by parts
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 17, 2016, 09:48:17 am
(http://uploads.tapatalk-cdn.com/20160516/a6eb482deff5136ea9b679311af571bc.jpg)

 got stuck after using integration by parts



Title: Re: 4U Maths Question Thread
Post by: amandali on May 17, 2016, 07:55:00 pm
(http://uploads.tapatalk-cdn.com/20160517/b10d17745781ddd5cdb9a417c3b51240.jpg)

i did part (i)  but not sure how it is used for part (ii)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 17, 2016, 08:56:12 pm
(http://uploads.tapatalk-cdn.com/20160517/b10d17745781ddd5cdb9a417c3b51240.jpg)

i did part (i)  but not sure how it is used for part (ii)

Title: Re: 4U Maths Question Thread
Post by: amandali on May 18, 2016, 11:53:05 pm
(http://uploads.tapatalk-cdn.com/20160518/5576cae1283ab996cac28240bd41a835.jpg)


not sure how to do part b)  part a is written on top
Title: Re: 4U Maths Question Thread
Post by: jakesilove on May 19, 2016, 09:18:48 am
(http://uploads.tapatalk-cdn.com/20160518/5576cae1283ab996cac28240bd41a835.jpg)


not sure how to do part b)  part a is written on top
Hey!

I've posted my solution below. This one was largely intuitive; once you've written out a couple expansions using your answer to a), it should be pretty obvious that b) is true.

(http://i.imgur.com/MGbu5St.png?1)

Jake
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 19, 2016, 01:06:26 pm
(http://uploads.tapatalk-cdn.com/20160518/5576cae1283ab996cac28240bd41a835.jpg)


not sure how to do part b)  part a is written on top

Judging by your writing to the side it would appear that you tried to work from bottom to top. If you wanted to do this approach instead then the bad idea was the choice to simplify the expression.

Title: Re: 4U Maths Question Thread
Post by: amandali on May 23, 2016, 07:57:09 pm
(http://uploads.tapatalk-cdn.com/20160523/610b5295ae58ef1f1167e411034acdde.jpg)

help with this ques thanks
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 23, 2016, 09:14:28 pm
(http://uploads.tapatalk-cdn.com/20160523/610b5295ae58ef1f1167e411034acdde.jpg)

help with this ques thanks




Title: Re: 4U Maths Question Thread
Post by: katherine123 on May 27, 2016, 12:20:19 am
what r the shortcut to simplify these
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on May 27, 2016, 11:51:02 am
what r the shortcut to simplify these

Hey Katherine! Do you mean what steps they used to get to the end of each line?

I'll pull the binomial-style term out of each one to show you the process. I'll just do the outline, I might skip a step here or there, let me know if anything doesn't make sense!  ;D

Here are the first two:



Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on May 27, 2016, 12:03:26 pm
what r the shortcut to simplify these

And here is the first term of the second two. Get a common term, factor out a sine on the top and a cosine on the bottom, you are left with -1 squared  :D



The second term of the second two will be very similar. Try it! Put everything over a common term, and you'll be able to simplify the numerator with a pythagorean identity. Then, you'll notice that there there will be cancellation with the numerator and denominator, after you factor out a cosine on the bottom. The result should follow right after  ;D hope this helped!
Title: Re: 4U Maths Question Thread
Post by: jakesilove on May 27, 2016, 12:03:52 pm
Hey Katherine! Do you mean what steps they used to get to the end of each line?

I'll pull the binomial-style term out of each one to show you the process. I'll just do the outline, I might skip a step here or there, let me know if anything doesn't make sense!  ;D

Here are the first two:





I was just posting a reply! Damn you and your superior LaTex skills
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on May 27, 2016, 01:22:26 pm
I was just posting a reply! Damn you and your superior LaTex skills

Damn sorry man! I think we need a system between Rui, you and myself  ;)
Title: Re: 4U Maths Question Thread
Post by: jakesilove on May 27, 2016, 02:41:33 pm
Damn sorry man! I think we need a system between Rui, you and myself  ;)

Aha no problem at all, usually we at least get out different methods, which can be useful for people learning the tricks of the trade!
Title: Re: 4U Maths Question Thread
Post by: amandali on May 29, 2016, 09:16:20 am
how do i find volume generated using cylindrical shell method
 when area bounded by y=1-x^2 and y=1-x is rotated About x axis
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 29, 2016, 10:28:00 am
how do i find volume generated using cylindrical shell method
 when area bounded by y=1-x^2 and y=1-x is rotated About x axis

(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpsmhzcmraq.png)





Title: Re: 4U Maths Question Thread
Post by: katherine123 on May 30, 2016, 10:39:45 pm
1. find the volume generated using cylindrical shells when region between y=2x^2 and y=x^4-2x^2 is revolved around y axis
whats the height?

2.find volume generated by using cylindrical shell method when area enclosed by x^2/25+ y^2/16=1 is rotated about x=8
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 31, 2016, 01:51:56 pm
1. find the volume generated using cylindrical shells when region between y=2x^2 and y=x^4-2x^2 is revolved around y axis
whats the height?

2.find volume generated by using cylindrical shell method when area enclosed by x^2/25+ y^2/16=1 is rotated about x=8

The height for the first one is just h=2x2 - (x4-2x2) by inspection of the graph

(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zps1yjgdqab.png)





Subject to inaccuracy
Title: Re: 4U Maths Question Thread
Post by: katherine123 on June 01, 2016, 11:45:10 pm
the region bounded by y=e^x, x=1 and y=1 is rotated about line x=1. Find the volume of solid

i used the slicing method
and got
(1-ln(y)) as the radius

v=integral of π(1-ln(y))^2 dy

is this correct or does cylindrical method only apply for this ques

the answer is 2π(e-2) units^3
Title: 4U Maths Question Thread
Post by: amandali on June 02, 2016, 12:34:08 am
(http://uploads.tapatalk-cdn.com/20160601/bdbc395aeadb78cf7cc7fd75c95009e4.jpg)

how to find the radius of smaller and bigger circle using slicing method.
the aim is to find the volume when rotated about y axis
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 02, 2016, 08:36:23 am
the region bounded by y=e^x, x=1 and y=1 is rotated about line x=1. Find the volume of solid

i used the slicing method
and got
(1-ln(y)) as the radius

v=integral of π(1-ln(y))^2 dy

is this correct or does cylindrical method only apply for this ques

the answer is 2π(e-2) units^3





Cylindrical shells may make it somewhat easier. If we just take the radius and height to be:
r=(1-x)
h=(ex)-(1)
This integral just requires one application of integration by parts to evaluate:



Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 02, 2016, 09:17:45 am
(http://uploads.tapatalk-cdn.com/20160601/bdbc395aeadb78cf7cc7fd75c95009e4.jpg)

how to find the radius of smaller and bigger circle using slicing method.
the aim is to find the volume when rotated about y axis
(http://i1164.photobucket.com/albums/q566/Rui_Tong/Screen%20Shot%202016-06-02%20at%208.51.42%20AM_zpslaf9bjd6.png)




(http://i1164.photobucket.com/albums/q566/Rui_Tong/Screen%20Shot%202016-06-02%20at%208.54.10%20AM_zpshsral4ue.png)







This is easily subject to inaccuracies. Especially given that I haven't been able to reconcile the answer with the shells method yet.

Edit: I have now. Hence this should be a fully accurate solution.
Title: Re: 4U Maths Question Thread
Post by: relativity1 on June 06, 2016, 02:09:18 pm
How do you find the restriction in this question
The hyperbola H has equation xy = 16.
(a) Sketch this hyperbola and indicate on your diagram the positions and coordinates
of all points at which the curve intersects the axes of symmetry.
(b) P(4p, ), where p > 0, and Q(4q, ), where q > 0, are two distinct arbitrary
points on H. Find the equation of the chord PQ.
(c) Prove that the equation of the tangent at P is x + p2y = 8p.
(d) The tangents at P and Q intersect at T. Find the coordinates of T.
(e) The chord PQ produced passes through the point N(0, 8).
(i) Find the equation of the locus of T.
(ii) Give a geometrical description of this locus.

I only need part ii) i think the restriction was x can only be between 0 and 4
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 06, 2016, 04:16:44 pm
How do you find the restriction in this question
The hyperbola H has equation xy = 16.
(a) Sketch this hyperbola and indicate on your diagram the positions and coordinates
of all points at which the curve intersects the axes of symmetry.
(b) P(4p, ), where p > 0, and Q(4q, ), where q > 0, are two distinct arbitrary
points on H. Find the equation of the chord PQ.
(c) Prove that the equation of the tangent at P is x + p2y = 8p.
(d) The tangents at P and Q intersect at T. Find the coordinates of T.
(e) The chord PQ produced passes through the point N(0, 8).
(i) Find the equation of the locus of T.
(ii) Give a geometrical description of this locus.

I only need part ii) i think the restriction was x can only be between 0 and 4

If you only needed part e) (ii) then you should've included the answers to the previous parts, as now I have to find an answer to a lof of the answers from scratch. Please remember to do so next time so my life is easier





(Note that p can't equal to q here or else the tangents at P and Q would coincide and thus T is not a strictly defined point.)


If one was to be more cautious we can take things a step further. (This is probably what you mean by restriction)






I highly doubt this is the fastest method, but it was all I could come up with for now.
________

The non calculus approach:



Proof of the AM-GM inequality for two terms should be known by every MX2 student.

Title: Re: 4U Maths Question Thread
Post by: birdwing341 on June 08, 2016, 05:36:19 pm
This is a more general question (with no specifics) - but its basically on how to do well at circle geometry.

Do you have any particular tips that could help in an exam situation when I'm stumped by the question. I've heard in general that there is a lot of cyclic quads, angles standing on the same arc etc. but I was wondering more if you had a particular process you would go through.

Thanks!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 08, 2016, 06:23:34 pm
This is a more general question (with no specifics) - but its basically on how to do well at circle geometry.

Do you have any particular tips that could help in an exam situation when I'm stumped by the question. I've heard in general that there is a lot of cyclic quads, angles standing on the same arc etc. but I was wondering more if you had a particular process you would go through.

Thanks!!

You should be decently skilled (if not already mastered) circle geometry at the Extension 1 level before attempting problems that target Extension 2 students.

At the Extension 1 level, I was already familiar with what my theorems "LOOKED LIKE". What does this mean? Examples:
- Alternate angles look like Z angles on parallel lines
- Angles standing on same arc theorem looked like a nice M shape to me
- Alternate segment theorem involves a tangent and a triangle.
However how you visualise it may differ.

When I tackle an Extension 2 question, I don't label 50 thousand things at once if it's way too irrelevant. I work in one direction.

That is, I look at what I am trying to prove (or find).

Then I try to either work forwards, or backwards. That is, I look at the LHS or the RHS, then I start looking for an angle/side that equals to THAT ONLY. THEN, I keep going.

At the Extension 2 level you should also be prepared to manipulate lots of things. Similar triangles and cyclic quadrilaterals are examples of cliches, however even base angles of isosceles triangle are important. Equal radii is something I always keep at the back of my head regardless of if I need to use it.

Progressively, I build up to the final answer. But in doing so, I only consider RELEVANT information, not EXTRANEOUS information that may be useful for another part but not this present part.

A rare trap is the occurrence of trigonometry in circle geometry. When that happens, always look out for any right angled triangles BEFORE you attempt sine/cosine rule.
Title: Re: 4U Maths Question Thread
Post by: birdwing341 on June 08, 2016, 06:38:30 pm
You should be decently skilled (if not already mastered) circle geometry at the Extension 1 level before attempting problems that target Extension 2 students.

At the Extension 1 level, I was already familiar with what my theorems "LOOKED LIKE". What does this mean? Examples:
- Alternate angles look like Z angles on parallel lines
- Angles standing on same arc theorem looked like a nice M shape to me
- Alternate segment theorem involves a tangent and a triangle.
However how you visualise it may differ.

When I tackle an Extension 2 question, I don't label 50 thousand things at once if it's way too irrelevant. I work in one direction.

That is, I look at what I am trying to prove (or find).

Then I try to either work forwards, or backwards. That is, I look at the LHS or the RHS, then I start looking for an angle/side that equals to THAT ONLY. THEN, I keep going.

At the Extension 2 level you should also be prepared to manipulate lots of things. Similar triangles and cyclic quadrilaterals are examples of cliches, however even base angles of isosceles triangle are important. Equal radii is something I always keep at the back of my head regardless of if I need to use it.

Progressively, I build up to the final answer. But in doing so, I only consider RELEVANT information, not EXTRANEOUS information that may be useful for another part but not this present part.

A rare trap is the occurrence of trigonometry in circle geometry. When that happens, always look out for any right angled triangles BEFORE you attempt sine/cosine rule.

Thanks for the advice! Prompt as usual :)
Title: Re: 4U Maths Question Thread
Post by: katherine123 on June 09, 2016, 01:03:26 pm
Ques: the region enclosed between the curve y=cos^-1(x) , the lines x=-1 and y=pie/2 is made to revolve about the axis y=-1 . find the volume of solid generated

im not sure what the height is
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 10, 2016, 07:01:11 am
Ques: the region enclosed between the curve y=cos^-1(x) , the lines x=-1 and y=pie/2 is made to revolve about the axis y=-1 . find the volume of solid generated

im not sure what the height is
(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpsq360p6ps.png)



Do you see why?

Title: Re: 4U Maths Question Thread
Post by: katherine123 on June 13, 2016, 11:37:47 am
the ans: 480

I did 3 cases 1 M, 2M and 3M :   so   5!+ 6P5/2! + 7P5/5! 
but it's not working
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 13, 2016, 01:15:43 pm
the ans: 480

I did 3 cases 1 M, 2M and 3M :   so   5!+ 6P5/2! + 7P5/5! 
but it's not working







Disclaimer: I thoroughly dislike combinatorics. My explanations for them are usually suboptimal to those for other topics. If the explanation is insufficient, please allow someone else to supplement my answer.
My explanations for combinatorics will improve after I do discrete maths, which may be a while.

I wasn't able to justify the method you used either.
Title: Re: 4U Maths Question Thread
Post by: amandali on June 15, 2016, 11:46:27 pm
need help with this ques
A ring of altitude 2h is generated by revolving about y-axis the area of the segment bounded by the circle x^2+ y^2=a^2 and the chord of length 2h that is parallel to the y-axis. By using cylindrical shell method, show volume is given by v=[4π(h)^3]/3


can you briefly explain the method of pairing cylindrical shell and when is it applicable?
where v=2πsA
s=distance from axis of rotation
A=area
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 16, 2016, 09:17:53 am
need help with this ques
A ring of altitude 2h is generated by revolving about y-axis the area of the segment bounded by the circle x^2+ y^2=a^2 and the chord of length 2h that is parallel to the y-axis. By using cylindrical shell method, show volume is given by v=[4π(h)^3]/3


can you briefly explain the method of pairing cylindrical shell and when is it applicable?
where v=2πsA
s=distance from axis of rotation
A=area

Also, whilst I've seen that formula before, I've never found a use for it in MX2 where they get you to derive everything from scratch. So you'll have to show me an example of a question+answer that your teacher said to use it on.
(This is because normally you can't just assume the area. You can only assume the height.)



(http://i1164.photobucket.com/albums/q566/Rui_Tong/Screen%20Shot%202016-06-16%20at%208.56.35%20AM_zpshy9dlpid.png)


Title: Re: 4U Maths Question Thread
Post by: amandali on June 16, 2016, 03:05:09 pm
(http://uploads.tapatalk-cdn.com/20160615/e6242109cab4f5b0451622052b73d6cd.jpg)
how do prove this thanks
Title: Re: 4U Maths Question Thread
Post by: jakesilove on June 16, 2016, 04:14:47 pm
(http://uploads.tapatalk-cdn.com/20160615/e6242109cab4f5b0451622052b73d6cd.jpg)
how do prove this thanks

Hey!

I'm terrible at inequality questions like this; I've had a bit of a crack, didn't get to an answer but got close. I can see the answer if we are give that


or something similar, but at the moment I don't THINK I have enough information. Just posting this here in case anyone can make sense of it and move forward from there!

(http://i.imgur.com/jB5v5Ni.png?1)

Jake
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 16, 2016, 07:48:44 pm





Can't take credit for this. Whilst I considered the AM-GM inequality several times it went way over my head how it could be used. And I can't spend forever on this question myself cause my exam is tomorrow.
Title: Re: 4U Maths Question Thread
Post by: amandali on June 16, 2016, 10:14:03 pm
got stuck on proving this ques(http://uploads.tapatalk-cdn.com/20160616/a93853910c20d07f8e845e6fc493b05d.jpg)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 17, 2016, 09:09:27 am
got stuck on proving this ques(http://uploads.tapatalk-cdn.com/20160616/a93853910c20d07f8e845e6fc493b05d.jpg)


Title: Re: 4U Maths Question Thread
Post by: katherine123 on June 18, 2016, 10:36:37 am
how do i do part (iii)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 18, 2016, 01:26:54 pm
how do i do part (iii)




Title: Re: 4U Maths Question Thread
Post by: birdwing341 on June 20, 2016, 02:10:39 pm
Hello!

I think I remember somewhere that one moderator had a small checklist that they went through after completing each question (to check for stuff like units, constants, dxs etc.).

Would you be able to remind me what that is?

Thanks!
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on June 20, 2016, 03:10:27 pm
Hello!

I think I remember somewhere that one moderator had a small checklist that they went through after completing each question (to check for stuff like units, constants, dxs etc.).

Would you be able to remind me what that is?

Thanks!

In a guide I wrote a while ago I had (and this is what I used):

SURD - Simplest Form, Units, Rationalised, Does it Make Sense?

However, I expanded on it in an article I wrote about the BIG mistakes students make in exams, which you can find here!

Integration has so many little traps, I should probably come up with a new acronym for it! I used SURD because I always forgot units, and I always forgot to rationalise  ;)

EDIT: How about SURDIC? Simplest Form, Units, Rationalised, Does it Make Sense, Integration Notation, Constants? Ahaha  ;D
Title: Re: 4U Maths Question Thread
Post by: amandali on June 20, 2016, 03:43:50 pm
ive found the equation chord for PQ shown below  and    Mpo * Mqo =-1
and im not sure how to progress
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 20, 2016, 04:53:38 pm
ive found the equation chord for PQ shown below  and    Mpo * Mqo =-1
and im not sure how to progress
If this is one of the further questions in the Cambridge textbook then it might not be so easy to do it. These questions do not reflect the difficulty of the HSC (even though they are valid in reflecting question type) as the question is not broken down into parts for you - you have to discover everything yourself.





Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on June 20, 2016, 05:04:18 pm
If this is one of the further questions in the Cambridge textbook then it might not be so easy to do it. These questions do not reflect the difficulty of the HSC (even though they are valid in reflecting question type) as the question is not broken down into parts for you - you have to discover everything yourself.







Damn, got about half as far as you did before you posted ;) these ellipse questions can actually be pretty nasty, I'm glad I didn't have to deal with them in 2014  :P
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 20, 2016, 05:17:08 pm
Damn, got about half as far as you did before you posted ;) these ellipse questions can actually be pretty nasty, I'm glad I didn't have to deal with them in 2014  :P
Thing is, when someone asks you a question and it's clearly out of Cambridge 4U, all you have to do is decipher the worked solutions :P
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on June 20, 2016, 05:38:04 pm

Thing is, when someone asks you a question and it's clearly out of Cambridge 4U, all you have to do is decipher the worked solutions :P

Okay, sorts it, I need to buy Cambridge Is that kind of the standard 4U text or are there other common ones?
Title: Re: 4U Maths Question Thread
Post by: amandali on June 21, 2016, 06:12:50 pm
1. the base of a solid is the circle x^2+y^2=8x and every plane section perpendicular to the x-axis is a rectangle whose height is one third of the distance of the plane of the section from the origin. Show that the volume of the solid is v=64π/3 units^3


2.A dome has a circular base of radius=10metres. Cross sections perpendicular to the base and one axis are parabolas who height is the same as the base width. Show that the area of the parabolic cross-sections is 8y^2/3 square metres
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 21, 2016, 06:37:05 pm
1. the base of a solid is the circle x^2+y^2=8x and every plane section perpendicular to the x-axis is a rectangle whose height is one third of the distance of the plane of the section from the origin. Show that the volume of the solid is v=64π/3 units^3


2.A dome has a circular base of radius=10metres. Cross sections perpendicular to the base and one axis are parabolas who height is the same as the base width. Show that the area of the parabolic cross-sections is 8y^2/3 square metres
(http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI21062016_zpsuwx6tey2.jpg)



Let me know if my handwriting is not legible.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 21, 2016, 07:10:18 pm
1. the base of a solid is the circle x^2+y^2=8x and every plane section perpendicular to the x-axis is a rectangle whose height is one third of the distance of the plane of the section from the origin. Show that the volume of the solid is v=64π/3 units^3


2.A dome has a circular base of radius=10metres. Cross sections perpendicular to the base and one axis are parabolas who height is the same as the base width. Show that the area of the parabolic cross-sections is 8y^2/3 square metres


(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpsw00xkwfr.png)
(The image makes the Y-intercept look like some number between 3 and 4. Just pretend that it's not there.)










This was a messy question and required some lateral thinking in considering the parabolic cross section in it's OWN X-Y plane. Tell me if something does not make sense.
Title: Re: 4U Maths Question Thread
Post by: amandali on June 22, 2016, 03:41:19 pm
(http://uploads.tapatalk-cdn.com/20160621/c5b8bebc5e795f6ef77ac064f25b0925.jpg)
(http://uploads.tapatalk-cdn.com/20160621/ab9bbc9e5e76c5a739da26fc95a4627d.jpg)

need help with these ques
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 22, 2016, 04:32:47 pm
(http://uploads.tapatalk-cdn.com/20160621/c5b8bebc5e795f6ef77ac064f25b0925.jpg)
(http://uploads.tapatalk-cdn.com/20160621/ab9bbc9e5e76c5a739da26fc95a4627d.jpg)

need help with these ques



(Note: The area of any ellipse is πAB, where A and B are the lengths of the semi-major and semi-minor axes respectively)

Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 22, 2016, 05:02:20 pm
(http://uploads.tapatalk-cdn.com/20160621/c5b8bebc5e795f6ef77ac064f25b0925.jpg)
(http://uploads.tapatalk-cdn.com/20160621/ab9bbc9e5e76c5a739da26fc95a4627d.jpg)

need help with these ques




__________________________________________



The more exhaustive method to integrating cos4(x) is to rewrite as (cos2(x))2, apply the double angle formula, expand the whole thing out, and reapply the double angle formula
Title: Re: 4U Maths Question Thread
Post by: katherine123 on June 23, 2016, 01:03:10 am
how do i do part (i)
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on June 23, 2016, 10:35:51 am
how do i do part (i)

I'll let Rui correct me if this isn't quite 4 Unit friendly, but this is actually very simple if we know the formula for a sphere in three dimensions (x,y,z):



At the cross section, the height above the xy plane (the z-value)  is h, so we can substitute:



What we notice here is that this is actually the formula for a circle. We've just found the formula for the outer circle of the cross section, with that term on the right being the square of the radius like normal. So:



Let me know if this makes sense! And Rui might come along and give a better method, I'm not sure if 3-dimensional surface equations are what would be the normal approach  ;D


Title: Re: 4U Maths Question Thread
Post by: katherine123 on June 23, 2016, 11:41:13 am
how do i do part iii)

i got
Part i) h=[(root3)*a]/2
part ii) ZY=xa/b
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 23, 2016, 01:23:54 pm
how do i do part iii)

i got
Part i) h=[(root3)*a]/2
part ii) ZY=xa/b


(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpsvdoj9neh.png)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 23, 2016, 01:49:22 pm
I'll let Rui correct me if this isn't quite 4 Unit friendly, but this is actually very simple if we know the formula for a sphere in three dimensions (x,y,z):



At the cross section, the height above the xy plane (the z-value)  is h, so we can substitute:



What we notice here is that this is actually the formula for a circle. We've just found the formula for the outer circle of the cross section, with that term on the right being the square of the radius like normal. So:



Let me know if this makes sense! And Rui might come along and give a better method, I'm not sure if 3-dimensional surface equations are what would be the normal approach  ;D
Haha don't even need a z-axis

(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zps7x0tokux.png)

Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on June 23, 2016, 02:14:51 pm
Haha don't even need a z-axis

Cheers Rui, that's much easier, too much 3-dimensional calculus has spoiled me  ;)
Title: Re: 4U Maths Question Thread
Post by: jakesilove on June 23, 2016, 03:45:57 pm
Cheers Rui, that's much easier, too much 3-dimensional calculus has spoiled me  ;)

Yeah, for better or worse you don't do any 3-dimensional work in Extension 2 (probably better, the course is hard enough)
 
Title: Re: 4U Maths Question Thread
Post by: amandali on June 24, 2016, 12:32:20 pm
(http://uploads.tapatalk-cdn.com/20160623/807dca2a1118738f9aa3aceccbbec6d9.jpg)

does this require comparison of areas
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 24, 2016, 01:25:19 pm
(http://uploads.tapatalk-cdn.com/20160623/807dca2a1118738f9aa3aceccbbec6d9.jpg)

does this require comparison of areas
Can't see what areas to compare. I would've just attacked it with an analysis of monotonic behaviour.

(Note that this proof is a bit informal. You may have to verify it's suitability with your teacher)




The other side does not look immediately obvious to me, so I will have a closer review of it later
_______________________________________
Edit: Suggestion

Pro tip (aka. Giveaway)
We're full of even powers here...
Title: Re: 4U Maths Question Thread
Post by: katherine123 on June 28, 2016, 01:32:30 am
questions for harder ex 1
1. For any real x,y,z,u   prove that  (x^2+y^2)(z^2+u^2)>and equal to (xz+yu)^2
2.for any real numbers x,y,z,a,b,c, prove that (x^2+y^2+z^2)(a^2+b^2+c^2)>and equal to (ax+by+cz)^2
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on June 28, 2016, 09:18:58 am
questions for harder ex 1
1. For any real x,y,z,u   prove that  (x^2+y^2)(z^2+u^2)>and equal to (xz+yu)^2
2.for any real numbers x,y,z,a,b,c, prove that (x^2+y^2+z^2)(a^2+b^2+c^2)>and equal to (ax+by+cz)^2

Hey Katherine! The first one is fairly okay once you see it:



This is a result from Extension 2 I believe, but you can move the RHS over to the left and obtain this using by factorisation if you like:



Either way, the result is proven  ;D

Your second question is very very similar to your first! Expand both sides, cancel like terms, then move everything over to one side and factorise:



Which is definitely true  ;D that last factorisation might be a little tricky to see, let me know if this doesn't make sense  ;D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 28, 2016, 09:43:13 am
Here is one massive take out from Jamon's answer:

Where possible, you should feel safe to work BACKWARDS. Sometimes trying to work forwards just doesn't produce an obvious result, and you must start from somewhere else.


@Jamon however, because we're working backwards we don't want to use \implies, because that's for working forwards! We can always use \iff wherever appropriate, however preferably use \impliedby (left sided implication)
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on June 28, 2016, 10:17:26 am
Here is one massive take out from Jamon's answer:

Where possible, you should feel safe to work BACKWARDS. Sometimes trying to work forwards just doesn't produce an obvious result, and you must start from somewhere else.

@Jamon however, because we're working backwards we don't want to use \implies, because that's for working forwards! We can always use \iff wherever appropriate, however preferably use \impliedby (left sided implication)

Aha mate, I just needed an arrow  ;)
Title: Re: 4U Maths Question Thread
Post by: amandali on June 28, 2016, 11:08:01 pm
help with this ques thanks
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 29, 2016, 09:37:12 am
help with this ques thanks





Title: Re: 4U Maths Question Thread
Post by: amandali on June 30, 2016, 12:22:02 am
1)for a,b>0  show that (a+b)/2 greater than and equal to [(a^2+b^2)/2]^1/2

2) if 0<t<1, show that 1/2<1/(1+t)<1

3) 1/2(x^3+y^3)greater than equal to [(x+y)/2]^3
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 30, 2016, 10:25:44 am
1)for a,b>0  show that (a+b)/2 greater than and equal to [(a^2+b^2)/2]^1/2

2) if 0<t<1, show that 1/2<1/(1+t)<1

3) 1/2(x^3+y^3)greater than equal to [(x+y)/2]^3



_________________



One way to set this one out would be:
t < 1 therefore t-1 < 0
0 < t therefore 0 < 2t
Therefore t-1 < 0 < 2t
t+1 < 2 < 2t+2
and etc.
___________________



Title: Re: 4U Maths Question Thread
Post by: amandali on July 03, 2016, 02:18:32 am
(http://uploads.tapatalk-cdn.com/20160702/bed5022433ec1c9394cc43c5d79dcb47.jpg)(http://uploads.tapatalk-cdn.com/20160702/a7a5b5d4b26453b72bfdf6813c8a4335.jpg)

need help with these inequality ques thanks 
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 03, 2016, 09:35:15 am
(http://uploads.tapatalk-cdn.com/20160702/bed5022433ec1c9394cc43c5d79dcb47.jpg)(http://uploads.tapatalk-cdn.com/20160702/a7a5b5d4b26453b72bfdf6813c8a4335.jpg)

need help with these inequality ques thanks


___________________________


Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 03, 2016, 09:37:04 am
(http://uploads.tapatalk-cdn.com/20160702/bed5022433ec1c9394cc43c5d79dcb47.jpg)(http://uploads.tapatalk-cdn.com/20160702/a7a5b5d4b26453b72bfdf6813c8a4335.jpg)

need help with these inequality ques thanks


Title: Re: 4U Maths Question Thread
Post by: amandali on July 03, 2016, 09:25:26 pm
(http://uploads.tapatalk-cdn.com/20160703/7333561a2dbd2be6d9cf862b8b57f7e5.jpg)(http://uploads.tapatalk-cdn.com/20160703/a085abe0cc473ed77d56a4a7b42ea586.jpg)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 04, 2016, 12:02:42 am
(http://uploads.tapatalk-cdn.com/20160703/7333561a2dbd2be6d9cf862b8b57f7e5.jpg)(http://uploads.tapatalk-cdn.com/20160703/a085abe0cc473ed77d56a4a7b42ea586.jpg)

I dare say that heaps of information is missing somewhere to make the 3 variable AM-GM inequality easy to prove.

(Once again I'm not taking credit for anything above the long line. I asked for help on this one so please don't mind how he used a, b, c instead of x, y, z)










It was considerably harder than expected to convert LaTeX (different packages) between two different sources.
__________________________

The purpose of introducing the three-variable AM-GM inequality is to also help do the first question.

Let:
For eqn 1: x=a, y=b, z=c
For eqn 2: x=a, y=b, z=d
For eqn 3: x=a, y=c, z=d
For eqn 4: x=b, y=c, z=d

Title: 4U Maths Question Thread
Post by: amandali on July 04, 2016, 02:36:41 pm
(http://uploads.tapatalk-cdn.com/20160703/9493d5f5a0b89e5fe76172b22b46a9f9.jpg)


is my starting working correct and how do i progress

for resistance force, do i normally include k as a constant


(http://uploads.tapatalk-cdn.com/20160703/0e9e456f8a2dc778045d4cda51b5deca.jpg)
Title: Re: 4U Maths Question Thread
Post by: jakesilove on July 04, 2016, 05:11:33 pm
(http://uploads.tapatalk-cdn.com/20160703/9493d5f5a0b89e5fe76172b22b46a9f9.jpg)


is my starting working correct and how do i progress

for resistance force, do i normally include k as a constant


(http://uploads.tapatalk-cdn.com/20160703/0e9e456f8a2dc778045d4cda51b5deca.jpg)

Hey! I don't have time for a full solution (I'm sure Rui will jump in), so I'll just give some quick pointers. The start of the first question looks fine, great job! What you want to do next is integrate between definite limits. So, the right hand side (dv) can be integrated between 30 and v, whilst the left hand side can be integrated between time 0 and T. Once you've done that, you know that the highest point will be when v is zero (ie. the object is stationary). Set v=0 and an answer should pop out neatly! Great job setting up the question.

As for the next question, your problem lies in your integration. I don't know how your school does it, but using indefinite integrals, and finding the +C term each time, is generally more difficult than just using definite integrals, like we did above. Can you see which limits to use?

Hope this helps!

Jake
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 04, 2016, 05:17:50 pm
Hey! I don't have time for a full solution (I'm sure Rui will jump in), so I'll just give some quick pointers. The start of the first question looks fine, great job! What you want to do next is integrate between definite limits. So, the right hand side (dv) can be integrated between 30 and v, whilst the left hand side can be integrated between time 0 and T. Once you've done that, you know that the highest point will be when v is zero (ie. the object is stationary). Set v=0 and an answer should pop out neatly! Great job setting up the question.

As for the next question, your problem lies in your integration. I don't know how your school does it, but using indefinite integrals, and finding the +C term each time, is generally more difficult than just using definite integrals, like we did above. Can you see which limits to use?

Hope this helps!

Jake

Oops. Didn't see. Thanks for the bump haha
_____________________________
In regards to Q14







Emphasis: FR=kv2 generally speaking, but you were given the value for k here.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 04, 2016, 05:46:23 pm
(http://uploads.tapatalk-cdn.com/20160703/9493d5f5a0b89e5fe76172b22b46a9f9.jpg)


is my starting working correct and how do i progress

for resistance force, do i normally include k as a constant


(http://uploads.tapatalk-cdn.com/20160703/0e9e456f8a2dc778045d4cda51b5deca.jpg)
In regards to Q21




Title: Re: 4U Maths Question Thread
Post by: amandali on July 07, 2016, 12:28:48 pm
(http://uploads.tapatalk-cdn.com/20160706/38e78a8db6661af362df6bf51d890da0.jpg)



how to do part (ii) for ques 4

(http://uploads.tapatalk-cdn.com/20160706/83dcb292ab8221de867e6b4aa4965444.jpg)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 07, 2016, 01:24:54 pm
(http://uploads.tapatalk-cdn.com/20160706/38e78a8db6661af362df6bf51d890da0.jpg)



how to do part (ii) for ques 4

(http://uploads.tapatalk-cdn.com/20160706/83dcb292ab8221de867e6b4aa4965444.jpg)


Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 07, 2016, 01:33:50 pm
(http://uploads.tapatalk-cdn.com/20160706/38e78a8db6661af362df6bf51d890da0.jpg)



how to do part (ii) for ques 4

(http://uploads.tapatalk-cdn.com/20160706/83dcb292ab8221de867e6b4aa4965444.jpg)


Title: Re: 4U Maths Question Thread
Post by: katherine123 on July 07, 2016, 04:17:01 pm





why is the boundary between 0 and U/3  for the second integration not  between U and U/3   since the initial speed is U 
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 07, 2016, 05:42:54 pm
Oh shoot

why is the boundary between 0 and U/3  for the second integration not  between U and U/3   since the initial speed is U

Title: Re: 4U Maths Question Thread
Post by: amandali on July 08, 2016, 07:24:12 pm
ive done part i but not sure how to do part ii (2nd picture)

(http://uploads.tapatalk-cdn.com/20160708/71756e4a3c2bdd84ebe34469ae7ced09.jpg)

(http://uploads.tapatalk-cdn.com/20160708/59912487b04846b167e9be2ffc06904b.jpg)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 08, 2016, 08:45:04 pm
ive done part i but not sure how to do part ii (2nd picture)

(http://uploads.tapatalk-cdn.com/20160708/71756e4a3c2bdd84ebe34469ae7ced09.jpg)

(http://uploads.tapatalk-cdn.com/20160708/59912487b04846b167e9be2ffc06904b.jpg)




Title: Re: 4U Maths Question Thread
Post by: amandali on July 12, 2016, 05:09:20 pm
(http://uploads.tapatalk-cdn.com/20160712/5b5d7d5ff255902628d390fbc182bf94.jpg)

ive done part i and showed that time taken to reach max height as
  t=[V*ln(1+  u/V)]/g for part ii
but for not sure how to do part iii  (pic below)

(http://uploads.tapatalk-cdn.com/20160712/643eeaca702f5075d4c70317f693f513.jpg)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 12, 2016, 07:52:17 pm




This is the kind of question where the previous parts are not necessary.
Title: Re: 4U Maths Question Thread
Post by: amandali on July 12, 2016, 11:11:22 pm
how to do part v) and vi)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 13, 2016, 08:53:33 am
how to do part v) and vi)





____________________


Title: Re: 4U Maths Question Thread
Post by: HKSMASO6 on July 13, 2016, 09:18:57 pm
i have problems with these types of questions especially these types.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 13, 2016, 09:38:55 pm
i have problems with these types of questions especially these types.
Bringing back memories of CSSA or something. This was a damn tough reduction formula.





More to come
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 13, 2016, 10:00:43 pm





Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 13, 2016, 10:19:23 pm



Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 13, 2016, 10:27:57 pm


OMG that took forever!!!
Title: Re: 4U Maths Question Thread
Post by: jakesilove on July 14, 2016, 09:05:39 am


OMG that took forever!!!

I just can't explain how much I envy your natural Mathematical ability. Love you matey :)
Title: Re: 4U Maths Question Thread
Post by: HKSMASO6 on July 14, 2016, 04:07:32 pm
I just can't explain how much I envy your natural Mathematical ability. Love you matey :)
[tex]\text{Finally, rewrite the expression being clever enough as to use what we have in (v)}\\ \text{that is to say, use what we know!
OMG that took forever!!!
In an exam, if you see this type of question what is the best way to know what to do and should i leave these types of questions last
Title: Re: 4U Maths Question Thread
Post by: HKSMASO6 on July 14, 2016, 04:09:49 pm
AND RUI THANKS SOOOOO MUCH. YOUR A LEGEND
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on July 14, 2016, 04:19:27 pm
In an exam, if you see this type of question what is the best way to know what to do and should i leave these types of questions last

Didn't do 4U, but I'm a big fan of leaving the super hard stuff till last. Get the easy marks first in my opinion, mark maximisation!!  ;D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 14, 2016, 04:21:45 pm
In an exam, if you see this type of question what is the best way to know what to do and should i leave these types of questions last
Pretty sure this question was the last question on the paper that year so that's saying something.

i) is just your ordinary proof. Not much can be said here. (Combining the fractions could've helped make it clearer though)

With ii) you try to start on the LHS, but when doing a reduction formula you always need to look at the final result. Most notably, whether you have In, In-2 and what not. If you want to manipulate the reduction formula, anything that doesn't "look right" needs to be eliminated somehow. Anything that looks right (or promising), stays.

iii) is actually quite common - evaluating a reduction formula over and over again. Factoring out 2n would've been the hard part here. But if you really look at your working out and compare it to what you're trying to prove, it shouldn't be as less obvious.

iv) was really hard mostly cause of the summation. It's a bit hard to tell that by the linearity of integrals you can just swap their order. That aside, your result in part (iii) was obviously looking somewhat similar to what Cn was in part (iv), so you have to play with the algebra to make it work.

Seeing that interchange of order is probably something new to add to your toolbox.

Also note how I broke that question up into bits and pieces and built up. That's what you do when things are becoming truly messy.

v) required you to note the similarity between the RHS and part (iv). Provided anything you do doesn't break maths, all you do is try to find the similarity, and then force it out. Like how I did it (introduce the square root, then the integral symbol).

vi) is an example of testing how well 4u students know their algebra. This comes with a lot of experience - algebraic manipulation.
Title: Re: 4U Maths Question Thread
Post by: Septemberblue on July 18, 2016, 06:03:52 pm
Is it ok if someone could tell me how to recognise  and also memorise the difference of integration in 'Volumes' between finding the volume respecting to dy  and finding the volume respecting to dx. Like the height and the radius, because i get coinfuse when it gets to dy, Thank you :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 18, 2016, 06:30:35 pm
Is it ok if someone could tell me how to recognise  and also memorise the difference of integration in 'Volumes' between finding the volume respecting to dy  and finding the volume respecting to dx. Like the height and the radius, because i get coinfuse when it gets to dy, Thank you :)
What do you mean?

In 4U, because you have to go through the more rigorous definition of washers/shells you always need to draw a diagram and set everything up. The only things that can be arbitrarily said are.

Washers are used when you rotate perpendicular across an axis, whereas shells are used when you rotate parallel to it.

(Also, take care that IF you use washers, and you use dy to do the question, you have to use dx for shells with that same question. And vice versa.)
Title: Re: 4U Maths Question Thread
Post by: MarkThor on July 18, 2016, 07:56:32 pm
Hi Jake,
I couldn't find a general Math Extension 2 forum, so I hope it's fine to post more general questions about the course here.
I was wondering if you could give me some tips on how to study for Extension 2 in the lead up to trials. Are Past Papers the best thing to be focusing on? Or is there something else that I should be focusing on like specifically just the harder areas of the course?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 18, 2016, 07:58:10 pm
Hi Jake,
I couldn't find a general Math Extension 2 forum, so I hope it's fine to post more general questions about the course here.
I was wondering if you could give me some tips on how to study for Extension 2 in the lead up to trials. Are Past Papers the best thing to be focusing on? Or is there something else that I should be focusing on like specifically just the harder areas of the course?
Yes. You cannot EVER have a resource better than past papers. And you can never do enough of them either.

There are most certainly harder areas of the course. Take the entire harder 3U topic as a basic example. But even if you want to shift your focus to doing, say, lots of textbook revision on it, you still cannot get better than a student who does heaps of papers.

Reason being? Only through past papers, can you be trained into the correct ways of thought required for the exam.
_____________________

That being said, however, for every individual and every subject the best studying methods and techniques will vary. One way is to reconsolidate everything - go through your textbook (unfortunately I am still in the midst of composing the Ext 2 notes to advertise them), and do some review questions whilst you're at it. Then, you can just work through past papers as though you were doing homework, OR you can try doing one open book when you're ready. The further you go, the more confident you should be and not require your book. Towards the end, try to replicate full exam conditions.

And never be scared to use the solutions if you're stuck. if you don't work through the solutions (copy them out for practice), how can you learn?
Title: Re: 4U Maths Question Thread
Post by: jakesilove on July 18, 2016, 08:01:26 pm
Hi Jake,
I couldn't find a general Math Extension 2 forum, so I hope it's fine to post more general questions about the course here.
I was wondering if you could give me some tips on how to study for Extension 2 in the lead up to trials. Are Past Papers the best thing to be focusing on? Or is there something else that I should be focusing on like specifically just the harder areas of the course?

Hey Mark,

Definitely post any questions you want here! Specific, course related stuff, or general, study related stuff.
I would just be doing past papers, as many as you can get your hands on. Do them over and over, the tough stuff and the stuff you're comfortable with. If you find yourself really failing to understand a specific section, definitely go ahead and try to learn it again, but generally just do past papers.

In Ext 2, I also wrote a list of stupid mistakes I often made, or things that I needed to make sure to remember to do. I would have a list of questions I couldn't figure out, and felt that I should, and every day go through them again until I was totally confident with them. Basically, as I was doing past papers, I identified weaknesses and tried as best I could to fix them.

At the end of the day, past papers are your best friend. Don't focus on the harder areas specifically, because there are the same number of marks allocated to the whole course. Spending 10 hours on harder three unit might get you an additional one mark, but spending 10 hours on complex numbers might give you full marks in that section. Just do past papers, over and over.

Also, post questions that you have here! Get your class into the forums, so you can all help each other out, and learn through teaching (would seriously recommend this).

Catch you around the forums :)

Jake
Title: Re: 4U Maths Question Thread
Post by: jakesilove on July 18, 2016, 08:01:54 pm
Yes. You cannot EVER have a resource better than past papers. And you can never do enough of them either.

There are most certainly harder areas of the course. Take the entire harder 3U topic as a basic example. But even if you want to shift your focus to doing, say, lots of textbook revision on it, you still cannot get better than a student who does heaps of papers.

Reason being? Only through past papers, can you be trained into the correct ways of thought required for the exam.
_____________________

That being said, however, for every individual and every subject the best studying methods and techniques will vary. One way is to reconsolidate everything - go through your textbook (unfortunately I am still in the midst of composing the Ext 2 notes to advertise them), and do some review questions whilst you're at it. Then, you can just work through past papers as though you were doing homework, OR you can try doing one open book when you're ready. The further you go, the more confident you should be and not require your book. Towards the end, try to replicate full exam conditions.

And never be scared to use the solutions if you're stuck. if you don't work through the solutions (copy them out for practice), how can you learn?

Standard Rui, sneaking in right before me
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 18, 2016, 08:02:24 pm
Hey Mark,

Definitely post any questions you want here! Specific, course related stuff, or general, study related stuff.
I would just be doing past papers, as many as you can get your hands on. Do them over and over, the tough stuff and the stuff you're comfortable with. If you find yourself really failing to understand a specific section, definitely go ahead and try to learn it again, but generally just do past papers.

In Ext 2, I also wrote a list of stupid mistakes I often made, or things that I needed to make sure to remember to do. I would have a list of questions I couldn't figure out, and felt that I should, and every day go through them again until I was totally confident with them. Basically, as I was doing past papers, I identified weaknesses and tried as best I could to fix them.

At the end of the day, past papers are your best friend. Don't focus on the harder areas specifically, because there are the same number of marks allocated to the whole course. Spending 10 hours on harder three unit might get you an additional one mark, but spending 10 hours on complex numbers might give you full marks in that section. Just do past papers, over and over.

Also, post questions that you have here! Get your class into the forums, so you can all help each other out, and learn through teaching (would seriously recommend this).

Catch you around the forums :)

Jake
Correction: You wrote a list of fucking do this shit 
Title: Re: 4U Maths Question Thread
Post by: jakesilove on July 18, 2016, 08:03:59 pm
Correction: You wrote a list of fucking do this shit

Look, fair call. If you want to see an example of the list I wrote, you can find it on our Facebook page here!
Title: Re: 4U Maths Question Thread
Post by: MarkThor on July 18, 2016, 08:07:34 pm
Hahahah thanks Rui and Jake, loving the banter btw
Title: Re: 4U Maths Question Thread
Post by: jakesilove on July 18, 2016, 08:08:36 pm
Hahahah thanks Rui and Jake, loving the banter btw

Feel free to join the banter, good fun on a Monday night
Title: Re: 4U Maths Question Thread
Post by: MarkThor on July 18, 2016, 08:43:48 pm
Too true. And out of curiosity what school did you go to? Like was it one that normally achieves quite high?
Title: Re: 4U Maths Question Thread
Post by: jakesilove on July 18, 2016, 08:47:00 pm
Too true. And out of curiosity what school did you go to? Like was it one that normally achieves quite high?

I went to Emanuel, a tiny little school in Randwick. It was a good school, that often did achieve highly, but not like top 10 or anything like that!
Title: Re: 4U Maths Question Thread
Post by: MarkThor on July 18, 2016, 08:50:18 pm
Ah ok cool to know, thanks
Title: Re: 4U Maths Question Thread
Post by: amandali on July 19, 2016, 08:00:23 pm
(http://uploads.tapatalk-cdn.com/20160719/fcb0de7d958f3fd102564ce662b28824.jpg)

is there always linear relationship for triangles/trapezium only
ie x=mh+c   or is using similar triangles better way to find x   
Title: Re: 4U Maths Question Thread
Post by: DUX99.95StateRanker on July 19, 2016, 08:51:49 pm
How much time should one ideally spend studying 4u maths per week/day?
Just a rough idea

Im in yr 11 currently and doing great in ext 1 maths. Im considering doing ext 2.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 19, 2016, 08:54:38 pm
How much time should one ideally spend studying 4u maths per week/day?
Just a rough idea

Im in yr 11 currently and doing great in ext 1 maths. Im considering doing ext 2.
Varies for person to person.

I did most of my 4u work during my free periods at school so I can't speak soundly enough from experience. I just did everything that was for homework (until exam time). Jake might have pointers here.
Title: Re: 4U Maths Question Thread
Post by: jakesilove on July 19, 2016, 08:57:02 pm
Varies for person to person.

I did most of my 4u work during my free periods at school so I can't speak soundly enough from experience. I just did everything that was for homework (until exam time). Jake might have pointers here.

Towards the start, it'll be about the same as Ext 1. Until exams come around, and all of your maths efforts are placed on Ext 2, it really isn't THAT different I think. You just need to be consistent with your effort; putting hours in every week, doing questions, because if you don't understand a topic when you learn it, you sure as hell won't be able to learn it right before the exam. I can't put a number to it, except to say that it will be the same amount of study that you do in Ext 1 now, on a great day (ie. the maximum amount you study for Ext 1), consistently, every week. Sorry I can't give you more of a pointer; if you enjoy Maths, do it. If you hate it after a term, drop it. Simple as that!
Title: Re: 4U Maths Question Thread
Post by: amandali on July 19, 2016, 11:07:09 pm
(http://uploads.tapatalk-cdn.com/20160719/ca9b1373ea48dff0148039a5c2eaf54e.jpg)

is ii) correct
isnt it supposed to be the positive portion on the right side of curve (x>0) reflected about y axis with x<0 discarded if there s absolute sign around x and inside the function
Title: Re: 4U Maths Question Thread
Post by: Spencerr on July 19, 2016, 11:10:10 pm
Yes ii is correct.

Since the absolute values are around x + 1 not just x itself.
Therefore we take the positive of x + 1 which is x > -1 and flip that around, disregarding the negative side.
What you are referring to is just ln(|x|)
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on July 19, 2016, 11:48:37 pm
Yes ii is correct.

Since the absolute values are around x + 1 not just x itself.
Therefore we take the positive of x + 1 which is x > -1 and flip that around, disregarding the negative side.
What you are referring to is just ln(|x|)

And remember you can always jump onto a Graphing Calculator to confirm if you are unsure amandali ;D
Title: Re: 4U Maths Question Thread
Post by: amandali on July 21, 2016, 12:53:52 am
(http://uploads.tapatalk-cdn.com/20160720/77c08216d84d59a1262f29b5416c78c7.jpg)


(http://uploads.tapatalk-cdn.com/20160720/f1db9488679f4bd31ca34803932dcfaa.jpg)

for lAst ques  is it possible to have 2 answers since you dont know whether the angle is obtuse or acute  and the location of the vectors?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 21, 2016, 09:38:41 am

Title: Re: 4U Maths Question Thread
Post by: Spencerr on July 21, 2016, 09:49:06 am
Hey rui could you show us how to do the last question?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 21, 2016, 09:52:33 am
for lAst ques  is it possible to have 2 answers since you dont know whether the angle is obtuse or acute  and the location of the vectors?
I have no clue why edit post wasn't working.

Anyway, because you don't know whether beta is acute or obtuse you definitely cannot tell if a major or minor arc is being traced out. But you do know that the arc traced out by z goes from z1 to z2 in a counter-clockwise direction no matter what. Also, if you draw any diagram you can always label beta as the angle formed between the vectors (z-z1) and (z-z2)
Title: Re: 4U Maths Question Thread
Post by: jazza47 on July 23, 2016, 11:49:47 am
Help with questions c and d please and thank you  ;D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 23, 2016, 12:19:16 pm
Help with questions c and d please and thank you  ;D



Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 23, 2016, 12:40:33 pm
Help with questions c and d please and thank you  ;D



(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpshjsxbkws.png)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 23, 2016, 12:54:40 pm
(http://uploads.tapatalk-cdn.com/20160719/fcb0de7d958f3fd102564ce662b28824.jpg)

is there always linear relationship for triangles/trapezium only
ie x=mh+c   or is using similar triangles better way to find x
You could go for the x=mh+c method but I'm too used to similar triangles myself

I remember seeing it before but I forgot where I saw it. It looked visibly more appealing
Bumping this. I figured out the x=mh+c method. Yes, there always is one.

It's a matter of preference as which you want to use.

Just be careful of how you fit the line.
Title: Re: 4U Maths Question Thread
Post by: amandali on July 24, 2016, 12:46:52 am
(http://uploads.tapatalk-cdn.com/20160723/26e3e2470d4ad2672c9dc2a71ccccc02.jpg)

how to do part iii)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 24, 2016, 01:19:24 am


(http://i1164.photobucket.com/albums/q566/Rui_Tong/Screen%20Shot%202016-07-24%20at%201.13.03%20AM_zpsfiuqsl6b.png)


Title: Re: 4U Maths Question Thread
Post by: amandali on July 24, 2016, 09:20:47 am
(http://uploads.tapatalk-cdn.com/20160723/2b90b86e42ceb3ccc60f85d487efa7f8.jpg)

what did i do wrong for part d)
(http://uploads.tapatalk-cdn.com/20160723/07a98e47d05cbddee2bf4ce9b80907f4.jpg)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 24, 2016, 09:34:21 am
(http://uploads.tapatalk-cdn.com/20160723/2b90b86e42ceb3ccc60f85d487efa7f8.jpg)

what did i do wrong for part d)
(http://uploads.tapatalk-cdn.com/20160723/07a98e47d05cbddee2bf4ce9b80907f4.jpg)
Nothing. Your answer is the exact same as what WolframAlpha said.
Title: Re: 4U Maths Question Thread
Post by: amandali on July 24, 2016, 07:37:25 pm

Nothing. Your answer is the exact same as what WolframAlpha said.
(http://uploads.tapatalk-cdn.com/20160724/8e9d77f1623770b5bc8dd09b0623f70b.jpg)

the ans is this  but it seems correct too
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 24, 2016, 07:42:54 pm



Edit: Slight breakdown
Title: Re: 4U Maths Question Thread
Post by: Spencerr on July 26, 2016, 01:20:14 pm
Hey there!

How can I prove inequalities with composite functions
e.g. How do i prove this: sin(x) - x + (x^2) / 2 > 0
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 26, 2016, 02:55:59 pm
Hey there!

How can I prove inequalities with composite functions
e.g. How do i prove this: sin(x) - x + (x^2) / 2 > 0
This proof uses a tiny bit of intuition, which is mostly trained at the university level. It also really tests if you know that a monotonic function satisfies the horizontal line test: it is one-to-one. It's perfectly understandable to a 4U student however calls on a bit above their skills required.




Let me know if something doesn't make sense here.
Title: Re: 4U Maths Question Thread
Post by: katherine123 on July 26, 2016, 06:05:23 pm
Can i have a quick rundown of the conical pendulum?

Im not sure where normal (reaction force) comes in?
is the upward force(opposite mg) always (N+Tcosα) for ques like 4
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 26, 2016, 06:37:11 pm
Can i have a quick rundown of the conical pendulum?

Im not sure where normal (reaction force) comes in?
is the upward force(opposite mg) always (N+Tcosα) for ques like 4
For these questions you need to analyse where the forces actually are.

Note that a normal reaction only exists if the particle is touching a surface. If the particle is suspended in the air then this force is non-existent.

The normal reaction force is always perpendicular to the edge of the surface.

Always draw a vector diagram. Going by memory can lead to massive consequences. (http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI26072016_0001_zpsizrq8xil.jpg)
However as you can see, if you resolved the vectors correctly for THIS question, you do have N + T cos 60o = mg

Does this vector diagram make sense to you? If yes, then to do the question just correctly substitute your values. (Note that r = sqrt(3)/2 from trigonometry)

Also, for Q7, period is related to angular velocity by the relationship Period = 2π/ω

Q7 is not a conical pendulum question. It's just your ordinary (however really weird) circular motion question.
Title: Re: 4U Maths Question Thread
Post by: willy-boy on July 27, 2016, 02:38:13 pm
Was wondering if anyone had a set of all the proofs we need to know to replicate? I'm fine with most eqns, but when we get those fundamentals questions, you really just have to know... ta
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 27, 2016, 04:02:54 pm
Was wondering if anyone had a set of all the proofs we need to know to replicate? I'm fine with most eqns, but when we get those fundamentals questions, you really just have to know... ta
Most of the time if you know your actual content you will be able to make up the proofs on the spot without roting them. However from memory here are some important ones

- Induction proof for de Moivre's Theorem
- Centripetal acceleration for circular motion (examined in 2003 or 2004)
- 1+w+w2+...+wn-1 = 0 for roots of unity
- PS+PS' = 2a for an ellipse (and as for hyperbola)
- Multiple root theorem corollary: alpha is a root of multiplicity k for P(x) => it is a root of multiplicity k-1 for P'(x)
Title: Re: 4U Maths Question Thread
Post by: willy-boy on July 27, 2016, 04:30:48 pm
Bless your cotton socks. I don't mind the ol' De Moivre//Mechanics, but that is a great list. Thank you kindly
Title: Re: 4U Maths Question Thread
Post by: jakesilove on July 27, 2016, 11:14:51 pm
Bless your cotton socks. Thank you kindly

Love how polite these forums are
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on July 27, 2016, 11:51:50 pm
Love how polite these forums are

Indubitably  8)
Title: Re: 4U Maths Question Thread
Post by: Spencerr on July 28, 2016, 03:16:19 am
Hey Rui, I have two questions

For reduction integrals, how do you know which one to put as u and which one to put as dv?Is there any general rule to choose? And also when the question usually asks e.g. Show I (n+1) = n/n+1 I(n) when would you recommend subbing in LHS or RHS or is it up to individual intuition.

Title: 4U Maths Question Thread
Post by: RuiAce on July 28, 2016, 07:25:10 am
Hey Rui, I have two questions

For reduction integrals, how do you know which one to put as u and which one to put as dv?Is there any general rule to choose? And also when the question usually asks e.g. Show I (n+1) = n/n+1 I(n) when would you recommend subbing in LHS or RHS or is it up to individual intuition.
Q1: Even with reduction formulae, you usually follow the rule of LIATE with integration by parts (assuming that IBP is necessary).

With In = sinn(x), tann(x) and secn(x) you just need to know how to do them though.

Q2: In+1 generally requires a bit more intuition. If I was given it out of the blue I would start on In+1, but if halfway through my working out, or by looking ahead I see something really dodgy, I will turn back and start using In.

The intuition method is that when you use In, you get something like xn-1 * x2 which becomes xn+1

Edit: as an extra rule of thumb, if you have nothing to integrate with LIATE just integrate the 1 into x like you have to for ln(x)
Title: Re: 4U Maths Question Thread
Post by: katherine123 on July 28, 2016, 11:46:27 am
why is the vertical component Ncosx=mg instead of N+Tcosx=mg
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 28, 2016, 12:01:00 pm
why is the vertical component Ncosx=mg instead of N+Tcosx=mg
For what question? In Q4 I got the exact same as you got.

For Q7 the two tensions are acting horizontally and x = 0o implies cos(x) = 1
Title: Re: 4U Maths Question Thread
Post by: amandali on July 29, 2016, 09:14:00 pm
(http://uploads.tapatalk-cdn.com/20160729/4967a683358303f5ec34f36f6c6a4267.jpg)
how to do part ii)

(http://uploads.tapatalk-cdn.com/20160729/9fb440162049e0890723552c5d27b503.jpg)

how to do part ii)  isnt  (x^4+1)(x^2+1)  irreducible over real fields



for the rectangular hyperbola, am i supposed to memorise the focus (+-root2*c, +-root2*c), directices x+y=-+ root2*c, vertex (c,c)    or is there a simple derivation method
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 29, 2016, 09:51:13 pm
(http://uploads.tapatalk-cdn.com/20160729/4967a683358303f5ec34f36f6c6a4267.jpg)
how to do part ii)

(http://uploads.tapatalk-cdn.com/20160729/9fb440162049e0890723552c5d27b503.jpg)

how to do part ii)  isnt  (x^4+1)(x^2+1)  irreducible over real fields



for the rectangular hyperbola, am i supposed to memorise the focus (+-root2*c, +-root2*c), directices x+y=-+ root2*c, vertex (c,c)    or is there a simple derivation method
I can't make sense out of that (ii) because mixing Z and omega around with z and w is doing my head in. If Z=z and omega=w then the expression would be false because the LHS has a term in Z2, Z3 and Z4 but the RHS does not. So those 1's would also have to be Z's.
___________________________


___________________________
You can informally prove it out of the fact that xy=1/2 a2 is just x2-y2=a2 rotated by 45o. But yeah I would just memorise it and take it for granted. Keep in mind that the eccentricity of a rectangular hyperbola is always sqrt(2).

If you ask me though, I have never seen a question examining that. I don't think it's within the scope of the syllabus technically speaking.
Title: Re: 4U Maths Question Thread
Post by: katherine123 on July 30, 2016, 09:34:50 pm
For what question? In Q4 I got the exact same as you got.

For Q7 the two tensions are acting horizontally and x = 0o implies cos(x) = 1


forgot to attach the ques
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 30, 2016, 09:46:05 pm

forgot to attach the ques

The diagram is a bit badly drawn because that should be a semi-circle shape, not the shape of a parabolic arc.


Title: Re: 4U Maths Question Thread
Post by: amandali on July 30, 2016, 09:57:08 pm
(http://uploads.tapatalk-cdn.com/20160730/2f3156eb9de64ae93a3e59e259b8919e.jpg)


how do to this ques?  am i required to know how to prove this?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 30, 2016, 10:11:15 pm
(http://uploads.tapatalk-cdn.com/20160730/2f3156eb9de64ae93a3e59e259b8919e.jpg)


how do to this ques?  am i required to know how to prove this?

Title: Re: 4U Maths Question Thread
Post by: katherine123 on August 03, 2016, 05:46:36 pm
for ques c is (0,0) excluded?
Title: Re: 4U Maths Question Thread
Post by: jakesilove on August 03, 2016, 06:02:26 pm
for ques c is (0,0) excluded?

For the first question, I would do this by deduction. We know that p cannot equal zero. This is because, if it did, the equation would be able to be reduced by difference of two cubes. This will produce one real root, and two unreal roots (but the question specifies that there are three real roots). Therefore, the answer cannot be C or D (allows p to be zero).

Now, personally I looked to the derivative. If we want there to be three real roots, there needs to be two real turning points. If we differentiate the function, we get



If we set this equal to zero, we know that



This will only have two real x solutions if p is negative. Therefore the answer is B.

Jake
Title: Re: 4U Maths Question Thread
Post by: jakesilove on August 03, 2016, 06:05:12 pm
for ques c is (0,0) excluded?

For your second question, I don't believe that z=0+0i needs to be excluded. The modulus of z is 0, and if you add z and its conjugate you get zero. So, the equation becomes



Which is true.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 03, 2016, 06:48:44 pm
for ques c is (0,0) excluded?
Yeah gonna back up Jake on the second one because



I guess whilst unnecessary, if I were to improve on his answer for the first question, well...

I think I would've done the same method.
Title: Re: 4U Maths Question Thread
Post by: Neutron on August 10, 2016, 01:35:46 am
Hi! I should be sleeping but,

With graphs like these x^2y^2 - x^2 + y^2=0 how would you know what the shape is? Like even after finding stationary points etc, thank you!

Title: Re: 4U Maths Question Thread
Post by: jakesilove on August 10, 2016, 01:59:16 am
Hi! I should be sleeping but,

With graphs like these x^2y^2 - x^2 + y^2=0 how would you know what the shape is? Like even after finding stationary points etc, thank you!

Personally, I would find the stationary points, any intercepts, and then make things up based on the question. If you can find limits, try doing that; if you can't, forget it. For a graph like this, where I would have no idea where to even begin, I would likely just plot a bunch of points until I feel comfortable sketching the graph. Let x=0, 1, -1 etc. and solve for y. Plot some points and sketch away! Sorry that I can't be of more use, but plotting points is a very valid method, even in Ext 2.

GO TO SLEEP
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 10, 2016, 08:20:03 am
Hi! I should be sleeping but,

With graphs like these x^2y^2 - x^2 + y^2=0 how would you know what the shape is? Like even after finding stationary points etc, thank you!
You should refrain from thinking about the shape of the graph because for something as ugly as that the shape does not have a name and in fact, it's virtually impossible to predict what it is.
Title: Re: 4U Maths Question Thread
Post by: Neutron on August 10, 2016, 11:38:58 am
Ahh yeah it came up in half yearlies, fingers crossed that it won't come up again :/ But anyhow, with complex loci such as:

arg(z+1)-arg(z-1)=pi/3

It looks like an arc of a circle, but how do you know whether to draw this arc above or below the x axis? Thanks ^^
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 10, 2016, 12:54:42 pm
Ahh yeah it came up in half yearlies, fingers crossed that it won't come up again :/ But anyhow, with complex loci such as:

arg(z+1)-arg(z-1)=pi/3

It looks like an arc of a circle, but how do you know whether to draw this arc above or below the x axis? Thanks ^^
For the locus arg(z-z1)-arg(z-z2)=α

I've found from experimentation that you always go in an anti-clockwise direction from z1 to z2

GeoGebra simulation to experiment with:
https://www.geogebra.org/m/AgXWc6mD
Title: Re: 4U Maths Question Thread
Post by: massive on August 13, 2016, 11:29:39 am
Hey guys for part iii of this question how do you get what b equals to. I've attached the worked solution as well but i don't understand where the third line comes from (how the p^4 just factorises out? :S). Thanks!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 13, 2016, 12:28:45 pm
Hey guys for part iii of this question how do you get what b equals to. I've attached the worked solution as well but i don't understand where the third line comes from (how the p^4 just factorises out? :S). Thanks!
b is the product of the roots. Note that the general quadratic Ax2+Bx+C has sum of roots α+β=-B/A, αβ=C/A

The p4 is pulled out by simply pulling p from the first bracket and p3 from the second.
Title: Re: 4U Maths Question Thread
Post by: massive on August 13, 2016, 04:17:44 pm
how do you do this guys?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 13, 2016, 04:21:15 pm
how do you do this guys?

Title: Re: 4U Maths Question Thread
Post by: massive on August 13, 2016, 05:20:31 pm
after proving that lhs=rhs do you use t formula? because it says "hence" :S
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 13, 2016, 05:22:39 pm
after proving that lhs=rhs do you use t formula? because it says "hence" :S

Title: Re: 4U Maths Question Thread
Post by: massive on August 13, 2016, 05:30:28 pm


WOAAAH that's so smart lol thnx!
Title: Re: 4U Maths Question Thread
Post by: massive on August 13, 2016, 07:28:51 pm
how do you get this guys??
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 13, 2016, 07:48:13 pm
how do you get this guys??

Title: Re: 4U Maths Question Thread
Post by: massive on August 13, 2016, 09:14:41 pm
how do you do this?
Title: Re: 4U Maths Question Thread
Post by: jakesilove on August 13, 2016, 09:37:00 pm
how do you do this?

This was just a brute-force question, hope the steps make sense!

(http://i.imgur.com/DkMZuK5.png)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 13, 2016, 09:49:32 pm
This was just a brute-force question, hope the steps make sense!

(http://i.imgur.com/DkMZuK5.png)
Woooaaaahhh Jake got a 4U question before me

Actually I was just studying for my discrete maths test.
Title: Re: 4U Maths Question Thread
Post by: jakesilove on August 13, 2016, 10:18:11 pm
Woooaaaahhh Jake got a 4U question before me

Actually I was just studying for my discrete maths test.

You're just too good at LaTeX mate, I don't stand a chance
Title: Re: 4U Maths Question Thread
Post by: massive on August 13, 2016, 10:24:44 pm
guys how do you do this?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 13, 2016, 10:46:04 pm
guys how do you do this?









Title: Re: 4U Maths Question Thread
Post by: massive on August 13, 2016, 11:17:58 pm


How do you get that?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 13, 2016, 11:23:52 pm
How do you get that?
Title: Re: 4U Maths Question Thread
Post by: massive on August 14, 2016, 07:51:46 am
Yo guys how do you do part iii. Rui I tried doing the method you did by taking it to I0 but i ended up with:
1/(In) = (n+2)(n+1), am i doing something wrong?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 14, 2016, 08:49:24 am
Yo guys how do you do part iii. Rui I tried doing the method you did by taking it to I0 but i ended up with:
1/(In) = (n+2)(n+1), am i doing something wrong?
I subbed in values for n on WolframAlpha. That question has a mistake; it is (n+1)(n+2)/2 as you said.
Title: Re: 4U Maths Question Thread
Post by: massive on August 14, 2016, 01:43:23 pm
hey can someone get this? i got the answer except with a 1/2 infront :S
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 14, 2016, 02:06:55 pm
hey can someone get this? i got the answer except with a 1/2 infront :S


Title: Re: 4U Maths Question Thread
Post by: massive on August 14, 2016, 02:29:39 pm
Thanks Rui!

Also how do do you do the q attached?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 14, 2016, 03:22:39 pm
Thanks Rui!

Also how do do you do the q attached?






Title: Re: 4U Maths Question Thread
Post by: birdwing341 on August 14, 2016, 09:49:54 pm
Hello!
Anyone got some top tips for how to deal with harder projectile motion questions that you don't know how to do?
Any help would be much appreciated!!
Thanks in advance :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 14, 2016, 09:56:13 pm
Hello!
Anyone got some top tips for how to deal with harder projectile motion questions that you don't know how to do?
Any help would be much appreciated!!
Thanks in advance :)
It is extremely rare that a projectile question should appear in the 4U exam. As far as my memory goes, the only time it was asked was sometime during 2003-2006 where they combined it with resisted motion.

Despite this, projectile motions are a simple matter of:
Early parts - Using relevant definitions as appropriate (e.g. y=0 for a time of flight) and just dealing with messy algebra.
Late parts - Expect to use anything you know. One can never predict what they will need on the day. (However, fact is after a truly sufficient amount of past papers, you'd be trained to adjust to the new ways of thought and thus be better equipped to handle the peculiar questions.)

The biggest tips with projectile motion are basically to know your fundamentals and then expect the unexpected. It's much easier to answer a question than to give generic tips, especially for questions on topics such as this.
Title: Re: 4U Maths Question Thread
Post by: massive on August 15, 2016, 09:59:45 am
hey guys is ln(x/2) the same as ln( (tanx/2 - 1) / (tanx/2) )? Also if it is true, how? Thanks!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 15, 2016, 10:04:10 am
hey guys is ln(x/2) the same as ln( (tanx/2 - 1) / (tanx/2) )? Also if it is true, how? Thanks!!
Nope.

Change the formatting a bit and still nope.
Title: Re: 4U Maths Question Thread
Post by: massive on August 15, 2016, 10:29:15 am
oh then how do you do the q attached. my ans was ln( (tanx/2 - 1) / (tanx/2) ) but the answer is ln(x/2). Thanks.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 15, 2016, 10:46:52 am
oh then how do you do the q attached. my ans was ln( (tanx/2 - 1) / (tanx/2) ) but the answer is ln(x/2). Thanks.


Title: Re: 4U Maths Question Thread
Post by: massive on August 15, 2016, 10:49:09 am
sweeeet thanks Rui!!

Also can i just ask another question. How do you do part ii for the q attached. Obviously you have to use part i because 'hence' but how? What i was thinking was to use by parts but I don't think that's gonna work.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 15, 2016, 10:57:06 am
sweeeet thanks Rui!!

Also can i just ask another question. How do you do part ii for the q attached. Obviously you have to use part i because 'hence' but how? What i was thinking was to use by parts but I don't think that's gonna work.
I think I've seen this question before. It's been tagged as faulty.

Yeah, IBP isn't going to work because to integrate the sqrt(x^2-4) we need an x in the numerator. But by introducing that x we just made the question substantially more messy.

Title: Re: 4U Maths Question Thread
Post by: massive on August 15, 2016, 11:25:12 am
It's been tagged as faulty.

So the questions wrong?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 15, 2016, 01:28:47 pm
So the questions wrong?

Yeah. Pretty sure it should ∫xLn(x2-4)/√(x2-4)dx
Title: Re: 4U Maths Question Thread
Post by: massive on August 15, 2016, 02:35:23 pm
how do you do this man ?
Title: Re: 4U Maths Question Thread
Post by: jakesilove on August 15, 2016, 03:03:16 pm
how do you do this man ?

Hey!

My solution is below. A bit of a trekk for an integration question, but really it was just brute force. I hope my working makes sense!

(http://i.imgur.com/5MQqhO3.png)
(http://i.imgur.com/VVEzX15.png)

Jake
Title: Re: 4U Maths Question Thread
Post by: massive on August 15, 2016, 04:03:35 pm
Thanks Jake!

Guys for part ii of the question attached. How do you do get the reduction In ?
Title: Re: 4U Maths Question Thread
Post by: jakesilove on August 15, 2016, 05:35:11 pm
Thanks Jake!

Guys for part ii of the question attached. How do you do get the reduction In ?

Hey,

I'm on the run so won't be able to write up a full solution right now, but I think you have to break up the integral into the thing in part i), and then whatever is left over (I think x^(n-1)). Then, integrate using i), differentiate, preso the answer should pop out!

Jake
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 15, 2016, 06:24:21 pm
Thanks Jake!

Guys for part ii of the question attached. How do you do get the reduction In ?


Title: Re: 4U Maths Question Thread
Post by: massive on August 15, 2016, 06:47:18 pm
Thanks Rui! yeah i eventually got it :P

Guys check out the integration q attached. Its pretty interesting never seen anything like it. how does one go about solving something like that. At first I thought partial fractions but i don't think that's going to work. Im guessing substitution, idek. Any thought??
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 15, 2016, 08:14:28 pm
Thanks Rui! yeah i eventually got it :P

Guys check out the integration q attached. Its pretty interesting never seen anything like it. how does one go about solving something like that. At first I thought partial fractions but i don't think that's going to work. Im guessing substitution, idek. Any thought??
(http://uploads.tapatalk-cdn.com/20160815/8fc5c3656b474dce408513c10d76ef78.jpg)

I briefly looked at it and first deduced that it had gone well beyond the scope of the HSC as any exponentiation on itself is tremendously hard to handle. I.e. x^x

But after spending 4 more minutes on it I realised wait no it's too ridiculous
Title: Re: 4U Maths Question Thread
Post by: massive on August 15, 2016, 08:18:47 pm
(http://uploads.tapatalk-cdn.com/20160815/8fc5c3656b474dce408513c10d76ef78.jpg)

I briefly looked at it and first deduced that it had gone well beyond the scope of the HSC as any exponentiation on itself is tremendously hard to handle. I.e. x^x

But after spending 4 more minutes on it I realised wait no it's too ridiculous

Thanks mate!!
Title: Re: 4U Maths Question Thread
Post by: massive on August 15, 2016, 08:28:02 pm
how do you do part ii guys?
Title: 4U Maths Question Thread
Post by: RuiAce on August 15, 2016, 08:29:52 pm
how do you do part ii guys?
Hint (since I'm having dinner right now)

Let x=tan(theta)

Inspiration: because the derivative of tan is sec squared and because 1+tan^2=sec^2 AND the boundaries were too convenient
Title: Re: 4U Maths Question Thread
Post by: massive on August 15, 2016, 08:42:45 pm
Hint (since I'm having dinner right now)

Let x=tan(theta)

Inspiration: because the derivative of tan is sec squared and because 1+tan^2=sec^2 AND the boundaries were too convenient

But then how do you get a reduction formula using that :S
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 15, 2016, 09:09:35 pm
(http://uploads.tapatalk-cdn.com/20160815/b5d444dde1be43097e949d2d97b75f26.jpg)
Title: Re: 4U Maths Question Thread
Post by: massive on August 15, 2016, 09:38:44 pm
far out thanks!!

Also how do you do part ii for this q?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 15, 2016, 09:55:40 pm
far out thanks!!

Also how do you do part ii for this q?


Title: Re: 4U Maths Question Thread
Post by: massive on August 15, 2016, 10:02:14 pm
Thank you Rui!!!

I have another question. Part ii - i don't get how you do it, i tried using part i but there's also a B, what do you with that?

Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 15, 2016, 10:07:48 pm
Thank you Rui!!!

I have another question. Part ii - i don't get how you do it, i tried using part i but there's also a B, what do you with that?

Title: Re: 4U Maths Question Thread
Post by: massive on August 15, 2016, 10:19:50 pm



Woaah thats so cool! This is probably gonna be a dumb q, but how do you know you're left with sin2nx, like how do you know that that term doesn't get cancelled out?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 15, 2016, 10:40:13 pm
Woaah thats so cool! This is probably gonna be a dumb q, but how do you know you're left with sin2nx, like how do you know that that term doesn't get cancelled out?
Because the last term has nothing to pair up and vanish with.

Neither does the first term. It just so happens that the first term was sin(0) though.
Title: Re: 4U Maths Question Thread
Post by: massive on August 15, 2016, 10:44:41 pm
Oh yeaah i get it!

Hey i just have one more question, last question for the night (hopefully). How do you do it? :P
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 15, 2016, 10:59:28 pm
Oh yeaah i get it!

Hey i just have one more question, last question for the night (hopefully). How do you do it? :P


Also, pretty sure that 1 in the boundary is supposed to be a t.
Title: Re: 4U Maths Question Thread
Post by: massive on August 16, 2016, 02:04:52 pm
Guys when doing volumes, when do you split the area into two sections and then find the volume of each and add the two separate volumes to get the total volume?
Title: Re: 4U Maths Question Thread
Post by: massive on August 16, 2016, 02:54:20 pm
how do you do this q guys?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 16, 2016, 06:02:51 pm
Guys when doing volumes, when do you split the area into two sections and then find the volume of each and add the two separate volumes to get the total volume?
Don't quite understand this question. I analyse what areas are relevant before I analyse the relevant volume. And the methods for area analysis should be familiar from 2U.
how do you do this q guys?
Before I have a go at this (lacking materials), have you drawn out a(n appropriate) diagram?
Title: Re: 4U Maths Question Thread
Post by: massive on August 16, 2016, 06:43:07 pm
.Before I have a go at this (lacking materials), have you drawn out a(n appropriate) diagram?

Yeah, but idk where the pi came from
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 16, 2016, 07:23:40 pm






If your diagram happened to be wrong and consequently you couldn't obtain the correct integral, you might want to read what I said at the start a few times. Because the π comes out of direct integration.
Title: Re: 4U Maths Question Thread
Post by: massive on August 16, 2016, 08:00:19 pm
OHHH mate thanks i get it!!

Also need help in the one attached. Wth is it even ?? :S
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 16, 2016, 08:34:34 pm
OHHH mate thanks i get it!!

Also need help in the one attached. Wth is it even ?? :S
This is ridiculous. I'll leave the proof of the volume of a square pyramid as an exercise. Note: The volume of any pyramid is V = 1/3 A h, where A is the area of the base and h is the vertical height.(http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI16082016_zpssm1jjg34.jpg)
Title: Re: 4U Maths Question Thread
Post by: massive on August 16, 2016, 08:48:21 pm




do we have to put this line for hsc (the line with the sigma notation), like do we lose marks if we don't, because in our school our teacher never told us to put this line in :S
Title: Re: 4U Maths Question Thread
Post by: massive on August 16, 2016, 08:49:17 pm
This is ridiculous. I'll leave the proof of the volume of a square pyramid as an exercise. Note: The volume of any pyramid is V = 1/3 A h, where A is the area of the base and h is the vertical height.(http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI16082016_zpssm1jjg34.jpg)


Also thanks heaps for this!!
Title: Re: 4U Maths Question Thread
Post by: jakesilove on August 16, 2016, 08:52:14 pm
do we have to put this line for hsc (the line with the sigma notation), like do we lose marks if we don't, because in our school our teacher never told us to put this line in :S

Nup it's fine if you go straight to the integral :)
Title: Re: 4U Maths Question Thread
Post by: massive on August 16, 2016, 09:00:35 pm
guys how do you do this :S
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 16, 2016, 11:42:42 pm
guys how do you do this :S






Title: Re: 4U Maths Question Thread
Post by: massive on August 17, 2016, 01:09:42 am
Thanks Rui!!

Guys the question attached is so simple but I can't seem to get it :/
The answer to part i is: 4pi.a.f(x).dx   How?? its so weird
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 17, 2016, 08:38:39 am
Thanks Rui!!

Guys the question attached is so simple but I can't seem to get it :/
The answer to part i is: 4pi.a.f(x).dx   How?? its so weird


Nup it's fine if you go straight to the integral :)
Are you sure?
Title: Re: 4U Maths Question Thread
Post by: massive on August 17, 2016, 10:53:18 am
Thanks Rui :D

You see for the q attached, i keep getting 32 for my ans, but the actual answer is 24 (part ii) :S
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 17, 2016, 11:25:25 am
Thanks Rui :D

You see for the q attached, i keep getting 32 for my ans, but the actual answer is 24 (part ii) :S
I also got 32
Title: Re: 4U Maths Question Thread
Post by: massive on August 17, 2016, 02:36:55 pm
Guys how do you do this badboy ?  :-\
(just part ii)
Title: Re: 4U Maths Question Thread
Post by: massive on August 17, 2016, 02:55:33 pm
wait part ii is easy, because you just use part a (which is attached)

But i have a new problem, how in the world do you do part a. ??
Title: Re: 4U Maths Question Thread
Post by: znaser on August 17, 2016, 03:20:52 pm
I tried following the answers but yh this question is something. Thanks in advance :)
Title: Re: 4U Maths Question Thread
Post by: massive on August 17, 2016, 04:13:56 pm
Guys how do you even do this question???? :S
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 17, 2016, 05:00:05 pm
wait part ii is easy, because you just use part a (which is attached)

But i have a new problem, how in the world do you do part a. ??





Use this GeoGebra simulation to help you.

I will leave the area computation as your exercise
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 17, 2016, 05:09:52 pm
I tried following the answers but yh this question is something. Thanks in advance :)
What's the confusion? Is there a particular part about the answers that makes no sense? (Sorry, but your question is generic and I don't know where to focus my response on.)

Line 1: Cosine double angle formula: cos(2x) = 2cos2(x) - 1 so cos(x) = 2cos2(x/2) - 1
Line 1 second part: Expand out the compound angle

Line 2: As with line 1 albeit with minus not plus

Line 3: Sums the above. This makes sense - we are just deriving something similar to product to sum. The inspiration to use squares and not to the power of 1 is simply because the result to prove involves squares.

Line 4: They tell you what they did. This isn't exactly classic in the HSC (classic for me) but you should anticipate it. Because how else do you turn an equation into an inequality

Line 5: Rearrange
_________

(ii) is using the inequality in (i) over and over again. This is not uncommon: The proof of the AM-GM inequality for 4 variables is just the AM-GM for 2 variables using the same method.

I.e. (x+y)/2 ≤ √(xy) and (a+b+c+d)/4 ≤ 4√(abcd)
_________

There's hints in (iii) to use part (ii). What was perhaps not obvious was that 1/3(a+b+c) also had to be there. But look at it closely - you proved something involving multiplying 4 different cosines. You had to bring it out somehow.

Their typeface is pretty tedious to read though; I'll give that.
Title: Re: 4U Maths Question Thread
Post by: massive on August 17, 2016, 08:55:55 pm
guys how do you do this, been stuck on it for ages
Title: Re: 4U Maths Question Thread
Post by: znaser on August 18, 2016, 12:14:47 pm
Sorry should've been more specific. But yeah it was how to tackle part 2. Thanks, really helped :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 18, 2016, 02:33:47 pm
Sorry should've been more specific. But yeah it was how to tackle part 2. Thanks, really helped :)
If part 2 was the hard part (which is understandable), I recommend you try attempting the example I provided. See if you can use a similar method to prove the 4 variable AM-GM.
Title: Re: 4U Maths Question Thread
Post by: massive on August 18, 2016, 06:30:16 pm
Btw did anyone manage to figure out the two volumes questions I posted previously, I still don't get them  :( . thanks!
Title: Re: 4U Maths Question Thread
Post by: jakesilove on August 18, 2016, 09:18:57 pm
Btw did anyone manage to figure out the two volumes questions I posted previously, I still don't get them  :( . thanks!

Just because of the volume (lol) of questions you've been asking, do you think you could show us some working out before we tackle the questions you post? Just so we can see where we can help you specifically
Title: Re: 4U Maths Question Thread
Post by: massive on August 19, 2016, 01:56:23 pm
Just because of the volume (lol) of questions you've been asking, do you think you could show us some working out before we tackle the questions you post? Just so we can see where we can help you specifically

Oh yeah fair enough, i'll start doing that from now on. Its just that for those 2 questions i posted I don't even know where to start  :-[ .
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 20, 2016, 07:25:32 pm
(http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI20082016_zps5khyvj5v.jpg)
I can't guarantee that this answer is correct.

I would like to know what is the source of these questions?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 20, 2016, 11:42:39 pm
Guys how do you even do this question???? :S
You need to draw diagrams.
(i)



(ii)
This is a question you should've done despite the difficulty of part (i) because it is much more standard.

(iii)


Not doing (iv). That's definitely not in the syllabus (not that slanted slicing ever was). But I reckon it can be described by visualising the slice in your head.
Title: Re: 4U Maths Question Thread
Post by: amandali on August 25, 2016, 02:48:12 pm
need help with these ques
a)a disc is free to rotate in a horizontal plane about an axis through its centre O.  A small object P is placed on disc so that OP=0.2m. Contact between the particle and the disc is rough and limiting friction of the particle can experience is 0.5 times the normal reaction between particle and disc. the disc then begins to rotate. Find angular v. of the disc when particle is about to slip.

b)to start a lawn mower, a green keeper keeper needs to pull a chord wound around a grooves wheel of diameter 12cm at speed 147cm/s. find angular velocity of wheel in revolutions per min
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 25, 2016, 05:24:30 pm
need help with these ques
a)a disc is free to rotate in a horizontal plane about an axis through its centre O.  A small object P is placed on disc so that OP=0.2m. Contact between the particle and the disc is rough and limiting friction of the particle can experience is 0.5 times the normal reaction between particle and disc. the disc then begins to rotate. Find angular v. of the disc when particle is about to slip.

b)to start a lawn mower, a green keeper keeper needs to pull a chord wound around a grooves wheel of diameter 12cm at speed 147cm/s. find angular velocity of wheel in revolutions per min
b) looks like a question that just involves tackling with units.

The radius is 6cm, so the angular velocity is given
ω = v/r = 147/6 rad s-1

Recall that 1 radian is only 1/(2π) of a revolution, and that there are 60 seconds in a minute. Hence, we multiply by 60/2π = 30/π to convert to rev/min

ω = 735/π rev min-1 (which you can plug into your calculator)
Title: Re: 4U Maths Question Thread
Post by: jakesilove on August 25, 2016, 05:53:45 pm
need help with these ques
a)a disc is free to rotate in a horizontal plane about an axis through its centre O.  A small object P is placed on disc so that OP=0.2m. Contact between the particle and the disc is rough and limiting friction of the particle can experience is 0.5 times the normal reaction between particle and disc. the disc then begins to rotate. Find angular v. of the disc when particle is about to slip.

b)to start a lawn mower, a green keeper keeper needs to pull a chord wound around a grooves wheel of diameter 12cm at speed 147cm/s. find angular velocity of wheel in revolutions per min

Hey!

For a) I understood this question to mean the following:

The particle is on spinning disk, at radius 0.2m
The friction keeping the particle from slipping is 0.5 the normal force

From these two pieces of information, we can easily equate forces. The force 'inwards' (ie. keeping the particle in place) is going to be the friction force. The force pushing the particle 'outwards' (ie. making the particle slip) is going to be centripetal force. Therefore, the particle will be just about to slip when the inwards force equals the outwards force.






Hopefully I interpreted the question correctly! I'm not sure what else you could do.

Jake
Title: Re: 4U Maths Question Thread
Post by: znaser on August 26, 2016, 09:25:31 pm
Hi. So i follow this question (with ofc additional workout since the solution skips a lot of steps) right up till this fraction: [ (a+b)^2 - [(a+b)^2]/2 ]/[...]
I get wats going on with the denominator since: (a^1/2 - b^1/2)^2 ≥ 0 ... (a^2 + b^2) ≥ 2ab ... [(a^2 + b^2)^2] / 4 ≥ (ab)^2
therefore [a^2 + b^2] / (ab)^2 ≥ (a^2 + b^2) / [(a^2 + b^2)^2 / 4] since the denominator of the former is smaller than the latter. But I don't get wats going on with the numerator of the first fraction I typed. Thanks :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 26, 2016, 09:34:39 pm
Hi. So i follow this question (with ofc additional workout since the solution skips a lot of steps) right up till this fraction: [ (a+b)^2 - [(a+b)^2]/2 ]/[...]
I get wats going on with the denominator since: (a^1/2 - b^1/2)^2 ≥ 0 ... (a^2 + b^2) ≥ 2ab ... [(a^2 + b^2)^2] / 4 ≥ (ab)^2
therefore [a^2 + b^2] / (ab)^2 ≥ (a^2 + b^2) / [(a^2 + b^2)^2 / 4] since the denominator of the former is smaller than the latter. But I don't get wats going on with the numerator of the first fraction I typed. Thanks :)

Title: Re: 4U Maths Question Thread
Post by: znaser on August 26, 2016, 09:41:57 pm



Thank you!  ;D
Title: Re: 4U Maths Question Thread
Post by: Mei2016 on August 27, 2016, 10:26:10 pm
Hi,

I'm stuck on a few questions from paper 1999
Q2c) why is Re(z)= -1 an empty circle?
Q2e) isn't turning anticlockwise x i, and turning clockwise x -i ?
So answer should be -->AP= -Z(1)i? [But ANS is -->AP =Z(1)i]

(I tried to upload the pics of the question, but it didn't work)

Mod edit: Merged posts. Try to keep everything together :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 27, 2016, 10:35:24 pm
Hi,

I'm stuck on a few questions from paper 1999
Q2c) why is Re(z)= -1 an empty circle?
Q2e) isn't turning anticlockwise x i, and turning clockwise x -i ?
So answer should be -->AP= -Z(1)i? [But ANS is -->AP =Z(1)i]

(I tried to upload the pics of the question, but it didn't work)

Mod edit: Merged posts. Try to keep everything together :)
Are we talking about the 1999 HSC exam? Because Q2c) of that does not have Re(z).

If the diagram failed to upload, what paper are you trying to refer to
Title: Re: 4U Maths Question Thread
Post by: Mei2016 on August 27, 2016, 11:12:11 pm
Yep, it's the HSC 1999 paper. So you have to sketch the region on the Argand diagram so the x-axis would be Re(z) and the y-axis would be Im(z) right?

(I'm new to the site, so this is taking me a while...)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 27, 2016, 11:17:03 pm
This is the 1999 paper

Q2c) says to sketch the region bound by |z-i|≤2 and 0≤arg(z+1)≤π/4
Title: Re: 4U Maths Question Thread
Post by: Mei2016 on August 27, 2016, 11:34:31 pm
sorry, that question seems so dumb now...
I get it now; just needed a quick scan of the textbook  :)....
sorry for wasting your time  :(
Title: Re: 4U Maths Question Thread
Post by: Mei2016 on August 27, 2016, 11:38:40 pm
What about Q2e) ?

Also, for Q6bvi) I followed through i to v) but when I got up to vi) what's U, U(o) and V? and also, T(o) is the total time if air resistance is ignored so this should be smaller than T (which if the total time with air resistance-I think) ? >>because an object falling with no air resistance will fall quicker than an object falling against air resistance. [But the ANS is T(o) > T]

Thanks.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 27, 2016, 11:50:06 pm
Hi,

I'm stuck on a few questions from paper 1999
Q2c) why is Re(z)= -1 an empty circle?
Q2e) isn't turning anticlockwise x i, and turning clockwise x -i ?
So answer should be -->AP= -Z(1)i? [But ANS is -->AP =Z(1)i]

(I tried to upload the pics of the question, but it didn't work)

Mod edit: Merged posts. Try to keep everything together :)
Regarding 2e) That is an anticlockwise rotation. Keep in mind the direction of the vector.

PA is a clockwise rotation.

However, AP is an anticlockwise rotation.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 27, 2016, 11:59:26 pm
What about Q2e) ?

Also, for Q6bvi) I followed through i to v) but when I got up to vi) what's U, U(o) and V? and also, T(o) is the total time if air resistance is ignored so this should be smaller than T (which if the total time with air resistance-I think) ? >>because an object falling with no air resistance will fall quicker than an object falling against air resistance. [But the ANS is T(o) > T]

Thanks.








Title: Re: 4U Maths Question Thread
Post by: Mei2016 on August 28, 2016, 09:04:42 pm
Thanks so much for your help Rui! That made the question a lot clearer. I get it now. Previously, I was really confused with all the Ts, Us, and Vs and I was visualising the fall of the object downwards only. So T which is with air resistance is 'pushed' against by the air downwards while it's travelling upwards resulting in a lower maximum height compared to T(o) which is without air resistance, thus allowing it to have a greater max height. So T(o)>T.

Thanks again.  :)
Title: Re: 4U Maths Question Thread
Post by: znaser on August 28, 2016, 09:28:07 pm
Hi, how do u solve this question?
In how many ways can 5 different red roses and 3 different white roses be planted in a straight line next to a fence so that no 2 white roses are together?
I personally did: 8! - 6! x 3! but this equates to 36 000 and the answer is 14 400. Can you please also tell me wat is wrong with my working out. Thanks :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 28, 2016, 09:33:24 pm
Hi, how do u solve this question?
In how many ways can 5 different red roses and 3 different white roses be planted in a straight line next to a fence so that no 2 white roses are together?
I personally did: 8! - 6! x 3! but this equates to 36 000 and the answer is 14 400. Can you please also tell me wat is wrong with my working out. Thanks :)
Can't exactly follow your logic how 6!3! eliminates the cases when the white roses are planted together in your working.



Title: Re: 4U Maths Question Thread
Post by: znaser on August 28, 2016, 11:00:32 pm
Can't exactly follow your logic how 6!3! eliminates the cases when the white roses are planted together in your working.





Thanks! Well I did 6! x 3! as the white roses being together as you would consider the 3 roses as a group and therefore there would be 6! ways of arranging them with the 5 roses and 3! ways of themselves being arranged in their own group. My wording is not probably the best but that's how I usually tackle these sorts of questions and I'm not sure why this method doesn't get me to the answer like it usually does. Do u know y that is the case?
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on August 28, 2016, 11:07:04 pm
Thanks! Well I did 6! x 3! as the white roses being together as you would consider the 3 roses as a group and therefore there would be 6! ways of arranging them with the 5 roses and 3! ways of themselves being arranged in their own group. My wording is not probably the best but that's how I usually tackle these sorts of questions and I'm not sure why this method doesn't get me to the answer like it usually does. Do u know y that is the case?

I follow you! That covers cases with all three white roses together, but if you think carefully, it actually doesn't cover two at a time. So, you aren't subtracting enough cases, which is why your answer is higher than it should be ;D

I love the method you use, and often it works really well, just not in this case, you need to be very careful with probability!
Title: Re: 4U Maths Question Thread
Post by: Mei2016 on August 28, 2016, 11:33:04 pm
Hi znaser,

I might be wrong, but I'm pretty sure that with your method, for the white flowers; it's just 6x3! as if you imagine the 3 flowers as a bunch, you can order them with only 6 choices and not 6! choices.

So instead, the answer should be 8!-3!x6= 40284, unless if I have done something horribly wrong :)
Title: Re: 4U Maths Question Thread
Post by: Mei2016 on August 28, 2016, 11:39:58 pm
But this doesn't align with Rui's answer (which I fully understand) so I'm pretty sure my working out is somehow horribly wrong....
(but I don't know why it doesn't work ...)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 29, 2016, 12:04:37 am
Hi znaser,

I might be wrong, but I'm pretty sure that with your method, for the white flowers; it's just 6x3! as if you imagine the 3 flowers as a bunch, you can order them with only 6 choices and not 6! choices.

So instead, the answer should be 8!-3!x6= 40284, unless if I have done something horribly wrong :)
Judging by your 6*3!, you probably forgot that all the red roses are different as well.

That's my guess.
Title: Re: 4U Maths Question Thread
Post by: znaser on August 29, 2016, 11:11:29 am
I follow you! That covers cases with all three white roses together, but if you think carefully, it actually doesn't cover two at a time. So, you aren't subtracting enough cases, which is why your answer is higher than it should be ;D

I love the method you use, and often it works really well, just not in this case, you need to be very careful with probability!

Thanks! I think i follow now. So that method would also include options like WRWWRRRR where 1 white rose may be separated but two may be together.
Title: Re: 4U Maths Question Thread
Post by: znaser on August 29, 2016, 11:16:51 am
Hi znaser,

I might be wrong, but I'm pretty sure that with your method, for the white flowers; it's just 6x3! as if you imagine the 3 flowers as a bunch, you can order them with only 6 choices and not 6! choices.

So instead, the answer should be 8!-3!x6= 40284, unless if I have done something horribly wrong :)

Hi Mei2016,
I think it would be 8! - 6! x 6 x 3! as that would remove the possibility for 1 white rose to be separated but 2 to be together. You get the answer but it may be a fluke 😅. But yh either way rui's method is probably better in this case as it's much more simpler.
Title: Re: 4U Maths Question Thread
Post by: Ali_Abbas on August 29, 2016, 12:14:33 pm
Can't exactly follow your logic how 6!3! eliminates the cases when the white roses are planted together in your working.





I'm no expert in permutations and combinations, but I believe Rui's answer is incorrect. To demonstrate this, we provide the following counterexample:

Suppose the red roses are fixed as Rui imposed. Then observe that another viable arrangement is:

_RRRR_R_

Here, there are only three blank spots to fill. Thus, Rui's calculation does not capture the entire spectrum of cases and I conjecture that to achieve the correct number of permutations using this method, one would have to apply brute force to obtain all possible arrangements of red roses that leave at most one blank spot in between any two that occur. This would prove to be an inefficient process.

I believe one approach that seems right, though I'm not entirely convinced that it is, is to group two white roses together, treating them as a single object, and evaluate the number of permutations this yields, then subtracting this off from the total number of arrangements, namely 8!. The important observation to make is that this will account for the case where all three white roses are together, inclusively. So we need not treat the two cases separately as requiring two independent calculations.

When two white roses are grouped together, we now have 7 distinct objects. These are arranged in 7! ways, however, the two white roses that are coupled can be chosen in 3P2 = 3! ways. So we thus have 7!3! ways this can happen.

The total number of arrangements where no two white roses are together is calculated as 8! - 7!3! = 10 080. However, this is lower than Rui's answer which I alluded to have understated the true answer since he only accounted for one special arrangement of the red roses. Hopefully, I still have the right idea in my method but just need to make a small alteration. Feel free to correct me where I have gone wrong.
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on August 29, 2016, 01:01:21 pm
I'm no expert in permutations and combinations, but I believe Rui's answer is incorrect. To demonstrate this, we provide the following counterexample:

Suppose the red roses are fixed as Rui imposed. Then observe that another viable arrangement is:

_RRRR_R_

Here, there are only three blank spots to fill. Thus, Rui's calculation does not capture the entire spectrum of cases and I conjecture that to achieve the correct number of permutations using this method, one would have to apply brute force to obtain all possible arrangements of red roses that leave at most one blank spot in between any two that occur. This would prove to be an inefficient process.

I believe one approach that seems right, though I'm not entirely convinced that it is, is to group two white roses together, treating them as a single object, and evaluate the number of permutations this yields, then subtracting this off from the total number of arrangements, namely 8!. The important observation to make is that this will account for the case where all three white roses are together, inclusively. So we need not treat the two cases separately as requiring two independent calculations.

When two white roses are grouped together, we now have 7 distinct objects. These are arranged in 7! ways, however, the two white roses that are coupled can be chosen in 3P2 = 3! ways. So we thus have 7!3! ways this can happen.

The total number of arrangements where no two white roses are together is calculated as 8! - 7!3! = 10 080. However, this is lower than Rui's answer which I alluded to have understated the true answer since he only accounted for one special arrangement of the red roses. Hopefully, I still have the right idea in my method but just need to make a small alteration. Feel free to correct me where I have gone wrong.

I believe that Rui's answer accounts for that; he has all possible places for the white roses to be placed. Taking his example:



We would put white roses in the first, fifth, and last positions to obtain your arrangement :)
Title: Re: 4U Maths Question Thread
Post by: Ali_Abbas on August 29, 2016, 02:52:50 pm
I believe that Rui's answer accounts for that; he has all possible places for the white roses to be placed. Taking his example:



We would put white roses in the first, fifth, and last positions to obtain your arrangement :)

Thanks Jamon :) I totally missed this fact before.

I believe I know what was wrong with my method (with some help from Rui). By pairing two of the white flowers and calculating the number of corresponding arrangements, I seemed to have double-counted the number of times where all three white flowers are together and hence, subtracted two multiples of the same thing giving a deficient answer. To understand why, we offer the following explanation:

Suppose we fix a pair of white flowers. Observe that no matter where this pair occurs, there is at least one neighbouring spot to be filled. This neighbouring spot has the option of being filled by the remaining white flower. When this occurs, we also have the option of interchanging the three whites, given that they are distinct. As we shift the paired flowers across one space at a time, the property of having at least one neighbouring spot is preserved. In particular, translating the pair across by a single space, we observe that all previous arrangements of the three whites occurring together can be reproduced simply by inputting the third white flower on the opposite side of the paired group to which it previously occupied, and again allowing all three to be interchanged. Visually, this can be represented as follows:

....(WW)W.... (this is where the third white is to the right of the pair of whites and before we shift them across)

--> ....W(WW).... (this is after translating the pair horizontally to the right by one space and inputting the third white on the opposite side, namely the left)

Clearly, an overlapping of arrangements occurs. We can be sure that each arrangement occurs exactly twice as applying a second translation in the same direction as the first, either the left-most or right-most space (depending on which direction we are traversing across) becomes excluded from the pact and a new space joins instead (to be made up of all three whites).

Thus, to maintain the correct number of permutations we desire, we must add a single multiple of the number of permutations of when all three flowers are together. This is easily calculated as 6!3!, as stated by other members.

Hence, the correct number of permutations using my method is:

8! - 7!3! + 6!3! = 14 400, as per Rui's answer.

QED

 
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 29, 2016, 03:02:53 pm
Ah so it does involve inclusion-exclusion.
Title: Re: 4U Maths Question Thread
Post by: Ali_Abbas on August 29, 2016, 03:04:49 pm
Ah so it does involve inclusion-exclusion.

I had to think about it for a while but yes, that was the underlying issue.
Title: Re: 4U Maths Question Thread
Post by: Mei2016 on August 30, 2016, 08:57:46 pm
Hi Mei2016,
I think it would be 8! - 6! x 6 x 3! as that would remove the possibility for 1 white rose to be separated but 2 to be together. You get the answer but it may be a fluke 😅. But yh either way rui's method is probably better in this case as it's much more simpler.

Hey znaser,

You mean I got the answer? Do you know what it is? Also, where did this question come from? (as it would normally be like 5 girls and 3 boys...)
I really like Rui's way of doing it, so if you have the answers, what source is if from?

Also, to continue with my method, (because I had previously neglected the white flowers being together case), I've come to:

All possibilities-all white tog-considering the case of 2 white flowers tog and one separate;
8!-6!x6x3!-5!x6x5x2= 7200 (which is exactly half of Rui's answer, and that's so interesting, but I don't know what's wrong with it)
Title: Re: 4U Maths Question Thread
Post by: Mei2016 on August 30, 2016, 09:05:21 pm
Hi,

This is Q6 iv) from the 1997 HSC paper. I just don't get the N.B part in the answers.
(i.e how the _< turned into a < sign)
Title: Re: 4U Maths Question Thread
Post by: znaser on August 30, 2016, 09:09:39 pm
Hey znaser,

You mean I got the answer? Do you know what it is? Also, where did this question come from? (as it would normally be like 5 girls and 3 boys...)
I really like Rui's way of doing it, so if you have the answers, what source is if from?

Also, to continue with my method, (because I had previously neglected the white flowers being together case), I've come to:

All possibilities-all white tog-considering the case of 2 white flowers tog and one separate;
8!-6!x6x3!-5!x6x5x2= 7200 (which is exactly half of Rui's answer, and that's so interesting, but I don't know what's wrong with it)

The question is from an S&G paper back in 2013. It's multiple choice and the answer is 14 400 (the same as Rui's). They did it the same way as him but their explanation was a bit confusing. Not sure if u've heard of their company but they're known for their rlly difficult papers relative to others.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 30, 2016, 09:16:46 pm
Hi,

This is Q6 iv) from the 1997 HSC paper. I just don't get the N.B part in the answers.
(i.e how the _< turned into a < sign)


Title: Re: 4U Maths Question Thread
Post by: Ali_Abbas on August 30, 2016, 09:19:18 pm
Hi,

This is Q6 iv) from the 1997 HSC paper. I just don't get the N.B part in the answers.
(i.e how the _< turned into a < sign)

As per the explanation given in the set of solutions, the key observation they have made is that the upper and lower bounds are irrational numbers, given that the number Pi is present in both cases. The intermediately-occurring sum of rationals will, by intuition, evaluate to another rational. The set of rational numbers and the set of irrational numbers are disjoint sets. That is, they do not have any common elements and so this facilitates their justification for replacing the "less than or equal to" symbols with strict inequalities as equality cannot hold between numbers that belong to non-overlapping sets.

EDIT: In case you are unsure about my wording, I will define some of the terms used.

Upper and lower bounds are the numbers that occur on the "outside" of the inequality statement. In this case, the lower bound is Pi/4, and the upper bound is Pi/4 + 1/1003.

An intermediate term is one which occurs in between, or roughly speaking, in the middle of the string of terms in the statement we are considering.

Again, disjoint simply means non-overlapping or mutually exclusive. Basically it's where nothing is common.

Strict inequalities are < and >.

Hopefully that alleviates any potential confusion.
Title: Re: 4U Maths Question Thread
Post by: Mei2016 on August 30, 2016, 10:43:52 pm




Hi Rui,

Thanks for your help. I know what a rational number is, I was just previously confused with the wording of their explanation.
In the test when we do this (taking the equality out) do we need to explain it in words like how the book does?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 30, 2016, 10:49:24 pm
Hi Rui,

Thanks for your help. I know what a rational number is, I was just previously confused with the wording of their explanation.
In the test when we do this (taking the equality out) do we need to explain it in words like how the book does?
If you're going to make an inequality strict you have to be explicit about why.

For something a bit more trivial like 1≤2 to 1<2 you can easily argue this by saying "because 1≠2".

Since what your given is far more complicated and the absence of equality is less obvious, you have to say something.


Your explanation doesn't have to be identical to the book's if that's your query. It just has to allow your logic and reasoning to flow.
Title: Re: 4U Maths Question Thread
Post by: Mei2016 on August 31, 2016, 07:17:35 am
Ok, thanks.
Title: Re: 4U Maths Question Thread
Post by: amandali on August 31, 2016, 03:39:03 pm
i(http://uploads.tapatalk-cdn.com/20160830/da2775400e4ebbd8ca652617d41b46ad.jpg)

i dont know how to do part iv) 
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 31, 2016, 06:57:19 pm
Hint for the time being:

The question just means to find the normal reaction the surface exerts on the particle.

Remember: According to Newton's Third Law of Motion, the normal reaction the surface exerts on the particle has the same magnitude as the force the particle exerts onto the bowl.

> > > Please do not double post < < <
This was the best I could come up with.

Basically just use a Pythagorean identity.
(http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI31082016_zpsgptzwfgp.jpg)
Title: Re: 4U Maths Question Thread
Post by: massive on August 31, 2016, 10:23:09 pm
guys quick question:
Find the equation of the normal to the ellipse at P (2cos(theta), sqrt(3)sin(theta)) and prove that it bisects angle SPS' where S and S' are the foci
I can get the normal but i just can't get the second part of the question, thanks for the help!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 31, 2016, 10:33:49 pm
guys quick question:
Find the equation of the normal to the ellipse at P (2cos(theta), sqrt(3)sin(theta)) and prove that it bisects angle SPS' where S and S' are the foci
I can get the normal but i just can't get the second part of the question, thanks for the help!
This is just asking you to prove the reflection property for the ellipse. It'll be in your textbook.

Some questions in past HSC papers (I think there was one in 2009) also made students prove the reflection property. They did it for the hyperbola, though, which tweaks the proof a little.
Title: Re: 4U Maths Question Thread
Post by: massive on August 31, 2016, 10:39:46 pm
This is just asking you to prove the reflection property for the ellipse. It'll be in your textbook.

Some questions in past HSC papers (I think there was one in 2009) also made students prove the reflection property. They did it for the hyperbola, though, which tweaks the proof a little.


The answers prove that PS/PS' = ST/S'T and therefore the normal bisects angle SPS' :S how does that work?? (BTW T is the x intercept of the normal)

Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 31, 2016, 10:55:50 pm

The answers prove that PS/PS' = ST/S'T and therefore the normal bisects angle SPS' :S how does that work?? (BTW T is the x intercept of the normal)
Check one of the past papers such as 2009. There's a small proof as to how you can then use similar triangles to induce your required result.

You're required to memorise up to PS/PS' = TS/TS' by yourself. That last bit you'd be told guided to prove separately if you had to.
Title: Re: 4U Maths Question Thread
Post by: katherine123 on September 01, 2016, 01:46:39 am
need help with these ques:
1.a particle of mass M is attached to one end of a light inelastic string of length `2a and the other end is fastened to a fixed point O on a smooth horizontal table. Another particle also of mass M is attached to the midpoint of the string. The system then rotates in a horizontal circle and in contact with the table at an angular velocity of w. Find the ratio of the tensions in each part of the string.

2.a find of mass 0.6kg is attached to a point P on a string AB of length 1.4m, where AP is 0.8m. The ends A and B are attached to 2 points 1.0m apart in a vertical line, A being above B. THe ring is made to travel in a horizontal circle with speed v m/s.
what is the smallest possible values of v if neither portion of string is slack?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 01, 2016, 04:57:42 pm
need help with these ques:
1.a particle of mass M is attached to one end of a light inelastic string of length `2a and the other end is fastened to a fixed point O on a smooth horizontal table. Another particle also of mass M is attached to the midpoint of the string. The system then rotates in a horizontal circle and in contact with the table at an angular velocity of w. Find the ratio of the tensions in each part of the string.

2.a find of mass 0.6kg is attached to a point P on a string AB of length 1.4m, where AP is 0.8m. The ends A and B are attached to 2 points 1.0m apart in a vertical line, A being above B. THe ring is made to travel in a horizontal circle with speed v m/s.
what is the smallest possible values of v if neither portion of string is slack?
I do not take credit for an answer that's not mine. All credit goes to the person i seek for help. (I honestly could not understand Q2 at all when I first saw it without an appropriate diagram.)
Quote





Apparently, the scenario to Q1 is similar to this one.
Again, the HSC would be more explicit about the forces involved.

Try to work through his help. Draw a relevant diagram, and upload anything you have thus far if you're stuck halfway through and I should be able to continue it.
Title: Re: 4U Maths Question Thread
Post by: massive on September 01, 2016, 10:20:02 pm
yo guys does anybody know what papers contain circular motion questions, I've been looking everywhere and can't seem to find any. thanks!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 01, 2016, 10:29:55 pm
yo guys does anybody know what papers contain circular motion questions, I've been looking everywhere and can't seem to find any. thanks!
That's because the HSC is highly unlikely to ask a question that's bluntly related to circular motion. It's virtually always on one of the applications. of it - conical pendulum or banked tracks.

Why? Because what is there to examine you on.

There may be some trial papers with a few multiple choice on them. Try browsing through THSC.
Title: Re: 4U Maths Question Thread
Post by: massive on September 01, 2016, 11:22:14 pm
Hey just quickly, do you guys know of any little pointers to remember for mechanics, i think it occurs mostly in conical pendulum. What i mean is for e.g. when N=0 the particle is no longer touching the surface and also if w^2>0 then the particle is moving. Does anybody know anything else to keep an eye out for?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 01, 2016, 11:33:16 pm
Hey just quickly, do you guys know of any little pointers to remember for mechanics, i think it occurs mostly in conical pendulum. What i mean is for e.g. when N=0 the particle is no longer touching the surface and also if w^2>0 then the particle is moving. Does anybody know anything else to keep an eye out for?
0≤θ<π/2 if you do it correctly. (If the angle is 0 then obviously you're at rest though, and that's basically the same thing as saying ω2=0 so you usually don't get that happening.)
You can't have a conical pendulum that goes inverted just by itself.

Maximal tension a string can cope with is something but you're usually guided as to how to approach this type of question.

And of course in general: string means tension, surface means normal reaction (and friction if appropriate - smooth surface implies no friction).

That's basically all off the top of my head I reckon.
Title: Re: 4U Maths Question Thread
Post by: jakesilove on September 04, 2016, 12:01:48 pm
For anyone looking to absolutely smash their HSC exam, understanding what the question actually expects of you is vital. Check our Rui's beast guide of Maths verbs HERE, and get an edge in your final exam! As always, thanks must go out to the legend himself, RuiAce; Improving Atars Since 2015.
Title: Re: 4U Maths Question Thread
Post by: katherine123 on September 06, 2016, 09:56:07 pm
how to do last part
Title: Re: 4U Maths Question Thread
Post by: jakesilove on September 06, 2016, 11:42:54 pm
how to do last part

Definitely a tough one. First, we know that if the particle is JUST at the point where it doesn't slide down the track, the force acting on it will be



Now, from i)



From ii)



Now, in the case that





Therefore, from above



For all real velocities, this will only be true when the velocity equals zero! Therefore, the particle will not travel down the slant for any value of v
Title: Re: 4U Maths Question Thread
Post by: amandali on September 08, 2016, 03:07:16 pm
(http://uploads.tapatalk-cdn.com/20160907/9f62dd8cc7b37f797025648a47ad2cee.jpg)

need help with this ques thanks
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 08, 2016, 06:07:43 pm
(http://uploads.tapatalk-cdn.com/20160907/9f62dd8cc7b37f797025648a47ad2cee.jpg)

need help with this ques thanks










Let me know if something doesn't makes sense or I got a wrong answer. Plus I also don't do physics anymore and I'm using a brain from a university student's level. Implicit differentiation isn't an obvious approach and I feel it's unnecessarily overcomplicating for 4U. I wanted to use circle geometry but the sine rule was speaking more loudly to me.
Title: Re: 4U Maths Question Thread
Post by: massive on September 10, 2016, 10:35:26 pm
Hey guys how do you do this ??
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 10, 2016, 11:26:08 pm
Hey guys how do you do this ??


Note that the things here are negative as:
a) The centre of the planet is defined as x=0
b) Gravity brings things BACK to the centre of the planet

And this is where I got lost. I kinda had to use HSC physics here... so I reckon this question is pushing the boundaries of the 4U course.





Can you please stop avoiding this question I've asked many times and tell us what is the source of these questions?
Title: Re: 4U Maths Question Thread
Post by: massive on September 10, 2016, 11:36:58 pm


Note that the things here are negative as:
a) The centre of the planet is defined as x=0
b) Gravity brings things BACK to the centre of the planet

And this is where I got lost. I kinda had to use HSC physics here... so I reckon this question is pushing the boundaries of the 4U course.





Can you please stop avoiding this question I've asked many times and tell us what is the source of these questions?
Hey thankyou so much. This particular question was from the catholic trial 1984...I don't usually do questions that are really old, it's just that i need to get good in maths HSC, theyre the only things pulling my atar up :L
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 10, 2016, 11:45:47 pm
Hey thankyou so much. This particular question was from the catholic trial 1984...I don't usually do questions that are really old, it's just that i need to get good in maths HSC, theyre the only things pulling my atar up :L
That was all you had to say though haha. That explains why these questions are extremely difficult - they're going beyond 1990.
Title: Re: 4U Maths Question Thread
Post by: Neutron on September 12, 2016, 07:41:43 pm
Hello y'all! I come with a complex numbers question (I have no bloody clue what's going on with this one):

Let the points A1,A2,...,An represent the nth roots of unity, w1, w2, ..., wn, and suppose P represents any complex number z such that lzl=1.

a) Prove that w1+w2+...+wn=0

b) Show that PA2i=(z-wi)(z (conjugate) - wi (conjugate) for i=1,2,...,n

c)Prove that:
Σ (from i=1 to n) PA2i=2n

Sorry it's really hard to type it out on this forum because of all the squares and subscripts and conjugate bars and sigmas and ah i hope it still makes sense because it doesn't to me! Thanks guys :)

Neutrons
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 12, 2016, 08:01:05 pm
Hello y'all! I come with a complex numbers question (I have no bloody clue what's going on with this one):

Let the points A1,A2,...,An represent the nth roots of unity, w1, w2, ..., wn, and suppose P represents any complex number z such that lzl=1.

a) Prove that w1+w2+...+wn=0

b) Show that PA2i=(z-wi)(z (conjugate) - wi (conjugate) for i=1,2,...,n

c)Prove that:
Σ (from i=1 to n) PA2i=2n

Sorry it's really hard to type it out on this forum because of all the squares and subscripts and conjugate bars and sigmas and ah i hope it still makes sense because it doesn't to me! Thanks guys :)

Neutrons


___________________________________________



___________________________________________



Title: Re: 4U Maths Question Thread
Post by: massive on September 16, 2016, 08:48:53 am
How do you do this?

If abs(a)>2abs(b), prove that 2abs(a-b)>abs(a)

thanks!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 16, 2016, 09:30:22 am
How do you do this?

If abs(a)>2abs(b), prove that 2abs(a-b)>abs(a)

thanks!


Title: Re: 4U Maths Question Thread
Post by: karenc. on September 19, 2016, 04:44:56 pm
Hi everyone can u please help me with this question, it was from my school's trial:
Title: Re: 4U Maths Question Thread
Post by: Ali_Abbas on September 19, 2016, 06:06:30 pm
Hi everyone can u please help me with this question, it was from my school's trial:





Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 26, 2016, 01:26:25 pm
Here is a question I was asked during the 3U lecture break. I will address it in the next post.
(http://uploads.tapatalk-cdn.com/20160926/0d2e6f8afe0be737393f3ee0a40a7d87.jpg)
Apologies for the poor quality photo
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 26, 2016, 02:00:06 pm
Here is a question I was asked during the 3U lecture break. I will address it in the next post.
(http://uploads.tapatalk-cdn.com/20160926/0d2e6f8afe0be737393f3ee0a40a7d87.jpg)
Apologies for the poor quality photo
Unfortunately, I was unable to generate a better proof than what I offered up on the day for a 2 mark question. I will give a full statement of the proof I had then. It may be worth mentioning that I considered an approach that utilises stationary points, however felt it was a bit overkill for a 2 mark question. The stationary points do not have a tidy form: \( x=q+\frac{q-r}{-1\pm \sqrt{\frac{p-r}{p-q}}} \)

(Note that this proof used a graph to get there, mostly to visualise the shape of the function, but doesn't properly involve the graph itself. I could not see how the partial fractions decomposition aided any further.)

So I'm basically welcoming a more elegant proof.

(Also, if you're the one who asked me this, sorry that I got the assumed inequality in the next line the wrong way around!)








Title: Re: 4U Maths Question Thread
Post by: amandali on September 26, 2016, 08:45:05 pm
(http://uploads.tapatalk-cdn.com/20160926/0ccb89568cb328c0c61d36a0e6e40d42.jpg)

for part d ii)  do i always need to divide 3! (factorial) because there are same number of people in all the groups




(http://uploads.tapatalk-cdn.com/20160926/04d8ba88c0f9b5da660ba5786b64748a.jpg)
how to do part c)
Title: Re: 4U Maths Question Thread
Post by: katherine123 on September 27, 2016, 03:48:55 am
how to do part iii)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 27, 2016, 08:51:46 am
how to do part iii)

(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zps8pu3m4jy.png)





Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 27, 2016, 08:55:57 am
(http://uploads.tapatalk-cdn.com/20160926/0ccb89568cb328c0c61d36a0e6e40d42.jpg)

for part d ii)  do i always need to divide 3! (factorial) because there are same number of people in all the groups




(http://uploads.tapatalk-cdn.com/20160926/04d8ba88c0f9b5da660ba5786b64748a.jpg)
how to do part c)
I think the division by 3! is because of the issue with ordering here.

Suppose you had persons A, B, C, ..., L
And you had groups 1, 2, 3

Group 3 having A B C D and group A having E F G H
is the same outcome as group 3 having E F G H and group A having A B C D.

Edit: On further thought, it's probably tied down with the fact there's the same people in each group as well.
____________________________________________________

I can't see the rest of the question (I know where it's headed though. Just equate real or imaginary parts for part (ii), picking whichever one is appropriate.)
Title: Re: 4U Maths Question Thread
Post by: xXCandyDXx on September 29, 2016, 01:32:15 pm
Hi everyone !!! I know this may sound a bit basic but I am kind of confused ... With relation to resisted motion in a horizontal line and vertically in Mechanics ... When do we use -mkv and just -kv??? I can see in some working out that mass does actually make a difference even though it cancels out in some cases ... I'm just confused as to when I'm supposed to add the m in -kv or -kv^2.... Yeah that's my question thanks !!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 29, 2016, 01:44:51 pm
Hi everyone !!! I know this may sound a bit basic but I am kind of confused ... With relation to resisted motion in a horizontal line and vertically in Mechanics ... When do we use -mkv and just -kv??? I can see in some working out that mass does actually make a difference even though it cancels out in some cases ... I'm just confused as to when I'm supposed to add the m in -kv or -kv^2.... Yeah that's my question thanks !!!
They either give it to you, or they will say something along the lines of "force per unit mass".

Generally, the HSC is nice enough to just give you it and you can easily guess which to use. But this doesn't happen in textbooks.

In general, when "per unit mass" is mentioned, you are using mkv.
Title: Re: 4U Maths Question Thread
Post by: xXCandyDXx on September 29, 2016, 02:28:58 pm

They either give it to you, or they will say something along the lines of "force per unit mass".

Generally, the HSC is nice enough to just give you it and you can easily guess which to use. But this doesn't happen in textbooks.

In general, when "per unit mass" is mentioned, you are using mkv.


Ohh.. Alright THANKS !! :)
Title: Re: 4U Maths Question Thread
Post by: katherine123 on October 01, 2016, 08:50:41 pm
how to do part a i)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 01, 2016, 10:30:49 pm
how to do part a i)




Answers to the 2008 HSC exam may be found here
Title: Re: 4U Maths Question Thread
Post by: Neutron on October 04, 2016, 05:22:49 pm
Hey!

Could you guys please help me with the last question of the 2011 paper? Thanks, I don't even understand the question haha

Neutron
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 04, 2016, 07:34:03 pm
Hey!

Could you guys please help me with the last question of the 2011 paper? Thanks, I don't even understand the question haha

Neutron






_____________________________________________________



Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 04, 2016, 07:46:50 pm



Title: Re: 4U Maths Question Thread
Post by: aoifera on October 04, 2016, 08:57:15 pm
Would I be able to get some help with this question:
"A stone of mass 6kg is tied at one end of a 3 metre long strong. The other end is fixed to a point O. Find the tension in the string when the mass is rotating at 40 revolutions per minute"
Thank you :)
Title: Re: 4U Maths Question Thread
Post by: jakesilove on October 04, 2016, 10:07:00 pm
Would I be able to get some help with this question:
"A stone of mass 6kg is tied at one end of a 3 metre long strong. The other end is fixed to a point O. Find the tension in the string when the mass is rotating at 40 revolutions per minute"
Thank you :)

Hey! You just need to find the force exerted by the stone. To do this, we calculate centripetal force, which has formula mv^2/r. We have all values except velocity, but this is something you can easily figure out by knowing the radius of the circle, and how many revolutions occur each minute! If you need any further help, post your working out and I'll see what I can do :)

Jake
Title: Re: 4U Maths Question Thread
Post by: nay103 on October 05, 2016, 11:33:26 pm
Hi! I'm a bit confused about this 2011 HSC question:
8(b) A bag contains seven balls numbered from 1 to 7. A ball is chosen at random
and its number is noted. The ball is then returned to the bag. This is done a total
of seven times.
(i) What is the probability that each ball is selected exactly once?
(ii) What is the probability that at least one ball is not selected?
(iii) What is the probability that exactly one of the balls is not selected?

