# ATAR Notes: Forum

## HSC Stuff => HSC Maths Stuff => HSC => HSC Mathematics Extension 1 => Topic started by: jakesilove on January 28, 2016, 07:02:15 pm

Post by: jakesilove on January 28, 2016, 07:02:15 pm
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 3U 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.

I got an ATAR of 99.80, and a mark of 98 in the Extension 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: 98 in 3U Maths: Ask me Anything!
Post by: Phillorsm on January 29, 2016, 07:43:03 pm
Hey Jake,
I'm struggling in binomial theorem. I can kinda half do questions but find it really difficult to finish them off. We've been doing it for quite a while now in class and I can't seem to get the hang of it. I was wondering if you had any specific tips to binomial theorem, i'm just finding it really overwhelming to try and wrap my head around it and yeah.
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: jakesilove on January 29, 2016, 08:29:32 pm
Hey Jake,
I'm struggling in binomial theorem. I can kinda half do questions but find it really difficult to finish them off. We've been doing it for quite a while now in class and I can't seem to get the hang of it. I was wondering if you had any specific tips to binomial theorem, i'm just finding it really overwhelming to try and wrap my head around it and yeah.

Hey Phillorsm!

I have to say, Binomial Theorum can really be one of the most difficult sections of the Extension One (and even Extension Two) course. There are those standard questions (find the expansion of [INSERT QUESTION HERE], find the greatest term in [INSERT SERIES HERE] etc), and then it jumps to an incredibly difficult level almost straight away (If you have an expansion X to the power of another expansion Y, what is the colour of the third term's eyes?).

The first thing to have a good understanding of is how to expand any series, quickly. Recall the standard expansion

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

From there, it's all about utilising the relevant properties of Binomial Theorum, all of which I can't really summarise in one post (Maybe I'll make a more comprehensive document at some point). Could you post specific questions you are having trouble with to this forum, so I can take you through the steps required to answer it?

This topic is all about practice, practice, practice. Once you've seen the methodology a few times, you'll be a natural!

Talk to you soon :)

Jake
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: Happy Physics Land on January 29, 2016, 09:48:47 pm
I'm struggling in binomial theorem. I can kinda half do questions but find it really difficult to finish them off. We've been doing it for quite a while now in class and I can't seem to get the hang of it. I was wondering if you had any specific tips to binomial theorem, i'm just finding it really overwhelming to try and wrap my head around it and yeah.

Hello Im also a year 12 student doing my maths extension I, and I have encountered similar issues beforehand as you, because binomial is very tedious and the sigma notations always scare me because I am unfamiliar with it. So I began doing questions first from maths in focus which provides easy questions on binomial theorem and then moved on to do more exercises in the cambridge book. It is very challenging however beneficial to do those development and extension questions as well because it is likely that your teacher will confront you with a similar style question.

Here are several tips that I think has helped me a lot with binomial:
1. It is very important to remember that the binomial co-efficients always starts with nC0, not nC1
2. When proving binomial identities, a lot of those can be related back to pascal's triangle. So if you are confused about whats the significance of the proof or what you are trying to prove, write down the first few rows of pascal's triangle to give you a better observation of what exactly you are trying to prove
3. DO NOT BE SCARE OF SIGMA NOTATIONS! A lot of my friends instantly give up as soon as they seen sigma notation because its such a weird representation of a series of numbers. Sometimes we dont think of sigma notation as normal maths, but rather, some "alien language". But its VERY CRUCIAL TO REMEMBER that SIGMA NOTATIONS ARE OUR FRIENDS. It is just A SERIES, nothing more, just A SERIES OF NUMBERS. It is helpful for us because instead of having to tediously look a long, boring chain of numbers, a simple sigma notation essentially summarise it for us in simple expressions. On the bottom of sigma notation there is r= some number or k = some number, this just means that for the expression next to the sigma sign, the initial variable is what r or k represents. E.g. for 3^r, r = 0, that means we start with 3^0. On the top of the sigma sign there is usually "n", which is indicative that the series terminates at r = n, whatever that n value maybe. E.g. for 3^r, we terminate at 3^n.
4. It is beneficial sometimes when solving binomial questions to expand the binomial out. If it is too long an expansion, just write out the first 3-4 terms and the last 3 terms. This helps us to find patterns that can help us to solve the question.
5. In Binomial questions associated with integration or differentiation, we almost always find a value for x (i.e. let x = something) to make our solution look more similar to what the question requires for us to prove/find. A sneaky tip is that HSC examiners would usually write the question in a way that students will let x = 0 or 1.
6. When we are proving an identity in binomial theorem, its not always compulsory to start with the side thats more complicated. This is counter-intuitive to what we have always been learning because we are always used to solving something thats looks more intimidating because there is a higher chance that we can somehow manipulate it to make it look more neat/tidy, and resemble the other side of the equation. In binomial theorem, this is not always the case. For example, consider the proof for "Sum of nCr from r=0 to r=n) = 2^n". Logically, we would begin with the left hand side because it is more complicated and we would hope for a neat result to come out in the end. However, if we begin with the right hand side it will be much easier because RHS = 2^n = (1+1)^n = sum of (nCr x 1^r) from r=0 to r=n. Since 1^r is always 1, we can effectively prove that 2^n = sum of nCr from r=0 to r=n.
7. It almost always helpful that when you are stuck on a binomial proof question to go back to the basics of expanding (1+x)^n, or remembering that (1+x)^n = the sum of (nCr x x^r ) from r= 0 to r=n.
8. When finding the constant term that involves expanding two binomials, expand both and select one term from each binomial expansion that will cancel each other's variable out when multiplied together, leaving us with just a number.
9. Transformations of (1+x)^n will always change the position of the greatest co-efficient in the expansion. (1+x)^n will have its greatest co-efficient at the centre, (1+3x)^n will have its greatest co-efficient shifted to the right and (1+5x)^n will have its greatest co-efficient shifted even further to the right. Adversely, (3+x)^n will have its greatest co-efficient shifted to the left and (5+x)^n will have its greatest co-efficient shifted even more to the left and so on.