I am fine with parts i and ii, but iii had me a bit confused, and I'm not really sure I quite understand the given answer. My answer was
7/7x6/7x5/7x4/7x3/7x2/7x6/7 -> i.e. probability of six balls being selected exactly once, then one of those six balls being selected again. I don't see what is incorrect with this, if you could explain that would be great!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 05, 2016, 11:47:18 pm
Hi! I'm a bit confused about this 2011 HSC question:
8(b) A bag contains seven balls numbered from 1 to 7. A ball is chosen at random
and its number is noted. The ball is then returned to the bag. This is done a total
of seven times.
(i) What is the probability that each ball is selected exactly once?
(ii) What is the probability that at least one ball is not selected?
(iii) What is the probability that exactly one of the balls is not selected?

I am fine with parts i and ii, but iii had me a bit confused, and I'm not really sure I quite understand the given answer. My answer was
7/7x6/7x5/7x4/7x3/7x2/7x6/7 -> i.e. probability of six balls being selected exactly once, then one of those six balls being selected again. I don't see what is incorrect with this, if you could explain that would be great!



Title: Re: 4U Maths Question Thread
Post by: nay103 on October 05, 2016, 11:53:00 pm





Ah, this makes a lot more sense now. Thank you!
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on October 06, 2016, 12:08:43 am


Love this question (and the amount of the questions that I chose appearing makes me think that I really smashed you guys in that MX1 Revision Lecture) ;)
Title: Re: 4U Maths Question Thread
Post by: katherine123 on October 06, 2016, 01:48:28 am
how to do part iv)
Title: 4U Maths Question Thread
Post by: RuiAce on October 06, 2016, 08:22:18 am
how to do part iv)
Evaluating:
P(0)=-8
P(1)=2
Hence P(x) must have some root that is real, between 0 and 1.

But there's more to it. Remember that a complex root comes with its complex conjugate. This is because the coefficients on P(x) are all real

Since there are 4 roots, we must have either:
No complex roots - Part iii says that this is impossible
2 complex roots
4 complex roots

So since we've just deduced that at least 1 of the roots is real, it can't have 4 complex roots either.

Hence there are exactly two complex roots.
Title: Re: 4U Maths Question Thread
Post by: massive on October 06, 2016, 04:23:27 pm
if 1<x<y , is x-y>0 ??
Title: Re: 4U Maths Question Thread
Post by: jakesilove on October 06, 2016, 04:51:47 pm
if 1<x<y , is x-y>0 ??

Clearly, y is greater than x, so x-y is going to be negative!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 06, 2016, 05:25:42 pm
if 1<x<y , is x-y>0 ??
1 < x < y

So x < y
So x - y < 0


Yeah, like Jake said, it's not positive; it's negative.
Title: Re: 4U Maths Question Thread
Post by: massive on October 06, 2016, 06:08:31 pm
thanks guys, also how do you do the q attached.

the question asks to solve for x
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 06, 2016, 06:19:08 pm
thanks guys, also how do you do the q attached.

the question asks to solve for x





Title: Re: 4U Maths Question Thread
Post by: massive on October 06, 2016, 07:55:46 pm







wow that was real neat. Thanks Rui. So whenever we have absolute values and we need to solve do we always use cases like you did?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 06, 2016, 09:01:46 pm
wow that was real neat. Thanks Rui. So whenever we have absolute values and we need to solve do we always use cases like you did?
Yeah. Whenever a course of action isn't obvious, split the absolute values into cases.
Title: Re: 4U Maths Question Thread
Post by: nay103 on October 06, 2016, 09:23:51 pm
Back with another question!
I'm super confused about the last part. Doesn't it essentially mean having to prove the existence of the circle in the very question??
Thanks :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 06, 2016, 10:02:09 pm
Back with another question!
I'm super confused about the last part. Doesn't it essentially mean having to prove the existence of the circle in the very question??
Thanks :)

i.e. If that circle is tangential to 3 sides, it must be tangential to the fourth as well. That's why the implication is reversed in comparison to part (i).
_________________________________________

(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpsewabyifr.png)






Answers
Title: Re: 4U Maths Question Thread
Post by: katherine123 on October 07, 2016, 01:34:23 am
how to do part iii)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 07, 2016, 09:50:01 am
how to do part iii)









Title: Re: 4U Maths Question Thread
Post by: kiwiberry on October 07, 2016, 06:35:22 pm
Hey, I'm new here so I'm not sure how this works haha
this is probably really simple but how would you graph
i) 1/f(x)
ii) [f(x)]^2

thank you :)
Title: Re: 4U Maths Question Thread
Post by: jakesilove on October 07, 2016, 06:42:35 pm
Hey, I'm new here so I'm not sure how this works haha
this is probably really simple but how would you graph
i) 1/f(x)
ii) [f(x)]^2

thank you :)

Hey! Welcome to the forums! You've done the exact right thing; post up any questions you've got, and we're here to answer them! You can also answer others' questions if you feel up to it :)

For these kinds of questions, you just need to think about what happens at critical points on the graph. So, let's start with 1/f(x).

We know that at x approaching infinity, y will approach 2. That means that, for 1/f(x), as x approaches infinity, y will approach 1/2. We also know that y equals -1 and x=2, therefore for 1/f(x), when x=2, y=1/(-1)=-1.

Now, we need to recognise that if we divide by zero, there will be an asymptote at that point. We know that at x=0, y=0. Therefore, on our new graph, there will be an asymptote at x=0!

On the left hand side, we know that as x reaches negative infinity, y reaches infinity. As 1/infinity=0, our new graph will reach zero as x reaches negative infinity, and reach infinity as x reaches zero from the left.

This is super hard to explain. I have to run, but hopefully someone can give a better explanation/show you the second graph before I come back!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 07, 2016, 06:46:08 pm
Hey, I'm new here so I'm not sure how this works haha
this is probably really simple but how would you graph
i) 1/f(x)
ii) [f(x)]^2

thank you :)






Jamon went through the effort of making a guide for 4U curve sketching. Consider reading it.
Title: Re: 4U Maths Question Thread
Post by: znaser on October 09, 2016, 10:03:04 am
Hi. I'm not sure of the working out for this question. The answer is A.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 09, 2016, 10:17:09 am
Hi. I'm not sure of the working out for this question. The answer is A.






A more brute force approach that would work is a case by case analysis. Note, however, that this requires significantly more effort (just because of how many cases there are), and thus we choose the complement for convenience.
Title: Re: 4U Maths Question Thread
Post by: znaser on October 09, 2016, 10:31:17 am






A more brute force approach that would work is a case by case analysis.

Thanks!
Title: Re: 4U Maths Question Thread
Post by: birdwing341 on October 09, 2016, 09:26:00 pm
Hello!

Another out-of-left-field question, but I was wondering how you would order the HSC Extension 2 papers in difficultly (from 2001 to 2015) or in rough equivalents?

Thanks again!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 09, 2016, 09:44:29 pm
Hello!

Another out-of-left-field question, but I was wondering how you would order the HSC Extension 2 papers in difficultly (from 2001 to 2015) or in rough equivalents?

Thanks again!
That's hard. I need to look at every single paper to give an answer to this one.
And now that I had a quick glance (I'm not spending 45 hours of my life doing all those papers...) this is what I reckon

From easiest to hardest

2014
2015
2012
2013
2010
2011
2007
2009
2004
2006
2005
2002
2003
2001

Reminder: This is very easily debatable/arguable.
Title: Re: 4U Maths Question Thread
Post by: kiwiberry on October 09, 2016, 09:55:11 pm






Jamon went through the effort of making a guide for 4U curve sketching. Consider reading it.
Hey! Welcome to the forums! You've done the exact right thing; post up any questions you've got, and we're here to answer them! You can also answer others' questions if you feel up to it :)

For these kinds of questions, you just need to think about what happens at critical points on the graph. So, let's start with 1/f(x).

We know that at x approaching infinity, y will approach 2. That means that, for 1/f(x), as x approaches infinity, y will approach 1/2. We also know that y equals -1 and x=2, therefore for 1/f(x), when x=2, y=1/(-1)=-1.

Now, we need to recognise that if we divide by zero, there will be an asymptote at that point. We know that at x=0, y=0. Therefore, on our new graph, there will be an asymptote at x=0!

On the left hand side, we know that as x reaches negative infinity, y reaches infinity. As 1/infinity=0, our new graph will reach zero as x reaches negative infinity, and reach infinity as x reaches zero from the left.

This is super hard to explain. I have to run, but hopefully someone can give a better explanation/show you the second graph before I come back!

Thanks so much guys :)
I had a read of Jamon's guide and it was really helpful!!
Here's my attempt at the questions
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 09, 2016, 10:01:29 pm
Thanks so much guys :)
I had a read of Jamon's guide and it was really helpful!!
Here's my attempt at the questions
Looking good. Well done :)
Title: Re: 4U Maths Question Thread
Post by: Justina Shehata on October 09, 2016, 10:05:33 pm
Do you lose marks in the HSC for not putting working out in your answers?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 09, 2016, 10:06:32 pm
Do you lose marks in the HSC for not putting working out in your answers?
Yes.

That's why some questions are worth 3 marks, not 1 mark each. You're actually AWARDED marks for working out.

If you magically write down an answer, people will think ok where did you get it from
Title: Re: 4U Maths Question Thread
Post by: birdwing341 on October 09, 2016, 10:20:13 pm
And now that I had a quick glance (I'm not spending 45 hours of my life doing all those papers...) this is what I reckon

From easiest to hardest

2014
2015
2012
2013
2010
2011
2007
2009
2004
2006
2005
2002
2003
2001

Reminder: This is very easily debatable/arguable.

Man you are a legend, thanks rui! Just did one of the past papers so wanted to see where it sat in terms of difficulty :)
Title: Re: 4U Maths Question Thread
Post by: Justina Shehata on October 09, 2016, 11:49:10 pm
Yes.

That's why some questions are worth 3 marks, not 1 mark each. You're actually AWARDED marks for working out.

If you magically write down an answer, people will think ok where did you get it from

would the working out in the sample answers be a good indication to how much they require?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 10, 2016, 08:40:53 am
would the working out in the sample answers be a good indication to how much they require?
Maths sample answers are somewhat superior to every other course because they answer the question. But they may or may not make sense and you should use Excel Success One or MANSW published books.

If it's too late into the HSC and you don't want to spend any more money then use the answers that can be found here.

That being said, you shouldn't have to ask to know how much you need to answer. You should just be able to do the question and LEAVE it there. Mark allocations only indicate a RELATIVE quantity of working out required, not absolute. Some 3 markers will require more effort than other 3 markers.
Title: Re: 4U Maths Question Thread
Post by: massive on October 10, 2016, 04:08:49 pm


How does this work? i've never really understood how do it and when to apply it.
Title: Re: 4U Maths Question Thread
Post by: jakesilove on October 10, 2016, 04:10:59 pm
How does this work? i've never really understood how do it and when to apply it.

Basically, you have 4 choices where to put the first person, 4 choices where to put the second person.... 4 choices where to put the sixth person, so

Title: Re: 4U Maths Question Thread
Post by: katherine123 on October 10, 2016, 11:43:29 pm
first ques: the answer assumes  0<=t<=1 for part ii)  and i dont get why thats the case
2nd ques; dont know how to do part iii)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 10, 2016, 11:58:07 pm
first ques: the answer assumes  0<=t<=1 for part ii)  and i dont get why thats the case
2nd ques; dont know how to do part iii)
You've already asked the second question and it was answered on the previous page.
_______________________




Title: Re: 4U Maths Question Thread
Post by: kiwiberry on October 12, 2016, 08:45:47 pm
how would you find the asymptotes for a parametric equation like
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 12, 2016, 09:29:27 pm
how would you find the asymptotes for a parametric equation like







Title: Re: 4U Maths Question Thread
Post by: kiwiberry on October 13, 2016, 04:28:52 pm








Sorry could you please explain why x≠2? Is it because as t --> infinity, x --> 2?
Title: Re: 4U Maths Question Thread
Post by: jakesilove on October 13, 2016, 04:37:35 pm
Sorry could you please explain why x≠2? Is it because as t --> infinity, x --> 2?

That's exactly right! x will never QUITE reach 2, it will always be slightly less than 2 :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 13, 2016, 04:39:13 pm


Title: Re: 4U Maths Question Thread
Post by: aoifera on October 14, 2016, 04:18:17 pm
Hey could I have a hand with this question? I'm not sure which formulae to use
Title: Re: 4U Maths Question Thread
Post by: jakesilove on October 14, 2016, 04:23:44 pm
Hey could I have a hand with this question? I'm not sure which formulae to use

If there are 10 revolutions per minute, then a particle on the circumference will move 10 circumferences per minute, or



So, it will move 100*pi cm per minute making the answer C
Title: 4U Maths Question Thread
Post by: RuiAce on October 14, 2016, 04:37:45 pm
Hey could I have a hand with this question? I'm not sure which formulae to use
i.e. (So this is just adding to Jake's answer)

This is your very standard mechanics formula v=rω

The only extra step is the conversion between "revolutions" to something more useful for ω. Which is standard for 4U so you must be aware of it.
Title: Re: 4U Maths Question Thread
Post by: nancy_cc on October 15, 2016, 04:58:15 pm
Could anyone have a look at this question I'm stuck on? Would you use slices or shells? Or is it a different method that I'm missing? Thanks :)
Title: Re: 4U Maths Question Thread
Post by: znaser on October 15, 2016, 05:07:44 pm
Hi. I can't seem to understand the working out of part (i).
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 15, 2016, 05:10:46 pm
Could anyone have a look at this question I'm stuck on? Would you use slices or shells? Or is it a different method that I'm missing? Thanks :)




__________________________________


Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 15, 2016, 05:34:18 pm
Hi. I can't seem to understand the working out of part (i).







Note: Consider an alternate source of answers to BOSTES' ones.

Title: Re: 4U Maths Question Thread
Post by: znaser on October 15, 2016, 05:40:24 pm







Note: Consider an alternate source of answers to BOSTES' ones.



Sweet! Thanks!
Title: Re: 4U Maths Question Thread
Post by: nancy_cc on October 15, 2016, 05:46:38 pm




__________________________________




Wow I didn't realise to use the bottom half and kept getting stuck! Thank you that really helped! :)
Title: Re: 4U Maths Question Thread
Post by: MarkThor on October 15, 2016, 07:47:33 pm
Hi guys,
Could you guys help me out with this probability multiple choice question
I always seem to get the harder probability questions wrong, does anyone have any advice for this topic that could help?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 15, 2016, 07:56:45 pm

Hi guys,
Could you guys help me out with this probability multiple choice question
I always seem to get the harder probability questions wrong, does anyone have any advice for this topic that could help?






_________________

Well this is the most annoying topic ever in the entire 4U course. Every question you can be hit by may be unique. Always break it down and choose a fastest route.
Title: Re: 4U Maths Question Thread
Post by: MarkThor on October 15, 2016, 08:24:48 pm
Cool thanks for the breakdown Rui
I absolutely dread probability  :'(
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 15, 2016, 08:27:26 pm
Cool thanks for the breakdown Rui
I absolutely dread probability  :'(
You should see me. I'm dying with my first year uni combinatorics.
Title: Re: 4U Maths Question Thread
Post by: MarkThor on October 16, 2016, 08:47:07 am
That sounds way too intense
Title: Re: 4U Maths Question Thread
Post by: massive on October 16, 2016, 10:37:46 am
Hey guys how do you part c, the answer is: Incorrect, n=1020 is one counter example

wth does this mean, could someone explain the q and answer please!

Thankss
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 16, 2016, 10:52:47 am
Hey guys how do you part c, the answer is: Incorrect, n=1020 is one counter example

wth does this mean, could someone explain the q and answer please!

Thankss





Title: Re: 4U Maths Question Thread
Post by: massive on October 16, 2016, 11:23:08 am







OHHH that makes more sense now, thankyou. But in an exam situation, how do you come up with a number like 1020 to prove by contradiction
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 16, 2016, 11:26:25 am
OHHH that makes more sense now, thankyou. But in an exam situation, how do you come up with a number like 1020 to prove by contradiction
For that particular question, the inspiration was that

If N was very large, then sqrt(N+1)-1 is about equal to sqrt(N).

And sqrt(1020) = 1010

But then because it's a proof, we always need to look at what we're trying to prove. We want to prove that it is greater than 1010. Quite fortunately enough, we actually had 2x1010 by our intuition.


With proof by contradiction, you need to look ahead. You need to try to think about how you would fail it. You firstly decide wait, does this make sense or not, before you try finding a counterexample.
Note that, for that question though, you could've picked say 10100 and you would still have a valid counterexample.
Title: Re: 4U Maths Question Thread
Post by: nancy_cc on October 16, 2016, 01:15:05 pm
Hey could someone help with this question please! :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 16, 2016, 01:18:30 pm
Hey could someone help with this question please! :)

Part a) was done recently here in post #533

Use a similar analysis to try part b) and c) first. Come back with your progress
Title: Re: 4U Maths Question Thread
Post by: massive on October 16, 2016, 02:27:19 pm
hey guys how do you part b, im so close yet so far :/ i keep getting (-1)1-k instead of (-1)k+1
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 16, 2016, 03:03:08 pm
hey guys how do you part b, im so close yet so far :/ i keep getting (-1)1-k instead of (-1)k+1
Title: Re: 4U Maths Question Thread
Post by: DaCoon on October 16, 2016, 06:15:42 pm
Just curious but is it possible to self learn 4U maths?

I've heard mixed rumors from people saying things such as "4U requires critical thinking", "you need a maths tutor for 4U or else you'll do bad", "yes 4U is self learnable" etc but the thing is I don't know what is right and what is wrong
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 16, 2016, 06:30:14 pm
Just curious but is it possible to self learn 4U maths?

I've heard mixed rumors from people saying things such as "4U requires critical thinking", "you need a maths tutor for 4U or else you'll do bad", "yes 4U is self learnable" etc but the thing is I don't know what is right and what is wrong
Things I successfully self learnt:

Integration: 100% of it
Graphs: 100% of it
Complex numbers: All of it except for the complex locus section
Conics: Mostly all of it
Polynomials: 100% of it
Harder 3U: Induction

Things I failed to self learn:

Everything else


So that sounded pretty arrogant admittedly, just saying I learnt this and that. But here's the moral of the story.

It varies from person to person. Some people will have it in them to be able to teach themselves a certain amount of the 4U course. Some may even be able to self learn the entire thing. Whereas it's not possible for others.

What I will say, however, is that you do not NEED tutoring. Whilst having one is most certainly beneficial, this "need" word is. I got my mark in 4U without ANY tutoring whatsoever. (I just offer tutoring simply because I have somewhat of a passion for teaching.)
Title: Re: 4U Maths Question Thread
Post by: massive on October 16, 2016, 07:53:05 pm
Guys how do you prove the attached question by mathematical induction :S
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 16, 2016, 08:18:26 pm
Guys how do you prove the attached question by mathematical induction :S




Title: Re: 4U Maths Question Thread
Post by: massive on October 16, 2016, 09:45:50 pm





thankyou <3
Title: Re: 4U Maths Question Thread
Post by: Blissfulmelodii on October 17, 2016, 03:46:49 pm
Just wondering if anyone has any tips on how to recognise which substitution to use in Integration for 4U.
In 3U integration is the easiest topic for me because majority of the time they give you the substitution to use and it just becomes a matter of algebra but with 4U clearly they don't and I always get stuck in recognising what substitution to use especially when trig is involved. Are there any little markers to recognise or patterns that i am just not seeing?
 
Title: Re: 4U Maths Question Thread
Post by: massive on October 17, 2016, 04:44:35 pm
Guys how do you do this?:

Prove by mathematical induction that the sum of the exterior angles of a n sided convex polygon is 360 degrees

Also in general is there any technique to do geometry induction proofs?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 17, 2016, 05:15:09 pm
Just wondering if anyone has any tips on how to recognise which substitution to use in Integration for 4U.
In 3U integration is the easiest topic for me because majority of the time they give you the substitution to use and it just becomes a matter of algebra but with 4U clearly they don't and I always get stuck in recognising what substitution to use especially when trig is involved. Are there any little markers to recognise or patterns that i am just not seeing?


_________________




Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 17, 2016, 05:17:41 pm



Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 17, 2016, 05:42:04 pm
Guys how do you do this?:

Prove by mathematical induction that the sum of the exterior angles of a n sided convex polygon is 360 degrees

Also in general is there any technique to do geometry induction proofs?
Let me say this now, there is a 0% chance of a geometric induction

See the mathsisfun link


__________________________________



__________________________________



Title: Re: 4U Maths Question Thread
Post by: Blissfulmelodii on October 17, 2016, 07:38:48 pm





OHH that makes so much sense, I had not thought of it in that way. Thank you!!
I've given up on maths for the night but I definitely will post up some questions tomorrow (cause there were a few where the worked solutions only left me more confused)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 17, 2016, 07:59:55 pm
OHH that makes so much sense, I had not thought of it in that way. Thank you!!
I've given up on maths for the night but I definitely will post up some questions tomorrow (cause there were a few where the worked solutions only left me more confused)
Post anytime you want and I'll be there as soon as I'm free (or Jake will) :)
Title: Re: 4U Maths Question Thread
Post by: massive on October 17, 2016, 10:10:20 pm
how does this line:   e1/(n+1) < 1 + 1/n < e1/n become this line: (1+1/n)n < e < (1+1/n)n+1
Like in the solutions there's no other lines in between, so i was just wondering how they got that second line :S
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 17, 2016, 10:11:55 pm
how does this line:   e1/(n+1) < 1 + 1/n < e1/n become this line: (1+1/n)n < e < (1+1/n)n+1
Like in the solutions there's no other lines in between, so i was just wondering how they got that second line :S
Hint: Split the inequality up and work on two sides at once. Then recombine them.
Title: Re: 4U Maths Question Thread
Post by: massive on October 17, 2016, 10:39:37 pm
Hint: Split the inequality up and work on two sides at once. Then recombine them.

genius, thanks i got it now!!
Title: Re: 4U Maths Question Thread
Post by: katherine123 on October 18, 2016, 05:08:01 pm
how to do part ii and iv
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 18, 2016, 05:54:55 pm
how to do part ii and iv






Note that there are other ways of approaching this question. Please consult past answers wherever possible.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 18, 2016, 06:12:37 pm




Title: Re: 4U Maths Question Thread
Post by: massive on October 18, 2016, 06:51:28 pm
guys how do you do part ii :S
Title: Re: 4U Maths Question Thread
Post by: Brenda0708 on October 18, 2016, 07:12:17 pm
Not sure how to do the first 2 questions. Thanks!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 18, 2016, 07:19:32 pm
guys how do you do part ii :S



Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 18, 2016, 07:36:27 pm
Not sure how to do the first 2 questions. Thanks!







I don't see how you're mistaken for part a) - There are only 22*15 favourable outcomes. Can you please post up the answers in the future? I'm not confident with combinatorics.
Title: Re: 4U Maths Question Thread
Post by: massive on October 18, 2016, 07:58:30 pm
guys what do the thetas in conics represent??
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 18, 2016, 09:00:58 pm

guys what do the thetas in conics represent??
The angle it makes with the x-axis at the origin, just like with a circle
Title: Re: 4U Maths Question Thread
Post by: massive on October 18, 2016, 11:50:04 pm
guys how do you do the attached q

ans: 8!/(4!x4!)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 18, 2016, 11:57:48 pm
guys how do you do the attached q

ans: 8!/(4!x4!)




Title: Re: 4U Maths Question Thread
Post by: massive on October 19, 2016, 12:02:11 am






DAMMNN, thanks fam!
Title: Re: 4U Maths Question Thread
Post by: massive on October 19, 2016, 01:28:02 pm
Hey guys whats the general solution to the following qs: (I put more than one coz they're pretty simple.. I think)

1) sin(x-pi/3) = cos2x

2) sin3x + sinx = 0

3) tan2x + cot3x =0

Also is the general solution for cosx = 0 , x= (2n+1)pi OR +/-pi/2 + 2npi ?

Thanks heaps for your help  ;D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 19, 2016, 02:17:48 pm
Hey guys whats the general solution to the following qs: (I put more than one coz they're pretty simple.. I think)

1) sin(x-pi/3) = cos2x

2) sin3x + sinx = 0

3) tan2x + cot3x =0

Also is the general solution for cosx = 0 , x= (2n+1)pi OR +/-pi/2 + 2npi ?

Thanks heaps for your help  ;D
The former is wrong. Look again.
I think you forgot a /2.

Only the latter is correct. (For the cos)
__________________
Hints for (i) and (ii):





I hope you do realise that practically 90% of the questions you've asked won't be on Friday. It's good seeing you tackle hard problems but I don't see the benefit of only doing these and not doing more recent and relevant past papers right now.
Title: Re: 4U Maths Question Thread
Post by: massive on October 19, 2016, 07:17:11 pm
thanks Rui, one more thing- can you quotes the formula for the chord of contact (in conics) or do you have to prove it?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 19, 2016, 07:21:28 pm
thanks Rui, one more thing- can you quotes the formula for the chord of contact (in conics) or do you have to prove it?
Prove it. You only quote things that they give you in the exam.
Title: Re: 4U Maths Question Thread
Post by: Brenda0708 on October 19, 2016, 09:03:04 pm
Hi, if you take the reciprocal of a function that has an oblique asymptote, what happens to the asymptoe? (HSC 2012 Q14bi) --> an oblique asymptote with equation y=2x-1 turned into an asymptote with equation y=0. How do you know? And is this the case for all oblique asymptotes? Thanks.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 19, 2016, 09:13:50 pm
Hi, if you take the reciprocal of a function that has an oblique asymptote, what happens to the asymptoe? (HSC 2012 Q14bi) --> an oblique asymptote with equation y=2x-1 turned into an asymptote with equation y=0. How do you know? And is this the case for all oblique asymptotes? Thanks.




_______________

_______________



_______________


_______________

If you can understand how things work, then it should be much more abundantly clear why f(x) having an oblique asymptote means that 1/f(x) has the horizontal asymptote y=0 (x-axis)
_______________

Title: Re: 4U Maths Question Thread
Post by: Brenda0708 on October 19, 2016, 09:30:17 pm
Oh my, thank you for your explanation. That makes so much more sense now :))). Could you please give an example of what you meant with y=a and y=1/a? I'm trying to make sense of it but I think Im just confusing myself :')
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 19, 2016, 09:31:31 pm
Oh my, thank you for your explanation. That makes so much more sense now :))). Could you please give an example of what you meant with y=a and y=1/a? I'm trying to make sense of it but I think Im just confusing myself :')
Title: Re: 4U Maths Question Thread
Post by: Brenda0708 on October 19, 2016, 09:39:10 pm
Thank you!!  ;D
Title: Re: 4U Maths Question Thread
Post by: massive on October 19, 2016, 09:57:51 pm
should we draw graphs and other diagrams in pen or pencil? I was just wondering because of the whole scanning process and them not being able to see your work unless its in black pen.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 19, 2016, 09:58:14 pm
should we draw graphs and other diagrams in pen or pencil? I was just wondering because of the whole scanning process and them not being able to see your work unless its in black pen.
I drew all of my graphs in pen last year
Title: Re: 4U Maths Question Thread
Post by: massive on October 19, 2016, 11:35:44 pm
guys how do you do this... i can't seem to get the table of values right :/
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 19, 2016, 11:43:33 pm
guys how do you do this... i can't seem to get the table of values right :/
Show us your working
Title: Re: 4U Maths Question Thread
Post by: massive on October 19, 2016, 11:51:10 pm
Show us your working
I don't really have working, this is what i have for the seven ordinates:
0, pi/8, 3pi/8, pi/2, 5pi/8, 6pi/8, pi
The only problem is these aren't evenly spaced :/
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 19, 2016, 11:53:12 pm
I don't really have working, this is what i have for the seven ordinates:
0, pi/8, 3pi/8, pi/2, 5pi/8, 6pi/8, pi
The only problem is these aren't evenly spaced :/
7 evenly spaced coordinates just mean that there should be 6 equally-sized 'intervals' between the ordinates.

Hence, take x=...
0, π/6, 2π/6, 3π/6, 4π/6, 5π/6, 6π/6

i.e.
0, π/6, π/3, π/2, 2π/3, 5π/6, π
Title: Re: 4U Maths Question Thread
Post by: massive on October 19, 2016, 11:54:46 pm
7 evenly spaced coordinates just mean that there should be 6 equally-sized 'intervals' between the ordinates.

Hence, take x=...
0, π/6, 2π/6, 3π/6, 4π/6, 5π/6, 6π/6

i.e.
0, π/6, π/3, π/2, 2π/3, 5π/6, π

Howdu figure that out :O
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 19, 2016, 11:57:23 pm
Howdu figure that out :O
Since there are 6 intervals, consider the difference in the boundaries and divide by 6

(π-0)/6 = π/6

Hence, start at 0 and increase by π/6, six times. It's actually a standard trick that 2U kids need to know, although Simpson's rule is, of course, extremely rare in 4U
Title: Re: 4U Maths Question Thread
Post by: massive on October 20, 2016, 12:41:53 am
Since there are 6 intervals, consider the difference in the boundaries and divide by 6

(π-0)/6 = π/6

Hence, start at 0 and increase by π/6, six times. It's actually a standard trick that 2U kids need to know, although Simpson's rule is, of course, extremely rare in 4U
that's so cool, thanks for sharing!
Title: Re: 4U Maths Question Thread
Post by: massive on October 20, 2016, 12:43:36 am
hey quickly, is the q attached wrong? because cos and sin aren't for hyperbolas, or can this q still be done?
Title: Re: 4U Maths Question Thread
Post by: amandali on October 20, 2016, 02:13:25 am
(http://uploads.tapatalk-cdn.com/20161019/4cc71cd3339cce81c19e5aca09c15b0f.jpg)
for answer to part c (below)  is it correct to have dotted line for the outer line of circle and the vertical line

(http://uploads.tapatalk-cdn.com/20161019/4dd7f38c176d1491867514836c3de400.jpg)

how to do part b

(http://uploads.tapatalk-cdn.com/20161019/33387b03a6d5b7d94071c5a5d7fd51c3.jpg)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2016, 08:13:29 am
hey quickly, is the q attached wrong? because cos and sin aren't for hyperbolas, or can this q still be done?
It can still be done. It's just extensively unconventional.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2016, 08:15:18 am
(http://uploads.tapatalk-cdn.com/20161019/4cc71cd3339cce81c19e5aca09c15b0f.jpg)
for answer to part c (below)  is it correct to have dotted line for the outer line of circle and the vertical line

(http://uploads.tapatalk-cdn.com/20161019/4dd7f38c176d1491867514836c3de400.jpg)

how to do part b

(http://uploads.tapatalk-cdn.com/20161019/33387b03a6d5b7d94071c5a5d7fd51c3.jpg)
Technically yes, the dotted line is correct for c). But I never drew it like that in the HSC. I just kept it blocked because of the ≤.

It's only technically correct because the outer regions are useless. But I believe the HSC will be more lenient on this one if you just draw it blocked and not dotted.
Title: 4U Maths Question Thread
Post by: RuiAce on October 20, 2016, 08:24:48 am
(http://uploads.tapatalk-cdn.com/20161019/4cc71cd3339cce81c19e5aca09c15b0f.jpg)
for answer to part c (below)  is it correct to have dotted line for the outer line of circle and the vertical line

(http://uploads.tapatalk-cdn.com/20161019/4dd7f38c176d1491867514836c3de400.jpg)

how to do part b

(http://uploads.tapatalk-cdn.com/20161019/33387b03a6d5b7d94071c5a5d7fd51c3.jpg)
b):





__________________







That's the MANSW book. That already has all the answers you need. If it's explaining an answer that doesn't make sense you're more than welcome to post that up as well but that should be all you need.
Title: Re: 4U Maths Question Thread
Post by: Brenda0708 on October 20, 2016, 12:18:24 pm
(http://uploads.tapatalk-cdn.com/20161019/50301deeb599b5b14a2ecf9661392f0e.jpg)(http://uploads.tapatalk-cdn.com/20161019/c55ed55960bd54bd986dd218bd5af0f1.jpg)

Hi, is there a way to do ii) without finding the equation of the ellipse? Is the locus saying the distance of z from the origin to (1+3i) + distance of z from the origin to (9+3i) = 10? I.e. PS + PS' = 10? For so then wouldn't the origin be a point on the ellipse?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2016, 12:20:04 pm
(http://uploads.tapatalk-cdn.com/20161019/50301deeb599b5b14a2ecf9661392f0e.jpg)(http://uploads.tapatalk-cdn.com/20161019/c55ed55960bd54bd986dd218bd5af0f1.jpg)

Hi, is there a way to do ii) without finding the equation of the ellipse? Is the locus saying the distance of z from the origin to (1+3i) + distance of z from the origin to (9+3i) = 10? I.e. PS + PS' = 10? For so then wouldn't the origin be a point on the ellipse?
(1+3i) and (9+3i) are the foci.

S and S' do NOT lie on the x-axis. So the origin doesn't have to play a role here.

And of course, PS+PS' = 10 implies that the length of the major axis is 10 (hence the length of the semi-major axis is 5)

So your interpretation is correct. But your interpretation does not mean that the origin has to lie on the ellipse.
_______________________
Title: Re: 4U Maths Question Thread
Post by: massive on October 20, 2016, 12:40:16 pm
guys is the equation for asymptotes y=+/-b/a x right?, but how come the answer to this q is +/4/3 x not +/-3/4 x :S. In conics does 'a' ALWAYS represent the number under x, i always thought a was the bigger number and b was the smaller no.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2016, 12:41:49 pm
guys is the equation for asymptotes y=+/-b/a x right?, but how come the answer to this q is +/4/3 x not +/-3/4 x :S. In conics does 'a' ALWAYS represent the number under x, i always thought a was the bigger number and b was the smaller no.

Title: Re: 4U Maths Question Thread
Post by: massive on October 20, 2016, 01:15:27 pm



wait so is 'a' always the one under x and 'b' the one under y? like when you use b^2=a^2(e^21), B^2 = 16 and a^2=9 right?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2016, 01:19:46 pm
wait so is 'a' always the one under x and 'b' the one under y? like when you use b^2=a^2(e^21), B^2 = 16 and a^2=9 right?
For x^2/a^2-y^2/b^2=1 yeah
Title: Re: 4U Maths Question Thread
Post by: massive on October 20, 2016, 01:44:44 pm
For x^2/a^2-y^2/b^2=1 yeah

is it the same for ellipses as well?
Title: Re: 4U Maths Question Thread
Post by: jakesilove on October 20, 2016, 02:00:18 pm
is it the same for ellipses as well?

Hey! Looks like you're mixing up your formulas a bit; I would take a look at this seriously comprehensive document HERE and come back if you have any more questions!
Title: Re: 4U Maths Question Thread
Post by: massive on October 20, 2016, 03:16:19 pm
Hey! Looks like you're mixing up your formulas a bit; I would take a look at this seriously comprehensive document HERE and come back if you have any more questions!
thanks dude
Title: Re: 4U Maths Question Thread
Post by: katherine123 on October 20, 2016, 05:38:55 pm
image 1: how to do part v)
image 2: how to do part ii) and iii)
Title: Re: 4U Maths Question Thread
Post by: Brenda0708 on October 20, 2016, 05:59:39 pm
(1+3i) and (9+3i) are the foci.

S and S' do NOT lie on the x-axis. So the origin doesn't have to play a role here.

And of course, PS+PS' = 10 implies that the length of the major axis is 10 (hence the length of the semi-major axis is 5)

So your interpretation is correct. But your interpretation does not mean that the origin has to lie on the ellipse.
_______________________


Oh thank you. But then how do you sketch the locus without finding the equation? As in how do you know it lies strictly in the first quadrant? And after finding 2a, how do you find 2b without the equation?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2016, 06:03:24 pm
Oh thank you. But then how do you sketch the locus without finding the equation? As in how do you know it lies strictly in the first quadrant? And after finding 2a, how do you find 2b without the equation?
You don't know everything. Some things you have to infer.

2a is of course, given.

But the distance between the two foci is 2ae. Recall that SS' = 2ae
You are given the foci in the question.

Since the foci are at (1,3) and (9,3), the focal length is 4.

Hence 4 = ae
So e=4/5

Then you can use b^2=a^2(1-e^2) to find the length of the minor axis.

Note also that the ellipse's centre is also at (5,3)
Given the centre, and the length of the major and minor axes, you should be able to sketch the ellipse.
Title: Re: 4U Maths Question Thread
Post by: Brenda0708 on October 20, 2016, 06:07:12 pm
You don't know everything. Some things you have to infer.

2a is of course, given.

But the distance between the two foci is 2ae. Recall that SS' = 2ae
You are given the foci in the question.

Since the foci are at (1,3) and (9,3), the focal length is 4.

Hence 4 = ae
So e=4/5

Then you can use b^2=a^2(1-e^2) to find the length of the minor axis.

Note also that the ellipse's centre is also at (5,3)
Given the centre, and the length of the major and minor axes, you should be able to sketch the ellipse.

ohhhh okay thanks :)
Title: Re: 4U Maths Question Thread
Post by: Brenda0708 on October 20, 2016, 06:10:22 pm
Haha here comes another question :'). For harder 3U, I don't quite get the logic behind proving f(x)>/= 0 for x>0. Can you please interpret this is more detail? Thanks.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2016, 06:11:26 pm
image 1: how to do part v)
image 2: how to do part ii) and iii)




Inequalities is the art of throwing out the garbage.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2016, 06:12:51 pm
Haha here comes another question :'). For harder 3U, I don't quite get the logic behind proving f(x)>/= 0 for x>0. Can you please interpret this is more detail? Thanks.
Just read those dot points again. Read them one at a time. And use a visual aid if you need it.

If it's still too hard, you might want to find a question. There are some that float around in past HSCs.



The basic idea is that if it's increasing, then it's always greater than something. So if you find the left endpoint, you'll know what it's always greater than.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2016, 06:24:31 pm
image 1: how to do part v)
image 2: how to do part ii) and iii)


Title: Re: 4U Maths Question Thread
Post by: Brenda0708 on October 20, 2016, 06:24:42 pm
(http://uploads.tapatalk-cdn.com/20161020/5e8255e0847472beeba2252a3a473974.jpg)
Can you please walk me through this question
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2016, 06:30:07 pm
(http://uploads.tapatalk-cdn.com/20161020/5e8255e0847472beeba2252a3a473974.jpg)
Can you please walk me through this question







Title: Re: 4U Maths Question Thread
Post by: jakesilove on October 20, 2016, 06:31:56 pm









I think it'd be easy to do this part graphically as well
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2016, 06:36:17 pm
I think it'd be easy to do this part graphically as well
Would they let that go in 4U? Cause people mess up drawing to scale graphs.

Also it's not a 1 marker
Title: Re: 4U Maths Question Thread
Post by: jakesilove on October 20, 2016, 06:38:52 pm
Would they let that go in 4U? Cause people mess up drawing to scale graphs.

Also it's not a 1 marker

I'd definitely back a graphical solution to an inequality as full marks, if it's draw to scale. Smart, easy solution; the sort of this 4U markers love. Your solution is kind of a formal 'proof' of the graphical solution; definitely better maths, but like bruh. Cmon. Easy answers = good answers.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2016, 06:40:30 pm
I'd definitely back a graphical solution to an inequality as full marks, if it's draw to scale. Smart, easy solution; the sort of this 4U markers love. Your solution is kind of a formal 'proof' of the graphical solution; definitely better maths, but like bruh. Cmon. Easy answers = good answers.
If you really wanted to be easy then "by inspection" is the best answer please.
Title: Re: 4U Maths Question Thread
Post by: massive on October 20, 2016, 06:46:15 pm
guys whats the difference between uniform and non-uniform circular motion??
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2016, 06:46:50 pm
guys whats the difference between uniform and non-uniform circular motion??
Non-uniform circular motion is not in the HSC 4U course.

It means that the angular velocity is not constant; it varies.
Title: Re: 4U Maths Question Thread
Post by: Brenda0708 on October 20, 2016, 06:55:44 pm









Ohhhhh so is it like this: if we didn't show that g(0)=0, then the curve could be below the x axis somewhere between 0<x<infinity. But seeing that is is =0 at 0 then after x=0 the curve must be >0. Which is what you pretty much said-omg took me a while. LMAO wow thank you!!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2016, 06:57:27 pm
Ohhhhh so is it like this: if we didn't show that g(0)=0, then the curve could be below the x axis somewhere between 0<x<infinity. But seeing that is is =0 at 0 then after x=0 the curve must be >0. Which is what you pretty much said-omg took me a while. LMAO wow thank you!!!
Lel glad that we figured out the issue :P
Title: Re: 4U Maths Question Thread
Post by: jakesilove on October 20, 2016, 07:09:19 pm
image 1: how to do part v)
image 2: how to do part ii) and iii)

Other than the equations given, we also know from basic trig that



Thus, subbing into the first equation,





This directly results in



Halfway there! Let's sub this new relationship into the second proven equation.




As required

Okay, onto the last part. We know that, if theta does not equal zero,



This last part is tricky to see, but obvious once you've seen it; we need to introduce an inequality, so I'll just recall that





As required! Lot's of subbing in, lot's of thinking, but not too difficult a question considering how annoying it first seems :)
Title: Re: 4U Maths Question Thread
Post by: Brenda0708 on October 20, 2016, 07:18:55 pm
(http://uploads.tapatalk-cdn.com/20161020/e3196b8c42fca4b4a1caf9a368362a1d.jpg)
Would this be the correct layout to part iii) of the previous question? :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2016, 07:35:03 pm
(http://uploads.tapatalk-cdn.com/20161020/e3196b8c42fca4b4a1caf9a368362a1d.jpg)
Would this be the correct layout to part iii) of the previous question? :)
Where you said since f"(x) > 0 for all x > 0 you might want to put in brackets (by part (i))

I think, for that one it's like a step by step process. Because we consider two derivatives here, not one.
Since f"(x) > 0 for all x > 0 we know that f'(x) is increasing.
Since f'(0) = 0 we know that f'(x) > 0 for all x > 0 as well now

So since f'(x) > 0, going further back up f(x) is increasing
And then finish it off with whatever you wrote at the end :)
Title: Re: 4U Maths Question Thread
Post by: amandali on October 20, 2016, 08:00:50 pm
(http://uploads.tapatalk-cdn.com/20161020/f34a06fb6272d9c3fc52bc4d93a233bb.jpg)

how to do part i) im confused with the angles for normal force
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2016, 08:32:07 pm
(http://uploads.tapatalk-cdn.com/20161020/f34a06fb6272d9c3fc52bc4d93a233bb.jpg)

how to do part i) im confused with the angles for normal force
Note that F is NOT perpendicular to N here. Here's some things done to the diagram to make it clearer. Note: Corresponding angles on parallel lines used.

The diagram is misleading because it goes against the norm.

(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpsbfurlm4e.png)
Title: Re: 4U Maths Question Thread
Post by: Neutron on October 20, 2016, 08:41:03 pm
(http://uploads.tapatalk-cdn.com/20161020/d8e0df8773cce450704d080e4194724c.jpg)

Yo! Can someone please help with the last question of 2015? I have no clue what's going on! Thank you
Title: Re: 4U Maths Question Thread
Post by: jakesilove on October 20, 2016, 09:25:20 pm
(http://uploads.tapatalk-cdn.com/20161020/d8e0df8773cce450704d080e4194724c.jpg)

Yo! Can someone please help with the last question of 2015? I have no clue what's going on! Thank you

Let's start with i)



Using the standard binomial expansion, we get



It might be a good idea to write out some more terms, I just can't be bothered using LaTeX.

Now, using standard DeMoivre's theorum,



If we equate just REAL parts, we get



As required (obvs add some more middle terms, just couldn't be bothered here)

Okay, clearly here we need to let




Therefore, from part i)



As required. Rui is coming along with the last part of the question :) Let me know if I can expand (haha) on anything :)

(Rui also fixed some of the \( \LaTeX \) up :) )
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2016, 09:35:26 pm
(http://uploads.tapatalk-cdn.com/20161020/d8e0df8773cce450704d080e4194724c.jpg)

Yo! Can someone please help with the last question of 2015? I have no clue what's going on! Thank you






I was so happy when I got this question out in the exam. I was fanboying for 10 seconds...
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2016, 10:18:25 pm




______________________________





_____________________________





_____________________________





Title: Re: 4U Maths Question Thread
Post by: katherine123 on October 21, 2016, 03:47:04 am
how to do part ii)
Title: Re: 4U Maths Question Thread
Post by: Ali_Abbas on October 21, 2016, 05:00:35 am
how to do part ii)

Expand the LHS using the Binomial Theorem and observe that the imaginary terms (which occur when the index is of odd parity) cancel out leaving twice the sum of the real terms. This 2 out the front cancels with the 2 on the RHS.

Next, simply note that if n is divisible by 4, then n/4 = k, k integer. That means that cos(n*pi/4) = cos(k*pi) = (-1)^k = (-1)^(n/4). Combining all these and you will arrive at the required identity.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 21, 2016, 05:09:45 am
how to do part ii)



________________________

Title: Re: 4U Maths Question Thread
Post by: karyn.mcdonald on October 21, 2016, 10:02:27 am
hello... just wondering if someone could help me... :P

I always struggle knowing whether to use radians or degrees in certain situations>? how do I distinguish coz i'd HATE to lose a mark just for not typing it into my calculator!!

thnx :D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 21, 2016, 10:11:07 am
hello... just wondering if someone could help me... :P

I always struggle knowing whether to use radians or degrees in certain situations>? how do I distinguish coz i'd HATE to lose a mark just for not typing it into my calculator!!

thnx :D
You ALWAYS use radians unless you see that circle o floating in the air. Only when that circle is there do you use degrees.

Never assume degrees when there's no symbols - radians is the standard in mathematics.
Title: Re: 4U Maths Question Thread
Post by: kb123 on October 25, 2016, 05:56:39 pm

I don't understand this question.... Help please??

 Show that | |z1|-|z2| | < |z1+z2|. State the condition for
     equality to hold.

Title: 4U Maths Question Thread
Post by: RuiAce on October 25, 2016, 06:03:10 pm
I don't understand this question.... Help please??

 Show that | |z1|-|z2| | < |z1+z2|. State the condition for
     equality to hold.
Is there a typo here. Equality never holds for what you gave and I know two possible questions you intended to give.

Edit actually it does hold if z2=0 but that makes me think the question has a typo more now
Title: Re: 4U Maths Question Thread
Post by: kb123 on October 25, 2016, 07:48:10 pm
Is there a typo here. Equality never holds for what you gave and I know two possible questions you intended to give.

Edit actually it does hold if z2=0 but that makes me think the question has a typo more now

its just supposed to be less than or equal to, but otherwise there aren't any typos
Title: Re: 4U Maths Question Thread
Post by: birdwing341 on October 25, 2016, 07:51:44 pm
Is there a typo here. Equality never holds for what you gave and I know two possible questions you intended to give.

Edit actually it does hold if z2=0 but that makes me think the question has a typo more now

Looks like equality holds if z2=-z1
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 25, 2016, 07:53:43 pm
its just supposed to be less than or equal to, but otherwise there aren't any typos


Looks like equality holds if z2=-z1
Oh right, true.

Equality if z2=-z1 or any one of z1, z2 being equal to 0
Title: Re: 4U Maths Question Thread
Post by: kb123 on October 25, 2016, 08:06:54 pm


Oh right, true.

Equality if z2=-z1 or any one of z1, z2 being equal to 0

I'm sorry I only started ext 2 maths last week, could you explain how you got from the second last step to the last step?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 25, 2016, 08:20:19 pm
Oh I see. Sorry, entirely my fault, must've been too hungry because I made a false assumption.





Title: Re: 4U Maths Question Thread
Post by: kb123 on October 25, 2016, 08:50:42 pm
Oh I see. Sorry, entirely my fault, must've been too hungry because I made a false assumption.







after you let a=z2 and b = z1 - z2

how come you got I z1 + z2 I in the second line??
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 25, 2016, 09:12:02 pm
I must be having a really bad day.

I just did it by paper. I knew what I was doing but I articulated it wrong every single time. Terrible impression.
(http://uploads.tapatalk-cdn.com/20161025/bb63fbd2d028a782e4f0ebcba704fc13.jpg)
Seriously so sorry for the hassle.
Title: Re: 4U Maths Question Thread
Post by: kb123 on October 26, 2016, 03:52:07 pm
I must be having a really bad day.

I just did it by paper. I knew what I was doing but I articulated it wrong every single time. Terrible impression.
(http://uploads.tapatalk-cdn.com/20161025/bb63fbd2d028a782e4f0ebcba704fc13.jpg)
Seriously so sorry for the hassle.

All good, my teacher explained it for me today :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 26, 2016, 08:57:21 pm
All good, my teacher explained it for me today :)
Awesome (y)

Sorry again though; feel a bit bad having a new MX2 student suffer the stupidity invoked by my bad day yesterday.
Title: Re: 4U Maths Question Thread
Post by: kiwiberry on October 29, 2016, 01:56:00 pm
how do you do this 😁

if a and b are two non-zero complex numbers, show that if a/b = ik,
(conjugate of a)b + (conjugate of b)a = 0

sorry I'm not sure how to type the conjugate sign hahah
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 29, 2016, 02:04:24 pm
how do you do this 😁

if a and b are two non-zero complex numbers, show that if a/b = ik,
(conjugate of a)b + (conjugate of b)a = 0

sorry I'm not sure how to type the conjugate sign hahah
If you're lazy with LaTeX just say conj(z)

Is k a constant? If so, is it real or complex?
Title: Re: 4U Maths Question Thread
Post by: kiwiberry on October 29, 2016, 02:13:31 pm
If you're lazy with LaTeX just say conj(z)

Is k a constant? If so, is it real or complex?

k is a real constant, oops should've typed the whole question
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 29, 2016, 02:19:15 pm



Title: Re: 4U Maths Question Thread
Post by: Yasminpotts1105 on November 17, 2016, 06:42:33 pm
If ODEF is a rhombus, where O, D and F represent complex numbers 0, 4 +7i, and 7 + 4i respectively, find
a) the number represented by the point E
b) the length of the diagonals
The area of the rhombus
Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 17, 2016, 06:59:34 pm
If ODEF is a rhombus, where O, D and F represent complex numbers 0, 4 +7i, and 7 + 4i respectively, find
a) the number represented by the point E
b) the length of the diagonals
The area of the rhombus

(http://i.imgur.com/nQV9tWF.png)


_____________________________



_____________________________

This is just the formula area = xy/2 where x and y are the lengths of the diagonal.
Title: Re: 4U Maths Question Thread
Post by: kiwiberry on November 21, 2016, 12:01:43 pm
how would you describe the locus of arg(z-i) - arg(1-iz) = pi/4 geometrically?
Title: Re: 4U Maths Question Thread
Post by: Ali_Abbas on November 21, 2016, 12:45:34 pm
how would you describe the locus of arg(z-i) - arg(1-iz) = pi/4 geometrically?

Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 21, 2016, 12:57:04 pm



I might just drop this question here for now. I'm probably gonna write a little guide on the arg(z-z1) = arg(z-z2) = theta locus. I hate it because it's too much effort to explain.
Title: Re: 4U Maths Question Thread
Post by: Yasminpotts1105 on November 21, 2016, 08:37:12 pm
Prove |z1z2| = |z1||z2|

Find |3-2i|
Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 21, 2016, 08:47:33 pm
Prove |z1z2| = |z1||z2|

Find |3-2i|

Note: Where applicable, you are permitted to use 'cis' notation, but I will never use it.


That second question is just finding the modulus of a given complex number. Is there anything problematic about using a formula?
(Answer is √13)

(Note: You do NOT need to use |z1z2|=|z1||z2| here. Just chuck straight into the formula.)
Title: Re: 4U Maths Question Thread
Post by: hinakamishiro on November 22, 2016, 04:35:39 pm
Hey guys could someone please help me with question 2? Thank you so much! :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 22, 2016, 04:49:09 pm
Hey guys could someone please help me with question 2? Thank you so much! :)
Pointers on part a), because you can't do the other two parts without it.

z1 is obviously any complex number, so put it wherever you want it to be put. Just don't put it at the origin or things become way less obvious for the later parts.

What is z2? The significant part is that z2 involves i times z1. What does multiplication by i do? Rotate the vector π/2 (i.e. 90 degrees) counterclockwise

And what does k do? Well that's just a constant, so it stretches (or shrinks) the vector. So it just means the length might be different. The important thing, is the rotation.

Post up working out if you want further advice.

Extra tip on part b) - lol the typo on the question though
Inevitably, you're gonna have to plot z1+z2
Extra tip on part c)
I suspect that a rectangle is involved... WHY?
Title: Re: 4U Maths Question Thread
Post by: Wales on December 03, 2016, 04:03:00 pm
Bit unsure how to do the question. I've expanded the binomial and now I'm a bit lost.

i) Use the binomial theorem to expand (CosX+iSinX)^3

ii) Use DeMoivre's theorem and your result from part (i) to prove that

    cos^3(X)=1/4Cos3X+3/4CosX

Where X is theta ( not sure how to type that :( )

Thanks, Wales
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 03, 2016, 04:25:32 pm
Bit unsure how to do the question. I've expanded the binomial and now I'm a bit lost.

i) Use the binomial theorem to expand (CosX+iSinX)^3

ii) Use DeMoivre's theorem and your result from part (i) to prove that

    cos^3(X)=1/4Cos3X+3/4CosX

Where X is theta ( not sure how to type that :( )

Thanks, Wales


Title: Re: 4U Maths Question Thread
Post by: Wales on December 03, 2016, 11:46:33 pm





Cheers, I didn't realise you had to equate them. Appreciate it :)
Title: Re: 4U Maths Question Thread
Post by: Wales on December 03, 2016, 11:47:29 pm
http://puu.sh/sCJ9I/789660a744.png

I keep struggling with (ii), I get i but ii gets me nowhere :( It's some drawn out proof...
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 03, 2016, 11:54:35 pm
http://puu.sh/sCJ9I/789660a744.png

I keep struggling with (ii), I get i but ii gets me nowhere :( It's some drawn out proof...

Title: Re: 4U Maths Question Thread
Post by: Yasminpotts1105 on December 11, 2016, 04:22:06 pm
What is the difference when graphing f(x2)) to [f(x)]2 ?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 11, 2016, 04:40:18 pm
What is the difference when graphing f(x2)) to [f(x)]2 ?
Careful with the typos in your BBC code there; you forgot to [/sup] the first one because you forgot the forward slash

The short answer is that the difference is what you tamper with. It comes back down to bracketing - remember, you always do what's INSIDE the brackets first. Not what's OUTSIDE.

For y=[f(x)]2, you're finding the function, and then squaring it AFTER you've done that computation.
For y=f(x2), you're squaring the function BEFORE you apply the function to it.
_____________________

For the longer answer, it is easier to illustrate with an example. Open up Desmos and consider the function f(x)=sin(x)
(Just because it's a bit clearer do illustrate what's going on, and isn't TOO laggy.)




________________________________



Title: Re: 4U Maths Question Thread
Post by: kiwiberry on December 12, 2016, 10:03:36 pm
Points P and Q are the endpoints of focal chord of the ellipse x^2/a^2 + y^2/b^2 = 1. if the parameters at P and Q are θ and α, show that the eccentricity is given by e = sin(θ-α)/(sinθ-sinα)

help please :-\
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 12, 2016, 10:16:39 pm
Points P and Q are the endpoints of focal chord of the ellipse x^2/a^2 + y^2/b^2 = 1. if the parameters at P and Q are θ and α, show that the eccentricity is given by e = sin(θ-α)/(sinθ-sinα)

help please :-\


Title: Re: 4U Maths Question Thread
Post by: Yasminpotts1105 on December 13, 2016, 06:53:55 pm
Sketch |w+3i| / |w-4i| = 1

I tried to let w = x + iy and realize the denominator but I don't understand how to do it with the three terms I end up with.
Title: Re: 4U Maths Question Thread
Post by: Wales on December 13, 2016, 06:58:59 pm
http://puu.sh/sNOEr/240e0f639a.jpg

I get part i but for part ii I don't quite get the arg(z1+z2) expansion. Do I throw in Z1 and Z2 and simplify the expression?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 13, 2016, 07:06:18 pm
Sketch |w+3i| / |w-4i| = 1

I tried to let w = x + iy and realize the denominator but I don't understand how to do it with the three terms I end up with.


Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 13, 2016, 07:26:31 pm
http://puu.sh/sNOEr/240e0f639a.jpg

I get part i but for part ii I don't quite get the arg(z1+z2) expansion. Do I throw in Z1 and Z2 and simplify the expression?



(http://i.imgur.com/Z8eObEX.png)



Title: Re: 4U Maths Question Thread
Post by: Yasminpotts1105 on December 15, 2016, 08:37:40 pm
If z^3 = 1, z cannot equal 1, show that (1+z)^5 = - z

I have tried using demoivre's theorem and mathematical induction or just substituting a value for z but I just can't get an answer. Any helpful hints as to how to approach this?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 15, 2016, 08:43:05 pm
If z^3 = 1, z cannot equal 1, show that (1+z)^5 = - z

I have tried using demoivre's theorem and mathematical induction or just substituting a value for z but I just can't get an answer. Any helpful hints as to how to approach this?



The question can be done by explicitly finding the two values for z, but that is a bit tedious and algebraic tricks save more time here.
Title: Re: 4U Maths Question Thread
Post by: Yasminpotts1105 on December 15, 2016, 09:09:03 pm
THANK YOU SO MUCH, you've just allowed me to actually sleep tonight. :))
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on December 15, 2016, 09:34:40 pm
That math above was a thing of beauty 8)
Title: Re: 4U Maths Question Thread
Post by: bluecookie on December 23, 2016, 11:58:49 am
Find a and b.
2+i=(1+31)/(a+bi)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 23, 2016, 12:02:58 pm
Find a and b.
2+i=(1+31)/(a+bi)
Is that 1+31 or 1+3i there on the top?

Also, this is just elementary complex numbers division. Have you been taught how to do this
Title: Re: 4U Maths Question Thread
Post by: Jakeybaby on December 23, 2016, 12:04:11 pm
Is that 1+31 or 1+3i there on the top?

Also, this is just elementary complex numbers division. Have you been taught how to do this
I'd assume that it's 1+3i, yet again, too quick for me
Title: Re: 4U Maths Question Thread
Post by: bluecookie on December 23, 2016, 12:22:55 pm
I'd assume that it's 1+3i, yet again, too quick for me


Yes its 1+3i.

Is that 1+31 or 1+3i there on the top?

Also, this is just elementary complex numbers division. Have you been taught how to do this

Yep. Aha, I just figured it out then! Thanks :)
Title: Re: 4U Maths Question Thread
Post by: bluecookie on December 23, 2016, 05:57:37 pm
Prove by induction.

a) Conjugate(z1+z2+...zn)=conjugate(z1)+conjugate(z2)+...conjugate(zn)
b) Conjugate(z1*z2*...zn)=conjugate(z1)*conjugate(z2)*...conjugate(zn)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 23, 2016, 10:08:13 pm
Prove by induction.

a) Conjugate(z1+z2+...zn)=conjugate(z1)+conjugate(z2)+...conjugate(zn)
b) Conjugate(z1*z2*...zn)=conjugate(z1)*conjugate(z2)*...conjugate(zn)
When typing in words, feel free to use conj(z1) for the conjugate




The addition question is very similar to this. I will leave it as your exercise to do.
Title: Re: 4U Maths Question Thread
Post by: bluecookie on December 24, 2016, 10:24:33 am
Thank you :)
Title: Re: 4U Maths Question Thread
Post by: bluecookie on January 15, 2017, 05:07:05 pm
7. A chord AB of a circle makes an angle theta with the diameter passing through A. If the area of the minor segment is one-quarter the area of the circle, show that sin2theta=pi/2 - 2theta. Sovle this equation graphically.
8. A chord AB of a circle subtends an angle theta at the centre of the circl. If the perimeter of the minor segment is one-half the circumference of the circle, show that 2*sin(theta/2)=pi-theta. Solve this equation graphically.
Title: Re: 4U Maths Question Thread
Post by: jakesilove on January 15, 2017, 07:23:16 pm
7. A chord AB of a circle makes an angle theta with the diameter passing through A. If the area of the minor segment is one-quarter the area of the circle, show that sin2theta=pi/2 - 2theta. Sovle this equation graphically.
8. A chord AB of a circle subtends an angle theta at the centre of the circl. If the perimeter of the minor segment is one-half the circumference of the circle, show that 2*sin(theta/2)=pi-theta. Solve this equation graphically.

First, make sure to draw out the relevant diagram, and label anything that's easy to figure out. Labeling the origin as 'O', and joining a line between O and B, we find that



As they are both the radius of the circle. As such,



As the base angles in an isosceles triangles are equal. Therefore, since the angle sum of a triangle is pi radians,



Using the area of a minor segment formula, we know that (for subtending angle alpha)



Subbing in the angle we found above, the area of the minor segment is going to be



As the area of the segment is a quarter of the area of the circle, we note that





Play around with this last line, and you'll quickly get the desired result. Then, plot the graph of sin(2theta) and pi/2 - 2theta. Find out where they intersect; it should be around 0.42.

I'm working on the second question now.

The second question is probably much easier; we know that the length of an arc will be



As we need the perimeter of the minor segment, so we also need the length of the chord AB. We can imagine that half the length is like a right angle triangle, with angle theta/2 and hypotenuse r. Therefore, the total length of the chord AB will be



The total perimeter is therefore



We know that the perimeter is half the circumference of the circle. Therefore,



Again, if you play around with this line, you'll get the desired result. Draw it graphically, you should get a number around 1.7
Title: Re: 4U Maths Question Thread
Post by: kiwiberry on January 21, 2017, 01:48:16 am
Hey, could I please get some help with this question :)

1) P(3secθ,2tanθ) and Q(3secθ,-2tanθ) are two points on the hyperbola x2/9 - y2/4 = 1. The normal at P meets the line OQ at R. Show that the locus of R is a concentric hyperbola

Thank you!!!
Title: Re: 4U Maths Question Thread
Post by: jakesilove on January 21, 2017, 09:16:11 am
Hey, could I please get some help with this question :)

1) P(3secθ,2tanθ) and Q(3secθ,-2tanθ) are two points on the hyperbola x2/9 - y2/4 = 1. The normal at P meets the line OQ at R. Show that the locus of R is a concentric hyperbola

Thank you!!!

Hey! Let's start by finding the equation for the normal. I'm going to use implicit differentiation here.





We want to find the normal, so



At P,



The equation of the normal will be



At the point P,







Okay, now let's find the line OQ.



So the equation is



Using simultaneous equations, we can find the point of intercept.



Jesus, this is getting complicated. I assume I made a mistake; I'll leave this here, and explain how the rest would work. Find the point of intersection (both x and y value), then figure out the locus (by subbing one into the other). I have to head out or I'd redo this; good luck!
Title: Re: 4U Maths Question Thread
Post by: kiwiberry on January 21, 2017, 02:27:36 pm
Hey! Let's start by finding the equation for the normal. I'm going to use implicit differentiation here.





We want to find the normal, so



At P,



The equation of the normal will be



At the point P,







Okay, now let's find the line OQ.



So the equation is



Using simultaneous equations, we can find the point of intercept.



Jesus, this is getting complicated. I assume I made a mistake; I'll leave this here, and explain how the rest would work. Find the point of intersection (both x and y value), then figure out the locus (by subbing one into the other). I have to head out or I'd redo this; good luck!

Thanks Jake I worked it out, you got the gradient for OQ the wrong way round haha :)
Title: Re: 4U Maths Question Thread
Post by: jakesilove on January 22, 2017, 06:11:18 pm
Thanks Jake I worked it out, you got the gradient for OQ the wrong way round haha :)

Glad that you got there! This is why you can't do 4U maths in a rush :D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 24, 2017, 12:45:22 pm
I know I've been missing but can I just say how happy I am seeing Jake answer all the 4U maths questions for once (without me beating him all the time) :P
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on January 26, 2017, 11:04:45 pm
Hi Could i please get some help on this question
Much appreciated
[(http://uploads.tapatalk-cdn.com/20170126/ba281edde4d2a25281757fd9281ad27e.jpg)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 26, 2017, 11:07:19 pm
Hi Could i please get some help on this question
Much appreciated
[(http://uploads.tapatalk-cdn.com/20170126/ba281edde4d2a25281757fd9281ad27e.jpg)

Title: Re: 4U Maths Question Thread
Post by: hanaacdr on January 27, 2017, 07:39:08 am




Thank you!
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on January 27, 2017, 09:19:07 am
Hi
could please help me out with these questions

thank you so much!
really appreciate it!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 27, 2017, 09:42:13 am
Hi
could please help me out with these questions

thank you so much!
really appreciate it!


Q9 shouldn't be as hard, unless it gets icky. Post up any progress in working.
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on January 27, 2017, 10:39:40 am


Q9 shouldn't be as hard, unless it gets icky. Post up any progress in working.

this really helps!
thank you!
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on January 27, 2017, 10:51:35 am
this really helps!
thank you!

for question 9,
i keep getting the wrong answer
i am not sure what im doing wrong

thanks
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on January 27, 2017, 11:51:10 am
could you also help me with this question
i have no idea how to do it,
thank you!
Title: Re: 4U Maths Question Thread
Post by: de on January 27, 2017, 11:53:56 am
for question 9,
i keep getting the wrong answer
i am not sure what im doing wrong

thanks

I assume you got to the point where you had from there ignore the 4 obviously and try dividing the numerator and the denominator by cos^2.
Then you're in the position to make another substitution.
Title: Re: 4U Maths Question Thread
Post by: de on January 27, 2017, 11:59:41 am
could you also help me with this question
i have no idea how to do it,
thank you!

For this

The second part is straightforward. The first try a similar trick to what I recommended before (divide numerator and denominator by sin^2).
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 27, 2017, 12:00:00 pm
I assume you got to the point where you had from there ignore the 4 obviously and try dividing the numerator and the denominator by cos^2.
Then you're in the position to make another substitution.

The integral of cosec-squared is known, and is negative cot. Just like how the integral of sec-squared is tan.
Title: Re: 4U Maths Question Thread
Post by: de on January 27, 2017, 12:02:23 pm

The integral of cosec-squared is known, and is negative cot. Just like how the integral of sec-squared is tan.
Didn't know what was assumed or not in HSC, apologies!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 27, 2017, 12:07:16 pm
could you also help me with this question
i have no idea how to do it,
thank you!
In general, you have to make the substitution t=tan(x/2)

Because of how that's written, i.e. a + b sin(x) where a and b are either 1 or -1, you can manipulate the Pythagorean trigonometric identity that de used.
Didn't know what was assumed or not in HSC, apologies!
All g
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on January 27, 2017, 01:37:57 pm
In general, you have to make the substitution t=tan(x/2)

Because of how that's written, i.e. a + b sin(x) where a and b are either 1 or -1, you can manipulate the Pythagorean trigonometric identity that de used.All g

thank you so much!
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on January 27, 2017, 01:38:47 pm
Didn't know what was assumed or not in HSC, apologies!

thank youuu
much appreciated!
Title: Re: 4U Maths Question Thread
Post by: cmadeleine on January 27, 2017, 11:42:29 pm
Hi! Could you please help me with this question: find the Cartesian equation of the locus of arg(z-4)-arg(z+4) = pi/4?

Thank you! :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 27, 2017, 11:59:55 pm
Hi! Could you please help me with this question: find the Cartesian equation of the locus of arg(z-4)-arg(z+4) = pi/4?

Thank you! :)
For now, I am going to redirect you to the discussion here.

Use the resources I have linked to.
Title: Re: 4U Maths Question Thread
Post by: cmadeleine on January 28, 2017, 12:01:58 am
For now, I am going to redirect you to the discussion here.

Use the resources I have linked to.

Thank you so much!!
Title: Re: 4U Maths Question Thread
Post by: kiwiberry on January 28, 2017, 09:01:49 pm
Hey could please I get some help on these :)

1) P(cp,c/p) and Q(-cp,-c/p) are points on the rectangular hyperbola xy=c2. The normal at P to the hyperbola meets the curve again at R. Find the coordinates of R, and hence show that \(\angle PQR\) is a right angle

2) P(3p,3/p) and Q(3q,3/q) are points on different branches of the hyperbola xy=9. The tangents at P and Q meet at T. Find the locus of T if the positions of P and Q vary so that the chord PQ passes through (0,4).
I managed to find the locus as x=9/2, but the answer says that it is x=9/2, y<0. I was wondering why y has to be less than 0? T is \(( \frac{6pq}{p+q} ,\frac{6}{p+q})\) for reference

Thank you!
Title: Re: 4U Maths Question Thread
Post by: jakesilove on January 28, 2017, 09:33:10 pm
Hey could please I get some help on these :)

1) P(cp,c/p) and Q(-cp,-c/p) are points on the rectangular hyperbola xy=c2. The normal at P to the hyperbola meets the curve again at R. Find the coordinates of R, and hence show that \(\angle PQR\) is a right angle

2) P(3p,3/p) and Q(3q,3/q) are points on different branches of the hyperbola xy=9. The tangents at P and Q meet at T. Find the locus of T if the positions of P and Q vary so that the chord PQ passes through (0,4).
I managed to find the locus as x=9/2, but the answer says that it is x=9/2, y<0. I was wondering why y has to be less than 0? T is \(( \frac{6pq}{p+q} ,\frac{6}{p+q})\) for reference

Thank you!

Hey!

Let's find the normal at P


So, the normal will be






So,



This meets the hyperbola again where y=c^2/x



Solve this, and you'll get



The first one is obvious (it's the point from which the normal is from!), but the second one is what we're looking for.

The y value will be



Now, you just need to find the equation of the lines from P to Q, and Q to R. Show that the two gradients are inverse negatives of each other, and bam! Q1 done.

I'll work on Q2 now.


Okay, so for Q2, I'm actually not sure! Might have to call in the infinite wisdom of RuiAce here
Title: Re: 4U Maths Question Thread
Post by: beau77bro on January 28, 2017, 11:39:25 pm
hey jake, or whoever decides to answer this. im having trouble with this question and its kinda put me off the whole of fitzpatrick ahahha. but i was just wondering how we are supposed to approach this? question 6b

(http://16325884_1357187997635286_1495004055_o.jpg)


Title: Re: 4U Maths Question Thread
Post by: kiwiberry on January 28, 2017, 11:49:00 pm
Hey!

Let's find the normal at P


So, the normal will be






So,



This meets the hyperbola again where y=c^2/x



Solve this, and you'll get



The first one is obvious (it's the point from which the normal is from!), but the second one is what we're looking for.

The y value will be



Now, you just need to find the equation of the lines from P to Q, and Q to R. Show that the two gradients are inverse negatives of each other, and bam! Q1 done.

I'll work on Q2 now.


Okay, so for Q2, I'm actually not sure! Might have to call in the infinite wisdom of RuiAce here

Thank you :-)
Title: Re: 4U Maths Question Thread
Post by: Shadowxo on January 28, 2017, 11:55:31 pm
hey jake, or whoever decides to answer this. im having trouble with this question and its kinda put me off the whole of fitzpatrick ahahha. but i was just wondering how we are supposed to approach this? question 6b

(http://16325884_1357187997635286_1495004055_o.jpg)




Just expand it out and go from there
x + xi + 2y - 3yi = 10
(x + 2y) + (x - 3y)i = 10
x + 2y = 10
x - 3y = 0
5y = 10
y = 2
x = 6
Title: Re: 4U Maths Question Thread
Post by: beau77bro on January 28, 2017, 11:56:01 pm
hello again i have a lot of questions about section 3.2 of 4u cambridge, mostly the parts involving multiple parametric points on ellipses or hyperbola. so the parts with chords between points: P (acosß, bsinß) and Q (acos∂, bsin∂). firstly, in the explanation/ working to show the lines equation and gradient there are a lot of steps missing and i cant understand how they get anywhere, mostly the gradients? secondly, how relevant/do we need to know this at all because my teacher doesnt even know how to do it really i dont think? and if we need to know, is it at a level of deriving, im gonna assume so because it cant hurt but still?

so yea if anyone wants to shoot up some kind of working for the gradients that would be great. thankyouuuuuuuu i hate this whole Ex.
p.s. im probably gonna reply with a couple more questions because there isnt a lot of explanations really, or i dont get them. but thankyou
Title: Re: 4U Maths Question Thread
Post by: kiwiberry on January 29, 2017, 12:10:07 am
hello again i have a lot of questions about section 3.2 of 4u cambridge, mostly the parts involving multiple parametric points on ellipses or hyperbola. so the parts with chords between points: P (acosß, bsinß) and Q (acos∂, bsin∂). firstly, in the explanation/ working to show the lines equation and gradient there are a lot of steps missing and i cant understand how they get anywhere, mostly the gradients? secondly, how relevant/do we need to know this at all because my teacher doesnt even know how to do it really i dont think? and if we need to know, is it at a level of deriving, im gonna assume so because it cant hurt but still?

so yea if anyone wants to shoot up some kind of working for the gradients that would be great. thankyouuuuuuuu i hate this whole Ex.
p.s. im probably gonna reply with a couple more questions because there isnt a lot of explanations really, or i dont get them. but thankyou

They've used the sums to products formulas to simplify the gradients
(http://www.sosmath.com/trig/prodform/img6.gif)
Title: Re: 4U Maths Question Thread
Post by: beau77bro on January 29, 2017, 12:20:28 am
Hey could please I get some help on these :)

1) P(cp,c/p) and Q(-cp,-c/p) are points on the rectangular hyperbola xy=c2. The normal at P to the hyperbola meets the curve again at R. Find the coordinates of R, and hence show that \(\angle PQR\) is a right angle

2) P(3p,3/p) and Q(3q,3/q) are points on different branches of the hyperbola xy=9. The tangents at P and Q meet at T. Find the locus of T if the positions of P and Q vary so that the chord PQ passes through (0,4).
I managed to find the locus as x=9/2, but the answer says that it is x=9/2, y<0. I was wondering why y has to be less than 0? T is \(( \frac{6pq}{p+q} ,\frac{6}{p+q})\) for reference

Thank you!

ok so see the photos for q2:

but pretty much y<0 because the point p and q are on separate branches of a standard xy = c^2, so if we say q is on the branch in the 3rd quadrant, then its coordinates are all negative Q (3q, 3/q) and hence q is negative.
 
and when u rearrange  3(p+q) = 4pq
into p + q = 4pq/3

and sub into y = 6/(p + q)
you get:  y = 9/2pq   

and since p is positive and q is negative, y cannot be greater than 0, and it cant equal 0 because there isnt a variable on the top of the fraction.
Title: Re: 4U Maths Question Thread
Post by: beau77bro on January 29, 2017, 12:22:53 am
They've used the sums to products formulas to simplify the gradients
(http://www.sosmath.com/trig/prodform/img6.gif)

uhhhhh i dont think ive learnt this, how do you get these equations? do we need to know how to derive?
Title: Re: 4U Maths Question Thread
Post by: kiwiberry on January 29, 2017, 12:27:39 am
ok so see the photos for q2:

but pretty much y<0 because the point p and q are on separate branches of a standard xy = c^2, so if we say q is on the branch in the 3rd quadrant, then its coordinates are all negative Q (3q, 3/q) and hence q is negative.
 
and when u rearrange  3(p+q) = 4pq
into p + q = 4pq/3

and sub into y = 6/(p + q)
you get:  y = 9/2pq   

and since p is positive and q is negative, y cannot be greater than 0, and it cant equal 0 because there isnt a variable on the top of the fraction.

omg that makes sense, thanks so much!! :)
Title: Re: 4U Maths Question Thread
Post by: kiwiberry on January 29, 2017, 12:33:35 am
uhhhhh i dont think ive learnt this, how do you get these equations? do we need to know how to derive?

I'm pretty sure you don't have to derive them to use them, but there's a proof in 3U year12 cambridge I believe
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 29, 2017, 12:45:39 am
uhhhhh i dont think ive learnt this, how do you get these equations? do we need to know how to derive?



Title: Re: 4U Maths Question Thread
Post by: QC on January 29, 2017, 02:30:55 am
Hi, can someone help me with this question?
I feel like I know how to do it and I just suck at algebra or something. I've tried both realising the denominator and just forming one fraction, cross multiplying then solving the complex and real components separately as real and imaginary components.
Thanks :)
Title: Re: 4U Maths Question Thread
Post by: kiwiberry on January 29, 2017, 08:20:30 am
Hi, can someone help me with this question?
I feel like I know how to do it and I just suck at algebra or something. I've tried both realising the denominator and just forming one fraction, cross multiplying then solving the complex and real components separately as real and imaginary components.
Thanks :)



you can equate real and imaginary parts from here :)
Title: Re: 4U Maths Question Thread
Post by: QC on January 29, 2017, 12:56:23 pm
Idk what to do, are you allowed to cross multiply since it isn't a proof?
Title: Re: 4U Maths Question Thread
Post by: jakesilove on January 29, 2017, 12:59:34 pm
Idk what to do, are you allowed to cross multiply since it isn't a proof?

You definitely can cross multiply! If you didn't want to do that, I would rationalise the denominator on the LHS, and see what pops out. Let me know if you want a full proof!

However, I think that if you multiply the LHS (both denom and numer) by (1+sin(theta)+icos(theta)), things should work out.
Title: Re: 4U Maths Question Thread
Post by: Shadowxo on January 29, 2017, 01:13:46 pm
Here's my solution if you're struggling :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 29, 2017, 01:46:05 pm
Idk what to do, are you allowed to cross multiply since it isn't a proof?

This is the clever way going about it. The standard way is, as mentioned above, realising the denominator.
Title: Re: 4U Maths Question Thread
Post by: QC on January 29, 2017, 03:10:44 pm

This is the clever way going about it. The standard way is, as mentioned above, realising the denominator.
WOW that sol is so nice, that's a pretty good trick factorising the i out at the start, i'll keep it in mind. Thanks !
Title: Re: 4U Maths Question Thread
Post by: beau77bro on January 29, 2017, 09:08:16 pm

This is the clever way going about it. The standard way is, as mentioned above, realising the denominator.

this is such a nice proof, but i was just wondering what did you do in the middle step where u got half angles for cis? what exactly were the steps/ method you used
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 29, 2017, 09:15:25 pm
this is such a nice proof, but i was just wondering what did you do in the middle step where u got half angles for cis? what exactly were the steps/ method you used
Title: Re: 4U Maths Question Thread
Post by: michaelalt on January 31, 2017, 10:33:18 pm
Hi! I've been struggling with complex numbers alot and was wondering if i could get some help with the questions below. I have tried to make sense of them, but just end up confused. Any help on any of them would be greatly appreciated!!
Title: Re: 4U Maths Question Thread
Post by: Shadowxo on January 31, 2017, 10:45:11 pm
I'll just do the first question
For this question, it's helpful to graph the points.
If point A represents alpha, alpha = cis(theta)
If B is A rotated anticlockwise 2pi/3 and halved, it equals 1/2(cis(theta + 3pi/2))
So it equals 1/2(cis(theta)*cis(3pi/2)) = 1/2(A*cis(3pi/2)) = A/2cis(3pi/2) = D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 31, 2017, 11:24:39 pm
Hi! I've been struggling with complex numbers alot and was wondering if i could get some help with the questions below. I have tried to make sense of them, but just end up confused. Any help on any of them would be greatly appreciated!!
With these questions though, you're not going to get anywhere without a diagram. For any question, post up a diagram with any progress, and we can finish it off.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 01, 2017, 09:46:42 am
Thanks for your help on 1, it does make sense. But I'm also stuck on 3. This is the diagram I have drawn so far.



___________________________

(http://i.imgur.com/QFnn6c6.png)

Title: Re: 4U Maths Question Thread
Post by: armtistic on February 01, 2017, 05:13:27 pm
Hey guys

I've been doing some revision questions on conics and I came across this one that I'm struggling with.

Any help would be appreciated  :D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 01, 2017, 05:54:22 pm
Hey guys

I've been doing some revision questions on conics and I came across this one that I'm struggling with.

Any help would be appreciated  :D

(http://i.imgur.com/DisGkJL.png)






______________________________

______________________________




Title: Re: 4U Maths Question Thread
Post by: QC on February 02, 2017, 06:11:16 pm
How do you systematically make sure you can answer any circle geo question or is it not possible. Also, do you have a checklist or something you run through when u see a circle geo question?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 02, 2017, 07:20:56 pm
How do you systematically make sure you can answer any circle geo question or is it not possible. Also, do you have a checklist or something you run through when u see a circle geo question?


(http://i.imgur.com/goYXwwyh.jpg)


Apologies for the somewhat messy handwriting

Edit: First line should say join GF
Title: Re: 4U Maths Question Thread
Post by: Rathin on February 03, 2017, 04:28:31 pm
Write the equation of the locus at point P that moves such that its distance from (3,0) is 4/5 as its distance from the line x=2.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 03, 2017, 04:45:41 pm
Write the equation of the locus at point P that moves such that its distance from (3,0) is 4/5 as its distance from the line x=2.




You can rearrange however you want.
Title: Re: 4U Maths Question Thread
Post by: armtistic on February 03, 2017, 08:52:42 pm

(http://i.imgur.com/DisGkJL.png)






______________________________

______________________________







Holy Crap

Thanks a lot Rui! That was really well explained  :D
Title: Re: 4U Maths Question Thread
Post by: armtistic on February 04, 2017, 05:59:28 pm
Hey guys

I know I've been asking a lot of questions here but I have a topic test soon and I want to make sure all my understanding is solid  :)

Could someone help with this conics question?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 04, 2017, 07:27:31 pm
Hey guys

I know I've been asking a lot of questions here but I have a topic test soon and I want to make sure all my understanding is solid  :)

Could someone help with this conics question?
Are you sure there hasn't been a typo anywhere?

I attempted this question, and got something catastrophically messy. Not to mention the bad grammar in the first sentence.
Title: Re: 4U Maths Question Thread
Post by: Rathin on February 05, 2017, 04:11:27 pm
I got the locus as x+8y=0 but apprantely there is a restriction to the domain..how do I find that out?
Title: Re: 4U Maths Question Thread
Post by: Shadowxo on February 05, 2017, 05:22:52 pm
I got the locus as x+8y=0 but apprantely there is a restriction to the domain..how do I find that out?

I'm not very familiar with those kinds of questions (not in VCE study design) but for an ellipse there is a domain restriction, in this case x can only be between -4 and 4 inclusive or else x2 would be more than 16, and the midpoint of the chord cannot be outside the circle, so x must be between -4 and 4
Does this help? :P
Title: Re: 4U Maths Question Thread
Post by: Rathin on February 05, 2017, 05:48:29 pm
Answer says -16/sqrt(17)<=x<=16/sqrt(17) for the restriction..have no idea how they got that.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 05, 2017, 06:07:09 pm
I got the locus as x+8y=0 but apprantely there is a restriction to the domain..how do I find that out?
As per this GeoGebra simulation, your locus is a correct one






Title: Re: 4U Maths Question Thread
Post by: Syndicate on February 05, 2017, 06:10:11 pm
Answer says -16/sqrt(17)<=x<=16/sqrt(17) for the restriction..have no idea how they got that.

Just adding...
For the chord to exist, the line must join two points at the circumference of ellipse, which means it cannot exist at certain points due to its gradient.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 05, 2017, 06:18:04 pm
Just adding...
For the chord to exist, the line must join two points at the circumference of ellipse, which means it cannot exist at certain points due to its gradient.
Pretty sure I addressed this.

Also, it is ok for the two points to coincide.
[url=https://ggbm.at/BEZYGzee]

Title: Re: 4U Maths Question Thread
Post by: Syndicate on February 05, 2017, 06:25:25 pm
Pretty sure I addressed this.

Also, it is ok for the two points to coincide.

Sorry, I didn't realise you had already posted. I was replying to Shadowxo
Title: Re: 4U Maths Question Thread
Post by: QC on February 07, 2017, 08:19:49 pm
Hey, so I was pretty happy with how much of this question I think I got but couldn't get the final result (this was in the last question of 2000 4u HSC). I got the first part then tried t-method, and tried subbing in the hint and equating imaginary parts but my algebra got super messy and I don't exactly know how to get the second part. Is there any faster way using the properties of Imaginary parts etc.? Also, what are your opinions on people not even bothering with these questions cos they would be very tough to get in the exam? My teacher says if you want a safe B6 get all the easy questions (all other than last/8) 100% and look through the rest if you have time. In 4u there is marginal benefit benefit to your atar going from a 90-92 to a 94-95 even though it requires far more effort. My teacher says your better off maximising your 3u or other subjects as there is a huge difference between 90-92 and 94-95 in english, chem, phys, eco etc. Like these questions are fun for me to do so I will probably try them anyway but I can see their point.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 07, 2017, 09:07:10 pm
Hey, so I was pretty happy with how much of this question I think I got but couldn't get the final result (this was in the last question of 2000 4u HSC). I got the first part then tried t-method, and tried subbing in the hint and equating imaginary parts but my algebra got super messy and I don't exactly know how to get the second part. Is there any faster way using the properties of Imaginary parts etc.? Also, what are your opinions on people not even bothering with these questions cos they would be very tough to get in the exam? My teacher says if you want a safe B6 get all the easy questions (all other than last/8) 100% and look through the rest if you have time. In 4u there is marginal benefit benefit to your atar going from a 90-92 to a 94-95 even though it requires far more effort. My teacher says your better off maximising your 3u or other subjects as there is a huge difference between 90-92 and 94-95 in english, chem, phys, eco etc. Like these questions are fun for me to do so I will probably try them anyway but I can see their point.






Maybe a few strategies with de Moivre's theorem, but otherwise what strategy. I just brute forced it and hoped for the best.

Also, yeah I agree with your teacher. Especially since this was before 2001 (even if it was barely before 2001) and no longer reflects the difficulty of the HSC. I only did the questions up to the new version of the HSC.
Title: Re: 4U Maths Question Thread
Post by: QC on February 07, 2017, 09:26:48 pm
oh dang I didn't apply de Moivres again with the (costheta+isintheta)^2 only the power n so i was getting random half powers and I couldn't get it. Thanks man, its pretty annoying this stuff isnt online
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 07, 2017, 09:32:27 pm
oh dang I didn't apply de Moivres again with the (costheta+isintheta)^2 only the power n so i was getting random half powers and I couldn't get it. Thanks man, its pretty annoying this stuff isnt online
You said this was a past HSC paper question?
Title: Re: 4U Maths Question Thread
Post by: hinakamishiro on February 10, 2017, 05:36:20 pm
Hey guys could i have some help with part b please? Thanks  :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 10, 2017, 06:03:09 pm
Hey guys could i have some help with part b please? Thanks  :)



Title: Re: 4U Maths Question Thread
Post by: Rathin on February 12, 2017, 03:22:34 pm
Find the equation of the tangents to the ellipse 4x^2+9y^2=36 given that these tangents are perpendicular to the line 3x+2y=5.

I got 2x-3y+6=0
Answer is 2x-3y+6sqrt(2)=0
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 12, 2017, 03:25:53 pm
Find the equation of the tangents to the ellipse 4x^2+9y^2=36 given that these tangents are perpendicular to the line 3x+2y=5.

I got 2x-3y+6=0
Answer is 2x-3y+6sqrt(2)=0
Judging by what the question asks it's likely you made a little mistake somewhere in your working, especially since you were only off by a factor. Can you please post it up?
Title: Re: 4U Maths Question Thread
Post by: Rathin on February 12, 2017, 03:39:20 pm
Judging by what the question asks it's likely you made a little mistake somewhere in your working, especially since you were only off by a factor. Can you please post it up?

I have done it completely wrong.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 12, 2017, 03:40:00 pm






Title: Re: 4U Maths Question Thread
Post by: Rathin on February 12, 2017, 03:57:29 pm








Thanks, that was very smart :)
I was trying to equate the implicitly differentiated ellipse and the given gradient (2/3) to come out with a point what the gradient of 2/3 is true and then find the eqn of it but that didn't work out.
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on February 12, 2017, 05:17:56 pm
Hi!!
could i get some help on this volumes question please,
i have done it but i keep getting the wrong answer
thank you!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 12, 2017, 05:42:36 pm
Hi!!
could i get some help on this volumes question please,
i have done it but i keep getting the wrong answer
thank you!
Which method do you want? Shells is easier unless you haven't been taught how to use it.
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on February 12, 2017, 05:58:54 pm
Yes i have been taught shells, but we were given this question to use the slicing method
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 12, 2017, 06:38:54 pm
(http://i.imgur.com/4H8bP8Q.png)


(http://i.imgur.com/8nKDIWf.png)



Title: Re: 4U Maths Question Thread
Post by: Rathin on February 17, 2017, 12:51:35 pm
The ellipse x^2/a^2 + y^2/b^ =1 meets the y=axis at C and D. Tangents drawn to C and D on the ellipse meet the tangent in (xcos(θ))/a + (ysin(θ))/b=1 at the points E and F respectively. Prove that CE*DF=a^2
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 17, 2017, 01:22:22 pm
The ellipse x^2/a^2 + y^2/b^ =1 meets the y=axis at C and D. Tangents drawn to C and D on the ellipse meet the tangent in (xcos(θ))/a + (ysin(θ))/b=1 at the points E and F respectively. Prove that CE*DF=a^2






Title: Re: 4U Maths Question Thread
Post by: kinky_khan on February 18, 2017, 06:14:29 pm
HEYYYY Jake, I was struggling to answer the following questions and I was wondering if you could help me with them? thank you SOOOO MUCHH!!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 18, 2017, 06:42:44 pm
HEYYYY Jake, I was struggling to answer the following questions and I was wondering if you could help me with them? thank you SOOOO MUCHH!!!
There are several questions here, and we will not do them all without seeing your attempts at them. I will provide the answer to a random choice of two.

____________________________________


GeoGebra simulation attached so you can play around with different values of a and b. This will be the diagram for reference.
A represents the complex number omega, and P represents any complex number z that satisfies the locus.




Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 18, 2017, 07:03:28 pm
HEYYYY Jake, I was struggling to answer the following questions and I was wondering if you could help me with them? thank you SOOOO MUCHH!!!







_______________________________________


Title: Re: 4U Maths Question Thread
Post by: hinakamishiro on February 18, 2017, 07:49:05 pm
Hi guys could i have some help with conics question 3 and 5b please? Thank you!
Title: Re: 4U Maths Question Thread
Post by: kiwiberry on February 18, 2017, 08:40:51 pm
Hi guys could i have some help with conics question 3 and 5b please? Thank you!

Hey! for 5b):
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 18, 2017, 08:49:10 pm
Hi guys could i have some help with conics question 3 and 5b please? Thank you!

Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 20, 2017, 08:59:59 am
any help would be greatly appreciated :) thanks
Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 20, 2017, 09:04:27 am
I also don't understand how to do these questions (3 a & b).
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on February 20, 2017, 09:51:17 am
any help would be greatly appreciated :) thanks

Hey hey! Here is the working for this question! ;D



You can then manipulate that answer any way that suits you - I hope that helps! ;D I'll get to your next question in a little bit ;D

Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 20, 2017, 10:56:57 am
Hey hey! Here is the working for this question! ;D



You can then manipulate that answer any way that suits you - I hope that helps! ;D I'll get to your next question in a little bit ;D

Hey Jamon!
Thank you so much for this, if i was trying to manipulate it to get the answer 11/10 +3/10i how would you go about getting it?
Would i substitute the value of z into the original equation and work it out that way?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 20, 2017, 11:02:14 am
Hey Jamon!
Thank you so much for this, if i was trying to manipulate it to get the answer 11/10 +3/10i how would you go about getting it?


Jamon Edit - Fixed er up ;D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 20, 2017, 11:04:51 am
I also don't understand how to do these questions (3 a & b).

Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on February 20, 2017, 11:07:53 am
Just came back to do those! Ta Rui ;D

Edit: On the answer above, I wrote the question wrong in the second step: I'll go back and fix the errors in my working and then change Rui's to match, sorry! ;D
Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 20, 2017, 12:01:49 pm
Just came back to do those! Ta Rui ;D

Edit: On the answer above, I wrote the question wrong in the second step: I'll go back and fix the errors in my working and then change Rui's to match, sorry! ;D

Thanks again Jamon + Rui! You guys are lifesavers.
Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 20, 2017, 12:25:10 pm
Hey hey! Here is the working for this question! ;D



You can then manipulate that answer any way that suits you - I hope that helps! ;D I'll get to your next question in a little bit ;D

Can i just ask- how did you get to this (iz=(3+2i)z-3-2i) step? I don't understand the continuation from the previous step to this one.
Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 20, 2017, 12:41:51 pm



This makes sense! However, I tried doing part (b.) and it didn't work out as well as I thought it would...
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on February 20, 2017, 12:52:56 pm
Can i just ask- how did you get to this (iz=(3+2i)z-3-2i) step? I don't understand the continuation from the previous step to this one.

Heres the step in between! It's just expansion and then re-factorisation (or partial expansion, if you will) ;D

Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 20, 2017, 12:54:34 pm
Heres the step in between! It's just expansion and then re-factorisation (or partial expansion, if you will) ;D



Ahhhh! I get it! Thank you Jamon!!
Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 20, 2017, 12:59:38 pm
Hey hey! Here is the working for this question! ;D



You can then manipulate that answer any way that suits you - I hope that helps! ;D I'll get to your next question in a little bit ;D



When you go on to this ((3+i)z=3+2i) step, where did the iz on the left hand side go?
Title: Re: 4U Maths Question Thread
Post by: kiwiberry on February 20, 2017, 01:20:00 pm
When you go on to this ((3+i)z=3+2i) step, where did the iz on the left hand side go?
Title: Re: 4U Maths Question Thread
Post by: kiwiberry on February 20, 2017, 01:33:44 pm
This makes sense! However, I tried doing part (b.) and it didn't work out as well as I thought it would...
From your second line:

Solve simultaneously from here :)
Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 20, 2017, 07:42:57 pm
From your second line:

Solve simultaneously from here :)

youre the best. it makes so much sense and seems so simple when you do it, thanks heaps!
Title: Re: 4U Maths Question Thread
Post by: smile123 on February 21, 2017, 12:48:18 pm
please help
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 21, 2017, 12:57:59 pm
please help


Title: Re: 4U Maths Question Thread
Post by: smile123 on February 21, 2017, 01:04:43 pm
thanks  :)
Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 21, 2017, 01:19:59 pm
How would you approach this question? I tried expanding it out but it got really messy and I just got confused :(
Title: Re: 4U Maths Question Thread
Post by: jakesilove on February 21, 2017, 01:30:10 pm
How would you approach this question? I tried expanding it out but it got really messy and I just got confused :(

Hey!

Start by writing everything out in cis form.




Then, use DeMoivre's




Finally, use cis rules to get the answer. Remember that when we divide one complex number in mod-arg form by another, we divide their modulus and subtract their argument. I'll leave that bit to you :)
Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 21, 2017, 01:46:29 pm
Hey!

Start by writing everything out in cis form.




Then, use DeMoivre's




Finally, use cis rules to get the answer. Remember that when we divide one complex number in mod-arg form by another, we divide their modulus and subtract their argument. I'll leave that bit to you :)

Hey Jake- thank you! That makes sense, however, I don't really understand what to do after this...
Title: Re: 4U Maths Question Thread
Post by: jakesilove on February 21, 2017, 01:56:24 pm
Hey Jake- thank you! That makes sense, however, I don't really understand what to do after this...

Just what I wrote above; when we divide two complex numbers in cis notation, we divide their modulus and subtract their argument. So,



You can use this to get the final answer
Title: Re: 4U Maths Question Thread
Post by: smile123 on February 21, 2017, 07:42:20 pm
please help
Title: Re: 4U Maths Question Thread
Post by: beau77bro on February 21, 2017, 11:00:18 pm
Heyyyy I keep forgetting how to do this question someone please help hahaha ty


(http://uploads.tapatalk-cdn.com/20170221/bdbe195a2dd1697db249517b4b54ede9.jpg)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 21, 2017, 11:02:44 pm
Heyyyy I keep forgetting how to do this question someone please help hahaha ty


(http://uploads.tapatalk-cdn.com/20170221/bdbe195a2dd1697db249517b4b54ede9.jpg)

Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 21, 2017, 11:04:58 pm
please help
Hint: Use trigonometry.

Use a right angled triangle and the fact cos is the adjacent side divided by the hypotenuse. Then, use the rule of ASTC (cosine is positive in the 4th quadrant)

(Alternatively, use the parametric definition of the circle)
Title: Re: 4U Maths Question Thread
Post by: Kle123 on February 22, 2017, 12:43:27 am
I don't know/forgot how to do this question. Please help. Doing 1part is sufficient (then i can do the other as both questions are similar,,,, probably?). Thank you
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 22, 2017, 01:20:35 am
I don't know/forgot how to do this question. Please help. Doing 1part is sufficient (then i can do the other as both questions are similar,,,, probably?). Thank you
Somewhat untidy.






Title: Re: 4U Maths Question Thread
Post by: Kle123 on February 22, 2017, 08:53:32 am
Thank you night owl Rui.  :D
Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 22, 2017, 09:14:51 am
Just what I wrote above; when we divide two complex numbers in cis notation, we divide their modulus and subtract their argument. So,



You can use this to get the final answer

Like this?
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on February 22, 2017, 09:21:52 am
Like this?

Looks perfect to me ;D well done!!
Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 22, 2017, 09:31:17 am
Looks perfect to me ;D well done!!

The answers say 3π/10 though so I don't know where I went wrong.
Title: Re: 4U Maths Question Thread
Post by: jakesilove on February 22, 2017, 10:25:55 am
The answers say 3π/10 though so I don't know where I went wrong.

Remember; on the Argand diagram, you can add or subtract 2*pi without it changing the answer! This is because the point goes 'full circle'. In fact, the HSC prefers that you write your arguments between -Pi and Pi. In this case, subtract 20*pi/10 (ie. 2*pi) to get the answer you're looking for :)
Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 22, 2017, 10:29:47 am
Remember; on the Argand diagram, you can add or subtract 2*pi without it changing the answer! This is because the point goes 'full circle'. In fact, the HSC prefers that you write your arguments between -Pi and Pi. In this case, subtract 20*pi/10 (ie. 2*pi) to get the answer you're looking for :)

OHH yea. How does it come so easily to you haha, thank you again Jake!
Title: Re: 4U Maths Question Thread
Post by: jakesilove on February 22, 2017, 10:31:01 am
OHH yea. How does it come so easily to you haha, thank you again Jake!

Don't worry buddy, once you've had enough practice it'll come easily to you as well! For now, this shit is hard.
Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 22, 2017, 01:20:36 pm
Is someone able to help me with one of these questions? I don't remember the pattern.
Title: Re: 4U Maths Question Thread
Post by: Kle123 on February 22, 2017, 05:29:43 pm
I don't see how part i) is related to ii). Soooo stressful. Please help, thanks again.
Title: Re: 4U Maths Question Thread
Post by: Kle123 on February 22, 2017, 06:27:59 pm
Is someone able to help me with one of these questions? I don't remember the pattern.
I've only shown working out in the first question. The rest are answers (if you were already given answers, this is quite useless). I don't think exam questions require working out, however i might be wrong. I've done all the questions as mostly those questions given are different and there is no specific pattern (not really a pattern).

side note: there are a set amount of locus sketches and if you learn them all this section of complex numbers should be a cruise
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 22, 2017, 06:39:01 pm
I don't see how part i) is related to ii). Soooo stressful. Please help, thanks again.




(http://i.imgur.com/PEwDX9O.png)


Title: Re: 4U Maths Question Thread
Post by: Kle123 on February 22, 2017, 07:01:57 pm



THANK YOU RUI. Your explanations are soooo easy to understand. You're the greatest!
Title: Re: 4U Maths Question Thread
Post by: Kle123 on February 24, 2017, 12:04:35 am
This questions seems easy enough but i just don't trust my proof (it's kinda dodgy). Could someone show me how it's done. Thank you.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 24, 2017, 12:36:25 am
This questions seems easy enough but i just don't trust my proof (it's kinda dodgy). Could someone show me how it's done. Thank you.



Title: Re: 4U Maths Question Thread
Post by: Kle123 on February 24, 2017, 07:31:57 am





Omg... I just realised this morning when I was thinking about it nonchalantly. Sleeep really does help! I used b^2=a^2(e^2-1) and didnt bother checking. Sorry for wasting your time rui.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 24, 2017, 11:10:09 am
Omg... I just realised this morning when I was thinking about it nonchalantly. Sleeep really does help! I used b^2=a^2(e^2-1) and didnt bother checking. Sorry for wasting your time rui.
Sleep is a beautiful thing. I had a lot of those moments over the last two years of my maths life.
Title: Re: 4U Maths Question Thread
Post by: VydekiE on February 26, 2017, 08:30:26 am
Hi,
I'm having trouble with this polynomial question. It would be wonderful if I could get some help on it.
1. Find the value of k such that y=x^2-4x+6 and y=k-6x-x^2 are tangential to each other at some point.
Thank you!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 26, 2017, 09:07:11 am
Hi,
I'm having trouble with this polynomial question. It would be wonderful if I could get some help on it.
1. Find the value of k such that y=x^2-4x+6 and y=k-6x-x^2 are tangential to each other at some point.
Thank you!!




The question is of a 3U calibre and applies a common 2U trick, but it's likely that it builds into something meant for only 4U students.
Title: Re: 4U Maths Question Thread
Post by: VydekiE on February 26, 2017, 09:16:21 am
Thank you so much!!  :)
Title: Re: 4U Maths Question Thread
Post by: VydekiE on February 26, 2017, 09:24:17 am
Hi, I have another polynomial question and it would be great if I could get some help on this
1. Find the values of k for which the polynomial p(x)=2x^3-9x^2+12x-k has
a) one distinct real root
b) repeated roots
Thank you!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 26, 2017, 10:05:24 am
Hi, I have another polynomial question and it would be great if I could get some help on this
1. Find the values of k for which the polynomial p(x)=2x^3-9x^2+12x-k has
a) one distinct real root
b) repeated roots
Thank you!!


GeoGebra simulation attached



Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 26, 2017, 04:39:10 pm
Does anyone know how to represent z^6=1 in mod arg form?
Title: Re: 4U Maths Question Thread
Post by: Sine on February 26, 2017, 04:43:33 pm
Does anyone know how to represent z^6=1 in mod arg form?

Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 26, 2017, 04:48:48 pm



I did get that one, thanks!
However the answers also had cis (+/- pi/3), cis (+/- 2pi/3) and cis pi, so I didn't understand how they got those values.
Title: Re: 4U Maths Question Thread
Post by: jakesilove on February 26, 2017, 04:56:51 pm
Does anyone know how to represent z^6=1 in mod arg form?

Hey! So, let's try to do this comprehensively. We're expecting six roots; called the six roots of unity

Let's create some complex number, z,



Now,



Clearly, r=1. So, we're left with




For all real, integer values of n. Now, we're expecting six roots, so we need six ns. It's easiest to use the plus/minus version of the lowest integers.



This gets us



So,



Divide through all theta values by 6, then then add/subtract 2*pi so that all thetas are within -pi and pi (the required range for arguments in 4U). That should get you the answers you're expecting!
Title: Re: 4U Maths Question Thread
Post by: VydekiE on February 26, 2017, 06:39:56 pm
Hi, it would be great if I could get some help on this question.
1.
a) If w is a seventh root of 1, w does not equal to 1, show that w^3+w^2+w+1+1/w+1/w^2+1/w^3=0
b) By letting z=w +1/w reduce this equation into a cubic equation in z.
Thank you!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 26, 2017, 06:46:15 pm
Hi, it would be great if I could get some help on this question.
1.
a) If w is a seventh root of 1, w does not equal to 1, show that w^3+w^2+w+1+1/w+1/w^2+1/w^3=0
b) By letting z=w +1/w reduce this equation into a cubic equation in z.
Thank you!!


____________________________________

Title: Re: 4U Maths Question Thread
Post by: VydekiE on February 26, 2017, 08:25:35 pm
Thank you!!  :)
Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 26, 2017, 10:03:25 pm
Thanks Jake! Could you possibly help with this one as well? Its Q. 18 and I've drawn the diagram, but I don't know how to find the diagonal.
Title: Re: 4U Maths Question Thread
Post by: jakesilove on February 26, 2017, 10:18:58 pm
Thanks Jake! Could you possibly help with this one as well? Its Q. 18 and I've drawn the diagram, but I don't know how to find the diagonal.

Hey! Weird question; you're not usually asked to find the cartesian equation of something on the argand diagram. Still, let's think about what we know.

Firstly, the line must be perpendicular to the diagonal AC. This is important, because we can find the gradient of the first diagonal, and use that to find the perpendicular gradient. Secondly, the line must bisect the diagonal AC. Now, we have a gradient and a point; all we need to find the equation!

Let's replace A and C with points on the cartesian plane; A(0, 3) and C(4, -5). The gradient of the line between them will be



So, the gradient of the perpendicular line will be 1/2.

Now, the midpoint of AC is (2, -1). So, the cartesian equation of the line perpendicular to the diagonal AC will be





So, we get the equation



Great! That's the first part down. Now, the second point wants us to show that



Let's let z=x+iy




And we're done!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 26, 2017, 10:20:33 pm
Thanks Jake! Could you possibly help with this one as well? Its Q. 18 and I've drawn the diagram, but I don't know how to find the diagonal.





Woah. Rare instance that Jake beat me.
Title: Re: 4U Maths Question Thread
Post by: jakesilove on February 26, 2017, 10:21:47 pm





Woah. Rare instance that Jake beat me.

Mate, I got one earlier today as well. Pick up your game son.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 26, 2017, 10:24:21 pm
Mate, I got one earlier today as well. Pick up your game son.
What did you want me to whilst I was dining out 8)
Title: Re: 4U Maths Question Thread
Post by: jakesilove on February 26, 2017, 10:26:07 pm
What did you want me to whilst I was dining out 8)

Ninja answer a 4U question under the table, obviously.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 26, 2017, 10:29:04 pm
Man... This guy...
Ninja answer a 4U question under the table, obviously.
Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 27, 2017, 09:01:11 am
Hey! Weird question; you're not usually asked to find the cartesian equation of something on the argand diagram. Still, let's think about what we know.

Firstly, the line must be perpendicular to the diagonal AC. This is important, because we can find the gradient of the first diagonal, and use that to find the perpendicular gradient. Secondly, the line must bisect the diagonal AC. Now, we have a gradient and a point; all we need to find the equation!

Let's replace A and C with points on the cartesian plane; A(0, 3) and C(4, -5). The gradient of the line between them will be



So, the gradient of the perpendicular line will be 1/2.

Now, the midpoint of AC is (2, -1). So, the cartesian equation of the line perpendicular to the diagonal AC will be





So, we get the equation



Great! That's the first part down. Now, the second point wants us to show that



Let's let z=x+iy




And we're done!

This makes perfect sense. Thank you so much! The only thing that confused me was when you substituted the gradient of the perpendicular line into the (y=mx+b) equation. You stated earlier that the gradient would be 1/2, however, when you substitute it into the (y=mx+b) equation, you've written it as -1/2.

Thank you so much otherwise Jake!
Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 27, 2017, 09:26:05 am
Also if anyone could help me with this question.. Should I using De Moivre's theorem?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 27, 2017, 09:50:34 am
This makes perfect sense. Thank you so much! The only thing that confused me was when you substituted the gradient of the perpendicular line into the (y=mx+b) equation. You stated earlier that the gradient would be 1/2, however, when you substitute it into the (y=mx+b) equation, you've written it as -1/2.

Thank you so much otherwise Jake!
Transcription error
Also if anyone could help me with this question.. Should I using De Moivre's theorem?
Title: Re: 4U Maths Question Thread
Post by: cutiepie30 on February 27, 2017, 09:34:38 pm
How would you do this question?

Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 27, 2017, 10:23:21 pm
How would you do this question?






_____
Title: Re: 4U Maths Question Thread
Post by: cutiepie30 on February 27, 2017, 11:51:30 pm




_____


Thanks alot RuiAce  :) !!

i get how you get equation 1 and 2

but i dont get how you added equations 1 and 2 and simplified from there , could you please do those steps in more depth  :)

Also for your last case is it meant to be pi/4 for all real number and not pi/2?
Title: Re: 4U Maths Question Thread
Post by: Shadowxo on February 28, 2017, 12:16:19 am
Thanks alot RuiAce  :) !!

i get how you get equation 1 and 2

but i dont get how you added equations 1 and 2 and simplified from there , could you please do those steps in more depth  :)

Also for your last case is it meant to be pi/4 for all real number and not pi/2?


Same denominator and boundaries , so it's just adding two fractions. The top is then equal to the bottom so it cancels out and equals 1.
Title: Re: 4U Maths Question Thread
Post by: cutiepie30 on February 28, 2017, 12:33:42 am

Same denominator and boundaries , so it's just adding two fractions. The top is then equal to the bottom so it cancels out and equals 1.

Thanks Shadowxo   :) :)

just wondering in the end it should be pi/4 for all cases in reference to rui ace's orignal solution ?
Title: Re: 4U Maths Question Thread
Post by: Shadowxo on February 28, 2017, 12:57:19 am
Thanks Shadowxo   :) :)

just wondering in the end it should be pi/4 for all cases in reference to rui ace's orignal solution ?

I believe so :)
Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 28, 2017, 08:58:24 am
Can someone help me with this? I don't know how to make it true for n=1
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 28, 2017, 12:49:08 pm
Can someone help me with this? I don't know how to make it true for n=1
Title: Re: 4U Maths Question Thread
Post by: cutiepie30 on February 28, 2017, 06:48:42 pm




_____


Hey ruiAce , just a question in regards to your solution?

 How did you know and get this trick ?
How did you figure it out / get to it?

Thanks
  :) !
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 28, 2017, 07:03:34 pm
Hey ruiAce , just a question in regards to your solution?

 How did you know and get this trick ?
How did you figure it out / get to it?

Thanks
  :) !


Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 28, 2017, 07:32:54 pm


Thanks Rui! You make it so simple haha.
I get really confused on the n=k+1 part, any help would be greatly appreciated! :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 28, 2017, 07:44:31 pm
Thanks Rui! You make it so simple haha.
I get really confused on the n=k+1 part, any help would be greatly appreciated! :)

Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 28, 2017, 08:00:50 pm



You are so helpful. Thank you so much for taking time to write it all out Rui!!
If you have time, would possibly be able to give me some help on this question as well? Thank you for all your help so far!!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 28, 2017, 08:11:48 pm
You are so helpful. Thank you so much for taking time to write it all out Rui!!
If you have time, would possibly be able to give me some help on this question as well? Thank you for all your help so far!!!
Hint: Part c) is the useful part in the worked example. Commence by \(\text{Let }x=\tan \theta\) then play around with the algebra until you can use the triple-angle identity like they did

Feel free to post any progress if you get a bit lost halfway through :)
Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 28, 2017, 08:30:06 pm
Hint: Part c) is the useful part in the worked example. Commence by \(\text{Let }x=\tan \theta\) then play around with the algebra until you can use the triple-angle identity like they did

Feel free to post any progress if you get a bit lost halfway through :)


OK... I've gotten abit lost. HAHA.
Title: Re: 4U Maths Question Thread
Post by: kiwiberry on February 28, 2017, 08:37:18 pm

OK... I've gotten abit lost. HAHA.

Nearly there! :) From your last line:
Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 28, 2017, 08:43:31 pm
Nearly there! :) From your last line:


ahh ! I get it haha, thanks heaps! :)
Title: Re: 4U Maths Question Thread
Post by: michaelalt on February 28, 2017, 09:05:44 pm
I have no idea how to do part b. and c. of this question. I've tried watching videos but the examples are all different, and theres no pattern to them so I'm having a hard time working them out...
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 28, 2017, 09:11:33 pm
I have no idea how to do part b. and c. of this question. I've tried watching videos but the examples are all different, and theres no pattern to them so I'm having a hard time working them out...





Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 28, 2017, 09:13:36 pm


Title: Re: 4U Maths Question Thread
Post by: cutiepie30 on February 28, 2017, 10:41:52 pm




Thankyou RuiAce  :) !! Makes sense now !!
Title: Re: 4U Maths Question Thread
Post by: Kle123 on March 01, 2017, 09:10:25 pm
Hey Rui, could you (or anybody) help me with this question. Thank you
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 01, 2017, 09:19:50 pm
Hey Rui, could you (or anybody) help me with this question. Thank you
I'm trying to look for any shortcuts right now. I'll edit this post later.

Chances are that this is an annoying brute-force question.
_____________________

Edit: Can't see anything, so here goes.
(http://i.imgur.com/8jSJiVW.png)





Try chucking all of that into the distance formula. Hopefully something good appears. I also worry that I made some algebra mistakes along the way.
Title: Re: 4U Maths Question Thread
Post by: Kle123 on March 02, 2017, 08:48:44 pm
Haha Rui, i stopped about the same place where you stopped too, where it was about to get really messy. Frankly, I think this is a question is a waste of time. Thanks again for your help Rui, i'll just leave this question incomplete because i think its a waste of time.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 02, 2017, 08:56:04 pm
I'm with you. Huge waste of time tbh
Title: Re: 4U Maths Question Thread
Post by: lsong on March 02, 2017, 09:44:41 pm
Hi,
I'm having trouble with taking the limits of an equation to find where the curve approaches. For example, in the question below, I know I could take points but I would still like to know how to take the limits of this question just for future reference.
Thanks in advanced!  :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 02, 2017, 09:51:45 pm
Hi,
I'm having trouble with taking the limits of an equation to find where the curve approaches. For example, in the question below, I know I could take points but I would still like to know how to take the limits of this question just for future reference.
Thanks in advanced!  :)


Your question is not clear. What exactly are you asking? I didn't test points here.
Title: Re: 4U Maths Question Thread
Post by: lsong on March 02, 2017, 09:58:09 pm
Ah, sorry about that.
Basically, when you get the asymptotes, how do you find out which region the curve lies in (i.e. where it approaches the asymptotes)?
Hopefully I was clearer this time .-.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 02, 2017, 10:05:18 pm
Methods of intuition take heaps of practice. I started sketching curves on GeoGebra and in my head randomly to develop the intuition needed to just "SEE" how exactly the asymptote is approached, i.e. from above or below.

Techniques I employ include splitting each region between vertical asymptotes up, and observing the presence of both x-intercepts and stationary points. Horizontal/Oblique asymptotes are used for any region not bound between two vertical asymptotes.

E.g. for something like y=1/(x^2-1), I know about the stationary point at x=0, and I also know that between the vertical asymptotes x=-1, x=1, there are no x-intercepts. So the stationary point (which happens to be a local min) is going to, in a way, deflect the curve back down, so that both asymptotes are approached from below.

I also note the general shape of some things. Quadratic/Quadratic and constant/quadratic look the same, but linear/quadratic looks different.

Again, HEAPS of practice needed to develop intuition.
Title: Re: 4U Maths Question Thread
Post by: lsong on March 02, 2017, 10:19:17 pm
Hmm, ok, thank you!
It seems I'll just have to practise more.
Btw, for equations which use two equations, like log [sin x], how would you recommend sketching them, because I've been considering asymptotes and mainly subbing in values. Surely there is a faster way to approach this?
Sorry I seem to be asking really basic questions.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 02, 2017, 10:21:23 pm
Hmm, ok, thank you!
It seems I'll just have to practise more.
Btw, for equations which use two equations, like log [sin x], how would you recommend sketching them, because I've been considering asymptotes and mainly subbing in values. Surely there is a faster way to approach this?
Sorry I seem to be asking really basic questions.
Sketch y=sin(x) first.

Read off the y-coordinates of that graph. What happens when you log, say 1. What happens when you "try" to log 0. What happens if sin(x) was negative the whole time?

And obviously you never have to consider when it's below -1 or above 1 because that's just defying the range of y=sin(x)

Compose the graph using the two graphs that build it.

Note - Inevitably you're going to "sub values". You're not supposed to get out of it. The important thing is realising WHICH values to sub in, and where possible just subbing in your head.
Title: Re: 4U Maths Question Thread
Post by: lsong on March 03, 2017, 07:20:14 am
Aah ok, it really looks like intuitive thinking. Thanks for the pointers!  :)
Title: Re: 4U Maths Question Thread
Post by: Wales on March 03, 2017, 04:39:10 pm
I've just started the Graphs topic and this question seems completely foreign to me. Could anyone please guide me through it? I've had a look at the solutions but nothing makes too much sense.

Cheers, Wales
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 03, 2017, 06:30:35 pm
I've just started the Graphs topic and this question seems completely foreign to me. Could anyone please guide me through it? I've had a look at the solutions but nothing makes too much sense.

Cheers, Wales

__________________________________________



Title: Re: 4U Maths Question Thread
Post by: hinakamishiro on March 04, 2017, 02:47:09 pm
Hey guys could I please have some help with this conics question?  :-\
Title: Re: 4U Maths Question Thread
Post by: jakesilove on March 04, 2017, 05:03:27 pm
Hey guys could I please have some help with this conics question?  :-\

Definitely a bloody tough question. I'm no good at drawing diagrams, so I'm gonna just say some diagram-related stuff and you can draw it out if it doesn't make conceptual sense.

First, let's let the tangent at P meet x=a and x=-a at Q, R respectively. QR can hit the y=-axis at C.

The tangent PR has equation



So, Q has coordinates



and R has coordinates



We can easily find the gradient of QS and RS, and we multiply them together to find whether they are at right angles to each other. Using our knowledge of eccentricity,




Now, recalling that



by definition, we find that the gradient of QS* the gradient of RS =-1. Thus, they are at right angles to each other. We only need to replace e with -e to prove the same relation for QS' and RS' (Potentially a good idea to write this out in full in an exam, but that's up to you).

As QR subtends a right angle at S and S', QSRS' are concyclic. By circle geometry rules, QR must be the diameter of the circle. As the y'axis must bisect the chord SS', the centre of the circle is the point C.

Now, we use our point.



on the hyperbola stated, then QR has equation



Clearly, the point C (where the line meets the y axis) will be



As




So we can find coordinates S




So, the circle will have equation



Subject to error here or there (I hate conics), but the method is correct!
Title: Re: 4U Maths Question Thread
Post by: cutiepie30 on March 04, 2017, 08:26:19 pm
how would you do this question?

Evaluate the following using integration by parts

Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 04, 2017, 09:10:45 pm
how would you do this question?

Evaluate the following using integration by parts


Title: Re: 4U Maths Question Thread
Post by: Kle123 on March 05, 2017, 07:51:57 am
Rui, could you help me out with part ii). Thank you!
Title: Re: 4U Maths Question Thread
Post by: cutiepie30 on March 05, 2017, 11:19:28 am




Thanks so much RuiAce  :) !!
Title: Re: 4U Maths Question Thread
Post by: cutiepie30 on March 05, 2017, 11:20:45 am
how would you do this question?

Evaluate the following using integration by parts

Title: Re: 4U Maths Question Thread
Post by: jakesilove on March 05, 2017, 11:34:57 am
how would you do this question?

Evaluate the following using integration by parts









Using integration by parts




Then, you can use a bunch of trig identities to simply down to



according to Wolfram Alpha. Don't necessarily think you need to do this in an exam situations. Also, I'm sure there are plenty of other ways to solve this question, some of which would have yielded the simplified answer above straight away.
Title: Re: 4U Maths Question Thread
Post by: cutiepie30 on March 05, 2017, 12:21:47 pm








Using integration by parts




Then, you can use a bunch of trig identities to simply down to



according to Wolfram Alpha. Don't necessarily think you need to do this in an exam situations. Also, I'm sure there are plenty of other ways to solve this question, some of which would have yielded the simplified answer above straight away.

THanks jakesilove  :)

but i dont get how you went from 



how did you get

Also how did you get the =
in one go ?
Title: Re: 4U Maths Question Thread
Post by: jakesilove on March 05, 2017, 12:40:18 pm
THanks jakesilove  :)

but i dont get how you went from 



how did you get



Also how did you get the =
in one go ?

Fair enough, I definitely didn't explain myself very well. We want to integrate dv, so



This integrates in the normal way



However, we can use trigonometric identities to get



as required.

For the last one, I just used the reverse chain rule. Recall that



Using this, and noting that



we can recognise that



And that's where I got it! If you can't do that in one go, that's no problem at all (I've had a lot of practice). Take your time, set out your working however you want. However, over time, integrations like this will become natural :)
Title: Re: 4U Maths Question Thread
Post by: cutiepie30 on March 05, 2017, 01:05:15 pm
Fair enough, I definitely didn't explain myself very well. We want to integrate dv, so



This integrates in the normal way



Thanks so much for

However, we can use trigonometric identities to get



as required.

For the last one, I just used the reverse chain rule. Recall that



Using this, and noting that



we can recognise that



And that's where I got it! If you can't do that in one go, that's no problem at all (I've had a lot of practice). Take your time, set out your working however you want. However, over time, integrations like this will become natural :)

Thanks so much for clarifying jakesilove  :) :) Makes sense now !!
Title: Re: 4U Maths Question Thread
Post by: cutiepie30 on March 05, 2017, 04:43:24 pm
How would you do this question?
Use a substitution and an integration by parts to evaluate each of the following
Title: Re: 4U Maths Question Thread
Post by: jakesilove on March 05, 2017, 05:32:27 pm
How would you do this question?
Use a substitution and an integration by parts to evaluate each of the following

I would use the substitution of theta=5x+2, then apply that throughout the integral, expand everything out and do integration by parts where necessary. Show us your working out, and if you're still stuck we can talk you through it!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 05, 2017, 07:58:40 pm
Rui, could you help me out with part ii). Thank you!
Fair warning: My activity is reduced for this month. I won't be able to get to everything and when I do it might take a while to get there.
(http://uploads.tapatalk-cdn.com/20170305/ac3f6bc091519bca0444bd2f94fee15f.jpg)
Supposedly it gets bizarre
Title: Re: 4U Maths Question Thread
Post by: Kle123 on March 05, 2017, 09:22:04 pm
Fair warning: My activity is reduced for this month. I won't be able to get to everything and when I do it might take a while to get there.
Thanks Rui.
NOOOOoooo...my life saver. Thankfully, its only for a month. Good luck with whatever your up to!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 05, 2017, 09:46:18 pm
Thanks Rui.
NOOOOoooo...my life saver. Thankfully, its only for a month. Good luck with whatever your up to!
Ahh I'll still be around, just not as regularly. Besides Jake's floating around too
Title: Re: 4U Maths Question Thread
Post by: Mahan on March 05, 2017, 11:07:16 pm
How many numbers between 1 and 2002 such that the sum of the digits is divisible by 5?
Title: Re: 4U Maths Question Thread
Post by: jakesilove on March 06, 2017, 09:13:53 am
How many numbers between 1 and 2002 such that the sum of the digits is divisible by 5?

Hey Mahan! Just before we answer these questions, just wondering, are they questions you struggle with? Are they homework questions? Because they all seem far beyond any curriculum I know of.
Title: Re: 4U Maths Question Thread
Post by: Mahan on March 06, 2017, 05:18:23 pm
Hey Mahan! Just before we answer these questions, just wondering, are they questions you struggle with? Are they homework questions? Because they all seem far beyond any curriculum I know of.
Hey Jakesilove! These questions are not directly out of HSC related textbooks but apparently in order to solve them we don't need any knowledge beyond HSC.I agree that they are tricky but after spending some time they should be doable.All the questions that I have posted are either come from other sources, Not HSC related but can be solved using only HSC knowledge, or I heard it from my friends.Sometimes I post them to share them with others or sometimes I'm curious to see whether someone comes up with a different idea or I might not know the answer to that question. For this particular one, I'm curious to see whether someone comes up with some interesting idea to solve it, because there are always multiple methods to solve counting questions, at the same time share some rather unusual and challenging question for those who like challenge.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 06, 2017, 06:31:28 pm
Please reserve the questions thread for those students require assistance with.their studies. There are challenge marathon threads for these types of questions that exist purely for the sake of testing students.

Note: For that one, I'd just use techniques I was taught in discrete maths because I don't have the time to figure out the clever way. Stars and bars + inclusion/exclusion principle. Long winded because we are considering the number of solutions to b+c+d = 4, 5, 9, 10, 14, 15, 19, 20, 24 or 25 though.
Title: Re: 4U Maths Question Thread
Post by: Kle123 on March 07, 2017, 08:06:26 pm
Hey, this question relates to the question in posted in the 3U thread. The answer 126deg 52min for part (V). How can we deduce the angle is obtuse?

Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 07, 2017, 09:49:46 pm
Hey, this question relates to the question in posted in the 3U thread. The answer 126deg 52min for part (V). How can we deduce the angle is obtuse?
Honestly, I could not see how this is possible unless you drew a diagram that was reasonably to scale. I simulated the scenario on GeoGebra and there was no way that a free-hand diagram would've implied it.
Title: Re: 4U Maths Question Thread
Post by: cutiepie30 on March 11, 2017, 01:24:27 pm
Use a substitution and an integration by parts to evaluate each of the following

For this question and answer attached where am i going wrong?
Why is my answer different to the supplied answer?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 11, 2017, 01:45:48 pm
Use a substitution and an integration by parts to evaluate each of the following

For this question and answer attached where am i going wrong?
Why is my answer different to the supplied answer?
It's not. They're the same answer.

They just simplified it further. Because cos(...) is a common factor so you can factorise it further
Title: Re: 4U Maths Question Thread
Post by: cutiepie30 on March 11, 2017, 05:53:10 pm
For the error in this question.(attached)

i said the working out forgot to add an integration constant(ie +c)
 when an +c is added the value of c=-1

therefore the equation becomes loge(|x|)=loge(|x|), which is true .

Would this be right/ enough ?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 11, 2017, 06:33:55 pm
For the error in this question.(attached)

i said the working out forgot to add an integration constant(ie +c)
 when an +c is added the value of c=-1

therefore the equation becomes loge(|x|)=loge(|x|), which is true .

Would this be right/ enough ?
Pretty much, +C is the cause of it. Because in indefinite integration, we always have to keep in mind that the statement is true to within a constant.

If you ask me though, "true to within a constant" is the key thing that needs to be stated. Of course, you can, by computation, find the value of C (which is -1 or 1, depending on if your C is on the LHS or RHS) to reunite the solutions if you wish.
Title: Re: 4U Maths Question Thread
Post by: VydekiE on March 11, 2017, 08:10:25 pm
Hi, it would be great if I could get some help on this question
1) Find the equations of the tangent and normal to the ellipse x=4cos theta, y=2sin theta at the point where theta= -pi/4
Thank you!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 11, 2017, 08:12:58 pm
Hi, it would be great if I could get some help on this question
1) Find the equations of the tangent and normal to the ellipse x=4cos theta, y=2sin theta at the point where theta= -pi/4
Thank you!!


Title: Re: 4U Maths Question Thread
Post by: jakesilove on March 11, 2017, 08:14:17 pm
Hi, it would be great if I could get some help on this question
1) Find the equations of the tangent and normal to the ellipse x=4cos theta, y=2sin theta at the point where theta= -pi/4
Thank you!!

First, we need an equation. We know that



Clearly,




We want to find the tangent at the point



Honestly Rui, I thought I'd get this one.

But your way is 100% faster.

Far out.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 11, 2017, 08:15:36 pm
First, we need an equation. We know that



Clearly,




We want to find the tangent at the point



Honestly Rui, I thought I'd get this one.

But your way is 100% faster.

Far out.
Aha.

Pro tip: If things are given to you parametrically, try to avoid Cartesian stuff
Title: Re: 4U Maths Question Thread
Post by: jakesilove on March 11, 2017, 08:17:51 pm
Aha.

Pro tip: If things are given to you parametrically, try to avoid Cartesian stuff

Wise words from a freakin' genius. Legend as always matey.
Title: Re: 4U Maths Question Thread
Post by: VydekiE on March 11, 2017, 09:02:17 pm




Thank you!!
Title: Re: 4U Maths Question Thread
Post by: Kle123 on March 12, 2017, 12:19:27 pm
Could someone please help me with this question
Title: Re: 4U Maths Question Thread
Post by: cutiepie30 on March 12, 2017, 01:04:11 pm
Pretty much, +C is the cause of it. Because in indefinite integration, we always have to keep in mind that the statement is true to within a constant.

If you ask me though, "true to within a constant" is the key thing that needs to be stated. Of course, you can, by computation, find the value of C (which is -1 or 1, depending on if your C is on the LHS or RHS) to reunite the solutions if you wish.

Thanks RuiAce  :)
Title: Re: 4U Maths Question Thread
Post by: jakesilove on March 12, 2017, 01:40:08 pm
Could someone please help me with this question

Hey! So,




First, let's use the sum of roots. This will equal -b/a, thus




Next, let's use product of roots. This will be equal to e/a, thus



Oh. This was stupid. If A= alpha + beta + gamma, then A is a root of the equation! So,




as required.

Now,



Hmmm... will keep working on this part inb4 Rui posts his answer before me

Okay, so

From the above step that I didn't really need to do (but turns out I did!), we know that



New tactic






I have to run; hopefully Rui can swoop in and save the day, otherwise I'll pick this up later.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 12, 2017, 06:45:14 pm

Subject to inaccuracies
Title: Re: 4U Maths Question Thread
Post by: cutiepie30 on March 12, 2017, 07:29:26 pm
How would you do this question?
Title: Re: 4U Maths Question Thread
Post by: Mahan on March 12, 2017, 07:45:31 pm
How would you do this question?
Note:



Title: Re: 4U Maths Question Thread
Post by: cutiepie30 on March 12, 2017, 08:25:16 pm
Note:




thanks Mahan  :)

but the question says find cosh(3x) in terms of cosh(x) and sinh(x), your answer only has cosh(x) not sinh(x)?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 12, 2017, 08:30:06 pm
thanks Mahan  :)

but the question says find cosh(3x) in terms of cosh(x) and sinh(x), your answer only has cosh(x) not sinh(x)?
Hmm.

What answer do they want you to get to? Because the convention is to stop at where he did.
Title: Re: 4U Maths Question Thread
Post by: jakesilove on March 12, 2017, 10:15:42 pm
Hmm.

What answer do they want you to get to? Because the convention is to stop at where he did.

Is sinh and cosh even part of the HSC?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 12, 2017, 10:28:25 pm
Is sinh and cosh even part of the HSC?
Nope.

At least it got defined for them though.
Title: Re: 4U Maths Question Thread
Post by: cutiepie30 on March 12, 2017, 11:46:15 pm
Hmm.

What answer do they want you to get to? Because the convention is to stop at where he did.

i am not sure myself Rui  ;), our teacher just gave us this question as an extension and said we should be able to do it from the info provided
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 13, 2017, 06:56:15 am
i am not sure myself Rui  ;), our teacher just gave us this question as an extension and said we should be able to do it from the info provided
But then again, if you get it in terms of only cosh it's still in terms of both cosh and sinh. This is because you're basically adding \(0\sinh x\) anyhow.
Title: Re: 4U Maths Question Thread
Post by: cutiepie30 on March 13, 2017, 08:38:40 pm
But then again, if you get it in terms of only cosh it's still in terms of both cosh and sinh. This is because you're basically adding \(0\sinh x\) anyhow.

Thanks RuiAce  :)

basically so mahans answer should be fine right?

is there a way you could get   cosh(3x) in terms of cosh(x) and sinh(x), with both sinh(x) and cosh(x) in the final answer?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 13, 2017, 08:40:22 pm
Thanks RuiAce  :)

basically so mahans answer should be fine right?

is there a way you could get   cosh(3x) in terms of cosh(x) and sinh(x), with both sinh(x) and cosh(x) in the final answer?
Should be, yeah.
Title: Re: 4U Maths Question Thread
Post by: cutiepie30 on March 13, 2017, 09:11:53 pm
Should be, yeah.


Thanks RuiAce  :)
Title: Re: 4U Maths Question Thread
Post by: chelseam on March 14, 2017, 08:57:25 pm
Hi! Can someone please help me with part 3 of this question :) I've read the board of studies solutions but I still don't get what I'm supposed to do!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 14, 2017, 10:19:47 pm
Hi! Can someone please help me with part 3 of this question :) I've read the board of studies solutions but I still don't get what I'm supposed to do!





_____________________________


Title: Re: 4U Maths Question Thread
Post by: chelseam on March 14, 2017, 10:56:10 pm
Thank you so much!! :D





_____________________________



Title: Re: 4U Maths Question Thread
Post by: Rathin on March 15, 2017, 06:13:53 pm
For the question: Factorise 4x^4+1 asa product of real polynomials.
I got (2x^2+1)(2x^2-1)
But the answer is (2x^2+2x+1)(2x^2-2x+1), why is it so?
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on March 15, 2017, 06:21:45 pm
For the question: Factorise 4x^4+1 asa product of real polynomials.
I got (2x^2+1)(2x^2-1)
But the answer is (2x^2+2x+1)(2x^2-2x+1), why is it so?

Hey Rathin! It's been a while ;D



I think you've used the difference of two squares shortcut when you shouldn't have, or otherwise misread the question?

I'll leave Rui/Jake to tag in with the actual method if you need it (I'm not sure of the 4U approach to this) ;D

Edit: Nevermind, you'd just complete the square (sort of, not the way you'd normally do) - Pretty sure you'd just do this ;D



I hope that helps ;D
Title: Re: 4U Maths Question Thread
Post by: Rathin on March 15, 2017, 06:29:09 pm
Hey Rathin! It's been a while ;D



I think you've used the difference of two squares shortcut when you shouldn't have, or otherwise misread the question?

I'll leave Rui/Jake to tag in with the actual method if you need it (I'm not sure of the 4U approach to this) ;D

Edit: Nevermind, you'd just complete the square (sort of, not the way you'd normally do) - Pretty sure you'd just do this ;D



I hope that helps ;D

That is probably the stupidest thing I have ever done lol
Thanks for the help :)
Title: Re: 4U Maths Question Thread
Post by: chelseam on March 15, 2017, 08:53:24 pm
Can someone please help me with the last two parts of this question? Thank you :D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 15, 2017, 09:08:09 pm
Can someone please help me with the last two parts of this question? Thank you :D
Part iv) needs the result of b), which you need to provide.





Title: Re: 4U Maths Question Thread
Post by: chelseam on March 15, 2017, 10:09:18 pm
Thank you! :) This is part b)

Part iv) needs the result of b), which you need to provide.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 16, 2017, 07:27:27 am
Thank you! :) This is part b)
Hint: in 3U, when you were doing Newton's method, you learnt that a root is between -1 and -1/2 if f(-1) and f(-1/2) change signs. The result of part b) is supposed to help make your life easier when you have to deal with f(-1/2).

There are many other solutions better than the BoS ones which you should consider reviewing. However if those don't make sense, post up your interpretation of whichever solutions you looked at and we will fill in the remainder of the puzzle.
Title: Re: 4U Maths Question Thread
Post by: bluecookie on March 16, 2017, 12:20:02 pm
a) Show that the roots of y^4+y^3+y^3+y+1=0 are y=cos(2kpi/5)+1sin(2kpi/5), k=1,2,3,4.
b) Hence deduce that cos36 deg = 1/2 + cos 72 deg.
^ I need help with part b.
Title: Re: 4U Maths Question Thread
Post by: bluecookie on March 16, 2017, 01:14:17 pm
Suppose that z^7=1, z is not equal to 1.
i) deduce that z^3+z^2+z+1+1/z+1/z^2+1/z^3=0.
ii) by letting x=z+1/z, reduce the equation in i to a cubic equation in x
iii) Hence deduce that cospi/7cos2pi/7cos3pi/7=1/8
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 16, 2017, 01:51:55 pm
Suppose that z^7=1, z is not equal to 1.
i) deduce that z^3+z^2+z+1+1/z+1/z^2+1/z^3=0.
ii) by letting x=z+1/z, reduce the equation in i to a cubic equation in x
iii) Hence deduce that cospi/7cos2pi/7cos3pi/7=1/8
First two parts addressed in post #885 albeit with different pronumerals



_________


_________



Title: Re: 4U Maths Question Thread
Post by: chelseam on March 16, 2017, 09:20:05 pm
Thank you so much! :)
Hint: in 3U, when you were doing Newton's method, you learnt that a root is between -1 and -1/2 if f(-1) and f(-1/2) change signs. The result of part b) is supposed to help make your life easier when you have to deal with f(-1/2).

There are many other solutions better than the BoS ones which you should consider reviewing. However if those don't make sense, post up your interpretation of whichever solutions you looked at and we will fill in the remainder of the puzzle.
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on March 17, 2017, 06:32:16 am
Hi
i am not sure how to find the asymptote
do you just do lim x approaches 0?
thank you
Title: Re: 4U Maths Question Thread
Post by: ellipse on March 17, 2017, 07:22:41 am
Hi
i am not sure how to find the asymptote
do you just do lim x approaches 0?
thank you
This is the equation of a hyperbola, so you can divide each sides by 4 to get it in the general form. Now the equation of the asymptotes are y= b/a or y=-b/a. In this hyperbola, a is 2 and b is 1. So the asymptotes are y=-1/2x and y=1/2x.
Alternatively, you can try factorising the LHS
so (x-2y)(x+2y)=4. As the RHS  is a non-zero constant, the LHS must not be equal to 0.
Therefore x-2y =/= 0 and x+2y =/= 0
So the asymptotes are y=1/2x and y=-1/2x
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on March 17, 2017, 05:12:39 pm
This is the equation of a hyperbola, so you can divide each sides by 4 to get it in the general form. Now the equation of the asymptotes are y= b/a or y=-b/a. In this hyperbola, a is 2 and b is 1. So the asymptotes are y=-1/2x and y=1/2x.
Alternatively, you can try factorising the LHS
so (x-2y)(x+2y)=4. As the RHS  is a non-zero constant, the LHS must not be equal to 0.
Therefore x-2y =/= 0 and x+2y =/= 0
So the asymptotes are y=1/2x and y=-1/2x

thank you!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 17, 2017, 11:04:18 pm
a) Show that the roots of y^4+y^3+y^3+y+1=0 are y=cos(2kpi/5)+1sin(2kpi/5), k=1,2,3,4.
b) Hence deduce that cos36 deg = 1/2 + cos 72 deg.
^ I need help with part b.
Your expression is wrong. cos 36deg = -1/2 - cos 72deg.
Title: Re: 4U Maths Question Thread
Post by: cutiepie30 on March 19, 2017, 02:56:47 am
How would you go about doing these questions ? (attached)

Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 19, 2017, 07:37:53 am
How would you go about doing these questions ? (attached)
I'm just gonna give this warning now.

Hyperbolic functions are NOT a part of the 4U course. In the previous case it was okay because they defined the functions for you and just made you do algebraic manipulation. I will not do any more questions that are beyond the scope of the course.


First image - Use similar identities to the trig ones.
Second image - Consider sinh for the first and cosh for the others
Title: Re: 4U Maths Question Thread
Post by: VydekiE on March 19, 2017, 08:52:09 am
Hi, it would be great if I could get some help on this question
1) Prove that if T (x0, y0) lies on a directrix on the hyperbola x2/a2 - y2/b2=1 then the chord of contact will be a focal chord
Thank you!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 19, 2017, 09:03:26 am
Hi, it would be great if I could get some help on this question
1) Prove that if T (x0, y0) lies on a directrix on the hyperbola x2/a2 - y2/b2=1 then the chord of contact will be a focal chord
Thank you!!



Title: Re: 4U Maths Question Thread
Post by: VydekiE on March 19, 2017, 09:25:33 am





Thank you!!
Title: Re: 4U Maths Question Thread
Post by: cutiepie30 on March 19, 2017, 12:38:37 pm
I'm just gonna give this warning now.

Hyperbolic functions are NOT a part of the 4U course. In the previous case it was okay because they defined the functions for you and just made you do algebraic manipulation. I will not do any more questions that are beyond the scope of the course.


First image - Use similar identities to the trig ones.
Second image - Consider sinh for the first and cosh for the others

yea i know they are not rui  :), but i think that maths is a very beautiful subject and so it doesn't hurt here and there to do questions outside of the scope of the 4U course  :)  ;)

next time i will post a disclaimer before posting these questions. :)

Just wondering rui for the second image, what sinh(u) substituition would i have to make, i have tryed x=sinh(u) for the 1st question of the 2nd image  , it didnt work ?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 19, 2017, 02:00:28 pm
yea i know they are not rui  :), but i think that maths is a very beautiful subject and so it doesn't hurt here and there to do questions outside of the scope of the 4U course  :)  ;)

next time i will post a disclaimer before posting these questions. :)

Just wondering rui for the second image, what sinh(u) substituition would i have to make, i have tryed x=sinh(u) for the 1st question of the 2nd image  , it didnt work ?
x=3sinh(u)

Treat it like a trigonometric substitution. You need a coefficient in front of the function. √9=3
Title: Re: 4U Maths Question Thread
Post by: Kle123 on March 20, 2017, 03:58:38 pm
Hey there is no solution let alone answer to this question within the past paper. Idk how to get an expression such as the ones in the choices, but i tried subbing each of the MC and i got C (working out may be wrong though). This method seems too long to me for a MC question. Could someone show me a more efficient way? THank you.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 20, 2017, 06:58:15 pm
Hey there is no solution let alone answer to this question within the past paper. Idk how to get an expression such as the ones in the choices, but i tried subbing each of the MC and i got C (working out may be wrong though). This method seems too long to me for a MC question. Could someone show me a more efficient way? THank you.


Title: Re: 4U Maths Question Thread
Post by: SmeagAL on March 21, 2017, 02:01:16 pm
Hi. It would be great if I could get some help on this question.


Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 21, 2017, 03:14:58 pm
Hi. It would be great if I could get some help on this question.
This question involves a Riemann sum. (Which is not 'explicitly' in the syllabus.)
(http://uploads.tapatalk-cdn.com/20170320/c6d1ad0eaf924edf54034f15885a1af2.jpg)
The whole foundation of integration is the idea of Riemann sums. By reducing the width of the rectangles, we assert that the area of the rectangles will eventually be EQUAL to the area under the curve. So since the limit is actually representative of an (infinite, aka. limiting) Riemann sum, it is equivalent to the ACTUAL area under the curve.

Do mention if my handwriting was too messy.
Title: Re: 4U Maths Question Thread
Post by: lsong on March 21, 2017, 09:04:33 pm
Hi, I can't seem to do part (i) by using the given statement. Any help would be great thanks!
Title: Re: 4U Maths Question Thread
Post by: SmeagAL on March 21, 2017, 10:10:24 pm
THANKS! Extremely helpful.

Never heard of the Riemann Sum before.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 22, 2017, 04:38:37 am
Hi, I can't seem to do part (i) by using the given statement. Any help would be great thanks!





Note: The question actually says x>0 and not x≥0. However, I have assumed otherwise because if x>0 then for part i), we should have > as well instead of ≥.
Title: Re: 4U Maths Question Thread
Post by: lsong on March 22, 2017, 10:24:37 am





Note: The question actually says x>0 and not x≥0. However, I have assumed otherwise because if x>0 then for part i), we should have > as well instead of ≥.

Thanks for the help! :)
Title: Re: 4U Maths Question Thread
Post by: VydekiE on March 22, 2017, 11:43:52 am
Hi,
It would be great if someone could help me with this question
1) The curve on the Argand diagram for which ||z-3| - |z+5||=4 is a hyperbola.
a) Find the eccentricity of this hyperbola
Thank you!!
Title: Re: 4U Maths Question Thread
Post by: theblackswan on March 22, 2017, 02:50:19 pm
Hey,
I'm kinda stuck on questions that require you to find the tangents to the ellipse or hyperbola that are parallel to a line. What's the best method for those types of questions? For example, find the tangents to the ellipse 16x^2 + 25y^2 =400 which are parallel to the line y = x+2. Thanks!!
Title: Re: 4U Maths Question Thread
Post by: jakesilove on March 22, 2017, 06:51:15 pm
Hi,
It would be great if someone could help me with this question
1) The curve on the Argand diagram for which ||z-3| - |z+5||=4 is a hyperbola.
a) Find the eccentricity of this hyperbola
Thank you!!

Hey! First, recall that |PS-PS'|=2a. It seems sort of intuitive that the focus of the hyperbola will be z=3 and z=5, right? If not, try think about some values for z that hold true, and the shape kinda pops out on its own. You can use the 'location' of the focus', plus the knowledge that a=2, to find the eccentricity.
Title: Re: 4U Maths Question Thread
Post by: jakesilove on March 22, 2017, 06:52:34 pm
Hey,
I'm kinda stuck on questions that require you to find the tangents to the ellipse or hyperbola that are parallel to a line. What's the best method for those types of questions? For example, find the tangents to the ellipse 16x^2 + 25y^2 =400 which are parallel to the line y = x+2. Thanks!!


If you are just given points, the implicit differentiation is the way to go (you can do this as normal, sovling for dy/dx and plugging your points in). If a given line is parallel (ie. tangents are involved etc.) then you can try to use the quadratic discriminant. Both Rui and I are seriously busy tonight, so hopefully we can expand on our answers either later or sometime tomorrow!
Title: Re: 4U Maths Question Thread
Post by: VydekiE on March 23, 2017, 10:00:57 am
Hey! First, recall that |PS-PS'|=2a. It seems sort of intuitive that the focus of the hyperbola will be z=3 and z=5, right? If not, try think about some values for z that hold true, and the shape kinda pops out on its own. You can use the 'location' of the focus', plus the knowledge that a=2, to find the eccentricity.

Thank you!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 23, 2017, 11:28:36 am
Hey,
I'm kinda stuck on questions that require you to find the tangents to the ellipse or hyperbola that are parallel to a line. What's the best method for those types of questions? For example, find the tangents to the ellipse 16x^2 + 25y^2 =400 which are parallel to the line y = x+2. Thanks!!





Please find a sample question in post #813, and one solution in post #816

Edit: Typo
Title: Re: 4U Maths Question Thread
Post by: smile123 on March 24, 2017, 08:47:15 am
PLEASE HELP
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 24, 2017, 09:04:27 am
PLEASE HELP
Three-dimensional curves are not a part of the HSC

Title: Re: 4U Maths Question Thread
Post by: smile123 on March 24, 2017, 03:42:08 pm
Thankyou
Title: Re: 4U Maths Question Thread
Post by: VydekiE on March 25, 2017, 09:06:22 am
Hi, it would be great if I could get some help on this question
1) P(x1,y1) is a variable point on the hyperbola x2/a2 - y2/b2 =1 such that the tangent at P cuts the asymptotes at M and N.
a) Find the equation of the tangent at P (which I've already done)
b) Find the coordinates of M and N
Thank you!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 25, 2017, 09:19:18 am
Hi, it would be great if I could get some help on this question
1) P(x1,y1) is a variable point on the hyperbola x2/a2 - y2/b2 =1 such that the tangent at P cuts the asymptotes at M and N.
a) Find the equation of the tangent at P (which I've already done)
b) Find the coordinates of M and N
Thank you!!


Title: Re: 4U Maths Question Thread
Post by: VydekiE on March 25, 2017, 09:27:10 am




Thank you!!
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on March 26, 2017, 01:49:31 pm

Hi does anyone know how to do question 2
I am not sure how to approach it
Thanks

(http://uploads.tapatalk-cdn.com/20170325/27fa092f784d778ae149c771e97576b7.jpg)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 26, 2017, 01:59:03 pm
Hi does anyone know how to do question 2
I am not sure how to approach it
Thanks

(http://uploads.tapatalk-cdn.com/20170325/27fa092f784d778ae149c771e97576b7.jpg)
C, because the complementary event is just having none of the windows open, which can only be done in 1 way.

Note obviously that the total number of arrangements is 2^4 = 16, which is where we get 16-1=15 from.
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on March 26, 2017, 02:10:41 pm
C, because the complementary event is just having none of the windows open, which can only be done in 1 way.

Note obviously that the total number of arrangements is 2^4 = 16, which is where we get 16-1=15 from.

ahhh true
thank you!
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on March 27, 2017, 11:27:52 am
Not sure how to do b part 1 and 2(http://uploads.tapatalk-cdn.com/20170326/b07a25248876ca85108791cb5b166db0.jpg)
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on March 27, 2017, 11:53:09 am
Also this question
It looks simple but i just cant get it
Thank you!


(http://uploads.tapatalk-cdn.com/20170326/7633d4393531b4f770eed32afca6c045.jpg)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 27, 2017, 12:42:52 pm
Also this question
It looks simple but i just cant get it
Thank you!


(http://uploads.tapatalk-cdn.com/20170326/7633d4393531b4f770eed32afca6c045.jpg)


________________________








Hint for the other one: There are 3 cases to consider. You have no E's in there, one E or two E's
Title: Re: 4U Maths Question Thread
Post by: cutiepie30 on March 28, 2017, 09:18:04 pm
How would you do this question? (attached)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 28, 2017, 10:10:15 pm
How would you do this question? (attached)
Title: Re: 4U Maths Question Thread
Post by: cutiepie30 on March 29, 2017, 12:28:59 am


Thanks RuiAce  :)

could you please write a solution for me, i am kind of lost on where to go  :(
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 29, 2017, 07:10:25 am

It's really just a standard by parts question.
Title: Re: 4U Maths Question Thread
Post by: chelseam on March 29, 2017, 08:10:43 pm
Hi! Can someone please help me with part b of this question? Thank you! :)
Title: Re: 4U Maths Question Thread
Post by: dux99.95 on March 29, 2017, 08:58:17 pm
Heey
I got part a by subbing z= i.
How do u solve part b?

You guys are awesome!!
Title: Re: 4U Maths Question Thread
Post by: ellipse on March 29, 2017, 09:11:07 pm
Heey
I got part a by subbing z= i.
How do u solve part b?

You guys are awesome!!

let the other root be 'a.'

By doing the sum of roots, a+i=(1+i)/(2-i)
By realising the RHS, we get 1/5+3i/5=a+i.
Now solving for 'a,' you get a=1/5(1-2i)
Title: Re: 4U Maths Question Thread
Post by: ellipse on March 29, 2017, 09:42:53 pm
Hi! Can someone please help me with part b of this question? Thank you! :)


Title: Re: 4U Maths Question Thread
Post by: jakesilove on March 30, 2017, 08:17:36 am


You absolute Mathematical legend
Title: Re: 4U Maths Question Thread
Post by: chelseam on March 30, 2017, 06:43:06 pm
Thank you so much!  :D

Title: Re: 4U Maths Question Thread
Post by: qwertycc on March 31, 2017, 09:19:08 pm
Hi can someone please help me find the time taken for this resisted motion question. thanks very much!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 01, 2017, 10:35:16 am
Hi can someone please help me find the time taken for this resisted motion question. thanks very much!!
In the future, please provide any progress.



Title: Re: 4U Maths Question Thread
Post by: chelseam on April 01, 2017, 10:04:17 pm
Hi! Is there a way to recognise when to draw a major/minor arc that's above or below the x-axis for the type of question
arg(z±(...)/z±(...)) = π/(...)?

Thank you :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 01, 2017, 10:14:12 pm
Hi! Is there a way to recognise when to draw a major/minor arc that's above or below the x-axis for the type of question
arg(z±(...)/z±(...)) = π/(...)?

Thank you :)


I keep forgetting why so I'll link to one of Eddie Woo's videos
Title: Re: 4U Maths Question Thread
Post by: chelseam on April 01, 2017, 10:27:52 pm
That's so helpful! Thank you so much! ;D


I keep forgetting why so I'll link to one of Eddie Woo's videos
Title: Re: 4U Maths Question Thread
Post by: qwertycc on April 02, 2017, 09:33:20 am
Rui Ace, thanks for your help!  :)
Title: Re: 4U Maths Question Thread
Post by: Kle123 on April 03, 2017, 01:02:31 pm
can anybody help me out with this question. thank you
Title: Re: 4U Maths Question Thread
Post by: Kle123 on April 03, 2017, 01:14:00 pm
In this question i got pi as the answer but i did it through sketching on the argand diagram. Is the answer pi? and is this the simplest method?
Title: Re: 4U Maths Question Thread
Post by: ellipse on April 03, 2017, 01:16:33 pm
can anybody help me out with this question. thank you

Title: Re: 4U Maths Question Thread
Post by: ellipse on April 03, 2017, 01:21:02 pm
In this question i got pi as the answer but i did it through sketching on the argand diagram. Is the answer pi? and is this the simplest method?

you could do it using argument rules
Title: Re: 4U Maths Question Thread
Post by: Kle123 on April 03, 2017, 02:00:32 pm
Thank you so much  ;D
Title: Re: 4U Maths Question Thread
Post by: ellipse on April 03, 2017, 02:05:12 pm
Thank you so much  ;D

No worries!
Title: Re: 4U Maths Question Thread
Post by: Kle123 on April 03, 2017, 02:23:57 pm
im gonna have alot questions for today  :'(. I don't know how to do part b) for part a) however i got A(0,-b/tan(theta)) B(0,(a^2+b^2)tan(theta)/b). Sorry I don't have any answers to this exam paper
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 03, 2017, 04:18:54 pm
im gonna have alot questions for today  :'(. I don't know how to do part b) for part a) however i got A(0,-b/tan(theta)) B(0,(a^2+b^2)tan(theta)/b). Sorry I don't have any answers to this exam paper
Hint: On scrap paper, start by considering a circle where AB IS the diameter of. Then, since A and B both lie on the y-axis, we know that the center also lies on the y-axis.

In fact, the coordinates of the center is not hard to find - it'll just be the midpoint of AB, which I will call M.

And we can easily find the length of MA, which is also the length of MB (as they are both radii). It falls to use the distance formula to check that MS is yet a third radius of the circle, where S is the point (ae, 0).
Title: Re: 4U Maths Question Thread
Post by: dux99.95 on April 04, 2017, 07:14:47 am
let the other root be 'a.'

By doing the sum of roots, a+i=(1+i)/(2-i)
By realising the RHS, we get 1/5+3i/5=a+i.
Now solving for 'a,' you get a=1/5(1-2i)


Omg!! THANK YOU :) You made it so simple :)
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on April 06, 2017, 11:08:09 am
Hi
Could i please get some help on this mechanics question
what is retardation?
thank youuu
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on April 06, 2017, 11:10:24 am
Hi
Could i please get some help on this mechanics question
what is retardation?
thank youuu

Perhaps the wrong screenshot? I'm seeing lots of Chemistry  :o
Title: Re: 4U Maths Question Thread
Post by: jakesilove on April 06, 2017, 11:53:50 am
Hi
Could i please get some help on this mechanics question
what is retardation?
thank youuu

I know this isn't what it's supposed to be, but MAKE SURE TO INCLUDE YOUR STATES WHEN YOU WRITE CHEMICAL EQUATIONS!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 06, 2017, 12:08:25 pm
I know this isn't what it's supposed to be, but MAKE SURE TO INCLUDE YOUR STATES WHEN YOU WRITE CHEMICAL EQUATIONS!
The man has a point
Title: 4U Maths Question Thread
Post by: hanaacdr on April 06, 2017, 10:35:36 pm
WOOPS sorry
Here it is,

(Question 5)
A particle...

(http://uploads.tapatalk-cdn.com/20170406/5c6c6ae7ba8c46153119ff303099de3f.jpg)
Title: Re: 4U Maths Question Thread
Post by: Shadowxo on April 06, 2017, 11:18:16 pm
WOOPS sorry
Here it is,

(Question 5)
A particle...

(http://uploads.tapatalk-cdn.com/20170406/5c6c6ae7ba8c46153119ff303099de3f.jpg)

Retardation is just deceleration. If you're still having trouble let us know :)
Title: Re: 4U Maths Question Thread
Post by: ellipse on April 07, 2017, 05:45:51 pm
If the tangent to the cruve x^2+2xy+y^5=4 is horizontal at P(X,Y), show that P is a unique solution to the equation X^5+X^2+4=0

what does unique solution mean?
Title: Re: 4U Maths Question Thread
Post by: wu345 on April 07, 2017, 07:23:16 pm
Unique means nothing else like it i.e it's the only solution to the equation.
This of course makes sense when you think about it geometrically - the tangent only touches the ellipse at one point so there will be only one intersection point (i.e one unique solution to the given equation)
Title: Re: 4U Maths Question Thread
Post by: ellipse on April 07, 2017, 07:26:58 pm
Unique means nothing else like it i.e it's the only solution to the equation

oh ok so I need to show that it is the only real solution (ie all other roots are complex)?

or is sufficient to just implicitly differentiate (you get x+y=0) and sub it into the curve to get that equation (x^5+x^2+4=0)?
Title: Re: 4U Maths Question Thread
Post by: wu345 on April 07, 2017, 07:39:54 pm
oh ok so I need to show that it is the only real solution (ie all other roots are complex)?

or is sufficient to just implicitly differentiate (you get x+y=0) and sub it into the curve to get that equation (x^5+x^2+4=0)?

I think it should be sufficient to differentiate and sub as long as you justify it (i.e tangent only intersects the ellipse once)
Title: Re: 4U Maths Question Thread
Post by: ellipse on April 07, 2017, 07:54:18 pm
I think it should be sufficient to differentiate and sub as long as you justify it (i.e tangent only intersects the ellipse once)

yup thanks
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 07, 2017, 07:58:20 pm
I think it should be sufficient to differentiate and sub as long as you justify it (i.e tangent only intersects the ellipse once)
There's no ellipse here. That's a 5th power on y there.

I'm not sure if there was a typo in the original question or not. Because if the 5 is replaced by two we have \(x^2+2xy+y^2=4\) which is apparently a pair of parallel lines



Edit: Actually it does work, but we can't use any conic theorems because of the \(y^5\)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 07, 2017, 08:15:18 pm



___________________________



Title: Re: 4U Maths Question Thread
Post by: hanaacdr on April 08, 2017, 10:32:35 am
Hi
Could i get some help on question 3b

thank you
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 08, 2017, 07:28:29 pm
Hi
Could i get some help on question 3b

thank you






Title: Re: 4U Maths Question Thread
Post by: hanaacdr on April 10, 2017, 12:03:07 pm
Hi,
could i get some help on this question please
Title: Re: 4U Maths Question Thread
Post by: jakesilove on April 10, 2017, 02:30:35 pm
Hi,
could i get some help on this question please

Hey! Firstly, since all the coefficients are real, we know that either ONE of the roots is real, or THREE of the roots are real (as imaginary roots will come in complex conjugates).

The easiest way to answer the first half is by differentiating the function, it proving that it is non-decreasing. So,



For a>0, this derivative is ALWAYS positive. Thus, there will only be one real root, as the graph will never 'turn around'. It's helpful to sketch an example of a non-decreasing function (ie. gradient is always positive), to prove to the marker that you know why this fact results in one real root.

Now, if two roots are equal, we can right them as



Using sum and product rules.



and



We know from above that



so




Now, subbing alpha into the original equation, we get





Not sure where we are going with this; let's keep going?
 
We are trying to prove that



Subbing it what we've found above;



Which looks like it equals zero! Bit of a round about way of getting there, but hey, it works!
Title: Re: 4U Maths Question Thread
Post by: ellipse on April 10, 2017, 04:08:46 pm
Hi,
could i get some help on this question please

Here's an alternate method for part ii
Title: Re: 4U Maths Question Thread
Post by: jakesilove on April 10, 2017, 04:10:06 pm
Here's an alternate method for part ii

That's a heaps better method. Cheers for posting it up!
Title: Re: 4U Maths Question Thread
Post by: ellipse on April 10, 2017, 04:20:06 pm
That's a heaps better method. Cheers for posting it up!

No problem :)
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on April 10, 2017, 07:04:09 pm
Hey! Firstly, since all the coefficients are real, we know that either ONE of the roots is real, or THREE of the roots are real (as imaginary roots will come in complex conjugates).

The easiest way to answer the first half is by differentiating the function, it proving that it is non-decreasing. So,



For a>0, this derivative is ALWAYS positive. Thus, there will only be one real root, as the graph will never 'turn around'. It's helpful to sketch an example of a non-decreasing function (ie. gradient is always positive), to prove to the marker that you know why this fact results in one real root.

Now, if two roots are equal, we can right them as



Using sum and product rules.



and



We know from above that



so




Now, subbing alpha into the original equation, we get





Not sure where we are going with this; let's keep going?
 
We are trying to prove that



Subbing it what we've found above;



Which looks like it equals zero! Bit of a round about way of getting there, but hey, it works!


THANK YOUU!
Title: Re: 4U Maths Question Thread
Post by: beau77bro on April 13, 2017, 09:46:03 am
(http://uploads.tapatalk-cdn.com/20170412/ae72a154ec188b771b9ad6e2288d5dc5.jpg)

Hey guys could someone help me with 3b and c, I can't quite work it out algebraically
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 13, 2017, 10:21:04 am
(http://uploads.tapatalk-cdn.com/20170412/ae72a154ec188b771b9ad6e2288d5dc5.jpg)

Hey guys could someone help me with 3b and c, I can't quite work it out algebraically

___________

Side note: It may be worth mentioning that the given questions are the inverse hyperbolic sine and tangent functions, i.e. \( \sinh^{-1}x\), \( \tanh^{-1}x\). What we are trying to find are the exponential forms of the original hyperbolic functions \(\sinh x\) and \(\tanh x\)
Title: Re: 4U Maths Question Thread
Post by: beau77bro on April 13, 2017, 12:00:02 pm
thanks rui, i didnt even see that you could just cancel out the y squared, rookie mistake sorry ahhaha
Title: Re: 4U Maths Question Thread
Post by: beau77bro on April 13, 2017, 12:02:47 pm
(http://uploads.tapatalk-cdn.com/20170413/1bd49af4c7a5ebef05864e2acdab9efb.jpg)

I hope this question is better, not sure how to do 8a, the rest is fine but just abit confused on whether its algebraic or word and where to go? Sub in value less than 0 and show it doesn't work because it x^2 +3x/2 has to be greater than 0?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 13, 2017, 03:28:22 pm

(http://uploads.tapatalk-cdn.com/20170413/1bd49af4c7a5ebef05864e2acdab9efb.jpg)

I hope this question is better, not sure how to do 8a, the rest is fine but just abit confused on whether its algebraic or word and where to go? Sub in value less than 0 and show it doesn't work because it x^2 +3x/2 has to be greater than 0?
The natural domain of the first is x>0 but for the second it's x>0 AND x<-3/2
Title: Re: 4U Maths Question Thread
Post by: VydekiE on April 13, 2017, 05:33:25 pm
Hi, it would be great if I could get some help on this integration question
1. Integrate 1/5+4x+x2 dx
Thank you!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 13, 2017, 06:44:12 pm
Hi, it would be great if I could get some help on this integration question
1. Integrate 1/5+4x+x2 dx
Thank you!!
Is there a bracket missing around the denominator?

(If yes, hint: complete the square)
Title: Re: 4U Maths Question Thread
Post by: VydekiE on April 13, 2017, 07:55:28 pm
Is there a bracket missing around the denominator?

(If yes, hint: complete the square)

Nope there's no brackets missing
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 13, 2017, 08:01:45 pm
Title: Re: 4U Maths Question Thread
Post by: VydekiE on April 14, 2017, 04:13:57 pm

Didn't think of that. Thank you!!
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on April 15, 2017, 07:17:52 am
Hi
could i please get some help on part b please
thanks
Title: Re: 4U Maths Question Thread
Post by: jakesilove on April 15, 2017, 02:48:21 pm
Hi
could i please get some help on part b please
thanks

We know that the net force, on the way down, will be












Also,









Subbing in what we found the first time,



And that should get you to the answer. Possible I messed up a sign/factor somewhere, but that's the general set-up :)
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on April 15, 2017, 08:28:39 pm
thank you!

We know that the net force, on the way down, will be












Also,









Subbing in what we found the first time,



And that should get you to the answer. Possible I messed up a sign/factor somewhere, but that's the general set-up :)
Title: Re: 4U Maths Question Thread
Post by: Rathin on April 16, 2017, 12:59:58 pm
Can someone check if my method is correct?

Q) I=∫=x^3*e^x^2dx

A)

u=x^3
u'=3x^2
v'=e^x^2
v=2xe^x^2

Therefore;
I=2x^4*e^x^2 - 6∫x^3*e^x^2dx
7I= 2x^4*e^x^2 +C
I= (2x^4*e^x^2)/7 +C
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 16, 2017, 01:13:33 pm
v'=e^x^2
v=2xe^x^2
Look here. You differentiated your v', instead of integrated it.
Title: Re: 4U Maths Question Thread
Post by: Rathin on April 16, 2017, 01:19:58 pm
Look here. You differentiated your v', instead of integrated it.


Omg these silly mistakes..
Title: Conics Question
Post by: beau77bro on April 18, 2017, 01:22:11 pm
(http://uploads.tapatalk-cdn.com/20170418/70e2141275b17432c4c1e8153aa2d60b.jpg)

Q15 bii, I don't really get it, how do we do simultaneous, i feel I missed something simple
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 18, 2017, 02:09:30 pm




Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 18, 2017, 02:12:16 pm
If you meant a) ii it's actually just rearranging.

Title: Re: 4U Maths Question Thread
Post by: Rathin on April 18, 2017, 10:06:50 pm
Help with this please
∫sin(x)sin(3x)dx
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 18, 2017, 10:16:01 pm
Help with this please
∫sin(x)sin(3x)dx

Title: Re: 4U Maths Question Thread
Post by: Rathin on April 19, 2017, 09:53:08 am



Hey, wouldn't sin(x)sin(3x)=1/2(cos(2x)-cos(4x))?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 19, 2017, 09:56:55 am
Hey, wouldn't sin(x)sin(3x)=1/2(cos(2x)-cos(4x))?
No idea why I wrote down sin again. Let me edit that
Title: Re: 4U Maths Question Thread
Post by: Rathin on April 20, 2017, 01:10:55 pm
No idea why I wrote down sin again. Let me edit that

Textbook answer says something else from my answer. I got 1/4(sin(2x)) - 1/8(sin(4x)) +c . Is this correct?
Title: Re: 4U Maths Question Thread
Post by: Rathin on April 20, 2017, 01:27:53 pm
By IBP I got I=(sin(3x)cos(x)-3cos(3x)sin(x))/8 +c which is what the textbook has as its answer.
Title: Re: 4U Maths Question Thread
Post by: jakesilove on April 20, 2017, 02:59:52 pm
By IBP I got I=(sin(3x)cos(x)-3cos(3x)sin(x))/8 +c which is what the textbook has as its answer.

Those two solutions are exactly the same :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 20, 2017, 04:46:40 pm
What Jake said. When integrating, you must ALWAYS consider the possibility that the final answers are no different to each other.
Title: Re: 4U Maths Question Thread
Post by: crazycodpro on April 20, 2017, 07:35:20 pm
 :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 20, 2017, 07:47:15 pm
:)



Title: Re: 4U Maths Question Thread
Post by: crazycodpro on April 20, 2017, 08:15:44 pm
this is a stupid question but RuiAce how did you find the Dv/dh
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 20, 2017, 08:16:17 pm
this is a stupid question but RuiAce how did you find the Dv/dh
Just differentiate it.

Title: Re: 4U Maths Question Thread
Post by: crazycodpro on April 20, 2017, 08:20:16 pm
that make sense, thanks RuiAca  :)
Title: Re: 4U Maths Question Thread
Post by: beau77bro on April 21, 2017, 01:46:07 pm
(http://uploads.tapatalk-cdn.com/20170421/439d26d3853f4d5a7918066085ae4c65.jpg)

Please help hahaha
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 21, 2017, 02:19:32 pm
Title: Re: 4U Maths Question Thread
Post by: beau77bro on April 23, 2017, 12:13:29 am
(http://uploads.tapatalk-cdn.com/20170422/fee98dccb1a6ec7f5c1bf8092883d003.jpg)

Please help me question 4c, I feel I've done it before, but a solution would be so helpful
Title: Re: 4U Maths Question Thread
Post by: Mahan on April 24, 2017, 12:48:36 am
(http://uploads.tapatalk-cdn.com/20170422/fee98dccb1a6ec7f5c1bf8092883d003.jpg)

Please help me question 4c, I feel I've done it before, but a solution would be so helpful

Using the first part,

Also as a reminder,
Note

Similarly

Then

This comes from the facts:

and

and
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on April 24, 2017, 12:40:58 pm
Hi
could i get some help on this question please
i feel like I have done it before but not getting the answer
thank you! :)
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on April 24, 2017, 12:43:34 pm
Hi
Sorry and this question please
I feel like it is very physicsy (hhaha not a word)
thank you!
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on April 25, 2017, 12:15:35 pm
hi
Could i please get some help on question 8 please
thank youuu
Title: Re: 4U Maths Question Thread
Post by: jakesilove on April 25, 2017, 12:42:37 pm
Hi
could i get some help on this question please
i feel like I have done it before but not getting the answer
thank you! :)

Hey! For this question, simply use the distance formula to find the distance between the points



Where the eccentricity takes its usual form. You'll find that the distance between the point and the 'positive' foci is



and the distance between the point and the 'negative' foci is



Adding this up gives us the required relation, 2a.

Hi
Sorry and this question please
I feel like it is very physicsy (hhaha not a word)
thank you!

I'm not sure that this is really a maths question; I think it wants you to say that the beam of light will again go through the first foci? Presumably, EVERY bounce will lead the light to one of the foci. Nothing to solve here


Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 25, 2017, 12:51:50 pm
Hi
Sorry and this question please
I feel like it is very physicsy (hhaha not a word)
thank you!
The ray is going to alternate through which focus it passes through. If you shine a light from, or directly through one focus, it will always be reflected to the other focus. This is the reflection property of the ellipse.

But yeah, it's probably more of a physics question.

Hey! For this question, simply use the distance formula to find the distance between the points



Where the eccentricity takes its usual form. You'll find that the distance between the point and the 'positive' foci is



and the distance between the point and the 'negative' foci is


It says Cartesian coordinates in the question. Give me a sec.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 25, 2017, 12:56:02 pm
Hi
could i get some help on this question please
i feel like I have done it before but not getting the answer
thank you! :)


Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 25, 2017, 01:01:37 pm
hi
Could i please get some help on question 8 please
thank youuu
I just simulated it on GeoGebra and the question is wrong. (Most likely a typo.)

Because ellipse ii is literally enclosed INSIDE ellipse i, a tangent to ellipse i can be shown to never touch ellipse ii.
Title: Re: 4U Maths Question Thread
Post by: chelseam on April 25, 2017, 10:15:08 pm
Hi! I'm confused about how to prove the reflection property for the hyperbola! Also, would the geometrical properties for the hyperbola also apply to rectangular hyperbolae (especially the type with its asymptotes being the x & y axis)? Thank you :D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 25, 2017, 11:16:26 pm
Hi! I'm confused about how to prove the reflection property for the hyperbola! Also, would the geometrical properties for the hyperbola also apply to rectangular hyperbolae (especially the type with its asymptotes being the x & y axis)? Thank you :D
A rectangular hyperbola is just a hyperbola with eccentricity √2, so any property with the hyperbola will always apply to the rectangular hyperbola.

That being said, the foci and directrices of \(xy=c^2\) are NOT required for the 4U course, so you should NOT have to expect to do much with them.

For a full proof of the reflection property, consider Q13b and Q13c of the 2012 paper as well as any good source of solutions. You may ask for any specific lines of working out to be explained.
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on April 26, 2017, 10:16:28 pm
I just simulated it on GeoGebra and the question is wrong. (Most likely a typo.)

Because ellipse ii is literally enclosed INSIDE ellipse i, a tangent to ellipse i can be shown to never touch ellipse ii.
Thank you!
Title: Re: 4U Maths Question Thread
Post by: JuliaPascale123 on April 27, 2017, 11:29:02 am
Could I please have the working out and answers to these questions
(http://i67.tinypic.com/ycq49.jpg)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 27, 2017, 12:26:59 pm
Could I please have the working out and answers to these questions
(http://i67.tinypic.com/ycq49.jpg)

Title: Re: 4U Maths Question Thread
Post by: beau77bro on April 27, 2017, 07:44:18 pm
Using the first part,

Also as a reminder,
Note

Similarly

Then

This comes from the facts:

and

and


omg thankyou soo much mahan, sorry such a late reply appreciate the make explanation
Title: Re: 4U Maths Question Thread
Post by: beau77bro on April 27, 2017, 07:46:08 pm
(http://uploads.tapatalk-cdn.com/20170427/48cf7999750683dbddda5b64f0288bab.jpg)

This is an absolute monster question. I think atleast ahahaha any help is ridiculously appreciated
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 27, 2017, 07:59:02 pm
(http://uploads.tapatalk-cdn.com/20170427/48cf7999750683dbddda5b64f0288bab.jpg)

This is an absolute monster question. I think atleast ahahaha any help is ridiculously appreciated


I don't see how that can be simplified any further and neither does Wolfram
Title: Re: 4U Maths Question Thread
Post by: legorgo18 on April 27, 2017, 09:54:07 pm
Hello, i need some help with the following

1) If x,y>0 show that 1/x + 1/y > or equal to 4/(x+y)
2) Using cauchy's inequality, prove that 2(a^3 + b^3 + c^3) greater than or equal to ab(a+b) + ac(a+c) + bc(b+c) greater than or equal to 6abc

Thank you :c
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 27, 2017, 11:30:09 pm
Hello, i need some help with the following

1) If x,y>0 show that 1/x + 1/y > or equal to 4/(x+y)
2) Using cauchy's inequality, prove that 2(a^3 + b^3 + c^3) greater than or equal to ab(a+b) + ac(a+c) + bc(b+c) greater than or equal to 6abc

Thank you :c


____________________

I have no idea what they teach there because the Cauchy-Schwarz inequality isn't even in the HSC...
I do not see how this benefits students when they can't even use it in the exam.


Remarks: A lot of algebra was rushed. You may consider working through the algebra more slowly.
For the linear algebra version |a.b| ≤ ||a|| ||b||, simply treat the terms of the sequence as the components of the vectors instead.

Haven't figured out how C-S can be used for the first half yet
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 28, 2017, 03:11:07 pm
Hello, i need some help with the following

1) If x,y>0 show that 1/x + 1/y > or equal to 4/(x+y)
2) Using cauchy's inequality, prove that 2(a^3 + b^3 + c^3) greater than or equal to ab(a+b) + ac(a+c) + bc(b+c) greater than or equal to 6abc

Thank you :c

I have absolutely no clue why it's called that though - that's wrong to me.

This means I'll have to look at 2) properly later to determine what to substitute into the 3-variable AM-GM inequality.

Edit: Found a convoluted proof for part 2





Title: Re: 4U Maths Question Thread
Post by: beau77bro on April 29, 2017, 01:11:25 pm



I don't see how that can be simplified any further and neither does Wolfram

Omg THANKYOU tho I don't know series. The answer is this if it helps. Sorry for such a random Q and it probably would've helped a lot if I'd shown where u actually need to get to sorrryy(http://uploads.tapatalk-cdn.com/20170429/79b9b702bd22d5b5028501de74152757.jpg)

Maybe this will help? All g if it doesn't just stumped
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 29, 2017, 04:35:20 pm
Omg THANKYOU tho I don't know series. The answer is this if it helps. Sorry for such a random Q and it probably would've helped a lot if I'd shown where u actually need to get to sorrryy(http://uploads.tapatalk-cdn.com/20170429/79b9b702bd22d5b5028501de74152757.jpg)

Maybe this will help? All g if it doesn't just stumped




\begin{align*}&\quad \left[(1-x\cos x)-i(x\sin x)\right][(1-x\cos x)+i(x\sin x)]\\ &= (1-x\cos x)^2+(x\sin x)^2\\ &= 1-2x\cos x + x^2(\cos^2x+\sin^2x)\\ &= 1-2x\cos x + x^2\end{align*}

Cis notation is ew.


I might've made an algebra mistake because I don't quite agree with what they got for b.
Title: Re: 4U Maths Question Thread
Post by: Kle123 on April 30, 2017, 10:43:53 am
HEY, could anyone help me out with this integration? THANK YOU. I edited the red minus in(forgot to rewrite it after erasing it). The answer is also attached.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 30, 2017, 11:06:25 am
HEY, could anyone help me out with this integration? THANK YOU. I edited the red minus in(forgot to rewrite it after erasing it). The answer is also attached.

The method of computing your integral can be found in post #1040. In your scenario, take R = 6
Title: Re: 4U Maths Question Thread
Post by: Kle123 on April 30, 2017, 01:08:47 pm
Sorry Rui, I dont understand how you got from the first to second step and how the integration of the first part of the 2nd step is equal to the first part of the 3rd step. Could you explain? Thanks
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 30, 2017, 01:10:48 pm
Sorry Rui, I dont understand how you got from the first to second step and how the integration of the first part of the 2nd step is equal to the first part of the 3rd step. Could you explain? Thanks


Title: Re: 4U Maths Question Thread
Post by: Kle123 on April 30, 2017, 01:40:38 pm
Sorry. Before I meant i didn't know how to get from the first line to the second line on the right hand side. Also instead i tried differentiating it to check the integration but i got different.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 30, 2017, 01:52:41 pm
Sorry. Before I meant i didn't know how to get from the first line to the second line on the right hand side. Also instead i tried differentiating it to check the integration but i got different.
Oh my bad, this whole time there were typos floating around.

Given the typo's I'm not too sure how my original answer came out nicely (unless it didn't).

That being said, these are 4U integration tricks that you are expected to know.
Title: Re: 4U Maths Question Thread
Post by: Kle123 on April 30, 2017, 02:25:59 pm
That being said, these are 4U integration tricks that you are expected to know.
I see. Thanks for the help Rui!
Title: Re: 4U Maths Question Thread
Post by: chelseam on May 01, 2017, 11:45:17 pm
Hi! Could someone please help me with this question? The exercise is about using the substitution t=tanθ/2. Thank you :)

Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 02, 2017, 06:55:40 am
Hi! Could someone please help me with this question? The exercise is about using the substitution t=tanθ/2. Thank you :)
What textbook is this? That substitution is a horrible idea; u = sin(x) would be a much better substitution.
Title: Re: 4U Maths Question Thread
Post by: chelseam on May 02, 2017, 07:00:59 pm
What textbook is this? That substitution is a horrible idea; u = sin(x) would be a much better substitution.
Yeah that's what I thought as well! It's from the old version of Fitzpatrick but it's in the same exercise in the new one as well. Thanks Rui :)
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on May 04, 2017, 07:09:32 pm
could i get some help on this question too please
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on May 04, 2017, 08:59:15 pm
could i get some help plz
thanks
Title: Re: 4U Maths Question Thread
Post by: jakesilove on May 04, 2017, 09:30:43 pm
could i get some help on this question too please


Hey! Check out my brief solution below

(http://i.imgur.com/4y1hhc1.jpg)
Title: Re: 4U Maths Question Thread
Post by: jakesilove on May 04, 2017, 09:32:43 pm
could i get some help plz
thanks


And here's the answer to your second question :)

(http://i.imgur.com/riSAvLa.png)
Title: Re: 4U Maths Question Thread
Post by: hinakamishiro on May 05, 2017, 08:58:33 pm
Hey guys can someone help me with part 3? Thanks!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 05, 2017, 09:09:47 pm
Hey guys can someone help me with part 3? Thanks!
Previous post deleted upon seeing part a)



Again, use graphing software such as Desmos to aid this process. And rememeber the dilation factor in the change of base law, or just directly smack in log2 if you wish.
Title: Re: 4U Maths Question Thread
Post by: beau77bro on May 06, 2017, 11:12:05 pm




\begin{align*}&\quad \left[(1-x\cos x)-i(x\sin x)\right][(1-x\cos x)+i(x\sin x)]\\ &= (1-x\cos x)^2+(x\sin x)^2\\ &= 1-2x\cos x + x^2(\cos^2x+\sin^2x)\\ &= 1-2x\cos x + x^2\end{align*}

Cis notation is ew.


I might've made an algebra mistake because I don't quite agree with what they got for b.

woweee. that is some simple but overly complicated algebra. i apologise for the late response but thankyou very much rui. I hope i never get anything like this in a test hahahah. you r a god thankyou
Title: Re: 4U Maths Question Thread
Post by: beau77bro on May 06, 2017, 11:14:41 pm
(http://uploads.tapatalk-cdn.com/20170506/b60a124d5b5956cf9eb1dbeb793dfd9b.jpg)

4b I feel like I'm Doing something wrong I'm getting 128, the answer is 16?

My working:
(http://uploads.tapatalk-cdn.com/20170506/5ab50ad3287d818f31aad2d40c72f21c.jpg)
Title: Re: 4U Maths Question Thread
Post by: chelseam on May 07, 2017, 04:46:56 pm
Hi! Do we need to know the condition of the focal chord for conics? Or is that only needed in 3u parametrics? Thank you :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 07, 2017, 04:55:34 pm
(http://uploads.tapatalk-cdn.com/20170506/b60a124d5b5956cf9eb1dbeb793dfd9b.jpg)

4b I feel like I'm Doing something wrong I'm getting 128, the answer is 16?

My working:
(http://uploads.tapatalk-cdn.com/20170506/5ab50ad3287d818f31aad2d40c72f21c.jpg)
Can't quite follow all of your working. I understand the use of 4a = length of latus rectum, but I don't understand how you found the area of the parabolic cross section to use as the volume of your slice.

Hi! Do we need to know the condition of the focal chord for conics? Or is that only needed in 3u parametrics? Thank you :)
Parabola: The tangents at the end of a focal chord intersect at right angles on the directrix

Ellipse/Hyperbola: The tangents at the end of a focal chord intersect on the directrix.

So simply put, if the chord is a chord of contact from the directrix, it is focal.
Title: Re: 4U Maths Question Thread
Post by: beau77bro on May 07, 2017, 05:18:43 pm
Can't quite follow all of your working. I understand the use of 4a = length of latus rectum, but I don't understand how you found the area of the parabolic cross section to use as the volume of your slice.

Oh so it's like a rectangle area of ax4a - like length of the directrix x hieght which is a, as it passes through the focus, then you minus the area under the curve so you just have the area of the cross section. then im not sure what to do in terms of working out the volume of the solid correctly, im not even sure the slice is correct. Sorry for unclear working and question, probably ignore my working as it is piss poor, but the help is very much appreciated.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 07, 2017, 05:27:32 pm
Unless I went blind, I just realised you wrote \(4a=\sqrt{36-9y^2}\) and then \(a=\sqrt{36-9y^2}\) without any division by 4?
Title: Re: 4U Maths Question Thread
Post by: chelseam on May 07, 2017, 07:45:17 pm
So simply put, if the chord is a chord of contact from the directrix, it is focal.
Thanks Rui :D
Title: Re: 4U Maths Question Thread
Post by: seventeenboi on May 08, 2017, 04:20:07 pm
HI,
just a conics question:
I can do everything else except q2b(gamma).. helppp!!!
Thanks so much :)))
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 08, 2017, 04:38:46 pm

HI,
just a conics question:
I can do everything else except q2b(gamma).. helppp!!!
Thanks so much :)))
Hints: Use PS=e*PM and PS'=ePM' to take care of the LHS.

If you need the full solution please provide the answers to the previous part.
Title: Re: 4U Maths Question Thread
Post by: seventeenboi on May 08, 2017, 04:48:21 pm
Hints: Use PS=e*PM and PS'=ePM' to take care of the LHS.

If you need the full solution please provide the answers to the previous part.


OMG THANK YOU SO MUCH
it worked out yayy
Title: Re: 4U Maths Question Thread
Post by: kiwiberry on May 09, 2017, 10:28:35 am
Hello, I'm having trouble starting off this volumes question, any help is appreciated ;D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 09, 2017, 03:00:03 pm
Hello, I'm having trouble starting off this volumes question, any help is appreciated ;D
(http://i.imgur.com/LPMpy9C.png)



* - Not fully accurate, but don't worry about that.
(http://i.imgur.com/dppN9cn.png)

______________________________________



Title: Re: 4U Maths Question Thread
Post by: kiwiberry on May 09, 2017, 04:26:17 pm
....

Ahhh thank you so much Rui, that helped a lot :)
Title: Re: 4U Maths Question Thread
Post by: Wales on May 12, 2017, 07:11:49 pm
Just dropping by with a integration question.

Integral of (1+x)/SQRT(1-x-x^2) dx

From Coroneos' 100 Integrals question 15.

Cheers, Wales
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 12, 2017, 08:03:43 pm
Just dropping by with a integration question.

Integral of (1+x)/SQRT(1-x-x^2) dx

From Coroneos' 100 Integrals question 15.

Cheers, Wales
Title: Re: 4U Maths Question Thread
Post by: chelseam on May 12, 2017, 10:41:20 pm
Hi! I'm having trouble integrating cot^3x, and cot^5x. Thank you :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 12, 2017, 11:03:43 pm
Hi! I'm having trouble integrating cot^3x, and cot^5x. Thank you :)
This integral works the exact same as \(\int \tan^3 x\, dx\). You should have a go at that one first.
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on May 13, 2017, 09:08:07 am
Hi
Could i get some help on this question
Title: Re: 4U Maths Question Thread
Post by: crazycodpro on May 13, 2017, 06:43:39 pm
hi
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 13, 2017, 06:48:09 pm
hi
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 13, 2017, 07:21:21 pm
Hi
Could i get some help on this question

A bit all over the place. Feel free to point out any areas of confusion
(http://i.imgur.com/FfE4rrr.jpg)
Title: Re: 4U Maths Question Thread
Post by: hanaacdr on May 14, 2017, 03:08:47 pm
A bit all over the place. Feel free to point out any areas of confusion
(http://i.imgur.com/FfE4rrr.jpg)

WOW!
thank you!
Title: Re: 4U Maths Question Thread
Post by: beau77bro on May 19, 2017, 10:11:18 am
(http://uploads.tapatalk-cdn.com/20170519/5042b959a17e098c0ceaa0192c2f81d1.jpg)

Please help
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 19, 2017, 11:48:04 am
(http://uploads.tapatalk-cdn.com/20170519/5042b959a17e098c0ceaa0192c2f81d1.jpg)

Please help

_________________



_________________


_________________
Title: Re: 4U Maths Question Thread
Post by: beau77bro on May 19, 2017, 02:19:18 pm
omg u r a god
Title: Re: 4U Maths Question Thread
Post by: Jyrgal on May 20, 2017, 08:16:14 pm
heyo, i have a question on the below (i think) inequality question. i got up to c)ii), i kinda fudged the induction process (doesnt rly make much sense) can anyone help me with step by step for that? (assume c)i) uve already done)

thanks :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 20, 2017, 08:33:04 pm
heyo, i have a question on the below (i think) inequality question. i got up to c)ii), i kinda fudged the induction process (doesnt rly make much sense) can anyone help me with step by step for that? (assume c)i) uve already done)

thanks :)
.


Title: Re: 4U Maths Question Thread
Post by: hanaacdr on May 21, 2017, 01:46:21 pm
hi could i please get some help on this question thanks
Title: Re: 4U Maths Question Thread
Post by: Mahan on May 21, 2017, 03:48:27 pm
hi could i please get some help on this question thanks

There are two methods:

Then just evaluate the integral in the interval.

Second method,


That implies,



Hence,



That changes the variable of the integration to t  and the limit of the integral is now between 0 and 1/(3^0.5)

evaluating the integral in the new interval will give you the same answer.
Title: Re: 4U Maths Question Thread
Post by: itssona on May 21, 2017, 07:16:15 pm
heey so for this, do we differentiate 3 times?? I got -9/8 is it right
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 21, 2017, 07:22:37 pm
heey so for this, do we differentiate 3 times?? I got -9/8 is it right



To finish off the question, the last root is easy to find. Just use the sum of roots formula.
Title: Re: 4U Maths Question Thread
Post by: itssona on May 21, 2017, 07:30:27 pm



To finish off the question, the last root is easy to find. Just use the sum of roots formula.

So I got -2 after eliminating a candidate point and now I use -b/a in the original thing- so the other root is -9/2?
Just making sure
Thank you so so so much Rui :D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 21, 2017, 07:32:55 pm
The triple root is indeed -2
Title: Re: 4U Maths Question Thread
Post by: itssona on May 21, 2017, 07:35:29 pm
The triple root is indeed 2

ah thank you :)
Title: Re: 4U Maths Question Thread
Post by: itssona on May 22, 2017, 08:26:28 am
Having trouble using the sum of roots method for this :/

they compare terms in (x+2)^2 (x^2 +ax +b)

but why did they write x^2 +ax +b instead of ax^2 + bx +c
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 22, 2017, 08:50:26 am

Having trouble using the sum of roots method for this :/

they compare terms in (x+2)^2 (x^2 +ax +b)

but why did they write x^2 +ax +b instead of ax^2 + bx +c
You could go for \(ax^2+bx+c\) but if you look carefully it should be beyond obvious that a=1 as the polynomial is monic (leading coefficient 1)
Title: Re: 4U Maths Question Thread
Post by: itssona on May 22, 2017, 09:41:52 am
You could go for \(ax^2+bx+c\) but if you look carefully it should be beyond obvious that a=1 as the polynomial is monic (leading coefficient 1)
ohhhhhhhhhhhhh
Title: Re: 4U Maths Question Thread
Post by: itssona on May 22, 2017, 05:04:46 pm
at y=f^2 (x)

why is the turning point squared of f(x)?? i dont get it :/
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 22, 2017, 05:12:03 pm

at y=f^2 (x)

why is the turning point squared of f(x)?? i dont get it :/
For the sake of disambiguity, are you referring to \(f(f(x))\) or \([f(x)]^2\)?
Title: Re: 4U Maths Question Thread
Post by: itssona on May 22, 2017, 05:16:35 pm
For the sake of disambiguity, are you referring to \(f(f(x))\) or \([f(x)]^2\)?
f(f(x))
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 22, 2017, 05:43:01 pm



In general, no rule immediately justifies how to find the values of x satisfying that criteria. These zeroes can potentially be hard to find
Title: Re: 4U Maths Question Thread
Post by: itssona on May 22, 2017, 05:54:03 pm



In general, no rule immediately justifies how to find the values of x satisfying that criteria. These zeroes can potentially be hard to find
oh that makes sense! :)

sorry to ask another thing, but does that mean:
if I graph y=(x^2 -1)^3 then I can see that f'(x) is 0 so we have x = + or -1
and then I make this function in terms of f(x) and see that x=0
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 22, 2017, 05:57:20 pm



Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 22, 2017, 05:58:40 pm
oh that makes sense! :)

sorry to ask another thing, but does that mean:
if I graph y=(x^2 -1)^3 then I can see that f'(x) is 0 so we have x = + or -1
and then I make this function in terms of f(x) and see that x=0
Not too sure what you mean here?
Title: Re: 4U Maths Question Thread
Post by: itssona on May 22, 2017, 06:10:38 pm
Not too sure what you mean here?
like you know how f(x) should =0 because f cube x 's derivative is f'(x) multiplied by derivate of f'f(x) (and this is made up of f(x) and f'(x))

so what I mean to say is that f(x) should be 0

soooo (x^2-1)^3 =0
and then x=0

Im probably wrong lol idk
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 22, 2017, 06:12:34 pm


Title: Re: 4U Maths Question Thread
Post by: itssona on May 22, 2017, 06:27:02 pm



ah I get you



ohh I get you! Thank you SOO much for explaining all this so well Rui :D
Title: Re: 4U Maths Question Thread
Post by: seventeenboi on May 23, 2017, 01:56:17 pm
Hi! pls help with the 'describe how the shape of this curve changes as λ increases from 1 towards 2' part
Thanks!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 23, 2017, 04:54:38 pm
Hi! pls help with the 'describe how the shape of this curve changes as λ increases from 1 towards 2' part
Thanks!






GeoGebra simulation attached
Title: Re: 4U Maths Question Thread
Post by: scienceislife on May 25, 2017, 07:12:55 am
Evaluate the integral between 0 and -1 sqrtx+2/sqrt1-x dx
Thank you!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 25, 2017, 09:01:59 am
Evaluate the integral between 0 and -1 sqrtx+2/sqrt1-x dx
Thank you!
For the sake of disambiguity, do you mean \( \frac{\sqrt{x+2}}{\sqrt{1-x}}\), \(\frac{\sqrt{x}+2}{\sqrt{1}-x} \) or \(\sqrt{x}+\frac{2}{\sqrt{1}}-x \)?

The way you have typed it (without bracketing) is implied to suggest that you mean the third one.
Title: Re: 4U Maths Question Thread
Post by: bluecookie on May 27, 2017, 12:06:00 am
What's the focus and directrix of a line? (with e=infinity)
Title: 4U Maths Question Thread
Post by: RuiAce on May 27, 2017, 09:05:25 am
What's the focus and directrix of a line? (with e=infinity)
In essence, we will have a pair of intersecting lines and not just a single line.

Consider what happens as the eccentricity is increased. As the eccentricity increases, the foci become further distanced and the directrices approaches the y-axis. When the eccentricity becomes infinite, the foci will be infinitely away and the directrices coincide on the y-axis.

This also makes intuitive sense, as PS/PM = e. As e->inf, we have either PS -> inf, or PM -> 0.

It may be worth considering that this case is the opposite of the circle, where the foci coincide at the origin to become the centre, and the directrices are infinitely far away
Title: Re: 4U Maths Question Thread
Post by: bluecookie on May 27, 2017, 12:59:00 pm
In essence, we will have a pair of intersecting lines and not just a single line.

Consider what happens as the eccentricity is increased. As the eccentricity increases, the foci become further distanced and the directrices approaches the y-axis. When the eccentricity becomes infinite, the foci will be infinitely away and the directrices coincide on the y-axis.

This also makes intuitive sense, as PS/PM = e. As e->inf, we have either PS -> inf, or PM -> 0.

It may be worth considering that this case is the opposite of the circle, where the foci coincide at the origin to become the centre, and the directrices are infinitely far away

Oh, thanks :)
Title: Re: 4U Maths Question Thread
Post by: beau77bro on May 29, 2017, 09:11:32 am
(http://uploads.tapatalk-cdn.com/20170528/4c8fbaeefbf0c308c274745c18beefd1.jpg)

Please help question 3
Title: Re: 4U Maths Question Thread
Post by: beau77bro on May 30, 2017, 09:18:59 pm
Hey guys can I grab some help with this whole question I'm seriously struggling (http://uploads.tapatalk-cdn.com/20170530/3cc03a64824ffd60ab8e131028ac200a.jpg)to get it
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 01, 2017, 07:28:10 pm
(http://uploads.tapatalk-cdn.com/20170528/4c8fbaeefbf0c308c274745c18beefd1.jpg)

Please help question 3
Honestly this question appears nonsensical.

The question says that the particle accelerates, until it reaches the maximum velocity. Then, once it is at the maximum velocity it starts travelling the distance S that it's supposed to.

And then they say \(T_1\) is the time taken for the car travel the full distance at maximum velocity. But it doesn't travel the full distance at maximum velocity?

If the question meant something else then it was worded horribly.
Title: 4U Maths Question Thread
Post by: RuiAce on June 01, 2017, 07:37:54 pm
Hey guys can I grab some help with this whole question I'm seriously struggling (http://uploads.tapatalk-cdn.com/20170530/3cc03a64824ffd60ab8e131028ac200a.jpg)to get it
The significance is in that it means the force acts against the direction of the position/velocity respectively.

In the future, where possible please avoid the sideways photos. This request is a bit selfish but it stems from the fact that I personally hate mechanics and get put off from doing it quite easily......(http://uploads.tapatalk-cdn.com/20170601/617ba62b13d717431bb7d9ee56ed6ef7.jpg)(http://uploads.tapatalk-cdn.com/20170601/c3aacaf262b4c38ae2f2384e11f2b964.jpg)(http://uploads.tapatalk-cdn.com/20170601/f7146d1d4807dd50b9d62bf8838cfae3.jpg)
Title: Re: 4U Maths Question Thread
Post by: kiwiberry on June 01, 2017, 09:16:09 pm
Hey, can someone please explain these perms and combs questions, I'm at a loss

1) Four sets of twins go to a party. In how many ways can these children be arranged in pairs so that no child is with his twin?

2) How many ways can three integers be chosen which are in arithmetic progression from the numbers 1 to 101?

Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 01, 2017, 10:11:54 pm
Hey, can someone please explain these perms and combs questions, I'm at a loss

1) Four sets of twins go to a party. In how many ways can these children be arranged in pairs so that no child is with his twin?

2) How many ways can three integers be chosen which are in arithmetic progression from the numbers 1 to 101?





__________________________________________________________________

Note: We could have used the second or third term as well. Provided the triplets are ordered, we have a reference point.




I'm not confident with my working out for the first one. Can I please have the answer?
Title: Re: 4U Maths Question Thread
Post by: kiwiberry on June 01, 2017, 10:31:59 pm
I'm not confident with my working out for the first one. Can I please have the answer?
Thanks so much omg. The first one is 60
Title: Re: 4U Maths Question Thread
Post by: beau77bro on June 02, 2017, 01:15:20 pm
Honestly this question appears nonsensical.

The question says that the particle accelerates, until it reaches the maximum velocity. Then, once it is at the maximum velocity it starts travelling the distance S that it's supposed to.

And then they say \(T_1\) is the time taken for the car travel the full distance at maximum velocity. But it doesn't travel the full distance at maximum velocity?

If the question meant something else then it was worded horribly.
The significance is in that it means the force acts against the direction of the position/velocity respectively.

In the future, where possible please avoid the sideways photos. This request is a bit selfish but it stems from the fact that I personally hate mechanics and get put off from doing it quite easily......(http://uploads.tapatalk-cdn.com/20170601/617ba62b13d717431bb7d9ee56ed6ef7.jpg)(http://uploads.tapatalk-cdn.com/20170601/c3aacaf262b4c38ae2f2384e11f2b964.jpg)(http://uploads.tapatalk-cdn.com/20170601/f7146d1d4807dd50b9d62bf8838cfae3.jpg)

thankyou soo much rui, i worked out the first question if you want me to post it? thankyou soo much.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 02, 2017, 01:17:48 pm
thankyou soo much rui, i worked out the first question if you want me to post it? thankyou soo much.

Up to you. Good idea though for the benefit of the others
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 02, 2017, 11:28:17 pm
Hey, can someone please explain these perms and combs questions, I'm at a loss

1) Four sets of twins go to a party. In how many ways can these children be arranged in pairs so that no child is with his twin?

2) How many ways can three integers be chosen which are in arithmetic progression from the numbers 1 to 101?
This Q1 took me ages to do and I had to call in my second year math friends to help me out. It dumbfounded me big time until I realised that I had mucked up the very last piece of my analysis...


__________________________________




___________________________________




__________________________________________________________________________________________________





___________________________________





___________________________________

Title: Re: 4U Maths Question Thread
Post by: kiwiberry on June 03, 2017, 11:37:04 am
This Q1 took me ages to do and I had to call in my second year math friends to help me out. It dumbfounded me big time until I realised that I had mucked up the very last piece of my analysis...

Oh wow. Thanks Rui, that question was much more complicated than I anticipated :O
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 03, 2017, 03:22:03 pm
That being said, the post was excessively long because Ext 2 students aren't expected to just know the inclusion-exclusion principle. I had to define it first before I could use it
Title: Re: 4U Maths Question Thread
Post by: Wales on June 04, 2017, 02:02:42 pm
Having a bit of trouble with conics. Part 2. Would be appreciated if you could guide me in it right direction.

Cheers, Wales
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 04, 2017, 03:07:29 pm
Having a bit of trouble with conics. Part 2. Would be appreciated if you could guide me in it right direction.

Cheers, Wales
ST should be easy to find because T is on the x-axis so the distance is horizontal.

Hint: Did you consider the statement PS = ePM (or equivalently PS/PM = e, the focus-directrix definition)?
Title: 4U Maths Question Thread
Post by: beau77bro on June 05, 2017, 08:25:37 pm
Absolute monster question: or atleast idk how to do it.

Help greatly appreciated(http://uploads.tapatalk-cdn.com/20170605/5ee6c4d840af4268c551a4a3e6adea96.jpg)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 05, 2017, 08:28:14 pm
Absolute monster question: or atleast idk how to do it.

Help greatly appreciated(http://uploads.tapatalk-cdn.com/20170605/5ee6c4d840af4268c551a4a3e6adea96.jpg)
Clarity needed. Do you mean the circled one? The arrowed one was never crossed out
Title: Re: 4U Maths Question Thread
Post by: beau77bro on June 05, 2017, 08:36:37 pm
Clarity needed. Do you mean the circled one? The arrowed one was never crossed out

oh yea sorry rui, the last one
sorry
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 05, 2017, 08:38:00 pm
Surely you just sub in V=25.7 and v=10?
Title: Re: 4U Maths Question Thread
Post by: beau77bro on June 05, 2017, 09:20:59 pm
Surely you just sub in V=25.7 and v=10?

but then isnt there a 0 in the denominator?....oh my god, u sub in 10.5. sorry rui i tricked myself into thinking this was hard my bad. THANKYOU
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 05, 2017, 09:23:13 pm
Whoops. Lol the 105% went over my head after I figured out what the terminal velocity actually was
Title: 4U Maths Question Thread
Post by: beau77bro on June 06, 2017, 09:33:34 am
Mod Edit: Posts merged, use the Modify function to add to your previous post instead of chainposting.
Yea same ahhaha
Thanks tho Rui
Hey I'm getting half of the answer for 22 a. and 4-3pi instead of 10-3pi for 22b idk what I'm doing wrong(http://uploads.tapatalk-cdn.com/20170605/3103a7177953b853fd3eaeac15843ec8.jpg)


Nevermind I got it
Title: Re: 4U Maths Question Thread
Post by: theblackswan on June 09, 2017, 08:27:02 pm
Hey! I was just wondering if you have tips on how to solve 4U maths inequalities - I've started with them and needless to say, I don't feel too enthusiastic about them - like where do you even start and how do you show your working? Thanks a bunch! :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 09, 2017, 08:39:41 pm
Hey! I was just wondering if you have tips on how to solve 4U maths inequalities - I've started with them and needless to say, I don't feel too enthusiastic about them - like where do you even start and how do you show your working? Thanks a bunch! :)
You should identify what technique you use to prove those inequalities, be it calculus, something elementary, induction or simple algebra etc. (Not sure what else you might need off the top of your head.)

For simple algebraic ones, there's usually some tricks to get you started. You may choose to work backwards on a spare sheet of paper and then copy down the proof the correct way in your exam booklet. You may consider moving everything to one side and proving LHS - RHS < 0 or > 0 instead. You should also be aware of the fact that squares can never be negative.

I'm not too sure with what you mean by "how you show your working" though - you just show it. Expand on that query if you need more assistance please.
Title: Re: 4U Maths Question Thread
Post by: Jyrgal on June 09, 2017, 08:40:52 pm
hello! for integration purposes, which way is preferable: an example makes this much easier
integrate (cosx/sin^2x)
U=sinx
du=dxcosx then sub in

OR
can i just immediately substitute dx with dsinx and get rid of cosx?
(change of variable) i find this method much easier to use especialyl for by parts cause u dont have to sub in u&v twice

Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 09, 2017, 08:44:55 pm


In regards to the question itself though - either is fine
Title: Re: 4U Maths Question Thread
Post by: chelseam on June 11, 2017, 12:22:35 am
Hi! Could someone please explain how to prove that x^2<xy<y^2 if 0<x<y? I've tried using the fact that (x-y)^2≥0 and I've multiplied the given inequality by y, but now I'm stuck :P Thank you! :D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 11, 2017, 12:24:51 am
Hi! Could someone please explain how to prove that x^2<xy<y^2 if 0<x<y? I've tried using the fact that (x-y)^2≥0 and I've multiplied the given inequality by y, but now I'm stuck :P Thank you! :D

Similarly for y
Title: Re: 4U Maths Question Thread
Post by: chelseam on June 11, 2017, 12:26:52 am

Similarly for y
Oh dear I can't believe I didn't think of that... Thanks so much Rui! ;D
Title: Re: 4U Maths Question Thread
Post by: chelseam on June 11, 2017, 04:44:59 pm
Hi, could I please get some help with 7b and 8? Would 7b relate to or use 7a's result? And I have no idea where to even start for 8! Thank you :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 11, 2017, 06:47:25 pm
Hi, could I please get some help with 7b and 8? Would 7b relate to or use 7a's result? And I have no idea where to even start for 8! Thank you :)


Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 12, 2017, 09:43:43 am
Hi, could I please get some help with 7b and 8? Would 7b relate to or use 7a's result? And I have no idea where to even start for 8! Thank you :)



I leave the formal structure of the proof as your exercise
Title: Re: 4U Maths Question Thread
Post by: ProfLayton2000 on June 12, 2017, 11:48:52 am
Help I don't even know where to begin with this one (or is this level of difficulty outside what the HSC expects of us?)
Title: Re: 4U Maths Question Thread
Post by: chelseam on June 12, 2017, 12:23:44 pm
I leave the formal structure of the proof as your exercise
Thank you so much for your help Rui! ;D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 12, 2017, 12:33:58 pm
Help I don't even know where to begin with this one (or is this level of difficulty outside what the HSC expects of us?)
The question looks wrong to me.
Title: Re: 4U Maths Question Thread
Post by: ProfLayton2000 on June 12, 2017, 01:30:42 pm
The question looks wrong to me.

Phew tysm for that!
Title: Re: 4U Maths Question Thread
Post by: johnk21 on June 13, 2017, 06:19:51 pm
Could someone please help me with this question.
Thanks in advance :)
Title: Re: 4U Maths Question Thread
Post by: jakesilove on June 13, 2017, 06:48:28 pm
Could someone please help me with this question.
Thanks in advance :)

Hey! We start with

Now, we make the substitution


noting that c is just a constant, so disappears when we differentiate. Now, we change the limits



Therefore,



However, we can change the arbitrary variable u to the arbitrary variable x



As required
Title: Re: 4U Maths Question Thread
Post by: chelseam on June 15, 2017, 10:39:11 pm
Hi! Could someone please explain the process of making a trig substitution (the kind that looks like the one in this question). Thank you so much! :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 15, 2017, 10:48:43 pm
Hi! Could someone please explain the process of making a trig substitution (the kind that looks like the one in this question). Thank you so much! :)
Most of it is like an ordinary substitution where you let x = f(u). This one isn't exactly any different so I'm not sure what you want me to explain

Title: Re: 4U Maths Question Thread
Post by: chelseam on June 15, 2017, 10:54:07 pm
Most of it is like an ordinary substitution where you let x = f(u). This one isn't exactly any different so I'm not sure what you want me to explain
I was confused about when and how to change the domain! Thanks Rui :D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 15, 2017, 11:01:06 pm
Ah. Usually when you have x = f(u) instead of u = f(x), you try to put a restriction on the domain of u (or in this case, theta) so that over that domain the function is invertible. (i.e. the function is "locally" invertible.) That gets out out of troubles, e.g. if you had to substitute x = u2 and your original boundaries were 4 and 9, you no longer have to worry about a ± and just use 2 and 3.
Title: Re: 4U Maths Question Thread
Post by: chelseam on June 15, 2017, 11:04:38 pm
Ah. Usually when you have x = f(u) instead of u = f(x), you try to put a restriction on the domain of u (or in this case, theta) so that over that domain the function is invertible. (i.e. the function is "locally" invertible.) That gets out out of troubles, e.g. if you had to substitute x = u2 and your original boundaries were 4 and 9, you no longer have to worry about a ± and just use 2 and 3.
Thanks again Rui! You've been so helpful ;D
Title: 4U Maths Question Thread
Post by: beau77bro on June 15, 2017, 11:23:32 pm
HEY GUYS I WAS WONDERING WHAT YOU THOUGHT (ESP MODERATORS) WHAT WAS THE BEST WAY TO PREPARE FOR TRIALS (OTHER THAN THAT BRILLIANT NEW TOPIC TEST FOR EXT 2 WHICH IVE ORDERED). So what is it? Textbook Qs all day or trial questions all the way? I personally did a tonne of trials before half yearlies and just didn't find them challenging in a good way... It's just a tonne of easy-ish questions with some half hard ones or ridiculously hard - I felt like I didn't cover much of the course either. And half of my study I spent quite uninterested/unmotivated (I realise trials will change that ahahahha and I need to revise- so they are a really good option). So yea what do you legends think. What's the way to go? Smash out textbooks for each topic or focus on trial papers? Or maybe something completely different hahaha just take notes?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 15, 2017, 11:43:22 pm
HEY GUYS I WAS WONDERING WHAT YOU THOUGHT (ESP MODERATORS) WHAT WAS THE BEST WAY TO PREPARE FOR TRIALS (OTHER THAN THAT BRILLIANT NEW TOPIC TEST FOR EXT 2 WHICH IVE ORDERED). So what is it? Textbook Qs all day or trial questions all the way? I personally did a tonne of trials before half yearlies and just didn't find them challenging in a good way... It's just a tonne of easy-ish questions with some half hard ones or ridiculously hard - I felt like I didn't cover much of the course either. And half of my study I spent quite uninterested/unmotivated (I realise trials will change that ahahahha and I need to revise- so they are a really good option). So yea what do you legends think. What's the way to go? Smash out textbooks for each topic or focus on trial papers? Or maybe something completely different hahaha just take notes?
By the time it was trials for 4U I honestly gave up textbooks. Only past papers can emulate what you're going to get on the day.

Yes, doing questions from here on is the way to go. But textbooks really don't reflect what you're going to get these days. The closest to the exam style questions would be the Syd Grammar questions, which I gather that you have but presumably you have covered quite a load of them already. You can probably still do some more in the meantime if your trials are far away, especially since motivation levels do spike up in preparation for trials, but once you get to that period where you have to study for trials it won't matter too much.

You might not need to worry too much about easy questions for now, but you're going to have to practice them seriously. Because if you don't practice the easy ones, you'll find you make an embarrassing amount of silly mistakes. On the contrary, seeing as though you look as though you have all the easy content covered, (provided you can juggle your other subjects) it's time to aim high. Those really hard questions aren't meant to be things you just look at and say "oh yep duh", even though every now and then you might. Unlike scraping, training yourself enough to score well into the 90s is going to require those questions be knocked down. You should be able to develop enough skills to be able to at least give it a shot, even if you don't get the final answer out. Remember, questions won't be worth only 1 mark unless they're trivial at the end of the paper.

Now, I need to make a comment on this:
I felt like I didn't cover much of the course either.
If you do a LOT of past papers, yes you will. No guarantee that you're going to cover every cornerstone of the course, but you're going to have covered quite a fair load.

Key word is a LOT though. Extension 2 is huge; it's not as restricted as Extension 1 in what they can ask. You're gonna have to go all over the place with them papers.
Title: Re: 4U Maths Question Thread
Post by: beau77bro on June 16, 2017, 12:02:52 am
Ok sweet, I see where your coming from and I agree completely - where from do u recommend and in what quantity - I can do a solid trial in under 2 hours usually. That's without accounting for tricky questions. Less if I skip the easier questions-however I do think I should start focusing on doing them properly Coz yes, Dumb mistakes are annoying(that's actually what my gf called me first day of school hahaha). I have 7 weeks till trials, what is a reasonable amount ( I will be in Europe for two of them and I will go ham when I get back)? Thanks so much for the advice Rui hahaha I'm genuinely taking notes on what uve said Ahahha
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 16, 2017, 09:32:23 am
Well honestly, look, it's not easy to simply 'recommend' an amount because everyone needs to put in amount that works for them. In my opinion, whatever constitutes a 'good' amount is an amount that allows you to get to the level you want to be, without impairing your performance in other subjects. Of course if you want to be like me and put in a tad too much effort into 4U that's possible, but you risk reducing your marks in the other subjects even more so think about if that's worth it.

There's nothing wrong with skipping some easy questions every now and then especially if they're becoming a bore and time-wasting. Just don't neglect them completely because yeah, silly mistakes.

(Them relationship struggles though...)

Unfortunately it's been too long and I don't remember exactly how I planned out my trial study plan, or I probably would've just told you about it
Title: Re: 4U Maths Question Thread
Post by: Wales on June 16, 2017, 01:22:51 pm
Well honestly, look, it's not easy to simply 'recommend' an amount because everyone needs to put in amount that works for them. In my opinion, whatever constitutes a 'good' amount is an amount that allows you to get to the level you want to be, without impairing your performance in other subjects. Of course if you want to be like me and put in a tad too much effort into 4U that's possible, but you risk reducing your marks in the other subjects even more so think about if that's worth it.

There's nothing wrong with skipping some easy questions every now and then especially if they're becoming a bore and time-wasting. Just don't neglect them completely because yeah, silly mistakes.

(Them relationship struggles though...)

Unfortunately it's been too long and I don't remember exactly how I planned out my trial study plan, or I probably would've just told you about it

I'm planning on picking up the Atarnotes 4u books and giving them a go. I'm doing that in combination with set sheets from my tutor. I just don't feel confident in doing the work.

I ALWAYS second guess my work, I'm always afraid to post solutions to 2U questions (even though I've sat the course) in the fear I'll get something wrong and I feel that it's affecting me here as well.

Do you have any advice for this? I've never been particularly talented in Maths but don't find myself struggling with the maths but more interpreting the question and being confident in approaching it.

Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 16, 2017, 01:40:44 pm
I'm planning on picking up the Atarnotes 4u books and giving them a go. I'm doing that in combination with set sheets from my tutor. I just don't feel confident in doing the work.

I ALWAYS second guess my work, I'm always afraid to post solutions to 2U questions (even though I've sat the course) in the fear I'll get something wrong and I feel that it's affecting me here as well.

Do you have any advice for this? I've never been particularly talented in Maths but don't find myself struggling with the maths but more interpreting the question and being confident in approaching it.
At uni, I'm always second guessing my work. I would write something down and doubting it because there was probably a silly mistake in there somewhere.

At the start, just roll with it. Literally, don't even bother checking it all and just move on. Do that until you've done every question you can easily do and all that's rest is the toughies that gets you a band E4.

THAT is when you go check. Focus on actually getting the question done, because if all your mistakes are arithmetic errors but your method is right, there's only so much they can deduct.

Of course, if you arrive at a contradiction you should know to backtrack. But if you reach an answer that makes sense and might only be off by a bit, just move on.

It's hard to be perfect at maths and just not make mistakes. Those that don't do that are who get state ranks, and that's only about what, 10 out of 3500 people? You're most going to need to let some mistakes happen.

The important thing is to get into the habit of moving on and accepting it for as long as you can. Go back only when it's the right time to in an exam.

With posting solutions on here, look being honest I make mistakes too. I try my best to make sure I do nothing wrong but more often than expected it just is plain wrong. And I can understand why some people are probably too shy to post solutions in front of me as well - I do have a habit of critiquing where the faults are as well. I get used to it after a bit though - people critique mine as well. That really just takes getting used to, I'd say.
Title: Re: 4U Maths Question Thread
Post by: beau77bro on June 16, 2017, 05:57:16 pm

I'm planning on picking up the Atarnotes 4u books and giving them a go. I'm doing that in combination with set sheets from my tutor. I just don't feel confident in doing the work.

I ALWAYS second guess my work, I'm always afraid to post solutions to 2U questions (even though I've sat the course) in the fear I'll get something wrong and I feel that it's affecting me here as well.

Do you have any advice for this? I've never been particularly talented in Maths but don't find myself struggling with the maths but more interpreting the question and being confident in approaching it.

Honestly man I have had the exact same problem. I'm sure tonnes of people have, especially this other guy in my class. And honestly, and this is what's so annoying about maths at least in my opinion, it's always easier then it looks. Like 95% of the time you know how to do it or have the ability. I found and so did this guy in my class that the hardest part is starting. The questions won't ask you something you can't do or that they haven't given you the information to do - it just becomes how you apply that, and you really can't see that or the sneaky tricks until your half way in (unless ur a god like Rui most likely). Idk if this is useful advice, but I'd say with the questions your not confident on, it's important just to start and see if you get somewhere because hopefully the tricks will become apparent once you get into it. AND HOPEFULLY YOU GET THERE. But if u don't that's fine too, and there's this forum and lots of worked solutions as well, so hopefully you can always work it out - and pick up the tricks they used. I really don't think there are many people that have gotten even 70% of the questions they've done throughout their 4u career.

My teacher always says this thing about Einstein. He said that Einstein didn't think he was that smart, it was just that he never gave up on a question - and I think that's really important here, at least to some extent because we have other things to do and it can be quite hard.

Sorry for the super long post, but hope it helps.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 16, 2017, 06:10:32 pm
With 4U, I too to this day still only see it halfway in every now and then.

Have faith though. 70% scrapes E4s occasionally; 72% quite often
Title: 4U Maths Question Thread
Post by: beau77bro on June 16, 2017, 06:13:45 pm
I just mean like with every single question you've ever done, especially at first try, not many people would've gotten 70-100% of those right. It's more than possible to get crazy high marks in the HSC, but of course mistakes as well just getting things plain wrong is half of the battle on the way there
Title: Re: 4U Maths Question Thread
Post by: Wales on June 17, 2017, 10:55:19 pm
At uni, I'm always second guessing my work. I would write something down and doubting it because there was probably a silly mistake in there somewhere.

At the start, just roll with it. Literally, don't even bother checking it all and just move on. Do that until you've done every question you can easily do and all that's rest is the toughies that gets you a band E4.

THAT is when you go check. Focus on actually getting the question done, because if all your mistakes are arithmetic errors but your method is right, there's only so much they can deduct.

Of course, if you arrive at a contradiction you should know to backtrack. But if you reach an answer that makes sense and might only be off by a bit, just move on.

It's hard to be perfect at maths and just not make mistakes. Those that don't do that are who get state ranks, and that's only about what, 10 out of 3500 people? You're most going to need to let some mistakes happen.

The important thing is to get into the habit of moving on and accepting it for as long as you can. Go back only when it's the right time to in an exam.

With posting solutions on here, look being honest I make mistakes too. I try my best to make sure I do nothing wrong but more often than expected it just is plain wrong. And I can understand why some people are probably too shy to post solutions in front of me as well - I do have a habit of critiquing where the faults are as well. I get used to it after a bit though - people critique mine as well. That really just takes getting used to, I'd say.

I've tried to take that advice on board, it's going to be difficult changing my mindset but I guess it's the best way. I constantly get surrounded by people who are just incredibly talented at Maths and it's hard not to feel inadequate. I've got friends complaining about 97%'s in their tests while I'm hoping for 75+. It's a bit ironic coming from a school where 5 people are doing 4u but even so, seeing my tutor effortlessly solve the questions at a glance (2016 HSC grad with 94 in 4U) and being my age doesn't help my confidence.

I feel like I need to stop lying to myself and believe that I'm somewhat capable in Maths relative to my peers. I just need that boost in confidence :\

Thanks to you both for the words, it has helped greatly.
Title: Re: 4U Maths Question Thread
Post by: beau77bro on June 18, 2017, 01:07:46 pm
(http://uploads.tapatalk-cdn.com/20170618/ea60f5392396eaace41e327deae8a61d.jpg)

Help pls what even is this??
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 18, 2017, 01:35:37 pm
(http://uploads.tapatalk-cdn.com/20170618/ea60f5392396eaace41e327deae8a61d.jpg)

Help pls what even is this??

I don't get what they have. I get -1/(n+1) not +1/(n+2)
Title: Re: 4U Maths Question Thread
Post by: beau77bro on June 18, 2017, 04:57:19 pm
Thanks Rui. I really don't think I can help u there AHHAHA I'll get back to u if I work anything out or if my teacher gets it. HOW DO U JUST KNOW HOW TO DO IT hahaha
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 18, 2017, 04:59:55 pm
Thanks Rui. I really don't think I can help u there AHHAHA I'll get back to u if I work anything out or if my teacher gets it. HOW DO U JUST KNOW HOW TO DO IT hahaha
Tbh it's pretty typical in 3U binomial theorem. Have you covered the whole topic?
Title: Re: 4U Maths Question Thread
Post by: beau77bro on June 18, 2017, 06:10:26 pm
Haha no not really. Not properly - could u post ur working, I think I have an idea of how u did it but not sure. Thanks rui
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 18, 2017, 06:34:49 pm



Note: Faster approach: Perform a definite integral from -1 to 0 instead.
Title: Re: 4U Maths Question Thread
Post by: beau77bro on June 18, 2017, 06:50:44 pm
oh i know where the mistake is from (the negative) you jsut times by negative across both and it fixes it.


I THINK HAHAH <-- coz i could be wrong hahah
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 18, 2017, 06:55:35 pm
Oops. I forgot how to do mental arithmetic. A positive answer also seems more legit :P

But yeah, in that case there's still a typo in that n+2 should be n+1
Title: Re: 4U Maths Question Thread
Post by: beau77bro on June 18, 2017, 07:04:11 pm
ok so does that fix it? omg yes my mate stresses so much about not being able to do his tutoring, im sure he appreciates this lots thanks rui!!!
Title: Re: 4U Maths Question Thread
Post by: beau77bro on June 20, 2017, 10:16:14 pm
(http://uploads.tapatalk-cdn.com/20170620/de50412b3a58aed1191806b585e1b9b4.jpg)

I seriously can't get question 1a. I think they did something weird I can't see or work out. Thankyou.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 21, 2017, 10:06:06 am
(http://uploads.tapatalk-cdn.com/20170620/de50412b3a58aed1191806b585e1b9b4.jpg)

I seriously can't get question 1a. I think they did something weird I can't see or work out. Thankyou.
If there's no velocity surely that means there's no centripetal force, and thus the only two forces in play are the normal reaction and the gravity?

Title: Re: 4U Maths Question Thread
Post by: beau77bro on June 21, 2017, 10:52:22 am
But there is still a frictional force or something, so u can't use components
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 21, 2017, 10:56:47 am
But there is still a frictional force or something, so u can't use components
Where's the friction if you aren't moving?
Title: Re: 4U Maths Question Thread
Post by: xboxer on June 21, 2017, 11:04:03 am
True
Title: Re: 4U Maths Question Thread
Post by: beau77bro on June 21, 2017, 11:16:10 am
Where's the friction if you aren't moving?

the reaction force will push it away and down the slope - the friction up the slope will mean it stays put, so you cant just disregard friction (i think - i couldnt get the answer doing that)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 21, 2017, 12:19:29 pm
Ok. I think this is the idea

Vertical resolution - Do it as normal
Horizontal resolution - The friction and normal reaction cancel out to 0 instead of resolve to give \(mr\omega^2\)

I'm about to go study for my exam coming up so I won't be able to prepare a solution just yet. Would appreciate if you could post up your working out if you're stuck though.
Title: Re: 4U Maths Question Thread
Post by: Jyrgal on June 21, 2017, 08:51:13 pm
heyo!

i was just wondering if there is any smart tips or tricks to do graph transformations for cos^-1(f(x)), sin^-1(f(x)) and tan^-1(f(x))?
so far, I've had to get domain, range, intercepts, then limits for x->0 & infinity, and then finally draw the general transformation, plugging into calculator to make sure each value is correct. this takes an absurd amount of time and i just wanna know some good tricks that i can apply to this transformation thatll make my life easier, like for reciprocal graphs when f(x) ->0+, 1/f(x) -> infinity+

Thanks  :D :D

Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 21, 2017, 08:59:51 pm
heyo!

i was just wondering if there is any smart tips or tricks to do graph transformations for cos^-1(f(x)), sin^-1(f(x)) and tan^-1(f(x))?
so far, I've had to get domain, range, intercepts, then limits for x->0 & infinity, and then finally draw the general transformation, plugging into calculator to make sure each value is correct. this takes an absurd amount of time and i just wanna know some good tricks that i can apply to this transformation thatll make my life easier, like for reciprocal graphs when f(x) ->0+, 1/f(x) -> infinity+

Thanks  :D :D


Whilst that list of things should always be looked out for, you're usually not going to need all of them for every single curve. It will depend on the exact example you're given which ones are of relevance.
Title: Re: 4U Maths Question Thread
Post by: dux99.95 on June 22, 2017, 11:33:50 pm
Where do i check for notifs? Is there a button that takes me directly to the exact post bc usually I go on the new replies and then scroll through the last few pages of the recent posts to find my post and see for any replies

2nd q
Am I able to copy and paste pictures here (it doesnt work when i copy from google docs)

Main q
Shouldn’t limits be -a and a instead of a and 0 when you do a volumes by cylindrical shells method question? (not sure if i've explained it right - i'd post a picture but not sure how to work this)


thanks so much :) your help is appreciated a lot!! 




Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 22, 2017, 11:53:49 pm
Main q
Shouldn’t limits be -a and a instead of a and 0 when you do a volumes by cylindrical shells method question? (not sure if i've explained it right - i'd post a picture but not sure how to work this)


thanks so much :) your help is appreciated a lot!! 
This question is arbitrary and only makes sense if we assume that the rotation is about the y-axis (the line x=0, as opposed to say x=1). Assuming this:

The method of cylindrical shells is designed based off the formula for an arbitrary cylinder and the area of just a cross section formed by cutting down the radius of the cylinder, not the entire diameter. We choose to rotate about an angle of 360 degrees to map out our cylinder.

If instead we chose to slice down the whole diameter, we would only have to rotate about an angle of 180 degrees, not 360 degrees. The formula is then no longer \(V=\int_0^a 2\pi r h\, dr\), but \(V=\int_{-a}^a \pi r h\, dr\). (There is a subtle link to even functions here.)

The reason this ends up being π instead of 2π is because intuitively speaking, rotation by 180 degrees only maps out half of the cylinder and not the entire thing.

In summary - this only happens because of how the method was designed to begin with

Questions related to how to use the forum should be posted in their respective sections. Before doing so, you should also read the article on how to use the forums before posting a question about it.
Title: Re: 4U Maths Question Thread
Post by: theblackswan on June 25, 2017, 11:10:00 am
Hi,
In regards to the arg (z) bit of complex numbers, how do you know if the locus of say, arg (z-2) - arg (z+2) = pi/2 is the semicircle above or below the x - axis? And for solving a question say, Re [(z-i)/(z+1)] how can you find arg (z) geometrically instead of algebraically? Thanks!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 25, 2017, 11:11:57 am
Hi,
In regards to the arg (z) bit of complex numbers, how do you know if the locus of say, arg (z-2) - arg (z+2) = pi/2 is the semicircle above or below the x - axis? And for solving a question say, Re [(z-i)/(z+1)] how can you find arg (z) geometrically instead of algebraically? Thanks!

Your second question is ambiguous. That's just an expression for the real part and not an equation, so nothing can be inferred about the argument.
Title: Re: 4U Maths Question Thread
Post by: itssona on June 25, 2017, 03:43:31 pm
Heey how would I graph
y^2=x^4 (4 + x)

Thank you :)
Title: 4U Maths Question Thread
Post by: RuiAce on June 25, 2017, 03:57:46 pm
If you're looking for what a graph would look like, you can do it online  :D I recommend Desmos Graphing Calculator

It would look like this...
(http://i.imgur.com/GUOOj1t.jpg)
Which really is not obvious at all.

I'll explain where it comes from later; was having a nap. Desmos only shows what it looks like, not where to actully get it which is important for 4U.
Title: Re: 4U Maths Question Thread
Post by: jakesilove on June 25, 2017, 04:08:56 pm
If you're looking for what a graph would look like, you can do it online  :D I recommend Desmos Graphing Calculator

It would look like this...
(http://i.imgur.com/GUOOj1t.jpg)

But thanks for the post! Always a good idea to have a graphing calculator on hand :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 25, 2017, 04:16:17 pm
Heey how would I graph
y^2=x^4 (4 + x)

Thank you :)


By now you should be familiar with techniques on drawing square root curves, e.g. Block out what's below the x-axis
Title: Re: 4U Maths Question Thread
Post by: itssona on June 25, 2017, 04:47:06 pm


By now you should be familiar with techniques on drawing square root curves, e.g. Block out what's below the x-axis

thank you Rui!! :D
Title: Re: 4U Maths Question Thread
Post by: beau77bro on June 25, 2017, 05:43:24 pm
(http://uploads.tapatalk-cdn.com/20170625/2fd1362ad5dfd16f62dba5b476377ca3.jpg)

Hey guys q2, the answer is:
 (g x cot(theta))/4(n x Pi)^2

But I'm getting tan? Just wondering if you guys could check it - can't see where I'm getting it wrong
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 25, 2017, 05:59:32 pm

(http://uploads.tapatalk-cdn.com/20170625/2fd1362ad5dfd16f62dba5b476377ca3.jpg)

Hey guys q2, the answer is:
 (g x cot(theta))/4(n x Pi)^2

But I'm getting tan? Just wondering if you guys could check it - can't see where I'm getting it wrong
Check that you didn't misplace your theta
(http://uploads.tapatalk-cdn.com/20170625/65702d4b984aa4a925d6104594b6e0ce.jpg)
Title: Re: 4U Maths Question Thread
Post by: itssona on June 25, 2017, 07:06:09 pm
Heey alsoo

Find the volume of rotation when the region bounded by the x and y axes, x=2 and the curve y= 1/(x^2 - 4x +13) is rotated by the y axis

I attempted it and got an answer of 4pi/3 arctan 2/3 + piln (9/13) but tbh got no idea

Thank you :)
Title: Re: 4U Maths Question Thread
Post by: bluecookie on June 25, 2017, 07:11:24 pm
For the circular motion around a circle of x^2+y^2=r^2, with x=rcostheta, y=rsintheta, x*=r(-sintheta)dtheta/dt. y*=rcosthetadtheta/dt. I don't get how the last two equations were derived.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 25, 2017, 07:11:33 pm
Heey alsoo

Find the volume of rotation when the region bounded by the x and y axes, x=2 and the curve y= 1/(x^2 - 4x +13) is rotated by the y axis

I attempted it and got an answer of 4pi/3 arctan 2/3 + piln (9/13) but tbh got no idea

Thank you :)

This is what WolframAlpha tells me.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 25, 2017, 07:12:50 pm
For the circular motion around a circle of x^2+y^2=r^2, with x=rcostheta, y=rsintheta, x*=r(-sintheta)dtheta/dt. y*=rcosthetadtheta/dt. I don't get how the last two equations were derived.

Title: Re: 4U Maths Question Thread
Post by: bluecookie on June 25, 2017, 07:15:55 pm


I don't get the second last step to the last step, why does the cos change to a -sintheta and why does the theta appear at the top of the derivative?
Title: Re: 4U Maths Question Thread
Post by: kiwiberry on June 25, 2017, 07:21:58 pm
I don't get the second last step to the last step, why does the cos change to a -sintheta and why does the theta appear at the top of the derivative?

Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 25, 2017, 07:23:24 pm
I don't get the second last step to the last step, why does the cos change to a -sintheta and why does the theta appear at the top of the derivative?
Like I said, implicit differentiation. This is a very standard 4U technique that you're expected to know.

Thanks for stepping it out kiwiberry
Title: Re: 4U Maths Question Thread
Post by: bluecookie on June 25, 2017, 07:29:44 pm
Thanks! Can I've some help with this question?

A particle P is attached by a light string of length l to a fixed point A, and P describes a horizontal circle with uniform angular velocity w around the vertical AC. Prove that if theta is the inclination of the string to the vertical, then h does not depend on l, and find a expression for h in terms of w.

Find the tension in the string when the particle P is rotating at 1 rev/s, the mass of P is 1kg and the length of the string is 35cm. Take pi^2=10.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 25, 2017, 07:31:46 pm
Thanks! Can I've some help with this question?

A particle P is attached by a light string of length l to a fixed point A, and P describes a horizontal circle with uniform angular velocity w around the vertical AC. Prove that if theta is the inclination of the string to the vertical, then h does not depend on l, and find a expression for h in terms of w.

Find the tension in the string when the particle P is rotating at 1 rev/s, the mass of P is 1kg and the length of the string is 35cm. Take pi^2=10.
What's h in this question? Is there a diagram that goes with the question that should be provided?
Title: Re: 4U Maths Question Thread
Post by: bluecookie on June 25, 2017, 07:48:30 pm
Yep, sorry!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 25, 2017, 08:00:58 pm
(http://uploads.tapatalk-cdn.com/20170625/07be8a998855e30bc8395646a966ee65.jpg)
Title: Re: 4U Maths Question Thread
Post by: itssona on June 25, 2017, 08:15:46 pm
Ooh also please explain this thank youuu :
Solve z^2 + (1+3i)z -8 -i =0
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 25, 2017, 08:22:56 pm
Ooh also please explain this thank youuu :
Solve z^2 + (1+3i)z -8 -i =0






Edit: Typo, forgot the square root
Title: Re: 4U Maths Question Thread
Post by: itssona on June 25, 2017, 09:06:37 pm




ahh i get it now, thank u :)
Title: Re: 4U Maths Question Thread
Post by: ProfLayton2000 on June 29, 2017, 03:08:32 pm
Hey for complex numbers how do we indicate the locus if it one of the axes? (Eg locus of Im(z) = |z|)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 29, 2017, 03:15:31 pm
Hey for complex numbers how do we indicate the locus if it one of the axes? (Eg locus of Im(z) = |z|)
You can just call it the y-axis (or imaginary axis) surely
Title: Re: 4U Maths Question Thread
Post by: chloe9756 on June 30, 2017, 03:41:26 pm

The combined air and road resistance of a car in motion is proportional to v2, where v is its speed. When the engine is disengaged the car moves down an incline making an angle arcsin(1/30) with the horizontal, with a velocity of 30m/s. Find the force required to drive the car up the incline with a steady speed of 24m/s, given that the mass of the car is 1200kg.

Answer = 656N
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 30, 2017, 04:36:29 pm

The combined air and road resistance of a car in motion is proportional to v2, where v is its speed. When the engine is disengaged the car moves down an incline making an angle arcsin(1/30) with the horizontal, with a velocity of 30m/s. Find the force required to drive the car up the incline with a steady speed of 24m/s, given that the mass of the car is 1200kg.

Answer = 656N
(http://uploads.tapatalk-cdn.com/20170630/afe8b8a0f064af9ef0b6367c09b2e695.jpg)An assumption had to be made that g=10 to make this work
Title: Re: 4U Maths Question Thread
Post by: beau77bro on July 03, 2017, 03:37:13 am
(http://uploads.tapatalk-cdn.com/20170702/5cc9639d16214eed5a3f4625ef72e8f6.jpg)

Ok so the answer is b. But it doesn't consider two yellows being selected - or somethings idk. I got 10/41 so yea I'm way off - help much appreciated
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 03, 2017, 08:53:02 am
(http://uploads.tapatalk-cdn.com/20170702/5cc9639d16214eed5a3f4625ef72e8f6.jpg)

Ok so the answer is b. But it doesn't consider two yellows being selected - or somethings idk. I got 10/41 so yea I'm way off - help much appreciated
Of course we can't have two yellows because we know that one jellybean has to be black




Title: Re: 4U Maths Question Thread
Post by: limtou on July 04, 2017, 01:57:20 pm
Hello! Could you please explain the solutions to these questions?
all of Q8 and part ii) for c)
Thanks :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 04, 2017, 04:09:30 pm
Hello! Could you please explain the solutions to these questions?
all of Q8 and part ii) for c)
Thanks :)
(Briefly recall that three lines being concurrent means that they intersect at a common point)


_____________________________


For the number of favourable outcomes, it helps to draw a diagram with 6 non-parallel non-concurrent lines.


_____________________________

Again, you should draw a diagram to visualise what's going on




I am not sure where the confusion is with the inequality proof. The solutions are quite self-explanatory. Please indicate where the problem is. Also please confirm that the stationary point obtained in the previous part is at x=c/2.
Title: Re: 4U Maths Question Thread
Post by: limtou on July 04, 2017, 05:03:53 pm
(Briefly recall that three lines being concurrent means that they intersect at a common point)
...

Thanks for the super long explanation for this question!
And don't worry about the inequality proof, I was momentarily confused at , but I see it now :)
Title: Re: 4U Maths Question Thread
Post by: bluecookie on July 12, 2017, 04:34:19 pm
Sketch:

(x^2+y^2)^2=2(x^2-y^2)
and x^2+y^2=3xy
Title: Re: 4U Maths Question Thread
Post by: seventeenboi on July 12, 2017, 04:42:36 pm
Hi :)

the ansewr to this question is C - i understand that there must be a vertical tangent at x, but how would you choose between a and c (concavity ??)
thanks :)
Title: Re: 4U Maths Question Thread
Post by: Shadowxo on July 12, 2017, 05:06:59 pm
Hi :)

the ansewr to this question is C - i understand that there must be a vertical tangent at x, but how would you choose between a and c (concavity ??)
thanks :)

For a question like that, I'd probably just test each of the quadrants. Quadrant 1, x and y are both positive so dy/dx is positive. Quadrant 2, y is positive and x is negative, so dy/dx is negative etc.
Then you just have to choose between c and d, when x approaches 0 dy/dx approaches positive/negative infinity so it's c

Hope this helps :)
Title: Re: 4U Maths Question Thread
Post by: seventeenboi on July 12, 2017, 06:12:00 pm
For a question like that, I'd probably just test each of the quadrants. Quadrant 1, x and y are both positive so dy/dx is positive. Quadrant 2, y is positive and x is negative, so dy/dx is negative etc.
Then you just have to choose between c and d, when x approaches 0 dy/dx approaches positive/negative infinity so it's c

Hope this helps :)

OHHH TRUE ok thankss :DDDD
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 12, 2017, 07:04:31 pm
Sketch:

(x^2+y^2)^2=2(x^2-y^2)
and x^2+y^2=3xy
These are just random implicit curves that authors throw 4U students when they're bored. Please indicate the source of these questions.

(My honest opinion though - not even worth attempting unless you've done polar curve sketching, which is not in the HSC course. Part b) can possibly be done by using the quadratic formula on \(y\) though.)
Title: Re: 4U Maths Question Thread
Post by: bluecookie on July 12, 2017, 10:48:01 pm
These are just random implicit curves that authors throw 4U students when they're bored. Please indicate the source of these questions.

(My honest opinion though - not even worth attempting unless you've done polar curve sketching, which is not in the HSC course. Part b) can possibly be done by using the quadratic formula on \(y\) though.)
Ohhh thanks, yeah I was wondering why the teacher was setting us so unseenly and obscure questions lol xP Thanks, I feel happier knowing I won't have to cover them at all XD They came from his holiday homework. We don't usually get this much, but since it's the holidays before the last school assessment he decided to give us extra :P
Title: Re: 4U Maths Question Thread
Post by: bluecookie on July 12, 2017, 11:18:57 pm
a) Find integral from 0 to pi of [dx/(5-4cosx)]
b) Un=integral from 0 to pi of [cosnxdx/(5-4cosx)], show that U(n+1)+U(n-1)-5/2*Un=0.
c) Calculte U0, U1, and hence find U2, U3.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 13, 2017, 08:33:02 am
a) Find integral from 0 to pi of [dx/(5-4cosx)]
b) Un=integral from 0 to pi of [cosnxdx/(5-4cosx)], show that U(n+1)+U(n-1)-5/2*Un=0.
c) Calculte U0, U1, and hence find U2, U3.


________________________________



________________________________

You should have no trouble using the recurrence formula to find U2 and U3.
Title: Re: 4U Maths Question Thread
Post by: seventeenboi on July 13, 2017, 08:14:33 pm
hiii :)
how would you do this question HELPPp
thankkkk
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 13, 2017, 08:22:50 pm
hiii :)
how would you do this question HELPPp
thankkkk


Title: Re: 4U Maths Question Thread
Post by: claudiarosaliaa on July 13, 2017, 08:39:30 pm
Can anyone solve this Volumes questions? Its Questions 15.a) from the 2015 James Ruse Trial Paper:

The diagram shows a segment of the circle X2 + Y2 = r2 which is rotated about the y axis to from a collar. This collar is thus a sphere was a symmetrical hole through it. Let the hole be of the height 2h as shown.

Use the method of cylidical shells to show that the volume of the material in the collar is given by the integral

4pi(integral between r and the square root of (r2-h2) of X multiplied by the square root of (r2-X2)dx)

Evaluate the integral to show that the volume of material in the collar is a function of h only independent of r.

Sidenote: How do I enter maths symbols into my posts?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 13, 2017, 08:49:18 pm
Can anyone solve this Volumes questions? Its Questions 15.a) from the 2015 James Ruse Trial Paper:

The diagram shows a segment of the circle X2 + Y2 = r2 which is rotated about the y axis to from a collar. This collar is thus a sphere was a symmetrical hole through it. Let the hole be of the height 2h as shown.

Use the method of cylidical shells to show that the volume of the material in the collar is given by the integral

4pi(integral between r and the square root of (r2-h2) of X multiplied by the square root of (r2-X2)dx)

Evaluate the integral to show that the volume of material in the collar is a function of h only independent of r.

Sidenote: How do I enter maths symbols into my posts?

Assuming a diagram was provided.




It should be clear why I took a positive square root.


Last bit subject to a bit of computational inaccuracy, but the method is still the same.

LaTeX guide
Title: Re: 4U Maths Question Thread
Post by: claudiarosaliaa on July 13, 2017, 09:11:16 pm
Can anyone explain why (D) is the correct answer?
Title: Re: 4U Maths Question Thread
Post by: claudiarosaliaa on July 13, 2017, 09:19:35 pm
Assuming a diagram was provided.




It should be clear why I took a positive square root.


Last bit subject to a bit of computational inaccuracy, but the method is still the same.

LaTeX guide

Thank you!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 13, 2017, 09:19:48 pm
Can anyone explain why (D) is the correct answer?



As an exercise, you should use either technology or by hand, attempt to sketch out what parts of the hyperbola would be present in the other parametrisations.
Title: Re: 4U Maths Question Thread
Post by: bluecookie on July 14, 2017, 02:12:46 pm

thanks :D
Title: Re: 4U Maths Question Thread
Post by: bluecookie on July 14, 2017, 02:46:26 pm
If In=integral tan^nx dx for n>=0 show that In=[1/(n-1)]*tan^(n-1)x-I(n-2) for n>=2. What's the setting out for proving n>=2.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 14, 2017, 02:59:41 pm
If In=integral tan^nx dx for n>=0 show that In=[1/(n-1)]*tan^(n-1)x-I(n-2) for n>=2. What's the setting out for proving n>=2.
The normal setting out.

Note that for the integral of tann[/sup(x) you should not be using by parts, but only a Pythagorean identity.
Title: Re: 4U Maths Question Thread
Post by: bluecookie on July 14, 2017, 03:21:05 pm
The normal setting out.

Note that for the integral of tann[/sup(x) you should not be using by parts, but only a Pythagorean identity.

Thanks!

Show that for In=integral sec^nxdx >=0, In=1/(n-1)tanxsec^n-2x+(n-2)/(n-1)*I(n-2) for n>=2.

For these type questions, how do we know which way to start? Because I can see several ways this question might work. (take out sec squared which gives sec^n-2 which matches the I(n-2) expression a little, or change it into secx*sec^n-1x and differentiate the sec^n-1x to get the I(n-2) (I tried this but got a really long expression that I was stuck unsure of what to do with) etc) but how do I know is the right method?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 14, 2017, 11:27:50 pm
Thanks!

Show that for In=integral sec^nxdx >=0, In=1/(n-1)tanxsec^n-2x+(n-2)/(n-1)*I(n-2) for n>=2.

For these type questions, how do we know which way to start? Because I can see several ways this question might work. (take out sec squared which gives sec^n-2 which matches the I(n-2) expression a little, or change it into secx*sec^n-1x and differentiate the sec^n-1x to get the I(n-2) (I tried this but got a really long expression that I was stuck unsure of what to do with) etc) but how do I know is the right method?
These really standard ones are ones you should memorise the method of.

For that one, you should consider \(\sec^2x\) and \(\sec^{n-2}x\) because of the relevant Pythagorean identity. But you will need integration by parts - Only questions that involve \(\tan^n x\) or can be made into that form by a u-substitution can fully avoid IBP as far as I know in 4U.
Title: Re: 4U Maths Question Thread
Post by: bluecookie on July 15, 2017, 01:03:43 pm
Thanks!
Title: Re: 4U Maths Question Thread
Post by: limtou on July 15, 2017, 01:24:41 pm
Hello, please help with d) iv)
Thanks :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 15, 2017, 01:50:23 pm
Hello, please help with d) iv)
Thanks :)
I had covered a bit of this during my lecture. The first steps are to let z=x+iy to reduce the expressions given.


One possible plan for the sketch:

1. Sketch \(x^2-y^2=3\) and \(xy = 2\) separately
2. Identify the regions satisfying \(x^2-y^2<3\) and \(xy<2\)
3. Add in the restriction \( 0 < x^2-y^2\) and \(0 < xy\) using your knowledge of the 4 quadrants. You may wish to do this separately from the rough sketches you had in step 2.
4. Combine the regions required.
Title: Re: 4U Maths Question Thread
Post by: bluecookie on July 15, 2017, 03:40:34 pm
If In=integral from 0 to 1 of x(1-x^3)^n for n>=0 show that In=[3n/(3n+2)]*I(n-1) for n>=1. Hence find an expression for In in terms of n for n>=0.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 15, 2017, 03:55:36 pm
If In=integral from 0 to 1 of x(1-x^3)^n for n>=0 show that In=[3n/(3n+2)]*I(n-1) for n>=1. Hence find an expression for In in terms of n for n>=0.

Title: Re: 4U Maths Question Thread
Post by: BradMate on July 16, 2017, 06:19:47 pm
Could someone please explain how to solve this polynomial question. Thanks so much.
Title: Re: 4U Maths Question Thread
Post by: limtou on July 16, 2017, 10:30:40 pm
I had covered a bit of this during my lecture. The first steps are to let z=x+iy to reduce the expressions given.


One possible plan for the sketch:

1. Sketch \(x^2-y^2=3\) and \(xy = 2\) separately
2. Identify the regions satisfying \(x^2-y^2<3\) and \(xy<2\)
3. Add in the restriction \( 0 < x^2-y^2\) and \(0 < xy\) using your knowledge of the 4 quadrants. You may wish to do this separately from the rough sketches you had in step 2.
4. Combine the regions required.

Thank you! I realised that I wasn't able to get the answer at first because I didn't consider the quadrants, but now I got it :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 16, 2017, 11:00:13 pm
Could someone please explain how to solve this polynomial question. Thanks so much.
No LaTeX because I'm at the gym, just a description: Edit; LaTeX in place now.

As the coefficients are integers, if a surd is a root then so must it's irrational conjugate. Here, that would be \(-\sqrt3\).

Then, as the polynomial is monic it must have leading coefficient is 1. By the product of roots, if the third root is \(\alpha\) we must have
\(\sqrt{3}\times -\sqrt3\times \alpha=-12 \implies \alpha=4\)

So the polynomial is
\(P(x)=(x-\sqrt3)(x+\sqrt3)(x-4)=\left(x^2-3\right)(x-4)\)

You may expand this out if you wish.
Thank you! I realised that I wasn't able to get the answer at first because I didn't consider the quadrants, but now I got it :)
Ah yep, that was most likely the hardest bit about that question.
Title: Re: 4U Maths Question Thread
Post by: Ali_Abbas on July 16, 2017, 11:54:43 pm
Could someone please explain how to solve this polynomial question. Thanks so much.

Here is my solution.

Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 17, 2017, 12:09:59 am
Here is my solution.


Whilst it looks nice I feel as though this is way too overkill when you can just use a simple algorithm and the irrational conjugate theorem for this type of question...
Title: Re: 4U Maths Question Thread
Post by: Ali_Abbas on July 17, 2017, 12:25:39 am
Whilst it looks nice I feel as though this is way too overkill when you can just use a simple algorithm and the irrational conjugate theorem for this type of question...

Overkill is my middle name
Title: Re: 4U Maths Question Thread
Post by: pikachu975 on July 17, 2017, 12:37:03 am
Whilst it looks nice I feel as though this is way too overkill when you can just use a simple algorithm and the irrational conjugate theorem for this type of question...

Is the irrational conjugate theorem in HSC? Never heard of it
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 17, 2017, 09:15:34 am
Is the irrational conjugate theorem in HSC? Never heard of it
Yes and it works exactly like the complex conjugate root theorem, which states that if a polynomial has real coefficients but has a complex root then another root must be its complex conjugate.

Also note that the HSC introduces theorems, but more often than not chooses not to give their names (e.g. intermediate value theorem subtly being placed in the 3U estimation of roots topic)
Title: Re: 4U Maths Question Thread
Post by: BradMate on July 17, 2017, 11:26:40 am
Thanks Rui and AA. Much appreciated.
Title: Re: 4U Maths Question Thread
Post by: chelseam on July 19, 2017, 11:42:43 pm
Hi! Could someone please help me with this question? I don't understand why the answers say that dV=2y(4-x)dx. I've tried to use similar triangles but I'm not sure if that's even relevant here... Thank you :)
Title: 4U Maths Question Thread
Post by: RuiAce on July 20, 2017, 12:36:16 am
Hi! Could someone please help me with this question? I don't understand why the answers say that dV=2y(4-x)dx. I've tried to use similar triangles but I'm not sure if that's even relevant here... Thank you :)


This is actually easier to explain if they taught 3D coordinate geometry with the x-y-z plane.


Can understand if that didn't make sense but I'm too sleepy now so if something doesn't puzzle up I'll attend to it tomorrow
Title: Re: 4U Maths Question Thread
Post by: chelseam on July 20, 2017, 12:50:23 am
Can understand if that didn't make sense but I'm too sleepy now so if something doesn't puzzle up I'll attend to it tomorrow
Thanks so much Rui, I think I get it now! ;D Is the 4 coming from the dotted line in the diagram?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 20, 2017, 09:10:17 am
Thanks so much Rui, I think I get it now! ;D Is the 4 coming from the dotted line in the diagram?
Pretty much that's one way of thinking about it
Title: Re: 4U Maths Question Thread
Post by: bluecookie on July 20, 2017, 09:40:10 am
In the diagram the circles XYPS and XYRQ intersect at the point X and Y, and PXQ, PYR, QSY, PST and QTR are straight lines.
(i) Explain why ∠STQ=∠YRQ+∠YPS.
(ii) Show that ∠YRQ+∠YPS+∠SXQ=pi.
(iii) Deduce that STQX is a cyclic quadrilateral.
(iv) Let ∠QPY=a and ∠PQY=b. Show that ∠STQ=a+b.

Can I have help with iv?
Title: Re: 4U Maths Question Thread
Post by: bluecookie on July 20, 2017, 09:42:45 am
Find V if
v=integral from 0 to h, of invcos(x/h)-(x/h)root(1-(x/h)^2) dx given that integral of invcos(theta)d(theta)=thetainvcos(theta)-root(1-theta^2) and let theta=x/h.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 20, 2017, 10:09:20 am
Find V if
v=integral from 0 to h, of invcos(x/h)-(x/h)root(1-(x/h)^2) dx given that integral of invcos(theta)d(theta)=thetainvcos(theta)-root(1-theta^2) and let theta=x/h.
This was difficult to read. In the future, when +'s and -'s are present please use an extra space so that the terms don't look clumped. Same with "thetainvcos" - a space is beneficial. Also, please use sqrt for the square root, and preferably "arccos" for invcos.

Additionally, at times like this please resort to the modify function instead of double posting.


Title: Re: 4U Maths Question Thread
Post by: seventeenboi on July 20, 2017, 12:26:18 pm
hiii :))
I'm 100% lost on how to do this question please help
thanks!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 20, 2017, 12:33:33 pm
In the diagram the circles XYPS and XYRQ intersect at the point X and Y, and PXQ, PYR, QSY, PST and QTR are straight lines.
(i) Explain why ∠STQ=∠YRQ+∠YPS.
(ii) Show that ∠YRQ+∠YPS+∠SXQ=pi.
(iii) Deduce that STQX is a cyclic quadrilateral.
(iv) Let ∠QPY=a and ∠PQY=b. Show that ∠STQ=a+b.

Can I have help with iv?
The diagram they offered was terrible in the fact that it did not make QSY appear to be a straight line.
(http://i.imgur.com/JBzf2K2.jpg)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 20, 2017, 12:34:27 pm
hiii :))
I'm 100% lost on how to do this question please help
thanks!
Hint: You will have to assume that the area of an ellipse is \(\text{Area}=\pi AB\), where A and B are the lengths of the major and minor axes.
Title: Re: 4U Maths Question Thread
Post by: Natmo243 on July 20, 2017, 07:53:04 pm
Could somebody please help me with this one? I'm trying to get an answer independent of a (their answer is sqrt(3)+pi/3). I've tried the substitution x=asinӨ and integration by parts. Any help would be much appreciated :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 20, 2017, 09:10:28 pm
Could somebody please help me with this one? I'm trying to get an answer independent of a (their answer is sqrt(3)+pi/3). I've tried the substitution x=asinӨ and integration by parts. Any help would be much appreciated :)
Just looking at this I can tell that the answer will not be independent of a, due to the upper bound being unpleasant. Presumably the upper bound on the integral is incorrect and should be \(a\).

Edit: According to Wolfram, if the upper bound is \(a\) then the answer is what's given, except with a minus.
Title: Re: 4U Maths Question Thread
Post by: Natmo243 on July 21, 2017, 06:58:33 am
Just looking at this I can tell that the answer will not be independent of a, due to the upper bound being unpleasant. Presumably the upper bound on the integral is incorrect and should be \(a\).

Edit: According to Wolfram, if the upper bound is \(a\) then the answer is what's given, except with a minus.
My teacher and I could only come to that conclusion after changing the limits. It's not often that a mistake exists both in the question and the answer. Thankyou very much!
Title: Re: 4U Maths Question Thread
Post by: limtou on July 21, 2017, 09:16:24 pm
Hey Rui,

Could you explain why y = log_e (e^x-2) has the asymptote y=x?

Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 21, 2017, 09:21:36 pm
Hey Rui,

Could you explain why

has the asymptote y=x?


That one isn't easy to explain mathematically. Consider a more intuitive approach




Also, LaTeX doesn't handle it well when you use double spaces.
Title: Re: 4U Maths Question Thread
Post by: limtou on July 21, 2017, 09:27:32 pm
That one isn't easy to explain mathematically. Consider a more intuitive approach




Also, LaTeX doesn't handle it well when you use double spaces.

Thanks!
The latex killed me :( Took me longer trying to get the latex than your explanation (and didn't even get the latex in the end).
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 21, 2017, 10:55:49 pm
Thanks!
The latex killed me :( Took me longer trying to get the latex than your explanation (and didn't even get the latex in the end).
Tbh, I think if you dropped all the spaces in your LaTeX (except for when splitting some functions up) it would've rendered properly.
Title: Re: 4U Maths Question Thread
Post by: beau77bro on July 23, 2017, 09:53:02 am
ok so im getting really close to the answer, using u=x-a substitution, but i can't understand how if f(x) is odd that we treat f(u) as odd) because aren't we technically shifting the function?

p.s. i never really understood dumby variables - rushed lesson so an explanation of anything that answers this would be really appreciated

thanks
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 23, 2017, 10:08:28 am


____________________________




Title: Re: 4U Maths Question Thread
Post by: beau77bro on July 29, 2017, 09:33:29 am
why dont we set one person in this and have to reduce the n part by another 1?

usually we say for n around a table it's (n-1)! because we set one person so they aren't the same. why don't we set one here? the answer is (n-4)!4! but what if n=4 then it's 4! and that's not right. is it?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 29, 2017, 10:03:55 am
why dont we set one person in this and have to reduce the n part by another 1?

usually we say for n around a table it's (n-1)! because we set one person so they aren't the same. why don't we set one here? the answer is (n-4)!4! but what if n=4 then it's 4! and that's not right. is it?




___________________________________________________________________


Compare the two cases. When n=4, what actually happens is that we've chosen everyone. This leaves nobody else left to use as, pretty much, a "basis" for the circular arrangement to handle the (n-1)! issue. So we actually have to cater for it manually.

Provided n>4, we have a way around it because there will always be one leftover guy. So as a result of the proof of the (n-1)! result, we address it in pretty much the way(s) explained above.

As far as 4U goes, it would probably have made sense to assume n>4 here. But pretty good point though.
Title: Re: 4U Maths Question Thread
Post by: beau77bro on July 29, 2017, 10:24:34 am




___________________________________________________________________


Compare the two cases. When n=4, what actually happens is that we've chosen everyone. This leaves nobody else left to use as, pretty much, a "basis" for the circular arrangement to handle the (n-1)! issue. So we actually have to cater for it manually.

Provided n>4, we have a way around it because there will always be one leftover guy. So as a result of the proof of the (n-1)! result, we address it in pretty much the way(s) explained above.

As far as 4U goes, it would probably have made sense to assume n>4 here. But pretty good point though.

OHHHH YEA I FORGOT WHEN YOU GROUP THEM AS 4, they are still treated a just 1 person, so n-3, THEN U SET ONE. ok i see my bad. thanks rui.
Title: Re: 4U Maths Question Thread
Post by: Kle123 on July 29, 2017, 09:09:47 pm
HI
It is true that (cis(2π/3))3=1 but why isn't it true when both sides are cube rooted? (cis2π/3≠1).
Thankss
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 29, 2017, 09:18:16 pm
HI
It is true that (cis(2π/3))3=1 but why isn't it true when both sides are cube rooted? (cis2π/3≠1).
Thankss
Simple reason: Cube rooting (or even square rooting) isn't that simple when you're dealing with the complex numbers, as opposed to the real numbers.

Recall when you computed elementary square, cube, fourth, fifth or whatever roots, you couldn't just pick one value of \( \theta\). e.g. To compute the fifth roots of \( i\), you couldn't just get away with \(5\theta = \frac{\pi}2\implies \theta = \frac{\pi}{10}\). You had to do the whole \(5\theta = \frac{\pi}2+2k\pi \) thing.
Title: Re: 4U Maths Question Thread
Post by: kiwiberry on July 31, 2017, 08:05:00 pm
Heya, could someone please explain how they found the number of terms? I understand the rest of the question :)
Title: Re: 4U Maths Question Thread
Post by: Opengangs on July 31, 2017, 09:18:36 pm
Heya, could someone please explain how they found the number of terms? I understand the rest of the question :)
Since we want to find when Alan eventually picks a black jellybean, we need to consider every case.

So, if Alan picks a black jellybean on his first go, it's simply 1/n
3rd go: (n - 1)/n * (n - 2)/(n - 1) * 1/(n - 2), since the first and second times, we don't want the black jellybean.
5th go: (n - 1)/n * (n - 2)/(n - 1) * (n - 3)/(n - 2) * (n - 4)/(n - 3) * 1/(n - 4), same reasoning.

So, we sum this up to get when Alan eventually picks a black jellybean.
P(Alan picks a black jellybean) = 1/n + (n - 1)/n * (n - 2)/(n - 1) * 1/(n - 2) + (n - 1)/n * (n - 2)/(n - 1) * (n - 3)/(n - 2) * (n - 4)/(n - 3) * 1/(n - 4) + ... + (n - 1)/n * (n - 2)/(n - 1) ... * 2/3 * 1/2 * 1

We can actually rewrite this as:
P(Alan picks a black jellybean) = 1/n + 1/n * (n - 1)/(n - 1) * (n - 2)/(n - 2) + 1/n * (n - 2)/(n - 1) * (n - 2)/(n - 2) * (n - 3)/(n - 3) * (n - 4)/(n - 4) + ... + 1/n * (n - 1)/(n - 1) ... * 3/3 * 2/2 * 1
= 1/n + 1/n + 1/n + ... 1/n

How do we know it's (n + 1)/2 terms?
Let's consider each term independently.

Case 1: 1st go -- 1/n
Case 2: 3rd go -- 1/n
Case 3: 5th go -- 1/n
Case 4: 7th go -- 1/n

We can develop a pattern here between the case number and the term we're interested in.
Case n: (2n - 1)th go -- 1/n - This is the general pattern.

Thus, the nth go is:
Case [(n + 1)/2]: nth go -- 1/n

This means that there are (n + 1)/2 terms.

Which means that the probability is (1/n) * (n + 1)/2 = (n + 1)/2n as required.
Title: Re: 4U Maths Question Thread
Post by: kiwiberry on July 31, 2017, 09:33:55 pm
Since we want to find when Alan eventually picks a black jellybean, we need to consider every case.
...
Omg thanks so much!!! I get it :D
Title: Re: 4U Maths Question Thread
Post by: Opengangs on July 31, 2017, 09:37:26 pm
Omg thanks so much!!! I get it :D
Glad I could help! ^-^
Title: Re: 4U Maths Question Thread
Post by: raymatar on August 02, 2017, 07:46:25 pm
Hey can someone please help me with these 2 questions. Thanks in advance.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 02, 2017, 09:48:08 pm
Hey can someone please help me with these 2 questions. Thanks in advance.



Are you having trouble with 7b or 7c?
Title: Re: 4U Maths Question Thread
Post by: beau77bro on August 03, 2017, 09:33:51 am
help please, i really don't understand the process to get here in the solution - maybe you have an alternative? or could u explain how u got there? should i add the solution?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 03, 2017, 09:47:38 am
help please, i really don't understand the process to get here in the solution - maybe you have an alternative? or could u explain how u got there? should i add the solution?
Why not?
Title: Re: 4U Maths Question Thread
Post by: beau77bro on August 03, 2017, 03:33:54 pm
idk ahhaha - just thought it wouldn't be necessary and i was hoping there was another way
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 03, 2017, 03:36:39 pm
idk ahhaha - just thought it wouldn't be necessary and i was hoping there was another way

That actually didn't cross my mind during the time I had to think about it. I'll probably explain it later tonight though (unless someone gets there before me) because of a few more classes I have today that I want to focus on.
Title: Re: 4U Maths Question Thread
Post by: beau77bro on August 03, 2017, 06:57:21 pm
why isnt the answer B? - if the centre is half way to the vertex Q, shoulds it be half OQ, which gives 1/2 (i+1)z
Title: Re: 4U Maths Question Thread
Post by: Ali_Abbas on August 03, 2017, 07:09:13 pm
why isnt the answer B? - if the centre is half way to the vertex Q, shoulds it be half OQ, which gives 1/2 (i+1)z


Unless I'm mistaken, I agree the answer should be B.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 03, 2017, 07:13:29 pm
why isnt the answer B? - if the centre is half way to the vertex Q, shoulds it be half OQ, which gives 1/2 (i+1)z

Well what did they say it was?

(I haven't forgotten about your binomial question; but I just got home and I'm having dinner first.)
Title: Re: 4U Maths Question Thread
Post by: beau77bro on August 03, 2017, 07:56:44 pm
Well what did they say it was?

(I haven't forgotten about your binomial question; but I just got home and I'm having dinner first.)

sorry they said it was C
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 03, 2017, 08:17:14 pm
sorry they said it was C
Wait... but isn't that correct? It goes from iz to z so it should be z - iz = (1-i)z

I feel like they made typos somewhere
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 03, 2017, 08:28:55 pm
idk ahhaha - just thought it wouldn't be necessary and i was hoping there was another way




Also, I believe they intended to write a2n-r in the place of ar but made a mistake in transcript




______________________________________



Title: Re: 4U Maths Question Thread
Post by: beau77bro on August 03, 2017, 09:22:28 pm



Also, I believe they intended to write a2n-r in the place of ar but made a mistake in transcript




______________________________________





ok so it's kind of a weird comparison/equating of co-efficients for different expansions, i can see what you did but I'm struggling with the reasoning. is it just re-denoting the question so that you can show a new relationship between a with the power? - seems liek they just gave u a question where u just use the answer to find something - i can't see a straightforward process other than changing the powers.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 03, 2017, 11:37:01 pm
ok so it's kind of a weird comparison/equating of co-efficients for different expansions, i can see what you did but I'm struggling with the reasoning. is it just re-denoting the question so that you can show a new relationship between a with the power? - seems liek they just gave u a question where u just use the answer to find something - i can't see a straightforward process other than changing the powers.
Isn't that what you're meant to do in a proof? Look at the answer and find something?

I did mention in my lecture that one of the key ingredients to proving is to actually know what you're trying to prove.
____________

Although, to be fair, I might've made my response sounded like I was really just overly using what I was trying to prove, so the explanation looked dodge.

The problem is that the standard method of actually writing out the coefficient of \(x^r\) and trying to invoke the identity \( \binom{N}{K}=\binom{N}{N-K}\) means we have to deal with the expansion of \( ((1+x)+x^2)^n \) and then the presence of the \(x^2\) makes things awkward. If you want an example of a question where they actually did that, look at the binomial theorem question of the HSC 3U 2013 paper.
Title: Re: 4U Maths Question Thread
Post by: johnk21 on August 04, 2017, 11:00:04 am
Could i please get some help with part i.
Thanks :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 04, 2017, 11:21:03 am
Could i please get some help with part i.
Thanks :)




Title: Re: 4U Maths Question Thread
Post by: johnk21 on August 04, 2017, 11:25:33 am






OOOH... obviously. I used the same method but failed in isolating for x1 and y1.
Thanks so much  :) :)
Title: Re: 4U Maths Question Thread
Post by: chelseam on August 13, 2017, 05:09:39 pm
Hi! Could someone please help me with proving this statement? I'm stuck on how to get the left hand side, but I managed to use (x-y)^2≥0 to get the right hand side (not sure if I'm meant to do that though)! Thank you  :D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 13, 2017, 05:17:42 pm
Hi! Could someone please help me with proving this statement? I'm stuck on how to get the left hand side, but I managed to use (x-y)^2≥0 to get the right hand side (not sure if I'm meant to do that though)! Thank you  :D
That question looks wrong to me (if we're to assume, as always, x and y are positive). Are you sure the fraction on the RHS isn't meant to be entirely under the square root?
Title: Re: 4U Maths Question Thread
Post by: chelseam on August 13, 2017, 05:20:09 pm
That question looks wrong to me (if we're to assume, as always, x and y are positive). Are you sure the fraction on the RHS isn't meant to be entirely under the square root?
Yes it is! I'm so sorry, I drew it wrong in the picture!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 13, 2017, 05:26:12 pm


I know you were at my lecture so i know you know what I'm talking about :P I wrote the bottom line first and worked my way up.
Title: Re: 4U Maths Question Thread
Post by: chelseam on August 13, 2017, 05:34:31 pm
Quote
I know you were at my lecture so i know you know what I'm talking about :P I wrote the bottom line first and worked my way up.
Thanks so much Rui! ;D The tips at your lecture have helped me like inequalities questions a lot more now knowing that there's different ways to approach them! :D
Title: Re: 4U Maths Question Thread
Post by: itssona on August 18, 2017, 06:33:00 pm
pls help
(x+iy)^2=7-24i
find x and y
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 18, 2017, 06:37:56 pm
pls help
(x+iy)^2=7-24i
find x and y



Title: Re: 4U Maths Question Thread
Post by: itssona on August 18, 2017, 06:54:28 pm




ohhh!
thank you Rui!! :)
Title: Re: 4U Maths Question Thread
Post by: bluecookie on August 26, 2017, 06:54:34 pm
If P(acosθ , bsinθ ) and Q(acos(-θ ), bsin(-θ )] are the eremities of the latus rectum x=ae of the ellipse x^2/a^2y^2/b^2=1.

Show that PQ has length 2b^2/a
____________________________

P(asecθ , btanθ ) lies on the hyperbola x^2/a^2-y^2/b^2=1 with foci S(ae,0) and S'(-ae,0).
a) Show that PS=a(esecθ -1) and PS'a(esecθ +1)
b) Deduce that |PS-PS'|=2a
____________________________

P(asecθ , btanθ ) and Q(asecphi, btanphi) lie on the hyperbola x^2/a^2-y^2/b^2=1. Use the result that the chord PQ has the equation (x/a)*cos[(θ -pi)/2]-(y/b)*sin[(θ +phi)/2]=cos[(θ +phi)/2] to show that if PQ is a focal chord, then tanθ /2tanphi/2 takes one of the values of (1-e)/(1+e) or (1+e)/(1-e).

P(2rt(3), 3rt(3)) is one extremity of a focal chord on the hyperbola x^2/3 - y^2/9=1. Find the coordinates of the other extremity Q.
____________________________

Show that cos4θ =8(cosθ )^4-8(cosθ )^2+1.
a) Solve the equation 8x^4-8x^2+1=0 and deduce the exact values of cospi/8 and cos5pi/8.
b) Solve the equation 16x^4-16x^2+1=0 and deduce the exact v alues of cospi/12 and cos5pi/12.

Moderator action: Posts merged. At times like these, please resort to the modify at the top right corner of a post, to refrain from multi-posting.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 26, 2017, 07:01:07 pm
If P(acosθ , bsinθ ) and Q(acos(-θ ), bsin(-θ )] are the eremities of the latus rectum x=ae of the ellipse x^2/a^2y^2/b^2=1.

Show that PQ has length 2b^2/a




Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 26, 2017, 07:22:12 pm
P(asecθ , btanθ ) lies on the hyperbola x^2/a^2-y^2/b^2=1 with foci S(ae,0) and S'(-ae,0).
a) Show that PS=a(esecθ -1) and PS'a(esecθ +1)
b) Deduce that |PS-PS'|=2a
____________________________

P(asecθ , btanθ ) and Q(asecphi, btanphi) lie on the hyperbola x^2/a^2-y^2/b^2=1. Use the result that the chord PQ has the equation (x/a)*cos[(θ -pi)/2]-(y/b)*sin[(θ +phi)/2]=cos[(θ +phi)/2] to show that if PQ is a focal chord, then tanθ /2tanphi/2 takes one of the values of (1-e)/(1+e) or (1+e)/(1-e).

P(2rt(3), 3rt(3)) is one extremity of a focal chord on the hyperbola x^2/3 - y^2/9=1. Find the coordinates of the other extremity Q.



______________________________________________________


Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 26, 2017, 07:59:21 pm
Show that cos4θ =8(cosθ )^4-8(cosθ )^2+1.
a) Solve the equation 8x^4-8x^2+1=0 and deduce the exact values of cospi/8 and cos5pi/8.
b) Solve the equation 16x^4-16x^2+1=0 and deduce the exact v alues of cospi/12 and cos5pi/12.

Moderator action: Posts merged. At times like these, please resort to the modify at the top right corner of a post, to refrain from multi-posting.





______________________________________



Title: Re: 4U Maths Question Thread
Post by: Kle123 on August 27, 2017, 12:29:20 pm
Could i get help with part 3. Thanks!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 27, 2017, 12:45:57 pm
Could i get help with part 3. Thanks!
In the future, please provide progress on the previous parts. Here, part i would've been useful.



________________________________________


Title: Re: 4U Maths Question Thread
Post by: Kle123 on August 27, 2017, 12:54:35 pm
THANKSSSS Rui. Ill make sure to include the working for previous parts in the post next time!
Title: Re: 4U Maths Question Thread
Post by: bluecookie on August 27, 2017, 03:52:50 pm





Thanks :D About the post mergings - I was actually considering doing that before I multi-posted (I think I've been told of for that before and know it's something generally not appreciated in this thread) but I ran into a problem where, if I spent time typing all those questions before, by the time I posted them all at once, there may have already been a solution offered to the first, you get what I mean? (Geezus I feel like I'm doing such a terrible job explaining sorry, but it's just that, if I post 1 problem, in the time it takes for me to type another, someone (like you) can read it and give me a solution - effectively saving time) so that's why I multi-posted. What should I do about this problem in the future?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 27, 2017, 05:28:43 pm
Thanks :D About the post mergings - I was actually considering doing that before I multi-posted (I think I've been told of for that before and know it's something generally not appreciated in this thread) but I ran into a problem where, if I spent time typing all those questions before, by the time I posted them all at once, there may have already been a solution offered to the first, you get what I mean? (Geezus I feel like I'm doing such a terrible job explaining sorry, but it's just that, if I post 1 problem, in the time it takes for me to type another, someone (like you) can read it and give me a solution - effectively saving time) so that's why I multi-posted. What should I do about this problem in the future?
Questions like these are very unlikely to fall under the umbrella of "quick" questions; their solutions require over 10 minutes to be typed up properly, making this unlikely. If you have a series of small questions there's nothing wrong with simply asking them all at once, otherwise preferably wait for one (or if it isn't overly excessive, maybe two) to be answered already before posting more.
Title: Re: 4U Maths Question Thread
Post by: bluecookie on August 28, 2017, 12:59:32 pm





______________________________________





Thank you :)

Questions like these are very unlikely to fall under the umbrella of "quick" questions; their solutions require over 10 minutes to be typed up properly, making this unlikely. If you have a series of small questions there's nothing wrong with simply asking them all at once, otherwise preferably wait for one (or if it isn't overly excessive, maybe two) to be answered already before posting more.

Okay, will do ^^
Title: Re: 4U Maths Question Thread
Post by: bluecookie on August 28, 2017, 01:46:18 pm
Show that cos5θ=15(cosθ)^5-20(cosθ)^3+5cosθ. Hence
a) Solve the equation 16x^5-20x^3+5x-1=0 and deduce the exact values of cos(2pi/5) and cos(4pi/5)
b) Solve the equaion 32x^5-40x^3+10x-1=0 and deduce that
i) cospi/15+cos7pi/15+cos13pi/15+cos19pi/15=-1/2
ii) cospi/15cos7pi/15cos13pi/15cos19pi/15=1/16.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 28, 2017, 03:22:19 pm
Show that cos5θ=15(cosθ)^5-20(cosθ)^3+5cosθ. Hence
a) Solve the equation 16x^5-20x^3+5x-1=0 and deduce the exact values of cos(2pi/5) and cos(4pi/5)
b) Solve the equaion 32x^5-40x^3+10x-1=0 and deduce that
i) cospi/15+cos7pi/15+cos13pi/15+cos19pi/15=-1/2
ii) cospi/15cos7pi/15cos13pi/15cos19pi/15=1/16.

If you have trouble replicating the same procedure you should post up your working instead for guidance/feedback.
_______________________________________




The next question is done via a similar method (with differences being that no exact values make it faster, and the equation to solve is weirder). To be honest, these methods are all in my 4U Notes book; you need to take a look at it.
Title: Re: 4U Maths Question Thread
Post by: bluecookie on September 04, 2017, 11:42:39 am

If you have trouble replicating the same procedure you should post up your working instead for guidance/feedback.
_______________________________________




The next question is done via a similar method (with differences being that no exact values make it faster, and the equation to solve is weirder). To be honest, these methods are all in my 4U Notes book; you need to take a look at it.

Thanks! Yeah I dunno why, I'm just not very good at those. For part b i) I got one of the roots as cos(-11pi/15), and the only one that doesn't match up to any of the other roots I got was cos19pi/15. I was wondering how you got from cos(-11pi/15) to cos(19pi/15).
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 04, 2017, 12:33:35 pm
Thanks! Yeah I dunno why, I'm just not very good at those. For part b i) I got one of the roots as cos(-11pi/15), and the only one that doesn't match up to any of the other roots I got was cos19pi/15. I was wondering how you got from cos(-11pi/15) to cos(19pi/15).


Title: Re: 4U Maths Question Thread
Post by: beau77bro on September 13, 2017, 04:43:03 pm
can someone explain this both in words and equations - struggle with dummy variables
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 13, 2017, 05:07:38 pm
can someone explain this both in words and equations - struggle with dummy variables



____________________________________________________



____________________________________________________


____________________________________________________
Title: Re: 4U Maths Question Thread
Post by: justwannawish on September 14, 2017, 01:18:49 pm
Hi!

I was wondering what are the best textbooks for 4U? I've heard that while cambridge is good for the harder questions, it isn't really realistic for the types of questions in the HSC?  (Any recs for 3U as well? Is Fitzpatrick 'better' than Cambridge?)

Thank you
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 14, 2017, 10:24:43 pm
Hi!

I was wondering what are the best textbooks for 4U? I've heard that while cambridge is good for the harder questions, it isn't really realistic for the types of questions in the HSC?  (Any recs for 3U as well? Is Fitzpatrick 'better' than Cambridge?)

Thank you
There is no such thing as best textbooks to rely on HSC questions/preparation, especially given that the ultimate focus is always past papers.

For some decently worthwhile mentions, if you're using the small Cambridge book that is only ever so decent. As for question types, they are either too easy or too hard.

On the other hand, things like Terry Lee's textbook are generally recommended for selective schools, due to the difficulty of the questions.

4U Fitzpatrick is decent, provided the newer version is chosen. The old version is quite useless.

What's generally regarded as a standout is the Sydney Grammar textbook, as it is most similar to the 3U Cambridge textbook. This textbook, however, must be purchased from the school. It was never published (presumably due to the new syllabus things), and comes as only a PDF.
Title: Re: 4U Maths Question Thread
Post by: beau77bro on September 15, 2017, 10:55:30 am
Hi!

I was wondering what are the best textbooks for 4U? I've heard that while cambridge is good for the harder questions, it isn't really realistic for the types of questions in the HSC?  (Any recs for 3U as well? Is Fitzpatrick 'better' than Cambridge?)

Thank you
I'm not sure if you are a year 12 or 11, if you are a year 12 then this isn't as helpful. but if u are going into year 12 i hope this helps u pick and choose.
3U:
Terry Lee - is very good for harder level questions, a lot are HSC esk, some are just ridiculous. overall one of the top recommends personally. Explanations for the exercises are usually not great, but the worked solutions are pretty good.
Cambridge - very good explanations (quite wordy) and a lot of questions to practice with. Some are just not HSC type Questions and are just tedious, strange, or just why? but they aren't irrelevant and all the questions are very good for drawing out some kind of understanding and for testing yourself (i personally think there are too many and some are unnecessary to do)
Fitzpatrick - is very well written and nicely set out, it's not as hard as the TL and Cambridge but it has some really good, tough questions (i don't use it that much but it's good). the explanations are very good and its a pretty book ngl
AtarNotes - is something i recommend for revision (topic specific which helps when u only need to study for a couple of topics for an exam)--> to me there aren't enough questions to practice on or learn stuff. but when it comes time for exams the questions are challenging, they cover almost all types of questions you can be asked and the solutions are so comprehensive that it's super easy to fix any holes.
Papers - are ultimately the best for revision holistically. HSC, or trial you are gonna be challenged and get some solid practice. if you do 30 papers you will undoubtably be exponentially better (but u need to stick with it through hard Qs).
NOTE: YOU WILL NEED TO REVISE CONSISTENTLY TO KEEP ON TOP OF THINGS SO WORK YOU DID DOESN'T GO TO WASTE

4U:
Terry Lee - is again amazing, and very very challenging. (some topics are better than others, so in 4u ull probably have to do a couple textbooks to work out the best)
Cambridge - i honestly don't use it much, not very appealing and the questions are weird (good challenge exercises and diagnostic tests). explanations are good but it's just not as good as others.
Sydney Grammar Notes - ARE EASILY THE BEST RESOURCE overall - explanations are amazing, and as rui said it's the closest to Cambridge 3U. There are lots of Questions, lots of chapters and they go from easy to medium to hard (i forgot what they are called) and if you had to do just one this is the one (in my opinion because it covers all topics pretty well). still i would say use this in conjunction with terry lee. we used these notes from my teacher throughout - but i think if u can do this for the whole course.
Fitzpatrick - is good, I don't use it much because it's a bit repetitive (for what i did), is definitely useful and has good questions. better for starting out than challenging urself, but it asks a lot of questions which is good.
Atarnotes - i haven't started doing, but from reading it they are good challenging questions for revision. if they are as good as the 3u ones then it is very worthwhile
Trials - are again just the best thing for revision and practice but u have to do a lot of them to cover everything since the course is massive. and u need to spend time working things out (dont be afraid to use the solutions just use them well dont copy)

For harder ext 1 I recommend a book by kinny lewis. like just lots of Questions and really good solutions. They are hard, but they cover most things which is amazing. still should do everything under the sun for this topic since its expected to be like 30% of the course by NESA.

yea so there you go. I'm sure there are more out there, but that's my top picks - can't go too wrong with any but it's good to use a couple in my opinion.
HOPE THIS HELPS.

Title: Re: 4U Maths Question Thread
Post by: statues on September 15, 2017, 11:14:12 pm
Hi -
I need a quick explanation of HSC 2016 q 15 b ii
So I can work to the answer without the limits of the integral (d and b) - I don't understand why they're there. I've read the suggested solutions posted by Rui. But I still don't get it. In other questions where I'm integrating to get an expression for time, I just integrate, solve for any constant and then substitute a point such as x in to get time. Why is it that in this question a definite integral is required?

Many thanks
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 15, 2017, 11:16:57 pm
Hi -
I need a quick explanation of HSC 2016 q 15 b ii
So I can work to the answer without the limits of the integral (d and b) - I don't understand why they're there. I've read the suggested solutions posted by Rui. But I still don't get it. In other questions where I'm integrating to get an expression for time, I just integrate, solve for any constant and then substitute a point such as x in to get time. Why is it that in this question a definite integral is required?

Many thanks
In the future, please provide a link or screenshot to the question.

Are these my solutions that I posted ages ago last year you're referring to?
Title: Re: 4U Maths Question Thread
Post by: statues on September 15, 2017, 11:24:38 pm
In the future, please provide a link or screenshot to the question.

Are these my solutions that I posted ages ago last year you're referring to?

Yup - the one here: https://atarnotes.com/forum/index.php?topic=168150.0
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 15, 2017, 11:36:13 pm
Yup - the one here: https://atarnotes.com/forum/index.php?topic=168150.0





________________________________________






Whilst it probably isn't required out of a 3U student, in general a 4U student is expected to be able to handle applications of definite integrals in the place of indefinite integrals as well.
Title: Re: 4U Maths Question Thread
Post by: statues on September 15, 2017, 11:46:40 pm
Thank you so much Rui!
Do you know what section it is in, in your MX2 notes, I can't seem to find it.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 15, 2017, 11:48:10 pm
Thank you so much Rui!
Do you know what section it is in, in your MX2 notes, I can't seem to find it.
Should be under 6.3.2 Vertical Resisted Motion
Title: Re: 4U Maths Question Thread
Post by: Shlomo314 on September 17, 2017, 03:35:03 pm
Could someone help me out with this integration using the t method please?

Integrate tanx/(1+cosx)

cheers
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 17, 2017, 04:03:15 pm
Could someone help me out with this integration using the t method please?

Integrate tanx/(1+cosx)

cheers



Subject to minor computation error
Title: Re: 4U Maths Question Thread
Post by: Shlomo314 on September 19, 2017, 06:55:46 pm
Hi guys, can someone help me on this question please?
Find the number of arrangements that can be made using the letters of the word PARTICULAR without altering the relative positions of vowels and consonants

I don't understand the part where it says the 'relative positions of vowels and consonants.

Cheers Atar team
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 19, 2017, 07:19:46 pm
Hi guys, can someone help me on this question please?
Find the number of arrangements that can be made using the letters of the word PARTICULAR without altering the relative positions of vowels and consonants

I don't understand the part where it says the 'relative positions of vowels and consonants.

Cheers Atar team


____________________________


Title: Re: 4U Maths Question Thread
Post by: atangirala1 on September 20, 2017, 05:47:34 pm
Hey RuiAce ,
Obviously I want to join ur atar notes forum but I'm currently unable to do so.
May u plz guide me and  let me join ur Ex2 math science group  :) ;)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 20, 2017, 05:54:33 pm

Hey RuiAce ,
Obviously I want to join ur atar notes forum but I'm currently unable to do so.
May u plz guide me and  let me join ur Ex2 math science group  :) ;)
What group is this?
Title: Re: 4U Maths Question Thread
Post by: justwannawish on September 20, 2017, 09:03:07 pm
I'm not sure if you are a year 12 or 11, if you are a year 12 then this isn't as helpful. but if u are going into year 12 i hope this helps u pick and choose.
3U:
Terry Lee - is very good for harder level questions, a lot are HSC esk, some are just ridiculous. overall one of the top recommends personally. Explanations for the exercises are usually not great, but the worked solutions are pretty good.
Cambridge - very good explanations (quite wordy) and a lot of questions to practice with. Some are just not HSC type Questions and are just tedious, strange, or just why? but they aren't irrelevant and all the questions are very good for drawing out some kind of understanding and for testing yourself (i personally think there are too many and some are unnecessary to do)
Fitzpatrick - is very well written and nicely set out, it's not as hard as the TL and Cambridge but it has some really good, tough questions (i don't use it that much but it's good). the explanations are very good and its a pretty book ngl
AtarNotes - is something i recommend for revision (topic specific which helps when u only need to study for a couple of topics for an exam)--> to me there aren't enough questions to practice on or learn stuff. but when it comes time for exams the questions are challenging, they cover almost all types of questions you can be asked and the solutions are so comprehensive that it's super easy to fix any holes.
Papers - are ultimately the best for revision holistically. HSC, or trial you are gonna be challenged and get some solid practice. if you do 30 papers you will undoubtably be exponentially better (but u need to stick with it through hard Qs).
NOTE: YOU WILL NEED TO REVISE CONSISTENTLY TO KEEP ON TOP OF THINGS SO WORK YOU DID DOESN'T GO TO WASTE

4U:
Terry Lee - is again amazing, and very very challenging. (some topics are better than others, so in 4u ull probably have to do a couple textbooks to work out the best)
Cambridge - i honestly don't use it much, not very appealing and the questions are weird (good challenge exercises and diagnostic tests). explanations are good but it's just not as good as others.
Sydney Grammar Notes - ARE EASILY THE BEST RESOURCE overall - explanations are amazing, and as rui said it's the closest to Cambridge 3U. There are lots of Questions, lots of chapters and they go from easy to medium to hard (i forgot what they are called) and if you had to do just one this is the one (in my opinion because it covers all topics pretty well). still i would say use this in conjunction with terry lee. we used these notes from my teacher throughout - but i think if u can do this for the whole course.
Fitzpatrick - is good, I don't use it much because it's a bit repetitive (for what i did), is definitely useful and has good questions. better for starting out than challenging urself, but it asks a lot of questions which is good.
Atarnotes - i haven't started doing, but from reading it they are good challenging questions for revision. if they are as good as the 3u ones then it is very worthwhile
Trials - are again just the best thing for revision and practice but u have to do a lot of them to cover everything since the course is massive. and u need to spend time working things out (dont be afraid to use the solutions just use them well dont copy)

For harder ext 1 I recommend a book by kinny lewis. like just lots of Questions and really good solutions. They are hard, but they cover most things which is amazing. still should do everything under the sun for this topic since its expected to be like 30% of the course by NESA.

yea so there you go. I'm sure there are more out there, but that's my top picks - can't go too wrong with any but it's good to use a couple in my opinion.
HOPE THIS HELPS.



I think I'll go with Fitzpatrick and Terry Lee for the moment. Thank you for your advice :)

There is no such thing as best textbooks to rely on HSC questions/preparation, especially given that the ultimate focus is always past papers.

For some decently worthwhile mentions, if you're using the small Cambridge book that is only ever so decent. As for question types, they are either too easy or too hard.

On the other hand, things like Terry Lee's textbook are generally recommended for selective schools, due to the difficulty of the questions.

4U Fitzpatrick is decent, provided the newer version is chosen. The old version is quite useless.

What's generally regarded as a standout is the Sydney Grammar textbook, as it is most similar to the 3U Cambridge textbook. This textbook, however, must be purchased from the school. It was never published (presumably due to the new syllabus things), and comes as only a PDF.

Thank you for your advice. Is the Sydney Grammar book available for students who don't go there?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 20, 2017, 09:12:56 pm
Thank you for your advice. Is the Sydney Grammar book available for students who don't go there?
Yes.

But like said, you have to order through your school.
Title: Re: 4U Maths Question Thread
Post by: Kle123 on September 28, 2017, 07:45:00 pm
Hi, Could I get help on this proof:
(n-1)Ck+(n-2)Ck+(n-3)Ck+...+kCk=nC(k+1)
Thankss
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 28, 2017, 07:47:23 pm
Hi, Could I get help on this proof:
(n-1)Ck+(n-2)Ck+(n-3)Ck+...+kCk=nC(k+1)
Thankss
This was one of the 3U questions in 2009.
Title: Re: 4U Maths Question Thread
Post by: raymatar on September 29, 2017, 10:16:48 am
Can anyone help with v please?

Thanks
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 29, 2017, 10:50:02 am
Can anyone help with v please?

Thanks

This is because if \(\beta\) or \(\gamma\) were chosen to be the right angle instead, the same argument would work. This falls from just the fact that \(\alpha,\beta,\gamma\) are just the angles of the triangle.

_________________________________

Remark: Internationally, cosec is denoted as "csc", and is what LaTeX renders. Hence it will be what I type.


Title: Re: 4U Maths Question Thread
Post by: raymatar on September 29, 2017, 12:12:48 pm

This is because if \(\beta\) or \(\gamma\) were chosen to be the right angle instead, the same argument would work. This falls from just the fact that \(\alpha,\beta,\gamma\) are just the angles of the triangle.

_________________________________

Remark: Internationally, cosec is denoted as "csc", and is what LaTeX renders. Hence it will be what I type.




How do you find the cot equation?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 29, 2017, 12:31:16 pm
How do you find the cot equation?
That was what you proved in iv.
Title: Re: 4U Maths Question Thread
Post by: itssona on September 29, 2017, 04:14:26 pm
halp pls
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 29, 2017, 07:46:11 pm
halp pls
Just looking at this, part b is clearly wrong, as \( w = z(\cos\theta+i\sin\theta)\) so \( w\overline{z} = z^2(\cos\theta-i\sin\theta)\)
Title: Re: 4U Maths Question Thread
Post by: itssona on September 30, 2017, 10:17:03 am
Just looking at this, part b is clearly wrong, as \( w = z(\cos\theta+i\sin\theta)\) so \( w\overline{z} = z^2(\cos\theta-i\sin\theta)\)
ohh okay thank you



help with this please

Mod edit: Post merge
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 30, 2017, 02:22:23 pm
help with this please
(https://i.imgur.com/7nqK4RQ.png)



___________________________________________



___________________________________________

Title: Re: 4U Maths Question Thread
Post by: itssona on October 01, 2017, 12:19:26 pm
hiii if i wanna sketch locus of arg (z+1/z-i)

then do i first see where its undefined and put open circles on the graph at those points? I think i should but i only ever heard of it once so I'm not sure.. thank you

edit: please show how to do this?  the solution is a semi circle but how
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 01, 2017, 02:29:53 pm
hiii if i wanna sketch locus of arg (z+1/z-i)

then do i first see where its undefined and put open circles on the graph at those points? I think i should but i only ever heard of it once so I'm not sure.. thank you

edit: please show how to do this?  the solution is a semi circle but how
I've never spent the time to fully understand the locus. I just remember how to draw it
Title: Re: 4U Maths Question Thread
Post by: seventeenboi on October 02, 2017, 03:56:42 pm
hellooo I don't understand this question or the solutions given pls help :)
thanks in advanced
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 02, 2017, 04:00:22 pm
hellooo I don't understand this question or the solutions given pls help :)
thanks in advanced



Title: Re: 4U Maths Question Thread
Post by: lsong on October 02, 2017, 10:02:34 pm
Hi all,

Just a general question if anyone knows.
If we copy down a diagram (e.g. circle geometry) and add a new element (e.g. a new point, or line), do we still need to write down what was added?

Thanks :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 02, 2017, 10:05:10 pm
Hi all,

Just a general question if anyone knows.
If we copy down a diagram (e.g. circle geometry) and add a new element (e.g. a new point, or line), do we still need to write down what was added?

Thanks :)
I would always state further constructions made, e.g produce AB to P. Else the reader might be confused.

But if you're in a rush for time you might get away without doing it.
Title: Re: 4U Maths Question Thread
Post by: lsong on October 02, 2017, 10:09:40 pm
I would always state further constructions made, e.g produce AB to P. Else the reader might be confused.

But if you're in a rush for time you might get away without doing it.

Noted, Thanks Rui :D
Title: Re: 4U Maths Question Thread
Post by: uusunny on October 03, 2017, 05:15:54 pm
How would I do this question?
A funfair game has the following set up: a player may toss two balls into any of k chutes and if both balls are returned via a single chute, the player wins. What is the probability that both balls enter via different chutes, but are returned via the same chute, where that chute is neither of the chutes by which any ball entered?

The answer is (k^2 - 3k +2)/ k^3

Thank you!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 03, 2017, 07:45:09 pm
How would I do this question?
A funfair game has the following set up: a player may toss two balls into any of k chutes and if both balls are returned via a single chute, the player wins. What is the probability that both balls enter via different chutes, but are returned via the same chute, where that chute is neither of the chutes by which any ball entered?

The answer is (k^2 - 3k +2)/ k^3

Thank you!



Title: Re: 4U Maths Question Thread
Post by: armtistic on October 04, 2017, 03:56:28 pm
Hey,
This implicit differentiation is simple but I didn't know that  dy/dx being a constant meant that the original relation represents two parallel lines.
So I was wondering if this is just something you have to know or if there is some explanation for how they concluded that.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 04, 2017, 04:04:03 pm
Hey,
This implicit differentiation is simple but I didn't know that  dy/dx being a constant meant that the original relation represents two parallel lines.
So I was wondering if this is just something you have to know or if there is some explanation for how they concluded that.


(where by extension, that constant is in fact the gradient of the line itself.)
______________________________________________


The full, proper reason why it's not a coincidence how we have a pair of parallel lines somewhat relies on a more formal understanding of either conics and/or graphs that the HSC doesn't teach. A pair of parallel lines is really one of the "degenerate" cases of the conic sections
______________________________________________

Title: Re: 4U Maths Question Thread
Post by: johnk21 on October 04, 2017, 10:35:46 pm
Can someone please help me with this probability q. Thanks
Title: Re: 4U Maths Question Thread
Post by: Opengangs on October 04, 2017, 10:49:37 pm
Can someone please help me with this probability q. Thanks
a) Total no. of outcomes is given by 9P3 = 504
Now, if it succeeds 400, the first number simply has to be greater than 3.
Thus, final probability = (6P1 * 8 * 7)/504 = 2/3

b) Again, total no. of outcomes is 9P3.
Now, if we were to randomly choose three numbers, there's only one way to arrange them.
So, the total number of arrangements we'll get is: 1 + 1 + 1 + 1 + 1 + ... + 1, since in each selection, there's only one way in arranging them in descending order.

The fact that there are 9C3 ways of arranging them is revealed when we count the fact that (1, 2, 3) arranged would be the same as (1, 3, 2) arranged, which is the same as (3, 2, 1) arranged.

This brings us to conclude that there is a probability of 9C3/9P3 = 1/6
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 04, 2017, 10:55:34 pm
b) Again, total probability is 9P3.
Now, if we were to randomly choose three numbers, there's only one way to do so.
So, the total number of arrangements we'll get is: 1 + 1 + 1 + 1 + 1 + ... + 1, since in each arrangement, there's only one way in arranging them in descending order.

The fact that there are 9C3 ways of arranging them is revealed when we count the fact that (1, 2, 3) arranged would be the same as (1, 3, 2) arranged, which is the same as (3, 2, 1) arranged.

This brings us to conclude that there is a probability of 9C3/9P3 = 1/6
Remark: Total outcomes is different to total probability. The sum of all probabilities is 1.

Actually, \( \binom{9}3 \) is the number of ways they can be chosen. Once the three numbers have been chosen, there is only one way for them to be arranged in descending order. We must not get mixed up between what a "selection" and an "arrangement" is in our reasoning.
Title: Re: 4U Maths Question Thread
Post by: Shlomo314 on October 05, 2017, 11:43:49 am
Hey Atar team,

Could someone give me a hand on 2016 ext 2 multi choice for 7,9, and 10 please?

Cheers
Title: Re: 4U Maths Question Thread
Post by: Natasha.97 on October 05, 2017, 11:51:07 am
Hey Atar team,

Could someone give me a hand on 2016 ext 2 multi choice for 7,9, and 10 please?

Cheers

Hi!
Rui has uploaded his explanations here :)
Title: Re: 4U Maths Question Thread
Post by: Shlomo314 on October 06, 2017, 09:22:56 am
Hi,
I was just view the 2016 solutions that RuiAce post and Q16 says I need to update 3rd party hosting. How do I do that

Cheers
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 06, 2017, 02:01:16 pm
Hi,
I was just view the 2016 solutions that RuiAce post and Q16 says I need to update 3rd party hosting. How do I do that

Cheers
At some point I had reached the limit for third party hosting with photobucket. I'll extract the images and put them on another image hosting site instead.
Title: Re: 4U Maths Question Thread
Post by: beau77bro on October 07, 2017, 09:45:29 pm
i dont get the jump from the equals to the inequality sign. 2011 HSC q8ci btw. but i only need the working explained not the question. it is basically the triangle inequality thing - if so could u explain both how it works and how the hell we were meant to see that


and how do we know its a Geometric Progression?!?!? because what does M represent - it doesn't really say. so do all the coefficients equal M? that doesn't make sense doe it?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 07, 2017, 10:05:59 pm
i dont get the jump from the equals to the inequality sign. 2011 HSC q8ci btw. but i only need the working explained not the question. it is basically the triangle inequality thing - if so could u explain both how it works and how the hell we were meant to see that


I gave a brief mention to this in my book. It's quite unfair and I couldn't get it out at the time even though it makes sense now.



_______________________________________

As for how the hell you were meant to see it?

Short answer: You don't. Unless you saw the next step


Title: Re: 4U Maths Question Thread
Post by: beau77bro on October 07, 2017, 10:19:56 pm
ok so i get the inequality. but wth is m. like is it the coefficient of every single term??? that seems so sus and impossible. how can u factorise it out if its the coeffiecient of each??
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 07, 2017, 10:39:36 pm
ok so i get the inequality. but wth is m. like is it the coefficient of every single term??? that seems so sus and impossible. how can u factorise it out if its the coeffiecient of each??


Visual example



Title: Re: 4U Maths Question Thread
Post by: beau77bro on October 08, 2017, 09:08:50 am
OHHHHH OK THANKYOU RUI.
Title: Re: 4U Maths Question Thread
Post by: seventeenboi on October 09, 2017, 11:12:38 am
hiii i need help with this question

thank you so much!!! :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 09, 2017, 11:15:56 am
hiii i need help with this question

thank you so much!!! :)
Already addressed in the compilation.
Title: Re: 4U Maths Question Thread
Post by: seventeenboi on October 09, 2017, 11:38:16 am
Hello, sorry I have another question -
what is the effect on a general hyperbola as e approaches infinity ???? i know that it approaches a 'pair of vertical lines' (directrices) but i don't understand why that is :))

thank you
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 09, 2017, 11:47:59 am
Hello, sorry I have another question -
what is the effect on a general hyperbola as e approaches infinity ???? i know that it approaches a 'pair of vertical lines' (directrices) but i don't understand why that is :))

thank you

(The directrices ultimately coincide thanks to the formula \(x=\pm \frac{a}{e}\) and that the negative of 0 is also 0.)


______________________


Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on October 09, 2017, 09:58:25 pm
hey! this is my very first post so please excuse me if this doesn't work out. anywho I wanted to ask about 4u. how hard is it? I reallyyy want to try it and I know I'm capable of doing it if I work for it but Ive been hearing things like this subject is killer and when you do it you'll be looking like a zombie from the lack of sleep, which has made me reconsider taking it but what do you recommend?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 09, 2017, 10:10:01 pm
hey! this is my very first post so please excuse me if this doesn't work out. anywho I wanted to ask about 4u. how hard is it? I reallyyy want to try it and I know I'm capable of doing it if I work for it but Ive been hearing things like this subject is killer and when you do it you'll be looking like a zombie from the lack of sleep, which has made me reconsider taking it but what do you recommend?
I may have forgotten to address these exact rumours, but they were basically what I cleared up at the start of the lecture.

Short answer to the validity of the rumours: NO. When people have this sort of problem, they're doing a course that they were never suited for anyway. Many high achieving students in MX2 have never had this problem.
Title: Re: 4U Maths Question Thread
Post by: Shlomo314 on October 12, 2017, 04:52:53 pm
Hey atar team, how would you solve this question?

Cheers
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 12, 2017, 05:33:07 pm
Hey atar team, how would you solve this question?

Cheers




Title: Re: 4U Maths Question Thread
Post by: Shlomo314 on October 13, 2017, 02:19:18 pm
Hi Atar team, Could someone help me with this question please?

Thanks
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 13, 2017, 02:21:56 pm
Hi Atar team, Could someone help me with this question please?

Thanks
This is just 1/2, because the other 1/2 is when the number 1 lies somewhere to the right of the number 2.

(Note that two numbers obviously cannot occupy the same position. So we only have two mutually exclusive cases - 1 is to the left of 2, or 1 is to the right of 2.)
Title: Re: 4U Maths Question Thread
Post by: itssona on October 16, 2017, 04:22:54 pm
heey im not sure if I'm getting the right answer for this,
basically I need to now integrate -1/(6 (2x+3)) and 1/(6 (2x+3)) for partial fractions,
is the answer -1/12 ln (2x+3) + 1/12ln (2x-3)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 18, 2017, 09:27:35 am
heey im not sure if I'm getting the right answer for this,
basically I need to now integrate -1/(6 (2x+3)) and 1/(6 (2x+3)) for partial fractions,
is the answer -1/12 ln (2x+3) + 1/12ln (2x-3)
That looks fine (although you should be using absolute values for ln|2x+3| and ln|2x-3|).
Title: Re: 4U Maths Question Thread
Post by: itssona on October 18, 2017, 02:23:41 pm
yo Rui can u tell the ext 2 methods? XD
find the values of a and b such that x^4 -x^3 +x^2 +ax +b is divisible by (x^2+4)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 18, 2017, 08:22:24 pm
yo Rui can u tell the ext 2 methods? XD
find the values of a and b such that x^4 -x^3 +x^2 +ax +b is divisible by (x^2+4)

where \( P(x) \) is the polynomial you were given, i.e. \( P(x) = x^4 - x^3 + x^2 + ax + b \)

Title: Re: 4U Maths Question Thread
Post by: julies on October 18, 2017, 11:58:22 pm
(http://)

hey not sure if this question's been asked already
but I don't understand how to do this q and the solutions don't make sense
thanks in advance!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 19, 2017, 07:12:17 am
()

hey not sure if this question's been asked already
but I don't understand how to do this q and the solutions don't make sense
thanks in advance!



Title: Re: 4U Maths Question Thread
Post by: Shlomo314 on October 20, 2017, 12:12:01 pm
Hey,
What are the best strategies for the next 3 days to maximise marks before the Ext 2 paper apart from past HSC papers?
Thankyou
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2017, 12:27:28 pm
Hey,
What are the best strategies for the next 3 days to maximise marks before the Ext 2 paper apart from past HSC papers?
Thankyou
I don't know of any better strategy (until either the night before the exam, where you should be resting, and the morning of the exam, where you should just reconsolidate all the methods/formulas/... you know).

If you don't feel like studying for maths anymore (ran out of papers, genuinely bored etc.) then it might be time to study for something else.

Basically, you stop worrying about past papers once you're at the last stretch. Until then, there is still no optimal strategy aside from them.
Title: Re: 4U Maths Question Thread
Post by: itssona on October 20, 2017, 07:39:02 pm
hiii
so if we know z^3=1
how do we evaluate (1+2w+3w^2)(1+2w^2+3w)

whats the way we do this?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2017, 07:40:21 pm

hiii
so if we know z^3=1
how do we evaluate (1+2w+3w^2)(1+2w^2+3w)

whats the way we do this?
I’m about to drive home but for reference, did you mean w^3=1?
Title: Re: 4U Maths Question Thread
Post by: beau77bro on October 20, 2017, 07:51:45 pm
ii and iii, it says "recorded", why don't we care about the arrangements in the answer, specifically ii?

also why can't we do iii, by going 1-probability of a draw and divide by 2?


Title: Re: 4U Maths Question Thread
Post by: itssona on October 20, 2017, 07:53:08 pm
I’m about to drive home but for reference, did you mean w^3=1?
ohh yeah I mean w^3 sorry
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2017, 08:28:59 pm
ohh yeah I mean w^3 sorry

__________________________________

\begin{align*}\therefore (1+2w+3w^2) (1+2w^2+3w) &= (w^2-1)(w-1)\\ &= w^3 - w^2 - w + 1\\ &+ w^3 - (w^2 - w - 1) + 2\\ &= 1 - 0 + 2\\ &= 3\end{align*}
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2017, 08:34:26 pm
ii and iii, it says "recorded", why don't we care about the arrangements in the answer, specifically ii?

also why can't we do iii, by going 1-probability of a draw and divide by 2?





___________________________

If you mean [1 - Pr(whole match is drawn) ] / 2 then yes you can. But if you do that, just make sure you've covered every single case for a draw.

Terry Lee's answers just chose not to do it.
Title: Re: 4U Maths Question Thread
Post by: beau77bro on October 20, 2017, 09:22:27 pm


___________________________

If you mean [1 - Pr(whole match is drawn) ] / 2 then yes you can. But if you do that, just make sure you've covered every single case for a draw.

Terry Lee's answers just chose not to do it.

then don't we need to consider the arrangements in the "total number of arrangements"? or something - otherwise isnt it just the probability of a win, law, draw, draw in any order?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2017, 09:28:10 pm
then don't we need to consider the arrangements in the "total number of arrangements"? or something - otherwise isnt it just the probability of a win, law, draw, draw in any order?

Why does the total number of arrangements matter?

What we've calculated is the probability of W-D-L-D in that order. If we multiply by the number of ways we can arrange WDLD like we did in part iii), then we would have the probability of WDLD in any order and not just that specific one.
Title: Re: 4U Maths Question Thread
Post by: Shlomo314 on October 21, 2017, 12:26:25 pm
Hey, how would I do this question?

Cheers
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 21, 2017, 12:32:37 pm

Hey, how would I do this question?

Cheers
A and D are integrals of odd functions so they are exactly equal to 0.

On the other hand B and C are integrals of even functions. But if we sketch the function being integrated in C we will find that the region is entirely BELOW the x-axis, so the integral is negative.

If you sketch the one for B, that one will always be above the x-axis. Hence that is the correct answer, as the integral will be positive as a consequence of this.
Title: Re: 4U Maths Question Thread
Post by: chelseam on October 21, 2017, 07:11:04 pm
Hi! Could someone please show me how to prove the triangle inequality? I'm confused about both the geometric and algebraic approaches! Thank you so much :D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 21, 2017, 07:33:32 pm
Hi! Could someone please show me how to prove the triangle inequality? I'm confused about both the geometric and algebraic approaches! Thank you so much :D
I'm fairly sure there is no algebraic proof that you're expected to know of in 4U. The first time I proved it algebraically was in first year uni after doing some linear algebra in advance. The only thing that can be proven algebraically is the generalised triangle inequality, and even then you have to assume that it's true for 2 complex numbers before you can prove it for n complex numbers.
(https://i.imgur.com/wLAIt4Z.png)



Title: Re: 4U Maths Question Thread
Post by: chelseam on October 21, 2017, 08:19:25 pm
I'm fairly sure there is no algebraic proof that you're expected to know of in 4U. The first time I proved it algebraically was in first year uni after doing some linear algebra in advance. The only thing that can be proven algebraically is the generalised triangle inequality, and even then you have to assume that it's true for 2 complex numbers before you can prove it for n complex numbers.
Thanks so much Rui! ;D
Title: Re: 4U Maths Question Thread
Post by: statues on October 22, 2017, 12:51:42 am
2001 HSC 5
(c) A class of 22 students is to be divided into four groups consisting of 4, 5, 6 and
7 students.
(i) In how many ways can this be done? Leave your answer in unsimplified
form.

The answer my book gives is 22C4x18C5x13C16
What I don't understand is why this isn't divided by 4! on account of the ways that the same groups can be selected in different orders, but still essentially be the same.
Many Thanks
Title: Re: 4U Maths Question Thread
Post by: Ali_Abbas on October 22, 2017, 02:24:33 am
Hi! Could someone please show me how to prove the triangle inequality? I'm confused about both the geometric and algebraic approaches! Thank you so much :D

Algebraic proof of the triangle inequality can be done in two ways.






Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 22, 2017, 09:12:43 am
2001 HSC 5
(c) A class of 22 students is to be divided into four groups consisting of 4, 5, 6 and
7 students.
(i) In how many ways can this be done? Leave your answer in unsimplified
form.

The answer my book gives is 22C4x18C5x13C16
What I don't understand is why this isn't divided by 4! on account of the ways that the same groups can be selected in different orders, but still essentially be the same.
Many Thanks

This is basically when to not use the identical groups problem I covered in my trial survival lectures.


Selecting the groups in another order was not what caused the problem; it was the actual groups themselves.
Title: Re: 4U Maths Question Thread
Post by: Shlomo314 on October 22, 2017, 09:57:04 am
Hi,
When finding the eccentricity for a conjugate hyperbola (y^2/b^2-x^2/a^2=1), is it e^2=1+a^2/b^2 or is it e^2=1+b^2/a^2
Thankyou
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 22, 2017, 10:01:16 am
Hi,
When finding the eccentricity for a conjugate hyperbola (y^2/b^2-x^2/a^2=1), is it e^2=1+a^2/b^2 or is it e^2=1+b^2/a^2
Thankyou
I always memorised it as this (because textbooks can use multiple conventions and it gets annoying)

So for \(\frac{y^2}{b^2}-\frac{x^2}{a^2} = 1\) it would be \( a^2=b^2(e^2-1) \)
Title: Re: 4U Maths Question Thread
Post by: Shlomo314 on October 22, 2017, 11:15:20 am
Hey Rui

For the 2016 question 9, I dont understand how you did h/4=x/6 with similar triangle method (I saw the 6 from finding the difference of the two lengths but how did you use similar triangles and got 6)?

Thanks
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 22, 2017, 11:25:57 am
Hey Rui

For the 2016 question 9, I dont understand how you did h/4=x/6 with similar triangle method (I saw the 6 from finding the difference of the two lengths but how did you use similar triangles and got 6)?

Thanks
(https://i.imgur.com/ZWXA6GU.png)
The front view is basically the trapezium itself, and by drawing in that blue line you get triangles and parallelograms appearing. Then, the proportional sides on similar triangles are useful to us. (We know that the triangles are similar as they are equiangular by inspection.)

This is a pretty fairly common technique that I wrote in my book
Title: Re: 4U Maths Question Thread
Post by: Shlomo314 on October 22, 2017, 11:44:08 am
Hey,
In the mechanics section, more specifically F=ma, how do you know when to include the 'm' when integrating? Is there a difference between 'Resistance' and 'Retardation' where you leave the m or take the m out of the equation?

Thanks
Title: Re: 4U Maths Question Thread
Post by: chelseam on October 22, 2017, 12:14:13 pm
Algebraic proof of the triangle inequality can be done in two ways.
Thank you so much! :D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 22, 2017, 12:25:19 pm
Hey,
In the mechanics section, more specifically F=ma, how do you know when to include the 'm' when integrating? Is there a difference between 'Resistance' and 'Retardation' where you leave the m or take the m out of the equation?

Thanks
Retardation is negative acceleration. Resistance is negative force.

As for the 'm' problem, many textbooks are full of confusion but the HSC usually clears up any disambiguity with whether or not the m is necessary. They will either say "proportional to the velocity/square of the velocity" or "proportional to the mass times the velocity/square of the velocity". So you will know what to do when that happens.

Besides, when in doubt, just look at what they want you to prove. If the m disappears, then the resistive force should be something like mkv. If the m stays, the resistive force should be something like kv.
Title: Re: 4U Maths Question Thread
Post by: beau77bro on October 22, 2017, 04:35:36 pm
HELPPP, i can't work out this question. and i am struggling with my reasoning. i have all the principles for Probability, but i can't apply them and im struggling, like in the last question how i didn't see how you didnt arrange them, well it was because it was an arrangement and was set. but now im sturggling with a lot of probability and was wondering how i could brush up or revise - like just arranging things properly or knowing when not to, or when it's just repeats or basic stuff i am over complicating.


i cant do the first part, i would like to attempt the second after you guys give me hints, thankyou.
Title: Re: 4U Maths Question Thread
Post by: beau77bro on October 22, 2017, 04:45:53 pm
and where can i find harder binomial theorem Qs, harder sequences and series? - not really major parts but i have major holes.
Title: Re: 4U Maths Question Thread
Post by: aryak on October 22, 2017, 04:50:27 pm
Hey,
How do you find the minor and major axis of a locus of an ellipse using complex numbers?
Ex: Iz-3I + Iz+3I = 12

Thanks
Title: Re: 4U Maths Question Thread
Post by: sudodds on October 22, 2017, 04:52:25 pm
HELPPP, i can't work out this question. and i am struggling with my reasoning. i have all the principles for Probability, but i can't apply them and im struggling, like in the last question how i didn't see how you didnt arrange them, well it was because it was an arrangement and was set. but now im sturggling with a lot of probability and was wondering how i could brush up or revise - like just arranging things properly or knowing when not to, or when it's just repeats or basic stuff i am over complicating.


i cant do the first part, i would like to attempt the second after you guys give me hints, thankyou.

hey! thought i'd give Rui a bit of a break and have a go aha :)

So this isn't too bad if you break it down a bit. So let's just put one of that pair in the left-most slot in the first rung (call them Player A). Let's look at the possibilities they verse Player B, depending on where Player B is.

If Player B is next to them in the same bracket (a 1 in 3 chance), the condition is already satisfied. So, 1/3 there. If Player B is in one of the other slots (a 2 in 3 chance), two things need to happen. Player A needs to win their game, and Player B does too. So multiply 2/3 by 1/2, and then 1/2 again, and you get 1/6.

Add these together!

1/3+1/6=1/2

(sorry I don't know how to use the fancy math text like Rui but hopefully this makes sense aha)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 22, 2017, 04:58:45 pm
hey! thought i'd give Rui a bit of a break and have a go aha :)

So this isn't too bad if you break it down a bit. So let's just put one of that pair in the left-most slot in the first rung (call them Player A). Let's look at the possibilities they verse Player B, depending on where Player B is.

If Player B is next to them in the same bracket (a 1 in 3 chance), the condition is already satisfied. So, 1/3 there. If Player B is in one of the other slots (a 2 in 3 chance), two things need to happen. Player A needs to win their game, and Player B does too. So multiply 2/3 by 1/2, and then 1/2 again, and you get 1/6.

Add these together!

1/3+1/6=1/2

(sorry I don't know how to use the fancy math text like Rui but hopefully this makes sense aha)
OI TRUUUUUUUU.

Now, hints for the second part:

(whereas for two we only had \(\binom42\binom22 \frac{1}{2!} = 3 \) )
Try splitting the cases this way before going into subcases:
Case 1 - The pairings are split something like (A,B) (C,D) || (E,F) (G,H), so A and C can meet by the second round. This will be the case that best reflects the computations in part (i) (although they won't be an exact replica)
Case 2 - The pairings are split something like (A,B) (E,F) || (C,D) (G,H), so A and C are forced to meet in the third round.

That allows you to generalise it upwards
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 22, 2017, 05:01:34 pm
and where can i find harder binomial theorem Qs, harder sequences and series? - not really major parts but i have major holes.

Your best bet is to try papers from before 2000 (or even 1990) if you want some of these ones. (Or some of the extension level questions in the Cambridge textbook; like they're not really examinable but they definitely are hard.)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 22, 2017, 05:08:05 pm
Hey,
How do you find the minor and major axis of a locus of an ellipse using complex numbers?
Ex: Iz-3I + Iz+3I = 12

Thanks


I presume you knew that the foci were at 3 and -3.



(Major axis = 12, minor axis = 6√3)
Title: Re: 4U Maths Question Thread
Post by: Checkmate123 on October 22, 2017, 05:33:57 pm
in finding the nth root of -1, all textbooks and solutions say it is cis(((2k+1)pi)/n). However isn't the general solution cis((2kpi +/- pi)/n)? Do we ignore the plus/minus for some reason?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 22, 2017, 05:38:41 pm
in finding the nth root of -1, all textbooks and solutions say it is cis(((2k+1)pi)/n). However isn't the general solution cis((2kpi +/- pi)/n)? Do we ignore the plus/minus for some reason?

You can check this by listing out a few cases.
For the ± case, you would have cis(π/n), cis(3π/n), cis(5π/n) and so on, and also the other way.
For the + case, you would also have cis(π/n), cis(3π/n), cis(5π/n) and the rest.

This happens more or less because
1. They both go both ways: you can just sub k=-1 instead of k=1.
2. The trig functions are periodic: \( \sin (2\pi + x ) = \sin x\) and \( \cos (2\pi+x) = \cos x\).
Title: Re: 4U Maths Question Thread
Post by: samanthachoy on October 22, 2017, 05:56:13 pm
Hey Rui!
Just wondering if you had any general tips for graphing functions when they're part of tan inverse ((tan^-1) f(x) )?
Title: Re: 4U Maths Question Thread
Post by: Checkmate123 on October 22, 2017, 05:58:04 pm

You can check this by listing out a few cases.
For the ± case, you would have cis(π/n), cis(3π/n), cis(5π/n) and so on, and also the other way.
For the + case, you would also have cis(π/n), cis(3π/n), cis(5π/n) and the rest.

This happens more or less because
1. They both go both ways: you can just sub k=-1 instead of k=1.
2. The trig functions are periodic: \( \sin (2\pi + x ) = \sin x\) and \( \cos (2\pi+x) = \cos x\).

Thanks. So for any question, I can safely use only + and sub in k=1?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 22, 2017, 05:59:16 pm
Hey Rui!
Just wondering if you had any general tips for graphing functions when they're part of tan inverse ((tan^-1) f(x) )?


If it's the limiting behaviour, i.e. as \(x\to \infty\) we had f(x) exploding, then we would end up with a horizontal asymptote
If it's just an asymptote, i.e. as \(x\to a\) we had f(x) exploding, anticipate a hole in the graph because that's gonna lead to a discontinuity.

Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 22, 2017, 06:00:51 pm
Thanks. So for any question, I can safely use only + and sub in k=1?
I've never found any problem with only using +

Although at the same time, I've never solved it by inspection in high school (unless it was a multiple choice). I preferred doing the long method just in case I'd get called out for skipping steps and thus lose marks.
Title: Re: 4U Maths Question Thread
Post by: samanthachoy on October 22, 2017, 06:19:30 pm
Thanks Rui, that really cleared it up for me! Also, would you be able to share any general tips on attacking 4u probability questions? (like how to answer certain types of questions - whether to use binomial probability, combinations, the box method or something else)
Title: Re: 4U Maths Question Thread
Post by: frog1944 on October 22, 2017, 06:22:20 pm
Now, hints for the second part:


Hi RuiAce, for the first round, what is the probability the players will meet? I understand that the number of ways to pick the 8 players is 105, but what is the number of ways that they will meet? I have the book, it says 1/7 but I'm not sure how to arrive at that using your method (I originally was attempting to do it this way)

Thanks
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 22, 2017, 06:27:38 pm
Thanks Rui, that really cleared it up for me! Also, would you be able to share any general tips on attacking 4u probability questions? (like how to answer certain types of questions - whether to use binomial probability, combinations, the box method or something else)
Not too sure what the box method is.

If you suspect a binomial probability then you're suspecting the same thing you saw in 3U. You're looking out for if you need to consider the number of ways k success can happen out of n trials, given a success probability p.

There's some extra counting techniques from the trial survival lectures in the notes section of this website
Title: Re: 4U Maths Question Thread
Post by: lsong on October 22, 2017, 06:35:01 pm
Hi,
For part (ii), are we allowed to use 'Similarly to argument in part (i)'? Or would we have to prove it separately? The sample answers allow the similarly argument but not sure if we should follow this.
Thanks.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 22, 2017, 06:38:55 pm
Hi RuiAce, for the first round, what is the probability the players will meet? I understand that the number of ways to pick the 8 players is 105, but what is the number of ways that they will meet? I have the book, it says 1/7 but I'm not sure how to arrive at that using your method (I originally was attempting to do it this way)

Thanks
EDIT I'M SORRY ALL OF THIS IS WRONG



Note that the identical groups will only occur for the other 4 people. The pairs that A and C must be with are immune from this because A and C must be in different groups, and that effectively forces all possible groups to be distinct.

The only difference is that we don't have (A,B)(C,D) || (E,F)(G,H) but rather (A,B)(D,E) || (C,F)(G,H). This doesn't actually change the number of arrangements though.

Sanity check: If we regroup the cases, we get 15+45+45 = 105 as we expected.
That method doesn't get to the correct final answer so I need to check something

Edit #2: I've found out what the theoretical values are by working backwards. I'm gonna try again and work out how we can get to them.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 22, 2017, 06:39:39 pm
Hi,
For part (ii), are we allowed to use 'Similarly to argument in part (i)'? Or would we have to prove it separately? The sample answers allow the similarly argument but not sure if we should follow this.
Thanks.
Similarly is allowed so long as if you replicate the exact same STEPS albeit with different angles/sides you will reach a very similar conclusion.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 22, 2017, 07:41:34 pm
EDIT I'M SORRY ALL OF THIS IS WRONG



Note that the identical groups will only occur for the other 4 people. The pairs that A and C must be with are immune from this because A and C must be in different groups, and that effectively forces all possible groups to be distinct.

The only difference is that we don't have (A,B)(C,D) || (E,F)(G,H) but rather (A,B)(D,E) || (C,F)(G,H). This doesn't actually change the number of arrangements though.

Sanity check: If we regroup the cases, we get 15+45+45 = 105 as we expected.
That method doesn't get to the correct final answer so I need to check something

Edit #2: I've found out what the theoretical values are by working backwards. I'm gonna try again and work out how we can get to them.
Ok scrap ALMOST ALL of that - I think I've worked out a way to figure out the initial starting positions properly.

___________________________________





Alternatively, I could've chosen the two extra pairs first: \( \binom62\binom42\frac{1}{2!}\) and then arranged the last two; one with A and one with C: \(2\times 1\). This still gives 90.
___________________________________

The reason I do this is because facing off in round 1 is different: (A,C) must form a pair. For ANY OTHER ROUND, (A,C) must NOT form a pair.






Title: Re: 4U Maths Question Thread
Post by: aryak on October 22, 2017, 09:36:32 pm
Hey for 2015, question 12 (d) I am very confused as to whether to use x or (3-x) for volumes by cylindrical shells. How do we know which one to use?
Please Help!
https://www.boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-maths-ext-2.pdf
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 22, 2017, 09:44:02 pm
Hey for 2015, question 12 (d) I am very confused as to whether to use x or (3-x) for volumes by cylindrical shells. How do we know which one to use?
Please Help!
https://www.boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-maths-ext-2.pdf
(https://i.imgur.com/roBd70E.png)
Title: Re: 4U Maths Question Thread
Post by: aryak on October 22, 2017, 09:48:59 pm
is it ok if you could explain this a bit more? I am still a bit confused
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 22, 2017, 09:49:59 pm
is it ok if you could explain this a bit more? I am still a bit confused

What part is the confusing part?

I essentially wrote down the two relevant lengths, and figured out which one was the radius of the cylinder.

Title: Re: 4U Maths Question Thread
Post by: Dragomistress on October 23, 2017, 07:55:47 pm
Would someone please do this. Solving quadratic equations with complex coefficients: finding the complex number equations.

4x^2-4(1+2i)x-(3-4i)=0
Title: Re: 4U Maths Question Thread
Post by: justwannawish on October 23, 2017, 08:36:46 pm
Hi, guys, do you use De Moivre's Thereom to solve:
z^5= -32
Thank you
Title: Re: 4U Maths Question Thread
Post by: beau77bro on October 23, 2017, 08:54:14 pm
What were the roots of the complex quadratic in the exam?
Title: Re: 4U Maths Question Thread
Post by: beau77bro on October 23, 2017, 09:33:33 pm
And what was 16ciii? I got 2(3)^5 -3
Title: Re: 4U Maths Question Thread
Post by: pikachu975 on October 23, 2017, 10:02:17 pm
And what was 16ciii? I got 2(3)^5 -3

Answer was 480
Title: 4U Maths Question Thread
Post by: beau77bro on October 23, 2017, 10:15:50 pm
I think I can see why, the arrangements of only 2 colours can be happen in two ways.

As in like changing which way they alternate - 3 different pairs x2 because flip alternating. Damn was cheering
Title: Re: 4U Maths Question Thread
Post by: paigek3 on October 24, 2017, 10:49:08 am
Yewww I can say I posted in 4u maths  8)

Helloooo, this is such a dumb question that has no relevance to any mathematical problems whatsoever, but I am so confused about the structure of 4u maths so I thought I might as well ask here since the exam is over  ;D Sooo I know you guys sit the exam the same time as Mathematics, meaning you don't do that exam, but do you do the Mathematics course still? Im 10/10 confused and always wanted to know the answer to this

- from your fellow 4 unit (English) student  ;)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 24, 2017, 11:39:01 am
Yewww I can say I posted in 4u maths  8)

Helloooo, this is such a dumb question that has no relevance to any mathematical problems whatsoever, but I am so confused about the structure of 4u maths so I thought I might as well ask here since the exam is over  ;D Sooo I know you guys sit the exam the same time as Mathematics, meaning you don't do that exam, but do you do the Mathematics course still? Im 10/10 confused and always wanted to know the answer to this

- from your fellow 4 unit (English) student  ;)
Perhaps the easiest way to put it

The 4U student is taught the 2U content as well, but then it doesn't get examined. It just becomes assumed knowledge for the 4U student
Title: Re: 4U Maths Question Thread
Post by: paigek3 on October 24, 2017, 11:48:20 am
Perhaps the easiest way to put it

The 4U student is taught the 2U content as well, but then it doesn't get examined. It just becomes assumed knowledge for the 4U student

Riiiight that makes sense!!! Ahhh thank you for answering this I have been trying to figure it out for so long :P
Title: Re: 4U Maths Question Thread
Post by: itssona on October 25, 2017, 08:47:25 am
solve x^2cos^2 theta + x sin2theta +1 =0
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 25, 2017, 08:47:01 pm
solve x^2cos^2 theta + x sin2theta +1 =0
Assuming that theta is not a typo, what are we trying to solve for here? Because if we're trying to solve for x, the easiest way out is really just the quadratic formula
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 25, 2017, 08:58:39 pm
Please note that these were demonstrated in the lectures. So I will not go into much depth with the solution.
Would someone please do this. Solving quadratic equations with complex coefficients: finding the complex number equations.

4x^2-4(1+2i)x-(3-4i)=0
\begin{align*}0&=4x^2-4(1+2i)x-(3-4i)\\ x&= \frac{4(1+2i)\pm \sqrt{16(1+2i)^2 +16(3-4i)}}{8}\tag{quadratic formula}\\ &= \frac{4+8i\pm \sqrt{16(1+4i-4)+16(3-4i)}}{8}\\ &= \frac{4+8i \pm \sqrt{0}}{8}\\ &= \frac{1}{2}+ i\end{align*}
Hi, guys, do you use De Moivre's Thereom to solve:
z^5= -32
Thank you
Handwriting as "cis" is ok



So our solutions are 2cis(pi/5), 2cis(3pi/5), ...

Remark: You could've taken k=-2,-1,0,1,2
Title: Re: 4U Maths Question Thread
Post by: itssona on October 25, 2017, 10:38:58 pm
lil question
say i have 2 complex roots which are conjugates of eachother. One is alpha and the other is beta.
how do we prove alpha^2=beta and beta^2=alpha

thank you :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 25, 2017, 10:40:02 pm
lil question
say i have 2 complex roots which are conjugates of eachother. One is alpha and the other is beta.
how do we prove alpha^2=beta and beta^2=alpha

thank you :)
That usually isn't true. Do they give more information?
Title: Re: 4U Maths Question Thread
Post by: itssona on October 25, 2017, 10:45:30 pm
That usually isn't tru6e. Do they give more information?

well alpha=(-1+root 3 i)/2
beta=(-1-root3 i)/2

verify that a^2=b, b^2=a, a^3=b^3=1, 1+a+b=0

maybe it's related to roots of unity? idk :/
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 25, 2017, 10:47:35 pm
well alpha=(-1+root 3 i)/2
beta=(-1-root3 i)/2

verify that a^2=b, b^2=a, a^3=b^3=1, 1+a+b=0

maybe it's related to roots of unity? idk :/

Lol.

Yes, I bet you those are the cube roots of unity
Title: Re: 4U Maths Question Thread
Post by: itssona on October 25, 2017, 10:48:42 pm
Lol.

Yes, I bet you those are the cube roots of unity
oh lmao how do I verify all that stuff tho
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 25, 2017, 10:57:15 pm
Tbh, whilst there might be a crafty solution, at a time like that I would just be lazy and actually do the multiplication by hand
Title: Re: 4U Maths Question Thread
Post by: itssona on October 25, 2017, 11:02:47 pm
Tbh, whilst there might be a crafty solution, at a time like that I would just be lazy and actually do the multiplication by hand
ooh okay thank you Rui
Title: Re: 4U Maths Question Thread
Post by: itssona on October 28, 2017, 08:05:54 pm
Heya, I tried doing this by subbing in x1+iy1 for w1 and x2+iy2 for w2 but im still not getting it :/

solve for w1, w2
2w1 + 3iw2 =0
(1-i)w1 + 2w2 = i-7

(when i write w1, i mean w subscript 1)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 28, 2017, 08:12:59 pm
Heya, I tried doing this by subbing in x1+iy1 for w1 and x2+iy2 for w2 but im still not getting it :/

solve for w1, w2
2w1 + 3iw2 =0
(1-i)w1 + 2w2 = i-7

(when i write w1, i mean w subscript 1)



Note that you will need to divide complex numbers.
Title: Re: 4U Maths Question Thread
Post by: itssona on October 28, 2017, 08:28:10 pm
Thank you so much Rui :)

Alsoo, kinda not getting anywhere with this despite using trig formulas and whatnot:
if z is cistheta
prove that
2/(1+z) = (1+costheta+isintheta)/(1+costheta) = (2cos^2theta - 2isin(theta/2) cos(theta/2))/ 2cos^2(theta/2) = 1-itan(theta/2)

or if you can point me in the general direction? thank you!!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 28, 2017, 08:38:26 pm
Thank you so much Rui :)

Alsoo, kinda not getting anywhere with this despite using trig formulas and whatnot:
if z is cistheta
prove that
2/(1+z) = (1+costheta+isintheta)/(1+costheta) = (2cos^2theta - 2isin(theta/2) cos(theta/2))/ 2cos^2(theta/2) = 1-itan(theta/2)

or if you can point me in the general direction? thank you!!!
\begin{align*}\frac{2}{1+z}&=\frac{2}{1+\cos\theta + i\sin \theta}\\ &= \frac{1}{\frac{1}{2}(1+\cos \theta) + \frac{1}2 i\sin \theta }\\ &= \frac{1}{\cos^2\frac\theta2 + i\cos\frac\theta2\sin\frac\theta2}\\ &= \frac{1}{\cos \frac\theta2} \times \frac{1}{\cos \frac\theta2 + i\sin \frac\theta2}\end{align*}
NOW try "realising" the denominator
Title: Re: 4U Maths Question Thread
Post by: itssona on October 28, 2017, 08:55:02 pm
\begin{align*}\frac{2}{1+z}&=\frac{2}{1+\cos\theta + i\sin \theta}\\ &= \frac{1}{\frac{1}{2}(1+\cos \theta) + \frac{1}2 i\sin \theta }\\ &= \frac{1}{\cos^2\frac\theta2 + i\cos\frac\theta2\sin\frac\theta2}\\ &= \frac{1}{\cos \frac\theta2} \times \frac{1}{\cos \frac\theta2 + i\sin \frac\theta2}\end{align*}
NOW try "realising" the denominator
ohh yes I got the final part!! Thank you!!
Buut wait,, how do I show that I got 1+costheta+isintheta/ 1+ costheta
?
edit: nvm, got it!!, thank you :D
Title: Re: 4U Maths Question Thread
Post by: itssona on October 28, 2017, 10:33:33 pm
hey for this, do i sub in x+iy later??

find x,y if
2z/(1+i) - 2z/i = 5/(2+i)
Title: Re: 4U Maths Question Thread
Post by: itssona on October 29, 2017, 02:04:05 pm
heey if my complex number has no real parts (e.g. it's in where a=0 and b=z) then how do I know my argument?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 29, 2017, 02:10:16 pm
hey for this, do i sub in x+iy later??

find x,y if
2z/(1+i) - 2z/i = 5/(2+i)

\begin{align*}\frac{2z}{1+i} - \frac{2z}{i}&=\frac{5}{2+i}\\ 2z \left(\frac{1}{1+i}-\frac{1}{i}\right)&= \frac{5}{2+i}\\ 2z\left(\frac{i-(1+i)}{i(1+i)}\right)&=\frac{5}{2+i} \\ \therefore z&=\frac{5}{2(2+i)} \times \frac{i(1+i)}{-1}\end{align*}
which you should be able to finish off.
heey if my complex number has no real parts (e.g. it's in where a=0 and b=z) then how do I know my argument?
If you draw the Argand diagram, you'll see any purely imaginary number has argument either \(\frac\pi2\) or \(-\frac\pi2\)

Exercise: Work out when it's positive and when it's negative.
Title: Re: 4U Maths Question Thread
Post by: itssona on October 29, 2017, 03:46:58 pm
thank you so much Rui oml
also can't figure this-
E+ni is a root for ax2+bx+c=0
where a and b and c are real
show that an^2 =aE^2 + bE +c

so I figured that the other root is E-ni but I've tried using product of roots, but can't seem to prove this :/
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 29, 2017, 10:23:00 pm
thank you so much Rui oml
also can't figure this-
E+ni is a root for ax2+bx+c=0
where a and b and c are real
show that an^2 =aE^2 + bE +c

so I figured that the other root is E-ni but I've tried using product of roots, but can't seem to prove this :/
Those letters E and n look ugly so I'm gonna replace them with Greek letters........ :P



Just use \( i^2=-1\) to finish it off from here
Title: Re: 4U Maths Question Thread
Post by: itssona on November 01, 2017, 04:30:20 am
prove by Induction that z1+z2+zn = z1 +z2 +zn

where the LHS has a big conjugate line on the top and RHS has small conjugate lines (lmao)

so I proved true for n=1 and assumed for n=k but im stuck on proving n=k+1

thank you!!! :D
Modify message
Title: Re: 4U Maths Question Thread
Post by: itssona on November 06, 2017, 05:45:34 pm
z^2-2(1+i)z+8i=0
when we solve we should get (1 + root 3) +i(1-root3) and this other root, but i cant seem to get this :/
Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 06, 2017, 09:38:49 pm
prove by Induction that z1+z2+zn = z1 +z2 +zn

where the LHS has a big conjugate line on the top and RHS has small conjugate lines (lmao)

so I proved true for n=1 and assumed for n=k but im stuck on proving n=k+1

thank you!!! :D
Modify message




Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 06, 2017, 09:46:36 pm
z^2-2(1+i)z+8i=0
when we solve we should get (1 + root 3) +i(1-root3) and this other root, but i cant seem to get this :/

____________________________________




You should be able to finish it off from here.
Title: Re: 4U Maths Question Thread
Post by: itssona on November 06, 2017, 09:53:12 pm




thank you rui :)
Title: Re: 4U Maths Question Thread
Post by: itssona on November 06, 2017, 10:02:51 pm

i dont get this part, moduli of complex numbers removes the i?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 07, 2017, 08:49:25 am
i dont get this part, moduli of complex numbers removes the i?
\begin{align*}|(x+iy)|^2 &= |-6i|\\ |x+iy|^2 &= 6\\ x^2+y^2&=6\end{align*}
Title: Re: 4U Maths Question Thread
Post by: Dragomistress on November 08, 2017, 07:29:45 am
Hello!
I would like to know if using cis is possible for complex numbers.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 08, 2017, 09:42:18 am
Hello!
I would like to know if using cis is possible for complex numbers.
Yeah for HSC MX2 this is fine
Title: Re: 4U Maths Question Thread
Post by: frog1944 on November 08, 2017, 03:07:54 pm
Hi Rui,

I read somewhere that apparently it causes issues later on at university using the cis notation? Is this correct?

Thanks
Title: Re: 4U Maths Question Thread
Post by: Dragomistress on November 08, 2017, 05:00:41 pm
May you also do this:
Solve the equation z^3=1 when z=x+yi using the De Moivre's Theorem to find the roots of complex numbers.
Title: Re: 4U Maths Question Thread
Post by: frog1944 on November 08, 2017, 05:26:57 pm
Title: Re: 4U Maths Question Thread
Post by: frog1944 on November 08, 2017, 05:31:06 pm
I've tried to post my solution, however, the formatting for LaTex doesn't seem to be working for me (any help what I did wrong)?
Title: Re: 4U Maths Question Thread
Post by: Dragomistress on November 08, 2017, 05:46:21 pm
Ha ha, thanks for trying though. Can you try just take a photo of it instead as I sort of need it very soon.
Title: Re: 4U Maths Question Thread
Post by: frog1944 on November 08, 2017, 07:32:30 pm
Sorry, it should be working now
Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 08, 2017, 08:39:07 pm
Hi Rui,

I read somewhere that apparently it causes issues later on at university using the cis notation? Is this correct?

Thanks

The "cis" notation is an extremely redundant notation and nobody uses it in the real world. Writing it as the complex exponential takes less time so you should always go about that instead. The reason why "cis" gets taught in high school is because to actually prove Euler's formula you need Taylor series, which will certainly not go into high school math.

Honestly a bit of a lame excuse, but there's not much we can do about it.
Title: Re: 4U Maths Question Thread
Post by: Jeeffffffffffffff on November 08, 2017, 08:57:48 pm
Hey guys, would anybody be keen to help me with some subset graphing?
the question says to illustrate the following subsets of C on the z-plane and one of them is:
Arg[z-(1-iroot3)]=2pi/3
I've been shown a couple ways of how to do these types of questions but I was just after an easy method that will work everytime
Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 08, 2017, 09:17:54 pm
Hey guys, would anybody be keen to help me with some subset graphing?
the question says to illustrate the following subsets of C on the z-plane and one of them is:
Arg[z-(1-iroot3)]=2pi/3
I've been shown a couple ways of how to do these types of questions but I was just after an easy method that will work everytime


So you'd have the ray drawn from \( 1-\sqrt3 i \) making an angle of \( \frac{2\pi}3 \) with the positive real axis.
Title: Re: 4U Maths Question Thread
Post by: Jeeffffffffffffff on November 08, 2017, 09:29:58 pm


So you'd have the ray drawn from \( 1-\sqrt3 i \) making an angle of \( \frac{2\pi}3 \) with the positive real axis.

Awesome, cheers Rui.
Title: Re: 4U Maths Question Thread
Post by: itssona on November 08, 2017, 10:19:46 pm
for any complex numbers z1,z2, show that |z1+z2|^2 + |z1-z2|^2 = 2(|z1|^2 + |z2|^2)
thank you :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 08, 2017, 10:35:36 pm
for any complex numbers z1,z2, show that |z1+z2|^2 + |z1-z2|^2 = 2(|z1|^2 + |z2|^2)
thank you :)


Title: Re: 4U Maths Question Thread
Post by: itssona on November 08, 2017, 11:21:03 pm



thank u Rui :D
Title: Re: 4U Maths Question Thread
Post by: Jeeffffffffffffff on November 09, 2017, 10:50:50 am
2i(y-1)>0

Can you divide by 2i and keep the inequality the same?

To give y>1

Or is there something I'm missng?
Title: Re: 4U Maths Question Thread
Post by: frog1944 on November 09, 2017, 11:46:09 am
for any complex numbers z1,z2, show that |z1+z2|^2 + |z1-z2|^2 = 2(|z1|^2 + |z2|^2)
thank you :)

An alternative method, without resorting to the modulus-argument form is;
Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 09, 2017, 11:46:40 am
2i(y-1)>0

Can you divide by 2i and keep the inequality the same?

To give y>1

Or is there something I'm missng?
Inequalities don't exist with complex numbers (because they don't make sense). You can do inequalities with the real/imag parts, the mod and the arg, but not the complex number itself. The fact that you got there means you most likely made a mistake along the way.
An alternative method, without resorting to the modulus-argument form is;

Lol right, that's what I failed to think up last night. I knew the linear algebra way but entirely forgot the complex analysis version.

@sssona09 please use this method instead :)
Title: Re: 4U Maths Question Thread
Post by: maria.micale on November 09, 2017, 05:02:59 pm
Hey is there a site where is can plug in things including arg and it will sketch it on complex plane? My textbook docent have answers to some of the questions.

I've used the complex plane on desmos but I can't or don't know how to type in arg.

If not then can you sketch for me:

 -π < arg < π

 arg (z - i) = 0

arg ( z + 2) = 3π/4

Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 09, 2017, 05:15:09 pm
Hey is there a site where is can plug in things including arg and it will sketch it on complex plane? My textbook docent have answers to some of the questions.

I've used the complex plane on desmos but I can't or don't know how to type in arg.

If not then can you sketch for me:

 -π < arg < π

 arg (z - i) = 0

arg ( z + 2) = 3π/4


If you want to use technology, consider feeding GeoGebra the following input:
Code: [Select]
arg(x+i*y+2)=3pi/4
Note that excluding -2 will be necessary. GeoGebra won't do this, but you will need to.
Title: Re: 4U Maths Question Thread
Post by: maria.micale on November 10, 2017, 05:48:16 pm
If you want to use technology, consider feeding GeoGebra the following input:
Code: [Select]
arg(x+i*y+2)=3pi/4
Note that excluding -2 will be necessary. GeoGebra won't do this, but you will need to.

Thanks Rui,

Could you please sketch for me -π < arg < π because I don't think geogebra can do it?
Or if not, would you sketch the whole plane excluding the real axis ( i = 0)?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 10, 2017, 06:09:25 pm
Thanks Rui,

Could you please sketch for me -π < arg < π because I don't think geogebra can do it?
Or if not, would you sketch the whole plane excluding the real axis ( i = 0)?
Well yeah -π < arg < π is essentially the whole plane.

But you would exclude only the negative real axis, and the point 0. Because anything on the positive real axis satisfies arg(z) = 0, which still lies between -π and π. Only those on the negative real axis satisfy arg(z) = π, which just barely lies outside that range.
Title: Re: 4U Maths Question Thread
Post by: Caleb Campion on November 11, 2017, 11:26:41 am
I just don't know where to start when attacking this question? Is sum of roots the right way to go? Would love if I could get a worked solution :)

cos(pi/7) = cos(2pi/7) + cos(4pi/7) + 1/2
Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 11, 2017, 11:31:11 am
I just don't know where to start when attacking this question? Is sum of roots the right way to go? Would love if I could get a worked solution :)

cos(pi/7) = cos(2pi/7) + cos(4pi/7) + 1/2
Sum of roots will indeed be necessary but you would certainly not be told to prove this in an exam without any guidance. Please provide the source of the question.
Title: Re: 4U Maths Question Thread
Post by: maria.micale on November 17, 2017, 05:50:39 pm
Hey how would you go about factorising z^5 + 3z^4 - z - 3?

I got the first 3 factors by finding factors of the constant term that sub into P(z) to get zero.

So far I got P(z) = (z-1)(z+1)(z+3)Q(z)

But I don't know the best way to go about finding Q(x).
Title: Re: 4U Maths Question Thread
Post by: Eric11267 on November 17, 2017, 05:57:59 pm
Hey how would you go about factorising z^5 + 3z^4 - z - 3?

I got the first 3 factors by finding factors of the constant term that sub into P(z) to get zero.

So far I got P(z) = (z-1)(z+1)(z+3)Q(z)

But I don't know the best way to go about finding Q(x).
I would do it by grouping terms

Edit: I don't know if this is complex numbers or whatever so I'll just leave it like this
Title: Re: 4U Maths Question Thread
Post by: maria.micale on November 18, 2017, 12:13:01 pm
I would do it by grouping terms

Edit: I don't know if this is complex numbers or whatever so I'll just leave it like this



Thankyou! Also,

How do I factorise this?

z^4 + 4z^3 + 3z^2 - 8z - 10 if I am given (z +2 - i) as one linear factor.

I know that (z -2 + i) would be another factor due to a theorem but I don't know what to do to get the other two factors.
Title: Re: 4U Maths Question Thread
Post by: Sine on November 18, 2017, 04:30:35 pm


Thankyou! Also,

How do I factorise this?

z^4 + 4z^3 + 3z^2 - 8z - 10 if I am given (z +2 - i) as one linear factor.

I know that (z -2 + i) would be another factor due to a theorem but I don't know what to do to get the other two factors.
If (z + 2 -i) is a factor then (z + 2 +i) is also a factor via the conjugate root theorem (probably a typo on your part) :)
From here I would expand the two factors that you have i.e. expand (z + 2 -i)(z + 2 +i) which would yield a real quadratic.From here you can long divide the quartic that you have with the quadratic to yield another quadratic then we can factorise this final quadratic using the techniques we know.
Title: Re: 4U Maths Question Thread
Post by: KT Nyunt on November 23, 2017, 02:37:35 pm
Hey  :)

I have a 4U assessment coming up and I have this question:

If given the graph of f(x), how would I sketch f^2(x)?

Is this the same as f(f(x))?


Thanks in advance :)

Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 23, 2017, 02:40:21 pm
Hey  :)

I have a 4U assessment coming up and I have this question:

If given the graph of f(x), how would I sketch f^2(x)?

Is this the same as f(f(x))?


Thanks in advance :)



You should consider what the source of your question has done and follow their rule.
Title: Re: 4U Maths Question Thread
Post by: maria.micale on November 24, 2017, 10:38:10 pm
Hi, Could someone please explain how to factorise 2z^2 - 2z +1 over C?

I used quadratic formula so that z = 1/2 + or - i/2
So I thought P(z) = (z - 1/2 + i/2)(z - 1/2 - i/2)

But the answer is P(z) = 2(z - 1/2 + i/2)(z - 1/2 - i/2) (s0 there is a 2 out the front, I thought this may have something to do with the polynomial being non-monic but I'm not sure)

Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 25, 2017, 06:54:03 am
Hi, Could someone please explain how to factorise 2z^2 - 2z +1 over C?

I used quadratic formula so that z = 1/2 + or - i/2
So I thought P(z) = (z - 1/2 + i/2)(z - 1/2 - i/2)

But the answer is P(z) = 2(z - 1/2 + i/2)(z - 1/2 - i/2) (s0 there is a 2 out the front, I thought this may have something to do with the polynomial being non-monic but I'm not sure)


Yeah. The quadratic formula gives you the roots under the assumption that the polynomial is monic. If you wish to factorise by quadratic formula, you must always pull out the leading coefficient first (and hence the 2 in front).

The reason why this happens is because the quadratic formula is derived from completing the square. But when completing the square, you ALWAYS pull out the leading coefficient.
Title: Re: 4U Maths Question Thread
Post by: KT Nyunt on November 26, 2017, 02:52:02 pm
Hi Rui, could you please help with this question?

Use De Moivre's Theorem to express cos3θ in terms of powers of cosθ

Thanks in advance :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 26, 2017, 02:57:05 pm
Hi Rui, could you please help with this question?

Use De Moivre's Theorem to express cos3θ in terms of powers of cosθ

Thanks in advance :)
The \(\cos 5\theta\) variant was covered in my lecture so I won't go in depth here - just provide the working out.

Title: Re: 4U Maths Question Thread
Post by: Jeeffffffffffffff on November 27, 2017, 06:50:32 pm
The letters from the word FIFTY are selected at random. Find the probability that the T is chosen somehwere between the 2 F's.
I don't even know where to start with this one, a step by step of the whole process would be greatly appreciated, cheers!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 28, 2017, 02:20:16 pm
The letters from the word FIFTY are selected at random. Find the probability that the T is chosen somehwere between the 2 F's.
I don't even know where to start with this one, a step by step of the whole process would be greatly appreciated, cheers!





Title: Re: 4U Maths Question Thread
Post by: radnan11 on December 04, 2017, 10:24:07 am
Can someone explain how to do part b and c?
Title: Re: 4U Maths Question Thread
Post by: Opengangs on December 04, 2017, 11:36:32 am
Can someone explain how to do part b and c?
Using part a), we know that:


----

Using the product of roots:


Multiplying these values together, we get:
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 04, 2017, 11:37:43 am
Can someone explain how to do part b and c?




________________________________________________





Title: Re: 4U Maths Question Thread
Post by: justwannawish on December 05, 2017, 11:47:32 am
Hey guys,
I'm not sure how to fnd the roots of this or find x and y. Am i supposed to sub z=1, because that only got me a "2i+ix-y-1=0" and I'm not sure how to simplify it. Also is that the traditional way of dealing with questions about roots
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 05, 2017, 11:49:39 am
Hey guys,
I'm not sure how to fnd the roots of this or find x and y. Am i supposed to sub z=1, because that only got me a "2i+ix-y-1=0" and I'm not sure how to simplify it. Also is that the traditional way of dealing with questions about roots

Indeed, if you're given information about one root but then lack any more information to do clever tricks, the best you can ever do is by literally substituting the root straight in. But at any time, you must never forget that "two complex numbers are equal by definition if they have the same real and imaginary parts", so the instance you form an equality there's nothing wrong with equating the real and imag parts.
Title: Re: 4U Maths Question Thread
Post by: maria.micale on December 09, 2017, 12:27:35 pm
How would I factorise x^4 + x^2 +1 over the real numbers. It has no real roots. The roots are 1/2 + or - √3i/2 and -1/2 + or - √3i/2

Thanks in advance.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 09, 2017, 12:33:46 pm
How would I factorise x^4 + x^2 +1 over the real numbers. It has no real roots. The roots are 1/2 + or - √3i/2 and -1/2 + or - √3i/2

Thanks in advance.


Or you could've used FOIL, but that would probably take longer.
Title: Re: 4U Maths Question Thread
Post by: RustyWasTaken on December 09, 2017, 06:11:45 pm
Hi. How do I know when my curve is able to cross the horizontal asymptote and when it isnt able ?
Also how do I know which part crosses ? On the diagram the left side crosses but the right side doesnt :(
 pls help
Cheers.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 09, 2017, 07:04:09 pm
Hi. How do I know when my curve is able to cross the horizontal asymptote and when it isnt able ?
Also how do I know which part crosses ? On the diagram the left side crosses but the right side doesnt :(
 pls help
Cheers.





Now, once you have this information, whether or not you cross the horizontal asymptote depends on common sense. (Because as far as this course goes, there will never be too many stationary points involved with sketching rational functions.)

As \(x\to +\infty\), \(y\to 1^-\). But as \(x \to 0^+\), \(y \to +\infty\). This was the nature of the vertical asymptote we had found earlier. So what happens to your graph?

On your graph, you start to come down from the asymptote at \(x=0\). You must keep going down, without ever cutting the \(x\)-axis (because you have no intercepts), and somehow you still need to approach the horizontal asymptote from BELOW. If you don't cut through this horizontal asymptote, then the only way you can approach it is from above, which contradicts what we've just shown. So in this case, it MUST go down.

The same argument is now much more handwavy as \(x\to -\infty\). Once again, as \(x\to -3^-\), \(y\to +\infty\). So you need to come out of the horizontal asymptote \(y=1\) approached from above, and go up to the vertical asymptote at \(x=3\). But there really is no way for you to deduce whether or not you have stationary points (or worse, oscillations) about \(y=1\).

The short answer here is that in general, it's safe to assume that no renegade stationary points like this will occur. The safety net is more or less ensured by the fact that the degree of the numerator is not strictly greater than that of the denominator (\(x^2+3\) and \(x(x+3)\) are both degree 2 polynomials).
Title: Re: 4U Maths Question Thread
Post by: Jeeffffffffffffff on December 21, 2017, 02:47:19 pm
Show that if a, b, c, d > 0, then:
a/b + b/c + c/a + d/a  >or= 4

Just started doing advanced 3 unit inequalities and I have no idea what to do with this question, any help would be appreciated,
cheers.
Title: Re: 4U Maths Question Thread
Post by: maria.micale on December 21, 2017, 02:53:17 pm
Hi, there is this inequality question I am having trouble with. Could someone please assist?

Using a^2 + b^2 + c^2 >or= ab + bc + ca,
Show that a^2 + b^2 + c^2 >or= 3*cube root(a^2*b^2*c^2)
Title: Re: 4U Maths Question Thread
Post by: mxrylyn on December 21, 2017, 05:41:30 pm
Hello

I am having trouble with seeing how (-2 plus/minus root 8 x i) over 2,
can be written as -1 plus/minus root 2 x i
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 21, 2017, 05:58:25 pm
Hello

I am having trouble with seeing how (-2 plus/minus root 8 x i) over 2,
can be written as -1 plus/minus root 2 x i


Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 21, 2017, 06:18:26 pm
Hi, there is this inequality question I am having trouble with. Could someone please assist?

Using a^2 + b^2 + c^2 >or= ab + bc + ca,
Show that a^2 + b^2 + c^2 >or= 3*cube root(a^2*b^2*c^2)
What is the source of this question? I know I'm quite braindead but I genuinely don't see how the first inequality is supposed to be of any use right now

(More or less because you're basically trying to prove the AM-GM inequality for three variables. But \(a^2+b^2+c^2 \ge 3\sqrt[3]{a^2b^2c^2} \) and \(ab+ac+bc \ge 3\sqrt[3]{a^2b^2c^2}\) are both direct applications of the AM-GM inequality anyway.)
Title: Re: 4U Maths Question Thread
Post by: maria.micale on December 21, 2017, 08:47:52 pm
Show that if a, b, c, d > 0, then:
a/b + b/c + c/a + d/a  >or= 4

Just started doing advanced 3 unit inequalities and I have no idea what to do with this question, any help would be appreciated,
cheers.


The question I asked was from Fitzpatrick. The question above is also from this textbook. Do you know how to solve either of them. If so could you please show working because I am stumped.

Thanks
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 21, 2017, 09:21:47 pm
Oh wow, I completely missed that there was another question on the previous page. I'll address your's soon.

Although, having said that,
Show that if a, b, c, d > 0, then:
a/b + b/c + c/a + d/a  >or= 4

Just started doing advanced 3 unit inequalities and I have no idea what to do with this question, any help would be appreciated,
cheers.
This is actually one of the questions covered in my 4U notes book. So all I will do is put the working out here.



___________________________________

Gonna just edit this post because there's no point making another one when I don't have a concrete answer.

The question I asked was from Fitzpatrick. The question above is also from this textbook. Do you know how to solve either of them. If so could you please show working because I am stumped.

Thanks
With that question you asked, the only logical thing to do is to somehow prove that \( ab+ac+bc\ge 3\sqrt[3]{a^2b^2c^2} \), because otherwise you aren't using the hence method. But to do that, you would have to make a very weird substitution to force certain terms to collapse (in your given inequality), and right now I can't see how that works. I won't say that the question is nonsensical yet, but it seems extremely peculiar. (They actually make you prove something using a HORRIBLE method.)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 21, 2017, 11:12:52 pm




Remark: In the HSC, you would be guided through it. This method is quite famous but you're not expected to know it off by heart.
Title: Re: 4U Maths Question Thread
Post by: MLov on December 21, 2017, 11:28:29 pm
Recall that for a rectangular prism, given a fixed volume, the sum of dimensions is minimised when it is a cube.
Consider a rectangular prism with sides a2,b2 and c2 and a cube with sides . It is not hard to check they have the same volume.
Therefore for any a,b,c: as required.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 22, 2017, 10:31:24 am
Recall that for a rectangular prism, given a fixed volume, the sum of dimensions is minimised when it is a cube.
Consider a rectangular prism with sides a2,b2 and c2 and a cube with sides . It is not hard to check they have the same volume.
Therefore for any a,b,c: as required.
This is really just bringing back memories of Lagrange multipliers...
Title: Re: 4U Maths Question Thread
Post by: itssona on December 26, 2017, 10:59:53 am


Mod edit: Was authorised to fix up the LaTeX
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 26, 2017, 11:14:27 am


Mod edit: Was authorised to fix up the LaTeX

Personally, however, I hate doing division of coordinates because it can get confusing. (In a way, I need to think of things upside down.) So I'll do this instead.

________________________________________________


So on GeoGebra, you can consider \( f(x) = x^2+x-2 \) and \( g(x) = 1/f(x) \).

On GeoGebra, you can consider \( h(x) = 2x^2-x-3 \)

And then just multiply ordinates.

Remark: Again, division is probably recommended, because it's faster. I just prefer the less brain-work involved with this longer approach
Title: Re: 4U Maths Question Thread
Post by: maria.micale on December 28, 2017, 07:33:32 pm
Hi, is there a way to find the zeros of polynomials with complex coefficients that have a degree higher than 2? Because I am having trouble solving theses types of polynomials. I'm not sure really where to start with them. Thanks!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 28, 2017, 07:37:41 pm
Hi, is there a way to find the zeros of polynomials with complex coefficients that have a degree higher than 2? Because I am having trouble solving theses types of polynomials. I'm not sure really where to start with them. Thanks!
There is no concrete rule of thumb. It always depends on a question-by-question basis.
Title: Re: 4U Maths Question Thread
Post by: maria.micale on December 28, 2017, 09:17:05 pm
There is no concrete rule of thumb. It always depends on a question-by-question basis.


But like its not like with polynomials with real coefficients wear you can check if factors of the constant term are zeros right? Like there is nothing like that right?

I have this question,

Find the real numbers k such that z=ki is a root of the equation z^3 + (2+i)Z^2 + (2+2i)z +4 = 0. Hence or otherwise, find the three roots of the equation.

I subbed in ki where there was a z and ended up getting -ik^3 + (-2-i)k^2 + (2i-2)k +4 =0 and I am not sure how to find k from here.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 28, 2017, 09:20:40 pm

But like its not like with polynomials with real coefficients wear you can check if factors of the constant term are zeros right? Like there is nothing like that right?

I have this question,

Find the real numbers k such that z=ki is a root of the equation z^3 + (2+i)Z^2 + (2+2i)z +4 = 0. Hence or otherwise, find the three roots of the equation.

I subbed in ki where there was a z and ended up getting -ik^3 + (-2-i)k^2 + (2i-2)k +4 =0 and I am not sure how to find k from here.

Hint: The fact that \(k\) is real means that you can do some rearranging, and then equate the real and imaginary parts. The real parts should give you a quadratic...
Title: Re: 4U Maths Question Thread
Post by: maria.micale on December 29, 2017, 07:18:30 am
Hint: The fact that \(k\) is real means that you can do some rearranging, and then equate the real and imaginary parts. The real parts should give you a quadratic...


Alright thanks heaps Rui!!! I got it an everything now from doing what you said but now Im wondering how would you go about solving that polynomial I had before with complex coefficients where I just subbed in ki where there was a z? Could you tell me how you would solve it because I don't think I have ever tried one with complex coefficients that wasn't a quadratic or quartic that could be reduced to a quadratic.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 29, 2017, 10:04:33 am

Alright thanks heaps Rui!!! I got it an everything now from doing what you said but now Im wondering how would you go about solving that polynomial I had before with complex coefficients where I just subbed in ki where there was a z? Could you tell me how you would solve it because I don't think I have ever tried one with complex coefficients that wasn't a quadratic or quartic that could be reduced to a quadratic.
Oh, I wouldn't have.

There's no giveaway at all when it's just some arbitrary cubic or quartic. The HSC always gives you some kind of hint to make sure everything else falls out nicely, but in general there's no 'clean' way of solving a quartic whatsoever (try Google searching the quartic formula, it's bizarre).
Title: Re: 4U Maths Question Thread
Post by: stesoo on December 29, 2017, 10:49:38 am
hi i have a question  ;D
how can I prove that (1+sintheta +icostheta)/(1+sintheta-icostheta) = sintheta + icostheta
thank you
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 29, 2017, 11:17:53 am
hi i have a question  ;D
how can I prove that (1+sintheta +icostheta)/(1+sintheta-icostheta) = sintheta + icostheta
thank you
\begin{align*}\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}&= \frac{1+i\text{ cis }(-\theta)}{1-i\text{ cis }\theta}\\ &= \frac{\text{ cis }\left(-\frac{\theta}{2}\right)}{\text{ cis }\frac{\theta}{2}}\times \frac{\text{ cis }\frac{\theta}{2}+i\text{ cis }\left(-\frac\theta2\right)}{\text{ cis }\left(-\frac\theta2\right)-i\text{ cis }\frac\theta2}\\ &= \text{cis }(-\theta)\times \frac{\cos \frac\theta2+i\sin \frac\theta2 + i \left(\cos \frac\theta2-i\sin \frac\theta2\right)}{\cos \frac\theta2-i\sin \frac\theta2-i\left(\cos\frac\theta2+i\sin \frac\theta2\right)}\\ &= \text{cis }(-\theta) \times \frac{(1+i)\left(\cos \frac\theta2+\sin \frac\theta2\right)}{(1-i)\left(\cos \frac\theta2+\sin \frac\theta2\right)}\\ &= (\cos\theta - i\sin \theta) \times i\\ &= \sin \theta + i \cos \theta\end{align*}
Title: Re: 4U Maths Question Thread
Post by: Jeeffffffffffffff on December 29, 2017, 01:30:20 pm
Find constants a,b,c such that the polynomial p(x) = x3-6x2+11x-13 is expressible as X3+aX+b where X=x-c. Hence show that the equation p(x)=0 has only one root.
I found a, b, and c quite easily but unsure of how to show that the equation has only one root?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 29, 2017, 01:46:39 pm
Find constants a,b,c such that the polynomial p(x) = x3-6x2+11x-13 is expressible as X3+aX+b where X=x-c. Hence show that the equation p(x)=0 has only one root.
I found a, b, and c quite easily but unsure of how to show that the equation has only one root?

You should be well aware of the fact that this is not 4U material as it requires you to pick the values for \(a, b\) and \(c\) yourself, without any guidance whatsoever.

________________________________________________________


Title: Re: 4U Maths Question Thread
Post by: maria.micale on December 29, 2017, 03:10:40 pm
Hi I am not sure where to begin with these questions.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 29, 2017, 03:39:59 pm
Deleted materials






The rest should be doable similarly (except maybe part d) - I'm not sure if that requires a different approach but I don't have time to think about it just yet). A remark that for part b), an assumption is made that none of the roots are 0. This won't be necessary for the other parts.




This approach should certainly be adaptable for part d). It is essentially equivalent to performing a \(u\)-substitution, where \(u = \frac1x\), and then working from there.
Title: Re: 4U Maths Question Thread
Post by: maria.micale on December 29, 2017, 07:51:25 pm

You should be well aware of the fact that this is not 4U material as it requires you to pick the values for \(a, b\) and \(c\) yourself, without any guidance whatsoever.

________________________________________________________




How do we know that p(x-c) is simply a horizontal translation of p(x)?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 29, 2017, 08:02:54 pm
How do we know that p(x-c) is simply a horizontal translation of p(x)?

And thus, the point \(( x_0, f(x_0) )\) lies on \(y=f(x)\).


Hence, the point \( (x_0+c, f(x_0) ) \) lies on \(y=f(x-c) \).

Title: Re: 4U Maths Question Thread
Post by: maria.micale on January 03, 2018, 11:47:10 am
Hi, I've done part a) and the answer is c = 1/27 a(9b -2a^2) but I really don't understand how to go about b). The answer is C - k = A/27 (9(B-m) -2A^2 ) fixed point has x-coordinate -A/3. Thanks in advanced!!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 03, 2018, 12:03:17 pm
Hi, I've done part a) and the answer is c = 1/27 a(9b -2a^2) but I really don't understand how to go about b). The answer is C - k = A/27 (9(B-m) -2A^2 ) fixed point has x-coordinate -A/3. Thanks in advanced!!!
I'll start you off. This certainly involves using part a) somehow.


Title: Re: 4U Maths Question Thread
Post by: maria.micale on January 04, 2018, 12:44:22 pm
I'll start you off. This certainly involves using part a) somehow.




Ok thanks Rui, that makes sense, any further hints?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 04, 2018, 01:05:34 pm
Ok thanks Rui, that makes sense, any further hints?
So that gives you the weird expression involving \(C-k\) and according to your answers we can stop there. Which is reasonable, because we now have something that links \(m\) and \(k\) together.

The second point is actually the fixed point. If you let the first, second and third points have \(x\)-coordinate \(\alpha-d, \alpha, \alpha+d\), you'll see that \(\alpha = -\frac{A}{3} \)
Title: Re: 4U Maths Question Thread
Post by: Jeeffffffffffffff on January 04, 2018, 05:46:20 pm
Use De Moivre’s theorem to express cos5theta and sin5theta in powers of costheta and sintheta. Hence, express tan5theta as a rational function of t where t=tantheta and deduce that tan(Pi/5)tan(2pi/5)tan(3pi/5)tan(4pi/5)=5.

I’ve gotten up to having
Tan5theta= (t5-10t3+5t)/(5t4-10t2+1)
But I don’t really know how I’m supposed to use that to deduce the last line in the question. Any help would be appreciated, thanks.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 04, 2018, 06:09:29 pm
Use De Moivre’s theorem to express cos5theta and sin5theta in powers of costheta and sintheta. Hence, express tan5theta as a rational function of t where t=tantheta and deduce that tan(Pi/5)tan(2pi/5)tan(3pi/5)tan(4pi/5)=5.

I’ve gotten up to having
Tan5theta= (t5-10t3+5t)/(5t4-10t2+1)
But I don’t really know how I’m supposed to use that to deduce the last line in the question. Any help would be appreciated, thanks.

The expression is basically your numerator from the previous part.

Note that this is safe, because we haven't divided by 0 at the "important" points. This is as a consequence of the fact that the numerator and denominator have no common factors.




Title: Re: 4U Maths Question Thread
Post by: RustyWasTaken on January 05, 2018, 08:13:04 pm
b) How do I identify which roots belong to z^6 + z^3 + 1 = 0
Cheers :D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 05, 2018, 10:37:34 pm
b) How do I identify which roots belong to z^6 + z^3 + 1 = 0
Cheers :D


These should be \(\text{ cis }\frac{2\pi}{9},\text{ cis }\frac{4\pi}{9},\text{ cis }\frac{8\pi}{9},\text{ cis }\frac{10\pi}{9},\text{ cis }\frac{14\pi}{9},\text{ cis }\frac{16\pi}{9} \)
Title: Re: 4U Maths Question Thread
Post by: maria.micale on January 06, 2018, 11:38:56 am
So that gives you the weird expression involving \(C-k\) and according to your answers we can stop there. Which is reasonable, because we now have something that links \(m\) and \(k\) together.

The second point is actually the fixed point. If you let the first, second and third points have \(x\)-coordinate \(\alpha-d, \alpha, \alpha+d\), you'll see that \(\alpha = -\frac{A}{3} \)

I think I’m confused, could you please explain further because I don’t know how to get to the answer still. Thanks :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 06, 2018, 11:42:01 am


Title: Re: 4U Maths Question Thread
Post by: maria.micale on January 06, 2018, 12:19:43 pm
Ok thanks heaps Rui!!!
Title: Re: 4U Maths Question Thread
Post by: Lefkiiii6 on January 06, 2018, 12:23:53 pm




Sorry I'm looking at this question too and I don't understand how to find that c-k = A/27 (9(B-m) - 2A^2)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 06, 2018, 12:41:48 pm
Sorry I'm looking at this question too and I don't understand how to find that c-k = A/27 (9(B-m) - 2A^2)
That part was addressed a few posts back when she first posted this question
Title: Re: 4U Maths Question Thread
Post by: Lefkiiii6 on January 06, 2018, 01:20:45 pm
That part was addressed a few posts back when she first posted this question

Yeah I read this but I am still not sure.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 06, 2018, 02:17:50 pm
Yeah I read this but I am still not sure.
Please elaborate further on the confusion.

(Pretty much all I did was consider what Maria had proven in part a), and demonstrated the link between that and part b). At that point, all that was left was basically just substituting a = A, b = B-m, c = C-k)
Title: Re: 4U Maths Question Thread
Post by: itssona on January 07, 2018, 06:04:25 pm
how would you go about graphing x^3 + y^3 =1
as in what process would you follow
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 07, 2018, 09:11:40 pm
how would you go about graphing x^3 + y^3 =1
as in what process would you follow


i.e. a S.P. at \( (0,1) \) and a point where the derivative is undefined at \( (1,0) \).

As opposed to the semicircle \( y=(1-x^2)^{1/2} \). Square rooting a negative number is problematic but cube rooting is not.

Think about it visually. If you graphed \(y=1\) it's just gonna be a horizontal line, because a constant doesn't change. If you graphed \(y=x^3\) however, that thing is gonna eventually blow up. Whatever blows up will have a much more significant impact on anything that does not blow up, as you let \(x\) become really really large.


Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 11, 2018, 02:01:46 pm
Hey Sona, I wanna come back to this one. I've found the formal way to prove where the asymptote is.
how would you go about graphing x^3 + y^3 =1
as in what process would you follow



Of course, it's also a bit advanced. But at least it's a legit proof now.
____________________________________________


____________________________________________



Note: \( \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}} \) and similarly \( \sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}\), which has been used here. Some steps could've been skipped if you're really good at tackling algebra.
Title: Re: 4U Maths Question Thread
Post by: RustyWasTaken on January 16, 2018, 01:05:36 pm
Cheers :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 16, 2018, 01:18:56 pm
Cheers :)



(https://i.imgur.com/aRtsfS1.png)
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on January 17, 2018, 07:26:05 pm
Hey Jake and all 4U lovers. Could I please get some help with some 4U graphing.
I'm quite confused on approaching the absolute value one (| |y|-|x| |=1) and I'd prefer to learn how to do the second one (y=x^3/(x^2-9)) without using calculus.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 17, 2018, 07:36:13 pm
Hey Jake and all 4U lovers. Could I please get some help with some 4U graphing.
I'm quite confused on approaching the absolute value one (| |y|-|x| |=1) and I'd prefer to learn how to do the second one (y=x^3/(x^2-9)) without using calculus.




This should be enough information to sketch the curve.
_______________________________________________

Recall key aspects of the graph of \(y = \frac{1}{f(x)} \). \(x\)-intercepts and vertical asymptotes interchange, the nature of stationary points are inverted and etc.

Of course, multiplication of ordinates will not guarantee a fully accurate graph. But it is the best compromise in only 4U methods.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 17, 2018, 07:52:30 pm

Definition




So in the first quadrant, you will sketch the pair of parallel lines \( y = x+1\) and \(y = x-1\).

So in the second quadrant, you will sketch the pair of parallel lines \(y=-x-1\) and \(y=-x+1\)

So in the third quadrant, sketch the same pair of parallel lines as in the first quadrant. i.e. \(y=x+1\) and \(y=x-1\).

Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on January 19, 2018, 08:33:36 pm
Wow Rui your breakdown of the absolute value one and treating them for 4 separate quadrants was really helpful, I actually didn't use this method but I realise now how working backwards from the definition can make it much easier. The other question I prefer your method of finding the reciprocal of the parabola and multiplying y=x^3 thats much easier to visualise and do on the spot. Thank you heaps this will be pretty darn useful so much appreciated.
Title: Re: 4U Maths Question Thread
Post by: philgee on January 20, 2018, 06:00:31 pm
Hey Rui! Not sure how to prove that Re(alpha)=1.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 20, 2018, 06:03:41 pm
Hey Jake! Not sure how to prove that Re(alpha)=1.



Note: \( z + \overline{z} = 2 \text{Re}(z) \) is a known result from complex numbers. It can easily be proven by letting \( z = x+iy\).
Title: Re: 4U Maths Question Thread
Post by: philgee on January 20, 2018, 06:27:27 pm
Thanks Rui! Your explanation was clear and concise!
Title: Re: 4U Maths Question Thread
Post by: philgee on January 20, 2018, 08:27:00 pm
Hi rui!
I couldnt find the roots for this question

Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 20, 2018, 09:56:50 pm
Hi rui!
I couldnt find the roots for this question


deleted materials
I got onto this later than when I first saw it because I was in the middle of an online meeting - soz

I'm still working on it, and having done some backtracking I realised that \(k=8\) and the roots are \(\pm \frac12, 1\pm \frac{i}2\). But all the ways that actually got me to the answer are pretty disastrous and not elegant, so I don't wanna post them just yet. Before I give up and just post something inelegant, can I get the source of the question?
I ended up finding a solution that's still not elegant, but at least it's less handwavy. I don't believe this was the intended approach but it will suffice.





From here, it's easy to finish the question off.

This particular approach is nice in that we don't have to find \(k\). But I'm somewhat unsure on if we're not meant to find \(k\) or if we actually were.
Title: Re: 4U Maths Question Thread
Post by: Jeeffffffffffffff on January 24, 2018, 01:42:37 pm
Use the fact that a2+b2>=2ab to prove that:
i) (ab+xy)(ax+by)>=4abxy
ii) ax+by<=1 if a2+b2=1 and x2+y2=1

Just starting out with this stuff still and I think our teacher is just as confused as us, any help would be appreciated, thanks :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 24, 2018, 02:03:51 pm
Use the fact that a2+b2>=2ab to prove that:
i) (ab+xy)(ax+by)>=4abxy
ii) ax+by<=1 if a2+b2=1 and x2+y2=1

Just starting out with this stuff still and I think our teacher is just as confused as us, any help would be appreciated, thanks :)

\begin{align*}(ab+xy)(ax+by)&=a^2bx+bxy^2+ab^2y+ayx^2\\ &= (a^2bx+bxy^2) + (ab^2y+ayx^2)\\ &= bx(a^2+y^2)+ay(b^2+x^2)\\ &\ge bx(2ay) + ay(2bx)\tag{using the given inequality}\\ &= 4abxy\end{align*}
Intuition behind this method - To use the identity, somehow I had to force out a bunch of squares. But it wasn't obvious what I could do until I had expanded it out in the first place.
__________________________________________

Note that I was only able to generate this proof after working backwards on scrap paper.


Title: Re: 4U Maths Question Thread
Post by: Dragomistress on January 24, 2018, 04:34:41 pm
May someone explain what has happened?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 24, 2018, 04:37:00 pm
May someone explain what has happened?
They just used the compound angle identities.
Title: Re: 4U Maths Question Thread
Post by: maria.micale on January 24, 2018, 06:10:54 pm
I've got this question from Fitzpatrick that I can't solve.

Given that x2 + y2 + z2 >= xy + yz + zx

Show that:

1) a2b2 + b2c2 + c2a2 >= abc(a +b +c)
2) (x + y + z)2 >= 3(xy + yz + zx)

Thanks in advanced!!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 24, 2018, 06:16:00 pm
I've got this question from Fitzpatrick that I can't solve.

Given that x2 + y2 + z2 >= xy + yz + zx

Show that:

1) a2b2 + b2c2 + c2a2 >= abc(a +b +c)
2) (x + y + z)2 >= 3(xy + yz + zx)

Thanks in advanced!!!


___________________________________

(this is actually a formula you used several times in the polynomials topic: \( \alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2-2(\alpha\beta+\alpha\gamma+\beta\gamma) )
Title: Re: 4U Maths Question Thread
Post by: RustyWasTaken on January 25, 2018, 12:31:40 pm
Cheeers  :'(
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 25, 2018, 01:05:51 pm
Cheeers  :'(
This question lacks information - it is not stated whether \(a\) and \(b\) are real or not.
Title: Re: 4U Maths Question Thread
Post by: philgee on January 25, 2018, 01:23:40 pm
Hello how would i go about finding the integral of the last question?

Title: Re: 4U Maths Question Thread
Post by: RustyWasTaken on January 25, 2018, 01:29:16 pm
This question lacks information - it is not stated whether \(a\) and \(b\) are real or not.

Oh. The answer from the textbook for a and b is real.
a = -2 and b =-2
I cant get the ans though :(
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 25, 2018, 02:02:09 pm
Hello how would i go about finding the integral of the last question?


This technically is not 4U material as improper integrals are not in the course. But having said that, many bits and pieces will be provided as it's quite easily doable regardless.
(https://i.imgur.com/jXTvbH3.png)



Extra: The function provided is actually the hyperbolic tan function \( f(x) = \tanh x\)
Oh. The answer from the textbook for a and b is real.
a = -2 and b =-2
I cant get the ans though :(




Title: Re: 4U Maths Question Thread
Post by: philgee on January 25, 2018, 02:18:15 pm
Dear Rui, i dont get why why it is 1- the original curve inside the integral.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 25, 2018, 02:23:32 pm
Dear Rui, i dont get why why it is 1- the original curve inside the integral.
Like mentioned, area between two curves.

If we just considered \(\int_0^\infty \frac{e^x+e^{-x}}{e^{x}+e^{-x}}\,dx \), we would've been considering the region that's merely below the curve. Not the region sandwiched between the curve and its asymptote.

(In fact, that integral would be infinite, because that area would just blow up.)
Title: Re: 4U Maths Question Thread
Post by: philgee on January 25, 2018, 02:26:52 pm
ohhhh okay thank you i understand now!
Title: Re: 4U Maths Question Thread
Post by: clovvy on January 25, 2018, 11:19:01 pm
This question is weird:
solve the equation t^4+8t^3-6t^2-8t=1 correct to 3 decimal places.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 25, 2018, 11:23:27 pm
This question is weird:
solve the equation t^4+8t^3-6t^2-8t=1 correct to 3 decimal places.
I’m at the gym right now but here’s some thoughts upon first glance.

Step 1. Divide both sides by t^2.
Step 2. Make the substitution u = t - 1/t. If that doesn’t work, try t + 1/t
Title: Re: 4U Maths Question Thread
Post by: clovvy on January 25, 2018, 11:31:57 pm
This question is weird:
solve the equation t^4+8t^3-6t^2-8t=1 correct to 3 decimal places.

God that's a typo, it is t^4+8t^3-6t^2-8t+1=0
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 26, 2018, 12:27:30 am
God that's a typo, it is t^4+8t^3-6t^2-8t+1=0

______________________




Note that the intuition behind the substitution \( u = t - \frac{1}{t} \) was essentially dependent on the symmetric nature of the coefficients (of the original equation). This kind of intuition isn't expected for a 4U student.
Title: Re: 4U Maths Question Thread
Post by: clovvy on January 26, 2018, 09:10:23 pm

______________________




Note that the intuition behind the substitution \( u = t - \frac{1}{t} \) was essentially dependent on the symmetric nature of the coefficients (of the original equation). This kind of intuition isn't expected for a 4U student.

I tried doing this again on paper and for some reason it didn't work..., the u sub I tried on the equation, I get the  u^2 but I did not get 8u...
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 26, 2018, 09:18:19 pm
I tried doing this again on paper and for some reason it didn't work..., the u sub I tried on the equation, I get the  u^2 but I did not get 8u...


Edit: Typo
Title: Re: 4U Maths Question Thread
Post by: Dragomistress on January 27, 2018, 09:46:36 pm
How should I do this?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 27, 2018, 10:01:41 pm
How should I do this?
Sketch the locus given and then deduce the max and min values of \( \arg(z) \). Recall that \( \arg(z) \) measures the angle made at 0+0i with respect to the positive real axis.

Note that the argument is optimised (i.e. maximised and minimised) whenever the ray drawn is tangential to the circle you sketched earlier.
Title: Re: 4U Maths Question Thread
Post by: clovvy on January 27, 2018, 10:04:38 pm
This is a locus question, the first thing you want to do is sketch the locus.... Iz-2iI=1... in this case let z=x+iy and then square root both sides than you will notice that it is a circle with radius 1 at the centre (0,2)...... and then to find arg(z)... Rui I can't remember exactly how to do that... -_-
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 28, 2018, 07:24:31 pm
Deleted a previous reply - Now that I have more time I've decided to just do this one. Having said that though, still use the 4U trial survival handout as a reference point.

It may be worth noting that if we had \( |z-2i|\le 1 \), for this type of question we would ignore the disc's interior. This is because the minimum and maximum of both \( |z| \) and \(\arg z\) can only be attained on the disc's boundary, i.e. the actual circle. It just so happens that for the argument, we only consider when it's tangential to the circle.
(https://i.imgur.com/19TcOK9.png)

Note, however, that the max can also be found from symmetry, once you know the min,


_________________________________________________


Title: Re: 4U Maths Question Thread
Post by: RustyWasTaken on January 29, 2018, 01:32:28 am
Cheers
  :'(
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 29, 2018, 09:22:04 am
Cheers
  :'(
Title: Re: 4U Maths Question Thread
Post by: Jeeffffffffffffff on January 31, 2018, 01:34:45 pm
Show that if x>0, y>0 then
1/x2+1/y2>=8/(x+y)2

Please help, fully worked solution would be appreciated.
Title: Re: 4U Maths Question Thread
Post by: maria.micale on January 31, 2018, 01:36:33 pm
Using expansion (x+ y + z)3, show that (x + y + z)3 >= 27xyz

I've written out expansion but cannot solve question. Any help appreciated.
Title: Re: 4U Maths Question Thread
Post by: Opengangs on January 31, 2018, 01:48:43 pm
Show that if x>0, y>0 then
1/x2+1/y2>=8/(x+y)2

Please help, fully worked solution would be appreciated.


Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 31, 2018, 01:49:06 pm
Show that if x>0, y>0 then
1/x2+1/y2>=8/(x+y)2

Please help, fully worked solution would be appreciated.





To provide some intuition:

Working backwards did not help at all here. As a more advanced technique, you should have proven \( a+b \ge 2\sqrt{ab} \) several times if you're up to questions of this calibre, so just assume that it's true and work with it. This was beneficial in collapsing the \(x^{-2}+y^{-2} \) into \( 2(xy)^{-1} \).

Now that we have something more manageable, we can work backwards again. It turns out that THIS time, working backwards actually got something relatively nice. So we use it.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 31, 2018, 01:54:16 pm
Using expansion (x+ y + z)3, show that (x + y + z)3 >= 27xyz

I've written out expansion but cannot solve question. Any help appreciated.

\begin{align*}&\quad (x+y+z)^3\\&= x^3+y^3+z^3 + 3(x^2y+x^2z+y^2x+y^2z+z^2x+z^2y) + 6xyz\\ &= (x^3+y^3+z^3) + 3z(x^2+y^2) + 3y(x^2+z^2) + 3x(y^2+z^2) + 6xyz\\ &\ge (x^3+y^3+z^3) + 6zxy + 6yxz + 6xyz + 6xyz\\ &= x^3+y^3+z^3 + 24xyz \end{align*}

This is really just the AM-GM inequality with three variables. The AM-GM inequality with four variables is easy to prove, and you can use that one to prove the version with three variables.


To provide some intuition: Since we have to expand a cube involving three terms to prove something with \(xyz\) in the RHS, it was already extremely likely that \( a+b+c\ge 3\sqrt[3]{abc} \) would be useful (by simply letting \(a = x^3\) and etc.). But this is quite tedious to prove, and there's still no guarantee that it'd do all the magic. So I started by considering every other term.

The terms \(x^2y + xy^2 + x^2z+xz^2+y^2z+yz^2 \) all took the form something-squared plus something. Just like in the above case, \(a+b \ge 2\sqrt{ab} \) could have been really useful. But I didn't want renegade square roots appearing because there's no square root at all in what I'm trying to prove.
Since a square and square root cancel each other out when dealing with positive numbers, it made sense to apply that on the squares instead. Hence, I used \(x^2+y^2\ge 2xy \) instead of \( x+y \ge 2\sqrt{xy} \).

But of course, I had to rearrange the terms and do some factorising before I could even think of using that formula.

It was from there, I saw that I suddenly had almost all of the \(xyz\) in place (I already had 24 of them, and only needed 3 more). Well, the last 3 takes effort to get at, but at least I know I could get there
Title: Re: 4U Maths Question Thread
Post by: Lefkiiii6 on January 31, 2018, 02:18:49 pm
Prove that, if a>0, b>0, a4 + b4 > a3b + ab3 by writing expansion (a-b)4

Using x3+y3>=(x/z+y/z)xyz and similar expressions for y3+z3 and z3+x3 to deduce that x3+y3+z3>=3xyz.


These are two questions I am having trouble with. Can anyone help?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 31, 2018, 02:24:20 pm
Prove that, if a>0, b>0, a4 + b4 > a3b + ab3 by writing expansion (a-b)4

Using x3+y3>=(x/z+y/z)xyz and similar expressions for y3+z3 and z3+x3 to deduce that x3+y3+z3>=3xyz.


These are two questions I am having trouble with. Can anyone help?

That second one is another famous way of proving the three-variable AM-GM inequality.




_______________________________________________________________



Title: Re: 4U Maths Question Thread
Post by: Lefkiiii6 on January 31, 2018, 09:14:40 pm
Ok thanks heaps for all that Rui and all your help in the past! I have another question. We didn't really cover cube roots of unity or roots of unity in general so I'm unsure about this stuff.

If 1, w1 and w2 are the cube roots of unity, prove that:

 w1 = conjugate of w2 = conjugate of (w2)2
w1 + w2 = -1
w1w2=1
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 01, 2018, 01:26:32 pm
Ok thanks heaps for all that Rui and all your help in the past! I have another question. We didn't really cover cube roots of unity or roots of unity in general so I'm unsure about this stuff.

If 1, w1 and w2 are the cube roots of unity, prove that:

 w1 = conjugate of w2 = conjugate of (w2)2 - Also, this bit is a mistake. It is equal to just \(w_2\,^2\); no conjugate.
w1 + w2 = -1
w1w2=1

Essentially, what the "cube roots of unity" means is just the "cube roots of 1". Hence the equation above.


Alternately, if we choose to use polynomials, and thus not explicitly evaluate w1 and w2



Also, you can just square \(w_2\) by hand, and you will see that it equals to \(w_1\).
________________________________________

(where \(w = w_1\))



Again, alternatively if we choose to use polynomials
This is easy to prove by just using the sum of roots.

The same goes for the product of roots for part 3.
The third part can be done by observing that \(w_1 w_2 = w w^2 = w^3 = 1\)

Edit: Tbh some bits of that might’ve been explained poorly. The thing to realise, however, is that w2^2 = w1 underlies a lot of the things we do with the cube roots of unity. This is why we usually never use w1 and w2, but prefer w and w^2 instead.
Title: Re: 4U Maths Question Thread
Post by: stesoo on February 01, 2018, 09:20:53 pm
Hi i need help with this q,
for a real number r, the polynomial 8x^3-4x^2-42x+45 is divisible by (x-r)^2. Find the value of r.

thanks
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 02, 2018, 02:43:45 pm
Hi i need help with this q,
for a real number r, the polynomial 8x^3-4x^2-42x+45 is divisible by (x-r)^2. Find the value of r.

thanks





Title: Re: 4U Maths Question Thread
Post by: Lefkiiii6 on February 02, 2018, 07:54:20 pm
Hi, what are the techniques used to graph y = f(x2)?

Thankyou!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 02, 2018, 08:57:21 pm
Hi, what are the techniques used to graph y = f(x2)?

Thankyou!!
When what's inside the \(f\) gets altered there usually isn't any strategy. I just rely on common sense.

Common sense here suggests that I break up the cases for positive \(x\) and negative \(x\) first. (We can put \(x=0\) with the positive ones or just do that one separately.)

I have \(y=f(x)\) sitting in front of me. I now do \(y=f(x^2)\). So if I put in \(x=2\), I don't get \(f(2)\) anymore, but rather \(f(4)\).
When I put in \(x=5\), I don't get \(f(5)\) anymore, but rather \(f(25)\). Effectively speaking, I'm getting something further away from the \(x\)-axis closer to it

So if \(x > 1\), I'm really just squeezing the graph inwards. But I'm squeezing it at a faster rate as \(x\) grows bigger.

Then, I could do a similar analogy with \( 0 < x < 1\). If i'm careful enough, I'll see that the graph expands outwards from the \(y\)-axis.

And of course, since at \(x=0\) we have \(f(x^2)=f(0)\) as well, and similarly for \(x=1\), nothing changes between \(y=f(x)\) and \(y=f(x^2)\) for these two specific values.

You won't be able to get a perfectly accurate sketch, because you're just squeezing things. But you should be able to see the idea after you do some simulations on GeoGebra/Desmos.

And lastly, for negative \(x\)? That is just a matter of reflecting whatever you've done for positive \(x\) over into the left of the \(y\)-axis as well. You should be able to see why that's the case.
Title: Re: 4U Maths Question Thread
Post by: RustyWasTaken on February 04, 2018, 03:51:47 pm
Cheers :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 04, 2018, 04:55:05 pm
Cheers :)
Hint: Just quote \( PS = ePM\), where in your case \(e = \frac35 \)
Title: Re: 4U Maths Question Thread
Post by: clovvy on February 04, 2018, 06:24:12 pm
Show that cos(2kpi/7), where k=1,2,3,4,5,6 are roots of the equation 64x^6-64x^5-48x^4-48x^3+8x^2+8x+1=0
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 06, 2018, 11:56:54 am
Show that cos(2kpi/7), where k=1,2,3,4,5,6 are roots of the equation 64x^6-64x^5-48x^4-48x^3+8x^2+8x+1=0
I just had a brief at this and they're not. \( \cos \frac{2\pi}7 \) does not go to 0 when I plug it into that thing.
Title: Re: 4U Maths Question Thread
Post by: clovvy on February 06, 2018, 01:27:14 pm
I just had a brief at this and they're not. \( \cos \frac{2\pi}7 \) does not go to 0 when I plug it into that thing.
there is a k after pi
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 06, 2018, 01:46:38 pm
there is a k after pi
Yeah.

I put \(k=1\) in, because it was a value of \(k\) listed. And it didn't work. (I haven't explicitly tested all of the other values of \(k\), but it is very likely that there will be more values of \(k\) that won't work either, just because of how these types of questions were designed in the first place.)
Title: Re: 4U Maths Question Thread
Post by: beau77bro on February 07, 2018, 11:30:22 pm
Trying to help a friend. This question(2a) is ridiculous tho. I've broken it up into two integral of negative inverse cot and inverse tan but I can't work out wat to use as the separating boundary. (1 works but I don't know why)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 08, 2018, 09:18:29 am
Trying to help a friend. This question(2a) is ridiculous tho. I've broken it up into two integral of negative inverse cot and inverse tan but I can't work out wat to use as the separating boundary. (1 works but I don't know why)



Or you could've skipped the "let y" bit and just recognised \(\tan^{-1}(\cot x)=\tan^{-1}\left(\tan \left(\frac\pi2-x\right)\right) \) and realised that \(0 < x < \frac\pi2\implies 0 < \frac\pi2-x < \frac\pi2 \) as well.
Title: Re: 4U Maths Question Thread
Post by: radnan11 on February 15, 2018, 01:43:51 pm
Being trying this question for the last two days, always getting it wrong

"Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region enclosed within the circle (x-1)2 + y2 = 1 about the y axis."

from Cambridge 4u
Title: Re: 4U Maths Question Thread
Post by: jazzycab on February 15, 2018, 02:43:23 pm
Being trying this question for the last two days, always getting it wrong

"Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region enclosed within the circle (x-1)2 + y2 = 1 about the y axis."

from Cambridge 4u

If we're rotating about the y-axis, the radius of each cylinder is the x-value.
When finding the volume using the formula for volume of a solid, it is the rotated volume between the curve and an axis. In this particular case, the region being rotated is the region within the circle (not between the curve an the y-axis, due to the translation from the base circle 1 unit right) - giving an annulus (donut-shape). However, the 'hole' of the annulus is technically infinitesimally small due to the fact that there is an x-intercept at the origin.
Thus, to find the volume required, we need to find the volume between the right-hand side of the circle and the y-axis and subtract the volume obtained between the left-hand side of the circle and the y-axis:

The left half of the circle has the equation

Where as the right half of the circle has the equation

The volume between the right half and the axis is (the terminals are the upper and lower limits of the range of the circle):

The volume between the left half and the axis is:

The total volume is therefore:

This integrand can be evaluated by using a trig substitution:


I'm Victorian, so I'm not familiar with the 4U HSC study design, so not sure how much of this you are expected to know how to do and whether you are taught other techniques or 'standard' integrals to evaluate the integral in this question.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 15, 2018, 03:03:34 pm
Being trying this question for the last two days, always getting it wrong

"Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region enclosed within the circle (x-1)2 + y2 = 1 about the y axis."

from Cambridge 4u
(https://i.imgur.com/qd0VaxHh.jpg)
Title: Re: 4U Maths Question Thread
Post by: jazzycab on February 15, 2018, 03:23:58 pm
Or RuiAce's solution. Much neater and more succinct than mine
Title: Re: 4U Maths Question Thread
Post by: radnan11 on February 15, 2018, 03:40:23 pm
Ahh i got most of RuiAce's solution but couldn't get the second last line of working. Thanks :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 15, 2018, 04:25:14 pm
Ahh i got most of RuiAce's solution but couldn't get the second last line of working. Thanks :)
If you meant the harder-to-see red marker, basically the first integral represents the integral of an odd function and the second represents the area of a semi-circle with radius 1. (Line above is just splitting the integral up in case I misinterpreted). Let me know if anything else was hard to see or etc
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on February 17, 2018, 03:41:50 pm
Hey Rui and all maths lovers i saw this integral online and im really trying to crack it but its just too complicated. Please give it a try if you can.
 Intergrate [ (ln (x) + 1/x) e^x ] dx
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 17, 2018, 03:50:40 pm
Hey Rui and all maths lovers i saw this integral online and im really trying to crack it but its just too complicated. Please give it a try if you can.
 Intergrate [ (ln (x) + 1/x) e^x ] dx

\begin{align*}\frac{d}{dx} e^x\ln x &= e^x \ln x + \frac{e^x}{x}\\ \therefore \int \left( \ln x + \frac{1}{x} \right)e^x\, dx &= e^x \ln x + C\end{align*}
________________________________________________________


I think this particular integral popped up quite recently in the MIT integration bee?
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on February 19, 2018, 03:17:07 am
Oh wow reversing from product rule made it look really simple. I'll be honest this question was from a qualifiers test for the mit intergration bee a friend showed me. I also noticed ,when looking at your 2nd working backwards, that expanding and rearranging is more insightful. Thanks heaps and you wont mind me asking for a solution for integral of (e^x sinx)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 19, 2018, 07:42:37 am
Sounds about right; I think I did all of those qualifier questions recently. Some MIT questions are a bit too far but this particular one could be done by a 4U student. Anyway, as for \( \int e^x\sin x\,dx \) you're gonna be expected to know how to do that one.
\begin{align*}\text{Let }I&=\int e^x\sin x\,dx \\ &= -e^x\cos x + \int e^x\cos x\,dx\\ &= -e^x\cos x + e^x\sin x - \int e^x\sin x\,dx\\ &= e^x(\sin x - \cos x) -
 I.\\ \therefore 2I &= e^x (\cos x - \sin x) + 2C\tag{*}\\ I &= e^x(\cos x - \sin x ) + C\end{align*}

Conventionally, the formal justification is just to say "statement is true to within a constant"
Title: Re: 4U Maths Question Thread
Post by: arii on February 22, 2018, 10:34:21 am
I need help with this question..
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 22, 2018, 10:43:30 am
I need help with this question..



Of course, this also means that \( g(x) \) is concave up, but we don't really need this.



Title: Re: 4U Maths Question Thread
Post by: arii on February 22, 2018, 11:14:39 am



Of course, this also means that \( g(x) \) is concave up, but we don't really need this.





Yeah there was a part before which I weirdly proved but I think your method might be better if I can get my head around the first bit.

Proved:
sinx > x
but as soon as you start rearranging...
sinx - x > 0
-(x-sinx) > 0
x-sinx < 0 (sign flips because you divide both sides by -1?)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 22, 2018, 11:17:38 am
Yeah there was a part before which I weirdly proved but I think your method might be better if I can get my head around the first bit.

Proved:
sinx > x
but as soon as you start rearranging...
sinx - x > 0
-(x-sinx) > 0
x-sinx < 0 (sign flips because you divide both sides by -1?)

Typo (I'll fix it now), sin x is definitely less than x.
Title: Re: 4U Maths Question Thread
Post by: arii on February 22, 2018, 07:17:33 pm
Hey Rui (and other math students),

Can you check if I did (a - first box) correctly and help me with (b - second box)?

Consider the graph of the function y=sqrtx
Spelling typo in picture: Curve not Cruve...
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 22, 2018, 08:08:43 pm
Hey Rui (and other math students),

Can you check if I did (a - first box) correctly and help me with (b - second box)?

Consider the graph of the function y=sqrtx
Spelling typo in picture: Curve not Cruve...

\( y\) increasing means \( y^\prime > 0\). \(y^{\prime\prime} > 0\) means \(y^\prime\) is increasing and \(y\) is concave up. So what you had to solve was \(\frac12 x^{-1/2} > 0 \).
___________________________________________
Firstly, two useful results.

The idea that \(f^\prime(x) \ge 0\) is only a way to actually prove that \(f(x)\) is increasing. It's not, bydefinition, what increasing actually means.






___________________________________________

Remark: This question could be done a tiny bit more quickly if it said "hence or otherwise". We can just sketch \( y = \sqrt{x} \) and draw a bunch of rectangles above it.
Title: Re: 4U Maths Question Thread
Post by: arii on February 22, 2018, 09:01:00 pm
Thank you so much Rui (you're very likely going to be saving my 4U life throughout the year).

I just wanted to ask you about the last section of your working where you start off as:
RHS = (given)
        = (how did you get this... the pro-numerals went from n to x a bit sudden)
        = MHS (given in the question)"
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 22, 2018, 09:10:17 pm
Thank you so much Rui (you're very likely going to be saving my 4U life throughout the year).

I just wanted to ask you about the last section of your working where you start off as:
RHS = (given)
        = (how did you get this... the pro-numerals went from n to x a bit sudden)
        = MHS (given in the question)"
Title: Re: 4U Maths Question Thread
Post by: arii on February 23, 2018, 05:12:37 pm
Not sure how to continue here...

Normally I'd try to factorise out sqrt{k+1} but it seems to complicate it even more...
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 23, 2018, 07:55:47 pm
Not sure how to continue here...

Normally I'd try to factorise out sqrt{k+1} but it seems to complicate it even more...

Working backwards




Title: Re: 4U Maths Question Thread
Post by: justwannawish on February 23, 2018, 10:06:30 pm
Hey,

any general types on drawing y=ln(e^f(x)) where a graph of f(x) is provided for you?

also drawing e^ln(f(x))
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 23, 2018, 10:12:06 pm
Hey,

any general types on drawing y=ln(e^f(x)) where a graph of f(x) is provided for you?

also drawing e^ln(f(x))
That's literally just going to be \( y = f(x) \) in both cases.

(Although the latter has an extra restriction - \( \ln (e^\alpha)= \alpha\) always, but \(e^{\ln \alpha} = \alpha\) only when \(\alpha > 0\), just because of the domain of \(\ln x\) being \( x > 0\))
Title: Re: 4U Maths Question Thread
Post by: amaher on February 24, 2018, 02:44:24 pm
Hi!!! I'm struggling with a conics q about ellipses -
M is the midpoint of PQ where P and Q lie on x2/a2 + y2/b2 = 1. O is the centre of the ellipse. Show that the tangents at P and Q intersect on OM produced.
Thank you to anyone that helps!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 24, 2018, 04:23:51 pm
Hi!!! I'm struggling with a conics q about ellipses -
M is the midpoint of PQ where P and Q lie on x2/a2 + y2/b2 = 1. O is the centre of the ellipse. Show that the tangents at P and Q intersect on OM produced.
Thank you to anyone that helps!!


_________________________________________________________


_________________________________________________________

Note that the intuition to do this was via working backwards. Here, the details in working backwards will be omitted.



Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on February 26, 2018, 01:00:09 am
Hey Rui please help with another conic question if you're free.
Two tangents to the hyperbola 4x^2-y^2=36 intersect at (0,4). Find the coordinates for the point(s) on the hyperbola for this to occur.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 26, 2018, 07:31:28 am
Hey Rui please help with another conic question if you're free.
Two tangents to the hyperbola 4x^2-y^2=36 intersect at (0,4). Find the coordinates for the point(s) on the hyperbola for this to occur.
I'll come back to this with a proper answer in a few hours time. Basically the idea is that you start by just finding the equation of the chord of contact from (0,4), then locating the points of intersection between the chord of contact and the hyperbola

Edit:
\[\text{The usual method to find the chord of contact}\\ \text{can be found in the previous lecture slides.}\]
(Or alternatively in the book.)


Note: The values of \(a\) and \(b\) were obtained because we can only infer them when the RHS of the equation is 1. So the equation of the hyperbola had to be rearranged into \( \frac{x^2}{9} - \frac{y^2}{36} = 1\)

And thus our points are \( \left(\pm\frac{3\sqrt{13}}2,-9 \right) \)
Title: Re: 4U Maths Question Thread
Post by: jazzycab on February 26, 2018, 11:20:54 am
Hey Rui please help with another conic question if you're free.
Two tangents to the hyperbola 4x^2-y^2=36 intersect at (0,4). Find the coordinates for the point(s) on the hyperbola for this to occur.
Let's say that the solution points are at \(\left(\alpha,\beta\right)\ \left(1\right)\). From the equation of the curve, we get:

Secondly, we can find the gradients of the tangents by implicitly differentiating the hyperbola:

Substituting in \(\left(1\right)\) and \(\left(2\right)\) gives the gradients:

The equation of each tangent line is given by \(y_T=m\left(x-\alpha\right)+\beta\). We can subsequently substitute in \(\left(1\right)\), \(\left(2\right)\) and \(\left(3\right)\):

We know that the tangent lines pass through \(\left(0,4\right)\):

Substituting \(\left(5\right)\) into \(\left(1\right)\) gives:

Given that I have squared and square rooted both sides of equations a significant number of times, we will need to check what combinations of \(\left(\alpha,\beta\right)\) are correct. Let's look at the tangent line equation, \(\left(4\right)\), which passes through \(\left(0,4\right)\):

From \(\left(7\right)\) we can see that \(\alpha=\frac{3\sqrt{13}}{2}\Rightarrow\beta=-9\). Now let's investigate for the other \(x\)-value:

Note that \(\left(9\right)\) is identical to \(\left(8\right)\) so it has the same solutions. Thus, \(\alpha=-\frac{3\sqrt{13}}{2}\Rightarrow\beta=-9\). Hence, the two points on the hyperbola whose tangents intersect at \(\left(0,4\right)\) are:
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on February 26, 2018, 08:08:18 pm
 Holy crap me, Jassycab, you broke it down so wonderfully, my only fear is that in an exam i may use the chord of contact method because its a bit simpler but forming a tangent and subbing in the values is pretty darn insightful.
Title: Re: 4U Maths Question Thread
Post by: jazzycab on February 26, 2018, 09:29:52 pm
Holy crap me, Jassycab, you broke it down so wonderfully, my only fear is that in an exam i may use the chord of contact method because its a bit simpler but forming a tangent and subbing in the values is pretty darn insightful.

We don't look at the chord of contact method in the VCE, so I just used what I could from the VCE Mathematics syallbi
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on February 27, 2018, 09:34:06 am
Hey maths lovers another conics question if you guys have time.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 27, 2018, 10:52:13 am
Hey maths lovers another conics question if you guys have time.

i.e. the points \( (1,3) \) and \( (9,3) \).

_______________________________________

_______________________________________



The length of the major axis is 10, and the length of the minor axis is 6.
(https://i.imgur.com/rge3jKM.png)
_______________________________________


For the ellipse, the boundary is pretty much just going to be the ellipse itself.


_______________________________________

(Note: We only care about the right-side of the x-axis. The reason we don't care about the left side is simply because there just is no ellipse there.)


More examples of types of problems like part iii)
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on February 28, 2018, 11:32:56 pm
Oh sweet the arg part i can see clearly it's angle made by the x and y axis to the +ve x-axis, the diagram makes it much clear.  I'd like to request help for 2 more questions though, again conics.
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on March 01, 2018, 12:01:40 am
Oh for question 12 i finally got a sorta solution please advise me if it's a preferable method.

first I prove a^2m^2+b^2=k^2, k is the y intercept of the tangent, (in the attached image)
and simply sub in a= 5, b=4 and m=1 and solve for k to put in eq. y=x+k
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 01, 2018, 12:49:52 am
Oh sweet the arg part i can see clearly it's angle made by the x and y axis to the +ve x-axis, the diagram makes it much clear.  I'd like to request help for 2 more questions though, again conics.
Just from memory I'm pretty sure you arrived at the best method for Q12. (I remember seeing that \(a^2m^2+b^2=k^2\) thing, or something very similar to it quite a lot.)


_____________________________________________________
\begin{align*}&\quad \frac{x^2}{a^2}-\frac{y^2}{b^2}\\ &= \frac14 \left[(\sec\theta+\sec\phi)^2 - (\tan\theta + \tan\phi)^2 \right]\\ &= \frac14 \big[ (\sec^2\theta-\tan^2\theta) + (\sec^2\phi-\tan^2\phi) \\ &\quad \quad+ 2(\sec\theta\sec\phi - \tan\theta\tan\phi)\big]\\ &= \frac12 \left[ 1 +\sec\theta\sec\phi-\tan\theta\tan\phi \right]\end{align*}
\begin{align*}&\therefore \frac{x^2}{a^2}-\frac{y^2}{b^2} - \frac{y}{b}\\ &= \frac12 (1+\sec\theta\sec\phi-\tan\theta\tan\phi-\tan\theta-\tan\phi)\\ &= \frac12 \left( \frac{\cos\theta\cos\phi+1-\sin\theta\sin\phi-\sin\theta\cos\phi-\cos\theta\sin\phi}{\cos\theta\cos\phi} \right)\\ &= \frac12 \left(\frac{\cos(\theta+\phi) +1- \sin(\theta+\phi)}{\cos\theta\cos\phi}\right)\\ &=\frac12 \left(\frac{\cos \frac\pi2 + 1 - \sin \frac\pi2}{\cos\theta\cos\phi} \right) \\ &= \frac12 \left( \frac{0 + 1 - 1}{\cos\theta\cos\phi} \right)\\ &= 0\end{align*}
\[ \therefore \boxed{\frac{x^2}{a^2}-\frac{y^2}{b^2} = \frac{y}{b}}\text{ as required.} \]
Note: You are not expected to solve locus problems for the ordinary hyperbola (and ellipse) for the HSC.
Title: Re: 4U Maths Question Thread
Post by: arii on March 01, 2018, 07:32:20 pm
If un = 2n+2 + 32n+1, show that un+1 = 2un + 7*32n+1. Hence, show that un is divisible by 7 for n≥1.
Title: Re: 4U Maths Question Thread
Post by: jazzycab on March 01, 2018, 09:01:46 pm
If un = 2n+2 + 32n+1, show that un+1 = 2un + 7*32n+1. Hence, show that un is divisible by 7 for n≥1.

If \(u_n=2^{n+2}+3^{2n+1}\) then:
. We can split the second term into \(2\times3^{2n+1}+7\times3^{2n+1}\):

For the second part, we can prove this using induction:
Title: Re: 4U Maths Question Thread
Post by: Lefkiiii6 on March 02, 2018, 08:41:28 pm
How do you graph y=ef(x) when given the graph of y=f(x). I think it is simply dealing what is below x-axis and then everything else is kind of higher up than original f(x) but thats just by looking at desmos and looking at some graphs

Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 02, 2018, 08:46:20 pm
How do you graph y=ef(x) when given the graph of y=f(x). I think it is simply dealing what is below x-axis and then everything else is kind of higher up than original f(x) but thats just by looking at desmos and looking at some graphs








These are just some examples of heaps of things you can consider.
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on March 03, 2018, 09:05:11 pm
A bit of help with complex polynomial division please.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 03, 2018, 09:11:17 pm
A bit of help with complex polynomial division please.

(https://i.imgur.com/7CZbk3i.png)
(For now, I'll presume you know how to finish off the remainder of the question.)

(Note that LaTeX "might've" done polynomial long division a bit differently to what you're used to, with the formatting.) Alternatively, if you view my lecture slides for the half yearly lectures, there's a different method I introduce that avoids polynomial long division altogether.
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on March 04, 2018, 01:39:52 pm
Ohh damn Rui i completely overlooked the fact that if the coefficients are real than the conjugate is also a root. I actually attempted long division with complex divisor (my teacher did something similar in class) but it was horrendously long.
Yikes i do feel a bit dumb now for forgetting a good amount of what I'd noted in the slides, thanks heaps for sharing em I realised the equalities & equating coefficients method is hella lot cleaner.
Title: Re: 4U Maths Question Thread
Post by: justwannawish on March 04, 2018, 08:36:41 pm
Hi, could you plese help me out with this question?

Given that arg(z+1)=pi/6 and arg(z-1)=2pi/3, write z in the form x+iy where x and y are real numbers


Thank you!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 04, 2018, 08:52:18 pm
Hi, could you plese help me out with this question?

Given that arg(z+1)=pi/6 and arg(z-1)=2pi/3, write z in the form x+iy where x and y are real numbers


Thank you!

(https://i.imgur.com/CDuZWhQ.png)



\begin{align*}\therefore z &= 1 \left( \cos \frac\pi3 + i \sin \frac\pi3\right)\\ &= \frac12 + \frac{\sqrt3}{2}i\end{align*}
Title: Re: 4U Maths Question Thread
Post by: Lefkiiii6 on March 05, 2018, 05:53:59 pm
Hi I came across this in the 2001 HSC paper and I don't know what to do and can't find solutions. Can anyone assist?



Hi, I'm also unsure about this question. Any help appreciated. Thankyou!  :)

Mod edit: Merged. Please use the "Modify" button at times like this to refrain from chain posting.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 05, 2018, 06:58:53 pm
Hi I came across this in the 2001 HSC paper and I don't know what to do and can't find solutions. Can anyone assist?



Hi, I'm also unsure about this question. Any help appreciated. Thankyou!  :)

Mod edit: Merged. Please use the "Modify" button at times like this to refrain from chain posting.
Please find the solution for the latter linked in the compilation. (The first one will be added shortly.)



Carefully note that the minus sign here can be safely ignored, because -1 is a factor of every integer and thus is not really relevant.



________________________




____________________________________________________________



________________________



________________________


____________________________________________________________

(Note that this deduce part is actually not linked, notation wise, to the above. We just happen to use the same letter.)

Essentially, case 1 is equivalent to saying (*) actually has a rational root.

________________________




____________________________________________________________
Title: Re: 4U Maths Question Thread
Post by: Dragomistress on March 06, 2018, 05:54:46 pm
How do I do this question?
There are 11 people and 2 circular tables, one has 5 seats and the other has 6 seats. How many arrangements are possible?
Title: Re: 4U Maths Question Thread
Post by: roygbivmagic on March 06, 2018, 06:17:25 pm
Hi, can you please explain how to do this question?
Prove that (x + y)^2 >= 4xy
Hence prove that 1/x^2 + 1/y^2 >= 4/(x^2 + y^2)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 06, 2018, 06:24:22 pm
How do I do this question?
There are 11 people and 2 circular tables, one has 5 seats and the other has 6 seats. How many arrangements are possible?


Alternatively, the answer is also \( \binom{11}{6}5! \binom55 4! \) if we choose the people for the second table first.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 06, 2018, 06:32:31 pm
Hi, can you please explain how to do this question?
Prove that (x + y)^2 >= 4xy
Hence prove that 1/x^2 + 1/y^2 >= 4/(x^2 + y^2)

_____________________________________________________________________________________



Title: Re: 4U Maths Question Thread
Post by: justwannawish on March 07, 2018, 06:08:44 pm

(https://i.imgur.com/CDuZWhQ.png)



\begin{align*}\therefore z &= 1 \left( \cos \frac\pi3 + i \sin \frac\pi3\right)\\ &= \frac12 + \frac{\sqrt3}{2}i\end{align*}

how would we do the question if the args didn't add up to pi/2 and we couldn't use the circle method
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 07, 2018, 07:58:52 pm
how would we do the question if the args didn't add up to pi/2 and we couldn't use the circle method
This is actually a lot harder in general. Whenever you encounter questions like this, usually there's a trick or two behind it all. In general, however, you may need to brute force it and actually find the equation of the ray, before going back to 2U methods for the point of intersection.
Title: Re: 4U Maths Question Thread
Post by: justwannawish on March 07, 2018, 09:42:17 pm
This is actually a lot harder in general. Whenever you encounter questions like this, usually there's a trick or two behind it all. In general, however, you may need to brute force it and actually find the equation of the ray, before going back to 2U methods for the point of intersection.

:/ That sounds tricky haha. If you don't mind, could you walk me through finding the equation of the ray? e.g. if the question was like arg(z-1)=pi/4, arg(z+1)=pi/3 find arg(z)

Also, what are some of the really common complex number qs that one must now. From looking at past trials, finding the roots of unity and factorising over complex/real fields, finding square roots and then solving a quadratic are often the easy marks at the beginning of the test. Can you think of any others, especially to do with vectors (e.g. finding locus)

Thank you!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 08, 2018, 12:06:59 am
:/ That sounds tricky haha. If you don't mind, could you walk me through finding the equation of the ray? e.g. if the question was like arg(z-1)=pi/4, arg(z+1)=pi/3 find arg(z)

Also, what are some of the really common complex number qs that one must now. From looking at past trials, finding the roots of unity and factorising over complex/real fields, finding square roots and then solving a quadratic are often the easy marks at the beginning of the test. Can you think of any others, especially to do with vectors (e.g. finding locus)

Thank you!
The method itself is actually not too bad. Take \( \arg (z-1) = \frac\pi4 \). The ray will just be one part of the line, through the point \( (1,0) \), with gradient \( m = \tan \frac\pi4 = 1 \). So using the point gradient form, you just have \( y = 1(x-1) \).

Having said that, whilst that's easy enough, if you did the same thing for \( \arg (z+1) = \frac\pi3 \) you would have \( y = \sqrt{3}(x + 1) \). And then you see the main problem - that \( \sqrt3\) is gonna be hella annoying. It's easy enough to go back to 2U here, and just do simultaneous equations to solve for the point of intersection. But would I expect a 4U student to want to deal with surds for this? Probably not.

Worse, they could move it away from -1 and 1 and give you a random complex number like \( -\sqrt{2} + i\). Or they could swap the angles into something which you'd have to do a bit of work to compute the exact values for.
________________________________________________________________________________

Quite a lot of the common ones are all included in my 4U notes book. Vector addition itself is uncommon altogether. For stuff like locus, there are many curves that you'd be expected to know how to draw. On top of that, it should be fairly easy to sketch regions in the Argand plane based off them.

A somewhat common culprit for the trials is to write down the minimum value of \( \arg z \), or \( |z| \), or the real part or whatever, given that you've already sketched its locus. Sometimes they'll swap that out for the maximum value as well. These were covered in my trial lectures last year and you should look at the notes section for a few examples on those.

A lot of the more popular ones tend to overlap into the polynomials topic as well.
Title: Re: 4U Maths Question Thread
Post by: arii on March 08, 2018, 10:45:11 pm
Hi Rui and other users, I'm not too sure how to do (b) and (c).. I feel like (b) might be related to cyclic quad through constructing another circle but I haven't had much luck doing it.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 08, 2018, 11:36:10 pm
Hi Rui and other users, I'm not too sure how to do (b) and (c).. I feel like (b) might be related to cyclic quad through constructing another circle but I haven't had much luck doing it.
Also with part a), you really should invoke that RASD is a cyclic quadrilateral first (reason: the angles you specified are supplementary), before you draw the circle around the points.


(In this case, to the right of \(DC\)).

Note that the theorem used here is essentially the converse of the "angles subtended to the circumference on the same arc are equal" theorem. It is also one of the reasons why I hate maths in focus, because they don't teach this theorem.

______________________________________________________________________

\begin{align*}\pi - \angle DST &= \angle DCT \tag{part ii}\\ &= \angle DAR \tag{ext. angle of cyclic quad}\\ &= \angle DSR \tag{part i}\end{align*}
Title: Re: 4U Maths Question Thread
Post by: arii on March 09, 2018, 12:11:39 am
Also with part a), you really should invoke that RASD is a cyclic quadrilateral first (reason: the angles you specified are supplementary), before you draw the circle around the points.


(In this case, to the right of \(DC\)).

Note that the theorem used here is essentially the converse of the "angles subtended to the circumference on the same arc are equal" theorem. It is also one of the reasons why I hate maths in focus, because they don't teach this theorem.

______________________________________________________________________

\begin{align*}\pi - angle DST &= \angle DST \tag{part ii}\\ &= \angle DAR \tag{ext. angle of cyclic quad}\\ &= \angle DSR \tag{part i}\end{align*}


Thank you for your reply. I wanted to ask about the third-to-second-last line for the solution to part (c). I don't understand how Angle DST = Angle DAR through the external angle of cyclic quad. DAR is the external angle to the cyclic quad, but DST is not one of the four corners in cyclic quad ABCD, it's kinda just floating in the middle... I think the external angle of cyclic quad means Angle DAR = Angle DCB?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 09, 2018, 08:23:53 am
Thank you for your reply. I wanted to ask about the third-to-second-last line for the solution to part (c). I don't understand how Angle DST = Angle DAR through the external angle of cyclic quad. DAR is the external angle to the cyclic quad, but DST is not one of the four corners in cyclic quad ABCD, it's kinda just floating in the middle... I think the external angle of cyclic quad means Angle DAR = Angle DCB?
Typo's from doing it before I went to sleep. Fixed
Title: Re: 4U Maths Question Thread
Post by: arii on March 10, 2018, 02:34:56 am
Hi Rui (and other users), so this one is a bit of a challenge.. I'm having some trouble actually visualising what is descriptively told to us so if someone can include a diagram, that'd be amazing.
Title: Re: 4U Maths Question Thread
Post by: jazzycab on March 10, 2018, 08:12:37 am
Hi Rui (and other users), so this one is a bit of a challenge.. I'm having some trouble actually visualising what is descriptively told to us so if someone can include a diagram, that'd be amazing.
This is what the image looks like:
(https://i.imgur.com/JcaKTti.png)
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on March 11, 2018, 06:46:29 pm
Hey there guys need some help with polynomials, in the first que i get that 2 roots are complex conugates as all coefficient are real and to have 2 turning points there must be a negative gradient (thus c<0) but the rest is tough to get.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 11, 2018, 07:06:14 pm
Hey there guys need some help with polynomials, in the first que i get that 2 roots are complex conugates as all coefficient are real and to have 2 turning points there must be a negative gradient (thus c<0) but the rest is tough to get.
What are the questions?
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on March 12, 2018, 02:53:37 am
Ohhhh so sorry i completely forgot the ques my bad. Here they are.

And please a bit more help with another hard conics question im having trouble with.
NP is the ordinate of a point P(x1,y1) on the hyperbola x^2/a^2 - y^2/b^2=1. The tangent at p meets the x axis at T. Prove thata (OT)(ON)=a^2, where O is the origin.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 12, 2018, 04:50:31 pm
Ohhhh so sorry i completely forgot the ques my bad. Here they are.

And please a bit more help with another hard conics question im having trouble with.
NP is the ordinate of a point P(x1,y1) on the hyperbola x^2/a^2 - y^2/b^2=1. The tangent at p meets the x axis at T. Prove thata (OT)(ON)=a^2, where O is the origin.
PDF attached. As stated, will get back to part a) of Q15 later. (I wasn't really able to use the internet at the time of writing up a response.)

I'm also not sure what the wording of your conics question suggests. The 'ordinate' is just the y-coordinate of a point, yet it states \(NP\), which is a line (more accurately a segment). The ordinate of the point \(P\) is just \(y_1\).
___________________________________________
EDIT




GeoGebra simulation



Note that this is guaranteed, because \( x^3-3x+k \) is guaranteed to never have a triple root.
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on March 12, 2018, 10:03:38 pm
oh i see what the k does it shifts the graph up and that restricts the the cubic to intersect the x axis more than once. Thank you heaps Rui deeply appreciate it even if you had no internet. The other conics que i asked is from sk patel (6E q8)  textbook, take your time and thank you for helping us out.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 12, 2018, 10:30:10 pm
oh i see what the k does it shifts the graph up and that restricts the the cubic to intersect the x axis more than once. Thank you heaps Rui deeply appreciate it even if you had no internet. The other conics que i asked is from sk patel (6E q8)  textbook, take your time and thank you for helping us out.
(No worries :) - Pretty much PDFs are a go-to option if I'm trying to answer it without access to a computer with internet, because I can still use the full LaTeX package. The other option is I pull out a whiteboard, which I didn't have at the time either)

I wonder what SK Patel was thinking when he wrote that question; that makes no sense in the literal manner. I've reinterpreted it to be something like this:




I feel as though the wording of the question is what's hard. Not the actual maths involved
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on March 14, 2018, 06:30:44 am
I'd probs agree with you that the question is very ambiguous (especially after seeing the working) but in mosts part i see it just finding the x intercept with the tangent (i learnt how to derive that before) and the x value of P. Thank you again, I usually do homework late at night any my friend are are probs sleeping so I really appreciate your help.
Title: Re: 4U Maths Question Thread
Post by: arii on March 14, 2018, 10:38:39 pm
I'm not sure how I'd find the gradient for this question:

Find the equation of the tangent to the curve sqrt(x)+sqrt(y)=sqrt(c) at the point P(a,b) on the curve.
Title: Re: 4U Maths Question Thread
Post by: Sine on March 14, 2018, 10:59:01 pm
I'm not sure how I'd find the gradient for this question:

Find the equation of the tangent to the curve sqrt(x)+sqrt(y)=sqrt(c) at the point P(a,b) on the curve.
Title: Re: 4U Maths Question Thread
Post by: arii on March 15, 2018, 10:33:37 pm
Thanks Sine. I managed to successfully figure out the rest.
Anyway, bumped into another question I wasn't sure how to do...

Prove that x^3+y^3=3xy is symmetrical about the line y=x
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 15, 2018, 11:26:05 pm
Thanks Sine. I managed to successfully figure out the rest.
Anyway, bumped into another question I wasn't sure how to do...

Prove that x^3+y^3=3xy is symmetrical about the line y=x
You proved something was symmetrical about y=0 (even function) by just showing that the replacement of x with -x didn't change the given equation. You do the same thing here by showing that if you replace x with y (and also y with x), it's symmetrical about the line y=x.

Which is obviously the case, because upon doing this replacement you get \( y^3 + x^3 = 3yx \) which rearranges quickly back into the original equation.

Nor is this a part of the 4U syllabus.
Title: Re: 4U Maths Question Thread
Post by: Dragomistress on March 16, 2018, 02:31:49 pm
How do you do this?
|2iy|≤4 (Not absolute value)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 16, 2018, 03:33:23 pm
How do you do this?
|2iy|≤4 (Not absolute value)
If that's not an absolute value then I actually don't know what that means. You can't have \(i\) appearing in an equality by itself.

(Modulus and absolute value mean the same thing if that's what you intended to mean)
Title: Re: 4U Maths Question Thread
Post by: Dragomistress on March 17, 2018, 08:38:12 am
The question is sketch |z-w|≤4 (w is conjugate of z)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 17, 2018, 09:25:38 am
The question is sketch |z-w|≤4 (w is conjugate of z)
That is the absolute value then. (As mentioned above, the modulus and the absolute value mean the same thing.)
\begin{align*}|2iy|&\le 4\\ 2y &\le 4\\ y&\le 2 \end{align*}
Title: Re: 4U Maths Question Thread
Post by: Dragomistress on March 17, 2018, 10:01:48 am
But apparently, it is -2≤y≤2
Title: Re: 4U Maths Question Thread
Post by: jazzycab on March 17, 2018, 12:04:28 pm
But apparently, it is -2≤y≤2


The solution to this is \(-2\le y\le 2\) (look at a graph of \(y=x^2-4\) to see why)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 17, 2018, 12:16:48 pm
But apparently, it is -2≤y≤2
Oh my bad. Technically speaking I should have left it as |y|≤2 Which solves to give -2≤y≤2 (it’s  an absolute value inequality).

In general, one can assume that |i|=1 but one can not assume |y|=y. This holds only when y is non negative, which is something we cannot confirm.
Title: Re: 4U Maths Question Thread
Post by: arii on March 23, 2018, 06:32:00 pm
Find the general solutions of the 1/cos(3x)=1/sin(2x). I got up to the line attached, but I'm not sure if I've lost solutions on the way. (Still haven't figured out how to input fractions and stuff despite Rui's guide)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 23, 2018, 06:59:07 pm
Find the general solutions of the 1/cos(3x)=1/sin(2x). I got up to the line attached, but I'm not sure if I've lost solutions on the way. (Still haven't figured out how to input fractions and stuff despite Rui's guide)
Code: [Select]
[tex]\frac{a}{b}[/tex]
Any further questions should be asked over in that thread
___________________________________________________


At no point anywhere else in the proof did you attempt to 'cancel things out', so that was the only instance that was risky.
Title: Re: 4U Maths Question Thread
Post by: arii on March 23, 2018, 07:50:30 pm


At no point anywhere else in the proof did you attempt to 'cancel things out', so that was the only instance that was risky.

I wasn't entirely sure how to continue from that last line in order to get to the answer they were expecting.

The answers had it as:


but technically, I only have the first two solutions...
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 23, 2018, 08:23:50 pm
I wasn't entirely sure how to continue from that last line in order to get to the answer they were expecting.

The answers had it as:


but technically, I only have the first two solutions...
Hint: Double/Triple angles were overkill. Just use the identity \( \cos \alpha = \sin \left( \frac\pi2 - \alpha \right) \)
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on March 25, 2018, 01:02:15 pm
Hey guys need some help with confusing trig questions.

Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 25, 2018, 01:22:12 pm
Hey guys need some help with confusing trig questions.

The question is to solve for x where 0< x < 2pi
Hints: the second one is a quadratic in tan(x) after expanding and dividing everything by cos^2(x). The last one is just 4sin^2(2x)=1.

I’ll get back to the first one when I’m at home, but if you’re stuck on any of the equations you should post up your progress thus far.
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on March 25, 2018, 01:46:52 pm
oh yeah i see how u used double angel formula for sin x.
And also thank you i got the quadratic after making the 5 cos^2x + sin^2x =3sin2x into 4 cos^2x +1= 6 sinx cosx  and like u said divide by cos^2x
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 25, 2018, 04:47:31 pm
(https://i.imgur.com/p1jMgd0l.jpg)

(Of course, when I say \(n\)-gon, I really mean a regular \(n\)-gon as stated in the question.)
Title: Re: 4U Maths Question Thread
Post by: clovvy on March 28, 2018, 03:55:15 pm
Question 13
Title: Re: 4U Maths Question Thread
Post by: jazzycab on March 28, 2018, 05:04:57 pm
Question 13

Where is Question 13?
Title: Re: 4U Maths Question Thread
Post by: clovvy on March 28, 2018, 05:43:56 pm
Lol just realised the number was covered, well the extension question where I have to proof by de moivre that sin(2n+1)= some nasty expression
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 28, 2018, 05:51:49 pm


Title: Re: 4U Maths Question Thread
Post by: arii on March 28, 2018, 10:14:02 pm
I need help with (c). I've put what I've found in (a) and (b) since the question says "Hence".
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 28, 2018, 10:41:28 pm
I need help with (c). I've put what I've found in (a) and (b) since the question says "Hence".




Title: Re: 4U Maths Question Thread
Post by: clovvy on March 29, 2018, 11:41:32 am
Question 13
The extension question have multiple parts, the sin(2n+1) is part a, part b with the cot^2 is easy, however I don't know how to do part c and d which is the last two questions
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 29, 2018, 03:30:53 pm



______________________________________________________


Title: Re: 4U Maths Question Thread
Post by: clovvy on March 30, 2018, 04:55:00 pm
Question 24 is insane
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 30, 2018, 05:08:18 pm
Question 24 is insane
Nor is it useful to do and not worth the time and effort.

A small remark on part b) though - the radius will approach infinity, and the locus ends up being a straight line. Uncoincidentally, that straight line is the perpendicular bisector of the interval joining \(z_1\) and \(z_2\).

If you're interested in the basic ideas behind the geometric method, refer to the Wikipedia page linked.
Title: Re: 4U Maths Question Thread
Post by: clovvy on March 30, 2018, 05:28:40 pm
Nor is it useful to do and not worth the time and effort.

A small remark on part b) though - the radius will approach infinity, and the locus ends up being a straight line. Uncoincidentally, that straight line is the perpendicular bisector of the interval joining \(z_1\) and \(z_2\).

If you're interested in the basic ideas behind the geometric method, refer to the Wikipedia page linked.

Is this question beyond the course?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 30, 2018, 06:20:28 pm
Is this question beyond the course?
No.

On the other hand, it takes far too long to do and would essentially be an 8-mark question in the HSC, and thus virtually pointless.
Title: Re: 4U Maths Question Thread
Post by: aadharmg on March 31, 2018, 01:43:01 pm
I'm really struggling with this, but how can I express 2cis(5pi/12) - 2cis(3pi/4) in mod-arg form? I don't know how to simplify this further. Please help.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 31, 2018, 01:58:15 pm
I'm really struggling with this, but how can I express 2cis(5pi/12) - 2cis(3pi/4) in mod-arg form? I don't know how to simplify this further. Please help.
It's easily doable once we decide to use compound angles but I'm very sceptic about how you had a \( \frac{5\pi}{12} \) in there. Was this the entire question? (If not, send the previous parts.)
Title: Re: 4U Maths Question Thread
Post by: aadharmg on March 31, 2018, 02:16:07 pm
It's easily doable once we decide to use compound angles but I'm very sceptic about how you had a \( \frac{5\pi}{12} \) in there. Was this the entire question? (If not, send the previous parts.)
This is the entire question. I was able to cruise through the first two parts. For the third part I tried to simplify cis5pi/12 to normal x + iy form and do the same for the other complex number, then add them and work backwards to get the total mod-arg form but that led me down an extremely complicated path. Must have done something wrong on the way.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 31, 2018, 02:32:37 pm
This is the entire question. I was able to cruise through the first two parts. For the third part I tried to simplify cis5pi/12 to normal x + iy form and do the same for the other complex number, then add them and work backwards to get the total mod-arg form but that led me down an extremely complicated path. Must have done something wrong on the way.

Carefully note the negative. This is because we're doing a clockwise rotation. If the rotation were anticlockwise, then we'd still use \( +\frac\pi3 \).
\begin{align*}z_2 - z_1 &= z_2 \left( \cos \frac\pi3 + i \sin \frac\pi3 \right)\\ &= 2\left( \cos \frac{5\pi}{12} + i\sin \frac{5\pi}{12} \right)\left( \cos \left(-\frac\pi3\right) + i\sin \left(-\frac\pi3\right) \right)\\ &= 2 \left( \cos \frac\pi{12} + i\sin \frac\pi{12} \right)\end{align*}
Note that there is a slight subtlety in that \(z_2-z_1\) actually points from \(z_1\) to \(z_2\). This contributes to why the rotation is clockwise.
Title: Re: 4U Maths Question Thread
Post by: aadharmg on March 31, 2018, 02:53:24 pm

Carefully note the negative. This is because we're doing a clockwise rotation. If the rotation were anticlockwise, then we'd still use \( +\frac\pi3 \).
\begin{align*}z_2 - z_1 &= z_2 \left( \cos \frac\pi3 + i \sin \frac\pi3 \right)\\ &= 2\left( \cos \frac{5\pi}{12} + i\sin \frac{5\pi}{12} \right)\left( \cos \left(-\frac\pi3\right) + i\sin \left(-\frac\pi3\right) \right)\\ &= 2 \left( \cos \frac\pi{12} + i\sin \frac\pi{12} \right)\end{align*}
Note that there is a slight subtlety in that \(z_2-z_1\) actually points from \(z_1\) to \(z_2\). This contributes to why the rotation is clockwise.
OHHHH. Okay that makes a lot more sense. I just

Carefully note the negative. This is because we're doing a clockwise rotation. If the rotation were anticlockwise, then we'd still use \( +\frac\pi3 \).
\begin{align*}z_2 - z_1 &= z_2 \left( \cos \frac\pi3 + i \sin \frac\pi3 \right)\\ &= 2\left( \cos \frac{5\pi}{12} + i\sin \frac{5\pi}{12} \right)\left( \cos \left(-\frac\pi3\right) + i\sin \left(-\frac\pi3\right) \right)\\ &= 2 \left( \cos \frac\pi{12} + i\sin \frac\pi{12} \right)\end{align*}
Note that there is a slight subtlety in that \(z_2-z_1\) actually points from \(z_1\) to \(z_2\). This contributes to why the rotation is clockwise.
Ohhhhh. Okay that makes a lot more sense. I guess I never saw it in terms of the rotations but kept thinking about the expression algebraically. Thank you so much!
Title: Re: 4U Maths Question Thread
Post by: mxrylyn on April 01, 2018, 06:57:18 pm
This is not so much a maths Q as it is an advice Q.

I have been studying mx2 for an exam on wednesday. The last 4 days i have spent 8 hours a day on time set aside to study the four topics i get tested on (complex #'s, polynomials, conic and graphs).

Nothing is clicking and i cant do even the esiest of questions (apart from in complex #'s) . I dont want to drop the subject but i know im going to fail, and i hate failing.

Does anyone have any tips on a new way to apprach study for this subject? (My test is on tuesday so i only have one more day to study for it, its also worth 30% of my grade)



Title: Re: 4U Maths Question Thread
Post by: jakesilove on April 01, 2018, 07:37:27 pm
This is not so much a maths Q as it is an advice Q.

I have been studying mx2 for an exam on wednesday. The last 4 days i have spent 8 hours a day on time set aside to study the four topics i get tested on (complex #'s, polynomials, conic and graphs).

Nothing is clicking and i cant do even the esiest of questions (apart from in complex #'s) . I dont want to drop the subject but i know im going to fail, and i hate failing.

Does anyone have any tips on a new way to apprach study for this subject? (My test is on tuesday so i only have one more day to study for it, its also worth 30% of my grade)





Hey!

If you're doing 4U maths, that means you're good at maths. If you get the complex numbers section, then that's a really good start to understanding the rest of the 4U maths curriculum.

The fact is that 4U is totally different to every other HSC course. Usually, you'll be walking into exams knowing close to 100% of the content you'll be assessed on. Sure, you'll need to work out how exactly to answer particular questions, but for the most part you're just regurgitating.

4U is nothing like that. When you study 4U maths, you resign yourself to never being quite sure if you'll be able to answer ANY questions, or even understand what the question is getting at. It's a hard bloody subject, no doubt about it. Everyone finds it freakin' difficult, so just know that you're not alone.

Past papers, and looking at the answers to questions and trying to establish patterns for particular classes of questions, is really the best way to study in my opinion. Once you have a basic grasp of the content itself, just do a billion past papers. It sounds like you're doing that at the moment, and it doesn't feel like it's sinking in. However, I promise you that it is helping, even if only a little bit per question.

My advice at this point is to think about the TYPES of questions you'll be getting in your exam, and writing out a brief 'structure' to answer such a question. This is possible for topics like conics and graphs. Failing that, you could even just write out a list of tips and techniques you pick up from the answers to past questions. It's like developing a 'cheat sheet', which you can use to do past papers and make your life a whole lot easier.

Everyone struggles with 4U, even those ridiculously good at Maths. It sucks to feel the way you're feeling, but honestly I'm 100% sure you're better than you think you are. Just keep slugging away, and do your best to answer every question in your exam. Seriously good luck in your study!
Title: Re: 4U Maths Question Thread
Post by: clovvy on April 01, 2018, 09:20:08 pm
Hey!

If you're doing 4U maths, that means you're good at maths. If you get the complex numbers section, then that's a really good start to understanding the rest of the 4U maths curriculum.

The fact is that 4U is totally different to every other HSC course. Usually, you'll be walking into exams knowing close to 100% of the content you'll be assessed on. Sure, you'll need to work out how exactly to answer particular questions, but for the most part you're just regurgitating.

4U is nothing like that. When you study 4U maths, you resign yourself to never being quite sure if you'll be able to answer ANY questions, or even understand what the question is getting at. It's a hard bloody subject, no doubt about it. Everyone finds it freakin' difficult, so just know that you're not alone.

Past papers, and looking at the answers to questions and trying to establish patterns for particular classes of questions, is really the best way to study in my opinion. Once you have a basic grasp of the content itself, just do a billion past papers. It sounds like you're doing that at the moment, and it doesn't feel like it's sinking in. However, I promise you that it is helping, even if only a little bit per question.

My advice at this point is to think about the TYPES of questions you'll be getting in your exam, and writing out a brief 'structure' to answer such a question. This is possible for topics like conics and graphs. Failing that, you could even just write out a list of tips and techniques you pick up from the answers to past questions. It's like developing a 'cheat sheet', which you can use to do past papers and make your life a whole lot easier.

Everyone struggles with 4U, even those ridiculously good at Maths. It sucks to feel the way you're feeling, but honestly I'm 100% sure you're better than you think you are. Just keep slugging away, and do your best to answer every question in your exam. Seriously good luck in your study!

I agree, 4U is totally different, even if you know the contents well they can somehow chuck in questions that are unpredictable at times....  Me personally going through past papers I have seen LOTS of unpredictable questions for the topics that I am currently doing...

Contrary to public opinion of 4U students in general, my basics are actually very very poor and not up to scratch when I start my hsc, so that also causes me to struggle more as well.. Nonetheless I did enjoy the subject
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 01, 2018, 11:54:03 pm
Hey!
JAKE IS ALIVE.
Title: Re: 4U Maths Question Thread
Post by: mxrylyn on April 06, 2018, 07:45:13 am
Hey!

If you're doing 4U maths, that means you're good at maths. If you get the complex numbers section, then that's a really good start to understanding the rest of the 4U maths curriculum.

The fact is that 4U is totally different to every other HSC course. Usually, you'll be walking into exams knowing close to 100% of the content you'll be assessed on. Sure, you'll need to work out how exactly to answer particular questions, but for the most part you're just regurgitating.

4U is nothing like that. When you study 4U maths, you resign yourself to never being quite sure if you'll be able to answer ANY questions, or even understand what the question is getting at. It's a hard bloody subject, no doubt about it. Everyone finds it freakin' difficult, so just know that you're not alone.

Past papers, and looking at the answers to questions and trying to establish patterns for particular classes of questions, is really the best way to study in my opinion. Once you have a basic grasp of the content itself, just do a billion past papers. It sounds like you're doing that at the moment, and it doesn't feel like it's sinking in. However, I promise you that it is helping, even if only a little bit per question.

My advice at this point is to think about the TYPES of questions you'll be getting in your exam, and writing out a brief 'structure' to answer such a question. This is possible for topics like conics and graphs. Failing that, you could even just write out a list of tips and techniques you pick up from the answers to past questions. It's like developing a 'cheat sheet', which you can use to do past papers and make your life a whole lot easier.

Everyone struggles with 4U, even those ridiculously good at Maths. It sucks to feel the way you're feeling, but honestly I'm 100% sure you're better than you think you are. Just keep slugging away, and do your best to answer every question in your exam. Seriously good luck in your study!

Hey,
I'd like to thank you so so so SO much for your advice.
I'm the only student in my Mx2 class and by the time I posted this I had already texted my teacher to tell him I didn't want to do the exam (because I'm dramatic af).
I ended up taking a break for the night and focusing on consolidating what i knew before trying to learn extra stuff and i eneded up with 80%, which is not where i wanted to be, but i got 40% last term to its a definite step forward.

THANK YOU!
Title: Re: 4U Maths Question Thread
Post by: radnan11 on April 07, 2018, 03:52:19 pm
I don't understand what part b is asking for
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 07, 2018, 10:11:27 pm
I don't understand what part b is asking for
Suggestions regarding what its asking: Draw the line through the point P and the focus S. Let P' be where that line meets the ellipse again.

Then PP' is a focal chord. Now show that the tangents from P and from P' intersect on the directrix.

(I haven't attempted this question myself yet - no time)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 08, 2018, 09:14:30 pm
Hi, I'm stuck on this rates of change question:
A ferris wheel has radius 50m and the carriages move at a constant speed of 0.3m/s without stopping. Suppose a carriage is C, the base of the ferris wheel is B and the centre is O, and z = angle <BOC. The position of C can be represented by: [50sin(z),50cos(z)] i.e. coordinates x=50sin(z) and y=50cos(z). Also, the speed of C is given by the formula = sqrt((dx/dt)^2 + (dy/dt)^2).
a) Find dz/dt in radians per second.
b) How long does it take for carriage C to make one complete revolution? (in minutes)
(https://i.imgur.com/HGpaZ7Wl.jpg)

Part b should now be easy. If it takes 1 second to travel \(0.0006\) radians, then how long does it take to travel \(2\pi\) radians?
Title: Re: 4U Maths Question Thread
Post by: Dragomistress on April 11, 2018, 08:49:07 am
I am having trouble visualising this question.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 11, 2018, 08:56:59 am
I am having trouble visualising this question.
This is just proving the reflection property.
(https://i.imgur.com/X687O2u.png)
What you need to prove is that the highlighted angles are equal
Title: Re: 4U Maths Question Thread
Post by: arii on April 11, 2018, 06:20:21 pm
Could I get some help with this question please?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 11, 2018, 07:01:06 pm
Could I get some help with this question please?


Title: Re: 4U Maths Question Thread
Post by: arii on April 12, 2018, 09:05:16 pm
Found another question I wasn't too sure about... Am I mean't to use double sigma? But then I wouldn't know what values to set it equal to since that's normally used when it says something like find the coefficient of x...
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 12, 2018, 09:15:25 pm
Found another question I wasn't too sure about... Am I mean't to use double sigma? But then I wouldn't know what values to set it equal to since that's normally used when it says something like find the coefficient of x...
This was in one of the older 3U papers. It has already been addressed in the corresponding compilation.
Title: Re: 4U Maths Question Thread
Post by: arii on April 12, 2018, 09:55:23 pm
The same exact question was in some 3U paper? Hmm, not sure if I see it.. Some links keep redirecting me to solutions I wrote up for others when its under the binomial heading..
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 12, 2018, 09:56:15 pm
The same exact question was in some 3U paper? Hmm, not sure if I see it.. Some links keep redirecting me to solutions I wrote up for others when its under the binomial heading..
From memory it was in 2001
Title: Re: 4U Maths Question Thread
Post by: Opengangs on April 12, 2018, 09:57:57 pm
The same exact question was in some 3U paper? Hmm, not sure if I see it.. Some links keep redirecting me to solutions I wrote up for others when its under the binomial heading..
The explanation can be found here. :)
Title: Re: 4U Maths Question Thread
Post by: arii on April 12, 2018, 10:10:39 pm
The explanation can be found here. :)

Life saver. Thank you  :)
Title: Re: 4U Maths Question Thread
Post by: arii on April 13, 2018, 06:59:16 pm
...and I'm back.
Another binomial question.. (idk if it's past hsc or not)

I started finding the coefficient, there's n cases for the double sigma thing (as shown in the table, after I did the binomial expansion) for the given LHS which is (x+1)^n *(x+1)^n. But I wasn't able to condense it into a closed form since I don't know how big n is.. There must be some way around it.

EDIT:
For the first and last row of the last column: nCn * nC1 = n, NOT 1.
Title: Re: 4U Maths Question Thread
Post by: Opengangs on April 13, 2018, 07:25:10 pm


You didn't have to use double sigma notation*









Title: Re: 4U Maths Question Thread
Post by: radnan11 on April 15, 2018, 10:15:56 pm
Are we required to know the proofs for the intersection of tangents and normals of a rectangular hyperbola. My teacher said yes but noted that it had never come in an exam before but my tutor said no since it was too tedious to ask as a question. Whose right? :-\
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 15, 2018, 10:42:18 pm
Are we required to know the proofs for the intersection of tangents and normals of a rectangular hyperbola. My teacher said yes but noted that it had never come in an exam before but my tutor said no since it was too tedious to ask as a question. Whose right? :-\
That should be a proof that you can wing on the spot, not one that you should memorise.

Although I do agree with your teacher in that I doubt it has been examined before. The reason for this is ironically what your tutor says - tediousness.
Starting the intersection of tangents
\begin{align*}x+p^2y&=2cp\\ x+q^2y&=2cq\end{align*}

Title: Re: 4U Maths Question Thread
Post by: aadharmg on April 21, 2018, 06:26:45 pm
I'm trying to teach myself a bit of Volumes and in the first exercise I've kind of stumbled upon this question (highlighted in image), I took multiple routes in order to get to the answer but I think I went wrong somewhere each time in terms of my interpretation of the solid. There are no worked solutions/examples, hence I require your help. Thank You.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 21, 2018, 07:28:50 pm
I'm trying to teach myself a bit of Volumes and in the first exercise I've kind of stumbled upon this question (highlighted in image), I took multiple routes in order to get to the answer but I think I went wrong somewhere each time in terms of my interpretation of the solid. There are no worked solutions/examples, hence I require your help. Thank You.
(https://i.imgur.com/GORo5QPl.jpg)
Title: Re: 4U Maths Question Thread
Post by: aadharmg on April 21, 2018, 07:47:39 pm
Thank you so much!
Title: Re: 4U Maths Question Thread
Post by: Dragomistress on April 28, 2018, 05:14:08 pm
For conics, is it worth trying to remember the many equations? Or should they all be derived on the spot?

Also, do people derive the chord of contact?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 28, 2018, 05:28:39 pm
For conics, is it worth trying to remember the many equations? Or should they all be derived on the spot?

Also, do people derive the chord of contact?
The exam will either tell you to derive it on the spot (including the chord of contact, yes), or they'll just give you the formula.
Title: Re: 4U Maths Question Thread
Post by: vikasarkalgud on May 06, 2018, 01:03:50 pm
hey, need help for this mechanics questions. very close to the solution they provided but not exactly it. just need help with part d)

A submarine traavels underwater with a constant driving force of mF (where m is the mass of the submarine). Water pressure exerts a resistance to its motion of v^2 per unit mass ( v is velocity and v > 0)

a) Show that the equation of motion is d^2x/dt^2 = F - v^2 --------------------------Done
b) Show that the terminal velocity is sqrt(F)-------------------------------Done
c) The cruising speed is half the terminal velocity. How far does the submarine need to travel to reach its cruising speed? (Ans: 1/2ln(4/3))------Done
d) How long does it take to reach its cruising speed? (So close!!!!!!)

answer is t = 1/(2sqrt(F)ln3) -------------------------I'm getting ln3 on numerator!!!!!!!!!!!! so triggered

Would appreciate if someone able to show their method if they get the solution, so I can compare and see where I might have gone wrong. Ty!!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 06, 2018, 01:30:52 pm
hey, need help for this mechanics questions. very close to the solution they provided but not exactly it. just need help with part d)

A submarine traavels underwater with a constant driving force of mF (where m is the mass of the submarine). Water pressure exerts a resistance to its motion of v^2 per unit mass ( v is velocity and v > 0)

a) Show that the equation of motion is d^2x/dt^2 = F - v^2 --------------------------Done
b) Show that the terminal velocity is sqrt(F)-------------------------------Done
c) The cruising speed is half the terminal velocity. How far does the submarine need to travel to reach its cruising speed? (Ans: 1/2ln(4/3))------Done
d) How long does it take to reach its cruising speed? (So close!!!!!!)

answer is t = 1/(2sqrt(F)ln3) -------------------------I'm getting ln3 on numerator!!!!!!!!!!!! so triggered

Would appreciate if someone able to show their method if they get the solution, so I can compare and see where I might have gone wrong. Ty!!!
Your answer is definitely correct.
Title: Re: 4U Maths Question Thread
Post by: vikasarkalgud on May 06, 2018, 01:39:53 pm
TY SO MUCH. relieved.
Title: Re: 4U Maths Question Thread
Post by: aadharmg on May 08, 2018, 12:07:35 am
Hey, had an issue with a mechanics question. I'm probably missing a very obvious part, but I've tried many times and I keep leading myself to these weeeird integrals. Has to be much easier. Someone please help.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 08, 2018, 07:51:12 am
Hey, had an issue with a mechanics question. I'm probably missing a very obvious part, but I've tried many times and I keep leading myself to these weeeird integrals. Has to be much easier. Someone please help.


Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on May 10, 2018, 03:41:36 am
A bit of help with an integration que please.
Title: Re: 4U Maths Question Thread
Post by: jazzycab on May 10, 2018, 12:03:52 pm
A bit of help with an integration que please.

You can evaluate that integral using integration by parts:


I'm not sure what you're expected to know in the HSC 4U syllabus as far as limits are concerned.
We have an additional condition that I haven't used as yet. That is, \(s\) is a positive constant. Thus, as \(N\) approaches \(+\infty, sN>0\Rightarrow e^{sN}>0\); \(e^{sN}\) is increasing.
So when \(N\) is large, as \(N\) increases \(e^{sN}\) increases a lot faster than \(sN+1\) increases. Essentially this tells us that \(\lim_{N\rightarrow\infty}\left(\frac{sN+1}{e^{sN}}\right)=0\) (we could show this readily using L'Hopital's rule if it is in the course).
So we end up with:
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on May 10, 2018, 03:57:45 pm
Dude Jazzycab you're actually a massive help, and i'm a bit weak with limits but i understood the denomitor is exponentially greater then numerator so the whole fraction approached 0. Thanks a lot man this is very nicely done.
Title: Re: 4U Maths Question Thread
Post by: vikasarkalgud on May 10, 2018, 07:50:27 pm
hey rui, another mechanics question,

A particle of unit mass moves in a straight line against a resistance kv^alpha (0<alpha<1), where v is the speed in m/s. It starts at a speed of 3000 m/s and this drops to 1500 m/s in one second. After a further 2 seconds the particle comes to rest. Find alpha, to 2 d.p (Answer: 0.42)

I attached where i got up to, giving about 0.86. I got 3 = (3000)^(1-alpha).Idk where i went wrong :(
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 10, 2018, 08:01:17 pm
hey rui, another mechanics question,

A particle of unit mass moves in a straight line against a resistance kv^alpha (0<alpha<1), where v is the speed in m/s. It starts at a speed of 3000 m/s and this drops to 1500 m/s in one second. After a further 2 seconds the particle comes to rest. Find alpha, to 2 d.p (Answer: 0.42)

I attached where i got up to, giving about 0.86. I got 3 = (3000)^(1-alpha).Idk where i went wrong :(
(https://i.imgur.com/GDfbWZN.png)
This transition wasn't justified. You multiplied out \(k (1-\alpha)\) but you didn't multiply that out to \(C\) as well.
Title: Re: 4U Maths Question Thread
Post by: vikasarkalgud on May 10, 2018, 09:17:42 pm
OMG, thankyou so much for helping with the +C common denominator error I had. I wasn't going to sleep if I couldn't get that solution.

Still a little fuzzy on the +C idea though, a little embarrassing as a 4u student, but was the problem here that I was trying to multiply +C by a variable/unknown term? coz its okay to do if its just a constant like "2", or "-e^0.1" since C represents any constant anyway?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 10, 2018, 09:22:51 pm
OMG, thankyou so much for helping with the +C common denominator error I had. I wasn't going to sleep if I couldn't get that solution.

Still a little fuzzy on the +C idea though, a little embarrassing as a 4u student, but was the problem here that I was trying to multiply +C by a variable/unknown term? coz its okay to do if its just a constant like "2", or "-e^0.1" since C represents any constant anyway?
You introduce a new constant, that depends on the old constant.

So say that your original constant was \(C\). Your new constant would be a new arbitrary constant, say \(C_0\) instead. The idea is that whilst \(C\) is arbitrary, you can't use the same \(C\) to represent two different things
-----------------------------------------------------
Basically what happened was that when you multiplied everything else by a term, namely the \(k(1-\alpha)\), you didn't multiply \(C\) to it at all. That's no longer mathematically correct.
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on May 19, 2018, 02:23:15 pm
Hey rui need some help. I've been given this volumes question and also the answer but am really confused how it works. I can do the second part but i'm not sure what to do for the first.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 19, 2018, 03:19:15 pm
Hey rui need some help. I've been given this volumes question and also the answer but am really confused how it works. I can do the second part but i'm not sure what to do for the first.

The thickness in your case is obviously just \( \delta x\).




The one thing to watch out for here is really just the height. Everything else is very routine in the method of cylindrical shells so if you have any problem with it you'll need to elaborate.
Title: Re: 4U Maths Question Thread
Post by: clovvy on May 23, 2018, 04:21:57 pm
I swear I see this equation all the time with implicit questions, but can't do it T_T
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 23, 2018, 08:10:08 pm
I swear I see this equation all the time with implicit questions, but can't do it T_T
This is a lot that you're throwing in all at once. What progress have you made on every part in question. Or alternatively would you like us to focus on certain bits.

There's no point in spoon-feeding every single bit.
Title: Re: 4U Maths Question Thread
Post by: clovvy on May 24, 2018, 06:13:23 am
This is a lot that you're throwing in all at once. What progress have you made on every part in question. Or alternatively would you like us to focus on certain bits.
Only the first part of c I could do..  Without knowing the intercepts and all that I cannot progress with it (I need to know how to get it), once I can recognize critical points or intercepts than perhaps I can work out there rest (I wasn't being specific there sorry)
There's no point in spoon-feeding every single bit.

The only thing I could do is differentiate but I cannot recognize the intercepts or anything like that..  It is a weird equation to work with
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 24, 2018, 08:27:56 am
For now, these are sketch solutions (which you can ask step by step for more details if necessary). Whilst you cannot tell the intercepts 'by inspection', they are very easy to find if we return to old techniques we were taught in Year 11. The question's difficulty really kicks in once you reach d) iii), with the very unusual deduction of a self intersecting curve.

Which certainly is not MX2 material. But this question wasn't meant to be usual MX2 difficulty anyway given that it was the last one in the Sydney Grammar book.

(https://i.imgur.com/B3CdsT2l.png)
(https://i.imgur.com/Obo1Wkol.png)
(https://i.imgur.com/lVluiQul.png)
(https://i.imgur.com/E7JFxXUl.png)
Title: Re: 4U Maths Question Thread
Post by: mxrylyn on May 24, 2018, 03:13:15 pm
Heyy

Just reading about the rule of LIATE for integration on my AN book.

It seems really useful, but I couldn't understand how to use it or weather it helps me pick the U or the V'

Edit:
After reading it a few more times I realised that it clearly says that the rule is to pick which one to differentiate, which is the U.

Sorry, & Thank you
Title: Re: 4U Maths Question Thread
Post by: arii on May 25, 2018, 06:56:43 pm
Harder 3U topic. Appeared in HSC 4-unit 1989 past paper.

I had a bit of trouble doing the questions from (c) onwards.
Title: Re: 4U Maths Question Thread
Post by: vikasarkalgud on May 25, 2018, 07:45:27 pm
Hey Rui, mechanics question - vertical motion

essentially, x = 1/2k (ln(g/(g-kv^2))
Terminal velocity, vT = sqrt(g/k)
I have to show distance particle must fall to reach velocity 1/2vT is x = 1/2k(ln(2/3))............I'm getting x = 1/2k(ln(4/3)) am I wrong?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 25, 2018, 08:02:00 pm
Harder 3U topic. Appeared in HSC 4-unit 1989 past paper.

I had a bit of trouble doing the questions from (c) onwards.
The first few parts only require 2U concepts.


_________________________________________________

_________________________________________________

(and is the main reason why it was in such an old paper, but would not be considered suitable exam content anymore.)



That it is tangential to \(PQ\) can be proven in a similar way, because the horizontal distance from the vertical line \(x=1\) to \(S(6,7)\) is also 5. The proper write-up is left as your exercise.
(https://i.imgur.com/v82Ku7u.png)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 25, 2018, 08:03:01 pm
Hey Rui, mechanics question - vertical motion

essentially, x = 1/2k (ln(g/(g-kv^2))
Terminal velocity, vT = sqrt(g/k)
I have to show distance particle must fall to reach velocity 1/2vT is x = 1/2k(ln(2/3))............I'm getting x = 1/2k(ln(4/3)) am I wrong?

1/2k is interpreted to mean \( \frac{1}{2} k\). Did you mean 1/(2k)?

Although regardless, I am also getting 4/3 assuming you do need \(v = \frac12 \sqrt{\frac{g}{k}} \). In fact \( \ln \frac23 \) would be weird, because that'd imply displacement was allowed to be negative for a 4U resisted motion question.
Title: Re: 4U Maths Question Thread
Post by: vikasarkalgud on May 25, 2018, 08:54:44 pm
1/2k is interpreted to mean \( \frac{1}{2} k\). Did you mean 1/(2k)?

Although regardless, I am also getting 4/3 assuming you do need \(v = \frac12 \sqrt{\frac{g}{k}} \). In fact \( \ln \frac23 \) would be weird, because that'd imply displacement was allowed to be negative for a 4U resisted motion question.

thanks, that's all i needed, source I use is often incorrect in their answers, similar thing happened couple of weeks ago and you verified for me. Thankyou so much, saves me a lot of time that might've been wasted trying to fix a correct answer :)
Title: Re: 4U Maths Question Thread
Post by: scienceislife on May 27, 2018, 04:02:09 pm
Why is the domain of cis(theta) restricted to between -pi and pi?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 27, 2018, 04:11:30 pm
Why is the domain of cis(theta) restricted to between -pi and pi?
It's not?

The value of \(\theta\) that satisfies \( -\pi < \theta \leq \pi\) is just the principal argument of a complex number
Title: Re: 4U Maths Question Thread
Post by: forrea099 on May 27, 2018, 09:06:39 pm
Hi,
If this was the graph of derivative function f'(x), what would the graph of f(x) look like?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 27, 2018, 09:12:00 pm
Hi,
If this was the graph of derivative function f'(x), what would the graph of f(x) look like?

It will look like:
- Between 0 and 1: A straight line of gradient 3
- Between 1 and 3: A parabolic shape that eventually flattens out as x approaches 3
- Between 3 and 5: The same parabolic shape, except now going downwards
- Between 5 and 7: A parabolic shape, with the gradient gradually diminishing from -3 to -2
- Between 7 and 8: A parabolic shape, flattening out a bit more quickly at 8
- Between 8 and 9: The same parabolic shape from 7 to 8, except now going upwards
- Between 9 and 10: If you think about that one, it will essentially look like what happens between 8 and 9, but inverted. This is because you want your parabolic curve to flatten out yet again at 10.
and it can start wherever you want it to because you haven't specified anything about a starting point.

Note that the actual shape of the graph is very hard to describe for such a graph because it's literally being composed by a LOT of straight lines. You would need to do a lot of work to determine the exact shape. So if you want a more descriptive answer, you should post your attempt at sketching the shape itself.
Title: Re: 4U Maths Question Thread
Post by: radnan11 on May 28, 2018, 11:46:08 am
are allowed to use euler's law to prove de moivre's theorem instead of induction?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 28, 2018, 11:58:25 am
are allowed to use euler's law to prove de moivre's theorem instead of induction?
No. Euler's formula is not a part of the course and therefore must never be used for a final answer.

In fact, attempting to use Euler's law to prove De Moivre's is actually assuming what you're trying to prove. Because you're essentially assuming that \(( e^{i\theta})^n = e^{ni\theta} \).
The index law \( (a^m)^n = a^{mn} \) is NOT a known fact over the complex numbers; De Moivre's law helps to prove that it's true.
Title: Re: 4U Maths Question Thread
Post by: justwannawish on May 31, 2018, 06:59:33 pm
Hey, I just have a polynomial question that I'm a bit confused about :/
 I've done part (i) but can't figure out the second part
 
It can be proven that cos6x =32cos6^(theta) - 48cos^4(theta) + 19cos^2(Theta) - 1
(i) Find the roots of P(x) = 32x^6 -48x^4 + 18x^2 - 1
Spoiler
My roots were x = cos(pi/12), cos(5pi/12), cos(7pi/12), cos (11pi/12), and positive and negative 1/root 2

(ii) Show that cos^2(pi/12)+cos^2(5pi/12)=1
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 31, 2018, 08:51:11 pm
Hey, I just have a polynomial question that I'm a bit confused about :/
 I've done part (i) but can't figure out the second part
 
It can be proven that cos6x =32cos6^(theta) - 48cos^4(theta) + 19cos^2(Theta) - 1
(i) Find the roots of P(x) = 32x^6 -48x^4 + 18x^2 - 1
Spoiler
My roots were x = cos(pi/12), cos(5pi/12), cos(7pi/12), cos (11pi/12), and positive and negative 1/root 2

(ii) Show that cos^2(pi/12)+cos^2(5pi/12)=1


Title: Re: 4U Maths Question Thread
Post by: mxrylyn on June 04, 2018, 11:40:44 am
Heyy

Im stuck on integrating x / the root of 1- x^2, with limits at 1/2 and - 1/2

 :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 04, 2018, 11:42:57 am
Heyy

Im stuck on integrating x / the root of 1- x^2, with limits at 1/2 and - 1/2

 :)
Hint: This is just a 3U integral, but of course in 4U you're not given the substitution. Try \( u = 1-x^2\).
Title: Re: 4U Maths Question Thread
Post by: mxrylyn on June 04, 2018, 03:34:06 pm
Hint: This is just a 3U integral, but of course in 4U you're not given the substitution. Try \( u = 1-x^2\).

Thank you!
Title: Re: 4U Maths Question Thread
Post by: vikasarkalgud on June 05, 2018, 04:05:02 pm
Hey Rui,
what would be the most efficient way of integrating I = (integral) sqrt((4+x)/(3-x)) dx ?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 05, 2018, 04:19:04 pm
Hey Rui,
what would be the most efficient way of integrating I = (integral) sqrt((4+x)/(3-x)) dx ?
\[ \int \sqrt{\frac{a+x}{b-x}}dx \text{ and similar forms are usually a little nasty.}\]

Of course, we all know that these integrals are still nasty. We have no choice but to then expand the bottom, complete the square and then split the top into two separate integrals (one which integrates to \( \ln \) and another which integrates to \( \sin^{-1} \)).

If you have some free time, you can try experimenting with this. This may or may not work, but you can perform the substitution \( u = x - \frac12\) to reduce it into \( \int \sqrt{\frac{\frac72 + u}{\frac72 - u}}du \) and then perform another substitution \( u = \frac72\cos2\theta\). You'll need double angles if you walk down this path.

Edit: But the idea is that it ultimately boils down to something like this: \( -14\int \cos^2\theta\,d\theta \). This might've taken longer than the other method tbh, but it's quite interesting to investigate.
Title: Re: 4U Maths Question Thread
Post by: Jeeffffffffffffff on June 05, 2018, 07:44:31 pm
Hi
I’ve got an integration with trig substitution question and I’ve got an answer for it but I’m not entirely confident in it and was wondering if you could check it for me please?
The question wanted the integral of the square root of 6x-x2-5 and I got
2sin-1((x-3)/2)+(((x-3)•sqrt(4-(x-3)2))/2) +C
Sorry for all the messy typing I tried uploading a photo but the photo was too big :///
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 05, 2018, 07:48:32 pm
Hi
I’ve got an integration with trig substitution question and I’ve got an answer for it but I’m not entirely confident in it and was wondering if you could check it for me please?
The question wanted the integral of the square root of 6x-x2-5 and I got
2sin-1((x-3)/2)+(((x-3)•sqrt(4-(x-3)2))/2) +C
Sorry for all the messy typing I tried uploading a photo but the photo was too big :///
Looks legit.

(Note that \( \sin^{-1} (-x) = -\sin^{-1}x\), so there's no discrepancy.)
Title: Re: 4U Maths Question Thread
Post by: mxrylyn on June 07, 2018, 09:50:46 am
Heyy

Which methods do I use to integrate (×+2) / (the root of: ×^2 - 1)

And also the root of: 9 -  x^2
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 07, 2018, 09:55:46 am
Heyy

Which methods do I use to integrate (×+2) / (the root of: ×^2 - 1)

And also the root of: 9 -  x^2
The second one should be screaming trig sub. Which I get is unpleasant, but still it should be obvious because of the \( \sqrt{a^2-x^2} \) form..

The former is just a usual substitution. As for \( \int \frac{1}{\sqrt{x^2-1}} \,dx\), if this question came from an HSC paper prior to 2016 then you could just use the table of standard integrals to write down the answer. Otherwise, you have no choice but to use a trig sub as well, because of the \( \sqrt{x^2-a^2} \) form. Subbing in \(x = \sec \theta\) is pretty disgusting though, admittedly.
Title: Re: 4U Maths Question Thread
Post by: vikasarkalgud on June 07, 2018, 02:35:52 pm
Hey Rui, got this question from a past paper. Not sure how to do. (attached to post)

ty for any help
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 07, 2018, 03:37:24 pm
Hey Rui, got this question from a past paper. Not sure how to do. (attached to post)

ty for any help


Title: Re: 4U Maths Question Thread
Post by: bdobrin on June 07, 2018, 06:53:46 pm
Hey Rui,

Im really stuck on this recurrence formulae question. Any help would be appreciated

"If In = integral (from pi/2 to 0) of sinnxcos2x dx for n >0, show that In = [(n-1)/(n+2)] In-2"

thanks
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 07, 2018, 07:38:16 pm
Hey Rui,

Im really stuck on this recurrence formulae question. Any help would be appreciated

"If In = integral (from pi/2 to 0) of sinnxcos2x dx for n >0, show that In = [(n-1)/(n+2)] In-2"

thanks

\begin{align*}I_n &= \int_0^{\pi/2} \sin^n x \cos^2 x \,dx\\ &= \int_0^{\pi/2} \left( \sin^{n-1}x \cos^2 x \right) \sin x \, dx\\ &= \underbrace{\left[ -\cos x \left( \sin^{n-1}x \cos^2 x \right) \right]_0^{\pi/2}}_{\text{this thing equals 0}} \\&\qquad+ \int_0^{\pi/2} \cos x \Big( \left( (n-1) \cos x \sin^{n-2}x \right) \cos^2 x - (2\cos x \sin x)\sin^{n-1}x \Big) dx \\ &= (n-1) \int_0^{\pi/2} \sin^{n-2}x \cos^2x \cos^2 x \,dx - 2\int_0^{\pi/2} \sin^{n}x\cos^2 x \,dx\\ &= (n-1) \int_0^{\pi/2} \sin^{n-2}x \cos^2x \left( 1 - \sin^2 x \right)dx - 2I_n \\ \therefore 3I_n &= (n-1) \int_0^{\pi/2} \sin^{n-2}x \cos^2 x \, dx - (n-1) \int_0^{\pi/2} \sin^n x \cos^2 x \, dx\\ &= (n-1) I_{n-2} - (n-1) I_n\\ \therefore (n+2) I_n &= (n-1) I_{n-2}\\ I_n &= \frac{n-1}{n+2}I_{n-2}\end{align*}
It may be slower, but it might be a lot easier to express \(I_n\) in terms of \(J_{n+2}\), where \(J_n = \int_0^{\pi/2} \sin^n x\,dx\), and then try working with that.
Title: Re: 4U Maths Question Thread
Post by: bdobrin on June 07, 2018, 07:55:21 pm
Hey Rui,

thanks for that - didn't realise how hard that question was

also, how did you know to change sinnx to sinn-1sinx ?
Title: Re: 4U Maths Question Thread
Post by: vikasarkalgud on June 07, 2018, 08:11:29 pm
Hey Rui, back again. 4u curve sketching question here and I'd like to know how exactly they got the integral graph to approach 2 in the final part, as x approaches negative infinity. Also, I had my own kind of way of doing f(e^x), was wondering if there is method to it coz mine is kind of dodgy. Thankyou very much.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 07, 2018, 09:03:39 pm
Hey Rui,

thanks for that - didn't realise how hard that question was

also, how did you know to change sinnx to sinn-1sinx ?
A bit of backtracking: I realised at an early onset that \( I_n = \int_0^{\pi/2} \sin^n x\,dx - \int_0^{\pi/2} \sin^{n+2} x \, dx\). But then I ignored that observation and kept trying to find smart ways to do this problem and surely enough, I was getting nowhere. So I started going back to my first idea and thought about how to handle the reduction formula for \(J_n = \int_0^{\pi/2} \sin^n x\,dx \). I recalled that for that one, I always had to split it into \( \sin x \sin^{n-1} x\) regardless. So (albeit with some reluctance) I decided to do that for this one as well. Thankfully it worked.
Hey Rui, back again. 4u curve sketching question here and I'd like to know how exactly they got the integral graph to approach 2 in the final part, as x approaches negative infinity. Also, I had my own kind of way of doing f(e^x), was wondering if there is method to it coz mine is kind of dodgy. Thankyou very much.
I don't see at all why the integral graph should approach 2. In fact, I don't know what it should approach at all - the only thing I know is that it must approach something greater than 0.

I would've done \(f(e^x)\) the same way I did \( f(x^2)\) tbh, which is also dodgy. When the transformation is under the brackets I usually don't rely on a toolkit, but rather visualise what's going on in my head. In a way, I visualise a "stretching" animation of the original graph.

Although, some things that are useful to note:
\( \lim_{x\to -\infty} f(e^x) = f(0) \), which is equal to 2 in this case so we expect our graph to asymptotically approach 2 towards the left
As \(x\to \infty\), \(f(e^x) \) is really just tending towards \( \lim_{x\to \infty}f(x) \), so we expect it to asymptotically approach 0 towards the right as well
Title: Re: 4U Maths Question Thread
Post by: itssona on June 10, 2018, 12:29:32 pm
ruii hey
would you happen to have a really simple proof that explains the rotation of a rectangular hyperbola? also if the hyperbola is in the form x^2-y^2=a^2, then do we just use the regular x=a/e for directrix right?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 10, 2018, 12:53:27 pm
ruii hey
would you happen to have a really simple proof that explains the rotation of a rectangular hyperbola? also if the hyperbola is in the form x^2-y^2=a^2, then do we just use the regular x=a/e for directrix right?

It’s a bit hidden because the post is old but I actually have talked about it here.

And yeah. In particular, \(e=\sqrt2\) in this case
Title: Re: 4U Maths Question Thread
Post by: itssona on June 10, 2018, 01:00:48 pm
It’s a bit hidden because the post is old but I actually have talked about it here.

And yeah. In particular, \(e=\sqrt2\) in this case
ah thank you!:)
Title: Re: 4U Maths Question Thread
Post by: justwannawish on June 10, 2018, 02:59:02 pm
Hey, I'm a bit confused on how you integrate this with t-formula:
1dx/(3+4sin(2x))
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 10, 2018, 03:22:42 pm
Hey, I'm a bit confused on how you integrate this with t-formula:
1dx/(3+4sin(2x))
This is really just a pointless grind and involves nothing clever, only heavily brute forced computations.
\begin{align*}\int \frac{dx}{3 + 4\sin 2x} &=\int \frac{ \frac{2}{1+t^2}}{3 + \frac{8t}{1+t^2}}\,dt\\ &= \int \frac{2}{3t^2 + 8t + 3}\,dt\\ &=\frac23 \int \frac{1}{\left(t + \frac43\right)^2 - \frac79}\,dt \\ &=\frac23 \int \frac{1}{\left(t + \frac{4+\sqrt7}{3} \right)\left(t + \frac{4-\sqrt7}{3} \right)}\,dt \\ &= \frac1{\sqrt7}\int \left( \frac{1}{t + \frac{4-\sqrt7}3}+\frac{1}{t - \frac{4+\sqrt7}3} \right)dt\\ &= \frac1{\sqrt7} \left( \ln \left|  t + \frac{4-\sqrt7}{3}\right| - \ln \left| t + \frac{4+\sqrt7}{3} \right|\right)+C\\ &= \frac1{\sqrt7} \ln \left| \frac{3\tan x + 4 - \sqrt7}{3 \tan x + 4 +\sqrt7} \right|+C\end{align*}
Title: Re: 4U Maths Question Thread
Post by: clovvy on June 10, 2018, 08:46:41 pm
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 10, 2018, 08:58:33 pm

Did you get this question off the meme? (It's final answer is actually quite disgusting.)
Title: Re: 4U Maths Question Thread
Post by: clovvy on June 10, 2018, 09:03:14 pm
Did you get this question off the meme? (It's final answer is actually quite disgusting.)
I did check the answer and it is disgusting, how do you guess that? (I mean you are right)- how do I approach this whole mess to get to the final answer?

Is this question beyond 4U?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 10, 2018, 09:30:37 pm
I think, once you're at that point, you might wanna try bashing it with a trig substitution. (Sadly hyperbolic substitutions are not taught in 4U)

In my opinion, it's a waste of time and you can be doing something way more productive than that. However if you have enough time to do it, make sure you include the boundaries and check with WolframAlpha as you go.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 11, 2018, 12:32:13 pm
Managed to finish it off but yeah I used a hyperbolic substitution, which is beyond 4U. The thought of subbing in \( u = \sec \theta\) horrified me.

But yeah, this is still a brute forced approach. I haven't been able to see anything elegant to bash this one with.

(https://i.imgur.com/xtjPdI7.png)
Title: Re: 4U Maths Question Thread
Post by: tiffany88 on June 13, 2018, 11:49:52 am
Hello,
Could I please get some help with part b of this question?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 13, 2018, 12:09:33 pm
Hello,
Could I please get some help with part b of this question?

The concept of Riemann sums are not involved in the MX2 course. Where is this question coming from?
Title: Re: 4U Maths Question Thread
Post by: tiffany88 on June 13, 2018, 12:35:16 pm
The concept of Riemann sums are not involved in the MX2 course. Where is this question coming from?

I got it from my tutor, who did these questions involving Riemann sums for integration. I'm glad its not part of the course because it seems difficult to understand :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 13, 2018, 12:45:11 pm
It's covered in first year uni because Riemann sums form the basis of the integration that we do (which is, incidentally enough, known as Riemann integration). The most that you'd be required to do with rectangles in 4U is use them to prove inequalities that involve integrals. The whole ideas of generalising this (replacing one rectangle with multiple rectangles, taking limits etc.) however is not important so I won't cover it here.

(However, if you want a solution anyway, you can find the first year uni math questions thread, repost it there and I'll be happy to address it.)
Title: Re: 4U Maths Question Thread
Post by: RustyWasTaken on June 13, 2018, 01:44:18 pm
Help please :p
Cheers
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 13, 2018, 03:46:30 pm
Help please :p
Cheers
(https://i.imgur.com/1Egcx0B.png)
Title: Re: 4U Maths Question Thread
Post by: clovvy on June 15, 2018, 04:22:31 pm
https://www.youtube.com/watch?v=F9ZJya5DK5w
Another way of approaching the meme integral
Title: Re: 4U Maths Question Thread
Post by: clovvy on June 16, 2018, 02:36:17 pm
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 16, 2018, 02:53:33 pm


\begin{align*}\int \frac{dx}{x+\sqrt{x}}&= \int \frac{dx}{\sqrt{x} \left(\sqrt{x}+1\right)}\\ &= 2\int \frac{du}{u+1}\\ &= 2\ln |u+1| + C\\ &= 2\ln \left( \sqrt{x} + 1\right) + C\end{align*}
Title: Re: 4U Maths Question Thread
Post by: vikasarkalgud on June 18, 2018, 10:12:52 pm
Error
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 18, 2018, 10:41:27 pm
Hey Rui, this is a different kind of problem but I don't know where else to ask, I'm doing a software major and have encountered a math problem. Using C# on Visual Studio, and for some reason, the square root is coming as square root of -23. It seems to be doing 4 - (3 * 9), instead of 3 * 9/169 which is 27/169 (what I'm meant to get since 3 - that, would give me a value I can square root.

With inputs:
a = 8
b = 5

Formula:

perimeter = (Math.PI) * (a + b) * (((1 + (3 * ((a - b) * (a - b) / (a + b) * (a + b)) / (10 + Math.Sqrt(4 - (3 * ((a - b) * (a - b) / (a + b) * (a + b)))))))));

the square root im dealing with is on 2nd half of that mess
I know, brackets at end are crazy but there seems to be that many since no syntactical error. During runtime, I placed a breakpoint and perimeter value returned 'NaN', or '0' when variable was hovered over. what is wrong? Also, you've probably realised that there is an approximation for the perimeter of the ellipse, assuming ive copied it right...

Any help would be much appreciated, ty
Fairly sure programming languages aren't in the HSC altogether, so if this is something more related to uni stuff then please make a new thread in the uni section of the forum.

Also, it's really not good practice at all to nest so many brackets. If you really wanna do that you should declare some new variables "e.g. sum = a + b and diff = a - b" instead. And maybe doing some divisions step by step. Because looking at that, I really can't tell what's going on
Title: Re: 4U Maths Question Thread
Post by: clovvy on June 24, 2018, 09:37:53 am
Hey guys,
how do I recognise horizontal or vertical tangents at a stat point or an asymptote, thanks
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 24, 2018, 09:44:33 am
Hey guys,
how do I recognise horizontal or vertical tangents at a stat point or an asymptote, thanks
Not sure what you mean by "tangents" at asymptotes.


Title: Re: 4U Maths Question Thread
Post by: envisagator on June 24, 2018, 10:17:00 am
Hey, I'm having trouble recognising how to draw a guide graph or how it helps when drawing a graph. This is a question that I didnt quite understand: By investigating the behaviour of the curve y^2=x^2(x-1) at the neighbourhood of x=1 and fr large values of x sketch the curve. And is it possible to tackle this question without calculus??
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 24, 2018, 10:42:00 am
Hey, I'm having trouble recognising how to draw a guide graph or how it helps when drawing a graph. This is a question that I didnt quite understand: By investigating the behaviour of the curve y^2=x^2(x-1) at the neighbourhood of x=1 and fr large values of x sketch the curve. And is it possible to tackle this question without calculus??



This should be unsurprising. What we are essentially sketching is \( y = \pm x \sqrt{x-1} \). For \(y = x\sqrt{x-1}\), \(y\to 1^+\). But for \(y = -x\sqrt{x-1}\), \(y\to 1^-\).


Intuitive explanation: It's very important to keep focusing on the point \( (1,0) \). We know that \(x \geq 1\). Suppose that as \(x \to 1^+\), \( \frac{dy}{dx}\) does not explode.

Edit: Earlier explanation wasn't sufficient. The most important criteria is that this graph needs to remain smooth. If we assume that \( \frac{dy}{dx}\) is finite, i.e. the curve is either increasing, decreasing or stationary, ultimately as we approach \(x\to 1^+\):
- The upper half of the graph, i.e. \( y = x\sqrt{x+1} \), approaches it at an angle or flat
- The lower half of the graph, i.e. \(y = -x\sqrt{x+1}\), does the same.

Either of these would lead to a corner or a cusp. These are sometimes feasible options, but they aren't here. This occurs more or less due to the fact we never introduce surds in our equation (until we manually take square roots), and all relevant functions \(y^2\), \(x^2\) and \(x+1\) are always differentiable everywhere along their respective domains: all real \(y\) and \(x \geq 1\).

So since we can't have a cusp or corner, everything above breaks, so our original assumption was wrong. \( \frac{dy}{dx} \) has to be infinite for the graph to smoothly bounce back off. (In fact, this is similar to just proving that \( \frac{dx}{dy} = 0\), but of course we wanted to avoid calculus here.)

Upon reflection, the intuitive method is a lot more catastrophic (at the 4U level) in contrast to just implicitly differentiating \(y^2 = x^3-x^2\).
Title: Re: 4U Maths Question Thread
Post by: mxrylyn on July 01, 2018, 12:04:34 pm
I am not sure how to integrate (x +2) / (x^2 -1)

And
1/ (3x^2 + 6x + 10)

Can I get the integral of 1/3x  and times 8t by the integral of 1/x and the integeral of 1/6x+10?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 01, 2018, 12:41:07 pm
I am not sure how to integrate (x +2) / (x^2 -1)

And
1/ (3x^2 + 6x + 10)

Can I get the integral of 1/3x  and times 8t by the integral of 1/x and the integeral of 1/6x+10?
You need to complete the square for the second one: \( \int \frac{1}{3x^2+8x+10}dx = \int \frac{1}{3(x+1)^2+7}dx \)
\[ \int \frac{x+2}{x^2-1} dx = \int \frac{x}{x^2-1}dx + \int \frac{2}{x^2-1}dx\\ \text{The first integral goes to }\frac12 \ln |x^2-1|\\ \text{The second requires partial fractions.} \]
I don't understand what that third one is supposed to be.
Title: Re: 4U Maths Question Thread
Post by: mxrylyn on July 01, 2018, 12:47:40 pm
You need to complete the square for the second one: \( \int \frac{1}{3x^2+8x+10}dx = \int \frac{1}{3(x+1)^2+7}dx \)
\[ \int \frac{x+2}{x^2-1} dx = \int \frac{x}{x^2-1}dx + \int \frac{2}{x^2-1}dx\\ \text{The first integral goes to }\frac12 \ln |x^2-1|\\ \text{The second requires partial fractions.} \]
I don't understand what that third one is supposed to be.

Thank you!
Yes I messed up with both explaining and typing the 3rd one
Title: Re: 4U Maths Question Thread
Post by: mxrylyn on July 01, 2018, 06:25:25 pm
What method do I use to integrate

 1/(x(3-x))

(1-x^2)/(1+x^2)

And (sec^2x)/(9 - tan^2 x)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 01, 2018, 06:30:37 pm
What method do I use to integrate

 1/(x(3-x))

(1-x^2)/(1+x^2)

And (sec^2x)/(9 - tan^2 x)

1. That is screaming partial fractions right there.
3. Looking at it again, after subbing \(u = \tan x\) you need partial fractions for this one as well.
\begin{align*}\int \frac{1-x^2}{1+x^2}\,dx &= -\int \frac{x^2-1}{x^2+1}\,dx\\ &= -\int \left( 1 - \frac{2}{x^2+1} \right)\,dx\\ &= -x +2\tan^{-1}x +C\end{align*}
Title: Re: 4U Maths Question Thread
Post by: mxrylyn on July 01, 2018, 10:20:33 pm
I am having trouble showing that 1/9-10sin^2 = sec^2 X / 9-tan^2 X
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 02, 2018, 09:29:23 am
I am having trouble showing that 1/9-10sin^2 = sec^2 X / 9-tan^2 X

Hint: After dividing top and bottom by \(\cos^2 x\), you need to use a Pythagorean identity involving \(\sec^2 x\) on the bottom
Title: Re: 4U Maths Question Thread
Post by: Caleb Campion on July 03, 2018, 05:12:10 pm
Hi there, I was wondering if there's a chance my version has been misprinted or all of them are like that, but both my HSC Mathematics Extension 1 and 2 TOPIC TESTS books from ATARNotes have significant mistakes in them. The issue is, that there are so many of them. Many of the answers answer a variant of the actual question, with different values etc. Some of them, like Q3 in the Volumes part of the MX2 tests book, have printed the question differently to the answer, such as the question deals with a function with (e^3 . x) whereas the answer answers a question that uses (e^(3x)).

There are several spelling errors in my other atarnotes books also.

I love these books, and the topic tests are helpful, but with this many mistakes, its embarrasing to my self and my teacher (who bought them), and stops me from reccommending these products.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 03, 2018, 05:15:40 pm
Hi there, I was wondering if there's a chance my version has been misprinted or all of them are like that, but both my HSC Mathematics Extension 1 and 2 TOPIC TESTS books from ATARNotes have significant mistakes in them. The issue is, that there are so many of them. Many of the answers answer a variant of the actual question, with different values etc. Some of them, like Q3 in the Volumes part of the MX2 tests book, have printed the question differently to the answer, such as the question deals with a function with (e^3 . x) whereas the answer answers a question that uses (e^(3x)).

There are several spelling errors in my other atarnotes books also.

I love these books, and the topic tests are helpful, but with this many mistakes, its embarrasing to my self and my teacher (who bought them), and stops me from reccommending these products.
Can you please compile the list of mistakes you have found and send them through to my inbox?

Whilst I can't comment on the MX2 TT's since I wasn't directly involved with them, a lot of things can easily get missed in either publication. Mistakes are things we all want to avoid, yet they still creep up without us noticing and it bugs all of us. I do not wish to disclose any information regarding my role in the MX1 TT's through online means - if any serious matters have come up I prefer to address them in person at the upcoming lecture series.

(Alternatively, can you just show me them at any one of the lectures?)


Edit: This list is not exhaustive (and is a bit lost in the forum as well), but here are some known typos in the MX2 publication.
Title: Re: 4U Maths Question Thread
Post by: clovvy on July 05, 2018, 08:19:17 pm
Having trouble with this integral
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 05, 2018, 08:32:17 pm

\begin{align*}\int_3^8 \frac{x}{x+1-\sqrt{x+1}}dx &= \int_2^3 \frac{u^2-1}{u^2 - u} 2u\,du\\ &= \int_2^3 \frac{2u(u-1)(u+1)}{u(u-1)}\,du\\ &= \int_2^3 (2u+2)\,du\\ &= [u^2+2u]_2^3\\ &= 7\end{align*}
Title: Re: 4U Maths Question Thread
Post by: clovvy on July 05, 2018, 08:37:38 pm

\begin{align*}\int_3^8 \frac{x}{x+1-\sqrt{x+1}}dx &= \int_2^3 \frac{u^2-1}{u^2 - u} 2u\,du\\ &= \int_2^3 \frac{2u(u-1)(u+1)}{u(u-1)}\,du\\ &= \int_2^3 (2u+2)\,du\\ &= [u^2+2u]_2^3\\ &= 7\end{align*}

Just implicit? God why did I forget implicit simply because this is integration haha.. thanks a lot tho
Title: Re: 4U Maths Question Thread
Post by: KT Nyunt on July 08, 2018, 06:51:42 pm
Hey Rui, I'm a bit confused about something in the ATAR Math ext. 2 topic notes...
With your LIATE rule, you said when we're integrating e^xsinx dx that we would differentiate sinx and integrate e^x.
But then in your solution, you integrated sinx and then differentiate e^x.

I feel like I'm missing something, but I don't know what... Pls help!

Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 08, 2018, 07:32:55 pm
Hey Rui, I'm a bit confused about something in the ATAR Math ext. 2 topic notes...
With your LIATE rule, you said when we're integrating e^xsinx dx that we would differentiate sinx and integrate e^x.
But then in your solution, you integrated sinx and then differentiate e^x.

I feel like I'm missing something, but I don't know what... Pls help!


My bad there, went against my own rule. I need to go fix that later.

Fortunately however, the order of the last two doesn't matter. I've found that with \( e^x\cos x\) I integrate cos, whereas with \(e^x \sin x\) I differentiate sin. But trying to say "LIAET" would be pretty hard lols
Title: Re: 4U Maths Question Thread
Post by: envisagator on July 09, 2018, 05:34:14 pm
Hei Rui, conics has been a topic I really struggle as i cant grasp what the question actually asks and the usefulness of sum of roots such as in this question: y=mx+1 is a chord of the hyperbola x^2 - 4y^2 = 16 , that passes through (0,1) for all real values of m. Find the locus of the midpoint of the chord as m varies. (This is a textbook question btw)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 09, 2018, 06:24:23 pm
Hei Rui, conics has been a topic I really struggle as i cant grasp what the question actually asks and the usefulness of sum of roots such as in this question: y=mx+1 is a chord of the hyperbola x^2 - 4y^2 = 16 , that passes through (0,1) for all real values of m. Find the locus of the midpoint of the chord as m varies. (This is a textbook question btw)
These questions aren't hard because of the fact they're conics. What makes them hard is trying to find the shortcut that makes the problem a lot simpler, and actually tying the concepts together. This one you have here is one of the classic ones. Note that drawing a diagram is highly recommended.


Which, upon expanding and rearranging, becomes a quadratic. The roots are, again, the points of intersection.

At this point, I remark on what's significant. The points of intersection between the chord and the hyperbola are simply the endpoints of the chord itself. Therefore, the MIDPOINT of the chord is actually just the midpoint of the points of intersection. So the \(x\)-coordinate of the midpoint, actually turns out to be \( x= \frac{\alpha+\beta}{2} \).

We can now kill this problem by realising that rewriting \(x = \frac{\alpha+\beta}{2} \) gives us \(x = \frac{\text{sum of roots}}{2} \). And then boom, the \(x\)-coordinate of the midpoint of the chord has been found.

(Now plug into \(y=mx+1\) for the corresponding \(y\)-coordinate, before attempting to eliminate the parameter. Which is \(m\) in this case.)

A remark: We could've actually solved the quadratic to find what \(x_1\) and \(x_2\) explicitly are. But then we still need to compute \( \frac{x_1+x_2}{2} \) anyway. The sum of roots is just a more convenient shortcut for this purpose.
Title: Re: 4U Maths Question Thread
Post by: clovvy on July 09, 2018, 08:04:33 pm
Hey, I am doing a volumes question where the equation y=x^3+1 between -1 to 0 is rotated about the x-axis and I have to use the slicing method.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 09, 2018, 08:31:00 pm
Hey, I am doing a volumes question where the equation y=x^3+1 between -1 to 0 is rotated about the x-axis and I have to use the slicing method.

Title: Re: 4U Maths Question Thread
Post by: clovvy on July 10, 2018, 07:33:08 am

Title: Re: 4U Maths Question Thread
Post by: Opengangs on July 10, 2018, 08:47:56 am

The volume is evaluated as:
\[ V = \pi \int_a^b y^2\,dx \]
Title: Re: 4U Maths Question Thread
Post by: clovvy on July 10, 2018, 09:04:22 am
The volume is evaluated as:
\[ V = \pi \int_a^b y^2\,dx \]
ah ok so I don't even need slicing.. make sense...
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 10, 2018, 09:56:56 am

The slicing formula is \( \delta V = \pi r^2 \times \text{thickness}\). The square is supposed to be there for the radius, which in your case is just \(r = x^3+1\). You should go back and revise the formula.

In general, slicing eventually gives you the same integral as if you did it the old way.
The volume is evaluated as:
\[ V = \pi \int_a^b y^2\,dx \]
They actually need to do a bit more work in MX2. I just skipped all the details in my earlier reply.
Title: Re: 4U Maths Question Thread
Post by: vikasarkalgud on July 10, 2018, 03:18:49 pm
Hey, Rui, for complex locus, I don't get when I'm meant to be putting in open circles for intersections and dotted lines...coz I thought dotted lines were only for when there is no =, hence it cannot exist there. So this question I've attached..

Thankyou for help
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 10, 2018, 05:31:15 pm
Hey, Rui, for complex locus, I don't get when I'm meant to be putting in open circles for intersections and dotted lines...coz I thought dotted lines were only for when there is no =, hence it cannot exist there. So this question I've attached..

Thankyou for help
Part i) shouldn't have any open circles for that region..? It doesn't look like there's any open circles there which should be fine, since nothing is dotted here.
Title: Re: 4U Maths Question Thread
Post by: vikasarkalgud on July 16, 2018, 06:24:12 pm
Hey Rui,

for another subject, I don't really know what area, perimeter and volume actually are categorised as. What I mean is, length, height, perpendicular height, radius are dimensions. Therefore, using these, u can get area, perimeter, volume, circumference.etc. But what are these things actually called? These results?

Thankyou in advance
Title: Re: 4U Maths Question Thread
Post by: envisagator on July 16, 2018, 08:50:56 pm
Hei Rui,can you help with this induction q, its a cambridge 4u textbook question. I'm not quite sure what exactly im supposed to assume. Thank You in advance!!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 16, 2018, 10:04:03 pm
Hei Rui,can you help with this induction q, its a cambridge 4u textbook question. I'm not quite sure what exactly im supposed to assume. Thank You in advance!!!
Fairly sure you can continue with this question once I give you what to assume.



Hey Rui,

for another subject, I don't really know what area, perimeter and volume actually are categorised as. What I mean is, length, height, perpendicular height, radius are dimensions. Therefore, using these, u can get area, perimeter, volume, circumference.etc. But what are these things actually called? These results?

Thankyou in advance
Length, height, perpendicular height and radius are 1-dimensional concepts.

Perimeter is also a 1-dimensional concept; it's just a special type of length (just like how perpendicular height and radius are). Area, however, is a 2-dimensional concept. Volume is a 3-dimensional concept.

The radian measure of an angle is a special case of a 0-dimensional concept.

Colloquially, the word "dimensions" typically just refers to lengths, especially when you're dealing with a volume. But in science (yes, this is actually more of a science question than a maths question), the dimensions refer to the SI base units that you require to build up a new concept. Volume involves 3 length-dimensions, because its SI unit is m^3, which is built from 3 SI base units. Force involves a mass dimension and a length dimension, PER 2 time-dimensions, because its SI unit is kg m s^-2.

(As an extra example, we can get a bit bizarre, and argue voltage involves a mass dimension and length dimension, PER 3 time-dimensions and current-dimension, because its SI unit is kg m s^-3 A^-1. But it gets a bit bizarre, because there's debate surrounding whether current is the dimension or electric charge is the dimension.)
Title: Re: 4U Maths Question Thread
Post by: envisagator on July 16, 2018, 10:29:25 pm
Fairly sure you can continue with this question once I give you what to assume.
[tex]\text{You know that }u_{k+1} = 4u_k - 3u_{k-1}. Is this statement just subbing in k+1 into the 'fact' that shows how each term in the relation is created?, and do you choose to sub in k+1 because you know you are going to use that to prove in the final step??. Thank You!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 16, 2018, 10:35:22 pm
Fairly sure you can continue with this question once I give you what to assume.
[tex]\text{You know that }u_{k+1} = 4u_k - 3u_{k-1}.
Is this statement just subbing in k+1 into the 'fact' that shows how each term in the relation is created?, and do you choose to sub in k+1 because you know you are going to use that to prove in the final step??. Thank You!!
Yeah well you're meant to prove it holds when \(n=k+1\) so you're trying to prove that \(u_{k+1} = 2+3^{k+1} \).

But the recurrence relation was defined to be \(u_n = 4u_{n-1} - 3u_{n-2}\). Of course you can assume what you've defined it to be in the first place
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on July 16, 2018, 11:44:39 pm
hey Rui a bit of help please with some harder inequalities.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 17, 2018, 06:46:18 am
hey Rui a bit of help please with some harder inequalities.
What are the conditions on a, b and c?
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on July 17, 2018, 11:05:57 am
Oh sorry a, b and c are all positive integers
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 17, 2018, 03:51:54 pm
Without any precursor assumptions to build on I really could not find any intuitive way of doing this, so I just resorted to working backwards.



Title: Re: 4U Maths Question Thread
Post by: envisagator on July 18, 2018, 04:38:25 pm
Hi Rui, is relying on Cauchy's inequalitity AM>= GM too much to prove inequalities a bad habit or should i considering using proofs previously done and expanding on it to proof another result, idk if you understand by what i meant. Plus, when I state i have used it is just saying AM>= GM fine??. Thanks in advance!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 19, 2018, 02:16:56 am
Hi Rui, is relying on Cauchy's inequalitity AM>= GM too much to prove inequalities a bad habit or should i considering using proofs previously done and expanding on it to proof another result, idk if you understand by what i meant. Plus, when I state i have used it is just saying AM>= GM fine??. Thanks in advance!!
Your questions at Terry Lee coaching are kinda rigged so that you have no choice but to use them. In the actual HSC you'll only need it if you prove it beforehand. Preferably, yeah, always use what you've already proven to prove new stuff.
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on July 19, 2018, 04:47:44 pm
Hey rui please some help with polynomials
solve 4x^5+x =0 over the complex field
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 19, 2018, 05:53:26 pm
Hey rui please some help with polynomials
solve 4x^5+x =0 over the complex field
\begin{align*}4x^5+x&=0\\ x(4x^4+1)&=0\\ x(2x^2+i)(2x^2-i)&=0\end{align*}
\[ \text{The obvious solution is }x=0\\ \text{We need to then solve }\boxed{x^2 = \pm \frac{i}{2}}\\ \text{which can be done using any one of the usual approaches.} \]
Rewrite \(x = r \text{cis}\theta\), so that \(x^2 = \frac{i}{2}\) can be rewritten as \( r^2 \text{cis}2\theta = \frac12 \text{cis} \frac\pi2 \). This gives the solutions \(x = \frac{1}{\sqrt2}\text{cis}\frac\pi4\), \(x=\frac{1}{\sqrt2}\text{ cis }\left(-\frac{5\pi}{4}\right) \). Then do the same thing for \(x^2 = -\frac{i}{2} \).

Alternatively, just solve \(4x^4+1=0\), i.e. \(x^4 = -\frac14\), using the same approach as above.
Title: Re: 4U Maths Question Thread
Post by: yammy on July 22, 2018, 02:02:30 am
Hey Rui,
Can you please help me with question 3 and 4?
Title: Re: 4U Maths Question Thread
Post by: clovvy on July 22, 2018, 07:38:06 am
Hey Rui,
Can you please help me with question 3 and 4?

With 3, sketch your locus first before you do anything with it.. with q4 you have to use by parts
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 22, 2018, 07:52:51 am
Hey Rui,
Can you please help me with question 3 and 4?
Sketch solution for Q3: That region isn't even gonna be a region. It's literally just gonna be a tiny segment of the ray. First, verify that the ray \( \arg z = \frac\pi3\) and the circle \( |z|=2\) intersect at \(z = 2 \text{cis} \frac\pi3\). (Recall from my lecture that we only need to consider the boundary, which is why we don't care too much about the interior of the circle; \(|z|<2\).)

Then, because you know the locus of \(z\), you also know the locus of \(z-i\). Reason being that \(-i\) is just a translation downwards by one unit.

Since the endpoints of the locus of \(z\) were at \(0\) with an open circle and \(2\text{cis}\frac\pi3\) with a closed circle, the endpoints of the locus of \(z-i\) are at \(-i\) and \(2\text{cis} \frac\pi3-i\). So you just have to find the arguments of these two complex numbers to know the extremities. Of course, you will need to expand out that cis into Cartesian form beforehand.


I will look at Q4 later, but my guess would to be start with \(U_n = \int \frac{dx}{(x^2+1)^{n-1}} \) and then take the IBP suggestion, differentiating \( \frac{1}{(x^2+1)^{n-1}} \) and integrating \(1\).
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 24, 2018, 11:12:23 am
Sorry, forgot to back to this. But I couldn't get what they got. Their a) is obviously wrong, because there's no reason for the power in the bottom to be \(n\), but it looks like everything in b) is inverted as well..
\begin{align*}U_{n-1} &= \int \frac{dx}{(x^2+1)^{n-1}}\\ &= \frac{x}{(x^2+1)^{n-1}}-\int x\cdot \frac{2(-n+1)x}{(x^2+1)^n}\\ &= \frac{x}{(x^2+1)^{n-1}}-\int \frac{2x^2(1-n)}{(x^2+1)^n}\,dx \end{align*}

Subject to some minor computational error
Title: Re: 4U Maths Question Thread
Post by: envisagator on July 24, 2018, 08:00:49 pm
Hi Rui, for the inequality proof question posted as an image below, I've been looking at it for a while, just cant see the actually logical reason to why the part highlighted in the box is true. Plus, when i do some questions, is it 'normal' just to sorta guess how some of your facts are bigger or smaller than another fact such that when you combine them you can get easily get to the desired result. I dont know whether that makes sense, also is it necessary to write equality iff - bla bla bla?? Thank You in advance, sorry for it being a mouthful to take in
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 25, 2018, 09:12:09 am
Hi Rui, for the inequality proof question posted as an image below, I've been looking at it for a while, just cant see the actually logical reason to why the part highlighted in the box is true. Plus, when i do some questions, is it 'normal' just to sorta guess how some of your facts are bigger or smaller than another fact such that when you combine them you can get easily get to the desired result. I dont know whether that makes sense, also is it necessary to write equality iff - bla bla bla?? Thank You in advance, sorry for it being a mouthful to take in
\begin{align*}a^2+2ab+b^2 &\leq 2a^2+2b^2\\ \frac{a^2+2ab+b^2}{4}&\leq \frac{a^2+b^2}{2}\\ \left( \frac{a+b}{2} \right)^2 &\leq \frac{a^2+b^2}{2}\end{align*}
Note that \( 2ab \leq a^2+b^2\) comes from rearranging \( \frac{a+b}{2} \geq \sqrt{ab} \).

I'm not sure if the lecture handout is uploaded yet but the basic facts that you can assume for inequality proofs are all there. There really isn't that many (surprisingly), but people tend to struggle because they don't know what they are, which is what I tried to address.
Title: Re: 4U Maths Question Thread
Post by: cthulu on July 26, 2018, 01:54:28 pm
Hello, having trouble with this one integration question.



The solution is:



But I am getting:

Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 26, 2018, 02:04:57 pm
Hello, having trouble with this one integration question.



The solution is:



But I am getting:



Title: Re: 4U Maths Question Thread
Post by: cthulu on July 26, 2018, 02:27:45 pm



Wow thanks, I did it a stupid way that is why I was getting a constant.

Another question:

Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 26, 2018, 03:27:12 pm
Wow thanks, I did it a stupid way that is why I was getting a constant.

Another question:


Whilst the question is doable by considering \(x+iy = \frac{t^2-1}{t^2+1} - \frac{2ti}{t^2+1}\), I find that it's only really viable for the equation of the locus, and becomes kinda useless when dealing with the restriction on \(t\). More or less because optimising \( f(t) = \frac{t^2-1}{t^2+1} \) and \(g(t) = \frac{-2t}{t^2+1}\) over \(0\leq t \leq 1\) is really handwavy unless we use calculus, but that's not just one, but two disgusting quotient rule bashes.



________________________________________________





________________________________________________



Note that \( (0,-1) \) is also included in the locus. But it was actually taken care of during the messier part of the question; it didn't need to be handled separately. May be a good idea to restate it as well just in case.
Title: Re: 4U Maths Question Thread
Post by: cthulu on July 26, 2018, 04:28:31 pm



Thanks for the reply, I was doing this a little differently with less working out:

Your response is very good, really helps explain everything!

Another question if you don't mind:

(https://scontent-syd2-1.xx.fbcdn.net/v/t1.15752-9/37820053_408870752968181_1265485472084262912_n.jpg?_nc_cat=0&oh=f69107dcbd32c7ad0860122c19e1c6f6&oe=5BC59D8C)
Title: Re: 4U Maths Question Thread
Post by: Ali_Abbas on July 26, 2018, 07:10:18 pm
Quote
Another question:



Rui's solution is a slight overkill, so I recommend the following solution as an (easier) alternative.








Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 26, 2018, 08:35:10 pm
The main reason why I avoid the arg when handling such problems is because arg properties are just hard to use and we cannot assume anything about atan2 in HSC maths. Use of arg relies exclusively on properties that don't always necessarily hold.

Plus, the concept of the principal argument isn't really well defined in the HSC. There's no real distinction between arg(z) and Arg(z) as there should be. That makes the properties of the argument more dangerous to me.

It is definitely at the expense of overkill, which is why I'm not necessarily the biggest fan of my own method. But I'm way more sure of what I'm doing without it.

Of course, as for the modulus, it's definitely easier to just compute \(|z|\) directly. I just didn't anticipate that it would actually be a unit circle by inspection; I was expecting something more bizarre.
Thanks for the reply, I was doing this a little differently with less working out:
  • Find |z| which is 1 so the locus lies on the unit circle.
  • Rationalise z and split it up so it is in the form z = x + iy
  • Use 0<=t<=1 with x and y to determine domain and arg(z) and then draw the locus. However when trying to find the arg I got 0<=arg(z)<= arctan(-2/0). Here was where I was stuck.

Your response is very good, really helps explain everything!

Another question if you don't mind:

(https://scontent-syd2-1.xx.fbcdn.net/v/t1.15752-9/37820053_408870752968181_1265485472084262912_n.jpg?_nc_cat=0&oh=f69107dcbd32c7ad0860122c19e1c6f6&oe=5BC59D8C)

I'll take a closer look at this soon but in the meantime is there a diagram attached to this? My hangry brain cannot visualise it right now

Edit: Actually, more importantly, are there any assumptions on the normal reaction forces of the particles? Or the masses of the particles? There's no indication that the particles actually travel with the same velocity here.
Title: Re: 4U Maths Question Thread
Post by: cthulu on July 27, 2018, 09:39:55 am
Removed

Oh yes, sorry I should have shown the first part too, my bad.

(https://imgur.com/sFy4bDZ.png)
Title: Re: 4U Maths Question Thread
Post by: hassrax on July 27, 2018, 09:55:37 am
Hi I need help with this question below on how to approach it and the logic behind the steps. Thank you
Sketch y=f(x)=(x^n)e^-x  for x>0, n>1
Thank you
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 27, 2018, 11:21:31 am
Oh yes, sorry I should have shown the first part too, my bad.

(https://imgur.com/sFy4bDZ.png)
This is just a brainstorm here:

Assume that the angular displacement goes on forever, i.e. after it reaches \(2\pi\) it'll keep going to \(4\pi\), \(6\pi\) and so on.

Using \(v = r\omega\) you can deduce that \(\omega = \sqrt{ \frac{g\tan \theta}{R} } \). Using \( \theta = \omega t\) you can then deduce that for each particle, \( \theta = t  \sqrt{ \frac{g\tan \theta}{R} } \). Note that since the particles start side by side, we just naturally assume that their initial angular displacement is \(\theta = 0\) for convenience.

By subbing in \(g=10\) along with both corresponding values of \(\tan\theta\) and \(R\), you can deduce which of the two particles is travelling faster.

Then, the first time the particles are next to each other should be when \( \boxed{ \theta_{faster} - \theta_{slower} = 2\pi}\). In fact, the second time they are next to each other should also be when this difference equals to 4*pi. Essentially, at each point where their angular displacements differ by some integer multiple of \(2\pi\), they are next to each other.

Hi I need help with this question below on how to approach it and the logic behind the steps. Thank you
Sketch y=f(x)=(x^n)e^-x  for x>0, n>1
Thank you

This question is simply screaming "multiplication of \(y\)-coordinates" and that's all there is to it.

Are you uncomfortable with the whole process of multiplying ordinates and require that logic to be expanded for your example?
Title: Re: 4U Maths Question Thread
Post by: scienceislife on August 18, 2018, 04:52:21 pm
Find the volume bounded by y = 10 and y = x + 16/x  about x = −2
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 18, 2018, 06:38:34 pm
Find the volume bounded by y = 10 and y = x + 16/x  about x = −2
Do you want to use cylindrical shells or washers (slicing)?
Title: Re: 4U Maths Question Thread
Post by: scienceislife on August 18, 2018, 06:55:03 pm
Either method would be great, because I've tried both but cannot seem to get a reasonable answer.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 18, 2018, 07:10:51 pm
(https://i.imgur.com/SeLU9OZ.png)
Title: Re: 4U Maths Question Thread
Post by: phunky on August 18, 2018, 11:24:32 pm
hey guys,
could someone please explain how to graph the last one? Nut quite sure how to do those types of questions!
lol figuring out how to paste an image in was such a struggle
(https://i.imgur.com/sIWXj2V.png)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 19, 2018, 09:51:13 am
hey guys,
could someone please explain how to graph the last one? Nut quite sure how to do those types of questions!
lol figuring out how to paste an image in was such a struggle
(https://i.imgur.com/sIWXj2V.png)
Firstly, similar to \(y = f(|x|) \) you block out the part of the curve to the left of the \(y\)-axis. This is because regardless of what \(x\) is, \(x^2\) will always be positive, so you're always taking \(f\) of a positive number.

In particular, it's somewhat similar to \( y = f(|x|) \) in that you need to reflect the right-portion of the curve onto the left. However the fact that it involves \(x^2\) means that the graph becomes somewhat distorted as well, because it's not an ordinary dilation like \( y = f(3x)\).

Furthermore, note that since \( (4, 3) \) lies on \( y = f(x)\), we should expect \( (2,3) \) AND \( (-2,3) \) to lie on \( y = f(x^2) \). Since \( \left( \frac74, 0 \right) \) lies on \(y = f(x)\), we should expect \( \left( \frac{\sqrt7}{2}, 0 \right) \) AND \( \left( -\frac{\sqrt7}{2}, 0 \right) \) to lie on \( y = f(x^2)\).

(Picture was way too late but edited in anyway for completeness)
(https://i.imgur.com/x4SnfPEl.png)
Title: Re: 4U Maths Question Thread
Post by: phunky on August 19, 2018, 04:37:14 pm
ahhh thanks, got it!   ;D
Title: Re: 4U Maths Question Thread
Post by: yammy on August 19, 2018, 06:08:11 pm
Hey Rui, can you please help me with Q15a) i and iii? -I don't understand the solutions provided
For i) why is p a double root if the hyperbola touches the ellipse at P?
For iii) what do they mean by the parameter at the point Q? Why is it -p? And how do they know that O is the midpoint of PQ?
Thank you in advance(:
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 19, 2018, 07:41:37 pm
Hey Rui, can you please help me with Q15a) i and iii? -I don't understand the solutions provided
For i) why is p a double root if the hyperbola touches the ellipse at P?
For iii) what do they mean by the parameter at the point Q? Why is it -p? And how do they know that O is the midpoint of PQ?
Thank you in advance(:
Because \(P\) is the point \( \left(cp, \frac{c}{p} \right) \), the parameter at \(P\) is \(p\). That's literally it - the parameter at a point is just the parameter that represents the coordinates of that point. The reason why \(Q\) must therefore be \(-p\) is because clearly the ellipse and the hyperbola intersect at only two distinct points. So because \( \left( cp, \frac{c}{p} \right) \) corresponds to \(P\), this leaves us with \( \left( -cp, -\frac{c}{p} \right) \) corresponding to \(Q\). So clearly \(-p\) must be the parameter representing the point \(Q\).

(It's the same as the parabola \(x^2 = 4ay\). If a point is marked \( P(2ap, ap^2)\), then the parameter at \(P\) is just \(p\).)

It is then easy to show by the midpoint formula that the midpoint of \(P\) and \(Q\) is the origin.



For example, the quadratic equation \( x^2 - 2x + 1 = 0\) has only one unique solution. (In particular, for that one it will be \(x = 1\).) That solution is also a double root of the equation.



It's quite abstract to argue formally, but essentially the idea is to generalise the intersection between a conic and a tangent line at the point of contact, to the intersection between two "touching" conics at their point of contact instead.
Title: Re: 4U Maths Question Thread
Post by: yammy on August 19, 2018, 08:48:54 pm
Because \(P\) is the point \( \left(cp, \frac{c}{p} \right) \), the parameter at \(P\) is \(p\). That's literally it - the parameter at a point is just the parameter that represents the coordinates of that point. The reason why \(Q\) must therefore be \(-p\) is because clearly the ellipse and the hyperbola intersect at only two distinct points. So because \( \left( cp, \frac{c}{p} \right) \) corresponds to \(P\), this leaves us with \( \left( -cp, -\frac{c}{p} \right) \) corresponding to \(Q\). So clearly \(-p\) must be the parameter representing the point \(Q\).

(It's the same as the parabola \(x^2 = 4ay\). If a point is marked \( P(2ap, ap^2)\), then the parameter at \(P\) is just \(p\).)

It is then easy to show by the midpoint formula that the midpoint of \(P\) and \(Q\) is the origin.



For example, the quadratic equation \( x^2 - 2x + 1 = 0\) has only one unique solution. (In particular, for that one it will be \(x = 1\).) That solution is also a double root of the equation.



It's quite abstract to argue formally, but essentially the idea is to generalise the intersection between a conic and a tangent line at the point of contact, to the intersection between two "touching" conics at their point of contact instead.

ahh i understand, thank youu!!
While you're at it, can you please help me with this question as well?
bi, ii and iv)

For i) i found f'(x) and let f'(x)=0 to find the stat point, but I cant really get an answer
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 19, 2018, 09:12:28 pm
ahh i understand, thank youu!!
While you're at it, can you please help me with this question as well?
bi, ii and iv)

For i) i found f'(x) and let f'(x)=0 to find the stat point, but I cant really get an answer
Hint for iv): You want \(n!\) to appear under the power. But you know that \( n! = 1\times 2\times \dots \times n\)....................

If you've correctly found the stationary point \( x = \frac{c}{n} \), you should then proceed to prove that it is a local minimum and argue that it is a global minimum. You're essentially doing the same things I did in my trial survival lecture regarding the global minimum.

Title: Re: 4U Maths Question Thread
Post by: yammy on August 20, 2018, 12:09:56 am
Hint for iv): You want \(n!\) to appear under the power. But you know that \( n! = 1\times 2\times \dots \times n\)....................

If you've correctly found the stationary point \( x = \frac{c}{n} \), you should then proceed to prove that it is a local minimum and argue that it is a global minimum. You're essentially doing the same things I did in my trial survival lecture regarding the global minimum.



ohhh thanks rui i get it  :D
Can you also help me with the 2nd part of di)?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 20, 2018, 08:51:10 am
ohhh thanks rui i get it  :D
Can you also help me with the 2nd part of di)?



Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on August 23, 2018, 12:19:32 pm
hey can someone please help me with q10
thank you

Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 23, 2018, 05:44:18 pm
hey can someone please help me with q10
thank you


I’ve attempted the question but because of its very vague wording there was very little that I could salvage from it. Please provide the final answer for me to confirm with before I post any solution.
Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on August 23, 2018, 09:09:04 pm
I’ve attempted the question but because of its very vague wording there was very little that I could salvage from it. Please provide the final answer for me to confirm with before I post any solution.


there is no solutions :(
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 23, 2018, 11:43:07 pm

there is no solutions :(
That's cool, but like was looking for just the final **answer**. Anyway here is a sketch solution.
(https://i.imgur.com/ZQhzONh.png)
Title: Re: 4U Maths Question Thread
Post by: yammy on August 23, 2018, 11:50:51 pm
hello, can someone please help me with d) - mathematical induction question and 16a)? i cant seem to get a solution and there are no solutions provided
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 24, 2018, 08:47:10 am
hello, can someone please help me with d) - mathematical induction question and 16a)? i cant seem to get a solution and there are no solutions provided
The induction question is already addressed in the compilation as it is a part of the 2016 paper.

That is a very long question. What have you attempted thus far?
Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on August 24, 2018, 02:53:15 pm
That's cool, but like was looking for just the final **answer**. Anyway here is a sketch solution.
(https://i.imgur.com/ZQhzONh.png)


why did you introduce another R?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 24, 2018, 03:52:20 pm

why did you introduce another R?
Because the radius of the circular motion isn’t necessarily the same as the radius of the actual hemisphere that makes up the bowl.

r is the radius of the hemisphere. But the particle would typically be at a point further below the rim of the hemisphere. It’ll be undergoing circular motion with a smaller radius because of the fact it’s at a lower depth.
Title: Re: 4U Maths Question Thread
Post by: Jeeffffffffffffff on August 29, 2018, 07:39:55 pm
Hey people, so just as I thought mechanics was going alright, good ol' fitzpatrick threw me a bit of a curveball.
A body, projected vertically upwards with speed U, returns with a speed V. Assuming constant gravity and air resistance proportional to the square of the speed, find the total time taken.
I just find it very vague and I'm not really sure how I'm meant to go about doing this, any help would be appreciated, thanks.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 29, 2018, 07:46:20 pm
Hey people, so just as I thought mechanics was going alright, good ol' fitzpatrick threw me a bit of a curveball.
A body, projected vertically upwards with speed U, returns with a speed V. Assuming constant gravity and air resistance proportional to the square of the speed, find the total time taken.
I just find it very vague and I'm not really sure how I'm meant to go about doing this, any help would be appreciated, thanks.
Your final answer should be in terms of \(U\) and \(V\). You should consider the cases where the particle is going up and going down separately. In the case of the particle going up you will have \( \ddot{x} = -g - kv^2\) and the boundary cases:
- Initially, \(t = 0\) and \(v = U\)
- At the max height, \(t = \text{what you want to find}\) and \( v = 0\).

Then when the particle is going down you will have \( \ddot{x} = g - kv^2\).
Title: Re: 4U Maths Question Thread
Post by: Jeeffffffffffffff on August 29, 2018, 08:51:11 pm
Your final answer should be in terms of \(U\) and \(V\). You should consider the cases where the particle is going up and going down separately. In the case of the particle going up you will have \( \ddot{x} = -g - kv^2\) and the boundary cases:
- Initially, \(t = 0\) and \(v = U\)
- At the max height, \(t = \text{what you want to find}\) and \( v = 0\).

Then when the particle is going down you will have \( \ddot{x} = g - kv^2\).
Aaaaahhh yes that makes a lot of sense, thanks for that Rui. Just out of curiosity though, I have two different formulas for integrating something with 1/(a2-x2) and the way to tell which one to use is that for one |x|<a and the other you use when |x|>a,but in this question a ended up being the square root of g/k and x was the velocity, which was just v, so how would I go about figuring out if |v| is larger than or less than the square root of g/k??
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 29, 2018, 08:59:13 pm
Aaaaahhh yes that makes a lot of sense, thanks for that Rui. Just out of curiosity though, I have two different formulas for integrating something with 1/(a2-x2) and the way to tell which one to use is that for one |x|<a and the other you use when |x|>a,but in this question a ended up being the square root of g/k and x was the velocity, which was just v, so how would I go about figuring out if |v| is larger than or less than the square root of g/k??
You really should be using partial fractions to integrate that expression instead of jump to \( \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right| \). But to compare \(v\) to something like \(\sqrt{\frac{g}{k}} \), you can exploit the fact that \( \ddot{x} > 0\) after you change the orientation to be the more convenient one.

(Equivalently, just set \( \ddot{x} = 0\) because \( \sqrt{\frac{g}{k}}\) can be shown to be the terminal velocity, which is always a cap for the velocity.)
Title: Re: 4U Maths Question Thread
Post by: aadharmg on August 31, 2018, 07:58:48 pm
Are we allowed to use the standard integral formulas in the HSC exam or not because of its omission from the paper? For example the ln(x^2 + sqrroot(x^2 +/- a^2) formula.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 31, 2018, 08:10:42 pm
Are we allowed to use the standard integral formulas in the HSC exam or not because of its omission from the paper? For example the ln(x^2 + sqrroot(x^2 +/- a^2) formula.
Not anymore in your actual exam. But the odds that they will examine that now are also extremely low.
Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on September 02, 2018, 04:32:13 pm
hey can someone please help me with this question?
-----

let a, b be positive real numbers so that a+b=1. prove: a/(1+a) + b/(1+b) is less that or equal to 2/3
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 02, 2018, 04:58:32 pm
hey can someone please help me with this question?
-----

let a, b be positive real numbers so that a+b=1. prove: a/(1+a) + b/(1+b) is less that or equal to 2/3



Title: Re: 4U Maths Question Thread
Post by: envisagator on September 02, 2018, 05:26:24 pm
Hi Rui, just a question on volumes of solids of revolution by slices.



Thank You!!!
Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on September 02, 2018, 05:32:27 pm





thank you so much. but shouldn't the second equation you boxed be 1>=4ab not < ?

is it ok if you can try part ii of this question? I can't seem to get the volume equation they have.

(https://scontent-syd2-1.xx.fbcdn.net/v/t1.15752-9/40622720_2119421968377060_115003800959320064_n.png?_nc_cat=0&oh=89517562bb0af26a4084b519943bc708&oe=5BFA3EDF)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 02, 2018, 06:30:32 pm
Hi Rui, just a question on volumes of solids of revolution by slices.



Thank You!!!
Fairly sure if you're using slicing the big radius is just \(1\). Not \(1 - \sin y\). (Which will give you \(\delta V = \pi (1 - \sin^2 y)\delta y \))
thank you so much. but shouldn't the second equation you boxed be 1>=4ab not < ?

is it ok if you can try part ii of this question? I can't seem to get the volume equation they have.

(https://scontent-syd2-1.xx.fbcdn.net/v/t1.15752-9/40622720_2119421968377060_115003800959320064_n.png?_nc_cat=0&oh=89517562bb0af26a4084b519943bc708&oe=5BFA3EDF)
Yeah I'll fix the typo shortly



Edit: Pictures
(https://i.imgur.com/vALEJr6.png)
Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on September 02, 2018, 06:33:17 pm



I don't get how u got x =10 + (10-h)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 02, 2018, 06:37:33 pm
I don't get how u got x =10 + (10-h)
See the edit. It's the exact same method I use in the book
Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on September 02, 2018, 06:42:22 pm
See the edit. It's the exact same method I use in the book

thank you so much!
Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on September 02, 2018, 08:03:19 pm
could we be asked hard simple harmonic questions and projectile questions in a 4u exam?


100 post!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 02, 2018, 08:15:31 pm
could we be asked hard simple harmonic questions and projectile questions in a 4u exam?


100 post!
They combined resisted motion with projectile motion back in 2003 so I wouldn't rule it out completely
Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on September 02, 2018, 08:36:33 pm
They combined resisted motion with projectile motion back in 2003 so I wouldn't rule it out completely

so its possible, but highly unlikely that they will?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 02, 2018, 08:42:36 pm
so its possible, but highly unlikely that they will?
Yeah pretty much
Title: Re: 4U Maths Question Thread
Post by: cthulu on September 06, 2018, 05:49:10 pm
Hey had this conics question I was stuck on:

Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 06, 2018, 05:55:38 pm
Hey had this conics question I was stuck on:


This is just the usual reflection property of the ellipse. You should work through Q4a of the 2009 paper to attempt this.
Title: Re: 4U Maths Question Thread
Post by: vikasarkalgud on September 09, 2018, 09:59:26 am
Hey Rui, got a conical pendulum question,

A smooth hollow cone with semi-vertex angle 'alpha' is placed with its axis vertical and vertex down.. A particle moves in a horizontal circle on its inner surface, making n revolutions per second. Find the radius of the circle of motion.

I got r = gtan(alpha)/(4n^2pi^2) but the answer has tan(alpha) in the denominator idk how.

Thanks for help
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 09, 2018, 10:27:31 am
Hey Rui, got a conical pendulum question,

A smooth hollow cone with semi-vertex angle 'alpha' is placed with its axis vertical and vertex down.. A particle moves in a horizontal circle on its inner surface, making n revolutions per second. Find the radius of the circle of motion.

I got r = gtan(alpha)/(4n^2pi^2) but the answer has tan(alpha) in the denominator idk how.

Thanks for help
It looks like I get \(\tan \alpha\) in the denominator as well.

(https://i.imgur.com/GnGMa9o.png)
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on September 14, 2018, 10:24:11 pm
hey Rui need some help with ext2 combinations que
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 15, 2018, 07:38:59 am
hey Rui need some help with ext2 combinations que


------------------------------------------------------------


Note now, that the fact that the last goal must be scored by Mary is important. That is a deadlocked position, because otherwise either the game is not finished, or we've started considering the situation where Ferdinand wins and thus their roles have been swapped. We don't want to do that - this was the purpose of the \(2!\) earlier.

Thus, in a typical arrangement of who wins each round, Mary constantly occupies four variable positions. In the following computations, I will drop off the fact Mary wins the last one.



Title: Re: 4U Maths Question Thread
Post by: itssona on September 19, 2018, 08:10:07 pm
Heya:)
from 2012 HSC paper, 16 c) iii
how do we simplify this expression
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 19, 2018, 08:42:49 pm
Heya:)
from 2012 HSC paper, 16 c) iii
how do we simplify this expression

Did you mean the second part? (If so, note that the RHS of your expression has a mistake in the denominator)
\begin{align*} \frac{(n-1)! k }{n^k (n-k)!} &\geq \frac{(n-1)!(k-1)}{n^{k-1} (n-k+1)!} \\ \frac{k}{n(n-k)!} &\geq \frac{k-1}{(n-k+1)!}\\ \frac{k}{n(n-k)!} &\geq \frac{k-1}{(n-k)! (n-k+1)}\\ k(n-k+1) &\geq n(k-1)\\ kn - k^2 + k & \geq kn - n\\ k^2 - k -n& \leq 0\end{align*}
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 28, 2018, 05:35:19 pm
Hi, there this question is not attached but it is the very last question of the 2005 hsc..... I can do every part but I am struggling to understand the solutions for the last part of the question..... If anyone could somehow explain?
See this recent add to the compilation (you will need to scroll down for the last part). The main reason I didn't address this sooner was because I wanted to cover the entire question as it is somewhat demanding, but I was also on vacation until today so I wasn't feeling bothered enough to do it all.
Title: Re: 4U Maths Question Thread
Post by: Dragomistress on September 29, 2018, 05:19:06 am
Hi,
For all cube root of unity questions do I do:
w^3=1
w^3-1=0
(w-1)(w^2+w+1)=0
w-1=0 and w^2+w+1=0 equations and use these to find solutions?
Title: Re: 4U Maths Question Thread
Post by: clovvy on September 29, 2018, 06:05:37 am
Hi,
For all cube root of unity questions do I do:
w^3=1
w^3-1=0
(w-1)(w^2+w+1)=0
w-1=0 and w^2+w+1=0 equations and use these to find solutions?
careful dude, for cube roots of unity w is a complex number, so w-1 cannot be one as it is not purely real, although w^3 is a real number that is one... this type of proof you need to be familiar with, and 1+w+w^2 has to be zero....
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 29, 2018, 11:44:50 am
Hi,
For all cube root of unity questions do I do:
w^3=1
w^3-1=0
(w-1)(w^2+w+1)=0
w-1=0 and w^2+w+1=0 equations and use these to find solutions?

Title: Re: 4U Maths Question Thread
Post by: envisagator on October 03, 2018, 06:20:08 pm
Hi Rui, just need help with this polynomial question.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 03, 2018, 06:39:41 pm
Hi Rui, just need help with this polynomial question.


Basically, if you get lost in the formulae, keep writing \( \gamma\) and \( \delta\) but remember to sub \(\gamma = \alpha\) and \(\delta = \beta\) later on.



Title: Re: 4U Maths Question Thread
Post by: zawszeyi on October 03, 2018, 08:55:24 pm
Ok so im a little worried doing the 4u maths past papers. Im getting around 65's at this point.
What kind of raw marks do I need for a band 6? About 70?
Thanks.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 03, 2018, 08:57:54 pm
Ok so im a little worried doing the 4u maths past papers. Im getting around 65's at this point.
What kind of raw marks do I need for a band 6? About 70?
Thanks.
Around there typically scrapes the E4 as you say.

I don't know how to help you any further without more info.
Title: Re: 4U Maths Question Thread
Post by: justwannawish on October 04, 2018, 11:36:50 am
Hey, how do you solve this integral:
(xe^x) (1+x)

I tried doing by parts using u=x/1+x but didn't really get anywhere
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 04, 2018, 11:51:40 am
Hey, how do you solve this integral:
(xe^x) (1+x)

I tried doing by parts using u=x/1+x but didn't really get anywhere
I’m reading this as \( \int xe^x(1+x)\,dx \) which is just \( \int (x^2+x) e^x\, dx \) and can be handled by two applications of integration by parts. Where are you getting that substitution from?
Title: Re: 4U Maths Question Thread
Post by: justwannawish on October 04, 2018, 12:07:23 pm
I’m reading this as \( \int xe^x(1+x)\,dx \) which is just \( \int (x^2+x) e^x\, dx \) and can be handled by two applications of integration by parts. Where are you getting that substitution from?

Sorry that was completely my fault!
It was meant to be xe^x/(1+x)
Title: Re: 4U Maths Question Thread
Post by: clovvy on October 04, 2018, 12:17:48 pm
Sorry that was completely my fault!
It was meant to be xe^x/(1+x)
I tried doing it with let u=1+x and it is a lot nicer than your substitution..  However integrals like e^x/x is a special property not in the syllabus...  I end up integrating e^(u-1)-e^(u-1)/u by using substitution and the final answer is a property that we don't even learn in 4U
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 04, 2018, 03:50:56 pm
Yeah basically that's a non elementary integral. It might be doable with appropriate boundaries but there is definitely no known antiderivative for it.

(https://i.imgur.com/foiv2PX.png)
Title: Re: 4U Maths Question Thread
Post by: justwannawish on October 04, 2018, 11:18:42 pm


Thanks guys! It was a question on one of the trial papers our school bought (our actual trial was made up of questions from 3~ independent papers) and our teacher briefly mentioned that the question (which was originally xe^x/(1+e^x) ) was wrong and that xe^x/(1+x) was meant to be the correct q. He might have just stuffed it up, but thank you guys for all your help!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 05, 2018, 09:13:20 am
Thanks guys! It was a question on one of the trial papers our school bought (our actual trial was made up of questions from 3~ independent papers) and our teacher briefly mentioned that the question (which was originally xe^x/(1+e^x) ) was wrong and that xe^x/(1+x) was meant to be the correct q. He might have just stuffed it up, but thank you guys for all your help!
Both of them are non-elementary actually. I do vaguely recall this of this debuckle actually, but I’m fairly sure whatever the integral was supposed to be should’ve had some boundaries attached to it. Check with your teacher what the integration boundaries were because I think your integrand may have been correct...?
Title: Re: 4U Maths Question Thread
Post by: justwannawish on October 05, 2018, 09:56:32 am
Both of them are non-elementary actually. I do vaguely recall this of this debuckle actually, but I’m fairly sure whatever the integral was supposed to be should’ve had some boundaries attached to it. Check with your teacher what the integration boundaries were because I think your integrand may have been correct...?

The boundaries were 1 to 0 if that helps :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 05, 2018, 12:39:04 pm
The boundaries were 1 to 0 if that helps :)
Had to consult with a friend just in case and even then it still looks not doable (damn). If you ever get to see the solutions on the independents paper please keep us updated haha
Title: Re: 4U Maths Question Thread
Post by: envisagator on October 07, 2018, 11:21:17 am
Hi Rui, can you help me with part (c) of the question. Thanks in advance
Title: Re: 4U Maths Question Thread
Post by: clovvy on October 07, 2018, 04:15:53 pm
Hi Rui, can you help me with part (c) of the question. Thanks in advance
Use the equaion from part (i)


Also I don't know how to do part (ii) so if anyone can help there that would be great..
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 07, 2018, 06:37:49 pm
Hi Rui, can you help me with part (c) of the question. Thanks in advance
Yeah pretty much see above. All that it really wanted you to interpret is that when \(v = V_1\), \(F = F_1\), and also when \(v = V_2\), \(F = F_2\). Then we just do the best thing we can (i.e. cancellation) to get rid of the \(m\), \(r\) and \(\cos \alpha\).
Use the equaion from part (i)


Also I don't know how to do part (ii) so if anyone can help there that would be great..
Then the friction is reversed and starts going upwards instead. Which makes sense because if the velocity is too small, the particle will start to slide down the banked track, so you wanna push it back up there.
Title: Re: 4U Maths Question Thread
Post by: wlam on October 09, 2018, 03:39:56 pm
Hi could someone pls help me w this question thanks
Title: Re: 4U Maths Question Thread
Post by: clovvy on October 09, 2018, 03:44:23 pm
Hi could someone pls help me w this question thanks
which parts do you need help with specifically?
Anyway, I'll start you off with part (i), expand the expression (cisx)^5 manually

Title: Re: 4U Maths Question Thread
Post by: wlam on October 09, 2018, 05:02:54 pm
which parts do you need help with specifically?
Anyway, I'll start you off with part (i), expand the expression (cisx)^5 manually


hi thanks!

i need help with part b
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 09, 2018, 08:09:26 pm
Hi could someone pls help me w this question thanks




Title: Re: 4U Maths Question Thread
Post by: wlam on October 10, 2018, 07:33:24 pm






thanks!!
Title: Re: 4U Maths Question Thread
Post by: A Selective Student on October 12, 2018, 10:24:09 am
Hey Rui can i get help with these 2 questions. One is circle geo and one is complex. Would need some help on the working out.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 12, 2018, 02:23:58 pm
Hey Rui can i get help with these 2 questions. One is circle geo and one is complex. Would need some help on the working out.
(https://i.imgur.com/fK6kTCk.png)




There is a geometric way around this question as well, and it's quite interesting. But I found I had to do a tiny bit more work than this algebraic method.
Title: Re: 4U Maths Question Thread
Post by: envisagator on October 12, 2018, 03:18:30 pm
Hi Rui, can I have help with this induction q: I was think assume n=k and n=k-1, but the answer's did n=k and n=k+1 just not sure.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 12, 2018, 03:35:30 pm
Hi Rui, can I have help with this induction q: I was think assume n=k and n=k-1, but the answer's did n=k and n=k+1 just not sure.

If it's just the issue of {assume n=k, n=k-1 and prove n=k+1} v.s. {assume n=k+1, n=k and prove n=k+2} either works. Perhaps the reason why they did the latter is because by doing the latter you let \(k\geq 1\), in accordance with what the question provides. Assuming the former makes the working out heaps tidier, just that you need to let \(k \geq 2\) instead.

Did you need help with the computations?
Title: Re: 4U Maths Question Thread
Post by: envisagator on October 12, 2018, 03:53:40 pm
If it's just the issue of {assume n=k, n=k-1 and prove n=k+1} v.s. {assume n=k+1, n=k and prove n=k+2} either works. Perhaps the reason why they did the latter is because by doing the latter you let \(k\geq 1\), in accordance with what the question provides. Assuming the former makes the working out heaps tidier, just that you need to let \(k \geq 2\) instead.

Did you need help with the computations?
Thats all I needed to know that you!!.
Title: Re: 4U Maths Question Thread
Post by: A Selective Student on October 12, 2018, 05:06:13 pm
(https://i.imgur.com/fK6kTCk.png)




There is a geometric way around this question as well, and it's quite interesting. But I found I had to do a tiny bit more work than this algebraic method.

Wow can't believe i didn't see those methods for both question o.O Thanks so much!
Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on October 20, 2018, 04:31:01 pm
hello

can someone help pleaseee because Im not getting the right answer.

thank you
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2018, 04:33:40 pm
hello

can someone help pleaseee because Im not getting the right answer.

thank you
\begin{align*} \int \frac{1}{1-\sin x}\,dx &= \int \frac{1 + \sin x}{1 - \sin^2 x}\,dx\\ &= \int \frac{1 + \sin x}{\cos^2 x}\,dx\\ &= \int \frac{1}{\cos^2 x} + \frac{\sin x}{\cos^2 x}\,dx\\ &= \int \sec^2 x + \sec x \tan x \,dx\\ &= \sec x +\tan x + C \end{align*}
Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on October 20, 2018, 04:39:19 pm
\begin{align*} \int \frac{1}{1-\sin x}\,dx &= \int \frac{1 + \sin x}{1 - \sin^2 x}\,dx\\ &= \int \frac{1 + \sin x}{\cos^2 x}\,dx\\ &= \int \frac{1}{\cos^2 x} + \frac{\sin x}{\cos^2 x}\,dx\\ &= \int \sec^2 x + \sec x \tan x \,dx\\ &= \sec x +\tan x + C \end{align*}

thank you!!
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on October 21, 2018, 06:39:36 am
hello people, need some help please, Q10 from 2014 hsc
Title: Re: 4U Maths Question Thread
Post by: clovvy on October 21, 2018, 07:20:42 am
hello people, need some help please, Q10 from 2014 hsc

So the answer is D, and pay attention at how I use u as a dummy variable for x...
Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on October 21, 2018, 09:29:03 pm
hello

I just wanted to ask,  up to what year/how far back should I go with the hsc past papers since the syllabus changed and the older papers are more difficult?

thank you
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 21, 2018, 09:38:38 pm
hello

I just wanted to ask,  up to what year/how far back should I go with the hsc past papers since the syllabus changed and the older papers are more difficult?

thank you
They do get harder the further back you go, however anything from 2001 onwards is still relevant, so as far as you can go back to there (factoring in that you also have other subjects to study for).
Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on October 21, 2018, 10:01:05 pm
sorry can someone help me with these questions please. with the volume question I'm confused because they did not specify which axis its rotated by so how am I supposed to know.
thank you
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 21, 2018, 11:05:30 pm
sorry can someone help me with these questions please. with the volume question I'm confused because they did not specify which axis its rotated by so how am I supposed to know.
thank you
Actually that is a fair call with the volumes question. That question is in theory not doable because they haven't specified the line about which the rotation is taken with respect to.

As this is a reasonably huge dump of strenuous questions I (more or less because I'll get weary :() won't go into complete depth with them. However it is worth mentioning that for the 3 variable AM-GM inequality you've mentioned, in the current HSC there would be hints that help you build up to that result, as opposed to being shoved straight into the deep end. This is one way of proving the AM-GM inequality for 3 variables given that you've already proven it for 4 variables, and to prove it for 4 variables you can work through my trial lecture handout solutions.

The angular velocity \( \omega\) is defined by \( \omega = \frac{\d\theta}{dt} \), i.e. the rate at which the angle the particle makes, at the origin in the positive \(x\)-axis, changes with respect to time. That derivation was examined in the 1981 paper (and from memory it should've been stated in my 4U notes book), but it essentially relies on you starting with \( x = r\cos \theta\), \(y = r\sin \theta\) and then using implicit differentiation to obtain results like \( \dot{x} = \frac{d\theta}{dt} \times -r\sin \theta = -r\omega \sin \theta\), etc.

That locus essentially describes the arc, with endpoints at \(z_1\) and \(z_2\), going in an anticlockwise direction from \(z_1\) to \(z_2\) and not including the points \(z_1\) and \(z_2\) themselves.
Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on October 21, 2018, 11:10:19 pm
The angular velocity \( \omega\) is defined by \( \omega = \frac{\d\theta}{dt} \), i.e. the rate at which the angle the particle makes, at the origin in the positive \(x\)-axis, changes with respect to time. That derivation was examined in the 1981 paper (and from memory it should've been stated in my 4U notes book), but it essentially relies on you starting with \( x = r\cos \theta\), \(y = r\sin \theta\) and then using implicit differentiation to obtain results like \( \dot{x} = \frac{d\theta}{dt} \times -r\sin \theta = -r\omega \sin \theta\), etc.

That locus essentially describes the arc, with endpoints at \(z_1\) and \(z_2\), going in an anticlockwise direction from \(z_1\) to \(z_2\) and not including the points \(z_1\) and \(z_2\) themselves.
the answers for the volumes question did it around the y axis.

1981 😵😵 does that mean its highly unlikely that they will ask this question, ever? and I don't understand what is the point of that question. how am I  supposed to know wtf they are talking about. I did not do that in school (coz I have a crappy teacher) so when I first saw it I was like wtffffff

for that locus that is what I did. but the answers had something weird 
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 21, 2018, 11:18:45 pm
the answers for the volumes question did it around the y axis.

1981 😵😵 does that mean its highly unlikely that they will ask this question, ever? and I don't understand what is the point of that question. how I sone supposed to know wtf they are talking about. I did not do that in school (coz I have a crappy teacher) so when I first saw it I was like wtffffff

for that locus that is what I did. but the answers had something weird 
Yeah ignore that volumes question. They didn't hint at all that you were meant to do that.

Yeah that mechanics question is somewhat famous (can be found via a google search), because it basically asked for the proof of all the circular motion formulae we've been taking for granted all our lives. I can't see it being examined again unless they help set it up for you (just like with that AM-GM inequality question), because the intuition behind thinking that up without any help whatsoever is a bit of a huge stretch.

Their diagram looks like it basically gave the proof for the locus. The reason why that locus gives the arc is because if you look at the diagram, using the exterior angle of a triangle we have \( \arg(z-z_1) = \beta + \arg(z-z_2)\). Using \( \arg z - \arg w = \arg \frac{z}{w}\), this rearranges to give the equation you started off with. The arc gets traced out because you can move \(z\) anywhere along that arc, and the same proof is still valid.
Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on October 21, 2018, 11:31:54 pm
Their diagram looks like it basically gave the proof for the locus. The reason why that locus gives the arc is because if you look at the diagram, using the exterior angle of a triangle we have \( \arg(z-z_1) = \beta + \arg(z-z_2)\). Using \( \arg z - \arg w = \arg \frac{z}{w}\), this rearranges to give the equation you started off with. The arc gets traced out because you can move \(z\) anywhere along that arc, and the same proof is still valid.
I don't understand what you mean :(
why is the "circle" midway of the quadrant and not touching the x axis like the normal ones like this (but ofc the angle wouldn't be 90, it would be beta)?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 21, 2018, 11:35:25 pm
I don't understand what you mean :(
why is the "circle" midway of the quadrant and not touching the x axis like the normal ones like this (but ofc the angle wouldn't be 90, it would be beta)?
(Yep, the angle would be \(\beta\).)

In that particular example, you didn't just take an arbitrary choice of \(z_1\) and \(z_2\). For your original question, \(z_1\) and \(z_2\) could've been anywhere you wanted it to. But for your newer question, you had specifically let \(z_1 = 2\), which represents the point \( (2,0)\) on the Argand plane, and you've also specifically let \(z_2 = -2\), which represents the point \( (-2,0)\) on the Argand plane. Both of these points specifically lie on the \(x\)-axis.

So pretty much, that newer example isn't a "normal" one. It's just a special case.
Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on October 21, 2018, 11:43:36 pm
(Yep, the angle would be \(\beta\).)

In that particular example, you didn't just take an arbitrary choice of \(z_1\) and \(z_2\). For your original question, \(z_1\) and \(z_2\) could've been anywhere you wanted it to. But for your newer question, you had specifically let \(z_1 = 2\), which represents the point \( (2,0)\) on the Argand plane, and you've also specifically let \(z_2 = -2\), which represents the point \( (-2,0)\) on the Argand plane. Both of these points specifically lie on the \(x\)-axis.

So pretty much, that newer example isn't a "normal" one. It's just a special case.

does that mean if I drew the original question like the answer I just sent, I would get it right (but again not 90, beta)?
Title: Re: 4U Maths Question Thread
Post by: UStoleMyBike on October 21, 2018, 11:44:39 pm
Hey, I'm new here

If
what is


and why?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 21, 2018, 11:48:53 pm
does that mean if I drew the original question like the answer I just sent, I would get it right (but again not 90, beta)?
In theory, I would give you the marks for it.

Although in practice, usually when you aren't told anything about what \(z_1\) and \(z_2\) are, I like to keep my answer 'as arbitrary as possible' in a sense. Placing \(z_1\) and \(z_2\) in places that almost look like you're just chucking them at random places in thin air feels less 'restricted' in a sense, as opposed to doing what that other one did. Copying what they did can unintentionally give the impression that \(z_1\) and \(z_2\) have to lie on the \(x\)-axis no matter what, when really they don't. So whilst it's technically not incorrect to do it that way, it probably isn't the best habit to go towards.

Especially since, potentially in the exam they could give you something like \( \arg \left( \frac{z-3+4i}{z+2-i} \right) = \frac\pi3\) instead.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 21, 2018, 11:51:13 pm
Hey, I'm new here

If
what is


and why?

Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on October 21, 2018, 11:52:42 pm
In theory, I would give you the marks for it.

Although in practice, usually when you aren't told anything about what \(z_1\) and \(z_2\) are, I like to keep my answer 'as arbitrary as possible' in a sense. Placing \(z_1\) and \(z_2\) in places that almost look like you're just chucking them at random places in thin air feels less 'restricted' in a sense, as opposed to doing what that other one did. Copying what they did can unintentionally give the impression that \(z_1\) and \(z_2\) have to lie on the \(x\)-axis no matter what, when really they don't. So whilst it's technically not incorrect to do it that way, it probably isn't the best habit to go towards.

Especially since, potentially in the exam they could give you something like \( \arg \left( \frac{z-3+4i}{z+2-i} \right) = \frac\pi3\) instead.

hmm true that makes sense. thank you!!!
Title: Re: 4U Maths Question Thread
Post by: justwannawish on October 22, 2018, 02:16:10 pm
Let w be a non real cube of unity. Show the other non-real cube root is w^2.

What would be the best method for this?
Title: Re: 4U Maths Question Thread
Post by: envisagator on October 22, 2018, 03:24:38 pm
Hi Rui, with the following question attached, can you please explain why you do the part highlighted and its significance. Thank You!!!

Title: Re: 4U Maths Question Thread
Post by: Opengangs on October 22, 2018, 03:29:51 pm
Let w be a non real cube of unity. Show the other non-real cube root is w^2.

What would be the best method for this?
For cube root of unity, we have \(z^3 - 1 = 0\), giving us \(z^3 = 1\). So clearly \(z = 1\) is a root. Note that \(z^3 - 1\) can be factorised in the form: \((z - 1)(z^2 + z + 1)\). Finding the roots of the irreducible quadratic is fairly straight forward. We have:
\[ \begin{align*}z &= \frac{-1 \pm \sqrt{3}i}{2}\end{align*}\]

Taking \(\omega\) to be \(\frac{-1 + \sqrt{3}i}{2}\) and squaring it gives us:
\[ \begin{align*}\omega^2 &= \left(\frac{-1 + \sqrt{3}i}{2}\right)^2 \\ &= \frac{1 - 2\sqrt{3}i - 3}{4} \\ &= \frac{-1 - \sqrt{3}i}{2}\end{align*}\]

which is the other cube root of unity.
Title: Re: 4U Maths Question Thread
Post by: Opengangs on October 22, 2018, 03:57:27 pm
Hi Rui, with the following question attached, can you please explain why you do the part highlighted and its significance. Thank You!!!
With multiple zeroes, we know that at least \(P(x) = P'(x) = 0\). To show that there are no multiple zeroes, we essentially want to show that for any \(x\), \(P(x) = 0 \neq P'(x)\). Notice that
\[P(x) = P'(x) + \frac{x^n}{n!}\],

which is also where the first highlighted part comes from. The goal now is to show that: \(\frac{x^n}{n!}\) is non-zero for \(n > 1\).

Suppose that, for some \(\alpha\), we have \(P(\alpha) = P'(\alpha)\). Then clearly that must occur if:
\[\frac{\alpha^n}{n!} = 0 \Rightarrow \alpha = 0\].

Then we have \(P(0) = P'(0) = 0\), which is equivalent to the second highlighted part.
Title: Re: 4U Maths Question Thread
Post by: justwannawish on October 22, 2018, 07:34:25 pm
For cube root of unity, we have \(z^3 - 1 = 0\), giving us \(z^3 = 1\). So clearly \(z = 1\) is a root. Note that \(z^3 - 1\) can be factorised in the form: \((z - 1)(z^2 + z + 1)\). Finding the roots of the irreducible quadratic is fairly straight forward. We have:
\[ \begin{align*}z &= \frac{-1 \pm \sqrt{3}i}{2}\end{align*}\]
Taking \(\omega\) to be \(\frac{-1 + \sqrt{3}i}{2}\) and squaring it gives us:
\[ \begin{align*}\omega^2 &= \left(\frac{-1 + \sqrt{3}i}{2}\right)^2 \\ &= \frac{1 - 2\sqrt{3}i - 3}{4} \\ &= \frac{-1 - \sqrt{3}i}{2}\end{align*}\]

which is the other cube root of unity.

Hey, I forgot to mention that the next part of the question was that using the above, prove w^2 + w + 1=0

So would I still be able to use this method and then requote it for part ii?

Thanks!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 22, 2018, 08:02:04 pm
Hey, I forgot to mention that the next part of the question was that using the above, prove w^2 + w + 1=0

So would I still be able to use this method and then requote it for part ii?

Thanks!



You can, of course, plug the value of \(\omega\) in there manually if you wish to. The computations should still fall out.
Title: Re: 4U Maths Question Thread
Post by: Jeeffffffffffffff on October 23, 2018, 10:44:15 am
Hey so uuuhh I need a lil help with question 7b)ii) from 2011 paper.
I got part i) easily, it was just a subsitution but then when I was looking at the solutions at the end of part i) they “relabelled” u as x so they could use it in their solution for part ii). Why is it that they can just relabel it as x straight after going through that whole process of substiting u=4-x?? Btw I tried to post screenshots of the question I was talking about but it said it was too large or something?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 23, 2018, 10:49:49 am
Hey so uuuhh I need a lil help with question 7b)ii) from 2011 paper.
I got part i) easily, it was just a subsitution but then when I was looking at the solutions at the end of part i) they “relabelled” u as x so they could use it in their solution for part ii). Why is it that they can just relabel it as x straight after going through that whole process of substiting u=4-x?? Btw I tried to post screenshots of the question I was talking about but it said it was too large or something?
With the screenshots being too large, when that happens you'll need to upload it to an image hosting website like imgur and extract the link instead.
\[ u\text{ and }x\text{ are ultimately, nothing more than 'dummy variables'}\\ \text{used for the sake of definite integration.} \]
\[ \text{In general, it is always true that }\int_a^b f(x)\,dx = \int_a^b f(u)\,du.\\ \text{As an example, you can try computing }\int_0^1 x^2\,dx\\ \text{and check that it does equal }\int_0^1 u^2\,du.\]
Note that the fact they're equal is facilitated by the fact that \(x\) and \(u\) are 'doomed' to disappear, because once you sub the boundary values in, you don't care at all about what the variables you had at the start anymore were. This is the reason for why they're called 'dummy' variables in integration. Note that you don't run into the same luck when dealing with indefinite integrals, because the variables never disappear then.
\[ \text{So for that one, by the same logic,}\\ \int_1^3 \frac{\sin^2 \frac{\pi u}{8}}{u(4-u)}\,du = \int_1^3 \frac{\sin^2 \frac{\pi x}{8}}{x(4-x)}\,dx\]
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on October 23, 2018, 03:57:10 pm
hey rui plz need some help, a friend gave me and some other mates this 2u que but I find it really hard.
When i first graphed x^2+xy+y^2=0, on desmos, it was just a dot on the origin.
So i just tried a blodge method by letting y=1 and used quadratic formula to get x=(1+sqrt[3] i)/2 and plugged in both values of x&y in the other eq.
After that i got it simplified to 2cos(2015pi/3)=1

But im not sure if this is a full proof method. Plz give it a try.
Title: Re: 4U Maths Question Thread
Post by: envisagator on October 23, 2018, 04:36:25 pm
Hi Rui, this was a question posted on here a few days ago. I'm not sure how to do part c of it, i know its got to do with product and sum of roots but just cant grasp wats happening. Thank You!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 23, 2018, 05:40:56 pm
hey rui plz need some help, a friend gave me and some other mates this 2u que but I find it really hard.
When i first graphed x^2+xy+y^2=0, on desmos, it was just a dot on the origin.
So i just tried a blodge method by letting y=1 and used quadratic formula to get x=(1+sqrt[3] i)/2 and plugged in both values of x&y in the other eq.
After that i got it simplified to 2cos(2015pi/3)=1

But im not sure if this is a full proof method. Plz give it a try.
Yeah I know the source of this question. I've tried to avoid their own solution here though:






Your method would only work if this were a multiple choice question and you just wanted to figure out the correct answer. Here, it's not ok, because until we got to the very end we could never confirm whether the value was constant, or if it depended on \(x\) and \(y\).
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 23, 2018, 05:47:50 pm
Hi Rui, this was a question posted on here a few days ago. I'm not sure how to do part c of it, i know its got to do with product and sum of roots but just cant grasp wats happening. Thank You!!

Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on October 24, 2018, 04:53:09 pm
hey guys some help here plz
Title: Re: 4U Maths Question Thread
Post by: Bruh01 on October 24, 2018, 06:49:16 pm
Hi guys, can anyone explain how to sketch im(z)=|z|. Thanks
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 24, 2018, 06:51:50 pm
Hi guys, can anyone explain how to sketch im(z)=|z|. Thanks


Title: Re: 4U Maths Question Thread
Post by: fkkiwi on October 24, 2018, 08:00:15 pm
What are some good time management strategies for the exam? (e.g. how long should I spend on each question, at what point should I move on from a question if it takes too long, what happens if I can't crack Q16)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 24, 2018, 08:05:03 pm
What are some good time management strategies for the exam? (e.g. how long should I spend on each question, at what point should I move on from a question if it takes too long, what happens if I can't crack Q16)
For me, 2 minutes blankly staring at a question is enough for me to say screw it for now. Or occasionally even 1 minute is too much. On the other hand, if it's taken me more than 2 minutes but it's because I've been constantly writing, I'd probably say 5 minutes of writing for one part is a healthy cut-off point.

If you can't crack Q16 immediately, too bad. Go back and check your working out for the rest of the paper and make sure you got all your marks there. But then after that, few time management strategies exist if it's only Q16 you have left. Just try to get essentially whatever marks you can get.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 24, 2018, 08:58:38 pm
hey guys some help here plz
Note that questions before 2001 do not reflect well the difficulty of the current HSC exams. Here are sketch solutions.

i) This is a classic case of recursion. We can split the cases depending on if the first throw was a head or a tail. If the first was a head, we'd still need \(r-1\) heads and \(s\) tails in their appropriate positions, giving \(P(r-1,s)\). And vice versa if a tail was first. Since the probability that the first throw is \( \frac12\) for a head or a tail, we get \( \frac12 P(r-1,s) +\frac12 P(r,s-1) \).

ii) We have \(P(2,3) = \frac12P(1,3) + \frac12P(2,2) \)
For \(P(1,3)\), we need one head before we reach three tails, so we'd have either \(H\), \(TH\) or \(TTH\).
For \(P(2,2)\), we need two heads before we reach two tails, so we'd have either \(HH\), \(THH\) or \(HTH\)

iii) For the induction, the base case at \(n=0\) would consequently imply that \(r+s-1 = 0\). Since \(s \geq 1\), the only case is \( \boxed{r=0, s=1} \). This should thus be an easy check.
Then assume that the statement holds for \(n=k\). This means that it holds for all choices of \(r\) and \(s\) that satisfy the property \(r+s-1=k\). We need to prove it for \(n=k+1\), i.e. all choices of \(r\) and \(s\) that satisfy the property \( r+s-1=k+1\).

If \(r=0\), we have \(s=k+2\). This case should be handled separately, as it should fall out immediately.
Otherwise, \(r \geq 1\) and we can take care of all of those in the one go with the inductive assumption.
\begin{align*}P(r,s) &= \frac12 P(r-1, s) + \frac12 P(r, s-1)\\ &=\frac12 \times \frac1{2^k} \left[ \binom{k}{0} + \binom{k}{1} +\dots +\binom{k}{s-2}+ \binom{k}{s-1} \right]\\ &\quad + \frac12 \times\frac1{2^k} \left[ \binom{k}{0} + \binom{k}{1} + \dots + \binom{k}{s-2} \right]\tag{assumption} \end{align*}
At this point, combine the brackets together, and make use of the property \( \binom{N+1}{K+1} = \binom{N}{K} + \binom{N+1}{K+1} \) and \( \binom{N}{0} = \binom{N+1}{0} = 1 \) to complete the proof of this case. And after this case is proven, the induction is complete.
Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on October 24, 2018, 10:02:25 pm
what are "dividers" in probability?

also thank you for the good luck story on ig. it made me feel calmer :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 24, 2018, 10:47:46 pm
what are "dividers" in probability?

also thank you for the good luck story on ig. it made me feel calmer :)
Glad it did :)

I've rarely seen the word "divider" come up in probability. Which question did you get that from?
Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on October 25, 2018, 08:50:03 am
Glad it did :)

I've rarely seen the word "divider" come up in probability. Which question did you get that from?

the question was "there are 10 coins to be placed in 4 boxes with max of 4 in each. find the number of ways this can happen"

and when I asked my friend she used "dividers" so what she did was there are 10 coins and 3 dividers. so 13 objects in total. so it'll be 13!, but since the dividers are identical you do 13!/3! but also the coins are identical so your final answer will be 13!/3!10!

and I don't understand what she did

thank you
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 25, 2018, 08:57:50 am
the question was "there are 10 coins to be placed in 4 boxes with max of 4 in each. find the number of ways this can happen"

and when I asked my friend she used "dividers" so what she did was there are 10 coins and 3 dividers. so 13 objects in total. so it'll be 13!, but since the dividers are identical you do 13!/3! but also the coins are identical so your final answer will be 13!/3!10!

and I don't understand what she did

thank you
Oh that. Typically that type of question won't appear in the HSC courses, but I have seen it happen. The technique is known as "stars and bars", and some people rename it to "dots and dividers" which basically means the same thing.

The "dividers" serve the role of splitting up where the coins go. For example, you could have something like this: X X X | X | X X | X X X X.
That would be a case of {3 in box 1}, {1 in box 2}, {2 in box 3}, {4 in box 4}.

What she did, was essentially count the number of arrangements of those dots (marked with a cross) and dividers. Which, using our knowledge of letter arrangements, is indeed \( \frac{13!}{3!10!} \).

However, and this is I'd say a huge however and why I don't anticipate this question to be in the exam. The fact that a max of 4 coins can be placed into each box doubles the difficulty of the question. The problem with the \( \frac{13!}{3!10!} \) is that this doesn't cater for that restriction. For a counterexample, an arrangement that would have been counted could've been this: X | X | X | X X X X X X X (1 in the first three boxes, but 7 in the last).

The usual approach to such a problem is to apply the inclusion exclusion principle on the complement. This gives \( \frac{13!}{3! 10!}\) minus number of arrangements when at least one box has 5 in there plus number of arrangements when at least two boxes have 5 in there. But of course, at this point we've gone beyond the scope of the HSC.

Remark: It turns out that if at least 5 are in any one box, then this becomes \( \frac{8!}{3!5!} \) and if there's at least 5 in any two boxes, there's only \( \frac{3!}{3!0!}=1\) case.

Also, I'd recommend not thinking too hard about this the morning of the exam either.
Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on October 25, 2018, 09:15:00 am
However, and this is I'd say a huge however and why I don't anticipate this question to be in the exam. The fact that a max of 4 coins can be placed into each box doubles the difficulty of the question. The problem with the \( \frac{13!}{3!10!} \) is that this doesn't cater for that restriction. For a counterexample, an arrangement that would have been counted could've been this: X | X | X | X X X X X X X (1 in the first three boxes, but 7 in the last).

The usual approach to such a problem is to apply the inclusion exclusion principle on the complement. This gives \( \frac{13!}{3! 10!}\) minus number of arrangements when at least one box has 5 in there plus number of arrangements when at least two boxes have 5 in there. But of course, at this point we've gone beyond the scope of the HSC.

Remark: It turns out that if at least 5 are in any one box, then this becomes \( \frac{8!}{3!5!} \) and if there's at least 5 in any two boxes, there's only \( \frac{3!}{3!0!}=1\) case.

I don't get what u meant by doing the 7 in one box and the "inclusion exclusion principle"

also, there was a similar question that said

"10 people arrive at the airport and there are only 4 counters opened. how many ways can 10 people line up in 4 lane queue?"

and my other friend did since there are 10 people so 10!. and there are 13 spots to put 3 dividers which divides them people into the 4 counters. so she just did 13c3 * 10!

and since its practically the same question as the coins, they both give very different answers. why?

Also, I'd recommend not thinking too hard about this the morning of the exam either.
I know I shouldn't be doing math, but yesterday after bio exam I came home sooo exhausted and couldn't do anything because I was so tired which is making me stressed because I feel like I haven't done enough questions, considering the difficulty of the exams !!!!!!!!!
Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on October 25, 2018, 09:15:35 am
Hi,
Apparently this is a De Moivre’s theorem question but I can’t seem to get this. Please just give me a hint so I can try it myself. I have tried everything. Also, can you tell me how to approach such questions where it says From this, deduce this. I.e am I allowed to work with the second line, the one we are trying to prove?

Thank you!!!
there is no question attached
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 25, 2018, 09:21:08 am
I don't get what u meant by doing the 7 in one box and the "inclusion exclusion principle"

also, there was a similar question that said

"10 people arrive at the airport and there are only 4 counters opened. how many ways can 10 people line up in 4 lane queue?"

and my other friend did since there are 10 people so 10!. and there are 13 spots to put 3 dividers which divides them people into the 4 counters. so she just did 13c3 * 10!

and since its practically the same question as the coins, they both give very different answers. why?
I know I shouldn't be doing math, but yesterday after bio exam I came home sooo exhausted and couldn't do anything because I was so tired which is making me stressed because I feel like I haven't done enough questions, considering the difficulty of the exams !!!!!!!!!
The other one - simply put:
1) people are distinguishable but coins aren't, justifying the multiplication by 10!.
2) there's no limit on how many people can line up in each lane, so no inclusion exclusion is necessary.

Essentially, the inclusion exclusion principle is the reason why I would not worry about this question, because it is not an HSC level technique.

(What it does is that it basically helps us count "OR" scenarios, when there's overlaps in the cases.)
Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on October 25, 2018, 09:33:07 am
Essentially, the inclusion exclusion principle is the reason why I would not worry about this question, because it is not an HSC level technique.
thank god!!!!

also, I was looking at your trial handout (thank you for that lecture btw it was an absolute lifesaver!!!! and haha thanks for the chocolates :) ) and I know this is dumb but I can't figure out how you got 4^6 I understood it at the time but now I can't remember what you said
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 25, 2018, 09:36:44 am
thank god!!!!

also, I was looking at your trial handout (thank you for that lecture btw it was an absolute lifesaver!!!! and haha thanks for the chocolates :) ) and I know this is dumb but I can't figure out how you got 4^6 I understood it at the time but now I can't remember what you said

For the total outcomes, we essentially throw out all the restrictions.

Without restriction, each of the 6 people can choose any of the 4 rooms to stay in. So we have \(4\times4\times4\times4\times4\times4\) (one for each of the six people), which is just \(4^6\)

(Haha glad you enjoyed the chocolate, it's become a known thing at my lectures ;))
Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on October 25, 2018, 10:07:35 am
For the total outcomes, we essentially throw out all the restrictions.

Without restriction, each of the 6 people can choose any of the 4 rooms to stay in. So we have \(4\times4\times4\times4\times4\times4\) (one for each of the six people), which is just \(4^6\)

(Haha glad you enjoyed the chocolate, it's become a known thing at my lectures ;))

ohhh it was 4 choices for each person. that makes sense, thank you!!!!
Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on October 25, 2018, 10:13:29 am
also should I take a math set, like protractor, compass or nah
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 25, 2018, 10:20:42 am
also should I take a math set, like protractor, compass or nah
Like, sure if you want. It helps to draw circles for circle geometry problems where you need to copy the diagram I guess
Title: Re: 4U Maths Question Thread
Post by: Maths4life on October 25, 2018, 11:06:25 am
From,
Z^6 +1= (z^2 -2zcospi/6 +1)(z^2 -2zcospi/2 +1)(z^2 -2zcos5pi/6 +1)

Deduce:
Cos3theta= 4(Costheta-cospi/6)(Costheta-cospi/2)(Costheta-cos5pi/6)

Hi, I wasn’t too sure how to do these sort of questions where it says from this deduce this. Like what are my boundaries. For example, am I allowed to work with the second equation (the one we have to deduce) or not?
Also apparently this is a De Moivres theorem question but I can’t seem to figure out why. If you could please give me a hint on where to start that would be great.
Thank you!!!!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 25, 2018, 03:15:44 pm
From,
Z^6 +1= (z^2 -2zcospi/6 +1)(z^2 -2zcospi/2 +1)(z^2 -2zcos5pi/6 +1)

Deduce:
Cos3theta= 4(Costheta-cospi/6)(Costheta-cospi/2)(Costheta-cos5pi/6)

Hi, I wasn’t too sure how to do these sort of questions where it says from this deduce this. Like what are my boundaries. For example, am I allowed to work with the second equation (the one we have to deduce) or not?
Also apparently this is a De Moivres theorem question but I can’t seem to figure out why. If you could please give me a hint on where to start that would be great.
Thank you!!!!!
You should only assume stuff from the first equation. You can work with one side at a time for what you wish to prove, but that probably was not the intended approach of the question since it says "deduce".

For this question, as a hint the first step is to identify that:
\begin{align*}\frac{z^6+1}{z^3} &= \frac{z^2-2z\cos \frac\pi6+1}{z} \times \frac{z^2-2z\cos\frac\pi2 + 1}{z} \times \frac{z^2-2z\cos\frac{5\pi}6 + 1}{z} \\ \therefore z^3 + z^{-3} &= \left(z - 2\cos \frac\pi6 + z^{-1} \right)\left(z - 2 \cos \frac\pi2 + z^{-1} \right)\left( z - 2\cos \frac{5\pi}6 + z^{-1}\right)\end{align*}
Now think about what you can do to throw away all the \(z\)'s and \(z^{-1}\)'s for a bunch of cosines. And yes, it is a de Moivre's question
Title: Re: 4U Maths Question Thread
Post by: hassrax on October 25, 2018, 04:17:35 pm
Thank you so much!!! Just got the answer by letting z=cisthtea.  I just had some other questions. How did you know to divide by z^3, did it have soething to do with the cos3theta? Also, is there a general approach to these questions or does it just take experience?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 25, 2018, 04:49:12 pm
Thank you so much!!! Just got the answer by letting z=cisthtea.  I just had some other questions. How did you know to divide by z^3, did it have soething to do with the cos3theta? Also, is there a general approach to these questions or does it just take experience?
Yeah. I saw that \( \cos \frac\pi6\) and the other three were already there so I wanted to leave them untouched. My initial idea was to somehow kill off the \(z^2+1\) with \(z=-i\sin\theta\) or something, but then I decided nope there's gonna be too many sine's floating about.

Then it hit me that \(\operatorname{cis}\theta + \operatorname{cis}(-\theta) = 2\cos\theta\), so I thought about if that was helpful. That's when I realised that division by \(z^3\) would probably lead somewhere, because as mentioned it forces the \(\cos 3\theta\) to appear when combined with de Moivre's. What was left was to figure out how I would fix up the RHS.
Title: Re: 4U Maths Question Thread
Post by: hassrax on October 25, 2018, 06:09:38 pm
Thank you for the detailed thought process and answer!! Legend!
Title: Re: 4U Maths Question Thread
Post by: 24221 on October 28, 2018, 10:48:43 pm
Hey! can someone pls help me with this question
thank you!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 29, 2018, 06:48:17 am
Hey! can someone pls help me with this question
thank you!!



______________________________________
\begin{align*}|z+w|^2 &= (z+w) \overline{(z+w)} \\ &= (z+w)(\overline{z} + \overline{w})\\ &= z\overline{z} + z\overline{w} + w\overline{z} + w\overline{w}\\ &= |z|^2 + z\overline{w} + \overline{z\overline{w}} + |w|^2\\ &= |z|^2 + 2 \operatorname{Re}(z\overline{w}) + |w|^2.\end{align*}
Title: Re: 4U Maths Question Thread
Post by: 24221 on October 29, 2018, 07:49:07 am
thank you!
Title: Re: 4U Maths Question Thread
Post by: terassy on October 31, 2018, 10:03:16 pm
Hello, could someone help me clarify something.

This is the question:
Show that the following equations represent hyperbolas and find their centre, focu and the equation of directrices and asymptotes.

1) (x-3)^2/64   -    (y+1)^2/36  = 1

For the first part of the question, in ‘showing that the equation is a hyperbola’ do I show that e > 1 or something else like |PS-PS’|=2a

Thanks  :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 31, 2018, 10:52:12 pm
Hello, could someone help me clarify something.

This is the question:
Show that the following equations represent hyperbolas and find their centre, focu and the equation of directrices and asymptotes.

1) (x-3)^2/64   -    (y+1)^2/36  = 1

For the first part of the question, in ‘showing that the equation is a hyperbola’ do I show that e > 1 or something else like |PS-PS’|=2a

Thanks  :)
Tbh, that is such a weird question. Normally in the HSC they'd let you assume it.

But yeah, you would either prove that \( \frac{PS}{PM} > 1 \) for a suitable choice of \(S\) and \(M\), or alternatively \( |PS - PS^\prime| = 2a\) for a suitable choice of \(S\) and \(S^\prime\) (well, and \(a\)).

If I were the examiner, I would've been happy with the fact that it is a translation of the curve \( \frac{x^2}{64} - \frac{y^2}{36} = 1\), which is clearly a hyperbola as it is a course assumption. But those methods you hinted are the more formal approaches. Where did this question come from?
Title: Re: 4U Maths Question Thread
Post by: terassy on October 31, 2018, 11:33:32 pm
Tbh, that is such a weird question. Normally in the HSC they'd let you assume it.

But yeah, you would either prove that \( \frac{PS}{PM} > 1 \) for a suitable choice of \(S\) and \(M\), or alternatively \( |PS - PS^\prime| = 2a\) for a suitable choice of \(S\) and \(S^\prime\) (well, and \(a\)).

If I were the examiner, I would've been happy with the fact that it is a translation of the curve \( \frac{x^2}{64} - \frac{y^2}{36} = 1\), which is clearly a hyperbola as it is a course assumption. But those methods you hinted are the more formal approaches. Where did this question come from?

From the old Fitzpatrick 4U textbook. Thanks
Title: Re: 4U Maths Question Thread
Post by: 3.14159265359 on November 04, 2018, 02:08:41 pm
Tbh, that is such a weird question. Normally in the HSC they'd let you assume it.

But yeah, you would either prove that \( \frac{PS}{PM} > 1 \) for a suitable choice of \(S\) and \(M\), or alternatively \( |PS - PS^\prime| = 2a\) for a suitable choice of \(S\) and \(S^\prime\) (well, and \(a\)).

If I were the examiner, I would've been happy with the fact that it is a translation of the curve \( \frac{x^2}{64} - \frac{y^2}{36} = 1\), which is clearly a hyperbola as it is a course assumption. But those methods you hinted are the more formal approaches. Where did this question come from?

wouldn't you show that it is in the form

xx       yy
---  -   ----  = 1
aa       bb

because that is how my teacher told me to do it
Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 04, 2018, 06:29:03 pm
wouldn't you show that it is in the form

xx       yy
---  -   ----  = 1
aa       bb

because that is how my teacher told me to do it
It's not of that form because it's not centred at the origin.
Title: Re: 4U Maths Question Thread
Post by: flacko on November 05, 2018, 04:47:30 pm
Hey been struggling with 2 questions from Patel.

"For the following, describe the locus of the complex number w, where z is restricted as indicated.

w=(z-2+i)/(z+2-i) , |z|=1

Thanks
Title: Re: 4U Maths Question Thread
Post by: hassrax on November 06, 2018, 10:38:41 pm
Hi i need help with this q, if z is a complex number, z/z-i  then show that it is imaginary. I let z=x+iy and tried doing it from there but i get stuck when i let re(z) =0 , thanks
Title: Re: 4U Maths Question Thread
Post by: terassy on November 10, 2018, 11:48:04 pm
Help please  :-\

P and Q are variable points on the rectangular hyperbola xy = c2. The tangents at P and Q meet at R. If PQ passes through the point (a,0), find the equation of the locus of R.

Okay,
So I've got the equation of the tangent at P:
x=2cp-p2y

and the tangent at Q
x=2cq-q2y

Simultaneously solve them and R is at:
x= 2cp/(p+q)
y=2c/(p+q)

And the equation of the chord PQ is:
x+pqy=c(p+q), but it passes through (a,0), so we get this:
a=c(p+q)

Now what do I do. I would usually try to form an equation by getting rid of the parameters (p+q), but then pq is left behind so what do I do?
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on November 11, 2018, 12:33:38 pm
Hi i need help with this q, if z is a complex number, z/z-i  then show that it is imaginary. I let z=x+iy and tried doing it from there but i get stuck when i let re(z) =0 , thanks

Hey! Super sorry about the delay in response - I'm not sure if everything is quite right with this question. Are we trying to prove that a complex number of the form \(\frac{z}{z-i}\) is imaginary for any \(z\)? That isn't quite true. If we're trying to prove that the solution to \(z=\frac{z}{z-i}\) is imaginary then that isn't quite true either (answers are \(z=0\) and \(z=1+i\)). Maybe I'm misinterpreting you too - Could you maybe share the question/source? :)


Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on November 11, 2018, 12:49:34 pm
Help please  :-\

P and Q are variable points on the rectangular hyperbola xy = c2. The tangents at P and Q meet at R. If PQ passes through the point (a,0), find the equation of the locus of R.

Okay,
So I've got the equation of the tangent at P:
x=2cp-p2y

and the tangent at Q
x=2cq-q2y

Simultaneously solve them and R is at:
x= 2cp/(p+q)
y=2c/(p+q)

And the equation of the chord PQ is:
x+pqy=c(p+q), but it passes through (a,0), so we get this:
a=c(p+q)

Now what do I do. I would usually try to form an equation by getting rid of the parameters (p+q), but then pq is left behind so what do I do?

Hey! So we're here for the coordinates of R (think you forgot a \(q\) in your message?):



And as you correctly get to, \(p+q=\frac{a}{c}\), so we can ditch that. Where do we ditch \(pq\)? So it's a bit of a trick - We don't need to! Take the y-coordinate:



That is the locus! It's a horizontal line, you don't need to bring the \(x\) at all. Our objective was to get an equation with no \(p\) or \(q\), and we have that here. The locus is the horizontal line \(y=\frac{2c^2}{a}\) ;D

Watch out for this in locus questions, sometimes you don't need to bring the \(x\) in :)
Title: Re: 4U Maths Question Thread
Post by: Livjane_2203 on November 11, 2018, 04:26:03 pm
Would appreciate some help please.

Use De Moivre's Theorem to solve the equation z^5=1. Show that the points representing the five roots of this equation on an Argand diagram form the vertices of a regular pentagon of area 5/2 sin 2pi/5 and perimeter 10 sin pi/5.

I understand how how to get the roots but not sure how to get to the area and perimeter.

Thanks.  :)
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on November 11, 2018, 04:31:44 pm
I understand how how to get the roots but not sure how to get to the area and perimeter.

Hey there! Nice job, you've done the hard bit! For the area and perimeter, pick two of the roots you found and the origin on the complex plane. It forms a triangle, and that triangle is the basis on which we'll get the area/perimeter.

- We can find the area of the triangle using \(A=\frac{1}{2}ab\sin{C}\), where \(a\) and \(b\) are the moduli of the two  roots (they'll be the same), and \(C\) is the angle at the origin (it is \(\frac{2\pi}{5}\), since it is a fifth of the full angle at the middle). Multiply that area by five to get the area of the pentagon.
- We can find the length of the side opposite the origin by using the cosine rule, same lengths/angles as above. Multiply that by five to get the perimeter.

There are definitely other ways, this is a geometrical approach which I think is pretty simple ;D hope this helps!
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on November 12, 2018, 09:42:09 pm
Hello, just a question on 4U Cambridge, can anyone please give me a detailed answer for question 7 and 8 in the link? Thanks :)
Title: Re: 4U Maths Question Thread
Post by: Divayth Fyr on November 13, 2018, 11:46:37 pm
Hello, just a question on 4U Cambridge, can anyone please give me a detailed answer for question 7 and 8 in the link? Thanks :)

Ah, these questions. I remember when I first saw them; it was not a pleasant discovery. I hope the solutions help.
Title: Re: 4U Maths Question Thread
Post by: g.xzhu on November 18, 2018, 12:32:26 pm
Hi,

Could I please get some help with these two maths induction questions?

For all positive intergers n, prove by maths induction that:
     1) |z1z2...zn| = |z1| |z2| ... |zn|
     2) arg(z1z2...zn) = arg(z1) + arg(z2) + ... + arg(zn)

Thank you!
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on November 18, 2018, 01:01:43 pm
Ah, these questions. I remember when I first saw them; it was not a pleasant discovery. I hope the solutions help.
Thanks for the help! :)
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on November 18, 2018, 05:37:58 pm
Hi,

Could I please get some help with these two maths induction questions?

For all positive intergers n, prove by maths induction that:
     1) |z1z2...zn| = |z1| |z2| ... |zn|
     2) arg(z1z2...zn) = arg(z1) + arg(z2) + ... + arg(zn)

Thank you!

Hey! Let's roll with (1), I'll assume you could prove it for \(n=1\) (it is self apparent), but we'll also want to prove it for \(n=2\) to use later.



We can prove this by letting \(z_1=x_1+iy_1\) and \(z_2=x_2+iy_2\), and substituting.



Find each of those moduli (expand out the LHS) and they will be the same. So you've proved the case for \(n=2\).

Now, the induction assumption for (1) is:



Now let's go with \(n=k+1\) and see what we get:



Let's break this into two pieces, \(a=z_1z_2...z_k\) and \(b=z_{k+1}\). We do this because we've already proven that \(|z_1z_2|=|z_1||z_2|\) above! So by splitting it in two, we can automatically therefore say that:



Now we use our induction assumption!!



And we are done!! Conclude as usual, and you're all set :) the second question is the exact same process, use the result for \(n=2\) to help you generalise it for \(n=k+1\) ;D
Title: Re: 4U Maths Question Thread
Post by: g.xzhu on November 18, 2018, 11:20:09 pm
Hey! Let's roll with (1), I'll assume you could prove it for \(n=1\) (it is self apparent), but we'll also want to prove it for \(n=2\) to use later.



We can prove this by letting \(z_1=x_1+iy_1\) and \(z_2=x_2+iy_2\), and substituting.



Find each of those moduli (expand out the LHS) and they will be the same. So you've proved the case for \(n=2\).

Now, the induction assumption for (1) is:



Now let's go with \(n=k+1\) and see what we get:



Let's break this into two pieces, \(a=z_1z_2...z_k\) and \(b=z_{k+1}\). We do this because we've already proven that \(|z_1z_2|=|z_1||z_2|\) above! So by splitting it in two, we can automatically therefore say that:



Now we use our induction assumption!!



And we are done!! Conclude as usual, and you're all set :) the second question is the exact same process, use the result for \(n=2\) to help you generalise it for \(n=k+1\) ;D

Thank you so much! I've done question no. 2 now using the method, and I just wanted to check if my working (esp. for S1) was the most efficient one?

Thanks!
Title: Re: 4U Maths Question Thread
Post by: jamonwindeyer on November 18, 2018, 11:42:24 pm
Thank you so much! I've done question no. 2 now using the method, and I just wanted to check if my working (esp. for S1) was the most efficient one?

Thanks!

Hey! Nicely done - Definitely looks cumbersome, but nothing immediately jumps to me as easier. If anyone reading this spots anything, give a shout ;D
Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 19, 2018, 09:00:40 am
Thank you so much! I've done question no. 2 now using the method, and I just wanted to check if my working (esp. for S1) was the most efficient one?

Thanks!
Hey! Nicely done - Definitely looks cumbersome, but nothing immediately jumps to me as easier. If anyone reading this spots anything, give a shout ;D
The quicker way was to let \(z = r \operatorname{cis} \theta\) instead of go for the Cartesian approach. That reduces the working out by a tiny bit for the modulus, but makes the argument just as fast as the modulus to deal with
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on November 21, 2018, 04:53:39 pm
Hello,
Can anyone please give me a detailed answer for 4 d) and 4 f) ASAP? Thanks 😁
Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 21, 2018, 05:38:14 pm
Hello,
Can anyone please give me a detailed answer for 4 d) and 4 f) ASAP? Thanks 😁
I don't see any attachment?
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on November 21, 2018, 08:26:21 pm
I don't see any attachment?
Sorry, my bad. Here's the attachment of the questions
Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 21, 2018, 08:31:34 pm

\begin{align*} \overline{5z} &= \overline{5(x+iy)}\\ &= \overline{5x+5iy}\\ &= 5x-5iy\\ &= 5(x-iy)\\ &= 5\overline{z} \end{align*}
_________________________________________________
\begin{align*} \overline{\left(\frac{1}{z}\right)}&= \overline{\left( \frac{1}{x+iy} \right)}\\ &= \overline{\frac{x-iy}{x^2+y^2}}\\ &= \frac{x+iy}{x^2+y^2}\\ &= \frac{1}{x-iy}\\ &= \frac{1}{\overline{z}} \end{align*}
The steps used are only common techniques in the MX2 course. If you require a step to be expanded on, please state that step.
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on November 21, 2018, 08:39:11 pm

\begin{align*} \overline{5z} &= \overline{5(x+iy)}\\ &= \overline{5x+5iy}\\ &= 5x-5iy\\ &= 5(x-iy)\\ &= 5\overline{z} \end{align*}
_________________________________________________
\begin{align*} \overline{\left(\frac{1}{z}\right)}&= \overline{\left( \frac{1}{x+iy} \right)}\\ &= \overline{\frac{x-iy}{x^2+y^2}}\\ &= \frac{x+iy}{x^2+y^2}\\ &= \frac{1}{x-iy}\\ &= \frac{1}{\overline{z}} \end{align*}
The steps used are only common techniques in the MX2 course. If you require a step to be expanded on, please state that step.
Thanks for the solutions :)
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on November 22, 2018, 08:24:57 pm
Hello,
Can anyone please give me a detailed answer for 9 d) and 9 c) ASAP? Thanks 😁
Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 22, 2018, 08:41:41 pm
Hello,
Can anyone please give me a detailed answer for 9 d) and 9 c) ASAP? Thanks 😁
I will only do the first one mentioned as they both follow the exact same procedure.
\[ \text{Note that }(1+i)^2 = 1+2i+i^2 = 2i. \]





\begin{align*} x &= \frac{2+2i \pm (4 - 4i)}{2i}\\ &= \frac{6-2i}{2i}, \frac{2-6i}{2i}\\ &=3+i, -1-3i\end{align*}
Title: Re: 4U Maths Question Thread
Post by: not a mystery mark on November 24, 2018, 06:54:33 pm
Heyyy. Currently stuck on a question and thought I'd ask some geniuses. Also does one even know the first step to take in questions like these?
haha thanks heaps guys!

If:
Prove that:
Title: Re: 4U Maths Question Thread
Post by: RuiAce on November 24, 2018, 07:38:19 pm
Heyyy. Currently stuck on a question and thought I'd ask some geniuses. Also does one even know the first step to take in questions like these?
haha thanks heaps guys!

If:
Prove that:
This particular question is more of a trick. It relies on whether you've seen ratios of lengths before. If you haven't, then the intuition to use a geometric approach becomes far less obvious.
Because the ratios of lengths are 3-4-5, this is hinting at some kind of Pythagoras' Theorem result.

(https://i.imgur.com/zS9MECt.png)
Whilst it turns out it doesn't look like that for this question, the important thing to recall is that \(z_2-z_1\) is just the vector from \(z_1\) to \(z_2\), and hence its length \(|z_2-z_1|\) just represents the remaining side in the triangle, whose vertices are \(0\), \(z_1\) and \(z_2\).





Note that the cases of plus/minus didn't matter as it gets made positive anyway after squaring.
Title: Re: 4U Maths Question Thread
Post by: not a mystery mark on November 24, 2018, 08:20:04 pm
That is amazing! Wow. Thank you heaps. I'll be on the look out for Pythagorean Triads next time haha. Thank you again!!!
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on December 02, 2018, 05:22:00 pm
Hello,
Can anyone please give me a detailed answer for Q4 to Q9 ASAP (I have included some of my working out, Q8 and Q9 I cannot do at all)?. Thanks :)
Title: Re: 4U Maths Question Thread
Post by: Opengangs on December 03, 2018, 11:23:33 pm
Hello,
Can anyone please give me a detailed answer for Q4 to Q9 ASAP (I have included some of my working out, Q8 and Q9 I cannot do at all)?. Thanks :)
Hey, don't usually go through the MX2 thread as I didn't do MX2 but considering no one's answered it, I'll give them a crack!

I'll provide hints for you and then a full solution to some of the questions in the spoiler! :)

Q4
Hint 1: If you want to prove that a triangle is right angled, then you may like to use Pythagoras' Theorem! Perhaps, try showing that:
\[ \left|OQ\right|^2 = \left|OP\right|^2 + \left|PQ\right|^2.\]

Q4 - full solution
As hinted, we'll try to employ Pythagoras' Theorem.
First note that given: \(z = x + iy\), we have \(\left|OP\right| = \sqrt{x^2 + y^2}\) and:
\[ \boxed{\left|OP\right|^2 = x^2 + y^2}\]

We also have:
\[\begin{align*}z + iz &= (x + iy) + i(x + iy) \\ &= (x - y) + i(x + y).\end{align*}\]
So we get:
\[ \boxed{\left|OQ\right|^2 = (x - y)^2 + (x + y)^2}\]

Finally, the line segment \(PQ\) is just \(Q - P = ((x - y) - x) + i((x + y) - y) = -y + ix.\)

So we get:
\[ \boxed{\left|PQ\right|^2 = y^2 + x^2}\]

Now, we know that:
\[ \begin{align*}\left|OQ\right|^2 &= (x - y)^2 + (x + y)^2 \\ &= x^2 - 2xy + y^2 + x^2 + 2xy + y^2 \\ &= (x^2 + y^2) + (x^2 + y^2) \\ &= \left|OP\right|^2 + \left|PQ\right|^2.\end{align*}\]

Q5
Q5 - full solution
Let:
\[z_1 = r\text{cis}\theta\] and \[z_2 = r\text{cis}\left(\theta + \frac{\pi}{3}\right)\]

We then have the following (using De Moivre's Theorem):
\[ z_1^2 = r^2\text{cis}(2\theta), \quad z_2^2 = r^2\text{cis}\left(2\theta + \frac{2\pi}{3}\right)\]
Adding these two gives us:
\[ \begin{align*}z_1^2 + z_2^2 &= r^2\text{cis}(2\theta) + r^2\text{cis}\left(2\theta + \frac{2\pi}{3}\right) \\ &= r^2\left(\text{cis}(2\theta) + \text{cis}\left(2\theta + \frac{2\pi}{3}\right)\right) \\ &= r^2\text{cis}\left(2\theta + \frac{\pi}{3}\right) \\ &= r^2\text{cis}\left(\theta + \left(\theta + \frac{\pi}{3}\right)\right) \\ &= r\text{cis}\theta \times r\text{cis}\left(\theta + \frac{\pi}{3}\right) \\ &= z_1z_2\end{align*}\]

Q6
Hint 1: You may like to take a geometric approach :) Try drawing up \(z_1\) on an Argand diagram and from the nose of \(z_1\), draw \(z_2\). The result gives us: \(z_1 + z_2\).
Hint 2: The triangle inequality (see Q8) states that:
\[ \left|x+y\right| \leq \left|x\right| + \left|y\right|\]
Q6 - full solution
We note that:
\[ \left|z_1\right| = \left|(z_1 - z_2) + z_2\right|\]
And by the triangle inequality, we see that:
\[ \left|z_1\right| \leq \left|z_1 - z_2\right| + \left|z_2\right|\]

By a similar argument, the same can be said for \(z_2\). That is:
\[ \left|z_2\right| \leq \left|z_1 - z_2\right| + \left|z_1\right|\]

We see that:
\[ \left|z_1\right| - \left|z_2\right| \leq \left|z_1 - z_2\right|, \quad \left|z_2\right| - \left|z_1\right| \leq \left|z_1 - z_2\right|\]
which implies that:
\[ \left|\left|z_1\right|-\left|z_2\right|\right| \leq \left|z_1 - z_2\right|.\]

Q8
Hint: Best bet would be to prove it by induction.
Q8 - full solution
Base case is just the simple triangle inequality.

When \(n = k\), we have:
\[ \left|z_1 + z_2 + z_3 + \dots + z_k\right| \leq \left|z_1\right| + \left|z_2\right| + \left|z_3\right| + \dots + \left|z_k\right|.\]

When \(n = k+1\), we have:
\[ \begin{align*}\left|z_1 + z_2 + z_3 + \dots + z_k + z_{k+1}\right| &\leq \left|z_1 + z_2 + z_3 + \dots + z_k\right| + \left|z_{k+1}\right| \tag{by our triangle inequality} \\ &\leq \left|z_1\right| + \left|z_2\right| + \left|z_3\right| + \dots + \left|z_{k+1}\right|\end{align*}\]
which completes the proof.

Q9
Q9 - full solution
If the triangles \(OBA\) and \(OID\) are similar, then it follows that:
\[ \frac{OD}{OA} = \frac{OI}{OB}\]

and thus, we have:
\[ \frac{OD}{OI} = \frac{OA}{OB} = \frac{z_1}{z_2} = OD.\]
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on December 04, 2018, 08:23:09 pm
Hey, don't usually go through the MX2 thread as I didn't do MX2 but considering no one's answered it, I'll give them a crack!

I'll provide hints for you and then a full solution to some of the questions in the spoiler! :)

Q4
Hint 1: If you want to prove that a triangle is right angled, then you may like to use Pythagoras' Theorem! Perhaps, try showing that:
\[ \left|OQ\right|^2 = \left|OP\right|^2 + \left|PQ\right|^2.\] Thank you for the help kind sir!

Q4 - full solution
As hinted, we'll try to employ Pythagoras' Theorem.
First note that given: \(z = x + iy\), we have \(\left|OP\right| = \sqrt{x^2 + y^2}\) and:
\[ \boxed{\left|OP\right|^2 = x^2 + y^2}\]

We also have:
\[\begin{align*}z + iz &= (x + iy) + i(x + iy) \\ &= (x - y) + i(x + y).\end{align*}\]
So we get:
\[ \boxed{\left|OQ\right|^2 = (x - y)^2 + (x + y)^2}\]

Finally, the line segment \(PQ\) is just \(Q - P = ((x - y) - x) + i((x + y) - y) = -y + ix.\)

So we get:
\[ \boxed{\left|PQ\right|^2 = y^2 + x^2}\]

Now, we know that:
\[ \begin{align*}\left|OQ\right|^2 &= (x - y)^2 + (x + y)^2 \\ &= x^2 - 2xy + y^2 + x^2 + 2xy + y^2 \\ &= (x^2 + y^2) + (x^2 + y^2) \\ &= \left|OP\right|^2 + \left|PQ\right|^2.\end{align*}\]

Q5
Q5 - full solution
Let:
\[z_1 = r\text{cis}\theta\] and \[z_2 = r\text{cis}\left(\theta + \frac{\pi}{3}\right)\]

We then have the following (using De Moivre's Theorem):
\[ z_1^2 = r^2\text{cis}(2\theta), \quad z_2^2 = r^2\text{cis}\left(2\theta + \frac{2\pi}{3}\right)\]
Adding these two gives us:
\[ \begin{align*}z_1^2 + z_2^2 &= r^2\text{cis}(2\theta) + r^2\text{cis}\left(2\theta + \frac{2\pi}{3}\right) \\ &= r^2\left(\text{cis}(2\theta) + \text{cis}\left(2\theta + \frac{2\pi}{3}\right)\right) \\ &= r^2\text{cis}\left(2\theta + \frac{\pi}{3}\right) \\ &= r^2\text{cis}\left(\theta + \left(\theta + \frac{\pi}{3}\right)\right) \\ &= r\text{cis}\theta \times r\text{cis}\left(\theta + \frac{\pi}{3}\right) \\ &= z_1z_2\end{align*}\]

Q6
Hint 1: You may like to take a geometric approach :) Try drawing up \(z_1\) on an Argand diagram and from the nose of \(z_1\), draw \(z_2\). The result gives us: \(z_1 + z_2\).
Hint 2: The triangle inequality (see Q8) states that:
\[ \left|x+y\right| \leq \left|x\right| + \left|y\right|\]
Q6 - full solution
We note that:
\[ \left|z_1\right| = \left|(z_1 - z_2) + z_2\right|\]
And by the triangle inequality, we see that:
\[ \left|z_1\right| \leq \left|z_1 - z_2\right| + \left|z_2\right|\]

By a similar argument, the same can be said for \(z_2\). That is:
\[ \left|z_2\right| \leq \left|z_1 - z_2\right| + \left|z_1\right|\]

We see that:
\[ \left|z_1\right| - \left|z_2\right| \leq \left|z_1 - z_2\right|, \quad \left|z_2\right| - \left|z_1\right| \leq \left|z_1 - z_2\right|\]
which implies that:
\[ \left|\left|z_1\right|-\left|z_2\right|\right| \leq \left|z_1 - z_2\right|.\]

Q8
Hint: Best bet would be to prove it by induction.
Q8 - full solution
Base case is just the simple triangle inequality.

When \(n = k\), we have:
\[ \left|z_1 + z_2 + z_3 + \dots + z_k\right| \leq \left|z_1\right| + \left|z_2\right| + \left|z_3\right| + \dots + \left|z_k\right|.\]

When \(n = k+1\), we have:
\[ \begin{align*}\left|z_1 + z_2 + z_3 + \dots + z_k + z_{k+1}\right| &\leq \left|z_1 + z_2 + z_3 + \dots + z_k\right| + \left|z_{k+1}\right| \tag{by our triangle inequality} \\ &\leq \left|z_1\right| + \left|z_2\right| + \left|z_3\right| + \dots + \left|z_{k+1}\right|\end{align*}\]
which completes the proof.

Q9
Q9 - full solution
If the triangles \(OBA\) and \(OID\) are similar, then it follows that:
\[ \frac{OD}{OA} = \frac{OI}{OB}\]

and thus, we have:
\[ \frac{OD}{OI} = \frac{OA}{OB} = \frac{z_1}{z_2} = OD.\]
Title: Re: 4U Maths Question Thread
Post by: myopic_owl22 on December 07, 2018, 11:02:22 pm
Hi there!
I was reading through Rui's notes when it's mentioned that the parallelogram method of adding complex number vectors is better than the 'tip to tail' method (which should be "avoided where possible in the MX2 course"). I was wondering what was the reason for this was, given my teacher's strong advocacy for the latter method.
Thanks a bunch!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 07, 2018, 11:53:49 pm
Hi there!
I was reading through Rui's notes when it's mentioned that the parallelogram method of adding complex number vectors is better than the 'tip to tail' method (which should be "avoided where possible in the MX2 course"). I was wondering what was the reason for this was, given my teacher's strong advocacy for the latter method.
Thanks a bunch!
I'm not a fan of tip-to-tail for half of the complex numbers proofs problems I see. The parallelogram makes it more clear that the main diagonal of the parallelogram is \(z+w\), but then illustrates how the other diagonal represents \(z-w\). It is also the more formal way of doing vector addition.

I've grown to appreciate tip-to-tail for more advanced math problems, because it's just intuitively easier to visualise in comparison (especially when you need more than one vector and thus you risk too many parallelograms). It makes more intuitive sense, so to speak. But the parallelogram has typically been able to give me more useful information in HSC 4U.
Title: Re: 4U Maths Question Thread
Post by: myopic_owl22 on December 08, 2018, 12:07:14 pm
Thanks Rui! Definitely clears things up. As a follow-up question, I was wondering how you'd go about approaching/ setting out an answer for a question like the one in 2017 - Q13(e)

(https://i.ibb.co/GV2YZ0m/Capture.png)

"By using vectors" - does that mean demonstrate geometrically?

Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 08, 2018, 02:55:33 pm
Thanks Rui! Definitely clears things up. As a follow-up question, I was wondering how you'd go about approaching/ setting out an answer for a question like the one in 2017 - Q13(e)

(https://i.ibb.co/GV2YZ0m/Capture.png)

"By using vectors" - does that mean demonstrate geometrically?


This question doesn't use vector addition directly but rather draws on a key consequence (in fact namely the \(z-w\) issue). Recognising that \(\overrightarrow{DC}\) represents \(c-d\) and etc. originate from the parallelogram, but it can be something that people just memorise.



Rearranging this equation gives what we wish to prove.
Title: Re: 4U Maths Question Thread
Post by: myopic_owl22 on December 08, 2018, 06:26:03 pm
Thanks again! Sometimes I find it difficult to know how to explain myself in an exam - good to see that a diagram and drawn vectors weren't required. :)
Title: Re: 4U Maths Question Thread
Post by: not a mystery mark on December 08, 2018, 08:27:28 pm
I feel like we'll get to know each other a lot on this forum haha.
I'm unsure if I'm way too tired or if this is hard:

Find the locus of z if

a) is purely real
b) is purely imaginary
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 08, 2018, 08:44:26 pm
I feel like we'll get to know each other a lot on this forum haha.
I'm unsure if I'm way too tired or if this is hard:

Find the locus of z if

a) is purely real
b) is purely imaginary
It is quite tedious.



Similarly if it is purely real, setting the imaginary part equal to 0 gives the straight line \(x-6y-2 = 0\), but again we must exclude the point \((2,0)\) on the line.
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on December 09, 2018, 06:19:16 pm
Hello,
Can anyone please give me a detailed answer for Q8 ASAP (I have included some of my working out)? Thanks :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 09, 2018, 06:34:40 pm
Hello,
Can anyone please give me a detailed answer for Q8 ASAP (I have included some of my working out)? Thanks :)
Please be mindful when using comments like this. It creates unnecessary pressure to rush people who are taking their time to help out.

There's no question attached.
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on December 09, 2018, 07:36:14 pm
Please be mindful when using comments like this. It creates unnecessary pressure to rush people who are taking their time to help out.

There's no question attached.
Sorry about the ASAP part Rui. Here is question 8 (although I have already written it down on the paper).
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 09, 2018, 08:15:30 pm
Sorry about the ASAP part Rui. Here is question 8 (although I have already written it down on the paper).




Title: Re: 4U Maths Question Thread
Post by: david.wang28 on December 09, 2018, 08:29:42 pm





Thank you for the help Rui! Much appreciated :)
Title: Re: 4U Maths Question Thread
Post by: Sr-1425 on December 12, 2018, 08:07:07 pm
Hi, I was wondering how one would go about proving that:
Sinx+sin2x+sin3x+......+sin(nx)=sin(x(n+1)/2)*sin(nx/2) / sin(x/2)

Knowing that De Moivre's theorem must be used.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 12, 2018, 09:25:37 pm
Hi, I was wondering how one would go about proving that:
Sinx+sin2x+sin3x+......+sin(nx)=sin(x(n+1)/2)*sin(nx/2) / sin(x/2)

Knowing that De Moivre's theorem must be used.
This particular one was addressed in this thread
Title: Re: 4U Maths Question Thread
Post by: Sr-1425 on December 12, 2018, 09:29:56 pm
Thank you.
Title: Re: 4U Maths Question Thread
Post by: Elias S on December 14, 2018, 06:47:34 pm
Hi :)
I have a problem that I can't seem to resolve in my head. Perhaps someone knows the answer. I want to compute the integral Of course, to an extension 2 student this is not too difficult, you just use the substitution However, when you eventually get to the line you encounter a problem, because However, everywhere I look I find the solution continue as Is it correct to continue like this, ignoring the absolute values? If yes, why? If not, how can we make it rigorous?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 14, 2018, 06:58:05 pm
Hi :)
I have a problem that I can't seem to resolve in my head. Perhaps someone knows the answer. I want to compute the integral Of course, to an extension 2 student this is not too difficult, you just use the substitution However, when you eventually get to the line you encounter a problem, because However, everywhere I look I find the solution continue as Is it correct to continue like this, ignoring the absolute values? If yes, why? If not, how can we make it rigorous?

The simple fix to make it rigorous is to impose the condition that \( -\frac\pi2 \leq \theta \leq \frac\pi2 \). This covers all values in the range of \(y=\sin \theta\), that is \(-1\leq y\leq 1\).

And then, note that for any angle \(\theta\) in the first or negative first quadrants, \(\cos\theta\) is positive. Therefore \( |\cos\theta| = \cos\theta\) for this case.

In practice, it just tends to be assumed. It's nicer to use, because now we can assume \(\theta = \sin^{-1}x\) without justification, thereby avoiding any general formula tediosity

Edit: You made a typo I think, but it was clear enough to me that you substituted in \(\boxed{x=\sin\theta}\)
Title: Re: 4U Maths Question Thread
Post by: Elias S on December 14, 2018, 08:00:12 pm
Edit: You made a typo I think, but it was clear enough to me that you substituted in \(\boxed{x=\sin\theta}\)

Ah, my apologies for the typo. Thanks for the reply!
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on December 15, 2018, 05:08:12 pm
Hello, I have a few problems that are bugging me. I have attempted some of my working out, but for a couple of the questions I don't know where to go. Can anyone please help me with Q3, Q4, Q8 and Q10? It would be greatly appreciated :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 15, 2018, 05:57:15 pm
Hello, I have a few problems that are bugging me. I have attempted some of my working out, but for a couple of the questions I don't know where to go. Can anyone please help me with Q3, Q4, Q8 and Q10? It would be greatly appreciated :)
Starting points:

Q3 and Q4 both represent standard loci that you are expected to memorise or identify on the spot.
- \( |z-1| = |z-i| \) is the perpendicular bisector of the points representing \(1\) and \(i\) (i.e. the perpendicular bisector of \((0,1) \) and \((1,0)\) Factoring in the inequality, we want the region cut off by this line that includes the point \((1,0)\).
- \( |z-2-2i| \leq 1 \) is the circle centred at \( (2,2)\) with radius \(1\), and the inequality suggests that we want its interior.
- \( \arg(z+3) = \frac\pi3 \) is the ray starting at \(z=-3\) (i.e. the point \((-3,0)\), inclined at an angle of \( \frac\pi3\) and excluding the point \((-3,0)\) itself. The equation of the ray itself is not of interest.

Note however that Q8 is a special locus. You should watch Eddie Woo's video to understand this locus if you were not introduced to it at school to get started on this question.

I will only do Q10 for now as this is still a bombardment of questions, however that particular one is not something taught at school and expected to be memorised.


As usual, keep in mind that we aim to eliminate any parameters when handling locus problems this way. Here, the parameter will be \(\theta\) for part a), and \(r\) for part b).




Title: Re: 4U Maths Question Thread
Post by: david.wang28 on December 15, 2018, 06:03:08 pm
Starting points:

Q3 and Q4 both represent standard loci that you are expected to memorise or identify on the spot.
- \( |z-1| = |z-i| \) is the perpendicular bisector of the points representing \(1\) and \(i\) (i.e. the perpendicular bisector of \((0,1) \) and \((1,0)\) Factoring in the inequality, we want the region cut off by this line that includes the point \((1,0)\).
- \( |z-2-2i| \leq 1 \) is the circle centred at \( (2,2)\) with radius \(1\), and the inequality suggests that we want its interior.
- \( \arg(z+3) = \frac\pi3 \) is the ray starting at \(z=-3\) (i.e. the point \((-3,0)\), inclined at an angle of \( \frac\pi3\) and excluding the point \((-3,0)\) itself. The equation of the ray itself is not of interest.

Note however that Q8 is a special locus. You should watch Eddie Woo's video to understand this locus if you were not introduced to it at school to get started on this question.

I will only do Q10 for now as this is still a bombardment of questions, however that particular one is not something taught at school and expected to be memorised.


As usual, keep in mind that we aim to eliminate any parameters when handling locus problems this way. Here, the parameter will be \(\theta\) for part a), and \(r\) for part b).





Thanks for the help kind sir! :)
Title: Re: 4U Maths Question Thread
Post by: 006896 on December 23, 2018, 08:42:22 pm
Hi,
I have a question about doing 4U Maths for my HSC.
I will graduate in 2019, so I've already started by Year 12 year. I also accelerated Maths 2U and 3U, in which I received full marks for both subjects. I also state ranked in 2U Maths (in the top 3). I love maths and am currently doing 4U maths, but I don't see myself getting better than a mid-top-band for 4U maths. So many people have been telling me that it's crazy for me to do 4U maths, because it will make my 2U marks redundant. So what should I do? Will state ranking in 2U help increase my ATAR, in comparison than, say, getting a 95 in 4U maths? What should I do to get the highest ATAR - should I drop 4U maths to keep my 2U marks, or keep 4U maths and ditch my 2U marks? To be honest, my aim is to get the highest ATAR, not to gain skills and learn more (sorry!).
I am doing 14 units right now, including my 3 units of maths, which I have already completed.
Any advice is greatly appreciated.
Thanks
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 23, 2018, 09:04:10 pm
Hi,
I have a question about doing 4U Maths for my HSC.
I will graduate in 2019, so I've already started by Year 12 year. I also accelerated Maths 2U and 3U, in which I received full marks for both subjects. I also state ranked in 2U Maths (in the top 3). I love maths and am currently doing 4U maths, but I don't see myself getting better than a mid-top-band for 4U maths. So many people have been telling me that it's crazy for me to do 4U maths, because it will make my 2U marks redundant. So what should I do? Will state ranking in 2U help increase my ATAR, in comparison than, say, getting a 95 in 4U maths? What should I do to get the highest ATAR - should I drop 4U maths to keep my 2U marks, or keep 4U maths and ditch my 2U marks? To be honest, my aim is to get the highest ATAR, not to gain skills and learn more (sorry!).
I am doing 14 units right now, including my 3 units of maths, which I have already completed.
Any advice is greatly appreciated.
Thanks
If your final result for 2U and 3U were literally 100 and 50 (100 if you do end up continuing 4U) then it becomes pretty hard to say. Because of course something like a 95 in 4U would be the minimum to have a chance of topping off 100 in 2U.

Yet at the same time it's still possible to get a 100 in 4U regardless if you've already managed a perfect score twice. Won't say that's easy obviously but it's still theoretically possible.

I'm still gonna stick to my same old advice of making the decision to drop after the half yearly. If after the half yearly you feel that there might not be a chance of gunning for your 95+, then you can go for it. But otherwise the gamble may pay off for all we know.

The only thing that takes me aback a bit is that last bit with the gain skills and learn more. If your maths potential is already that great and you have a passion for it, there's no reason for you not to. Whilst it's not as great as rank 3, I know someone who came rank 15 in 2U maths and still made the decision to carry through 4U anyway. But because I can understand the desire to get the maximum ATAR, you really should consider still learning the 4U content but then dropping it before, say, the trials. You don't need to necessarily put effort into 4U, so that your ATAR doesn't get hit by anything, but at least you come out with the skills set everyone else did. (Trying to put myself in your shoes, if I were having the debate over keeping my 2U mark and somewhat favouring that option, that's what I would've done.)

Don't blame you at all for your predicament, especially since you have 14 units in total, but again that's probably my suggestion. Take as you will, or ask more if you wish to
Title: Re: 4U Maths Question Thread
Post by: terassy on January 07, 2019, 08:49:31 pm
Question from Terry Lee's textbook:

The normal at an end P(x1,y1) of the latus rectum of the ellipse x^2/a^2+y^2/b^2=1 meets the y-axis in M, and PN is the abscissa of P (i.e. PN is perpendicular to the y-axis). Prove that MN=a.

I know how to do this but I'm getting confused whether to use the ± sign. For example when finding the x coordinate of P it would be ±ae right? But the answers only use ae. Same when finding the y coordinate of P, which I substitute ae into x^2/a^2+y^2/b^2=1 to find the y value which would be ±b2/a. But again, the answers only use b2/a. Please help :) sorry to disturb your holidays
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 07, 2019, 09:05:25 pm
Question from Terry Lee's textbook:

The normal at an end P(x1,y1) of the latus rectum of the ellipse x^2/a^2+y^2/b^2=1 meets the y-axis in M, and PN is the abscissa of P (i.e. PN is perpendicular to the y-axis). Prove that MN=a.

I know how to do this but I'm getting confused whether to use the ± sign. For example when finding the x coordinate of P it would be ±ae right? But the answers only use ae. Same when finding the y coordinate of P, which I substitute ae into x^2/a^2+y^2/b^2=1 to find the y value which would be ±b2/a. But again, the answers only use b2/a. Please help :) sorry to disturb your holidays
Well, most likely in the exam they'd give you a diagram so that you only have to consider one of the endpoints of one particular latus rectum.

But for the textbook question, you're right that in theory you should consider the four cases of \( P\left(\pm ae, \pm a(1-e^2) \right) \) and \(P\left( \pm ae, \mp a(1-e^2) \right) \). And you could do this and it should work in every case (provided you can juggle the plus/minus and the minus/plus, else you'd have to do all 4 cases by hand). But at the same time I could argue via the symmetry of the ellipse that this will work for all four cases?

(Remark: you could rewrite the two subcases as \( \left(ae, \pm a(1-e^2)\right) \) and \(\left(-ae, \pm a(1-e^2) \right) \), but you may still bump into some minus/plus symbols.)

So I could say something like, "suppose without loss of generality that \(P\) is the endpoint \( (ae, a(1-e^2) ) \) of the right-hand latus rectum", then do yadayadaya, and justify by symmetry why it must also hold for the other three endpoints as well.

But again, in the exam most likely they wouldn't leave it so ambiguous.
Title: Re: 4U Maths Question Thread
Post by: DrDusk on January 09, 2019, 05:27:27 am
I was so happy to be done with 4u maths, now starting at UNSW I found out just some time ago I need to do math1131 or math1141 which covers 4u in apparently like a month. Ah RIP to me
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 09, 2019, 07:07:59 am
I was so happy to be done with 4u maths, now starting at UNSW I found out just some time ago I need to do math1131 or math1141 which covers 4u in apparently like a month. Ah RIP to me
Maybe post this under the UNSW section next time.

Anyway, whilst I'd say MATH1141 is mildly harder than 4u, I personally reckon MATH1131 is equal in difficulty overall. There are some learning curves (e.g. definition of a limit) that you do need to wrap your head around and it takes a fair while, just as harder 3u inequalities does. But other than that a lot of the stuff is just coverage of the basics in 4u. (Note that proofs in conics and volumes are not examinable in MATH1131 anymore - only the fact that the ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) and similarly for the hyperbola is necessary.)

The biggest overlap in MATH1131 and 4u is really just complex numbers and polynomials.

I also feel like MATH1231 (the follow up course in term 2) is easier
Title: Re: 4U Maths Question Thread
Post by: louisaaa01 on January 11, 2019, 02:48:47 pm
Just in relation to the ATAR Notes MX2 Topic Tests, I came across this question in Complex Numbers:

"The complex number z satisfies |z-i| = 1. What is the greatest distance that z can be from the point (1,0) in the Argand diagram?"

Intuitively, I thought that the line from (1,0) to the circle |z-i| = 1 with greatest length must pass through the centre of the circle, since the diameter is the longest 'chord' in a circle.

I got the right answer, but I was just wondering whether this was a valid conclusion or not? Would this kind of logic occur in every case or was I just lucky in getting the answer right? The worked solutions have two alternative methods, none of which use this reasoning.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 11, 2019, 02:57:49 pm
Just in relation to the ATAR Notes MX2 Topic Tests, I came across this question in Complex Numbers:

"The complex number z satisfies |z-i| = 1. What is the greatest distance that z can be from the point (1,0) in the Argand diagram?"

Intuitively, I thought that the line from (1,0) to the circle |z-i| = 1 with greatest length must pass through the centre of the circle, since the diameter is the longest 'chord' in a circle.

I got the right answer, but I was just wondering whether this was a valid conclusion or not? Would this kind of logic occur in every case or was I just lucky in getting the answer right? The worked solutions have two alternative methods, none of which use this reasoning.
That's interesting; I was surprised to see that that method is not there. Your method is certainly also valid.
Title: Re: 4U Maths Question Thread
Post by: _Himani_ on January 15, 2019, 04:20:56 pm
Hi
I'm fairly new to integration and got a bit stuck with this question type: ∫(x-1)/(x^2+1) dx. Should I be integrating these question with partial fractions or can I do this with inverse trig?
PS: This is my first post on the forum!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 15, 2019, 04:42:14 pm
Hi
I'm fairly new to integration and got a bit stuck with this question type: ∫(x-1)/(x^2+1) dx. Should I be integrating these question with partial fractions or can I do this with inverse trig?
PS: This is my first post on the forum!
\[ \text{Seeing as though the quadratic in the denominator is irreducible}\\ \text{i.e. it cannot be factored any further without complex numbers}\\ \text{some kind of splitting may be necessary.} \]
\[ \text{In your case, you may wish to write the integral as}\\ \int \frac{x}{x^2+1} - \frac{1}{x^2+1}\,dx.\\ \text{You should be able to identify the inverse trig in the second integral.} \]
Partial fractions would usually be the go-to if you found that the fraction on the denominator could be factorised, for example \( \frac{x-1}{x^2-4}\) instead.

Note: If you look at it closely, the "split" is always done to force a log to appear, whenever there's \(x\) in the top and \(x^2\) in the bottom.
Title: Re: 4U Maths Question Thread
Post by: Aviator_13 on January 23, 2019, 05:31:27 pm
Need some help with this question.
Find the Cartesian equation of the locus of z such that arg(z-2)=arg(z2). Describe the locus geometrically, noting any restrictions.

I used a geometric approach to do this question where i made arg(z) = theta therefore arg(z-2) = 2arg(z) = 2 theta. This essentially makes the locus of z a circle. |z-2| = 4. The restriction being that z cannot equal 0 and 2. But in the answers there was another answer which said when y=0....x>2 therefore making x>2 (ray) also a solution. Can someone explain this in a bit more detail?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 23, 2019, 07:46:51 pm
I briefly discussed it with a friend and we kinda agree that you're supposed to just "see" the ray. But there may be a slight bit of intuition as to why it's at least there.

First, note that \( \arg(z-z_1) - \arg(z-z_2) = \theta\) represents a major arc, when \( 0 < \theta < \frac\pi2\). But as \(\theta \to 0\), the arc keeps expanding and expanding until when \(\theta = 0\), we end up with a pair of diametrically opposed rays.

But a significant thing to note is that we ended up with a pair of rays. Usually the scenario of \( \arg z = \arg w\) results in two separate components. It just so happened that for the specific case of \( \arg (z-z_1) = \arg(z-z_2)\), the two rays were what happened.

Whereas here, more or less that \(z^2\) is why one of them is a circle. But at least now we have some intuition behind why there's two relevant components here.

With that settled, I still think we more or less need to "see" the ray. Our equation was \( \arg (z- 2) = \arg (z^2)\), but there is a particular edge case involved. (This usually happens when we deal with the argument just because it isn't always so nice.)
The idea is that here, the special edge case is when \(z-2\) and \(z^2\) are both strictly positive real numbers. Note that under those conditions, \( \arg (z-2) = 0\) and \(\arg (z^2) = 0\).

So the solutions just considered when \(z^2 > 0\) and \(z-2\) simultaneously hold, having treated \(z\) as a real number. This is, of course, when \( z > 2\).

What's the intuition behind seeing it? I'm actually not too sure how to put it in concrete high school terms at this stage. Can't see anything that would make it super obvious.

Remark: From doing some research I've found that \( \arg z = \arg w \) loci can be handled algebraically. But I won't go into that unless you request it, more or less because I'm trying to keep things reasonably geometric here.
Title: Re: 4U Maths Question Thread
Post by: terassy on January 27, 2019, 06:19:39 pm
Please help with the second part of this question.

The points P( cp , c/p ) and Q( cq , c/q ), c > 0, lie on different branches of the hyperbola with equation xy = c2. The tangent at Q is parallel to the tangent at P.

(i) Show that the equation of the tangent of P is y = 2c/p - x/p2.                     (2 marks)
(ii) Deduce that q = -p                                             (1 mark)
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on January 27, 2019, 07:00:44 pm
Hey there!

For the first question, you can simply rearrange the equation and differentiate, before subbing in x=cp and then subbing into the point gradient form of a line.



And obviously substituting x=cp gets you the gradient -1/p^2, and you should be able to do the rest to get it to match up to the equation in the question.

For the second question, if the gradients are equal for both P and Q, this means that -1/p^2=-1/q^2. You can rearrange to p^2=q^2. From this, you should be able to see that p is equal to plus or minus q. But since they are on different branches, and thus not the same point, p has to equal -q and thus q=-p.

Hope this helps :)
Title: Re: 4U Maths Question Thread
Post by: terassy on January 27, 2019, 08:09:06 pm
Thank you.

I'm going through a topic test and got this question wrong:

The line PN is the normal to the ellipse x2/25 + y2/9 = 1 at P(x0, y0) and S1 and S2 are the foci of the ellipse. Angle NPS1 = alpha and Angle NPS2 = beta. Show that alpha = beta.

(https://i.imgur.com/TJYMg5T.jpg)

I've been doing this question for so long and can't come up with an answer. My teacher said to use the angle between two lines formula. I found that mPN= 25y0/9xo and mPS1=y0/x0  - 4 and mPS2=y0/x0  + 4. But it doesn't work out.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 27, 2019, 10:15:00 pm
Thank you.

I'm going through a topic test and got this question wrong:

The line PN is the normal to the ellipse x2/25 + y2/9 = 1 at P(x0, y0) and S1 and S2 are the foci of the ellipse. Angle NPS1 = alpha and Angle NPS2 = beta. Show that alpha = beta.

(https://i.imgur.com/TJYMg5T.jpg)

I've been doing this question for so long and can't come up with an answer. My teacher said to use the angle between two lines formula. I found that mPN= 25y0/9xo and mPS1=y0/x0  - 4 and mPS2=y0/x0  + 4. But it doesn't work out.
In the future, please use brackets to indicate what's going on. When reading your gradients, the gradient of PS1 is interpreted as \( m_{PS_1}= \frac{y_0}{x_0} - 4 \), not the intended \( m_{PS_1} = \frac{y_0}{x_0 - 4}  \)
\[ \text{Note that since }(x_0,y_0)\text{ lies on the ellipse we have}\\ \frac{x_0^2}{25}+ \frac{y_0^2}{9} = 1.\\ \text{Multiplying both sides of the equation by 225 gives}\\ \boxed{9x_0^2 + 25y_0^2 = 225} \]
\[ \text{Now subbing into the angle between two lines formula we have}\\ \begin{align*} \tan \alpha &= \left| \frac{\frac{25y_0}{9x_0} - \frac{y_0}{x_0-4}}{1+ \frac{25y_0}{9x_0}\, \frac{y_0}{x_0-4}} \right| \\ &= \left| \frac{\frac{25y_0(x_0-4) - 9x_0y_0}{9x_0 (x_0-4)}}{\frac{9x_0(x_0-4) + 25y_0^2}{9x_0(x_0-4)}} \right|\tag{combining fractions}\\ &= \left| \frac{25x_0y_0-100y_0-9x_0y_0}{9x_0^2-36x_0+25y_0^2} \right|\\ &= \left|\frac{16x_0y_0-100y_0}{225-36x_0}\right|\tag{using boxed result}\\ &= \left|\frac{4y_0(4x_0-25)}{-9(4x_0-25)} \right|\\ &= \frac49 |y_0|\end{align*} \]
You should now be able to prove that \(\tan \beta\) equals the same expression, so \(\tan \alpha = \tan \beta\). And since \(\alpha\) and \(\beta\) are both acute angles, this implies that \(\alpha=\beta\).
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on January 30, 2019, 04:58:27 pm
Hello,
I'm having trouble with this question in the ink below. I tried to use an argument approach, but I was stuck midway. Can anyone please help me with this question? Thanks :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 30, 2019, 05:03:43 pm
Hello,
I'm having trouble with this question in the ink below. I tried to use an argument approach, but I was stuck midway. Can anyone please help me with this question? Thanks :)
This question came from the 2016 HSC. Fairly sure it was Q16. Should be in the compilation
Title: Re: 4U Maths Question Thread
Post by: terassy on January 31, 2019, 08:19:43 pm
Hi, could someone help me solve part (iii) of this question:

(https://i.gyazo.com/d211669d58aa2d3139e74bf1e0574792.png)

I know how to show the angle of inclination between the normal and semi-minor axis of the ellipse is equal, through creating an isosceles triangle. But I don't know how to show the tangents to the circle and the ellipse at the point of intersection is equal to the angle of inclination between the normal.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 31, 2019, 09:00:20 pm
Hi, could someone help me solve part (iii) of this question:

(https://i.gyazo.com/d211669d58aa2d3139e74bf1e0574792.png)

I know how to show the angle of inclination between the normal and semi-minor axis of the ellipse is equal, through creating an isosceles triangle. But I don't know how to show the tangents to the circle and the ellipse at the point of intersection is equal to the angle of inclination between the normal.
Hint: The whole point of the previous part was to show that \(N\) actually lies on the circle. Does this diagram seem to evoke the alternate segment theorem?

(https://i.imgur.com/KM4fSBA.png)

Remark: You may need to look more closely to see the triangle.
Title: Re: 4U Maths Question Thread
Post by: not a mystery mark on February 06, 2019, 09:29:09 am
Hey! My teacher and I are completely stuck on this Polynomials question. We can complete part i), but part ii) & iiI) is a mystery.

If w=x+x^(-1)
i)
ii)
iii) Show that the roots of are the four complex roots of and deduce that
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 06, 2019, 10:15:48 am
Hey! My teacher and I are completely stuck on this Polynomials question. We can complete part i), but part ii) & iiI) is a mystery.

If w=x+x^(-1)
i)
ii)
iii) Show that the roots of are the four complex roots of and deduce that
\begin{align*} x^4+x^3+x^2+x+1&= x^2(x^2+x+1+x^{-1}+x^{-2})\\ &= x^2 (\omega^2-2 + \omega + 1)\\ &= x^2(\omega^2 + \omega - 1)\end{align*}
\[ \text{The roots of }\omega^2+\omega - 1 = 0\\ \text{are }\omega = \frac{-1\pm \sqrt{5}}{2}\text{ by quadratic formula}\\ \text{so we have}\\ \begin{align*} x^2(\omega^2+\omega - 1) &= x^2\left(\omega - \frac{-1-\sqrt{5}}{2} \right)\left( \omega - \frac{-1+\sqrt{5}}{2} \right)\\ &= x^2 \left(x + \frac12 (1+\sqrt5) + x^{-1} \right)\left( x+ \frac12 (1-\sqrt5) + x^{-1}\right)\\ &= \left(x^2+ \frac12 x (1+\sqrt5) + 1 \right) \left(x^2 + \frac12x (1-\sqrt5)+1 \right)\end{align*} \]
__________________________________________

Sketch solution:
\begin{align*} x^4+x^3+x^2+x+1&=0\\ \implies \frac{x^5-1}{x-1}&=0 \tag{geometric series}\\ \implies x^5-1 &= 0\\ \implies x^5 &= 1\end{align*}
Note that \(36^\circ = \frac\pi5\) and \(72^\circ = \frac{2\pi}{5} \).
\[ \text{Using De Moivre's theorem and the usual approach}\\ \text{the roots of }x^5 = 1\text{ are}\\ x = 1, \, x = \operatorname{cis} \left( \pm \frac{2\pi}{5}\right), \, x = \operatorname{cis} \left( \pm\frac{4\pi}{5} \right) \]
\[ \text{But of course, since }x=1\text{ is not a root of }x^4+x^3+x^2+x+1=0\\ \text{and we know that the quartic equation has four solutions,}\\ \text{it follows that the remaining four roots, i.e. }\operatorname{cis} \left( \pm \frac{2\pi}5 \right)\text{ and }\operatorname{cis}\left( \pm \frac{4\pi}{5} \right)\\ \text{are the solutions of }x^4+x^3+x^2+x+1=0.\]
What requires a bit more work now is figuring whether \( \operatorname{cis} \left( \pm \frac{2\pi}5 \right) \) are the roots of \(x^2 + \frac12 x(1+\sqrt5) +1=0\), or the roots of \(x^2 + \frac12x(1-\sqrt5) - 1 = 0\). Note that because all coefficients are real even in the quadratics, non-real roots come in conjugate pairs.
\[ \text{The quadratic whose roots are }\alpha = \operatorname{cis} \left( -\frac{2\pi}{5} \right) \text{ and }\beta = \operatorname{cis} \left( \frac{2\pi}{5} \right)\\ \text{has the sum of roots }\alpha + \beta = 2\cos \frac{2\pi}{5}.\\ \text{Since }\frac{2\pi}{5}\text{ is a first quadrant angle, }\cos \frac{2\pi}{5} > 0\\ \text{so it follows that }\alpha + \beta > 0. \]
\[ \text{The quadratic }x^2+ \frac12 x(1-\sqrt5) + 1 = 0\\ \text{has sum of roots }\alpha_1+\beta_1 = \frac{\sqrt5 - 1}{2}\\ \text{which is positive}. \]
\[ \text{However the quadratic }x^2+ \frac12 x(1+\sqrt5) + 1 = 0\\ \text{has sum of roots }\alpha_2+\beta_2 = -\frac{\sqrt5 + 1}{2}\\ \text{which is negative.}\]
\[ \text{Hence }x=\operatorname{cis} \left( \pm \frac{2\pi}{5} \right)\text{ must be the two roots of}\\ x^2 + \frac12 x (1-\sqrt5) + 1 = 0\\ \text{and further by the sum of roots,}\\ 2\cos \frac{2\pi}{5} = \frac{\sqrt5 - 1}{2} \implies \boxed{\cos \frac{2\pi}{5} = \frac{\sqrt5 - 1}{4}}\\ \text{as required.}\]
\[ \text{Similarly }2\cos \frac{4\pi}{5} = -\frac{ \sqrt5 + 1}{2}\\ \text{However since }\cos (\pi - x) = -\cos x\text{, we then have}\\ -2 \cos \frac\pi5 = -\frac{\sqrt5+1}{2} \implies \boxed{\cos \frac\pi5 = \frac{\sqrt5 + 1}{2}}\\ \text{as required.}\]
Title: Re: 4U Maths Question Thread
Post by: not a mystery mark on February 06, 2019, 10:47:23 am
\[ \text{Hence }x=\operatorname{cis} \left( \pm \frac{2\pi}{5} \right)\text{ must be the two roots of}\\ x^2 + \frac12 x (1-\sqrt5) + 1 = 0\\ \text{and further by the sum of roots,}\\ 2\cos \frac{2\pi}{5} = \frac{\sqrt5 - 1}{2} \implies \boxed{\cos \frac{2\pi}{5} = \frac{\sqrt5 - 1}{4}}\\ \text{as required.}\]
\[ \text{Similarly }2\cos \frac{4\pi}{5} = -\frac{ \sqrt5 + 1}{2}\\ \text{However since }\cos (\pi - x) = -\cos x\text{, we then have}\\ -2 \cos \frac\pi5 = \frac{\sqrt5+1}{2} \implies \boxed{\cos \frac\pi5 = \frac{\sqrt5 + 1}{2}}\\ \text{as required.}\]

Oh my - you are an absolute god! Thank you!
Title: Re: 4U Maths Question Thread
Post by: louisaaa01 on February 06, 2019, 10:17:12 pm
Can someone please help with this question?

On an Argand diagram the points P and Q represent the numbers z1 and z2 respectively. OPQ is an equilateral triangle. Show that z12 + z22 = z1z2

Thank you!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 06, 2019, 10:48:43 pm
Can someone please help with this question?

On an Argand diagram the points P and Q represent the numbers z1 and z2 respectively. OPQ is an equilateral triangle. Show that z12 + z22 = z1z2

Thank you!
\[ \text{Because of }\triangle OPQ\text{ being isosceles}\\ \text{it follows that }\overrightarrow{OQ}\text{ must be some }\frac\pi3\text{-rotation of }\overrightarrow{OP}. \]
\[ \therefore \boxed{z_2 = z_1 \operatorname{cis} \left(\pm \frac\pi3 \right)} \]
\begin{align*} \therefore z_1^2 + z_2^2 &= z_1^2 \left( 1 + \left[\operatorname{cis} \left( \pm \frac{\pi}{3} \right)^2\right] \right)\\ &= z_1^2 \left(1+ \operatorname{cis}\left( \pm \frac{2\pi}{3} \right) \right)\\ &= z_1^2 \left(1 + \left(-\frac12 \pm \frac{\sqrt3}{2}i \right) \right)\\ &= z_1^2 \left( \frac12 \pm \frac{\sqrt3}{2}i \right)\\ &= z_1 \left[z_1 \operatorname{cis} \left( \pm \frac\pi3\right) \right]\\ &= z_1z_2 \end{align*}
Title: Re: 4U Maths Question Thread
Post by: Fatemah.S on February 13, 2019, 11:21:54 am
Hey!
Needed help with this integration by parts question. I tried letting u equal tan^n-2 x and v' equal tan^2x but I wasn't able to make enough progress. Pic attached :) (Cambridge 5.5 question 15)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 13, 2019, 11:25:41 am
That one doesn't use by parts. (Any integral that can be converted to \(I_n = \int \tan^n u \,du\) through substitution doesn't use by parts either.)
\begin{align*} I_n&= \int \tan^n x\,dx\\ &= \int \tan^{n-2}x\tan^2x\,dx\\ &= \int \tan^{n-2}x\sec^2 x\,dx - \int \tan^{n-2}x\,dx \tag{Pythagorean identity}\\ &= \frac{\tan^{n-1}x}{n-1} - I_{n-2} \end{align*}
Title: Re: 4U Maths Question Thread
Post by: Fatemah.S on February 13, 2019, 11:50:15 am
That one doesn't use by parts. (Any integral that can be converted to \(I_n = \int \tan^n u \,du\) through substitution doesn't use by parts either.)
\begin{align*} I_n&= \int \tan^n x\,dx\\ &= \int \tan^{n-2}x\tan^2x\,dx\\ &= \int \tan^{n-2}x\sec^2 x\,dx - \int \tan^{n-2}x\,dx \tag{Pythagorean identity}\\ &= \frac{\tan^{n-1}x}{n-1} - I_{n-2} \end{align*}

Oh that makes sense. Would you please be able to explain how you got from the third to the fourth row? Thank you so much!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 13, 2019, 12:01:26 pm
Oh that makes sense. Would you please be able to explain how you got from the third to the fourth row? Thank you so much!
That \(I_{n-2}\) bit should be clear. As for the first integral I basically just reversed the chain rule. However if you're uncomfortable with this, observe that the substitution \(u=\tan x \implies du = \sec^2 x\,dx\) gives
\[ \int \tan^{n-2}x\sec^2 x\,dx = \int u^{n-2}\,du = \frac{u^{n-1}}{n-1}+C = \frac{\tan^{n-1}x}{n-1}+C \]
Title: Re: 4U Maths Question Thread
Post by: Fatemah.S on February 13, 2019, 12:06:53 pm
That \(I_{n-2}\) bit should be clear. As for the first integral I basically just reversed the chain rule. However if you're uncomfortable with this, observe that the substitution \(u=\tan x \implies du = \sec^2 x\,dx\) gives
\[ \int \tan^{n-2}x\sec^2 x\,dx = \int u^{n-2}\,du = \frac{u^{n-1}}{n-1}+C = \frac{\tan^{n-1}x}{n-1}+C \]

Got it! Thanks
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on February 16, 2019, 11:07:44 am
Hey rui please help with these two questions, I have no absolute idea on how to attempt it except making the denominator 2 for the first one
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 16, 2019, 11:17:29 am
Hey rui please help with these two questions, I have no absolute idea on how to attempt it except making the denominator 2 for the first one
Where did this come from? It looks like competition level maths instead of 4U. But what vexes me more is that a quick check on Wolfram shows that \( \sqrt{\frac{2009\times2010\times2011\times2012+1}{4}} \) is actually a rational number, and the 50th digit after the decimal is (not coincidentally) zero.

One approach of starting Q2 using competition level mathematics
Edit: Later found out that I made a small mistake at the start. I think the expression was \(\sum_{k=0}^\infty \frac{F_{k+1}}{2^{k+2}} \) instead. The final answer should end up being \(1\), not \( \frac12\). Won't bother editing this though - I think all the working out should be reasonably adaptable.
\[ \text{Let }\phi = \frac{1+\sqrt5}{2}, \, \psi = \frac{1-\sqrt5}{2}\\ \text{and recall that }F_n = \frac{\phi^n - \psi^n}{\sqrt{5}} \]
\begin{align*} \sum_{k=0}^\infty \frac{F_k}{2^{k+2}} &= \frac{1}{4\sqrt5} \sum_{k=0}^\infty \frac{\phi^k - \psi^k}{2^k}\\ &= \frac1{4\sqrt5}\left(\sum_{k=0}^\infty \left( \frac\phi2\right)^k - \sum_{k=0}^\infty \left( \frac\psi2 \right)^k \right)\\ &= \frac{1}{4\sqrt5}\left(\frac{1}{1-\frac\phi2} - \frac{1}{1-\frac\psi2} \right) \end{align*}
because each of the sums are (infinite) geometric series. It can be checked through various non-calculator approaches that the common ratio is certainly less than 1.
One approach to commence proving the rationality of the expression in Q1
\[ \text{Let }x=2010+\frac12.\text{ Then,} \]
\begin{align*}\sqrt{\frac{2009\times2010\times2011\times2012+1}4}&= \sqrt{\frac{\left(x-\frac32\right)\left(x-\frac12\right)\left(x+\frac12\right)\left(x+\frac32\right)+1}4}\\ &= \sqrt{\frac{\left(x^2-\frac94\right)\left(x^2-\frac14\right)+1}{4}}\\ &= \sqrt{\frac{(4x^2-9)(4x^2-1) + 16}{64}}\\ &= \frac{\sqrt{16x^4-40x^2+25}}8\\ &= \frac{\sqrt{(4x^2-5)^2}}{8}\\ &= \frac{4x^2-5}{8} \end{align*}
noting that we expect our answer to remain positive.
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on February 16, 2019, 01:07:26 pm
I actually found it on this book website, i was just scrolling and these ques were really interesting but they didnt have answers.https://www.scribd.com/document/99166274/153-TG-Math-Problems-E-book
and theres one with the answers but i need a paid subscription for it
https://www.scribd.com/doc/188084024/153-TG-Math-Problems
Title: Re: 4U Maths Question Thread
Post by: louisaaa01 on February 17, 2019, 05:28:20 pm
Can someone please help with this question?

"Find the Cartesian equation of the locus of z such that arg(z-2) = arg(z2). Describe the locus geometrically, noting any restrictions"

I was okay with finding the locus, but I'm stuck on how to determine the restrictions.

Thanks!
Title: Re: 4U Maths Question Thread
Post by: goodluck on February 17, 2019, 05:40:50 pm
Hey, I'm not sure if this is part of the syllabus but my teacher gave us some extension questions to do (apparently our school doesn't teach harder 3u saying we ought to have already have learned it...) but could someone help me with this:

By considering the cases x ≥ 0, −1 < x < 0 and x ≤ −1 (or otherwise) prove that 1+x+x^2+x^3+x^4 is always positive.
Title: Re: 4U Maths Question Thread
Post by: AlphaZero on February 17, 2019, 06:59:17 pm
Hey, I'm not sure if this is part of the syllabus but my teacher gave us some extension questions to do (apparently our school doesn't teach harder 3u saying we ought to have already have learned it...) but could someone help me with this:

By considering the cases x ≥ 0, −1 < x < 0 and x ≤ −1 (or otherwise) prove that 1+x+x^2+x^3+x^4 is always positive.

I'm not too familiar with the HSC courses yet (I'll be reading through the study designs so I can help in the HSC boards in the future). Nonetheless, here's a start to your question.

Case 1: \(x\geq 0\)
This is the simplest of the three cases. For all \(x\geq0\), we have \(x^n\geq 0\) for all natural numbers \(n\).
\[\text{So, }\ 1+x+x^2+x^3+x^4\geq 1+0+0+0+0>0.\]

Case 2: \(-1<x<0\)
This case is a little trickier, but the result becomes obvious to prove if we write \begin{align*}1+x+x^2+x^3+x^4&=(1+x)+x^2(1+x)+x^4\\
&=(1+x)(1+x^2)+x^4.\end{align*}For \(-1<x<0\), we have \(0<x+1<1\) with \(x^2+1>1\) and \(x^4>0\). \[\text{So, }\ 1+x+x^2+x^3+x^4>0\times 1+0=0.\]

Case 3: \(x\leq -1\).
There are a few ways to solve this case. I'll let you try this one on your own.

If you still need help, don't hesitate to ask :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 17, 2019, 08:32:29 pm
Can someone please help with this question?

"Find the Cartesian equation of the locus of z such that arg(z-2) = arg(z2). Describe the locus geometrically, noting any restrictions"

I was okay with finding the locus, but I'm stuck on how to determine the restrictions.

Thanks!
If you've managed to find both the circle and the ray, the restrictions are that the circle only excludes \( (0,0) \), but the ray is essentially \( \arg (z-2) = 0 \)

I'll let someone else go into more depth though - don't quite have the time right now to figure this locus out again yet sadly

Note
The relevant circle is centred at \( (2,0) \) with radius \(r=2\).
Title: Re: 4U Maths Question Thread
Post by: AlphaZero on February 17, 2019, 09:17:49 pm
Hey, I'm not sure if this is part of the syllabus but my teacher gave us some extension questions to do (apparently our school doesn't teach harder 3u saying we ought to have already have learned it...) but could someone help me with this:

By considering the cases x ≥ 0, −1 < x < 0 and x ≤ −1 (or otherwise) prove that 1+x+x^2+x^3+x^4 is always positive.

Alrighty... I'll give the explanation a go, but it's been a very long time since I've any complex analysis this difficult. (VCE complex numbers is so much easier lol).

I'm assuming that you've already described the locus. Mainly,
> circle of radius 2 centered at \(2+0i\)
> line \(\text{Im}(z)=0\) (essentially the real axis)

All the points on the circle are fine except one, and we'll see why this is in a second.

Let's consider individually different sections of the line \(\text{Im}(z)=0\). For our convenience, let's write \(z=a+0i\).

If  \(a<2\)  (\(a-2<0\)),  then we have  \(\arg(z-2)=\arg(a-2)=\pi\)  and  \(\arg(z^2)=\arg(a^2)=0\)  (since \(a^2>0\)).

Note that obviously we cannot have \(z=2+0i\) since \(\arg(0)\) is undefined.

If \(a>2\)  (\(a-2>0\)),  then we have  \(\arg(z-2)=\arg(a-2)=0\)  and  \(\arg(z^2)=\arg(a^2)=0\)  (since \(a^2>0\)).

Therefore, what is essentially left of the line is the ray given by  \(\arg(z-2)=0\).


Altogether, the locus of \(z\) is given by \[\left\{z\mid \arg(z-2)=0\right\}\cup\left\{z\mid\ |z-2|=2\right\}\setminus\{0\},\] which can be read as "the set of values \(z\) that satisfy  \(\arg(z-2)=0\)  or  \(|z-2|=2\),  and  \(z\neq0\)"
Title: Re: 4U Maths Question Thread
Post by: goodluck on February 18, 2019, 07:42:51 pm
Hey, could someone explain to me how to do this:
Prove using vector methods that the midpoints of the sides of a convex quadrilateral form a parallelogram.

I'm not getting the same results for both the sides to prove it's parallel and I don't really get the internet explanations, I think I might need it to simplified a lot!

Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 18, 2019, 07:49:54 pm
Hey, could someone explain to me how to do this:
Prove using vector methods that the midpoints of the sides of a convex quadrilateral form a parallelogram.

I'm not getting the same results for both the sides to prove it's parallel and I don't really get the internet explanations, I think I might need it to simplified a lot!


Is this supposed to be under complex numbers? No other use of vectors appears in the current MX2 course.
Title: Re: 4U Maths Question Thread
Post by: goodluck on February 18, 2019, 08:21:04 pm
Is this supposed to be under complex numbers? No other use of vectors appears in the current MX2 course.

I'm not quite sure, my substitute teacher gave us a bunch of maths questions to do, saying her son said was appropriate for us (I think some of it is uni maths since there were some she marked wasn't in our syllabus). I thought it was just vector addition like in physics, but I'm not sure how to go about it
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 18, 2019, 08:38:09 pm
I'm not quite sure, my substitute teacher gave us a bunch of maths questions to do, saying her son said was appropriate for us (I think some of it is uni maths since there were some she marked wasn't in our syllabus). I thought it was just vector addition like in physics, but I'm not sure how to go about it
I'm not convinced that they understand the current 4U syllabus. It's actually doable within the boundaries of the new MX1 syllabus but not with any of the current syllabuses. Although if you're really interested in a solution, you can post it in the first year uni maths question thread and I'm happy to address it there.

It requires slightly more than the tip-to-tail stuff you learn in physics.
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on February 19, 2019, 04:57:23 pm
Hello,
I have partially done the working out, but I don't know what to do after. Can anyone please help me with this question? Thanks :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 19, 2019, 05:00:49 pm
Hello,
I have partially done the working out, but I don't know what to do after. Can anyone please help me with this question? Thanks :)
Follow the usual partial fractions approach. Sub \(x=1\), then \(x=2\) and then \(x=3\) to find your coefficients \(A\), \(B\) and \(C\).
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on February 19, 2019, 05:52:13 pm
Follow the usual partial fractions approach. Sub \(x=1\), then \(x=2\) and then \(x=3\) to find your coefficients \(A\), \(B\) and \(C\).
Thank you!
Title: Re: 4U Maths Question Thread
Post by: veronicaaat on February 22, 2019, 06:57:19 pm
Does anyone know how to attempt the answer to q19 and 20?
Thank you so much !
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 22, 2019, 07:05:42 pm
Does anyone know how to attempt the answer to q19 and 20?
Thank you so much !
Considering you've posted two potentially long questions, what attempts have you made so far? Please provide any understanding relevant.
Title: Re: 4U Maths Question Thread
Post by: goodluck on February 25, 2019, 10:29:43 am
Hi, I have an induction question that I'm a bit stuck on

Let A1=1 and for n>1, let A(n+1) (the n+1 is a subscript sorry if it's not clear) = sqrt(1+A(n)). let G=(1+sqrt(5))/2

a) Prove that if A(n) < G, then A(n+1) <G
b) Prove that for all n, A(n+1) > A(n)

I'm a bit confused on whether I should use induction for the first part or for the second part, or whether there's an easier way of doing the question?

Thanks for your help!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 25, 2019, 11:04:20 am
Hi, I have an induction question that I'm a bit stuck on

Let A1=1 and for n>1, let A(n+1) (the n+1 is a subscript sorry if it's not clear) = sqrt(1+A(n)). let G=(1+sqrt(5))/2

a) Prove that if A(n) < G, then A(n+1) <G
b) Prove that for all n, A(n+1) > A(n)

I'm a bit confused on whether I should use induction for the first part or for the second part, or whether there's an easier way of doing the question?

Thanks for your help!
You can actually use induction for both. However, the key thing to remember for both parts is that if \(n \geq 1\), then by definition, \(\boxed{A_{n+1} = \sqrt{1 + A_n}} \). Note - You wrote \(n >1 \), however because of how you defined it I'm actually convinced that you should've written \(n \geq 1\). Because otherwise we cannot compute what \(A_2\) is.
\[ \text{For the second one, we can check that when }n=1,\\ A_2 = \sqrt{1+A_1} = \sqrt{1+1} = \sqrt{2} > 1 = A_1\\ \text{as required.} \]
\[ \text{Now assume that for any integer }k\geq 1,\\ \boxed{A_k > A_{k-1}} \]
\[ \text{Then we have}\\ \begin{align*} A_{k+1} &= \sqrt{1+A_k} \tag{by definition}\\ &>\sqrt{1+A_{k-1}} \tag{inductive assumption}\\ &= A_k \tag{by definition} \end{align*}\\ \text{and hence the statement holds when }n=k+1\text{ as well.}\]
Note that we had used the definition twice.
________________________________________________________
\[ \text{For the first, we can check that when }n=1,\\ A_1 = 1 = \frac{1+\sqrt1}{2} < \frac{1+\sqrt5}{2} = G\\ \text{as required.}\]
\[ \text{Now assume that for any integer }k\geq 1,\\ \boxed{A_k < G}\]
\[ \text{Then we have}\\ \begin{align*} A_{k+1} &= \sqrt{1+A_k}\tag{by definition}\\ &< \sqrt{1+G}\tag{inductive assumption}\\ &= \sqrt{1+\frac{1+\sqrt5}{2}} \\ &= \sqrt{\frac{3+\sqrt5}{2}}\\ &= \sqrt{\frac{6+2\sqrt5}{4}}\\ &= \sqrt{\frac{1+2\sqrt5+5}{4}} \tag{completing the square}\\ &= \sqrt{\frac{(1+\sqrt5)^2}4} \\ &= \frac{1+\sqrt5}{2}\\ &= G\end{align*}\\ \text{as required, having taken positive root}\\ \text{noting that }\frac{1+\sqrt5}{2}\text{ is positive.}\]
Title: Re: 4U Maths Question Thread
Post by: goodluck on February 25, 2019, 03:19:34 pm
You can actually use induction for both. However, the key thing to remember for both parts is that if \(n \geq 1\), then by definition, \(\boxed{A_{n+1} = \sqrt{1 + A_n}} \). Note - You wrote \(n >1 \), however because of how you defined it I'm actually convinced that you should've written \(n \geq 1\). Because otherwise we cannot compute what \(A_2\) is.
\[ \text{For the second one, we can check that when }n=1,\\ A_2 = \sqrt{1+A_1} = \sqrt{1+1} = \sqrt{2} > 1 = A_1\\ \text{as required.} \]
\[ \text{Now assume that for any integer }k\geq 1,\\ \boxed{A_k > A_{k-1}} \]
\[ \text{Then we have}\\ \begin{align*} A_{k+1} &= \sqrt{1+A_k} \tag{by definition}\\ &>\sqrt{1+A_{k-1}} \tag{inductive assumption}\\ &= A_k \tag{by definition} \end{align*}\\ \text{and hence the statement holds when }n=k+1\text{ as well.}\]
Note that we had used the definition twice.
________________________________________________________
\[ \text{For the first, we can check that when }n=1,\\ A_1 = 1 = \frac{1+\sqrt1}{2} < \frac{1+\sqrt5}{2} = G\\ \text{as required.}\]
\[ \text{Now assume that for any integer }k\geq 1,\\ \boxed{A_k < G}\]
\[ \text{Then we have}\\ \begin{align*} A_{k+1} &= \sqrt{1+A_k}\tag{by definition}\\ &< \sqrt{1+G}\tag{inductive assumption}\\ &= \sqrt{1+\frac{1+\sqrt5}{2}} \\ &= \sqrt{\frac{3+\sqrt5}{2}}\\ &= \sqrt{\frac{6+2\sqrt5}{4}}\\ &= \sqrt{\frac{1+2\sqrt5+5}{4}} \tag{completing the square}\\ &= \sqrt{\frac{(1+\sqrt5)^2}4} \\ &= \frac{1+\sqrt5}{2}\\ &= G\end{align*}\\ \text{as required, having taken positive root}\\ \text{noting that }\frac{1+\sqrt5}{2}\text{ is positive.}\]

I'll ask the teacher about the greater than and equal to sign, since the greater than sign was what was written on our worksheet. But if it's any comfort, I completely agree ;)

I have another from a tutoring centre and I'm not sure whether it's just a bit advanced for the course or where it fits in exactly into harder 3u (i think it's applications of inequalities)

Considering 1/x (take x>0), explain why 1/n+1 < the integral from (n+1) to n of dx/x < 1/n
As the integral can be simplified to ln(1 +1/x), do we just graph the intervals to prove that it works (I checked on desmos and it fits nicely)

b) hence or otherwise, deduce (1+1/n)^n < e < (1 + 1/n)^(n+1)
         Do we separate the two sides from part a) and then raise it to e (e.g. from 1+1/k to e^ (1/k+1)) and multiply by the two sides and somehow manipulate the result? I'm not sure on how to get the lone e

c) this one is an extension question but it goes like this:
((1+n)^n)/(e^n) <n! < ((1+n)^(1+n))/(e^n)

Would that be relevant for this course or not?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 25, 2019, 06:04:19 pm
I'll ask the teacher about the greater than and equal to sign, since the greater than sign was what was written on our worksheet. But if it's any comfort, I completely agree ;)

I have another from a tutoring centre and I'm not sure whether it's just a bit advanced for the course or where it fits in exactly into harder 3u (i think it's applications of inequalities)

Considering 1/x (take x>0), explain why 1/n+1 < the integral from (n+1) to n of dx/x < 1/n
As the integral can be simplified to ln(1 +1/x), do we just graph the intervals to prove that it works (I checked on desmos and it fits nicely)

b) hence or otherwise, deduce (1+1/n)^n < e < (1 + 1/n)^(n+1)
         Do we separate the two sides from part a) and then raise it to e (e.g. from 1+1/k to e^ (1/k+1)) and multiply by the two sides and somehow manipulate the result? I'm not sure on how to get the lone e

c) this one is an extension question but it goes like this:
((1+n)^n)/(e^n) <n! < ((1+n)^(1+n))/(e^n)

Would that be relevant for this course or not?
Like yeah it's just harder 3U. Since you posted it on the 4U thread that's fine. I'm gonna swap out all the \(n\)'s for \(k\)'s, except for in the last part.

Yeah so for part i), just sketch the right branch of \(y= \frac1x\) and plot the points \( \left(k, \frac1k\right) \) and \(\left( k+1, \frac{1}{k+1} \right)\) on it. Note that the area under \(y = \frac{1}{x}\) from \(k\) to \(k+1\) has area given by \( \int_k^{k+1} \frac{1}{x}\,dx \). Note that a rectangle with length \(1\) and breadth \( \frac{1}{k+1} \) can be drawn right below it, so comparing areas we have \( \frac{1}{k+1} < \int_k^{k+1} \frac{1}{x}\,dx\). The other part is done similarly.

And yep. Upon evaluating, \( \int_k^{k+1} \frac{1}{x}\,dx = \ln \left( 1 + \frac1{k}\right)\), so you can rearrange by chopping the three-sided inequality into 2 two-sided inequalities first.
\[ \text{With your third part, note that rewriting the inequality in part ii) we have}\\ \boxed{ \left(\frac{k+1}{k}\right)^k < e < \left(\frac{k+1}{k}\right)^{k+1} } \]
\[ \text{Now observe that by subbing in }k=1,2,3,\dots, n\text{ we obtain}\\ \left(\frac21\right)^1 < e < \left(\frac21\right)^2\\ \left(\frac32\right)^2 < e < \left(\frac32\right)^2\\ \left( \frac43\right)^2 < e < \left(\frac43\right)^3\\ \vdots\\ \left( \frac{n+1}{n}\right)^n < e < \left(\frac{n+1}{n}\right)^{n+1}\]
\[ \text{Upon multiplying all of that out, we obtain}\\ \boxed{\left(\frac21\right)^1 \left(\frac32\right)^2 \left( \frac43\right)^3 \dots \left( \frac{n+1}{n}\right)^n < e^n < \left( \frac21\right)^2\left(\frac32\right)^3\left(\frac43\right)^4\dots\left(\frac{n+1}n\right)^{n+1}}\]
\[ \text{On the LHS, observe that}\\ 2^1\text{ in the top cancels out with }2^2\text{ in the bottom to yield }\frac12.\\ 3^2\text{ in the top cancels out with }3^3\text{ in the bottom to yield }\frac13\\ \text{and this cancellation process continues. Ultimately}\\ \text{we're just left with }\frac{(n+1)^n}{n!}\text{ after doing all the cancellations.} \]
Note that the \( (n+1)^n\) in the top has nothing in the bottom to be cancelled out with.
\[ \text{An essentially similar thing occurs on the RHS and leaves us }\frac{(n+1)^{n+1}}{n!}\\ \text{So the whole thing boils down to}\\ \boxed{\frac{(n+1)^n}{n!} < e^n < \frac{(n+1)^{n+1}}{n!}}.\]
This immediately rearranges to produce the required result.
Title: Re: 4U Maths Question Thread
Post by: Fatemah.S on February 27, 2019, 09:16:14 pm
Hey!
The question I have attached is finding volumes using cylindrical/shell method. I am able to get the diagram and what I think is the right volume integration. However, I am struggling to progress with it.  (6.2 Question 7 Cambridge)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 27, 2019, 09:31:48 pm
Hey!
The question I have attached is finding volumes using cylindrical/shell method. I am able to get the diagram and what I think is the right volume integration. However, I am struggling to progress with it.  (6.2 Question 7 Cambridge)
Your integral should look something like \(V = 2\pi \int_0^2 2x\sqrt{1-(x-1)^2}\,dx\). For this integral, first sub \(u=x-1\). Then, split the integral up, and use the area of a semi-circle and properties of odd functions to help out.
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on March 01, 2019, 05:43:48 pm
Hello,
I am stuck on one integration by parts question in the link below (my working out is below as well). Can anyone please help me with this question? Thanks :)
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on March 01, 2019, 06:17:11 pm


Skipped a few steps, but happy to explain further :) Hope this helps
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on March 01, 2019, 06:31:23 pm


Skipped a few steps, but happy to explain further :) Hope this helps
I don't understand how you got from the second step to the third step.
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on March 01, 2019, 07:14:25 pm
Sorry, my integral was actually wrong! (but the answer is the same :) )

Going back to fix that right now, but basically it involves a bit of side working considering the integral of the function u^2 e^2u. You do integration by parts twice. Have a shot at it yourself while I fix the code!

You should get down to something similar to 1/2 e^2u(u^2-u+1/2) (in factorised form).

EDIT: Original answer has been fixed.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 02, 2019, 10:37:51 am
That one can be done using only integration by parts if we're cautious with it. (But it definitely needs two applications of it.) Note that according to the rule of LIATE, we wish to differentiate the L, i.e. the logarithm.
\begin{align*}\int_1^e x(\ln x)^2\,dx &= \left[\frac{x^2}{2} (\ln x)^2\right]_1^e - \int_1^e \frac{x^2}{2} \left( \frac{2\ln x}{x} \right)\,dx\\ &= \frac{e^2}{2} - \int_1^e x\ln x\,dx\\ &= \frac{e^2}{2} - \left[ \frac{x^2}{2}\ln x\right]_1^e + \int_1^e \left(\frac{x^2}{2}\right)\left(\frac1x\right)\,dx \\ &= \frac{e^2}{2} - \frac{e^2}{2} + \left[ \frac{x^2}{4} \right]_1^e\\ &= \frac14 (e^2-1)\end{align*}
Title: Re: 4U Maths Question Thread
Post by: spnmox on March 04, 2019, 10:09:55 pm
Hi, I'm having trouble with this inequality question:
Prove that, if a,b,c and d are any four positive numbers, then ab+cd <= sqr((a^2+c^2)(b^2+d^2))
 I worked backwards and got that (a^2+c^2)(b^2+d^2) >=4abcd, and I need to prove that (ab+cd)^2<=4abcd. I don't know how to manipulate the LHS to do that, and also is this the right method?

Also, just generally: what's the best way to approach these kinds of inequality questions?
What are some useful inequality identities to know eg. a^2+b^2>=2ab or are you expected to derive them from the beginning?

I really appreciate you answering my questions!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 04, 2019, 10:19:17 pm
Hi, I'm having trouble with this inequality question:
Prove that, if a,b,c and d are any four positive numbers, then ab+cd <= sqr((a^2+c^2)(b^2+d^2))
 I worked backwards and got that (a^2+c^2)(b^2+d^2) >=4abcd, and I need to prove that (ab+cd)^2<=4abcd. I don't know how to manipulate the LHS to do that, and also is this the right method?

Also, just generally: what's the best way to approach these kinds of inequality questions?
What are some useful inequality identities to know eg. a^2+b^2>=2ab or are you expected to derive them from the beginning?

I really appreciate you answering my questions!
When you write out your answer, you must start from the beginning, which is either the basic fact that \( (a-b)^2 \geq 0\), or some equation they've given to you. Of course, if there's a part i) or something, you can assume the result of that for part ii).

Working backwards is good to help determine how you would write it up in a forwards direction. But note that your final solution needs to be written the correct way no matter what.
\[ \text{With your problem, }\textbf{further}\text{ working backwards from}\\ (ab+cd)^2 \leq 4abcd\text{ will have provided the path.} \]
\begin{align*}(ab-cd)^2 &\geq 0\\ a^2b^2 - 2abcd + c^2d^2 &\geq 0\\ a^2b^2 + 2abcd + c^2d^2 &\geq 4abcd\\ (ab+cd)^2 &\geq 4abcd \end{align*}
Alternatively, if you've proven \( (x+y)^2 \geq 4xy\) using \( (x-y)^2 \geq 0\) from an earlier part, you can just sub \(x=ab\) and \(y=cd\) without regurgitating the same proof from scratch.

HOWEVER:
Later, when I worked backwards on what you were originally trying to prove, I obtained this.
\begin{align*}ab+cd&\leq \sqrt{(a^2+b^2)(c^2+d^2)}\\ a^2b^2+2abcd+c^2d^2 &\leq a^2b^2+a^2d^2+c^2b^2+c^2d^2\\ a^2d^2-2abcd+c^2b^2 &\geq 0\\ (ad-bc)^2 &\geq 0 \end{align*}
So to prove that, I would've written that proof out backwards.
\begin{align*}(ad-bc)^2 &\geq 0\\ a^2d^2-2abcd + b^2c^2 &\geq 0\\ a^2b^2+a^2d^2+c^2b^2+c^2d^2 &\geq a^2b^2+2abcd+c^2d^2\\ (a^2+c^2)(b^2+d^2) &\geq (ab+cd)^2\\ \therefore ab+cd &\leq \sqrt{(a^2+c^2)(b^2+d^2)} \end{align*}
\[ \text{taking positive roots, since }ab+cd > 0\text{ from the given information.} \]
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on March 06, 2019, 08:20:31 pm
Hello,
I haven't got a clue how to do this question in the link below. However, I have attached a separate attachment based on my thoughts of doing this question. Can anyone please help me with this question? Thanks :)
Title: Re: 4U Maths Question Thread
Post by: wlam on March 07, 2019, 09:01:57 pm
help thanks
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 07, 2019, 09:10:09 pm
Hello,
I haven't got a clue how to do this question in the link below. However, I have attached a separate attachment based on my thoughts of doing this question. Can anyone please help me with this question? Thanks :)
The question's really ambiguous with it's wording (in particular when it said "ordinate"). Can you please provide the value of the answer so I know what I'm aiming for?
help thanks
Elaborate on what you need help with please?
Title: Re: 4U Maths Question Thread
Post by: wlam on March 07, 2019, 09:20:16 pm
The question's really ambiguous with it's wording (in particular when it said "ordinate"). Can you please provide the value of the answer so I know what I'm aiming for?Elaborate on what you need help with please?

i cant get the answer for part b
Title: Re: 4U Maths Question Thread
Post by: wlam on March 07, 2019, 09:21:41 pm
heres a better pic
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 07, 2019, 10:08:18 pm
heres a better pic
Would this diagram be helpful?
(https://i.imgur.com/wvwYVNy.png)

If you managed to figure out everything on the diagram and it's still not coming out, preferably post up your working. Otherwise give it a try with the diagram drawn. (Also note that the diagram has some right-angles that aren't displayed on it.)
Title: Re: 4U Maths Question Thread
Post by: wlam on March 07, 2019, 11:19:27 pm
thanks! i just misinterpreted the question
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on March 08, 2019, 03:30:09 pm
Hey rui need some of your maths experties plz
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 08, 2019, 06:06:29 pm
Hey rui need some of your maths experties plz
After tracking down the second question I realise that as it was stated, \(x\) and \(y\) were intended to be real. Of course, this wasn't specified in the question, but apparently it was assumed by the author for some unknown reason.

With that added knowledge, the \(x^2+y^2\) bit hints at the modulus of a complex number. Hence, let \(z=x+iy\).
\begin{align*}x + \frac{3x-y}{x^2+y^2}&=3\tag{1}\\ y-\frac{x+3y}{x^2+y^2}&=0\tag{2} \end{align*}
\[ \text{So upon doing }(1)+i\times (2)\text{ we obtain}\\ \begin{align*} x+iy + \frac{3x-y-ix-3iy}{x^2+y^2}&=3\\ x+iy + \frac{3(x-iy)-i(x-iy)}{x^2+y^2}&=3\\ \therefore z+\frac{(3-i)\overline{z}}{z\overline{z}}&=3\\ z^2-3z+(3-i)&=0\\ \therefore x+iy &= \frac{3\pm \sqrt{9-(12-4i)}}{2} \tag{quadratic formula}\\ &= \frac{3\pm \sqrt{-3+4i}}2\\ &= \frac{3\pm \sqrt{1 + 4i - 4}}2 \tag{completing the square}\\ &= \frac{3\pm \sqrt{1+4i+4i^2}}2 \\ &= \frac{3\pm \sqrt{(1+2i)^2}}2\\ &= \frac{3\pm (1+2i)}2\\ &= 2+i, \, 1-i\end{align*} \]
Note that it is a common MX2 formula that \( z\overline{z} = |z|^2\), and it was used here.
\[ \text{Hence equating real/imaginary parts,}\\ (x,y) = (2,1)\text{ or }(x,y) = (1,-1) \]
Regardless though, the first question still requires Taylor series, and hence should be put in the first year uni maths questions thread.
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on March 08, 2019, 08:37:05 pm
The question's really ambiguous with it's wording (in particular when it said "ordinate"). Can you please provide the value of the answer so I know what I'm aiming for?Elaborate on what you need help with please?
Hi Rui, the answer is 7√2➗16 + (3➗16) Ln(1+√2).
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 08, 2019, 09:03:19 pm
Hi Rui, the answer is 7√2➗16 + (3➗16) Ln(1+√2).
The thing that completely horrifies me with this question is how they expect you to know how to sketch \(x^{2/3} = y^{2/3} + 1\), and the awkwardness of the wording. Sketching that curve should, in theory, be a completely separate question on its own, guided using stuff like implicit differentiation and identifying the domain of the curve.

The wording of the question is also ambiguous, as it could represent the region below. They should have specified that they specifically want the region in the first quadrant.

Nevertheless, the question becomes doable with those assumptions put into place (albeit with some awkwardness). The idea is that they want you to find the green area shown below, but with the aid of the blue area. I will start it off.
(https://i.imgur.com/RZVb6De.png)
\[ \text{Note that when }x=2\sqrt{2}, \, y = 1.\\ \text{The combined area of the blue and green region is therefore just }2\sqrt{2}.\]
\[ \text{Note that the equation given rearranges to }x = \left( y^{2/3}+1\right)^{3/2}.\\ \text{The area of the blue region is therefore equal to}\\ \int_0^1 \left( y^{2/3}+1\right)^{3/2}\,dy\]
\[ \text{Let }y = tan^3\theta\implies dy = 3\tan^2\sec^2\theta.\text{ Then}\\ \begin{align*}\int_0^1 \left( y^{2/3} +1\right)^{3/2}\,dy &= \int_0^{\pi/4} \left( \tan^2\theta + 1 \right)^{3/2} (3\tan^3\theta \sec^2\theta)\,d\theta\\ &= 3\int_0^{\pi/4} \sec^2\theta (\sec^2\theta-1)\sec^2\theta\,d\theta\\ &= 3\left( \int_0^{\pi/4} \sec^6\theta\,d\theta + \int_0^{\pi/4} \sec^4\theta\,d\theta \right) \\ &= 3\left( \left[I_6\right]_0^{\pi/4} + \left[I_4\right]_0^{\pi/4} \right)\\ &= 3\left( \frac{\sec^4\frac\pi4 \tan \frac\pi4}{5} - \frac{\sec^4 0 \tan 0}5 - \frac45 \left[I_4\right]_0^{\pi/4} + \left[ I_4 \right]_0^{\frac\pi4}\right)\\ &= 3\left( \frac45 + \frac15 \left[ I_4\right]_0^{\pi/4}\right)\end{align*}\]
The remaining use of the reduction formulae and the final subtraction at the end is left as your exercise.

Subject to minor computational error.
Title: Re: 4U Maths Question Thread
Post by: spnmox on March 08, 2019, 09:40:36 pm
Hi, please could I have some help with proving this absolute value inequality:
l (a+b+c) l = lal + lbl + lcl
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 08, 2019, 09:52:47 pm
Hi, please could I have some help with proving this absolute value inequality:
l (a+b+c) l = lal + lbl + lcl
Were you trying to use \( \leq \) there?

(Note: If so, then this is just the usual triangle inequality. First prove that \( |z_1|+|z_2| \geq |z_1+z_2| \) using the usual geometric approach, and then prove by induction that \( |z_1|+|z_2|+\dots + |z_n| \geq |z_1+z_2+\dots+z_n| \). And then sub \(n=3\), with \(z_1 = a\), \(z_2 = b\), \(z_2 = c\).)
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on March 08, 2019, 10:18:42 pm
The thing that completely horrifies me with this question is how they expect you to know how to sketch \(x^{2/3} = y^{2/3} + 1\), and the awkwardness of the wording. Sketching that curve should, in theory, be a completely separate question on its own, guided using stuff like implicit differentiation and identifying the domain of the curve.

The wording of the question is also ambiguous, as it could represent the region below. They should have specified that they specifically want the region in the first quadrant.

Nevertheless, the question becomes doable with those assumptions put into place (albeit with some awkwardness). The idea is that they want you to find the green area shown below, but with the aid of the blue area. I will start it off.
(https://i.imgur.com/RZVb6De.png)
\[ \text{Note that when }x=2\sqrt{2}, \, y = 1.\\ \text{The combined area of the blue and green region is therefore just }2\sqrt{2}.\]
\[ \text{Note that the equation given rearranges to }x = \left( y^{2/3}+1\right)^{3/2}.\\ \text{The area of the blue region is therefore equal to}\\ \int_0^1 \left( y^{2/3}+1\right)^{3/2}\,dy\]
\[ \text{Let }y = tan^3\theta\implies dy = 3\tan^2\sec^2\theta.\text{ Then}\\ \begin{align*}\int_0^1 \left( y^{2/3} +1\right)^{3/2}\,dy &= \int_0^{\pi/4} \left( \tan^2\theta + 1 \right)^{3/2} (3\tan^3\theta \sec^2\theta)\,d\theta\\ &= 3\int_0^{\pi/4} \sec^2\theta (\sec^2\theta-1)\sec^2\theta\,d\theta\\ &= 3\left( \int_0^{\pi/4} \sec^6\theta\,d\theta + \int_0^{\pi/4} \sec^4\theta\,d\theta \right) \\ &= 3\left( \left[I_6\right]_0^{\pi/4} + \left[I_4\right]_0^{\pi/4} \right)\\ &= 3\left( \frac{\sec^4\frac\pi4 \tan \frac\pi4}{5} - \frac{\sec^4 0 \tan 0}5 - \frac45 \left[I_4\right]_0^{\pi/4} + \left[ I_4 \right]_0^{\frac\pi4}\right)\\ &= 3\left( \frac45 + \frac15 \left[ I_4\right]_0^{\pi/4}\right)\end{align*}\]
The remaining use of the reduction formulae and the final subtraction at the end is left as your exercise.

Subject to minor computational error.
Thank you Rui! But surely this question won't be in an Extension 2 exam?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 08, 2019, 11:26:33 pm
Thank you Rui! But surely this question won't be in an Extension 2 exam?
If it were, it would be chopped up into multiple parts.
Title: Re: 4U Maths Question Thread
Post by: spnmox on March 09, 2019, 10:11:34 am
Show that sinx<x for x>0.

Here's my working so far:
Let f(x)=sinx-x
f'(x)=cosx-1 =0 for stat pts
cosx=1
x=0,360
x>0 so x=360 degrees
f''(x)=-sinx
f''(360)=0

so apparently there's a horizontal point of inflexion at x=360
but don't you have to show that it's the absolute max turning point?
Title: Re: 4U Maths Question Thread
Post by: spnmox on March 09, 2019, 10:30:29 am
Consider the function f(x) =e^x(1-x/10)^10.  Sketch the curve and from your graph, deduce that e^x <= (1-x/10)^-10 for x<10

I'm having trouble finding the asymptotes and sketching the graph. I've already found that turning points (),1) and (10,0), and I think that as x-->infinity, y--> infinity and x--> -infinite y--> -infinity, hopefully that's right?

I'm not sure how to deduce that inequality either.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 09, 2019, 10:30:39 am
Show that sinx<x for x>0.

Here's my working so far:
Let f(x)=sinx-x
f'(x)=cosx-1 =0 for stat pts
cosx=1
x=0,360
x>0 so x=360 degrees
f''(x)=-sinx
f''(360)=0

so apparently there's a horizontal point of inflexion at x=360
but don't you have to show that it's the absolute max turning point?
Firstly, for the sake of calculus, never use degrees.

Secondly, \(f^{\prime\prime}(x) = 0\) doesn't always imply that \(x\) is a point of inflexion. The theorem says that if \(x\) is a point of inflexion, then \(f^{\prime\prime}(x) = 0\). But the converse (i.e. what you get when you flip that statement around) does not hold. An easy example is \( h(x) = x^4\) - we can prove that \( h^{\prime\prime}(0) = 0\), but \(x=0\) is certainly not a point of inflexion.
To test if it is actually a point of inflexion, you need to do test both sides for concavity changes. So since you've computed that \(h^{\prime\prime}(2\pi) = 0\), you could plug \(2\pi - 0.01\) and \(2\pi + 0.01\) into \(f(x)\), to determine if there is a concavity change. You'll see that there is not, so the stationary point you found will not be a horizontal point of inflexion.

Finally, as for the original question itself, hint - you've overcomplicated it. Verifying that \(f^{\prime}(x) \leq 0\) for all real \(x\) is the starting point. Recall that if \(f^\prime(x) \leq 0\) for all real \(x\), i.e. the curve is decreasing (albeit not strictly decreasing), then we have the conclusion
\[ \text{If }x_1 \geq x_2\\ \text{Then }f(x_1) \leq f(x_2) \]
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on March 12, 2019, 08:57:37 am
Hello,
I am stuck on an integration question. The question and my working out are attached below. Can anyone please help me out? Thanks :)
Title: Re: 4U Maths Question Thread
Post by: AlphaZero on March 12, 2019, 09:37:48 am
Hello,
I am stuck on an integration question. The question and my working out are attached below. Can anyone please help me out? Thanks :)
\begin{align*}\int_{-1}^1 \frac{x^2}{e^x+1}dx&=\int_0^1\left(\frac{x^2}{e^x+1}+\frac{x^2}{\frac{1}{e^x}+1}\right)dx\\
&=\int_0^1\left(\frac{x^2}{e^x+1}+\frac{x^2e^x}{1+e^x}\right)dx\\
&=\int_0^1\frac{x^2(e^x+1)}{e^x+1}dx\\
&=\qquad \vdots \end{align*}
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on March 12, 2019, 12:10:26 pm
Ahhh k, got the answer. Thank you!  :D
Title: Re: 4U Maths Question Thread
Post by: wlam on March 13, 2019, 10:00:39 pm
im lost on part b...
thanks :-\
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on March 13, 2019, 10:46:32 pm


From here it should be evident that since (x1,y1) or (x2,y2) lie on the chord, we can just sub them in. x1y1 or x2y2 should cancel out and you should have the given equation.

Hope this helps :)
Title: Re: 4U Maths Question Thread
Post by: spnmox on March 15, 2019, 10:58:49 pm
If f(x) = x/sqr(1+x^2), prove by mathematical induction for n>=2 that f(f(...(f (x))...)) = x/sqr(1+nx^2), if there are n number of letter f's on the LHS.

Please help! I don't understand what f of f(x) means. What exactly am I subbing in when I am trying to prove true for n=2 and assuming true for n=k?

Thank you.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 15, 2019, 11:11:45 pm
If f(x) = x/sqr(1+x^2), prove by mathematical induction for n>=2 that f(f(...(f (x))...)) = x/sqr(1+nx^2), if there are n number of letter f's on the LHS.

Please help! I don't understand what f of f(x) means. What exactly am I subbing in when I am trying to prove true for n=2 and assuming true for n=k?

Thank you.
To help you get started, say you have \(h(x) = x-1\). Then, \(h(h(x)) = h(x-1)\). Of course, clearly this simplifies to \( h(h(x)) = x-2\), but it illustrates the idea - you sub the expression the function is equal to, back into itself.

The notation is standard - it is just composition of functions \(g(f(x))\), except \(g(x)\) and \(f(x)\) are taken to be the same thing here. Note we also learnt to use the chain rule in 2U differentiation for composite functions - this let us establish that \( \frac{d}{dx} g(f(x)) = f^\prime(x) \cdot g^\prime(f(x)) \).
\[ \text{So for your question, just for the sake of starting you off,}\\ \text{If }f(x) = \frac{x}{\sqrt{1+x^2}} \]
\[\text{Then}\\ \begin{align*} f(f(x)) &= f\left( \frac{x}{\sqrt{1+x^2}} \right)\\ &= \frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1 + \left( \frac{x}{\sqrt{1+x^2}}\right)^2}}\end{align*} \]
For the base case, you wish to prove that that expression equals to \(\frac{x}{\sqrt{1+2x^2}} \).
Title: Re: 4U Maths Question Thread
Post by: spnmox on March 15, 2019, 11:36:05 pm
To help you get started, say you have \(h(x) = x-1\). Then, \(h(h(x)) = h(x-1)\). Of course, clearly this simplifies to \( h(h(x)) = x-2\), but it illustrates the idea - you sub the expression the function is equal to, back into itself.

The notation is standard - it is just composition of functions \(g(f(x))\), except \(g(x)\) and \(f(x)\) are taken to be the same thing here. Note we also learnt to use the chain rule in 2U differentiation for composite functions - this let us establish that \( \frac{d}{dx} g(f(x)) = f^\prime(x) \cdot g^\prime(f(x)) \).
\[ \text{So for your question, just for the sake of starting you off,}\\ \text{If }f(x) = \frac{x}{\sqrt{1+x^2}} \]
\[\text{Then}\\ \begin{align*} f(f(x)) &= f\left( \frac{x}{\sqrt{1+x^2}} \right)\\ &= \frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1 + \left( \frac{x}{\sqrt{1+x^2}}\right)^2}}\end{align*} \]
For the base case, you wish to prove that that expression equals to \(\frac{x}{\sqrt{1+2x^2}} \).

Thank you so much you actual genius! I figured it out and I'm so proud.

I also have another question:

If the sums of the first three terms of an AP and a GP are equal and non-zero, the common difference of the AP and the common ratio of the GP are equal and the ratio of the AM to the GM is 1:-2, find the common ratio of the GP. Also, find the relationship between their first terms.

So far I've just defined all the terms and apparently you need to solve them simultaneously but not sure how to do that.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 16, 2019, 12:04:06 am
Thank you so much you actual genius! I figured it out and I'm so proud.

I also have another question:

If the sums of the first three terms of an AP and a GP are equal and non-zero, the common difference of the AP and the common ratio of the GP are equal and the ratio of the AM to the GM is 1:-2, find the common ratio of the GP. Also, find the relationship between their first terms.

So far I've just defined all the terms and apparently you need to solve them simultaneously but not sure how to do that.
I can look at this one tomorrow later today if you want but mind if I grab the final answer off you so I know what to aim for?
Title: Re: 4U Maths Question Thread
Post by: spnmox on March 16, 2019, 11:49:59 am
I can look at this one tomorrow later today if you want but mind if I grab the final answer off you so I know what to aim for?

Sure:
d=-1/2, 4A-G=2
d=-2, A-G=2
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 16, 2019, 01:03:59 pm
Note that the actual words "AM" and "GM" are not a part of the HSC syllabus. Technically they should not be used unless they have been defined. (Note that harder 3U inequalities make students prove these results without actually explaining what they are.)

The question is also slightly ambiguous in that they should mention the AM of the terms in the A.P., and similarly the GM for the terms of the G.P., as opposed to letting the reader assume that for themselves.
\[ \text{To follow the convention of the answers, the common ratio}\\ \text{is also denoted by }d\text{ for this question.} \]
\[ \text{At times like these it helps to be a bit cleverer.}\\ \text{Let the first three terms in the AP be }\boxed{a-d, \, a, \, a+d} \\ \text{and the first three terms in the GP be }\boxed{\frac{b}{d}, \, b, \, bd}.\]
\[ \text{The ratio of the AM and GM here tell us that}\\ \begin{align*} \frac{\frac{(a-d)+a+(a+d)}3}{\sqrt[3]{\left(\frac{b}{d}\right) b (bd)}} &= -\frac12\\ \therefore \frac{a}{b} &= -\frac12\\ b&=-2a \end{align*} \]
\[ \text{The sum of the terms in the AP and GP being the same tell us that} \\ 3a = b \left( \frac{1}{k}+ 1 + k \right).\]
Pretty much everything is set up from here. You'll be able to get the two values of \(d\) that you've stated. Then for, say, the case of \(d=-\frac12\), if you let \(A = a-k\) and \(G = \frac{b}{k}\) you should be able to reach that relationship you have stated, using simultaneous equations.
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on March 19, 2019, 05:26:38 pm
Hello,
I am stuck on an integration question which I have partially done. Can anyone please help me out, especially with the second part? Thanks :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 19, 2019, 06:03:05 pm
Hello,
I am stuck on an integration question which I have partially done. Can anyone please help me out, especially with the second part? Thanks :)

\[ \text{Using the recurrence, we thus have}\\ \begin{align*}\int_0^1 x^4\sqrt{1-x^2}\,dx&= I_4\\ &= \frac{3}{6}I_2\\ &= \frac{3}{6} \, \frac{1}{4} I_0\\ &= \frac18 \int_0^1 \sqrt{1-x^2}\,dx\\ &= \frac18 \times \frac\pi4 \tag{1/4 area of a circle with radius 1}\\ &= \frac\pi{32} \end{align*}\]
Haven't checked your working for the first part but it should be right given that you got the right answer.
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on March 19, 2019, 06:20:36 pm
Thanks for the help! :)
Title: Re: 4U Maths Question Thread
Post by: 006896 on March 19, 2019, 06:40:30 pm
Hi,
Could you help me with a graphing question attached?
I just don't understand why when 1/x is multiplied by f(x)approaching y= infinity, when x approaches infinity on the new graph, the new graph doesn't approach y=0?
Thanks
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 19, 2019, 06:48:46 pm
Hi,
Could you help me with a graphing question attached?
I just don't understand why when 1/x is multiplied by f(x)approaching y= infinity, when x approaches infinity on the new graph, the new graph doesn't approach y=0?
Thanks
It is not true to assume things like \( \infty\times 0 = 0\), or that \( \frac{\infty}{\infty} = 0\).
\[ \text{For this particular graph, it is not too hard to see why.}\\ \text{The reason is because }f(x)\text{ is a cubic.} \]
More specifically, you could guess that \( f(x) = (x+3)(x-1)^2\). But the following explanation shows that for any cubic \(f(x) = ax^3+bx^2+cx+d\), as \(x\to \infty\), \(y\to \infty\) as well.
\[\text{If we take any cubic } f(x) = ax^3+bx^2+cx+d\\ \text{Then }f(x) \times \frac{1}{x} = ax^2+bx+c+\frac{d}{x}. \]
\[ \text{As }x\to \infty,\\ \text{whilst it is certainly true that }\frac{d}{x}\to 0,\\ \text{the terms }ax^2\text{ and }bx\text{ still tend to }\infty\text{ as well.}\]
\[ \text{And the sum of something tending to infinity, plus another tending to infinity,}\\ \text{plus of course the constant and the 0 term}\\ \text{should still also tend to infinity, as required.}\]
Note that this explanation can be generalised for all polynomials with degree at least 2. If you multiplied something linear, i.e. \(f(x)=ax+b\) to \(\frac1x\), then this does not happen. But for any higher powers, this will always be the case.

Note also that if we multiplied \(f(x) = ax^3+bx^2+cx+d\) to \( \frac{1}{x^3}\), then as \(x\to \infty\), \(y \to a\). And if we multiply \(f(x)\) to \( \frac{1}{x^n}\), where \(n > 3\), then \(y\to 0\).

If you are genuinely interested in some intuition behind why
The reason more or less is because \(x^3\) grows at a much more significant rate than \(x\) grows. Therefore \(x^3\) grows much faster than \( \frac{1}{x}\) actually decays.

The growth of \(x^3\) is said to dominate the decay of \( \frac{1}{x} \). Hence as \(x\to \infty\), \(y\to \infty\) as well.
Title: Re: 4U Maths Question Thread
Post by: Aaron Lillis on March 20, 2019, 06:52:34 am
Hey guyssss, im stuck on these integration questions, could someone please help me?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on March 20, 2019, 08:37:23 am
Hey guyssss, im stuck on these integration questions, could someone please help me?

In both cases, you need to factor a 3 from the denominator. For example, for the second, you will obtain \( \int_0^{1/3} \frac{dx}{3\sqrt{\frac49-x^2}}\). Did you try this?
Title: Re: 4U Maths Question Thread
Post by: Aaron Lillis on March 20, 2019, 09:13:40 am
In both cases, you need to factor a 3 from the denominator. For example, for the second, you will obtain \( \int_0^{1/3} \frac{dx}{3\sqrt{\frac49-x^2}}\). Did you try this?

thanks SO MUCHHHHH Rui.
would you be able to show me the same application on this integral? i tried factoring the (5)^1/2 but im just not getting the integral out.
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on March 20, 2019, 04:27:33 pm
Hello,
I have trouble with the question in the link below. Can anyone please help me out? Thanks :)
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on March 20, 2019, 05:29:46 pm


Hope this helps :)
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on March 20, 2019, 05:32:29 pm


Hope this helps :)
Don't worry; already did it with 1 page of working. Thanks for the input though :)
Title: Re: 4U Maths Question Thread
Post by: goodluck on April 08, 2019, 02:50:59 pm
hey how do you do this one:

h/(x+h) < ln(x+h) - ln(x) <h/x . (it's meant to be greater than and equal to btw)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 08, 2019, 03:13:57 pm
hey how do you do this one:

h/(x+h) < ln(x+h) - ln(x) <h/x . (it's meant to be greater than and equal to btw)
Hint: You can use a sketch to verify the following inequality.
\[ h \times \frac{1}{x+h} < \int_x^{x+h} \frac{1}{t}dt < h\times \frac{1}{x} \]
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on April 13, 2019, 10:01:55 pm
Hello,
I am stuck on Example 7 part a) in the link below. I have also attached my working out. Can anyone please help me out? Thanks :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 13, 2019, 10:40:29 pm
Hello,
I am stuck on Example 7 part a) in the link below. I have also attached my working out. Can anyone please help me out? Thanks :)
Two things:
\[ \text{Firstly, you've correctly noted that }A = \frac12 x^2 \tan \alpha.\\ \text{So your volume of a cross section should just be }\boxed{\delta V = \frac12 x^2 \tan \alpha\, \delta y}\\ \text{i.e. }\boxed{\delta V = \frac12 (r^2-y^2)\tan \alpha\, \delta y} \]
Not sure why you multiplied \(x^2\) to another \(r^2-y^2\) term in your expression. By doing so, because of the fact that \(x^2+y^2=r^2\), you're basically trying to deal with \(x^4\), or equivalently \((r^2-y^2)^2\)
\[ \text{Secondly, what is }a?\\ \text{The boundaries of integration (and also the sum) are just }-r\text{ and }r \]
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on April 15, 2019, 03:25:52 pm
Two things:
\[ \text{Firstly, you've correctly noted that }A = \frac12 x^2 \tan \alpha.\\ \text{So your volume of a cross section should just be }\boxed{\delta V = \frac12 x^2 \tan \alpha\, \delta y}\\ \text{i.e. }\boxed{\delta V = \frac12 (r^2-y^2)\tan \alpha\, \delta y} \]
Not sure why you multiplied \(x^2\) to another \(r^2-y^2\) term in your expression. By doing so, because of the fact that \(x^2+y^2=r^2\), you're basically trying to deal with \(x^4\), or equivalently \((r^2-y^2)^2\)
\[ \text{Secondly, what is }a?\\ \text{The boundaries of integration (and also the sum) are just }-r\text{ and }r \]
I realised where I went wrong; thanks :)
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on April 24, 2019, 09:49:37 pm
Hello,
I am stuck on a volumes question in the attachment below (especially the proving part). Can anyone please help me out? Thanks :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 25, 2019, 06:17:54 pm
Hello,
I am stuck on a volumes question in the attachment below (especially the proving part). Can anyone please help me out? Thanks :)
Sorry about the delay with this one. Been ages since I used the linear equation method for the volume of a frustrum so I wanted to be more alert when revising it. I just attached the full solution here.

It's essentially based off the exact same method you used. Just with some arbitrary variables, instead of given constants.

(https://i.imgur.com/NPmtuYtl.png)

(https://i.imgur.com/WU2u3B5l.png)
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on April 25, 2019, 09:08:04 pm
Sorry about the delay with this one. Been ages since I used the linear equation method for the volume of a frustrum so I wanted to be more alert when revising it. I just attached the full solution here.

It's essentially based off the exact same method you used. Just with some arbitrary variables, instead of given constants.

(https://i.imgur.com/NPmtuYtl.png)

(https://i.imgur.com/WU2u3B5l.png)
Wow, this seems top-notch. Thank you! :)
Title: Re: 4U Maths Question Thread
Post by: diggity on April 26, 2019, 10:27:38 pm
Hey Rui! Following the MX2 lecture today at UTS (was awesome btw!), I've decided to try this thread for help. I think my worst area is Applications of Polynomials (locuses and all that). Here's my question from that topic (from textbook):

a) Show that the equation of the normal to the parabola y = x^2 at the point P(t, t^2) may be written as


b) Hence, if there exist three normals passing through P(x_0, y_0) then using the result of part (b) in question 7 above show that .

Question a) is easy, just added it for context. The real head-scratcher is part b), and I don't even know where to start. (side note; is there any way to do LaTeX inline?)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 26, 2019, 10:56:14 pm
Hey Rui! Following the MX2 lecture today at UTS (was awesome btw!), I've decided to try this thread for help. I think my worst area is Applications of Polynomials (locuses and all that). Here's my question from that topic (from textbook):

a) Show that the equation of the normal to the parabola y = x^2 at the point P(t, t^2) may be written as


b) Hence, if there exist three normals passing through P(x_0, y_0) then using the result of part (b) in question 7 above show that .

Question a) is easy, just added it for context. The real head-scratcher is part b), and I don't even know where to start. (side note; is there any way to do LaTeX inline?)
Glad you enjoyed it! :)

I just had a look at this question. I'm hesitant to comment right now though because I actually got it to work easily with the cubic discriminant formula, Will probably spend more time on it later, but for my reference which textbook is this from?
Title: Re: 4U Maths Question Thread
Post by: diggity on April 26, 2019, 11:05:05 pm
Terry Lee's "Advanced mathematics; A complete HSC MX2 Course" (exercise 3.4 q8). I've never heard of the cubic discriminant formula before? Edit: and that's okay! take all the time you need
Title: Re: 4U Maths Question Thread
Post by: diggity on April 26, 2019, 11:09:17 pm
I just noticed that I did not include the result mentioned in the question! It's; if f(x) is in the form:


then,


(I'm not actually sure how much information from the previous question is needed, but this is the result obtained)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 26, 2019, 11:38:44 pm
I just noticed that I did not include the result mentioned in the question! It's; if f(x) is in the form:


then,


(I'm not actually sure how much information from the previous question is needed, but this is the result obtained)
Ah cool, the question is definitely doable now aha! That above result is quite important because it holds only when \(f(x)\) has 3 distinct \(x\)-intercepts.
\[ \text{The idea is the following.}\\ \text{Firstly, suppose that the normal at }P\text{ passes through }(x_0,y_0).\\ \text{Then, from subbing in we have}\\ \boxed{t^3 + \left( \frac{1-2y_0}{2}\right)t + \frac{-x_0}{2} =0}. \]
\[ \text{But then if }\textbf{three}\text{ normals pass through }(x_0,y_0)\\ \text{that means we have }\textbf{three distinct choices}\text{ for the point }P\\ \text{that will make the boxed equation hold true.}\]
\[ \text{Which is then equivalent to saying that}\\ \text{there are three distinct solutions for }t\\ \text{that make the equation true.}\]
It is because of that why we're able to use the result of the previous question. (The tough part of this question was translating what they tell you from words into mathematics.)
\[ \text{So using the previous question's result, here we have }\\\boxed{27\left(- \frac{x_0}{2} \right)^2 + 4\left( \frac{1-2y_0}{2} \right) ^3< 0} \]
\[ \text{Now we just do battle with algebra.}\\ \begin{align*} 4\left( \frac{2y_0-1}{2} \right)^3 &> 27\left( \frac{x_0}{2} \right)^2\\ \left( y_0-\frac12 \right)^3 &> 27 \left( \frac{x_0}{4} \right)^2\\ y_0 - \frac12 &> 3\left( \frac{x_0}{4} \right)^{2/3}\\ y_0 &= 3\left( \frac{x_0}{4} \right)^{2/3}+\frac12\end{align*} \]
Code: (Also, in regards to inline LaTeX, it works like this) [Select]
\( stuff goes here \)
Title: Re: 4U Maths Question Thread
Post by: diggity on April 27, 2019, 12:15:17 am
OH! That makes so much sense! I didn't recognise the equation of the normal was in that form, and then being able to use that inequality to achieve the answer. You're right, a huge part of this topic is the translation from words to mathematics. I see now that this question is quite simple when boiled down to its fundamental components. Major props!
Title: Re: 4U Maths Question Thread
Post by: diggity on April 27, 2019, 02:28:39 pm
Okay, I have another question (from the same exercise):

6 Assume \(cos3\theta = 4cos^3\theta - 3cos\theta \), find in terms of \(\pi\) the exact values of the solutions of
a) \(x^3-3x+1=0\)

The answer in the book begins quite simply, with \(x=kcos\theta\), \(k^3cos^3\theta - 3kcos\theta + 1 = 0\). They then mention that:
"We want \(\frac{k^3}{3k} = \frac{4}{3}\), ∴ \(k^2=4\), ∴ \(k=2\)."
From this, they attain the equation:

... and being able to solve for theta. What I'm confused at is why they wanted that original equation that gave them k. What I originally tried to do was something like:

Why did they divide the two equations? I see that there is no solution for my two equations, which is where I got stuck. Help!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 27, 2019, 03:00:30 pm
Okay, I have another question (from the same exercise):

6 Assume \(cos3\theta = 4cos^3\theta - 3cos\theta \), find in terms of \(\pi\) the exact values of the solutions of
a) \(x^3-3x+1=0\)

The answer in the book begins quite simply, with \(x=kcos\theta\), \(k^3cos^3\theta - 3kcos\theta + 1 = 0\). They then mention that:
"We want \(\frac{k^3}{3k} = \frac{4}{3}\), ∴ \(k^2=4\), ∴ \(k=2\)."
From this, they attain the equation:

... and being able to solve for theta. What I'm confused at is why they wanted that original equation that gave them k. What I originally tried to do was something like:

Why did they divide the two equations? I see that there is no solution for my two equations, which is where I got stuck. Help!
Your equations are essentially a bit too restrictive.

Your claim is that the coefficient of \(\cos^3\theta\) must be \(4\), and the coefficient of \(\cos\theta\) must be \(3\). But we do not need to go that far.

For example, the coefficients could've been, say, \(16\) and \(12\). Then we could've still use the formula, except we would've just had \(16\cos^3\theta - 12\cos\theta+1 = 0\), which can still be simplified into \( 4\cos3\theta+1=0\) anyway!
\[ \text{The idea is, only the }\textbf{ratio}\text{ of the coefficients need}\\ \text{to be }4:3.\\ \text{The actual coefficients themselves don't have to follow that rule}\\ \text{because we don't }\textbf{need}\text{ to aim for }\cos3\theta\text{ by itself.}\\ \text{Rather, we can aim for a constant multiplied to }\cos3\theta\text{ instead.} \]
\[ \text{This is where they get }\frac{k^3}{3k} = \frac{4}{3}\text{ from.} \]
It just so turns out that \(16\) and \(12\) weren't correct, but rather \(8\) and \(6\) were. Which of course, are still in the ratio 4:3, but don't actually equal 4 and 3 respectively!
Title: Re: 4U Maths Question Thread
Post by: pmcnair on April 29, 2019, 05:11:22 pm
Hello everyone,
Just stuck on this question, which has a weird sequence I can't really figure out. Any help with part (iii) would be very much appreciated.
It's from the 2003 HSC, question 6
Title: Re: 4U Maths Question Thread
Post by: RuiAce on April 29, 2019, 08:09:02 pm
Hello everyone,
Just stuck on this question, which has a weird sequence I can't really figure out. Any help with part (iii) would be very much appreciated.
It's from the 2003 HSC, question 6
\[ \text{Since this is a second order recurrence relation}\\ \text{we require a stronger version of induction that requires two base cases.} \]
\[ \text{When }n=1, \, s_1 =1 \geq \sqrt{1!}\text{ as required.}\\ \text{When }n=2, \, s_2 = 2 \geq \sqrt{2!}\text{ as required.} \]
\[ \text{With two base cases, now assume that the statement holds}\\ \text{when }n=k\text{ and when }n=k-1.\\ \text{i.e. }\boxed{s_k \geq \sqrt{k!}}\text{ and }\boxed{s_{k-1}\geq \sqrt{(k-1)!}} \]
\[\text{Then when }n=k+1,\\ \begin{align*} s_{k+1} &= s_k + k s_{k-1} \tag{definition}\\ &\geq \sqrt{k!} + k \sqrt{(k-1)!} \tag{assumption} \\ &= \sqrt{k (k-1)!} + k \sqrt{(k-1)!} \tag{classic factorial trick}\\ &= \sqrt{k}\sqrt{(k-1)!} + k \sqrt{(k-1)!} \tag{split product inside sqrt}\\ &= \left( \sqrt{k} + k\right) \sqrt{(k-1)!}\tag{factorise}\\ &\geq \sqrt{k(k+1)}\sqrt{(k-1)!}\tag{part ii)}\\ &= \sqrt{(k+1)k(k-1)!}\tag{recombine product into sqrt}\\ &= \sqrt{(k+1)!}\tag{classic factorial trick} \end{align*}\\ \text{as required.} \]
Note: The factorial trick used was that \(n! = n(n-1)!\).
Title: Re: 4U Maths Question Thread
Post by: pmcnair on April 29, 2019, 08:43:20 pm
Thanks very much Rui! I haven't seen a second order recurrence relation before but it does make sense why you need it.
Also gotta love the classic factorials
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on May 07, 2019, 05:57:57 pm
Hello,
I am stuck on this mechanics question in the link below. Can anyone please help me out? Thanks :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 07, 2019, 08:38:38 pm
Hello,
I am stuck on this mechanics question in the link below. Can anyone please help me out? Thanks :)
\[ \text{Your displacement bit looks fudgy.}\\ \text{I'm with you up to and including this step:}\\ \frac{1}{2k} \left[ \ln (c^2+v^2)\right]_c^0 = -x .\]
I can't see why \(c^2\) should equal 1 here - not sure where your explanation comes from. The next step here is to do what you would normally do in 2U, i.e. sub in the boundaries.
\begin{align*} \frac1{2k} \left[ \ln (c^2) - \ln (c^2+c^2) \right] &= -x\\ \frac{1}{2k} \ln \left(\frac{c^2}{c^2+c^2}\right)&=-x \tag{log laws}\\ \frac{1}{2k}\ln \frac12 &=-x\\ \therefore x &= -\frac1{2k}\ln \frac12\\ &= \frac1{2k}\ln 2. \end{align*}
\[ \text{Also I think there's a mistake with the velocity.}\\ \text{From }\int -dt = \frac1k\int \frac{dv}{c^2+v^2}\text{ you should've obtained }\boxed{-t = \frac{1}{k} \left( \frac{1}{c}\tan^{-1} \frac{v}{c}\right)+C} .\]
Note that whilst it is true that \( \int \frac{x}{a^2+x^2}dx = \frac12 \ln (a^2+x^2)+C \), when the numerator is instead a constant you have \( \int \frac{1}{a^2+x^2}\,dx = \frac1a \tan^{-1} \frac{x}a \). The first is a \( \int \frac{f^\prime(x)}{f(x)}\,dx \) pattern, but the second is a 3U standard integral from the inverse trig topic.
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on May 07, 2019, 08:50:10 pm
\[ \text{Your displacement bit looks fudgy.}\\ \text{I'm with you up to and including this step:}\\ \frac{1}{2k} \left[ \ln (c^2+v^2)\right]_c^0 = -x .\]
I can't see why \(c^2\) should equal 1 here - not sure where your explanation comes from. The next step here is to do what you would normally do in 2U, i.e. sub in the boundaries.
\begin{align*} \frac1{2k} \left[ \ln (c^2) - \ln (c^2+c^2) \right] &= -x\\ \frac{1}{2k} \ln \left(\frac{c^2}{c^2+c^2}\right)&=-x \tag{log laws}\\ \frac{1}{2k}\ln \frac12 &=-x\\ \therefore x &= -\frac1{2k}\ln \frac12\\ &= \frac1{2k}\ln 2. \end{align*}
\[ \text{Also I think there's a mistake with the velocity.}\\ \text{From }\int -dt = \frac1k\int \frac{dv}{c^2+v^2}\text{ you should've obtained }\boxed{-x = \frac{1}{k} \left( \frac{1}{c}\tan^{-1} \frac{v}{c}\right)+C} .\]
Note that whilst it is true that \( \int \frac{x}{a^2+x^2}dx = \frac12 \ln (a^2+x^2)+C \), when the numerator is instead a constant you have \( \int \frac{1}{a^2+x^2}\,dx = \frac1a \tan^{-1} \frac{x}a \). The first is a \( \int \frac{f^\prime(x)}{f(x)}\,dx \) pattern, but the second is a 3U standard integral from the inverse trig topic.
After a moment of refining my working out, I got the answer for the first part. I have trouble understanding and doing the second part, can you help me out please? Thanks :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 07, 2019, 09:39:40 pm
After a moment of refining my working out, I got the answer for the first part. I have trouble understanding and doing the second part, can you help me out please? Thanks :)
(There's a slight typo in that I wrote \(x\) instead of \(t\) for that last boxed expression. Fixed that now.)

When \(t=0\), \(v=c\), so \( C = -\frac{1}{ck}\tan^{-1}\frac{c}{c} = -\frac{1}{ck}\times\frac\pi4\). Then just sub \(v=0\). I'm not sure what else you want elaboration with?
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on May 07, 2019, 11:18:25 pm
(There's a slight typo in that I wrote \(x\) instead of \(t\) for that last boxed expression. Fixed that now.)

When \(t=0\), \(v=c\), so \( C = -\frac{1}{ck}\tan^{-1}\frac{c}{c} = -\frac{1}{ck}\times\frac\pi4\). Then just sub \(v=0\). I'm not sure what else you want elaboration with?
The part which says 'Find its velocity when it reaches a point C, having travelled for a time pi/12ck from A. Find the distance AC'. The very bottom part.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 08, 2019, 08:22:30 am
The part which says 'Find its velocity when it reaches a point C, having travelled for a time pi/12ck from A. Find the distance AC'. The very bottom part.
Completely missed that bit oops.
\[ \text{After showing that}\\ \boxed{t = \frac1{ck} \left( \frac\pi4 - \tan^{-1} \frac{v}{c} \right)}\\ \text{if we sub }t = \frac\pi{12ck}\text{ we have:} \]
\begin{align*}\frac\pi{12ck} &= \frac1{ck} \left( \frac\pi4 - \tan^{-1} \frac{v}{c} \right)\\ \frac\pi{12} &= \frac\pi4 - \tan^{-1} \frac{v}{c}\\ \tan^{-1} \frac{v}{c} &= \frac\pi4 - \frac\pi{12}\\ &= \frac\pi6\\  \therefore \frac{v}{c} &= \frac1{\sqrt3}\\ v &= \frac{c}{\sqrt3} \end{align*}
From here, we should be able to just sub straight into \( x= -\frac1{2k} \ln (c^2+v^2) + \frac1{2k} \ln (2c^2) \). The computations look a bit nasty but I don't see any curveball in this part either.
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on May 08, 2019, 05:57:04 pm
Completely missed that bit oops.
\[ \text{After showing that}\\ \boxed{t = \frac1{ck} \left( \frac\pi4 - \tan^{-1} \frac{v}{c} \right)}\\ \text{if we sub }t = \frac\pi{12ck}\text{ we have:} \]
\begin{align*}\frac\pi{12ck} &= \frac1{ck} \left( \frac\pi4 - \tan^{-1} \frac{v}{c} \right)\\ \frac\pi{12} &= \frac\pi4 - \tan^{-1} \frac{v}{c}\\ \tan^{-1} \frac{v}{c} &= \frac\pi4 - \frac\pi{12}\\ &= \frac\pi6\\  \therefore \frac{v}{c} &= \frac1{\sqrt3}\\ v &= \frac{c}{\sqrt3} \end{align*}
From here, we should be able to just sub straight into \( x= -\frac1{2k} \ln (c^2+v^2) + \frac1{2k} \ln (2c^2) \). The computations look a bit nasty but I don't see any curveball in this part either.
Ok, thanks for the help :)
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on May 11, 2019, 10:38:37 am
Hello,
I am stuck on both parts of this question in the attachment below. I have done part of the question but I can't seem to integrate the function. Can anyone please help me out? Thanks :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 12, 2019, 10:01:20 am
Hello,
I am stuck on both parts of this question in the attachment below. I have done part of the question but I can't seem to integrate the function. Can anyone please help me out? Thanks :)
Sorry must've missed this. Here's the integration bit done for you:
\begin{align*} \int_U^0 \frac{v}{V+v}\,dv &= \int_U^0 \frac{V+v}{V+v} - \frac{V}{V+v}\,dv\\ &= \int_U^0 1 - \frac{V}{V+v}\,dv\\ &= \left[v - V \ln(V+v) \right]_U^0 \end{align*}
(Basically the classic 'add something and subtract the same thing' trick)
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on May 12, 2019, 09:53:37 pm
Sorry must've missed this. Here's the integration bit done for you:
\begin{align*} \int_U^0 \frac{v}{V+v}\,dv &= \int_U^0 \frac{V+v}{V+v} - \frac{V}{V+v}\,dv\\ &= \int_U^0 1 - \frac{V}{V+v}\,dv\\ &= \left[v - V \ln(V+v) \right]_U^0 \end{align*}
(Basically the classic 'add something and subtract the same thing' trick)
Ok, thanks :)
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on May 14, 2019, 10:51:35 am
Hello,
I am confused with this question in the attachment below. I know how to find the area of the triangle (stated in my working out), but I don't know why Area (OLM) is independent of the point P. Can anyone please explain this thoroughly? Thanks :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 14, 2019, 11:04:31 am
Hello,
I am confused with this question in the attachment below. I know how to find the area of the triangle (stated in my working out), but I don't know why Area (OLM) is independent of the point P. Can anyone please explain this thoroughly? Thanks :)
The point \(P\) depends entirely on its parameter \(t\). Since your expression for the area does not have this parameter appearing, i.e. it is independent of whatever \(t\) is, equivalently speaking it is independent of the point \(P\).

(Note that \(c\) is assumed to be a constant, not some parameter that is allowed to vary.)
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on May 14, 2019, 11:10:28 am
The point \(P\) depends entirely on its parameter \(t\). Since your expression for the area does not have this parameter appearing, i.e. it is independent of whatever \(t\) is, equivalently speaking it is independent of the point \(P\).

(Note that \(c\) is assumed to be a constant, not some parameter that is allowed to vary.)
Ahhh, I get it now. I didn't quite get this concept since the book does not clearly mention the parameter stuff with that question. Thanks Rui!
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on May 21, 2019, 05:28:32 pm
Hello,
I am stuck on Q22 in the link below. I have done some of the question, but I don't get why PRQS is a rhombus unless t^2 = 1 (see attachment). Can anyone please help me out with the remaining part of the question? Thanks :)
Title: Re: 4U Maths Question Thread
Post by: esteban on May 22, 2019, 06:53:48 am
The midpoint of RS and  the midpoint of PQ are both (ct,c/t). If the diagonals of a quadrilateral are perpendicular bisectors of each other, then that quadrilateral is a rhombus.
Title: Re: 4U Maths Question Thread
Post by: not a mystery mark on May 22, 2019, 06:47:59 pm
Currently stuck on these recurrence relation questions.
Any ideas?

Cheers! :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 22, 2019, 07:00:11 pm
Currently stuck on these recurrence relation questions.
Any ideas?

Cheers! :)
Remember to also use the add/subtract or multiply/divide same thing trick for recurrence relations, not just integration by parts. The first question is relatively similar to what I did in the April lecture.
\begin{align*}I_n &= \int_0^1 (1-x^r)^n \,dx\\ &= \left[x (1-x^r)^n \right]_0^1 - \int_0^1 x\cdot -rx^{r-1} n (1-x^r)^{n-1}\,dx\\ &= - nr \int_0^1 -x^r(1-x^r)^{n-1} \\ &= -nr \int_0^1 \left[ (1-x^r) - 1 \right] (1-x^r)^{n-1}\\ &= -nr \int_0^1 (1-x^r)^n - (1-x^r)^{n-1}\,dx\\ &= -nr \left( I_n - I_{n-1} \right)\\ \therefore (nr+1) I_n &= nr I_{n-1}\end{align*}
Title: Re: 4U Maths Question Thread
Post by: not a mystery mark on May 22, 2019, 07:41:03 pm
Remember to also use the add/subtract or multiply/divide same thing trick for recurrence relations, not just integration by parts. The first question is relatively similar to what I did in the April lecture.
\begin{align*}\therefore (nr+1) I_n &= nr I_{n-1}\end{align*}

Oh my god, you're an absolute weapon. This logic makes sense. I knew I shouldn't have missed the April lectures.

Just a few questions, where did the [x(1-x^r)^n] bit go? and, How do you know when to use the add/subtract trick?

Thank you so much!!
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on May 22, 2019, 07:52:45 pm
The [x(1-x^r)^n] bit equates to zero when subbing in the boundaries, so it basically just disappears :)

Add/subtract trick comes down mostly to intuition. In this case, seeing that you have an nr out the front of the integral and xr just hanging around, with part of In-1 should tell you that you should manipulate this in some shape or form to the original, especially since you have an nr+1 coefficient for In. Basically, you're looking for an nr x (In-1 - In) to manipulate the integral to find the result. A good way of thinking about it is that if you have a result, think about what you're working towards and think about how you might get to that result. It really just comes down to practice and intuition :)
Title: Re: 4U Maths Question Thread
Post by: not a mystery mark on May 22, 2019, 07:57:12 pm
The [x(1-x^r)^n] bit equates to zero when subbing in the boundaries, so it basically just disappears :)

Ahh cheers, man! Can't wait for you to state rank this course haha. Also an absolute weapon.
Title: Re: 4U Maths Question Thread
Post by: louisaaa01 on May 24, 2019, 06:40:36 pm
Can someone please help with this question? Thanks!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 24, 2019, 07:02:19 pm
Can someone please help with this question? Thanks!
What attempts have you made so far at this problem? And what is the answer we are aiming for?
Title: Re: 4U Maths Question Thread
Post by: louisaaa01 on May 24, 2019, 07:24:19 pm
The answer aims for: α = tan-12

So far, I've considered similar triangles in Solid A, discovering that the area of the hollowed square = a2-x2

However, I'm slightly confused as to how to find the area of the slice in Solid B- if you could please help with this, that would be much appreciated!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 24, 2019, 07:31:26 pm
(https://i.imgur.com/sM4iQSj.png)

I considered the side view briefly but realised it was getting me nowhere. The top view became the nicer one apparently, because you can draw the radii of the circle wherever you feel is convenient.
Title: Re: 4U Maths Question Thread
Post by: not a mystery mark on May 29, 2019, 11:21:13 am
Heyy, back with a volume question. Still trying to get my head around cylindrical shell since we haven't done it in class yet.

The area bounded by the curve y = e^(x^(2)), the lines x=1 and y=1 is rotated about the y-axis. Use the method of cylindrical shells to calculate the volume of the solid of revolution formed.

I got the answer pi(0.5e^2 -e-1) but I am pretty sure that isn't right since it returns a negative value...
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 29, 2019, 12:02:16 pm
Heyy, back with a volume question. Still trying to get my head around cylindrical shell since we haven't done it in class yet.

The area bounded by the curve y = e^(x^(2)), the lines x=1 and y=1 is rotated about the y-axis. Use the method of cylindrical shells to calculate the volume of the solid of revolution formed.

I got the answer pi(0.5e^2 -e-1) but I am pretty sure that isn't right since it returns a negative value...
You should've arrived at \(\delta V =2\pi x\left(e^{x^2} - 1 \right)\delta x\) and \(V = 2\pi \int_0^1 xe^{x^2} - x\,dx\). (If you got that then you probably made an accident with the actual integration)
Title: Re: 4U Maths Question Thread
Post by: goodluck on June 02, 2019, 10:43:08 am
Hey, I'm a bit stuck over this integration question.

Prove the integral of 0 to 1 (x^m)(1-x)^n dx = (m!n!)/(m+n+1)! Hint: think of binomial theorem.

I tried using integration by parts but got stuck.  I'm not sure how to use binomial theorem since we haven't really covered that in 3U yet, even though it looks like the binomial theorem formula from my textbook.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 02, 2019, 03:54:08 pm
Hey, I'm a bit stuck over this integration question.

Prove the integral of 0 to 1 (x^m)(1-x)^n dx = (m!n!)/(m+n+1)! Hint: think of binomial theorem.

I tried using integration by parts but got stuck.  I'm not sure how to use binomial theorem since we haven't really covered that in 3U yet, even though it looks like the binomial theorem formula from my textbook.
What textbook is this? This is usually proven via a double-recurrence formula which is outside of the scope of the MX2 course.
Title: Re: 4U Maths Question Thread
Post by: martinstran on June 03, 2019, 08:04:56 pm
Hey everyone, I'm quite stuck on this Harder 3U inequalities question:
a²+b²≥ab+a+b-1
I know that a²+b²≥2ab from the AM≥GM inequality, so I've attempted to prove that  2ab≥ab+a+b-1 hence, ab-a-b+1≥0, but I can't prove that either.
There are no restrictions on the values of a and b. Any help is much appreciated :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 03, 2019, 08:53:58 pm
Hey everyone, I'm quite stuck on this Harder 3U inequalities question:
a²+b²≥ab+a+b-1
I know that a²+b²≥2ab from the AM≥GM inequality, so I've attempted to prove that  2ab≥ab+a+b-1 hence, ab-a-b+1≥0, but I can't prove that either.
There are no restrictions on the values of a and b. Any help is much appreciated :)
For this question I took on a bit of an unorthodox method. I'm not too sure at this stage how I'd approach it in a sense similar to AM-GM. For now I'll post what I've seen, but it may not be the intended approach.
\[ \text{In my approach, I first note that the statement is equivalent to}\\ a^2+b^2-ab-a-b+1 \geq 0.\\ \text{Which is equivalent to}\\ a^2-a(b+1) + (b^2-b+1) \geq 0. \]
\[ \text{Therefore I consider the quadratic expression}\\ a^2-(b+1)a+(b^2-b+1)\\ \text{as a quadratic in terms of }a. \]
Note that the quadratic could've been re-expressed in terms of \(b\), and the following proof should still work upon adapting.
\[ \text{The leading coefficient is 1, which is positive.}\\ \text{The discriminant of this quadratic is}\\ \begin{align*}\Delta &= [-(b+1)]^2 - 4(b^2-b+1)\\ &= b^2+2b+1 - 4b^2+4b-4\\ &= -3b^2+6b-3\\ &= -3(b-1)^2\\ &\leq 0\end{align*} \\ \text{for }\textbf{all}\text{ real }b.\]
\[ \text{Therefore the expression will never have two distinct real roots}\\ \text{for }a,\text{ in terms of }b.\\ \text{So since the leading coefficient is positive,}\\ \text{it follows that }a^2-(b+1)a+(b^2-b+1)\geq 0\text{ for all }a,b. \]
Which rearranges to produce the desired result.
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on June 11, 2019, 11:59:55 am
Hello,
I am stuck on Q 23 in the link below. I got the first part and attempted the second part, but I don't know where to go. Can anyone please help me out? Thanks :)
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on June 11, 2019, 04:52:52 pm
Hello,
I have another question on 4u, this time on integration. The question is evaluate the integral ∫(cos6x*cos2x)dx. My working out is shown, but I can't seem to get the answer using substitution. Can anyone please help me out? Thanks :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 11, 2019, 05:23:54 pm
Hello,
I have another question on 4u, this time on integration. The question is evaluate the integral ∫(cos6x*cos2x)dx. My working out is shown, but I can't seem to get the answer using substitution. Can anyone please help me out? Thanks :)
It's way easier to use the product-to-sum identities (which should be derived) for integrals of that form. Here, we derive:
\begin{align*}
\frac12 [\cos (A-B) + \cos(A+B)] &= \frac12[(\cos A \cos B + \sin A \sin B)-(\cos A \cos B -  \sin A \sin B)]\\
&= \cos A \cos B
\end{align*}
\begin{align*}
\therefore \int \cos 6x \cos 2x\,dx &= \frac12 \int \cos 4x + \cos 8x\,dx\\
&= \frac{\sin 4x}{8} + \frac{\sin 8x}{16}+C
\end{align*}
The method is outlined in my April lecture slides.
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on June 11, 2019, 05:25:10 pm
There's also a way without sums-to-products, albeit longer:



Hope this helps :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 11, 2019, 10:40:01 pm
Hello,
I am stuck on Q 23 in the link below. I got the first part and attempted the second part, but I don't know where to go. Can anyone please help me out? Thanks :)
Just taking a look at this now:

With b) (i), it's basically relating back to where banked tracks originate from. The idea is that the velocity you're travelling at is faster than the velocity that gives you the optimal banking angle. The train therefore has a tendency to skid up the track.

To ensure this does not happen, the lateral force must bring the train somehow downwards. The lateral force must therefore be directed inwards to the centre of the circle of motion.

This can only be done by the rail that is further outwards. The outer rail effectively 'pushes' the train back inwards.

With b) (iii), if you draw your force diagram correctly you should now arrive at \(N\cos \theta -F\sin \theta = mg\) and \(N\sin\theta + F\cos\theta = \frac{mv^2}{500} \). Hint: Multiply both sides of the first equation by \(\sin\theta\), and both sides of the second by \(\cos\theta\)...
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on June 11, 2019, 10:45:05 pm
Just taking a look at this now:

With b) (i), it's basically relating back to where banked tracks originate from. The idea is that the velocity you're travelling at is faster than the velocity that gives you the optimal banking angle. The train therefore has a tendency to skid up the track.

To ensure this does not happen, the lateral force must bring the train somehow downwards. The lateral force must therefore be directed inwards to the centre of the circle of motion.

This can only be done by the rail that is further outwards. The outer rail effectively 'pushes' the train back inwards.

With b) (iii), if you draw your force diagram correctly you should now arrive at \(N\cos \theta -F\sin \theta = mg\) and \(N\sin\theta + F\cos\theta = \frac{mv^2}{500} \). Hint: Multiply both sides of the first equation by \(\sin\theta\), and both sides of the second by \(\cos\theta\)...
Thank you Rui! :)
Title: Re: 4U Maths Question Thread
Post by: Shaynell01 on June 12, 2019, 03:25:18 pm
Can someone please help me rearrange to form an integratable expression.
cosinverse(pi*x) * sin(pi*x)
Title: Re: 4U Maths Question Thread
Post by: goodluck on June 14, 2019, 03:16:32 pm
What textbook is this? This is usually proven via a double-recurrence formula which is outside of the scope of the MX2 course.

This was from tutoring but I asked both my tutor and my class teacher and they both say it's outside the scope, so I shouldn't worry about it!
I have some other questions though (from past assessments from my school).

1.  Explain why multiplying a complex number z by cistheta rotates z anticlockwise about the origin, through an angle of theta.
    I get the theory behind this, but I'm confused on how to set this out. Would it be best to let z= rcis(delta) and then use the multiplication of polar numbers:
    so you get that: z*cistheta= rcis(delta+theta)? But I'm still a bit confused on how to prove it shifts anticlockwise

2. Assume w is a nth root of infinity. Using geometric sums, prove that w+w^2+w^3+...w^n=0
    When I tried doing this, I got this: w(1-w^n)/(1-w)= w(1-w^n)/w(1/w-1)                           
                                                                   = (1-w^n)/(1/w-1)
             was I meant to use smth to do with the property 1+w+w^2=0??

3. If the mod of z=1, prove |a+bz|= |az+b|. You may assume a and b are elements of the Real numbers,.
What I did was:
     square both sides: |a+bz|^2= |az+b|^2 . [[I'm just going to let the conjugates = capitalised version of the letter to make it easier to read]
   (a+bz)(conjugate(a)+conjugate(bz))= (az+b)conjugate(az)+conjugate(b))
(a+bz)(A+BZ)= (az+b)(AZ+B)
aA+Abz+aBZ+BbZz=aAzZ+AZb+aBz+Bb (but Zz=1)
LHS= |a| +Abz+ aBZ+|b|
RHS= |a|+AZb+aBz +|b|

but these aren't the same? if Zz=1, isn't Z=1/z? So I'm not sure how to get the answer.

Thank you!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 14, 2019, 03:32:52 pm
This was from tutoring but I asked both my tutor and my class teacher and they both say it's outside the scope, so I shouldn't worry about it!
I have some other questions though (from past assessments from my school).

1.  Explain why multiplying a complex number z by cistheta rotates z anticlockwise about the origin, through an angle of theta.
    I get the theory behind this, but I'm confused on how to set this out. Would it be best to let z= rcis(delta) and then use the multiplication of polar numbers:
    so you get that: z*cistheta= rcis(delta+theta)? But I'm still a bit confused on how to prove it shifts anticlockwise

2. Assume w is a nth root of infinity. Using geometric sums, prove that w+w^2+w^3+...w^n=0
    When I tried doing this, I got this: w(1-w^n)/(1-w)= w(1-w^n)/w(1/w-1)                           
                                                                   = (1-w^n)/(1/w-1)
             was I meant to use smth to do with the property 1+w+w^2=0??

3. If the mod of z=1, prove |a+bz|= |az+b|. You may assume a and b are elements of the Real numbers,.
What I did was:
     square both sides: |a+bz|^2= |az+b|^2 . [[I'm just going to let the conjugates = capitalised version of the letter to make it easier to read]
   (a+bz)(conjugate(a)+conjugate(bz))= (az+b)conjugate(az)+conjugate(b))
(a+bz)(A+BZ)= (az+b)(AZ+B)
aA+Abz+aBZ+BbZz=aAzZ+AZb+aBz+Bb (but Zz=1)
LHS= |a| +Abz+ aBZ+|b|
RHS= |a|+AZb+aBz +|b|

but these aren't the same? if Zz=1, isn't Z=1/z? So I'm not sure how to get the answer.

Thank you!!

If they have an answer I'm genuinely curious to see tbh, because I can't see how the coefficients work nicely when the binomial theorem is used.
__________________________________________________

1.That is because the angle is taken with respect to the positive real axis (i.e. positive \(x\)-axis).

As the argument increases, the more we go counterclockwise, because it's always up and outwards from the positive \(x\)-axis. Or else you cannot, in some sense, make the 'angle' 'bigger'.

Clockwise shifts occur when the argument is decreased.

2. Because \(\omega\) is an nth root of unity (watch out for the typo), by definition it means that \( \boxed{\omega^n = 1} \). Which once you sub that in, the expression becomes 0.

3. Firstly, never assume what you're trying to prove, or else you will receive 0 marks. You've started from the final result and worked backwards from it, but that can only be done when you know the final result is true, not when you are trying to prove it.

To work around this issue, you should only work on one side of the equation at a time. Not both sides simultaneously.
\begin{align*} LHS&=|a+bz|^2 \\&= (a+bz)\overline{(a+bz)}\\ &= (a+bz)(a+b\overline{z}) \tag{since a and b are real}\\ &= a^2 + abz + ab\overline{z} + b^2z\overline{z}\\ &= a^2+ab(z+\overline{z})+b^2 \tag{given |z|=1} \end{align*}
\begin{align*} RHS&=|az+b|^2\\ &= (az+b)\overline{(az+b)}\\ &= (az+b)(a\overline{z}+b) \tag{again, a and b are real}\\ &= az\overline{z} + ab\overline{z}+abz+b^2\\ &= a|z|^2 +abz+ ab\overline{z} + b^2\\ &= a^2+ ab(z+\overline{z})+b^2\\ & =LHS \end{align*}
However it is worthwhile to note that yes, in general \(z^{-1} = \frac{\overline{z}}{|z|^2}\). But when \(|z|=1\), the denominator basically collapses, and we arrive at \( z^{-1}=\overline{z}\)
Title: Re: 4U Maths Question Thread
Post by: goodluck on June 14, 2019, 08:58:26 pm


Took me ages bc Word crashed but here's my "latexed" version of their solution

Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 14, 2019, 09:14:15 pm
Took me ages bc Word crashed but here's my "latexed" version of their solution


That's basically the double recurrence relation method actually. It's not the binomial theorem.

(Double recurrences are outside the scope of MX2 but as seen in that example, they're handled the same way as usual recurrence formulas. i.e. with integration by parts. But I feel as though the question is misleading to mention the binomial theorem in that case.)
Title: Re: 4U Maths Question Thread
Post by: Livjane_2203 on June 15, 2019, 02:05:36 pm
I'm struggling with a volumes question help would be great thanks.

A hemisphere has radius r. By considering cross-sections parallel to the base of the hemisphere, show the volume is given by V=2/3(pi)r^3.


Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on June 15, 2019, 02:48:27 pm
Hey there!

Consider on a cartesian plane the semicircle

Then, as you're taking slices as it rotates from front to back, the slices will be semicircles that have radius x for 0<=x<=r. They'll go from zero, reach a radius of r, then go back down to zero. ie

Then, since the semicircle has radius r, and when you rotate it into a hemisphere, it will have depth 2r (from r to -r), your integral just becomes


Hope this helps :)
Title: Re: 4U Maths Question Thread
Post by: luckystrike826 on June 15, 2019, 06:51:29 pm
Hello! I have a question from class when doing the harder 3u section.

Given that d/dx (lnx) = 1/x,

Prove that:

Thanks a lot!
Title: Re: 4U Maths Question Thread
Post by: goodluck on June 15, 2019, 11:05:11 pm
That's basically the double recurrence relation method actually. It's not the binomial theorem.

(Double recurrences are outside the scope of MX2 but as seen in that example, they're handled the same way as usual recurrence formulas. i.e. with integration by parts. But I feel as though the question is misleading to mention the binomial theorem in that case.)

Yeah, I think it's like some of those recurrence formulas were, which is probably why they got away with it. would it be something that we could expect?

I'm stuck on some other problems (Is it just me or is complex numbers a more complicated area that everyone seems to treat it as? I've heard a lot of people saying it's super easy, which makes me feel awful)

1. Let z= (i+ix)/(1-ix). Show that:
                                                 if x is less than or equal to 1, arg(z)= pi/2+2arctan(x)
                                                 if x is greater than 1, arg(z)= -3pi/2+2arctan(x)
Challenge q: Are there interesting patterns you can spot as x approaches plus and minus infinity

2. Find the real and imaginary part of (1+cos2theta+isin2theta)/(1+cos2theta-isin2theta). I tried to use the 2cos^2theta-1 for it but it didn't come out nicely at all
Title: Re: 4U Maths Question Thread
Post by: kaustubh.patel on June 16, 2019, 08:52:39 am
I'm struggling with a volumes question help would be great thanks.

A hemisphere has radius r. By considering cross-sections parallel to the base of the hemisphere, show the volume is given by V=2/3(pi)r^3.


Umm what i did was imagine a QUATER of a semi circle and found the volume of that, you can simply multiply that by 2 to get the volume of the semi circle.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 16, 2019, 09:50:22 am
Yeah, I think it's like some of those recurrence formulas were, which is probably why they got away with it. would it be something that we could expect?

I'm stuck on some other problems (Is it just me or is complex numbers a more complicated area that everyone seems to treat it as? I've heard a lot of people saying it's super easy, which makes me feel awful)

1. Let z= (i+ix)/(1-ix). Show that:
                                                 if x is less than or equal to 1, arg(z)= pi/2+2arctan(x)
                                                 if x is greater than 1, arg(z)= -3pi/2+2arctan(x)
Challenge q: Are there interesting patterns you can spot as x approaches plus and minus infinity

2. Find the real and imaginary part of (1+cos2theta+isin2theta)/(1+cos2theta-isin2theta). I tried to use the 2cos^2theta-1 for it but it didn't come out nicely at all
Nah, double recurrence relations are actually not examinable in the HSC. Can't remember where I was first told of this but it is an official statement.
________________________________________________________________________

Just before I jump into 1, I want to confirm that you meant \( z = \frac{i+ix}{1-ix} \) as opposed to \( z = \frac{1+ix}{1-ix} \)? Haven't tried it yet, just wanted to double check first.
\[ \text{Double angle formulas should be fine for the second one.}\\ \text{It's just that even more work is necessary after you use it.} \\ \begin{align*} \frac{1+\cos2\theta+i\sin2\theta}{1+\cos2\theta-i\sin2\theta} &= \frac{2\cos^2\theta+2i\sin\theta\cos\theta}{2\cos^2\theta-2i\sin\theta\cos\theta}\\ &= \frac{\cos\theta+i\sin\theta}{\cos\theta-i\sin\theta}\\ &= \frac{\cos\theta+i\sin\theta}{\cos(-\theta) + i\sin(-\theta)}\\ &= \frac{\operatorname{cis}\theta}{\operatorname{cis}(-\theta)}\\ &= \operatorname{cis}2\theta\\ &= \cos2\theta+i\sin2\theta \end{align*} \]
Note that the complex numbers questions you have given are on the upper end of the spectrum - they tend to rock up in Q14-16 instead of Q11-13 of the exam. But at the E4 range, recognising things like \( \operatorname{cis}(-\theta) = \cos\theta - i\sin\theta \) becomes quite common, as you're expected to have a full mastery of trigonometric identities. And of course, the fact that division of complex numbers in mod-arg form involves subtracting their arguments.
Title: Re: 4U Maths Question Thread
Post by: goodluck on June 16, 2019, 10:27:19 am
Nah, double recurrence relations are actually not examinable in the HSC. Can't remember where I was first told of this but it is an official statement.
________________________________________________________________________

Just before I jump into 1, I want to confirm that you meant \( z = \frac{i+ix}{1-ix} \) as opposed to \( z = \frac{1+ix}{1-ix} \)? Haven't tried it yet, just wanted to double check first.
\[ \text{Double angle formulas should be fine for the second one.}\\ \text{It's just that even more work is necessary after you use it.} \\ \begin{align*} \frac{1+\cos2\theta+i\sin2\theta}{1+\cos2\theta-i\sin2\theta} &= \frac{2\cos^2\theta+2i\sin\theta\cos\theta}{2\cos^2\theta-2i\sin\theta\cos\theta}\\ &= \frac{\cos\theta+i\sin\theta}{\cos\theta-i\sin\theta}\\ &= \frac{\cos\theta+i\sin\theta}{\cos(-\theta) + i\sin(-\theta)}\\ &= \frac{\operatorname{cis}\theta}{\operatorname{cis}(-\theta)}\\ &= \operatorname{cis}2\theta\\ &= \cos2\theta+i\sin2\theta \end{align*} \]
Note that the complex numbers questions you have given are on the upper end of the spectrum - they tend to rock up in Q14-16 instead of Q11-13 of the exam. But at the E4 range, recognising things like \( \operatorname{cis}(-\theta) = \cos\theta - i\sin\theta \) becomes quite common, as you're expected to have a full mastery of trigonometric identities. And of course, the fact that division of complex numbers in mod-arg form involves subtracting their arguments.

Whoops, I made a typo!: the original statement was i(1+ix)/(1-ix) so I just multiplied it in. So it should equal (i-x)/(1-ix). Sorry about that!

Also for the previous question, when you divided the fraction by costheta, do we need to say smth like costheta can't equal 0? or is that not required?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 16, 2019, 04:05:09 pm
Whoops, I made a typo!: the original statement was i(1+ix)/(1-ix) so I just multiplied it in. So it should equal (i-x)/(1-ix). Sorry about that!

Also for the previous question, when you divided the fraction by costheta, do we need to say smth like costheta can't equal 0? or is that not required?
With the double angle formula, technically speaking you should, and then check the solutions for \(\cos x = 0\), i.e. \( x=2n\pi \pm \frac\pi2\) separately. There is another method that completely avoids this issue, but it's a bit harder to see. You can request to see that method if you wish in a later post.
\[ \text{Looking at that question now, the question would be split into multiple parts}\\ \text{should it appear in an exam.}\\ \text{The nature of the question looks very assignment-like.} \]
\[ \text{An assumption is made that }x\text{ is real.} \]
\[ \text{The initial expression given is not very manageable, so we're tempted to simplify it.}\\ \text{This should be done in the classic way of division by complex numbers.}\\ \begin{align*} \frac{i(1+ix)}{1-ix} &= \frac{i(1+ix)^2}{(1+ix)(1-ix)}\\ &= \frac{i(1+2ix-x^2)}{1^2+x^2}\\ &= \frac{i(1+2ix-x^2)}{1+x^2}\\ &= \frac{-2x}{1+x^2} + \frac{1-x^2}{1+x^2}. \end{align*} \]
\[ \text{Personally I feel there should be more cases - the case }x\leq 1\text{ looks a bit too powerful}\\ \text{at first glance. But I could be wrong.}\\ \text{For now, here's the solution to the second case:}\]
It might be fine if it is replaced by \( 0 < x \leq 1\) though, but then the bit about \(-\infty\) would be bad.
\[ \text{Suppose now that }x > 1. \\ \text{Then, note that }-\frac{2x}{1+x^2} < 0\\ \textbf{and }\frac{1-x^2}{1+x^2} < 0.\\ \text{Therefore, the complex number will be in the third quadrant.}\]
\[ \text{Upon drawing the diagram, we see that the related angle }\theta\text{ satisfies}\\ \tan \theta = \frac{x^2-1}{2x}.\\ \text{Therefore the argument will be}\\ \arg z = -\pi + \tan^{-1} \left(  \frac{x^2-1}{2x}\right) \]
___________________________________
\[ \text{So we've boiled the problem down to proving that}\\ \boxed{-\pi + \tan^{-1} \frac{x^2-1}{2x} = -\frac{3\pi}{2} + 2\tan^{-1}x}. \]
Of course, this may not be the recommended approach. But if the source of your question hasn't given any hints, anything goes. Here, I will now treat this problem similarly to a harder 3U question. On a separate sheet of paper, I've done some working backwards to make sure that this proof works. But I only present my answer with the working in a forwards, logical manner. (The intuition is that proving this statement is equivalent to proving that \( \frac\pi2 - 2\tan^{-1}x = \tan^{-1} \frac{1-x^2}{2x} \) for \(x > 1\).)
\[ \text{Observe that}\\\begin{align*} \tan \left( \frac\pi2 - 2\tan^{-1}x \right) &= \cot \left(2\tan^{-1}x \right)\\ &= \frac{1}{\tan (2\tan^{-1}x)}\\ &= \frac{1}{\left( \frac{2\tan(\tan^{-1}x)}{1-\tan^2(\tan^{-1}x)} \right)}\\ &= \frac{1}{\left( \frac{2x}{1-x^2} \right)}\\ &= \frac{1-x^2}{2x}\end{align*}\]
\[ \text{Hence taking inverse tan on both sides,}\\ \begin{align*} \frac\pi2 - 2\tan^{-1}x &= \tan^{-1} \frac{1-x^2}{2x}\\ \therefore \frac\pi2 - \tan^{-1} \frac{1-x^2}{2x} &= 2\tan^{-1}x\\ \frac\pi2 + \tan^{-1} \frac{x^2-1}{2x} &= 2\tan^{-1}x\\ \therefore -\pi + \tan^{-1} \frac{x^2-1}{2x} &= -\frac{3\pi}{2} + 2\tan^{-1}x \end{align*} \]
There's one huge subtlety here - I did not justify why we could cancel out the tan-inverse with the tan on the LHS. In general, \( \tan^{-1}(\tan x) = x \) only when \( -\frac\pi2 < x < \frac\pi2\).

Indeed, this situation holds here. Note that if \(x > 1\), then \( \frac\pi4 < \tan^{-1}x < \frac\pi2\). This can then be rearranged to give \( -\frac\pi2 < \frac\pi2 - 2\tan^{-1}x < 0 \), which satisfies the criteria that \( -\frac\pi2 < \frac\pi2 - 2\tan^{-1}x < \frac\pi2 \) regardless, so the cancellation is justified here.
Title: Re: 4U Maths Question Thread
Post by: Ollierobb1 on June 19, 2019, 12:28:17 pm
Four letters are chosen from the letters of the word TELEGRAPH.
These four letters are then placed alongside one another to form a four letter arrangement. Find the number of distinct four letter arrangements which are possible, considering all choices. (2 marks)

Help would be much appreciated
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 19, 2019, 01:08:34 pm
Four letters are chosen from the letters of the word TELEGRAPH.
These four letters are then placed alongside one another to form a four letter arrangement. Find the number of distinct four letter arrangements which are possible, considering all choices. (2 marks)

Help would be much appreciated
Hint: You will need to consider three separate cases:
- 2 E's among the four letters
- 1 E among the four letters
- 0 E's among the four letters

This is because E is the only repeated letter present. Note that the problem quickly becomes catastrophic when we have more repeated letters.

You should post any progress you have if you require further help.
Title: Re: 4U Maths Question Thread
Post by: DrDusk on June 19, 2019, 06:50:06 pm
Hello! I have a question from class when doing the harder 3u section.

Given that d/dx (lnx) = 1/x,

Prove that:

Thanks a lot!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 19, 2019, 07:34:18 pm
Hello! I have a question from class when doing the harder 3u section.

Given that d/dx (lnx) = 1/x,

Prove that:

Thanks a lot!
With this question, I will say that I heavily dislike this question as it has been provided without any context whatsoever, especially since the supposed result to be proved is commonly taken as a definition. Without the source of the problem, or any extra parts that were provided, I fail to see how this is appropriate even at the MX2 level.

In saying that, here is another approach. The intuition relies on the fact that the derivative and the limit are more or less related by the first principles formula.
\begin{align*} \frac{1}{y} &= \frac{d}{dy}\ln y\\ &= \lim_{h\to 0}\frac{\ln (y+h) - \ln y}{h}\\ &= \lim_{h \to 0} \ln \frac1h \left(1 - \frac{h}{y} \right) \tag{difference log law}\\ &= \lim_{h\to 0}\ln \left( 1-\frac{h}{y}\right)^{1/h}\\ &= \lim_{n\to \infty}\ln \left( 1 - \frac{1}{ny} \right)^n \end{align*}
\[ \text{where we perform the substitution }n=\frac1h.\\ \text{Furthermore, we sub }x=\frac1y\text{ to obtain}\\ x = \lim_{n\to \infty} \ln \left( 1 - \frac{x}{n} \right)^{n}. \]
\[ \text{Due to the continuity of }\ln(\cdot), \text{ the limit can be moved inside the logarithm to obtain}\\ x = \ln \left( \lim_{n\to \infty} 1-\frac{x}{n} \right).\\ \text{Which immediately rearranges to }\boxed{e^x = \lim_{n\to \infty} \left( 1-\frac{x}{n}\right)^n} \]
Title: Re: 4U Maths Question Thread
Post by: joelg on June 20, 2019, 04:39:18 pm
Hey guys, I'm struggling with part iv. (see photo) help would be greatly appreciated

Thanks!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 20, 2019, 04:49:15 pm
Hey guys, I'm struggling with part iv. (see photo) help would be greatly appreciated

Thanks!
Hint: You can try to show that the interval \(QR\) subtends angles of equal size at \( P\) and \(S\), which are both on the same side of \(QR\). That'll help you prove that the points are concyclic.

To do so, you may need \(b^2=a^2(1-e^2)\), but it's likely you'll also need the angle between two lines formula to find \( \tan \angle QPR \) and \(\tan \angle QSR\). I haven't tried the problem yet, but I wouldn't be surprised by some nasty algebra.
Title: Re: 4U Maths Question Thread
Post by: Ollierobb1 on June 21, 2019, 02:00:40 pm
Hint: You can try to show that the interval \(QR\) subtends angles of equal size at \( P\) and \(S\), which are both on the same side of \(QR\). That'll help you prove that the points are concyclic.

To do so, you may need \(b^2=a^2(1-e^2)\), but it's likely you'll also need the angle between two lines formula to find \( \tan \angle QPR \) and \(\tan \angle QSR\). I haven't tried the problem yet, but I wouldn't be surprised by some nasty algebra.

Damn, im still struggling through the algebra of this question. Maybe i haven't gotten the right y coordinate for R, -sinΘ(a^2-b^2)/b?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 21, 2019, 03:57:38 pm
Damn, im still struggling through the algebra of this question. Maybe i haven't gotten the right y coordinate for R, -sinΘ(a^2-b^2)/b?
\[ \angle QPR = 90^\circ\\ \text{because the normal is perpendicular to the tangent by definition.} \\ \text{So we just have to show that }\angle QSR = 90^\circ. \]
\[ \begin{align*} m_{QS} &= \frac{0 - \frac{b}{\sin\theta}}{ae - 0}\\ &= -\frac{b}{ae\sin\theta}\\ m_{RS} &= \frac{0 + \frac{(a^2-b^2)\sin\theta}{b}}{ae-0} \\ &= \frac{(a^2-b^2)\sin\theta}{abe}\end{align*} \]
\begin{align*}
\therefore m_{QS} m_{RS} &= -\frac{b}{ae\sin\theta} \cdot \frac{(a^2-b^2)\sin\theta}{abe}\\
&= -\frac{a^2-b^2}{a^2e^2}\\
&= -\frac{a^2-a^2(1-e^2)}{a^2e^2}\\
&= -\frac{a^2e^2}{a^2e^2}\\
&= -1
\end{align*}
Hence \( QS \perp RS\) as required. You should now be able to puzzle it all together.
Title: Re: 4U Maths Question Thread
Post by: classof2019 on June 23, 2019, 06:02:28 pm
Can someone please help with the attached q? Thanks!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 23, 2019, 06:06:50 pm
Hint for the time being: Apply differentiation from first principles to an appropriate function \(f(y)\). What you're computing is just \( f^\prime \left(\frac\pi4\right)\) of that function.
Title: Re: 4U Maths Question Thread
Post by: not a mystery mark on June 24, 2019, 07:31:35 pm
Hey! Our class is currently freaking out over this Complex Number - Vector question.
Does anyone have a definitive answer?  ;D

Cheers!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 25, 2019, 03:27:15 pm
Hey! Our class is currently freaking out over this Complex Number - Vector question.
Does anyone have a definitive answer?  ;D

Cheers!!
\[ \text{As }M\text{ is the midpoint of }AB,\\ \overrightarrow{OM} = \frac{z_1+z_2}{2}.\]
Note that I shouldn't really say they're "equal". I should say that \( \overrightarrow{OM} \) represents \( \frac{z_1+z_2}{2} \). But here I'll just be lazy.
\[ \text{Then,}\\ \begin{align*} \overrightarrow{MA} &= \overrightarrow{MO} + \overrightarrow{OA}\\ &= -\frac{z_1+z_2}{2} + z_1\\ &= \frac{z_1-z_2}{2} \end{align*} \]
\[ \text{Since }\overrightarrow{MQ}\text{ is a }90^\circ\text{ anti-clockwise rotation of }\overrightarrow{MA},\\ \overrightarrow{MQ} = i\left( \frac{z_1-z_2}{2} \right) \]
\[ \text{Then finally,}\\ \begin{align*}\overrightarrow{OQ} &= \overrightarrow{OM} + \overrightarrow{MQ}\\ \therefore \omega &= \frac{z_1+z_2}{2} + i\left( \frac{z_1-z_2}{2}\right).\end{align*} \]
Hence D.

It's very easy to get lost in the vectors so try approaching it somewhat systematically. Use tip-to-tail addition purely for mental purposes as much as you can. Also, I compare MQ to MA as opposed to QM to AM, because things are more obvious to me when they start from the same point.
Title: Re: 4U Maths Question Thread
Post by: not a mystery mark on June 25, 2019, 05:10:27 pm
It's very easy to get lost in the vectors so try approaching it somewhat systematically. Use tip-to-tail addition purely for mental purposes as much as you can. Also, I compare MQ to MA as opposed to QM to AM, because things are more obvious to me when they start from the same point.

This is super clear! Thanks heaps for making 4U feel really easy!!
Title: Re: 4U Maths Question Thread
Post by: Ollierobb1 on June 26, 2019, 05:55:27 pm
Hey guys, could anyone help me with this? Part i is fine and i found the coordinates of R but i just cant finish it off. Thanks
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on June 26, 2019, 06:09:59 pm
Hey there!

I assume that you've found that the co-ordinates of R are


Hope this helps :)

EDIT: for some insights into this approach, I've learned from my teacher to have a look at what you're working towards: basically the locus of R has no x1 or y1 involved anywhere; this tells me that since x1 and y1 are points on the locus we should look to cancel them out somehow using x2 - y2 = a2. Noticing that there's only x1 and y1 and not those terms squared in the co-ordinates, it makes sense to square them. It all comes out properly from there :)
Title: Re: 4U Maths Question Thread
Post by: luckystrike826 on June 30, 2019, 12:44:46 am
Hey Guys, can anyone help with the last part of this question? It's from Sydney Grammar 2014 Trials and I don't really understand their solutions.

The answer to i) is (r^2)/2     * (sin alpha + sin beta + sin gamma).
The answer to ii) a) is 1/3 (alpha + beta + gamma), 1/3 ( sin alpha + sin beta + sin gamma).
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 30, 2019, 08:03:48 am
Hey Guys, can anyone help with the last part of this question? It's from Sydney Grammar 2014 Trials and I don't really understand their solutions.

The answer to i) is (r^2)/2     * (sin alpha + sin beta + sin gamma).
The answer to ii) a) is 1/3 (alpha + beta + gamma), 1/3 ( sin alpha + sin beta + sin gamma).

The first one is simply applying \( A = \frac{1}{2}ab\sin \theta \) three times, once for each of the smaller triangles. Note that the angles are all made at the centre of the circle, hence the lengths \(a\) and \(b\) in the area formula equal the length of the radius \(r\). So we just get \( \frac12 r^2 \sin \alpha +\frac12r^2\sin \beta + \frac12 r^2\sin\gamma\).

Part ii) (alpha) is not examinable material in the HSC and likely Sydney Grammar being Sydney Grammar, purely because the "centroid" is not something HSC students are expected to memorise. I don't really see how part i) is helpful for this part either - it looks more helpful for the next part. In the HSC, you would be given the definition that the centroid really reflects the arithmetic mean of all the points you're considering. (Note that the arithmetic mean is just the usual average.)

So averaging the \(x\)-coordinates we get \( \frac{\alpha+\beta+\gamma}{3} \). Similarly for the \(y\)-coordinates.
Title: Re: 4U Maths Question Thread
Post by: luckystrike826 on June 30, 2019, 08:19:16 am
Sorry, I must have phrased my question wrong. I get the first two parts, but not sure about the last part which is the part beta. Thanks!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 30, 2019, 09:39:08 am
\[ \text{Intuition suggests that the values }\alpha,\,\beta\text{ and }\gamma\text{ should somehow}\\ \text{correspond to the same ones given in the previous part.}\\ \text{Else it does not make sense.} \]
\[ \text{How the answers start off is basically relying on part i) for some simplifications.}\\ \text{In part i), it is clear that the angles all meet at a point, which gives the}\\ \alpha+\beta+\gamma = 2\pi\text{ bit.} \]
Which is just 3 times the expression for the \(x\)-coordinate of the centroid. So the centroid must have \(x\)-coordinate \( \frac{2\pi}{3} \).
\[ \text{But on one hand, the area of }\triangle TUV\text{ was }\frac12 r^2 (\sin\alpha+\sin\beta+\sin\gamma)\\ \text{whilst the }y\text{-coordinate of the centroid is }\frac{\sin\alpha+\sin\beta+\sin\gamma}{3}. \]
\[ \text{So pretty much a bit of rearranging is what's needed here to show that}\\ \text{the centroid has }y\text{-coordinate }\frac{2\times \operatorname{Area}_{\triangle TUV}}{3r^2}. \]
The answers write the area as \(|\triangle TUV|\) instead. This isn't really common in the HSC but it does get used at times in the real world, so I'll use the absolute value brackets too.

Of course, that was the easy mark to get. By considering the link between the two parts we've more or less reinterpreted the problem. What we wished to do is to maximise \( |\triangle TUV| \), but we see that the \(y\)-coordinate of the centroid of \(\triangle ABC\) happens to be a scalar multiple of it. (Note of course that \(r\) is constant.)

So if we maximise the \(y\)-coordinate of said centroid, we achieve the same goal. This is a much harder mark to get. Naively, we'd be tempted to go back to maximising \(\sin\alpha+\sin\beta+\sin\gamma\), but this problem is still not viable because we only have the one constraint \(\alpha+\beta+\gamma = 2\pi\). This is where we get the intuition that something about the curve \(y=\sin x\) must be useful - we want to do this maximisation with a completely different argument, not via calculus in optimisation.

To my knowledge, concavity hasn't been examined much at all in the HSC - in fact, I only know of one question in 2018 or 2017 that did. This is the idea.
____________________________________________________________________________
\[ \text{Given two points }x_1\text{ and }x_2,\text{ we say that a curve is concave down for all }x_1 \leq x \leq x_2\\ \text{if for every }0\leq t \leq 1,\text{ it is true that}\\ f(tx_1 + (1-t)x_2) \geq tf(x_1) + (1-t)f(x_2).\]
\[ \text{This definition is awkward to work with, so we clarify what is going on.}\\ \text{Essentially, if your curve is concave down over a region, what you can do is draw a straight line segment}\\ \text{from one endpoint of the region, to the other.} \]
\[ \text{A curve is concave }\textbf{down}\text{ if that line segment}\\ \text{is guaranteed to lie }\textbf{below}\text{ the curve about that region.} \]
Note similarly that for a curve concave up over a certain interval, the line segment drawn between the endpoints should lie above the curve.
\[ \text{This is illustrated in the diagram. For example, between the points }A\text{ and }C\\ \text{the curve is concave down.}\\ \text{The interval }AC\text{ is also, in fact, under the curve.} \]
Similarly, between the points \(A\) and \(B\) the curve is concave down, and the line segment \(AB\) is below the curve. Same goes for \(BC\).

The answers claim that this will always be the case, regardless of what \(A\), \(B\) and \(C\) are. This is because the \(x\)-coordinates of these points are respectively \(\alpha\), \(\beta\) and \(\gamma\), which in part (i) we're told are always between \(0\) and \(\pi\). For a MX2 Q16 problem, it should be safe to just assume without proof that \(y=\sin x\) is always concave down for \(0 < x< \pi\) - this can easily be proven using the second derivative, if a graph is not adequately convincing.

(Aside: The second derivative is used as a theorem to prove concavity. The definition, however, is the clunky thing I mentioned earlier.)
\[ \text{So because we know that }AB, \, AC\textbf{ and }BC\text{ lie under the curve }y=\sin x,\\ \text{it follows that }\triangle ABC\text{ also lies under it too.} \]
And after all, since the centroid is just the intersection of the triangles' medians, that is consequently forced under the curve as well.
____________________________________________________________________________
\[ \text{However, the crucial thing to note now is that }\textbf{only}\text{ the }y\text{-coordinate}\\ \text{influences }|\triangle TUV|. \text{ Recall that the }x\text{-coordinate}\\ \text{of the centroid is }\textbf{constant}.\text{ It is always equal to }\frac{2\pi}{3}. \]
This is where advanced intuition becomes necessary. The idea is that we know the centroid can never float above \(y=\sin x\). Rather, the \(y\)-coordinate of the centroid is bounded above by the curve \(y=\sin x\).

So the natural question to ask her is whether or not this bound is attainable. That is, is there a configuration of points \(A\), \(B\) and \(C\) that would ensure that the centroid must lie on the curve as well. This is useful, because if we know it can lie on the curve, and also know that it can't go above it, then we've deduced that the maximum possible \(y\)-coordinate IS when it is on the curve.
\[ \text{The answer to these questions is usually yes,}\\ \text{when we allow for a thing called }\textbf{degeneration}. \]
\[ \text{The idea is to }\textit{collapse }\triangle ABC\text{ somehow.}\\ \text{Without loss of generality, let's start by collapsing }BC. \]
\[ \text{What we are going to do is let }B\text{ and }C\text{ become the same point.}\\ \text{That is, }C=B.\\ \text{Then the concept of }\triangle ABC\text{ no longer exists.} \]
\[ \text{However what happens is that if we gradually bring }C\text{ closer to }B,\\ \text{what happens geometrically is that }\triangle ABC\\ \text{will gradually turn into just another line segment, namely the interval }AB! \]
Once we're left with the interval \(AB\), the "centroid" of \(\triangle ABC\) will be nothing more than a point on said interval. (To be somewhat specific, it will be the point that divides \(AB\) internally in the ratio 1:2, I believe. It could be 2:1 though.)

But the point is this is not enough. We know that \(AB\) is still under the curve \(y=\sin x\), so if the centroid lies on \(AB\), it is also under the curve.
\[ \text{This is why we need to think about the most }\textit{borderline}\text{ case.}\\ \text{This is when we make all }\textbf{three}\text{ points coincide.}\\ \text{That is, set }A = B = C\text{, as the answers have stated.} \]
The intuition behind this is a general rule of mathematics - borderline cases tend to influence optimality (be it minimality or maximality). This can prove useful at times, but trying to recall this in an exam is hard.
____________________________________________________________________________
\[ \text{And the thing is, we see that this is good!}\\ \text{When }A=B=C\text{, we're saying that all 3 points are now the same point}\\ \text{and hence }\triangle ABC\text{ has also collapsed down into a single point.} \]
\[ \text{Consequently, the centroid has also been collapsed to a single point.}\\ \text{But the key idea is, the centroid now has nowhere to run.}\\ \text{Since }A,\, B\text{ and }C\text{ are the same point, the centroid is forced to being the same point as well.} \]
\[ \text{The reason why this is optimal is because we know that }A,\, B\text{ and }C\text{ are on the curve.}\\ \text{Hence the centroid is }\textbf{also}\text{ on the curve }y=\sin x\text{ now!} \]
Thus, we've shown that the centroid both cannot lie above the curve, but can also lie ON the curve. This must therefore give the maximum \(y\)-coordinate.

Basically, the argument was pretty much fully geometric. But in a specific way.
____________________________________________________________________________
\[ \text{Once this bit of the puzzle is cracked, everything is easy.}\\ \text{We know that }A = B = C\text{ automatically imposes the condition}\\ \text{that their corresponding }x\text{-coordinates are equal.} \]
\[ \text{Therefore }\alpha = \beta = \gamma.\\ \text{So by subbing back into }\alpha+\beta+\gamma = 2\pi\\ \text{we can conclude that in fact, }\\ \alpha = \beta = \gamma = \frac{2\pi}{3}.\\ \text{Which, of course, can now be plugged back into the area formula.} \]
Title: Re: 4U Maths Question Thread
Post by: Ollierobb1 on June 30, 2019, 10:38:21 am
Could anyone help me with this question? Thanks in advanced

A triangle has sides a, b & c
If a^2 + b^2 + c^2 = ab + bc + ca show that triangle ABC is equilateral
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on June 30, 2019, 11:42:45 am


Hope this helps :)
Title: Re: 4U Maths Question Thread
Post by: Ollierobb1 on June 30, 2019, 01:25:26 pm


Hope this helps :)

Ahhhh, it was only worth one mark so I thought maybe there was a quick way but this makes sense, thanks a lot!
Title: Re: 4U Maths Question Thread
Post by: luckystrike826 on June 30, 2019, 02:37:52 pm
\[ \text{Intuition suggests that the values }\alpha,\,\beta\text{ and }\gamma\text{ should somehow}\\ \text{correspond to the same ones given in the previous part.}\\ \text{Else it does not make sense.} \]
\[ \text{How the answers start off is basically relying on part i) for some simplifications.}\\ \text{In part i), it is clear that the angles all meet at a point, which gives the}\\ \alpha+\beta+\gamma = 2\pi\text{ bit.} \]
Which is just 3 times the expression for the \(x\)-coordinate of the centroid. So the centroid must have \(x\)-coordinate \( \frac{2\pi}{3} \).
\[ \text{But on one hand, the area of }\triangle TUV\text{ was }\frac12 r^2 (\sin\alpha+\sin\beta+\sin\gamma)\\ \text{whilst the }y\text{-coordinate of the centroid is }\frac{\sin\alpha+\sin\beta+\sin\gamma}{3}. \]
\[ \text{So pretty much a bit of rearranging is what's needed here to show that}\\ \text{the centroid has }y\text{-coordinate }\frac{2\times \operatorname{Area}_{\triangle TUV}}{3r^2}. \]
The answers write the area as \(|\triangle TUV|\) instead. This isn't really common in the HSC but it does get used at times in the real world, so I'll use the absolute value brackets too.

Of course, that was the easy mark to get. By considering the link between the two parts we've more or less reinterpreted the problem. What we wished to do is to maximise \( |\triangle TUV| \), but we see that the \(y\)-coordinate of the centroid of \(\triangle ABC\) happens to be a scalar multiple of it. (Note of course that \(r\) is constant.)

So if we maximise the \(y\)-coordinate of said centroid, we achieve the same goal. This is a much harder mark to get. Naively, we'd be tempted to go back to maximising \(\sin\alpha+\sin\beta+\sin\gamma\), but this problem is still not viable because we only have the one constraint \(\alpha+\beta+\gamma = 2\pi\). This is where we get the intuition that something about the curve \(y=\sin x\) must be useful - we want to do this maximisation with a completely different argument, not via calculus in optimisation.

To my knowledge, concavity hasn't been examined much at all in the HSC - in fact, I only know of one question in 2018 or 2017 that did. This is the idea.
____________________________________________________________________________
\[ \text{Given two points }x_1\text{ and }x_2,\text{ we say that a curve is concave down for all }x_1 \leq x \leq x_2\\ \text{if for every }0\leq t \leq 1,\text{ it is true that}\\ f(tx_1 + (1-t)x_2) \geq tf(x_1) + (1-t)f(x_2).\]
\[ \text{This definition is awkward to work with, so we clarify what is going on.}\\ \text{Essentially, if your curve is concave down over a region, what you can do is draw a straight line segment}\\ \text{from one endpoint of the region, to the other.} \]
\[ \text{A curve is concave }\textbf{down}\text{ if that line segment}\\ \text{is guaranteed to lie }\textbf{below}\text{ the curve about that region.} \]
Note similarly that for a curve concave up over a certain interval, the line segment drawn between the endpoints should lie above the curve.
\[ \text{This is illustrated in the diagram. For example, between the points }A\text{ and }C\\ \text{the curve is concave down.}\\ \text{The interval }AC\text{ is also, in fact, under the curve.} \]
Similarly, between the points \(A\) and \(B\) the curve is concave down, and the line segment \(AB\) is below the curve. Same goes for \(BC\).

The answers claim that this will always be the case, regardless of what \(A\), \(B\) and \(C\) are. This is because the \(x\)-coordinates of these points are respectively \(\alpha\), \(\beta\) and \(\gamma\), which in part (i) we're told are always between \(0\) and \(\pi\). For a MX2 Q16 problem, it should be safe to just assume without proof that \(y=\sin x\) is always concave down for \(0 < x< \pi\) - this can easily be proven using the second derivative, if a graph is not adequately convincing.

(Aside: The second derivative is used as a theorem to prove concavity. The definition, however, is the clunky thing I mentioned earlier.)
\[ \text{So because we know that }AB, \, AC\textbf{ and }BC\text{ lie under the curve }y=\sin x,\\ \text{it follows that }\triangle ABC\text{ also lies under it too.} \]
And after all, since the centroid is just the intersection of the triangles' medians, that is consequently forced under the curve as well.
____________________________________________________________________________
\[ \text{However, the crucial thing to note now is that }\textbf{only}\text{ the }y\text{-coordinate}\\ \text{influences }|\triangle TUV|. \text{ Recall that the }x\text{-coordinate}\\ \text{of the centroid is }\textbf{constant}.\text{ It is always equal to }\frac{2\pi}{3}. \]
This is where advanced intuition becomes necessary. The idea is that we know the centroid can never float above \(y=\sin x\). Rather, the \(y\)-coordinate of the centroid is bounded above by the curve \(y=\sin x\).

So the natural question to ask her is whether or not this bound is attainable. That is, is there a configuration of points \(A\), \(B\) and \(C\) that would ensure that the centroid must lie on the curve as well. This is useful, because if we know it can lie on the curve, and also know that it can't go above it, then we've deduced that the maximum possible \(y\)-coordinate IS when it is on the curve.
\[ \text{The answer to these questions is usually yes,}\\ \text{when we allow for a thing called }\textbf{degeneration}. \]
\[ \text{The idea is to }\textit{collapse }\triangle ABC\text{ somehow.}\\ \text{Without loss of generality, let's start by collapsing }BC. \]
\[ \text{What we are going to do is let }B\text{ and }C\text{ become the same point.}\\ \text{That is, }C=B.\\ \text{Then the concept of }\triangle ABC\text{ no longer exists.} \]
\[ \text{However what happens is that if we gradually bring }C\text{ closer to }B,\\ \text{what happens geometrically is that }\triangle ABC\\ \text{will gradually turn into just another line segment, namely the interval }AB! \]
Once we're left with the interval \(AB\), the "centroid" of \(\triangle ABC\) will be nothing more than a point on said interval. (To be somewhat specific, it will be the point that divides \(AB\) internally in the ratio 1:2, I believe. It could be 2:1 though.)

But the point is this is not enough. We know that \(AB\) is still under the curve \(y=\sin x\), so if the centroid lies on \(AB\), it is also under the curve.
\[ \text{This is why we need to think about the most }\textit{borderline}\text{ case.}\\ \text{This is when we make all }\textbf{three}\text{ points coincide.}\\ \text{That is, set }A = B = C\text{, as the answers have stated.} \]
The intuition behind this is a general rule of mathematics - borderline cases tend to influence optimality (be it minimality or maximality). This can prove useful at times, but trying to recall this in an exam is hard.
____________________________________________________________________________
\[ \text{And the thing is, we see that this is good!}\\ \text{When }A=B=C\text{, we're saying that all 3 points are now the same point}\\ \text{and hence }\triangle ABC\text{ has also collapsed down into a single point.} \]
\[ \text{Consequently, the centroid has also been collapsed to a single point.}\\ \text{But the key idea is, the centroid now has nowhere to run.}\\ \text{Since }A,\, B\text{ and }C\text{ are the same point, the centroid is forced to being the same point as well.} \]
\[ \text{The reason why this is optimal is because we know that }A,\, B\text{ and }C\text{ are on the curve.}\\ \text{Hence the centroid is }\textbf{also}\text{ on the curve }y=\sin x\text{ now!} \]
Thus, we've shown that the centroid both cannot lie above the curve, but can also lie ON the curve. This must therefore give the maximum \(y\)-coordinate.

Basically, the argument was pretty much fully geometric. But in a specific way.
____________________________________________________________________________
\[ \text{Once this bit of the puzzle is cracked, everything is easy.}\\ \text{We know that }A = B = C\text{ automatically imposes the condition}\\ \text{that their corresponding }x\text{-coordinates are equal.} \]
\[ \text{Therefore }\alpha = \beta = \gamma.\\ \text{So by subbing back into }\alpha+\beta+\gamma = 2\pi\\ \text{we can conclude that in fact, }\\ \alpha = \beta = \gamma = \frac{2\pi}{3}.\\ \text{Which, of course, can now be plugged back into the area formula.} \]

Thank you so much!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on June 30, 2019, 05:37:48 pm
Ahhhh, it was only worth one mark so I thought maybe there was a quick way but this makes sense, thanks a lot!
Note: There is if you have previous parts to help you out.
Title: Re: 4U Maths Question Thread
Post by: Ollierobb1 on July 01, 2019, 12:37:42 pm
The first 3 sections were fine and i havnt tried the second part of iv, but for finding SP i used the locus definition of a hyperbola and got the result but i ended up with a y still hanging around. Help would be much appreciated
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on July 01, 2019, 05:31:22 pm
The first 3 sections were fine and i havnt tried the second part of iv, but for finding SP i used the locus definition of a hyperbola and got the result but i ended up with a y still hanging around. Help would be much appreciated

Hey there!

The absence of any y-value indicates to me that you should be substituting it away in terms of x1 in some way ie.


If you still need help with the second part of iv), would be glad to help :)
Title: Re: 4U Maths Question Thread
Post by: wlam on July 17, 2019, 09:58:36 pm
Hi!!

Can somebody work out how to do part (a)?

Thanks  :)
Title: Re: 4U Maths Question Thread
Post by: AlphaZero on July 18, 2019, 12:41:40 am
Hi!!

Can somebody work out how to do part (a)?

Thanks  :)

If it helps, think about this in two dimensions.

What is the length of a chord of a circle of radius \(r\) situated a perpendicular distance of \(y\) from the centre of the circle?

The area of the square will be the square of this quantity.

In the image below, you are looking down the line segment \(BOA\).

(https://i.imgur.com/uUctRJ7.png)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 19, 2019, 04:06:55 pm
Hi!!

Can somebody work out how to do part (a)?

Thanks  :)
In the future, please provide any relevant understanding/working you've already attempted, or a final result to give a direction to head towards.

(https://i.imgur.com/dTQksfg.png)

The key bit was perhaps identifying AlphaZero's diagram. It's basically the quick tip I (tried to) mention in my July lecture regarding perspectives.
Title: Re: 4U Maths Question Thread
Post by: diggity on July 28, 2019, 03:16:27 pm
Hi! Was working through some old trials my school gave me and encountered this:
(https://cdn.discordapp.com/attachments/426624789084635149/604901810729123863/IMG_20190728_150150.jpg)
I'm infamously bad with these sorts of questions, so I approached it the best I could, forming two equations;



However, I was unable to get the desired result. After going back later and looking at the given solution, they seemed to completely ignore F and got these equations:



Which they could easily get the answer from. My question is, why use these equations instead of the one I initially used? Why is F ignored; is it because of the phrase "no tendency to slip sideways"?
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on July 28, 2019, 04:21:53 pm
Why is F ignored; is it because of the phrase "no tendency to slip sideways"?

Pretty much answered your own question :)

Basically the force F acting down the slope is going to be some form of motion resistance (basically friction every time). Since you're told that there's no tendency for the particle to slip up and down the slope this implies there is no such force F that exists ie. the only forces you consider are the weight force, and the normal force.

Hope this helps :)
Title: Re: 4U Maths Question Thread
Post by: diggity on July 28, 2019, 06:05:15 pm
Pretty much answered your own question :)

Basically the force F acting down the slope is going to be some form of motion resistance (basically friction every time). Since you're told that there's no tendency for the particle to slip up and down the slope this implies there is no such force F that exists ie. the only forces you consider are the weight force, and the normal force.

Hope this helps :)

So then the F in the diagram is essentially a red herring?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on July 28, 2019, 06:17:30 pm
So then the F in the diagram is essentially a red herring?
For that question, or at least that specific part, yes.

In other questions, the \(F\) may be important.
Title: Re: 4U Maths Question Thread
Post by: diggity on July 28, 2019, 06:27:39 pm
For that question, or at least that specific part, yes.

In other questions, the \(F\) may be important.

Yeah, the other question has the same deal. Those darn exam writers >:( Thank you both though!
Title: Re: 4U Maths Question Thread
Post by: louisaaa01 on August 07, 2019, 07:50:57 pm
Can someone please help with this question?

P(2p,2/p) and Q(2q, 2/q) are two points on the rectangular hyperbola xy=4. M is the midpoint of PQ.

Find the equation of the locus of M if PQ is a tangent to the parabola y2=4x

Essentially, I determined the locus to be y2=-x/4 (not sure if this is right as I don't have solutions). However, I'm conscious of the fact that there may be restrictions, but I can't seem to find what these may be. How do I determine the restrictions, if any?

Thank you.
Title: Re: 4U Maths Question Thread
Post by: not a mystery mark on August 07, 2019, 08:38:43 pm
Weird math question my teacher brought up today.
A numbers, 1, 2, 3, and 4 are rearranged to create a 4-digit number.

a) How many total numbers can you create if no numbers are repeated.
b) What is the sum of all possible 4-digit numbers?

I guess for (b) you can just list them all, but I was wondering if there was a quicker, more "mathematical" way of solving this.


**Edited spelling.
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on August 07, 2019, 09:23:23 pm
Weird math question my teacher brought up today.
A numbers, 1, 2, 3, and 4 are rearranged to create a 4-digit number.

a) How many total numbers can you create if no number are repeated.
b) What is the sum of all possible 4-digit numbers?

I guess for (b) you can just list them all, but I was wondering if there was a quicker, more "mathematical" way of solving this.

All possible 4-digit numbers would range from 1111 to 4444. As such there 256 numbers, and there is 1/4 chance any single digit appears in a certain place value. ie the answer will be 64 x (4000+3000+2000+1000+400+300+200+100+40+30+20+10+4+3+2+1).

Hope this helps :)
Title: Re: 4U Maths Question Thread
Post by: not a mystery mark on August 07, 2019, 09:44:38 pm
All possible 4-digit numbers would range from 1111 to 4444. As such there 256 numbers, and there is 1/4 chance any single digit appears in a certain place value. ie the answer will be 64 x (4000+3000+2000+1000+400+300+200+100+40+30+20+10+4+3+2+1).

Hope this helps :)

For this question the numbers aren't repeating, such returning number such as 1234, 1243, 1324... such returning 4! or 24 numbers.
But, I've used the logic from your answer because I'm pretty sure it still works here to get.
6 x (4000+3000+2000+1000+400+300+200+100+40+30+20+10+4+3+2+1)

Thanks heaps. It did help haha.
Title: Re: 4U Maths Question Thread
Post by: Aaron Lillis on August 07, 2019, 11:27:16 pm
Can someone please help with this question?

P(2p,2/p) and Q(2q, 2/q) are two points on the rectangular hyperbola xy=4. M is the midpoint of PQ.

Find the equation of the locus of M if PQ is a tangent to the parabola y2=4x

Essentially, I determined the locus to be y2=-x/4 (not sure if this is right as I don't have solutions). However, I'm conscious of the fact that there may be restrictions, but I can't seem to find what these may be. How do I determine the restrictions, if any?

Thank you.

That locus is correct, I think i did that question in a cssa paper. Does the question ask to specify restrictions? I don't see what the restrictions would be as you've eliminated the parameters
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 08, 2019, 09:33:58 am
Can someone please help with this question?

P(2p,2/p) and Q(2q, 2/q) are two points on the rectangular hyperbola xy=4. M is the midpoint of PQ.

Find the equation of the locus of M if PQ is a tangent to the parabola y2=4x

Essentially, I determined the locus to be y2=-x/4 (not sure if this is right as I don't have solutions). However, I'm conscious of the fact that there may be restrictions, but I can't seem to find what these may be. How do I determine the restrictions, if any?

Thank you.
That locus is correct, I think i did that question in a cssa paper. Does the question ask to specify restrictions? I don't see what the restrictions would be as you've eliminated the parameters
My working out and GeoGebra are claiming \(y^2 = -\frac{x}{2}\). Note that the line(s) demonstrating that \(PQ\) is a tangent to \(y^2=4x\) is hidden, but you can put it back in there.
\[m_{PQ} = \frac{\frac2q-\frac2p}{2q-2p}=-\frac1{pq}\\ \text{so }PQ\text{ has equation }x+pqy=2(p+q).\]
\[ \text{For this to be tangential to }y^2=4x\text{, upon substitution we'd require}\\ \text{the point of intersection to satisfy}\\ \frac{y^2}{4}+pqy-2(p+q)=0. \]
\[ \text{The tangent will only have one point of intersection with the parabola}\\ \text{and hence this quadratic in }y\text{ has only one solution.}\\ \text{Setting the discriminant equal to 0 gives the constraint}\\ \boxed{p^2q^2+2(p+q) = 0}. \]
\[ \text{The coordinates of }M\text{ satisfy}\\ \begin{align*}x&= \frac{2p+2q}{2}=p+q\\ y&= \frac{\frac2p+\frac2q}2 = \frac{q+p}{pq}\end{align*} \]
\[ \text{Hence to find the locus of }M\text{, we first see that}\\ y=\frac{x}{pq} \implies pq = \frac{x}{y}. \]
\[ \text{Subbing this and }x=p+q\text{ into the constraint found above gives}\\ \frac{x^2}{y^2}+2x=0 \implies\boxed{ y^2 = -\frac{x}{2}} \]
Restrictions on this locus are not so trivially found. First note that the relationship between \(p\) and \(q\), when treated as a quadratic in \(q\), gives \( q=\frac{-1\pm \sqrt{1-2p^3}}{p^2}\). The expression under the square root shows that \(p < 2^{-1/3}\).

Close examination of the GeoGebra output shows that the restrictions on the locus appear to differ, depending on if the positive or negative root was taken as the correct one. In general, restrictions do apply as the locus will not include every point on its defining equation.

Here, taking the positive root appears to force the condition that \(y > 0\), i.e. we only consider the upper branch of the parabola \(y^2=-\frac{x}{2}\).

Taking the negative root gives a condition very difficult to describe. There are subcases to consider. When \(p < 0\), it can be shown that the locus is restricted to \(y <- 2^{-1/3}\). The significance of \(-2^{-1/3}\) is that it is the point of intersection between the locus \(y^2=-\frac{x}{2}\) and the original hyperbola \(xy=4\). On the other hand when \(0 < p < 2^{1/3}\), it instead appears as though the locus will always be on the upper branch of the parabola, with a new condition that \(x < -2^{-1/3} \) instead.

So overall (combining the two), the locus would be \(y^2=-\frac{x}{2}\), subject to the restriction that the parabolic arc from \( \left(-2^{5/3}, -2^{1/3}\right) \) to the origin \((0,0)\) is excluded. The exclusion includes the origin, but not the other endpoint.

The main thing making this so awkward is that sometimes \(P\) and \(Q\) will be in the same quadrant, and other times they won't be. Furthermore, each value of \(p\) corresponds to two values of \(q\), and it appears as though the algebra works out differently depending on which is taken. For that reason, if you were asked to find the restrictions in an exam, you would be given hints. In general, excluded points on the locus do not need to be found unless the examination has specifically requested for it.
Title: Re: 4U Maths Question Thread
Post by: diggity on August 10, 2019, 03:09:42 pm
Hello!

For this question:

"A ball of mass m is projected vertically upwards with speed u. The acceleration acting against the ball is proportional to it speed. Find the time (t) taken to reach the greatest height."

I'm simply confused on how to resolve forces. I resolved them like this: "ma = -mg - kv",  but the answers resolved them as "a = -g - kv"

Why is that the correct resolution? Thanks in advance!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 10, 2019, 03:53:43 pm
Hello!

For this question:

"A ball of mass m is projected vertically upwards with speed u. The acceleration acting against the ball is proportional to it speed. Find the time (t) taken to reach the greatest height."

I'm simply confused on how to resolve forces. I resolved them like this: "ma = -mg - kv",  but the answers resolved them as "a = -g - kv"

Why is that the correct resolution? Thanks in advance!
Considering they said acceleration acting against the ball instead of the "force", \(a=-g-kv\) would be the correct answer.
Title: Re: 4U Maths Question Thread
Post by: blyatman on August 10, 2019, 10:15:41 pm
Hello!

For this question:

"A ball of mass m is projected vertically upwards with speed u. The acceleration acting against the ball is proportional to it speed. Find the time (t) taken to reach the greatest height."

I'm simply confused on how to resolve forces. I resolved them like this: "ma = -mg - kv",  but the answers resolved them as "a = -g - kv"

Why is that the correct resolution? Thanks in advance!

They're both correct. However, the force balance equation should always be the first line in any of these types of problems. The equations of motion are always derived from the force balance equation. In this case, "ma = -mg - kv", which means that a = -g-(k/m)*v. Note that k/m is still just a proportionality constant, so you could replace k/m with another constant, called k, for simplicity.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 11, 2019, 08:48:17 am
They're both correct. However, the force balance equation should always be the first line in any of these types of problems. The equations of motion are always derived from the force balance equation. In this case, "ma = -mg - kv", which means that a = -g-(k/m)*v. Note that k/m is still just a proportionality constant, so you could replace k/m with another constant, called k, for simplicity.
Although this makes perfect sense in the real world, from what I have seen they are far more specific about things in the HSC (in what they're looking for). The questions are worded so that students will always go along the exact same line of thinking, with the exact same constant \(k\) in mind. (As opposed to swapping it out for another.)

If we want to follow the rule that the force resolution should be the start, if anything they would like you to turn the resistance into \(mkv\). And then write \(ma = -mg-mkv\).
Title: Re: 4U Maths Question Thread
Post by: blyatman on August 11, 2019, 10:56:35 am
Although this makes perfect sense in the real world, from chat I have seen they are far more specific about things in the HSC (in what they're looking for). The questions are worded so that students will always go along the exact same line of thinking, with the exact same constant \(k\) in mind. (As opposed to swapping it out for another.)

If we want to follow the rule that the force resolution should be the start, if anything they would like you to turn the resistance into \(mkv\). And then write \(ma = -mg-mkv\).

Yeh I'm aware of that, but in the question it doesn't state what the propotionality constant is, so it should be perfectly valid to use k/m and then combine that into one final constant. They shouldn't lose marks for it - I remember doing this back in the day and it was fine, according to my teacher as well as my friends teachers.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 18, 2019, 11:26:07 am
Yeh I'm aware of that, but in the question it doesn't state what the propotionality constant is, so it should be perfectly valid to use k/m and then combine that into one final constant. They shouldn't lose marks for it - I remember doing this back in the day and it was fine, according to my teacher as well as my friends teachers.
What concerns me about this is that whilst it may have been fine once upon a time, marking guidelines may fluctuate in how strict they are. (For example, stuff like L’Hopitals rule is instantly penalised now.)

It feels too much like gambling on chance to rely on these handwaves. Although I can respect what teachers have told you, and that they probably still are examiners, this would be something I want to hear from this year’s examiners first-hand as well. I’ve never been comfortable recommending tricks that could potentially be taken as a stretch unless I have first-hand permission granted with it.
Title: Re: 4U Maths Question Thread
Post by: blyatman on August 18, 2019, 01:13:53 pm
What concerns me about this is that whilst it may have been fine once upon a time, marking guidelines may fluctuate in how strict they are. (For example, stuff like L’Hopitals rule is instantly penalised now.)

It feels too much like gambling on chance to rely on these handwaves. Although I can respect what teachers have told you, and that they probably still are examiners, this would be something I want to hear from this year’s examiners first-hand as well. I’ve never been comfortable recommending tricks that could potentially be taken as a stretch unless I have first-hand permission granted with it.
Fair enough. I wasn't even aware that there was ever a time where L'Hopitals rule was acceptable lol.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on August 18, 2019, 01:27:44 pm
Fair enough. I wasn't even aware that there was ever a time where L'Hopitals rule was acceptable lol.
Oddly enough, once upon a time it was ok. But the stance now is strictly against it and I think it's a good call by NESA here.

Title: Re: 4U Maths Question Thread
Post by: david.wang28 on August 25, 2019, 08:47:48 pm
Hello,
I am stuck on an integration question in the attachment below. I solved it, but I can't seem to find errors in my working out even though the answer required is different. Can anyone please help me out? Thanks, David.
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on August 26, 2019, 09:11:15 am
Hey there!

Firstly, I think the reason you're not getting to the correct solution is because your dv is incorrect!! Remember that adding one to the power, dividing by the new power and the derivative doesn't actually work unless you have a linear function (which is what I think you did there.)

An alternative solution is below:



If there's anything you don't understand, please ask further :)

Hope this helps :)
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on August 30, 2019, 06:00:23 pm
Hey there!

Firstly, I think the reason you're not getting to the correct solution is because your dv is incorrect!! Remember that adding one to the power, dividing by the new power and the derivative doesn't actually work unless you have a linear function (which is what I think you did there.)

An alternative solution is below:



If there's anything you don't understand, please ask further :)

Hope this helps :)
Thank you so much! Sorry for the late reply; I had trials.
Title: Re: 4U Maths Question Thread
Post by: classof2019 on September 13, 2019, 07:06:38 pm
Can someone please help?

The numbers 5,6,7,8 and 9 are chosen without replacement and arranged to form a five-digit number. Find the probability that the number formed has the three odd digits in increasing numerical order.

This is what I've attempted to reason (though it's wrong).

Since the 5, 7, and 9 are increasing numerical order, let them be fixed. We're essentially arranging the 6 and 8 around them. Let crosses denote possible positions for 6 and 8.

I ended up with: XX5XX7XX9XX

I then said that the number of arrangements was 8P2 as we're choosing two X's then arranging them around the 5, 7 and 9 - then divide by the total arrangements (5!) to get the answer. I've used similar logic in other questions and have gotten them right, however in this case this is wrong.

Just wondering, why am I wrong/ how do I go about getting the right answer?

Thank you.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 13, 2019, 08:39:28 pm
Can someone please help?

The numbers 5,6,7,8 and 9 are chosen without replacement and arranged to form a five-digit number. Find the probability that the number formed has the three odd digits in increasing numerical order.

This is what I've attempted to reason (though it's wrong).

Since the 5, 7, and 9 are increasing numerical order, let them be fixed. We're essentially arranging the 6 and 8 around them. Let crosses denote possible positions for 6 and 8.

I ended up with: XX5XX7XX9XX

I then said that the number of arrangements was 8P2 as we're choosing two X's then arranging them around the 5, 7 and 9 - then divide by the total arrangements (5!) to get the answer. I've used similar logic in other questions and have gotten them right, however in this case this is wrong.

Just wondering, why am I wrong/ how do I go about getting the right answer?

Thank you.
The mistake was in how you wrote it: xx5xx7xx9xx.

Consider the following. On one hand, we can do 6x58x7xx9xx. This would give us the number 65879.

On another hand, we can do x65x87xx9xx. This is a different arrangement in how you did 8P2, but observe that it also gives 65879! Hence the number 65879 has been double counted.

The flaw was that you focused too much on the possibility of something like 68579, which actually escapes any double counting. But it only does so because the even numbers got grouped between two odd numbers. It doesn’t work when the even numbers are split.

Instead, a better option would be the following. First arrange the 6 by picking one of the 4 crosses in: x5x7x9x.

Then arrange the 8. Note that now that the 6 has been arranged, we’d have something like x6x5x7x9x. Therefore we pick one out of 5 crosses for the 8.

In total, we therefore have \(4\times5=20\) favourable arrangements, giving a probability of \(\frac{20}{120}=6\).

Note: there is also a clever approach here. Recall the problem of arranging the letters ABCDEFG, if say the E must come before the F. But you can match each arrangement with E before F with another one with F before E. (For example, match BCDEFGA with BCDFEGA.) So exactly half of the arrangements have E before F, I.e. \(\frac{7!}{2}\) arrangements.

Here, the idea is that we can group any configuration of the odd numbers among each other. Take for example 65879. We can match it with:
- 65897
- 67859
- 67895
- 69875
- 69857

So each number with 5,7,9 in order can be matched with exactly five other numbers, without changing the position of the 6 and the 8, but worn 5,7,9 not in order. Hence only one sixth of arrangements have 5,7,9 in order.


Nevertheless, that was a nice idea with the stars and bars like idea. Just executed with a flaw!
Title: Re: 4U Maths Question Thread
Post by: classof2019 on September 13, 2019, 08:57:19 pm
The mistake was in how you wrote it: xx5xx7xx9xx.

Consider the following. On one hand, we can do 6x58x7xx9xx. This would give us the number 65879.

On another hand, we can do x65x87xx9xx. This is a different arrangement in how you did 8P2, but observe that it also gives 65879! Hence the number 65879 has been double counted.

The flaw was that you focused too much on the possibility of something like 68579, which actually escapes any double counting. But it only does so because the even numbers got grouped between two odd numbers. It doesn’t work when the even numbers are split.

Instead, a better option would be the following. First arrange the 6 by picking one of the 4 crosses in: x5x7x9x.

Then arrange the 8. Note that now that the 6 has been arranged, we’d have something like x6x5x7x9x. Therefore we pick one out of 5 crosses for the 8.

In total, we therefore have \(4\times5=20\) favourable arrangements, giving a probability of \(\frac{20}{120}=6\).

Note: there is also a clever approach here. Recall the problem of arranging the letters ABCDEFG, if say the E must come before the F. But you can match each arrangement with E before F with another one with F before E. (For example, match BCDEFGA with BCDFEGA.) So exactly half of the arrangements have E before F, I.e. \(\frac{7!}{2}\) arrangements.

Here, the idea is that we can group any configuration of the odd numbers among each other. Take for example 65879. We can match it with:
- 65897
- 67859
- 67895
- 69875
- 69857

So each number with 5,7,9 in order can be matched with exactly five other numbers, without changing the position of the 6 and the 8, but worn 5,7,9 not in order. Hence only one sixth of arrangements have 5,7,9 in order.


Nevertheless, that was a nice idea with the stars and bars like idea. Just executed with a flaw!

Thank you so much! This really helps.

Just a thought, and I'm not sure whether this is purely coincidental, but could we potentially say that all of these 5 digit numbers have some sort of arrangement of 5,7, and 9. Now, 5,7 and 9 can be arranged in 3! = 6 ways. Only one such arrangement is them in ascending order (5,7,9). So the probability is 1/6.

Is this a mathematically sound justification or a coincidence?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 13, 2019, 09:10:14 pm
Thank you so much! This really helps.

Just a thought, and I'm not sure whether this is purely coincidental, but could we potentially say that all of these 5 digit numbers have some sort of arrangement of 5,7, and 9. Now, 5,7 and 9 can be arranged in 3! = 6 ways. Only one such arrangement is them in ascending order (5,7,9). So the probability is 1/6.

Is this a mathematically sound justification or a coincidence?
That is perfectly mathematically sound. It’s just the more cleaner way of expressing my quicker way :)
Title: Re: 4U Maths Question Thread
Post by: Ollierobb1 on September 15, 2019, 10:19:01 am
Given that  zn+z-n=2cosnx, find the solutions for 2z4+3z3+5z2+3z+2=0. Help would be much appreciated :)
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on September 15, 2019, 12:42:24 pm
Hey there!

In this sort of situation, we want to use the result given in some shape or form, so we try to manipulate any expressions to 'match up' with the given result.
For 2z4+3z3+5z2+3z+2=0, note that  z2(2z2+3z+5+3z-1+2z-2)=0.
Then, we have z2(4cos2x+6cosx+5)=0, using the result that zn+z-n=2cosnx where z=cisx. Clearly, from the original expression, z cannot equal zero, so we essentially just have a quadratic in cos x ie. 8cos2x+6cosx+1=0. Solving for cosx, we have that


Hope this helps :)
Title: Re: 4U Maths Question Thread
Post by: worldno1 on September 17, 2019, 08:13:58 pm
Hey guys,

For this question:  An urn contains n red balls, n white balls and n blue balls. Three balls are drawn at random from the urn, one at a time, without replacement. What is the probability  that the three balls are all the same colour?

Why is the sample space (3N)C3, instead of (3N)C1 * (3N-1)C1 * (3N-2)C1?

My logic was that, since each ball was taken out one at a time, the total number of balls available would decrease by one each time. Why is this not so?

Thanks in advance! :) Also, sorry, I don't know how LaTex works.
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on September 17, 2019, 10:55:59 pm


Basically you have when you're picking for the second and third colour you're still allowing yourself to pick from colours that you can no longer pick from ie. with the 3n-1. You're still after all just picking 3 balls from 3n balls.

Hope this helps :)
Title: Re: 4U Maths Question Thread
Post by: mani.s_ on September 18, 2019, 09:17:36 pm
Will you guys make a new Maths Extension 2 course notes for the new syllabus???
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 18, 2019, 09:19:37 pm
Will you guys make a new Maths Extension 2 course notes for the new syllabus???
Yep.

Sadly, they won't be out in time for the October head-start lectures due to other commitments I have with the company. :(
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on September 20, 2019, 05:19:31 pm
Hello, I have trouble working out this integral. My working out is attached, the answer is 4𝜋/27 - √3 /9. Can anyone please help me out? Thanks :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 20, 2019, 05:55:59 pm
Hello, I have trouble working out this integral. My working out is attached, the answer is 4𝜋/27 - √3 /9. Can anyone please help me out? Thanks :)
Your mistake was pretty deep in actually. It was in the transition from line 7 to line 8, next to the blue text.
\[ \text{Notice how you wrote}\\ \int_{-\frac1{\sqrt3}}^{\frac1{\sqrt3}} \frac{dx}{1+9x^2} = \int_{-\frac1{\sqrt3}}^{\frac1{\sqrt3}} \frac{dx}{\frac19 + x^2}\\ \text{without appropriately adjusting in the numerator as well!} \]
\[ \text{The correct change is}\\ \int_{-\frac1{\sqrt3}}^{\frac1{\sqrt3}} \frac{dx}{1+9x^2} = \int_{-\frac1{\sqrt3}}^{\frac1{\sqrt3}} \frac{\frac19\, dx}{\frac19 + x^2}\]
Title: Re: 4U Maths Question Thread
Post by: david.wang28 on September 20, 2019, 06:40:31 pm
Your mistake was pretty deep in actually. It was in the transition from line 7 to line 8, next to the blue text.
\[ \text{Notice how you wrote}\\ \int_{-\frac1{\sqrt3}}^{\frac1{\sqrt3}} \frac{dx}{1+9x^2} = \int_{-\frac1{\sqrt3}}^{\frac1{\sqrt3}} \frac{dx}{\frac19 + x^2}\\ \text{without appropriately adjusting in the numerator as well!} \]
\[ \text{The correct change is}\\ \int_{-\frac1{\sqrt3}}^{\frac1{\sqrt3}} \frac{dx}{1+9x^2} = \int_{-\frac1{\sqrt3}}^{\frac1{\sqrt3}} \frac{\frac19\, dx}{\frac19 + x^2}\]
Ahhh I see, thank you :)
Title: Re: 4U Maths Question Thread
Post by: louisaaa01 on September 27, 2019, 10:18:36 pm
Can someone please help with part (iii) for this question? Thank you.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on September 28, 2019, 07:09:23 am
Can someone please help with part (iii) for this question? Thank you.
Edit: At later glance I realised I slightly messed up the base.
\[ \text{When }n=0,\\ \text{the only configurations for }(r,s)\text{ are }(1,0)\text{ and }(0,1).\\ \text{So we can just check each individually.}\]
\[ \text{For the }(0,1)\text{ case, we have }P(0,1)=0.\\ \text{Here }s=1\text{, so we verify that}\\ \frac{1}{2^1}\left[ \binom10 + \binom11 \right] =\frac22 = 1 \]
\[ \text{For the }(1,0)\text{ case, we have }P(1,0) = 0.\\ \text{Here }s=0\text{, which is outside the valid range }s\geq 1.\\ \text{Hence this does not contradict the inductive assumption anyhow.} \]
So in both cases, the base case \(n=0\) is valid.
_________________________________________________________________________________
\[ \text{Now assume the statement holds }\textbf{for all}\text{ configurations }(r,s)\\ \text{such that }r+s-1=k. \\ \text{We then need to prove the statement holds for all configurations }(r,s)\\ \text{such that } r+s-1=k+1. \]
That was perhaps the most confusing part of the induction - figuring out the goal of the inductive step.
\[ \text{Firstly, from the definition of the recurrence,}\\ P(r,s) = \frac12 P(r-1, s) + \frac12 P(r,s-1).  \]
\[ \text{Now if }r+s-1 = k+1\text{, we know that}\\ (r-1)+s-1 = k\text{ and } r + (s-1) - 1= k.\\ \text{Hence we may use the inductive assumption on }\textit{both}\text{ terms:} \]
\begin{align*}
P(r,s) &=  \frac12 P(r-1, s) + \frac12 P(r,s-1)\\
&= \frac12 \frac1{2^n} \left[ \binom{k}0 + \binom{k}1 + \dots + \binom{k}{s-2}+ \binom{k}{s-1} \right] + \frac12 \frac1{2^n} \left[ \binom{k}0 + \binom{k}1 + \dots + \binom{k}{s-2} \right] \\
&= \frac1{2^{n+1}} \left[ \binom{k}0 + \binom{k}1 + \dots + \binom{k}{s-2}+ \binom{k}{s-1} \right] +\frac1{2^{n+1}} \left[ \binom{k}0 + \binom{k}1 + \dots + \binom{k}{s-2} \right]
\end{align*}
Ensure that you understand why the first sum terminates at \(s-1\), whilst the second terminates at \(s-2\).
\[ \text{Now to finish, the tricks here are similar to the induction proof of the binomial theorem.}\\ \text{We need to recall that firstly, }\binom{k}{0} = \binom{k+1}{0} = 1.\\ \text{Also, we need Pascal's identity }\boxed{\binom{N+1}{K+1} = \binom{N}{K} + \binom{N}{K+1}}.\\ \text{With }\textit{both}\text{ properties in mind, we proceed to cleverly rearrange the terms.} \]
\begin{align*}
P(r,s) &= \frac1{2^{n+1}} \left[ \binom{k}{0} + \left[ \binom{k}0 + \binom{k}1\right] + \left[ \binom{k}1+ \binom{k}2 \right]+ \dots + \left[\binom{k}{s-2}+\binom{k}{s-1} \right] \right]\\
&= \frac1{2^{n+1}} \left[ \binom{k+1}{0} + \binom{k+1}{1}+\binom{k+1}{2} + \dots + \binom{k+1}{s-1} \right].
\end{align*}
And this is exactly what we wished to prove, so we are done.

*I did this in a bit of a rush so I may have given too little clarification. Let me know if there's any bit you want me to elaborate on
Title: Re: 4U Maths Question Thread
Post by: louisaaa01 on September 28, 2019, 11:56:42 am
\[ \text{When }n=1,\\ \text{the only configurations for }(r,s)\text{ are }(1,0)\text{ and }(0,1).\\ \text{In both cases, we know that }P(r,s) = 1\text{ from the question.} \]
\[\text{And also, }\\ \frac{1}{2^1}\left[ \binom10 + \binom11\right] = \frac22 = 1.\\ \text{Hence the statement holds for }n=1. \]
_________________________________________________________________________________
\[ \text{Now assume the statement holds }\textbf{for all}\text{ configurations }(r,s)\\ \text{such that }r+s-1=k. \\ \text{We then need to prove the statement holds for all configurations }(r,s)\\ \text{such that } r+s-1=k+1. \]
That was perhaps the most confusing part of the induction - figuring out the goal of the inductive step.
\[ \text{Firstly, from the definition of the recurrence,}\\ P(r,s) = \frac12 P(r-1, s) + \frac12 P(r,s-1).  \]
\[ \text{Now if }r+s-1 = k+1\text{, we know that}\\ (r-1)+s-1 = k\text{ and } r + (s-1) - 1= k.\\ \text{Hence we may use the inductive assumption on }\textit{both}\text{ terms:} \]
\begin{align*}
P(r,s) &=  \frac12 P(r-1, s) + \frac12 P(r,s-1)\\
&= \frac12 \frac1{2^n} \left[ \binom{k}0 + \binom{k}1 + \dots + \binom{k}{s-2}+ \binom{k}{s-1} \right] + \frac12 \frac1{2^n} \left[ \binom{k}0 + \binom{k}1 + \dots + \binom{k}{s-2} \right] \\
&= \frac1{2^{n+1}} \left[ \binom{k}0 + \binom{k}1 + \dots + \binom{k}{s-2}+ \binom{k}{s-1} \right] +\frac1{2^{n+1}} \left[ \binom{k}0 + \binom{k}1 + \dots + \binom{k}{s-2} \right]
\end{align*}
Ensure that you understand why the first sum terminates at \(s-1\), whilst the second terminates at \(s-2\).
\[ \text{Now to finish, the tricks here are similar to the induction proof of the binomial theorem.}\\ \text{We need to recall that firstly, }\binom{k}{0} = \binom{k+1}{0} = 1.\\ \text{Also, we need Pascal's identity }\boxed{\binom{N+1}{K+1} = \binom{N}{K} + \binom{N}{K+1}}.\\ \text{With }\textit{both}\text{ properties in mind, we proceed to cleverly rearrange the terms.} \]
\begin{align*}
P(r,s) &= \frac1{2^{n+1}} \left[ \binom{k}{0} + \left[ \binom{k}0 + \binom{k}1\right] + \left[ \binom{k}1+ \binom{k}2 \right]+ \dots + \left[\binom{k}{s-2}+\binom{k}{s-1} \right] \right]\\
&= \frac1{2^{n+1}} \left[ \binom{k+1}{0} + \binom{k+1}{1}+\binom{k+1}{2} + \dots + \binom{k+1}{s-1} \right].
\end{align*}
And this is exactly what we wished to prove, so we are done.

*I did this in a bit of a rush so I may have given too little clarification. Let me know if there's any bit you want me to elaborate on

Thank you!
Title: Re: 4U Maths Question Thread
Post by: not a mystery mark on October 01, 2019, 11:36:18 am
Hey, would somebody be able to explain why the answer to this question is C? I feel myself going insane and hope somebody can assist me in my time of lunacy. Thanks heaps!
Title: Re: 4U Maths Question Thread
Post by: louisaaa01 on October 01, 2019, 11:48:56 am
Hey, would somebody be able to explain why the answer to this question is C? I feel myself going insane and hope somebody can assist me in my time of lunacy. Thanks heaps!

Hey not a mystery mark!

So basically the way we approach this question is to split our F and N vectors into vertical and horizontal components.

Firstly, we know that F is inclined at θ to the horizontal, thus the horizontal component of F is Fcosθ and vertical component is Fsinθ.

On the other hand, N is at an angle of θ to the vertical (applying geometry - N is perpendicular to F). So the horizontal component of N is Nsinθ and the vertical component is Ncosθ

Now, we consider the net horizontal and vertical forces on C.

Since the particle isn't going anywhere vertically, we can conclude that all vertical forces are balanced. Since F and N are both pointing upwards, this means that: Fsinθ + Ncosθ = mg.

Horizontally, the particle is undergoing uniform circular motion - the net force is mrw2. Since F and N are in opposite directions, labelling the direction towards the centre as positive, we can say that:

Fcosθ - Nsinθ = mrw2

Which gives us C.

Hope this helps - if you need any further clarification, let me know!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 01, 2019, 11:54:31 am
Hey, would somebody be able to explain why the answer to this question is C? I feel myself going insane and hope somebody can assist me in my time of lunacy. Thanks heaps!
Above explanation should sum it up pretty much. Here's a diagram to go with it.

(https://i.imgur.com/tPY4uLb.jpg)
Title: Re: 4U Maths Question Thread
Post by: not a mystery mark on October 01, 2019, 12:31:52 pm
Above explanation should sum it up pretty much. Here's a diagram to go with it.

I don't know how, but you made that so much clearer. Thank you so much. I never understood it in class but this has diamond clarity now.
Thanks so much!
Title: Re: 4U Maths Question Thread
Post by: duncand on October 02, 2019, 03:36:25 pm
hi, I am not sure the answers for this trial paper for part ii)
 went from the 4th last line to the last line
(or from the 2nd tick to the third tick)
thanks. 8)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 02, 2019, 06:37:13 pm
hi, I am not sure the answers for this trial paper for part ii)
 went from the 4th last line to the last line
(or from the 2nd tick to the third tick)
thanks. 8)
This requires a somewhat combinatorial argument to justify.
\[ \text{Firstly, using the AM-GM inequality}\]\begin{align*}
 \sum_{1\leq i < j \leq n} 2\sqrt{\binom{n}{i}}\sqrt{\binom{n}{j}}  &= \sum_{1\leq i<j\leq n} 2\sqrt{\binom{n}i\binom{n}j}\\
&= \sum_{1\leq i < j \leq n} \left[\binom{n}{i} + \binom{n}{j} \right]
\end{align*}
\[ \text{When expanding this sum, we obtain}\\ \left[\binom{n}1+\binom{n}2 \right]+ \left[\binom{n}1+\binom{n}3 \right] + \left[\binom{n}1+\binom{n}4\right]+ \cdots + \left[ \binom{n}1+\binom{n}n\right]\\ \qquad \qquad\qquad+ \left[\binom{n}2+\binom{n}3 \right] + \left[\binom{n}2+\binom{n}4\right]+ \cdots + \left[ \binom{n}2+\binom{n}n\right] \\
\qquad \qquad\qquad\qquad \qquad\qquad+ \left[\binom{n}3+\binom{n}4\right]+ \cdots + \left[ \binom{n}3+\binom{n}n\right]\\ \qquad \qquad\qquad\qquad \qquad\qquad+\ddots \\
\qquad \qquad\qquad\qquad \qquad\qquad\qquad \qquad\qquad\qquad \qquad\qquad+ \left[\binom{n}{n-1}+ \binom{n}{n}\right]\]
Basically, we had to sub (i,j) = (1,2), (1,3), (1,4), ..., (1,n),
and then (i,j) = (2,3), (2,4), ..., (2,n),
and then (i,j) = (3,4), ..., (3,n), and so on.
\[ \text{Now exactly how many times does each term appear?}\\ \text{It is hopefully clear that }\binom{n}{1}\text{ appears exactly }n-1\text{ times}\\ \text{because it's only ever in the first row, so we just count the number of pairs in the first row.} \]
\[ \text{But then, observe how }\binom{n}{2}\text{ appears once in the first row}\\ \text{followed by }n-2\text{ times in the second row.}\\ \text{And then, observe how }\binom{n}3\text{ appears once in the first and second row}\\ \text{followed by }n-3\text{ times in the third row.} \]
\[ \text{In general, every term }\binom{n}{k}\text{ will appear}\\ \text{once for each of the preceding }k-1\text{ rows}\\ \text{and then }n-k\text{ times in the }k\text{-th row.}\\ \text{(And never again).} \]
\[ \text{The consequence:}\\ \textbf{Each }\binom{n}k\text{ term always appears }\textbf{exactly }n-1\text{ times.} \]
This is the transition from the 4th last to the 3rd last line.
_________________________________________________________________________________
\[ \text{Then, following identity is used:}\\ \binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\cdots + \binom{n}{n} = 2^n.\\ \text{This is a classic MX1 question, proven by subbing }x=1\\\text{into the binomial theorem.} \]
Note: At the MX2 level, some trial papers treat this as easy to prove. And hence there are no hints provided on when it can be used.
\[ \text{But in the sum we have from pulling out the }n-1,\\ \text{our sum actually starts at }\binom{n}{1}\text{ instead.}\\ \text{Hence it effectively evaluates to }2^n - \binom{n}{0}\\ \text{and it is well known that }\binom{n}{0}=1\text{ always.} \]
And this is the transition from the 3rd last tot he 2nd last line.

The final one should be obvious.
Title: Re: 4U Maths Question Thread
Post by: duncand on October 03, 2019, 03:54:14 pm
.
wow, would of never thought of this
thanks!! :D :D :D
Title: Re: 4U Maths Question Thread
Post by: classof2019 on October 07, 2019, 03:54:58 pm
Can someone please help with part (ii) of this question?

 A group of 12 people is to be divided into discussion groups.
(i) In how many ways can the discussion groups be formed if there are
8 people in one group, and 4 people in another?
(ii) In how many ways can the discussion groups be formed if there are
3 groups containing 4 people each?

More specifically, why do we need to divide our answer by 3! instead of just doing 12C8 x 8C4 x 4C4 + why doesn't a similar division apply for part (i)?

Cheers.
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on October 07, 2019, 04:05:50 pm
Hey there!

Consider 12 people A, B, C, D, E, F, G, H, I , J, K, L.
Dividing these people into a group of 8 and a group of 4 can only be done in one way ie. 12C8 x 4C4.
However, when we divide into 3 groups of 4, we need to account for the equal group sizes ie. we can't differentiate between say ABCD EFGH IJKL and EFGH ABCD IJKL. Essentially, dividing by 3! removes the ordering of identical groupings, as it doesn't matter whether we choose a certain four people in a group first, second or third. This doesn't occur with the first case because neither group is identical to the other.

Hope this makes sense :)
Title: Re: 4U Maths Question Thread
Post by: louisaaa01 on October 10, 2019, 04:47:09 pm
Hi,

I'm wondering if anyone can provide a more efficient way to solve this question? (multiple choice Q10 from 2013 HSC)

A hostel has four vacant rooms. Each room can accommodate a maximum of four people.
In how many different ways can six people be accommodated in the four rooms?
The way I approached it was a little tedious. I basically listed the possible ways of separating 6 people into four groups (allowing groups of zero):

4, 1, 1, 0
4, 2, 0, 0
3, 1, 1, 1
3, 2, 1, 0
3, 3, 0, 0
2, 2, 2,0
2, 2, 1, 1

Then I determined the number of ways of selecting people to occupy these groups and 'arranging' these groups into the four different hotel rooms - I wound up with 4020 which is the correct answer.

However, I'm aware that a multiple-choice question should take about 1.5 minutes, so I was wondering if there's a better way to go about this question?

Thanks!
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on October 10, 2019, 05:11:49 pm
Hey there!

You're right in that there is a quicker way; usually, if you're listing all these possibilities and the exhaustive list gets a bit long, using the complement is a better idea.

What the restriction actually does is ensure you can't have 5 or 6 people in a single room ie. the combinations 5/1/0/0 and 6/0/0/0 are invalid. Noting that the total number of combinations is just 46, we subtract (4C1x6C5x3C1x1C1 + 4C1x6C6) (which are just the total number of ways of the above combinations respectively) to find the total with the restriction.

Hope this helps :)
Title: Re: 4U Maths Question Thread
Post by: louisaaa01 on October 10, 2019, 05:48:50 pm
Hey there!

You're right in that there is a quicker way; usually, if you're listing all these possibilities and the exhaustive list gets a bit long, using the complement is a better idea.

What the restriction actually does is ensure you can't have 5 or 6 people in a single room ie. the combinations 5/1/0/0 and 6/0/0/0 are invalid. Noting that the total number of combinations is just 46, we subtract (4C1x6C5x3C1x1C1 + 4C1x6C6) (which are just the total number of ways of the above combinations respectively) to find the total with the restriction.

Hope this helps :)

Thanks so much!
Title: Re: 4U Maths Question Thread
Post by: classof2019 on October 19, 2019, 01:43:26 pm
Hi, does anyone have any sort of tips they could offer when attempting the harder/ more abstract questions in the 4U paper (e.g. in Q16)? Especially those involving Harder 3U topics. Cheers.
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on October 19, 2019, 02:31:35 pm
Hey there!

I think people by the time they get to Q16 will have a decent amount of time to weigh up how to attack each question in Q16. Therefore, there should be time to make a few considerations :)

1. Try and identify what the question is asking you to do
A tip I've restated here on the forums a lot and that I've found really handy is to think about what you're working towards. This usually works for when you have to prove a result of some sort, and this can be done in many ways, whether it's working backwards, or converting something into a more familiar form.

2. Explore your options
If you're stuck on circle geo, the best thing you can do sometimes is just mess around and chase angles. A similar principle applies; like have a go and see where an approach takes you, and with the time you have you should have ample opportunity to try a few things.

Outside of this, it's best to just get in loads of practice for end questions, and perhaps take a look at them during reading time instead of solving all the MC in your head. Identifying what you need to do relatively early then seeing it again makes things a lot easier more often than not. As well as this, you don't actually lose much time for not solving MC in reading time, since MC (especially Q1-6/7) are usually quite quick :)

Hope this helps :)
Title: Re: 4U Maths Question Thread
Post by: louisaaa01 on October 20, 2019, 02:07:45 pm
Hi!

Can someone please help with part (v) of this question (from the 1993 MX2 HSC paper)?

Also, just a question, I've heard (but I'm not entirely sure) that the question types in the MX2 HSC changed after 2001 - what exactly does this mean? I know they introduced MC in 2012 but I heard something about a change in 2001 as well.

Thanks  :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 20, 2019, 08:08:25 pm
Hi!

Can someone please help with part (v) of this question (from the 1993 MX2 HSC paper)?

Also, just a question, I've heard (but I'm not entirely sure) that the question types in the MX2 HSC changed after 2001 - what exactly does this mean? I know they introduced MC in 2012 but I heard something about a change in 2001 as well.

Thanks  :)
The main thing is that from 2001 onwards, mark allocation for each part of a question was promised.

But apart from that, it's especially true with 4U that many past papers liked to test some perhaps common concepts, but in absurdly weird ways in the final exam. (And of course, they were notably more time consuming.) For a question like this, the notation is somewhat peculiar, and the use of graphing in ii) and iii) seems a bit out of place to me. It turns out that part v) is just a recursive argument, but I'm not sure if I've ever seen that kind of pattern in recent papers.
\[ \textbf{1993 HSC 4U Additional} \]
Note that \(n\) is a fixed integer for this question.
\[ \text{By immediately using our existence assumption that }\\ x_n = \cot \theta_n\text{ for some }0<\theta_n<\pi,\text{ we obtain} \\ \boxed{\cot \theta_1 = \cot \theta_{n+1}}. \]
\[ \text{Now upon applying the recursive relationship once, we obtain}\\ \boxed{\cot\theta_1 = \cot 2\theta_n}.\\ \text{But applying it again, we obtain}\\ \boxed{\cot \theta_1 = \cot 2^2\theta_{n-1}}.\\ \text{And just to make it clearer, explicitly applying it once more gives}\\ \boxed{\cot \theta_1= \cot 2^3 \theta_{n-2}}. \]
\[ \text{The idea is that upon further repeated use of this recursion,}\\ \text{we will eventually arrive at }\boxed{\cot \theta_1 = \cot 2^n\theta_1 }. \]
One way of tracking the required power of 2 is to observe that the power plus the index on \(\theta\) must equal to \(n+1\).
\[ \text{At this point, it is now perfectly fine for us to use the general solution}\\ \text{in an attempt to track down all solutions.}\\ \text{Solving }\cot (2^n \theta_1) = \cot\theta_1 \text{ gives}\\ 2^n\theta_1 = k\pi + \theta_1 \implies \boxed{\theta_1 = \frac{k\pi}{2^n-1}}\\ \text{where }k\text{ is an integer.} \]
Note: If you put the \(k\pi\) on the other side, you'd end up with \( \theta_1 = \frac{-k\pi}{2^n-1}\), But that's all good - just sub out \(\ell = -k\) and work with \(\ell\) instead.
\[ \text{But we need to track down which values of }k\text{ give values for }\theta_1\text{ that are not out of bounds.}\\ \text{Recall that we still require }0< \theta_1 < \pi. \]
\[ \text{Subbing into that expression gives}\\ 0 < \frac{k\pi}{2^n-1} < \pi \implies \boxed{0 < k < 2^n -1} \]
Note: Of course, we're assuming that \(2^n -1 > 0\), since \(n \geq 1\). This assumption on \(n\) is not explicitly stated in the question, but is implicitly assumed to be the lowest index.
\[ \text{Hence our valid values for }k\text{ are}\\ k = 1, 2, 3, \dots, 2^n - 2.\\ \text{Which means our final answers are}\\ x_1 = \cot \left( \frac{\pi}{2^n - 1} \right), \, \cot \left( \frac{2\pi}{2^n - 1}\right), \dots, \cot \left( \frac{ (2^n-2)\pi}{2^n -1} \right). \]

Aside: For the interested reader, this question basically shows one special class of consequences when Newton's method is applied on a function that has no roots to begin with. Newton's method is designed with the intention of converging to a root once used often enough, so if we don't have a root, well where does it converge to?

More often than not, at each iteration we just get some debatably randomised value appearing. In other words, it definitely does not converge. But this question basically asks - is it possible for the values to start cycling at some point? That is, could Newton's method somehow exhibit periodic behaviour? The question more or less answers yes to this question, but challenges us by asking what initial values must we start with, to exhibit a period of \(n\).
Title: Re: 4U Maths Question Thread
Post by: classof2019 on October 22, 2019, 03:04:33 pm
How do you go about this question?

What is the total number of different combinations that can be made using 3 of the letters of the word MADAGASCAR?

Thanks
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on October 22, 2019, 03:37:07 pm
Hey there!

When we make a selection of letters then look at combinations, it's important to note that we need to choose the letters we arrange first, before arranging them to form combinations. The issue with this question is there are duplicate A's, so we need to consider cases in terms of how many A's we get. What does make it easier is that though there are four A's, there are only ever three letters chosen, which reduces the number of cases.

First note that there is ever only one way to pick any number of A's, since they're all the same. We just need to pick the 6 other filler letters depending on how many A's we have out of three ie. if there is one A, we need two filler letters.

Thus, the number of ways will just be the number of ways to pick the filler letters, then rearrange the whole set itself - 6C3 x 3! + 6C2 x 3! + 6C1 x 3 + 6C0 x 1.

Hope this helps :)
Title: Re: 4U Maths Question Thread
Post by: rh45_21 on October 24, 2019, 12:23:12 pm
Just a question that I came across that even stumped my teacher, would appreciate the help. Its pt. iv) of the first and ii) of the second

Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 24, 2019, 01:24:42 pm
Just a question that I came across that even stumped my teacher, would appreciate the help. Its pt. iv) of the first and ii) of the second


For the first part, we just use a double angle formula on \(\cos \frac\pi5 \).
\begin{align*}
\cos\frac\pi5 \sin \frac\pi{10} &= \frac14\\
\left(1 - 2\sin^2 \frac\pi{10}\right)\sin \frac\pi{10} &= \frac14\\
\sin \frac\pi{10} - 2sin^3 \frac\pi{10} &= \frac14\\
4\sin \frac\pi{10} - 8\sin^3 \frac\pi{10} &= 1
\end{align*}
\[ \text{Hence }8\sin^3 \frac\pi{10} - 4\sin \frac\pi{10} + 1 = 0 \]
By definition, this literally means that \(x=\frac\pi{10}\) is a root of the equation \(8x^3-4x+1=0\).
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on October 24, 2019, 03:03:41 pm
Hey there!

The diagram in the spoiler might come in handy :)
Spoiler
(https://i.imgur.com/7R0MnNb.png?1)

For i), note that at height h, we have the area being equal to \(\int_{-\sqrt{h}}^{\sqrt{h}} (h-x^2) \ dx \). This will equal the value in given in the question :)

For ii), we clearly want to find the volume as x changes, and as such we need to somehow relate h to x, since the variable slice is in terms of h. Refer to the diagram in the spoiler! From similar triangles, we can see that h=x/a + 1, and thus we can evaluate the volume of the solid as below:
Title: Re: 4U Maths Question Thread
Post by: mani.s_ on October 28, 2019, 07:19:42 pm
Hi can someone help me with Question 14???

Thanks :)
Title: Re: 4U Maths Question Thread
Post by: DrDusk on October 28, 2019, 07:38:56 pm
Hi can someone help me with Question 14???

Thanks :)
Title: Re: 4U Maths Question Thread
Post by: mani.s_ on October 28, 2019, 08:22:55 pm

Hi Thanks so much for your reply!!! I just had one question, how did you deduce that a = 0 from a^2 + ai / a^2 + b^2 + 1 ???
Title: Re: 4U Maths Question Thread
Post by: louisaaa01 on October 28, 2019, 08:32:21 pm
Hi Thanks so much for your reply!!! I just had one question, how did you deduce that a = 0 from a^2 + ai / a^2 + b^2 + 1 ???

Hi!

So just following up from DrDusk's solution - we are able to deduce that a = 0 since the question specifies that the given expression, z/(z-i) is purely real.

This essentially implies that the imaginary part of the expression is zero. After realising the denominator, we look for the coefficient of i and let this equal zero. In this case, the coefficient of i (the imaginary part) is; ai / (a2 +b2 +1). Since the denominator cannot equal zero, this means that a = 0.
Title: Re: 4U Maths Question Thread
Post by: mani.s_ on October 28, 2019, 08:37:47 pm
Hi!

So just following up from DrDusk's solution - we are able to deduce that a = 0 since the question specifies that the given expression, z/(z-i) is purely real.

This essentially implies that the imaginary part of the expression is zero, that is, after realising the denominator, the coefficient of i. Since the denominator cannot equal zero, this means the coefficient of i on the numerator, in this case 'a' = 0.
I'm so sorry, but I still don't understand how we can say a =0??? I'm sorry for troubling you guys :(
Title: Re: 4U Maths Question Thread
Post by: mani.s_ on October 28, 2019, 08:46:22 pm
Hi!

So just following up from DrDusk's solution - we are able to deduce that a = 0 since the question specifies that the given expression, z/(z-i) is purely real.

This essentially implies that the imaginary part of the expression is zero. After realising the denominator, we look for the coefficient of i and let this equal zero. In this case, the coefficient of i (the imaginary part) is; ai / (a2 +b2 +1). Since the denominator cannot equal zero, this means that a = 0.
Wait I think i understand it. So basically we're solving a / a^2 + b^2 +1  =  0, since the imaginary part is 0. from this we get a = 0
Title: Re: 4U Maths Question Thread
Post by: louisaaa01 on October 28, 2019, 08:47:33 pm
Wait I think i understand it. So basically we're solving a / a^2 + b^2 +1  =  0, since the imaginary part is 0. from this we get a = 0

Yes, that's exactly right! We let the imaginary part = 0, since, if a number has no imaginary part, it must be real.
Title: Re: 4U Maths Question Thread
Post by: mani.s_ on October 28, 2019, 09:03:00 pm
I'm sorry for bothering you guys again, but can someone help me with this question?

THANK YOU SO MUCH FOR YOUR HELP!!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 28, 2019, 09:07:43 pm
I'm sorry for bothering you guys again, but can someone help me with this question?

THANK YOU SO MUCH FOR YOUR HELP!!!
\begin{align*}
z^2 &= \overline{z}^2\\
z^2 - \overline{z}^2 &= 0\\
(z+\overline{z})(z-\overline{z} )&= 0\tag{diff. of two squares}
\end{align*}
\[ \text{Let }z=x+iy.\\ \text{If }z+\overline{z}=0\text{, then }(x+iy)+(x-iy)=0\\ \text{and hence }2x = 0 \implies x = 0 \implies z\text{ is purely imaginary.} \]
The other case is handled similarly.

Alternatively, just sub \(z=x+iy\) at the start and deduce that \(2xy = 0\), so either \(x=0\) or \(y=0\).

In the future, please post any attempts you have also made in the question. Although there's nothing wrong with us helping, it's quite redundant if we're always answering the whole thing - it doesn't show us where the gap in learning is, and doesn't necessarily help it sink in either.
Title: Re: 4U Maths Question Thread
Post by: mani.s_ on November 06, 2019, 09:41:54 pm
How would I solve this question? I know that if you multiply a complex number by i, you get a rotation of pi/2. I also know that arg(zw)=arg(z)+arg(w), but how would I use these to show the working out??

Any help is appreciated :)
Title: Re: 4U Maths Question Thread
Post by: louisaaa01 on November 07, 2019, 08:30:08 pm
How would I solve this question? I know that if you multiply a complex number by i, you get a rotation of pi/2. I also know that arg(zw)=arg(z)+arg(w), but how would I use these to show the working out??

Any help is appreciated :)

Hi Mani.s_

Speaking in general terms, when you multiply a complex number by (cosθ + isinθ), you're really rotating it anticlockwise about the origin by θ radians. Indeed, a rotation by i will produce an anticlockwise rotation of π/2 since i in mod/arg form is really (cos(π/2) + isin(π/2)).

For this question, you are essentially given a value of θ. What you must do is convert cosθ + isinθ (the mod/arg form) into the Cartesian form - more generally expressed as x + iy. To do so, you use exact trigonometric values, e.g. for b, cos(π/3) + isin(π/3) becomes 1/2 + i√3/2.

I hope this helps!
Title: Re: 4U Maths Question Thread
Post by: mani.s_ on November 08, 2019, 11:05:48 pm
Hi Mani.s_

Speaking in general terms, when you multiply a complex number by (cosθ + isinθ), you're really rotating it anticlockwise about the origin by θ radians. Indeed, a rotation by i will produce an anticlockwise rotation of π/2 since i in mod/arg form is really (cos(π/2) + isin(π/2)).

For this question, you are essentially given a value of θ. What you must do is convert cosθ + isinθ (the mod/arg form) into the Cartesian form - more generally expressed as x + iy. To do so, you use exact trigonometric values, e.g. for b, cos(π/3) + isin(π/3) becomes 1/2 + i√3/2.

I hope this helps!
Thank you so much!!!
Title: Re: 4U Maths Question Thread
Post by: mani.s_ on November 08, 2019, 11:06:42 pm
Hi, can someone explain what fractals are???

Thanks :)
Title: Re: 4U Maths Question Thread
Post by: mani.s_ on November 08, 2019, 11:17:05 pm
Hi Mani.s_

Speaking in general terms, when you multiply a complex number by (cosθ + isinθ), you're really rotating it anticlockwise about the origin by θ radians. Indeed, a rotation by i will produce an anticlockwise rotation of π/2 since i in mod/arg form is really (cos(π/2) + isin(π/2)).

For this question, you are essentially given a value of θ. What you must do is convert cosθ + isinθ (the mod/arg form) into the Cartesian form - more generally expressed as x + iy. To do so, you use exact trigonometric values, e.g. for b, cos(π/3) + isin(π/3) becomes 1/2 + i√3/2.

I hope this helps!
Thank you so much!!!
Title: Re: 4U Maths Question Thread
Post by: shekhar.patel on December 18, 2019, 10:38:17 pm
I need help in this question, and at the end the messed up letter is meant be be alpha conjucate
Title: Re: 4U Maths Question Thread
Post by: DrDusk on December 18, 2019, 11:17:40 pm
I need help in this question, and at the end the messed up letter is meant be be alpha conjucate
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 18, 2019, 11:22:19 pm

Did you justify why the bar vanishes off the coefficients \(a\), \(b\) and \(c\)?
Title: Re: 4U Maths Question Thread
Post by: DrDusk on December 18, 2019, 11:23:31 pm
Did you justify why the bar vanishes off the coefficients \(a\), \(b\) and \(c\)?
I knew this was coming hahaha  ;)

Title: Re: 4U Maths Question Thread
Post by: shekhar.patel on December 19, 2019, 03:48:47 pm


Oh, I see. But how do we know that we need to sub alpha at the start. Thanks
Title: Re: 4U Maths Question Thread
Post by: RuiAce on December 19, 2019, 03:51:03 pm
Oh, I see. But how do we know that we need to sub alpha at the start. Thanks
From the fact that \(\alpha\) is by definition a root of \(ax^2+bx+c=0\).

You're not being asked to show that \(\alpha\) is a root; you're given it. (What has to be shown is that the same result holds when \(\overline{\alpha}\) is subbed in.)
Title: Re: 4U Maths Question Thread
Post by: shekhar.patel on December 19, 2019, 03:54:14 pm
From the fact that \(\alpha\) is by definition a root of \(ax^2+bx+c=0\).

You're not being asked to show that \(\alpha\) is a root; you're given it. (What has to be shown is that the same result holds when \(\overline{\alpha}\) is subbed in.)

Got it. Cheers
Title: Re: 4U Maths Question Thread
Post by: shekhar.patel on December 19, 2019, 04:06:51 pm
I have this question, and my attempted answer is below, I am not completely sure if it is the right approach.
Any assistance would be great.
Title: Re: 4U Maths Question Thread
Post by: DrDusk on December 19, 2019, 04:27:57 pm
I have this question, and my attempted answer is below, I am not completely sure if it is the right approach.
Any assistance would be great.
You can't do that because the conjugate doesn't apply to a, b and c IF they are real. We haven't proved this so you can't just use it in your proof like that. Instead you should do this:


Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on December 19, 2019, 04:30:05 pm
Hey there!

Could you possibly explain your line of thinking when writing 'since the conjugate roots only applies to imaginary components we can say a, b and c are real?'. I'm a little confused - we could properly correct/understand/explain better if you clarified this :)

Perhaps a more clear approach (alternative to the above) would be to let \(\alpha = x+iy\) and \(\bar{\alpha} = x-iy\), thus leading to the fact that both
\(a(x+iy)^2+b(x+iy)+c=0\) and \(a(x-iy)^2+b(x-iy)+c=0\) are true. Adding these up, we see that we have \(2a(x^2-y^2)+2bx+c=0\) - given a, x and y are real, b and c must also be real.

Hope this helps :)
Title: Re: 4U Maths Question Thread
Post by: shekhar.patel on December 19, 2019, 04:54:51 pm
Hey there!

Could you possibly explain your line of thinking when writing 'since the conjugate roots only applies to imaginary components we can say a, b and c are real?'. I'm a little confused - we could properly correct/understand/explain better if you clarified this :)

Perhaps a more clear approach (alternative to the above) would be to let \(\alpha = x+iy\) and \(\bar{\alpha} = x-iy\), thus leading to the fact that both
\(a(x+iy)^2+b(x+iy)+c=0\) and \(a(x-iy)^2+b(x-iy)+c=0\) are true. Adding these up, we see that we have \(2a(x^2-y^2)+2bx+c=0\) - given a, x and y are real, b and c must also be real.

Hope this helps :)

Hi there, I was trying to just say that the conjugate only affects the imaginary part for example, x= a+ib then x bar = a- ib.
But yes, I see your way of solving the question. Many thanks for the help  :)
Title: Re: 4U Maths Question Thread
Post by: shekhar.patel on December 20, 2019, 12:42:16 pm
In the Vector question, I am getting a different answer to the book's. My answer and book's answer is attached. Would you please help me where I went wrong or if the book answer is wrong.
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on December 20, 2019, 02:38:12 pm
Hi there, I was trying to just say that the conjugate only affects the imaginary part for example, x= a+ib then x bar = a- ib.
But yes, I see your way of solving the question. Many thanks for the help  :)

This makes things clearer - but as with your original statement 'since the conjugate roots only applies to imaginary components we can say a, b and c are real' - the jump from the first part to the second part is very awkward; the logic that takes you from A to B is very flawed. I can't quite see how or why that was assumed?Perhaps explaining your line of thinking could help us better your understanding :) Though it's true in this case, you can't assume this every time, and you certainly shouldn't be assuming things in the first place - it's much safer to use one of the methods described above :)

In the Vector question, I am getting a different answer to the book's. My answer and book's answer is attached. Would you please help me where I went wrong or if the book answer is wrong.

The book's answer is wrong. In future, a more helpful check (when given the value of each of the complex numbers!) is to actually compute what z-w is: in this case it's \(\left(1-\frac{1}{\sqrt{2}}\right) + \left(\sqrt{3}+\frac{1}{\sqrt{2}}\right)i\) - which is clearly in the first quadrant, not on the imaginary axis as the book's answer suggests. In actuality, the book's answer and your answer is identical - the book likely hasn't taken care in drawing a diagram properly, as well as positioning the origin of the ray at the actual origin.

Hope this helps :)
Title: Re: 4U Maths Question Thread
Post by: shekhar.patel on December 22, 2019, 02:06:17 pm
This makes things clearer - but as with your original statement 'since the conjugate roots only applies to imaginary components we can say a, b and c are real' - the jump from the first part to the second part is very awkward; the logic that takes you from A to B is very flawed. I can't quite see how or why that was assumed?Perhaps explaining your line of thinking could help us better your understanding :) Though it's true in this case, you can't assume this every time, and you certainly shouldn't be assuming things in the first place - it's much safer to use one of the methods described above :)

The book's answer is wrong. In future, a more helpful check (when given the value of each of the complex numbers!) is to actually compute what z-w is: in this case it's \(\left(1-\frac{1}{\sqrt{2}}\right) + \left(\sqrt{3}+\frac{1}{\sqrt{2}}\right)i\) - which is clearly in the first quadrant, not on the imaginary axis as the book's answer suggests. In actuality, the book's answer and your answer is identical - the book likely hasn't taken care in drawing a diagram properly, as well as positioning the origin of the ray at the actual origin.

Hope this helps :)

Hi, yes the reasoning behind checking the answer is helpful. Thanks
Title: Re: 4U Maths Question Thread
Post by: shekhar.patel on December 22, 2019, 02:44:00 pm
I need help in this question, I tried to rotate vector by +- i but I am getting the wrong answer.
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on December 22, 2019, 03:01:09 pm
Hey there!

Note that while \(\vec{AB}\) represents the same complex number as \(\vec{OC}\), you can't just use that to find \(\vec{OB}\), since \(\vec{OB}\) clearly does not equal \(\vec{OC}\) (one is the diagonal of the square, the other is a side!). What you've actually done is rotate  \(\vec{OA}\) clockwise 90 degrees to give \(\vec{OC}\). 

There are two ways to do this:
a) Consider that \(\vec{OB}=\vec{OA}+\vec{AB}\) - you've got both values; the first given in the question and the second you've just calculated as equal to  \(\vec{OC}\). Add them up and you'll have \(\vec{OB}\).

b) Consider that \(\vec{AB}\) is a rotation anti-clockwise of \(\vec{AO}\) by 90 degrees. Recalling that  \(\vec{AO}\) = \(-\vec{OA}\), we find that \(\vec{AO}=-2-i\). Rotating to find \(\vec{AB}\), the computation then becomes identical to the one above via addition of vectors.

There technically is also a 'cheat' way of doing it by inspection; it's a relatively simple square and some people might pick up that B is 1+3i straightaway, but there's no working out involved; for the most part this is only helpful for checking your answer.

Hope this helps :)
Title: Re: 4U Maths Question Thread
Post by: shekhar.patel on December 22, 2019, 03:20:46 pm
Hey there!

Note that while \(\vec{AB}\) represents the same complex number as \(\vec{OC}\), you can't just use that to find \(\vec{OB}\), since \(\vec{OB}\) clearly does not equal \(\vec{OC}\) (one is the diagonal of the square, the other is a side!). What you've actually done is rotate  \(\vec{OA}\) clockwise 90 degrees to give \(\vec{OC}\). 

There are two ways to do this:
a) Consider that \(\vec{OB}=\vec{OA}+\vec{AB}\) - you've got both values; the first given in the question and the second you've just calculated as equal to  \(\vec{OC}\). Add them up and you'll have \(\vec{OB}\).

b) Consider that \(\vec{AB}\) is a rotation anti-clockwise of \(\vec{AO}\) by 90 degrees. Recalling that  \(\vec{AO}\) = \(-\vec{OA}\), we find that \(\vec{AO}=-2-i\). Rotating to find \(\vec{AB}\), the computation then becomes identical to the one above via addition of vectors.

There technically is also a 'cheat' way of doing it by inspection; it's a relatively simple square and some people might pick up that B is 1+3i straightaway, but there's no working out involved; for the most part this is only helpful for checking your answer.

Hope this helps :)

Hi there. Yes I see the way of doing it, I think a) is much easier for me to follow, but I don't get how OB=OA + AB, and the fact that we added OA after rotating is not quite clear. Would you please explain me why that is. Thanks
Title: Re: 4U Maths Question Thread
Post by: shekhar.patel on December 22, 2019, 05:48:04 pm
I just need some clarification here about why the following cannot be one of the possible positions.
Thanks
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on December 22, 2019, 06:02:52 pm
Hi there. Yes I see the way of doing it, I think a) is much easier for me to follow, but I don't get how OB=OA + AB, and the fact that we added OA after rotating is not quite clear. Would you please explain me why that is. Thanks

\(\vec{OB}=\vec{OA}+\vec{AB}\) because in general, for any complex numbers X, Y and Z, \(\vec{XZ}=\vec{XY}+\vec{XZ}\) - consider the following diagram:
Click for the diagram!
(https://i.imgur.com/cOHW9hF.png)
Also, consider the square we have as well, as per the diagram below:
Click for the next diagram!
(https://i.imgur.com/LfLxdc9.png)
When we rotate \(\vec{OA}\), we have \(\vec{OC}\), which you'll note is identical to \(\vec{AB}\). From the above, we must then add to have \(\vec{OB}\). Note that here \(\vec{OB}\) is the diagonal of the square; you'll need to add the two sides to get the value of the complex number.

I just need some clarification here about why the following cannot be one of the possible positions.
Thanks

Before I begin assisting you with this question, have you tried using the Argand diagram to help you in working out the answer? It's very helpful, and may make you realise why the answers are correct :)

Hope this helps :)
Title: Re: 4U Maths Question Thread
Post by: shekhar.patel on December 22, 2019, 06:35:14 pm
\(\vec{OB}=\vec{OA}+\vec{AB}\) because in general, for any complex numbers X, Y and Z, \(\vec{XZ}=\vec{XY}+\vec{XZ}\) - consider the following diagram:
Click for the diagram!
(https://i.imgur.com/cOHW9hF.png)
Also, consider the square we have as well, as per the diagram below:
Click for the next diagram!
(https://i.imgur.com/LfLxdc9.png)
When we rotate \(\vec{OA}\), we have \(\vec{OC}\), which you'll note is identical to \(\vec{AB}\). From the above, we must then add to have \(\vec{OB}\). Note that here \(\vec{OB}\) is the diagonal of the square; you'll need to add the two sides to get the value of the complex number.

This is clear now. I understand it. Thanks
Title: Re: 4U Maths Question Thread
Post by: Vxncent288 on December 23, 2019, 02:01:46 am
Hi,
I was wondering what resources I could use to study for the new 3U and 4U courses? Are there certain websites I can find past HSC papers? I'm a bit worried about studying for mathematics with the new syllabuses having come in so any advice/recommendations would be greatly appreciated!
Title: Re: 4U Maths Question Thread
Post by: shekhar.patel on December 23, 2019, 12:38:35 pm
Hi,
I was wondering what resources I could use to study for the new 3U and 4U courses? Are there certain websites I can find past HSC papers? I'm a bit worried about studying for mathematics with the new syllabuses having come in so any advice/recommendations would be greatly appreciated!


I mean there are only so many changes in the new syllabus. You can use the new Cambridge 3 unit and 4 unit textbooks. They contain really great questions, and relevant to the new syllabus. If you are doing past hsc papers ( find them on THSC) and you feel that you haven’t quite learnt the content to do the question, check the NESA syllabus and see if that topics is listed there. If it’s not just chill. Having a good understanding and knowing your syllabus is definitely helpful. Maybe print out the major topics and outcomes of the syllabus and stick it on your study desk.
Other than that the resources that are available are your teachers, tutor, textbooks, syllabus, past papers and your creativity.
Hope that helps  :)
Title: Re: 4U Maths Question Thread
Post by: shekhar.patel on January 07, 2020, 10:43:34 pm
Hi, I am not sure how to do this question, any hints to start off will be much appreciated .
Thanks
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 07, 2020, 10:48:06 pm
Hi, I am not sure how to do this question, any hints to start off will be much appreciated .
Thanks
For \(y^2 = f(x)\) curves, the curve is always symmetric about the \(x\)-axis. Hence the area under the \(x\)-axis should equal the area above it.

The upper curve is \( y = x^{\frac{3}{2}} \). (Recall that square rooting is equivalent to taking powers of a half.)

So the area under the upper curve will be \( \int_0^3 x^{\frac32}\,dx \). The combined area will be double this.
Title: Re: 4U Maths Question Thread
Post by: shekhar.patel on January 07, 2020, 10:51:12 pm
Ok, I see. Got the answer thanks.
Title: Re: 4U Maths Question Thread
Post by: shekhar.patel on January 07, 2020, 10:55:17 pm
In this question we need to find the area of the curve bounded by the abscissas. So we need to make x the subject. I have many operations but I can't find a way of making x the subject.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 07, 2020, 11:02:52 pm
In this question we need to find the area of the curve bounded by the abscissas. So we need to make x the subject. I have many operations but I can't find a way of making x the subject.
\begin{align*}y &= \frac{x}{\sqrt{x^2+5}}\\ y^2 &= \frac{x^2}{x^2+5}\\ y^2(x^2+5) &= x^2\\ 5y^2 &= x^2 - x^2y^2\\ 5y^2 &= x^2(1-y^2)\\ x^2 &= \frac{5y^2}{1-y^2} \\ x &= \frac{\sqrt{5}y}{\sqrt{1-y^2}}\end{align*}
Note that the positive square root was taken here because we are working in the first quadrant. In the first quadrant, \(x \geq 0\).

(Note also that this question could be done without remaking \(x\) the subject. One way is to consider the area of a rectangle, minus the area of some region by the \(x\)-axis. But both methods would've relied on the same integration technique.)
Title: Re: 4U Maths Question Thread
Post by: shekhar.patel on January 08, 2020, 12:07:46 pm
\begin{align*}y &= \frac{x}{\sqrt{x^2+5}}\\ y^2 &= \frac{x^2}{x^2+5}\\ y^2(x^2+5) &= x^2\\ 5y^2 &= x^2 - x^2y^2\\ 5y^2 &= x^2(1-y^2)\\ x^2 &= \frac{5y^2}{1-y^2} \\ x &= \frac{\sqrt{5}y}{\sqrt{1-y^2}}\end{align*}
Note that the positive square root was taken here because we are working in the first quadrant. In the first quadrant, \(x \geq 0\).

(Note also that this question could be done without remaking \(x\) the subject. One way is to consider the area of a rectangle, minus the area of some region by the \(x\)-axis. But both methods would've relied on the same integration technique.)

Oh, that makes sense. Thanks.
Title: Re: 4U Maths Question Thread
Post by: shekhar.patel on January 25, 2020, 07:33:43 pm
Hi, I have a question. Would both these answers be considered valid.

Convert -4i into polar form
4 cis (3pi/2)
4 cis (-pi/2)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on January 26, 2020, 11:24:17 am
Hi, I have a question. Would both these answers be considered valid.

Convert -4i into polar form
4 cis (3pi/2)
4 cis (-pi/2)
They're both correct, but ideally try to use the principle argument where possible. Here, \( \operatorname{Arg}(z) = -\frac\pi2\).
Title: Re: 4U Maths Question Thread
Post by: shekhar.patel on January 27, 2020, 01:58:39 pm
They're both correct, but ideally try to use the principle argument where possible. Here, \( \operatorname{Arg}(z) = -\frac\pi2\).
Ok. Thanks
Title: Re: 4U Maths Question Thread
Post by: milie10 on January 31, 2020, 11:49:25 pm
Hi!

(https://i.imgur.com/ICZD0f3.jpg)
Is my working out right for this? it says answers may vary, but how did they get the answer below?:
(https://i.imgur.com/kLfoqSQ.jpg)

Also, I'm really struggling with these questions:
(https://i.imgur.com/PzSmz03.jpg)

(https://i.imgur.com/gNT8YEr.jpg)

Thanks!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 01, 2020, 02:55:17 am
Hi!

(https://i.imgur.com/ICZD0f3.jpg)
Is my working out right for this? it says answers may vary, but how did they get the answer below?:
(https://i.imgur.com/kLfoqSQ.jpg)

Also, I'm really struggling with these questions:
(https://i.imgur.com/PzSmz03.jpg)

(https://i.imgur.com/gNT8YEr.jpg)

Thanks!
Your working makes perfect sense to me. But just take note that \( \frac{\lambda}{\frac12 \lambda - \frac32} \) reads like a fraction: "lambda over (half lambda - 3 halves)". Try to keep to vector notation: \( \begin{pmatrix} \lambda\\ \frac12 \lambda - \frac32\end{pmatrix} \).
\[ \text{Their working out looks as though they parametrised }y\text{ instead.}\\ \text{That is, they set }y=\lambda.\quad (\lambda \in \mathbb{R}) \]
\[ \text{Then }x=3+2y\text{, so }x=3+2\lambda.\\ \text{Continue with the same approach to get to their answer.} \]
In general, it doesn't matter which variable you choose to parametrise.
___________________________________________________________________________

You're expected to know and understand that in the vector equation \( \vec{u} = \begin{pmatrix}a\\b\end{pmatrix} + \lambda \begin{pmatrix}x\\y\end{pmatrix} \) that \( \begin{pmatrix}a\\b\end{pmatrix} \) is the position vector of a point on the line. With that in mind, for both questions 3 and 4 you should be able to figure out the \( \begin{pmatrix}a\\b\end{pmatrix} \) vector.

As for the direction vector \( \begin{pmatrix}x\\y\end{pmatrix} \):

- For Q3, you've already begun to use \(m=\tan\theta\). In general, if \(m\) is the gradient of the line, then a direction vector for the line is \( \begin{pmatrix}1\\m\end{pmatrix}\). Think about why! (Or if you can't think about why intuitively, prove it by finding the vector equation of the line itself.)
 
- For Q4, note that the angle is not made with respect to the positive \(x\)-axis this time. Rather, it's perpendicular to some random line, in particular the line \(x+3y+1=0\). Using only vector techniques, you would probably want to convert this line into its vector equation form.

Once you've done this, apply the scalar product formula \( \vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta \), where \(\theta = 45^\circ\), to find the possible direction vectors for the lines you seek. Set \(\vec{a}\) to be the direction vector of \(x+3y+1=0\), and in the same spirit as Q3, set \(\vec{b} = \begin{pmatrix}1\\m\end{pmatrix} \).

(The syllabus is still new, so I'm not aware of any concrete rules of thumb in doing these problems. I'm relying on what makes sense based off what I know, and what the syllabus allows for.)
___________________________________________________________________________

The last problem is the shortest distance between a point in a line problem, but necessitating vector methods. When vectors are required for shortest distances, always think projections.

Suppose that you want the shortest distance from a point \( P \) to a line. Let \( \vec{p}\) be the position vector of the point. Furthermore, suppose that the equation of the line is \( \vec{u}=\vec{a}+\lambda\vec{b}\).

1. Then the vector \( \vec{a} - \vec{p}\) is a vector that joins the point \(P\), to some random point on the line.
2. Now, \( \operatorname{proj}_{\vec{b}} (\vec{a}-\vec{p}) \) represents the projection of the vector you found above, onto this line.
3. Therefore, \( \operatorname{proj}_{\vec{b}} (\vec{a}-\vec{p}) - \vec{p}\) is another vector joining \(P\), to a point on the line. However, this vector is special, in that it is perpendicular to the line.
4. The shortest distance between a point and a line is always the perpendicular distance between them. Hence, the answer you seek can be found by taking the magnitude of this vector, i.e. \(|\operatorname{proj}_{\vec{b}} (\vec{a}-\vec{p}) - \vec{p}|\).

Note: All of the above is hard to visualis in your head. You should use a diagram when going through all of that.

In your case, \( \vec{p}= \begin{pmatrix}2\\3\end{pmatrix}\). You have to convert your line into its vector equation to write down your \(\vec{a}\) and \(\vec{b}\).
Title: Re: 4U Maths Question Thread
Post by: shekhar.patel on February 12, 2020, 06:19:59 pm
Hi. I am not quite sure why we integrate with the following limits.
Thanks.
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on February 12, 2020, 09:59:17 pm
Hi. I am not quite sure why we integrate with the following limits.
Thanks.

Hey there!

One of the ideas that was stressed to me as a student was 'look at what you're working towards'. It's become a really handy hint - note that in the result, the logarithm does not use 'n', rather, it uses '1/n'. Since we know the integral of a linear function is a logarithm, and we can directly see the connection between the given result and the result we're working towards (particularly in the middle term), it makes sense to try out 1/n as an upper limit. As a result of this, this means the right hand side will become 1/n, meaning everything will be multiplied by n, leaving a power on the logarithm. This kind of foresight often reduces the amount of work that needs to be done trialling different limits amongst other things - attempt to use it wherever you can.

Hope this helps :)
Title: Re: 4U Maths Question Thread
Post by: milie10 on February 15, 2020, 01:11:59 am
Hi!

Another vector question I'm struggling with:
"Find the shortest distance between the line through the points (1, 3, 1) and (1, 5, -1) and the line through the points (0, 2, 1) and (1, 2, -3)"

how do you find the equations using vector methods? what would the subsequent steps be?

Thanks!!!
Title: Re: 4U Maths Question Thread
Post by: shekhar.patel on February 15, 2020, 11:29:31 am
Hey there!

One of the ideas that was stressed to me as a student was 'look at what you're working towards'. It's become a really handy hint - note that in the result, the logarithm does not use 'n', rather, it uses '1/n'. Since we know the integral of a linear function is a logarithm, and we can directly see the connection between the given result and the result we're working towards (particularly in the middle term), it makes sense to try out 1/n as an upper limit. As a result of this, this means the right hand side will become 1/n, meaning everything will be multiplied by n, leaving a power on the logarithm. This kind of foresight often reduces the amount of work that needs to be done trialling different limits amongst other things - attempt to use it wherever you can.

Hope this helps :)

Ok thanks.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 16, 2020, 05:49:00 pm
Hi!

Another vector question I'm struggling with:
"Find the shortest distance between the line through the points (1, 3, 1) and (1, 5, -1) and the line through the points (0, 2, 1) and (1, 2, -3)"

how do you find the equations using vector methods? what would the subsequent steps be?

Thanks!!!
This is actually an extremely demanding question given the scope of MX2. The shortest distance between two lines usually requires knowledge of planes in 3D space or the cross product, both of which aren’t in the syllabus. What is the source of this question?
Title: Re: 4U Maths Question Thread
Post by: milie10 on February 17, 2020, 08:10:00 pm
This is actually an extremely demanding question given the scope of MX2. The shortest distance between two lines usually requires knowledge of planes in 3D space or the cross product, both of which aren’t in the syllabus. What is the source of this question?

I think my coaching taught us some out of syllabus content for vectors- they said that for completeness of this topic, learning the cross product is necessary. I'd still love to know how it would be done though, but I'm not too fussed about it since it won't be tested :)

Also, could someone explain this proofs question: Q10d from the cambridge textbook? I've forgotten how roots work- am I meant to use the sum and product of roots?
(https://i.imgur.com/Pkpxczz.png)

thanks so much :D
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on February 17, 2020, 09:26:26 pm
I'll leave the vector question for a more competent user while I continue getting accustomed to them myself, but I'll answer your second question :)

There are a few ways to prove to yourself this is true:
- You can prove to yourself that they are one and the same by doing the sum of roots and product of roots of both those quadratics
- You can also consider the roots as per the quadratic equation for each quadratic, inverting them and rationalising the denominator
- Also, consider what happens when you have \(f(\alpha)\) in the first quadratic, and factor out \(\alpha ^2\), and vice versa

Hopefully this will start to make sense! :)
Title: Re: 4U Maths Question Thread
Post by: milie10 on February 17, 2020, 09:32:00 pm
I'll leave the vector question for a more competent user while I continue getting accustomed to them myself, but I'll answer your second question :)

There are a few ways to prove to yourself this is true:
- You can prove to yourself that they are one and the same by doing the sum of roots and product of roots of both those quadratics
- You can also consider the roots as per the quadratic equation for each quadratic, inverting them and rationalising the denominator
- Also, consider what happens when you have \(f(\alpha)\) in the first quadratic, and factor out \(\alpha ^2\), and vice versa

Hopefully this will start to make sense! :)

Thank you so much!!
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 17, 2020, 09:32:48 pm
I think my coaching taught us some out of syllabus content for vectors- they said that for completeness of this topic, learning the cross product is necessary. I'd still love to know how it would be done though, but I'm not too fussed about it since it won't be tested :)

Also, could someone explain this proofs question: Q10d from the cambridge textbook? I've forgotten how roots work- am I meant to use the sum and product of roots?
(https://i.imgur.com/Pkpxczz.png)

thanks so much :D
Yeah, their perspective is understandable, but definitely out of the syllabus. Regardless, with the cross product:

1. Take one point on each line and find a direction vector through these points. So for example, (1,3,1) to (0,2,1). This vector is then a vector going from one of the lines, to the other. (Doesn't matter which line it goes from and to, since ultimately we'll be considering a distance, and hence ignore the direction.) Call this vector \( \vec{a}\).

2. Find the direction vectors of each of the lines.

3. Take the cross product of the vectors in step 2 to find a vector \( \vec{n}\) perpendicular to both of the lines.

4. The projection of the vector joining the two lines \(\vec{a}\), onto the normal vector \(\vec{n}\), that is the vector
\[ \operatorname{proj}_{\vec{n}}\vec{a}, \]
will have the shortest distance between the two lines.
________________________________________________________

Within the scope of the new syllabus, sums and products of roots would've been the first thing I thought of. You should find that the equivalence is indeed true, but I only prove the reverse implication. The forward implication is left as your exercise.
\[ \text{Let }\frac1\alpha\text{ and }\frac1\beta\text{ be the roots of}\\ cx^2 + bx + a. \]
\[ \text{Then it follows from the sums and products of roots that}\\ \begin{align*} \frac{\alpha+\beta}{\alpha\beta} = \frac1\alpha+\frac1\beta &= -\frac{b}{c},\\ \frac{1}{\alpha\beta} &= \frac{a}{c} \end{align*} \]
\[ \text{The second equation immediately rearranges to }\alpha\beta = \frac{c}{a}.\\ \text{Subbing into the first equation,}\\ \frac{\alpha+\beta}{\frac{c}{a}} = -\frac{b}{c} \implies \alpha+\beta = -\frac{b}{a}. \]
\[ \text{Thus a quadratic equation with roots }\alpha\text{ and }\beta\text{ is}\\ \begin{align*}(x-\alpha)(x-\beta) &= 0\\
\implies x^2-(\alpha+\beta)x + \alpha\beta &= 0\\
\implies x^2 + \frac{b}{a}x + \frac{c}{a} &= 0\\
\implies ax^2+bx+c &= 0 \end{align*}\\ \text{as required.} \]
Title: Re: 4U Maths Question Thread
Post by: milie10 on February 17, 2020, 10:48:47 pm
Hi, back again :D

Yeah, their perspective is understandable, but definitely out of the syllabus. Regardless, with the cross product:
...
Thanks heaps again Rui!

This proofs topic is so confusing for me (it's such a new concept!). Hopefully it will get a lot easier with practice :'(

(https://i.imgur.com/xKFU7xX.png)
Last one for the day- I asked a few of my friends this and we are all quite mind boggled by this question (cambridge enrichment question). Does anyone know how to do this?

Thanks :)
Title: Re: 4U Maths Question Thread
Post by: RuiAce on February 18, 2020, 04:29:39 pm
Hi, back again :D
Thanks heaps again Rui!

This proofs topic is so confusing for me (it's such a new concept!). Hopefully it will get a lot easier with practice :'(

(https://i.imgur.com/xKFU7xX.png)
Last one for the day- I asked a few of my friends this and we are all quite mind boggled by this question (cambridge enrichment question). Does anyone know how to do this?

Thanks :)
Part i:
From (4) there exists a passenger that earns exactly 3 times as much as the guard. Therefore immediately from (3) Mr Ward cannot be this passenger, because the guard would have to earn $33333.3333333... a year which is impossible (because money is only correct to the nearest cent.)

Therefore the passenger, who's neighbours with the guard from (4), must be Dr Pender or Mr Sadler. However in order to be neighbours with the guard, they must live halfway between Melbourne and Sydney as well, from (2). This eliminates Mr Sadler because of (1).

Conclusion: Dr Pender lives halfway between Melbourne and Sydney.

Part ii:
There is a passenger who lives in Melbourne from (6). Well we know it instantly ain't Mr Sadler because that would contradict (1). But we just singled out where Dr Pender lives.

Conclusion: Mr Ward is the passenger who lives in Melbourne. Consequently, we now know that Ward is the guard, again using (6).

Part iii:
We're told from (5) that apparently Pender beats the fireman at pool. But the only roles left are now the fireman and the driver. If Pender were the fireman, he'd have to beat himself in pool?!

Conclusion: Pender is the driver, because he'd have to be a different person to the fireman.
__________________________________________________________________________________________

Alright. This took me a while because I'm not the best at logic puzzles. Honestly didn't expect one to show up in the Cambridge textbook even as enrichment! Perfectly understandable because the enrichment section is known to push students to think beyond syllabus expectations.

I could argue that it does lie in the syllabus expectations because you are expected to know how logic works, and use a sequence of implications. But there were lots of implications to think through here; all of them requiring you to think beyond the basic facts given in the question.

To be totally honest, I'm not 100% convinced by my own solution yet. I checked and it does match the answer given, but I would advise you to repeatedly scan each step and be able to say "I can agree with your logic" at each point. Only if you can confidently say you agree with all of the logic, can the solution be regarded as very likely correct. (If not, well better ask haha.) ;)

Also, note that this may not be the most optimal route, even if it ends up being correct!
Title: Re: 4U Maths Question Thread
Post by: milie10 on March 05, 2020, 12:24:24 pm
Hi!!

I have reached the dreaded topic of proving ~inequalities~ :')

I have a general question about working out layout:
e.g. RTP \( (a+b)( \frac{1}{a} + \frac{1}{b} ) \ge 4\)
I like to work backwards by assuming that the proof is correct, shuffling it around until it gives me a property that I know is correct (e.g. in this case I ended up with \( (a-b)^2 \ge 0 \)) and then writing the proof out backwards in my final answer- if I do this, where should I put my original assumption? I know we're not meant to write it as an inequalities since that means that you're assuming that it's true
I was thinking about doing LHS= and RHS= but then that doesn't let me move numerals from either side

orrr is there a better method to solving these that doesn't involve me working backwards?

Thanks so much  ;D
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on March 05, 2020, 01:15:35 pm
Hey there!

Working backwards is often the best way to do this - after you're done working backwards to something you know is true (ie. an identity!), literally write your working backwards. It sounds confusing, but essentially working backwards twice gets your working to go forwards from the identity to the expression you wanted to prove in the first place. The identity usually goes at the top. You can't put 'your original assumption' anywhere - assuming something is true then proving it is a huge no-no. This should only be a tool to point you in the correct direction! :)

In this case, we'd have that:



which is our identity.

Now, we then use this identity such that \(x = \sqrt{\frac{a}{b}} \text{ and } y = \sqrt{\frac{b}{a}}\) to have that

Adding two to both sides we have

Then factorising the left hand side, we have our expression.

You can do LHS = xxxx and RHS = yyyy and make them meet in the middle (making sure you only work on one side at a time!) but this is often tacky and is best used as a last resort.

Hope this makes sense!
Title: Re: 4U Maths Question Thread
Post by: esteban on March 06, 2020, 06:55:20 am
Hi!!

I have reached the dreaded topic of proving ~inequalities~ :')

I have a general question about working out layout:
e.g. RTP \( (a+b)( \frac{1}{a} + \frac{1}{b} ) \ge 4\)
I like to work backwards by assuming that the proof is correct, shuffling it around until it gives me a property that I know is correct (e.g. in this case I ended up with \( (a-b)^2 \ge 0 \)) and then writing the proof out backwards in my final answer- if I do this, where should I put my original assumption? I know we're not meant to write it as an inequalities since that means that you're assuming that it's true
I was thinking about doing LHS= and RHS= but then that doesn't let me move numerals from either side

orrr is there a better method to solving these that doesn't involve me working backwards?

Thanks so much  ;D

You can also just start with the LHS and manipulate it directly. This doesn't involve writing down any inequalities that you have not yet proven.

Eg/

LHS = (a+b)(1/a+1/b) = (a+b)^2/ab = ((a-b)^2+4ab)/ab >= 4ab/ab = 4 = RHS, where the inequality follows from non-negativity of squares.
Title: Re: 4U Maths Question Thread
Post by: milie10 on March 07, 2020, 01:24:10 am
hii back again

(https://i.imgur.com/qc6b3Q4.jpg)

really struggling with these two - I made a start on the first one but not sure where to go from there

thank you!
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on March 08, 2020, 07:23:37 pm
Hey there!

Just notice for the first question when transitioning from the second line to the third, you've made a small computational error: the third line should read \(\frac{1}{n} < \ln (n) - \ln (n-1) < \frac{1}{n-1}\) instead of what you've got. From here, we can manipulate the expressions with log laws amongst other things to obtain the result in the question. Perhaps work from these lines here:



For the second question, consider a similar idea where instead of upper rectangles and lower rectangles, we have a lower rectangle, an integral and a trapezium.

Hope this helps :)
Title: Re: 4U Maths Question Thread
Post by: milie10 on March 19, 2020, 09:48:03 pm
hi!

(https://i.imgur.com/IQiuA4Z.png)
I did this using X~bin(24, 1/3) - does this method work?

(https://i.imgur.com/lpu1KhH.png)
These are the answers and I'm not sure how they got the 16/81

Thank you!
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on March 19, 2020, 10:51:44 pm
Hey there!

'I did this using X~bin(24, 1/3) - does this method work?' - yes, for X=0, since we want there to be no tagged sheep.

However, doing parts (a) and (b) are supposed to lead you on to the answer given! Multi-step questions are supposed to lead you into discovering a result - make sure you look to incorporate earlier results in the later parts of the question :)

Having said that, from (a) you would have found that the probability of having no tagged sheep on a day is \(\left(\frac{2}{3}\right)^4 = \frac{16}{81}\), which is where that comes from. They simply did it on a per day basis, which makes more intuitive sense given they asked for no sheep on six consective days, not 24 consecutive selections.

Hope this makes sense :)
Title: Re: 4U Maths Question Thread
Post by: shekhar.patel on March 21, 2020, 10:13:07 am
Hi. How do I do Q7     
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on March 21, 2020, 05:37:24 pm
Hey there!

If I'm not mistaken, an assumption should be made that \(a, b \in \mathbb{Z}\). Notice that if a and b are integers, and they have an even difference, they must both be odd or both be even. Consider now what happens to the sum of a and b, and thus the difference of the squares.

Hope this helps :)
Title: Re: 4U Maths Question Thread
Post by: mrsc on May 24, 2020, 01:13:13 am
Hey guys, just needed help with this complex number question part (d). Thanks
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on May 24, 2020, 06:50:26 am
Hey there!

Consider what happens if you take \((\omega + \omega ^8)(\omega ^2 + \omega ^7)(\omega ^4+ \omega ^5)\), where \(\omega = \cos \frac{2\pi}{9} +i\sin \frac{2\pi}{9}\).

We have that \((\omega + \omega ^8)(\omega ^2 + \omega ^7)(\omega ^4+ \omega ^5) = (2\cos \frac{2\pi}{9})(2\cos \frac{4\pi}{9})(-2\cos \frac{\pi}{9})\).

We also have that by expanding the brackets and simplifying using part (b), \((\omega + \omega ^8)(\omega ^2 + \omega ^7)(\omega ^4+ \omega ^5) = -1\).

Hence, we have that \((2\cos \frac{2\pi}{9})(2\cos \frac{4\pi}{9})(-2\cos \frac{\pi}{9}) = -1 \implies \cos \frac{2\pi}{9} \cos \frac{4\pi}{9} \cos \frac{\pi}{9} = 1\).

What's evident in this question is that you occasionally need to look to manipulate the question into a nicer form - here, the solution only really became obvious for me after manipulating the result to \((2\cos \frac{2\pi}{9})(2\cos \frac{4\pi}{9})(-2\cos \frac{\pi}{9}) = -1\). From this sort of position, it's a lot simpler and much less time consuming to explore possible solutions, especially given a number of results from previous parts of the question. Try this line of thinking with similar questions!

Hope this helps :)
Title: Re: 4U Maths Question Thread
Post by: mrsc on May 24, 2020, 11:41:38 am
Thanks a lot for your help!!! I get it now. Also stuck with part (c) of another question. For these types of questions, I know how to solve the LHS but for the RHS I'm not sure. I tried changing the roots from the equation in part (b) into mod-arg form first and then dividing by z^3 but it didn't work.
Title: Re: 4U Maths Question Thread
Post by: RuiAce on May 24, 2020, 08:09:07 pm
Thanks a lot for your help!!! I get it now. Also stuck with part (c) of another question. For these types of questions, I know how to solve the LHS but for the RHS I'm not sure. I tried changing the roots from the equation in part (b) into mod-arg form first and then dividing by z^3 but it didn't work.
Dividing by \(z^3\) before subbing in \(z = \cos\theta+i\sin\theta\) seems to be the required approach here.
\begin{align*}
\frac{RHS}{z^3} &= \frac{(z^2+1)(z^2-\sqrt3 z+1)(z^2+\sqrt3 z + 1)}{z^3}\\
&= \left(\frac{z^2+1}{z} \right) \left(\frac{z^2-\sqrt3z+1}{z}\right)\left(\frac{z^2+\sqrt3z+1}{z}\right)\\
&= \left( z +\frac1z\right) \left(z + \frac1z - \sqrt3\right) \left( z + \frac1z + \sqrt3\right).
\end{align*}
When \(z = \cos\theta+i\sin\theta\), it can be shown that \(z + \frac{1}{z} = 2\cos \theta\). Therefore
\[ \frac{RHS}{z^3} = (2\cos\theta) \left(2\cos\theta - \sqrt3\right) \left(2\cos\theta + \sqrt3\right).  \]
You should be able to finish things from here, noting that \(\sqrt3 = 2\cos\frac\pi6\).
Title: Re: 4U Maths Question Thread
Post by: ursa on May 31, 2020, 09:12:26 pm
Hello! First time I'm posting on an ATAR Notes forum, so hope I'm doing this right   8)
In class, we've just been finishing looking at operations on the complex plane. Long story short, I was having difficulty with a question, and the worked solutions provided by the textbook didn't seem to sufficiently answer the question... I spent the whole lesson working with my teacher and the other kid in my ext. 2 class, but we didn't really come up with anything satisfactory.
The question is:
 Prove that the vectors representing the complex numbers u, v and (u - iv)/(1-i) form a right-angled triangle.
Would really appreciate seeing someone's method of solving it! Sorry if my formating is confusing, I couldn't figure out how to attach an image.......
Hope that makes sense, thank you!  :D
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on May 31, 2020, 11:08:43 pm
Welcome to the forums!

In future, you can use an image hosting service like imgur to upload your picture, then copy a link through to the forum to post up a picture. You've definitely got the post right, don't worry about that :)

I think you seem to have some ideas, so instead of the answer, I'm going to drop a few hints:

Geometric approach
- What does the vector \(\frac{u-iv}{1-i}\) actually denote? Can you express it in terms of some other more familiar/simpler vectors?
- Construct a parallelogram
- Consider the properties of a parallelogram and a kite
- Draw a diagram of all of the above!

Algebraic approach
- Consider the arguments of the vectors \(u-\frac{u-iv}{1-i}\) and \(v-\frac{u-iv}{1-i}\). What do the values of the arguments tell you?

If there's anything else, feel free to ask in this thread  :)
Title: Re: 4U Maths Question Thread
Post by: vernburn on June 28, 2020, 02:53:52 pm
Hey there!

Consider what happens if you take \((\omega + \omega ^8)(\omega ^2 + \omega ^7)(\omega ^4+ \omega ^5)\), where \(\omega = \cos \frac{2\pi}{9} +i\sin \frac{2\pi}{9}\).

We have that \((\omega + \omega ^8)(\omega ^2 + \omega ^7)(\omega ^4+ \omega ^5) = (2\cos \frac{2\pi}{9})(2\cos \frac{4\pi}{9})(-2\cos \frac{\pi}{9})\).

We also have that by expanding the brackets and simplifying using part (b), \((\omega + \omega ^8)(\omega ^2 + \omega ^7)(\omega ^4+ \omega ^5) = -1\).

Hence, we have that \((2\cos \frac{2\pi}{9})(2\cos \frac{4\pi}{9})(-2\cos \frac{\pi}{9}) = -1 \implies \cos \frac{2\pi}{9} \cos \frac{4\pi}{9} \cos \frac{\pi}{9} = 1\).

What's evident in this question is that you occasionally need to look to manipulate the question into a nicer form - here, the solution only really became obvious for me after manipulating the result to \((2\cos \frac{2\pi}{9})(2\cos \frac{4\pi}{9})(-2\cos \frac{\pi}{9}) = -1\). From this sort of position, it's a lot simpler and much less time consuming to explore possible solutions, especially given a number of results from previous parts of the question. Try this line of thinking with similar questions!

Hope this helps :)

Alternatively, we know that:
\(\begin{aligned}1+z +z ^2+...+z^8 &= \prod_{k=1}^4\left(z- cis\frac{2k\pi}{9}\right)\left(z-cis\left(-\frac{2k\pi}{9}\right)\right)\\
&=\prod_{k=1}^4\left(z^2-2\cos\frac{2k\pi}{9}z+1\right)\end{aligned}\)

letting \(z=i\):
\(\begin{aligned}\prod_{k=0}^4\left(-2i\cos\frac{2k\pi}{9}\right)&=1\\
16\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}\cos\frac{2\pi}{3}\cos\frac{8\pi}{9}&=1\\
-\frac{1}{2}\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}\cos\frac{8\pi}{9}&=\frac{1}{16}\\
-\cos\frac{\pi}{9}\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}&=-\frac{1}{8}\\
\implies \cos\frac{\pi}{9}\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}&=\frac{1}{8}\end{aligned}\)
Title: Re: 4U Maths Question Thread
Post by: mrsc on October 02, 2020, 02:26:51 pm
Hi, just wondering what steps you would take to solve these types of questions. I know the graphical method but not sure about the algebraical. Thanks
Title: Re: 4U Maths Question Thread
Post by: Paradoxica on October 02, 2020, 10:12:42 pm
We can safely assume z²≠1, as the problem itself is undefined if that were true.

Arguments disregard the modulus of the number, so consider the equivalent number upscaled by the positive number The argument of this being equal to π/4 is the same as saying the real and imaginary parts are equal and positive. Hence:This is the cartesian equation of a circle of radius √2 and centre (0,-1).

Now we have to go back and restrict the circle for positive real and imaginary parts.

From the imaginary part -2y, we must have y<0.

Now for the real part, x²+y²>1, we begin by parametrising the circle as This is equivalent to Looking at the parametrisation of we see that this is exactly the same as saying y<0.

Hence, the locus is the arc of the circle with negative imaginary part.
Title: Re: 4U Maths Question Thread
Post by: Husky on October 18, 2020, 03:55:45 pm
Hey I was wondering if I could get some help with this question from the NESA HSC sample paper for Ext 2. It's question 14 of the paper but It feels harder to understand than the final question, I just can't wrap my head around the strange intervals and use of the pigeonhole principle, even after looking at the solutions.
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on October 18, 2020, 06:53:39 pm
Hey there :)

Imagine you have a number line from 0 to 1 (including 0, not including 1 - the question is kinda dodgy here, since adding one doesn't modify \(\{\alpha\}\)) divvied up into N intervals, each of length 1/N (draw it out!). Counting the numbers they give you, there are N+1 objects and you've drawn up N intervals. You can assign them one by one and there will be an interval that has two numbers in it - since we know that the interval length is 1/N, we can deduce that the difference is less than the length of the interval. Note also that if we put a number at each endpoint of every interval, this would still be the case (as 1, or N/N is excluded - this is why this was key in the first place!).

If you're struggling to visualise it still, I recommend a few things:
- draw it out
- try simpler PHP questions then apply the above to that, see if you can replicate the same ideas here
- come back to it later
- don't worry about it too much, it's only going to stress you out this close to the exam
Title: Re: 4U Maths Question Thread
Post by: Husky on October 18, 2020, 07:59:35 pm
Hey there :)

Imagine you have a number line from 0 to 1 (including 0, not including 1 - the question is kinda dodgy here, since adding one doesn't modify \({\alpha}\)) divvied up into N intervals, each of length 1/N (draw it out!). Counting the numbers they give you, there are N+1 objects and you've drawn up N intervals. You can assign them one by one and there will be an interval that has two numbers in it - since we know that the interval length is 1/N, we can deduce that the difference is less than the length of the interval. Note also that if we put a number at each endpoint of every interval, this would still be the case (as 1, or N/N is excluded - this is why this was key in the first place!).

If you're struggling to visualise it still, I recommend a few things:
- draw it out
- try simpler PHP questions then apply the above to that, see if you can replicate the same ideas here
- come back to it later
- don't worry about it too much, it's only going to stress you out this close to the exam

Ah okay that makes much more sense than the NESA solutions which are too brief IMO. Would you mind also explaining the third part of the question?
Title: Re: 4U Maths Question Thread
Post by: fun_jirachi on October 18, 2020, 09:03:22 pm
Note that as we can represent \(\alpha\) as \(k + \{\alpha\}\). Hence, for some integer n, we can express \(n\alpha\) as \(k + \{n\alpha\}\).
Now, we choose integers a and b in the range [0, N] such that \(\{a\alpha\}\) and \(\{b\alpha\}\) differ by less than 1/N ie. \(0 < |\{a\alpha\} - \{b\alpha\}| < \frac{1}{N}\).

From the above, we can express \(a\alpha\) and \(b\alpha\) as \(k + \{a\alpha\}\) and \(l + \{b\alpha\}\) respectively, where k and l are also integers. We can rearrange to have that \(\{a\alpha\} = a\alpha - k\) and \(\{b\alpha\} = b\alpha - l\), and combining the two, we have that \(\{a\alpha\} - \{b\alpha\} = a\alpha - b\alpha - k + l = (a-b)\alpha - (k - l)\) - but we've noted that a, b, l, and k are integers, and hence subtraction, addition and multiplication using these numbers will definitely result in other integers. Just for the sake of the question, we can let q = a-b and p = k-l, and replace this in the equation I gave in the first paragraph, which matches up with the result given in the question.

When doing questions like this, a few things to note:
- The value of the question in question; it's one mark, so you've definitely already done the hard work beforehand. Look to deduce things from previous parts instead of diving in headfirst. Head first, you will lose as opposed to head first can't lose
- I've reiterated this many times as my teacher reiterated to me; look at what you're working towards. It always helps to have a result in front of you, and it's similar to seeing an answer that's too big, too small, got the wrong sign or the wrong units. Keep yourself on the right track
- Don't stress too much, these are good questions to leave and come back to, especially if you can't see them right away (especially if as you say you find these harder to understand than the questions further back)

Hope this helps :)
Title: Re: 4U Maths Question Thread
Post by: ti_tirrr on October 24, 2021, 12:50:57 pm
Hi,

I needed some advice on what to do for my subjects in Year 12. Currently, I have started year 12 term 1 at my school and my subjects are as follows:

4U English
3U Math
Chemistry
Business Studies

I've been content with these subjects and am doing relatively well in them at school. We're starting week 4 at school and I've been thinking; is it too late to consider studying math ext 2? I love doing math but I don't know if I have the right to make that decision yet as Year 12 has already started and so has term classes for math ext 2. My friends in 4U math have already had 2 classes so far. Could anyone give me some advice on what to do about this?
Title: Re: 4U Maths Question Thread
Post by: RuiAce on October 24, 2021, 02:08:54 pm
Hi,

I needed some advice on what to do for my subjects in Year 12. Currently, I have started year 12 term 1 at my school and my subjects are as follows:

4U English
3U Math
Chemistry
Business Studies

I've been content with these subjects and am doing relatively well in them at school. We're starting week 4 at school and I've been thinking; is it too late to consider studying math ext 2? I love doing math but I don't know if I have the right to make that decision yet as Year 12 has already started and so has term classes for math ext 2. My friends in 4U math have already had 2 classes so far. Could anyone give me some advice on what to do about this?
I don't think it's too late at all. A more important question would be if your teachers would let you. If they deny you entry, that's a barrier out of your control.

On the other hand, catching up is entirely in your control. You might have to deal with a busy 1-3 weeks with the catch-up, but remember that's 1-3 weeks out of around 50 weeks before your HSC exams start. (And remember, you can always drop later if it turned out to be a mistake.)