These are all just some of my tips that l found very helpful to know. Im not sure how much this will help you but yeah good luck in everything this year!

Best Regards

Happy Physics Land
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: jakesilove on January 29, 2016, 09:58:03 pm
Hello Im also a year 12 student doing my maths extension I, and I have encountered similar issues beforehand as you, because binomial is very tedious and the sigma notations always scare me because I am unfamiliar with it. So I began doing questions first from maths in focus which provides easy questions on binomial theorem and then moved on to do more exercises in the cambridge book. It is very challenging however beneficial to do those development and extension questions as well because it is likely that your teacher will confront you with a similar style question.

Here are several tips that I think has helped me a lot with binomial:
1. It is very important to remember that the binomial co-efficients always starts with nC0, not nC1
2. When proving binomial identities, a lot of those can be related back to pascal's triangle. So if you are confused about whats the significance of the proof or what you are trying to prove, write down the first few rows of pascal's triangle to give you a better observation of what exactly you are trying to prove
3. DO NOT BE SCARE OF SIGMA NOTATIONS! A lot of my friends instantly give up as soon as they seen sigma notation because its such a weird representation of a series of numbers. Sometimes we dont think of sigma notation as normal maths, but rather, some "alien language". But its VERY CRUCIAL TO REMEMBER that SIGMA NOTATIONS ARE OUR FRIENDS. It is just A SERIES, nothing more, just A SERIES OF NUMBERS. It is helpful for us because instead of having to tediously look a long, boring chain of numbers, a simple sigma notation essentially summarise it for us in simple expressions. On the bottom of sigma notation there is r= some number or k = some number, this just means that for the expression next to the sigma sign, the initial variable is what r or k represents. E.g. for 3^r, r = 0, that means we start with 3^0. On the top of the sigma sign there is usually "n", which is indicative that the series terminates at r = n, whatever that n value maybe. E.g. for 3^r, we terminate at 3^n.
4. It is beneficial sometimes when solving binomial questions to expand the binomial out. If it is too long an expansion, just write out the first 3-4 terms and the last 3 terms. This helps us to find patterns that can help us to solve the question.
5. In Binomial questions associated with integration or differentiation, we almost always find a value for x (i.e. let x = something) to make our solution look more similar to what the question requires for us to prove/find. A sneaky tip is that HSC examiners would usually write the question in a way that students will let x = 0 or 1.
6. When we are proving an identity in binomial theorem, its not always compulsory to start with the side thats more complicated. This is counter-intuitive to what we have always been learning because we are always used to solving something thats looks more intimidating because there is a higher chance that we can somehow manipulate it to make it look more neat/tidy, and resemble the other side of the equation. In binomial theorem, this is not always the case. For example, consider the proof for "Sum of nCr from r=0 to r=n) = 2^n". Logically, we would begin with the left hand side because it is more complicated and we would hope for a neat result to come out in the end. However, if we begin with the right hand side it will be much easier because RHS = 2^n = (1+1)^n = sum of (nCr x 1^r) from r=0 to r=n. Since 1^r is always 1, we can effectively prove that 2^n = sum of nCr from r=0 to r=n.
7. It almost always helpful that when you are stuck on a binomial proof question to go back to the basics of expanding (1+x)^n, or remembering that (1+x)^n = the sum of (nCr x x^r ) from r= 0 to r=n.
8. When finding the constant term that involves expanding two binomials, expand both and select one term from each binomial expansion that will cancel each other's variable out when multiplied together, leaving us with just a number.
9. Transformations of (1+x)^n will always change the position of the greatest co-efficient in the expansion. (1+x)^n will have its greatest co-efficient at the centre, (1+3x)^n will have its greatest co-efficient shifted to the right and (1+5x)^n will have its greatest co-efficient shifted even further to the right. Adversely, (3+x)^n will have its greatest co-efficient shifted to the left and (5+x)^n will have its greatest co-efficient shifted even more to the left and so on.

These are all just some of my tips that l found very helpful to know. Im not sure how much this will help you but yeah good luck in everything this year!

Best Regards

Happy Physics Land

Happy Physics Land, this is a behemoth of helpful tips and incredibly succinct ways to solve often difficult questions. I just wanted to personally thank you for your contribution, and encourage anyone with knowledge to always step in and lend others a hand. By teaching others, you are just proving to yourself that you understand the content inside and out.

I'd love to incorporate your tips into a future document of tips for Binomial theorum, including diagrams and practice questions. For now, though, be proud that you will be helping the literally thousands of students who will check this forum daily.

For anyone else that wants to help contribute to this forum, both by asking questions and by answering them, 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: 98 in 3U Maths: Ask me Anything!
Post by: Happy Physics Land on January 29, 2016, 10:04:23 pm
Happy Physics Land, this is a behemoth of helpful tips and incredibly succinct ways to solve often difficult questions. I just wanted to personally thank you for your contribution, and encourage anyone with knowledge to always step in and lend others a hand. By teaching others, you are just proving to yourself that you understand the content inside and out.

I'd love to incorporate your tips into a future document of tips for Binomial theorum, including diagrams and practice questions. For now, though, be proud that you will be helping the literally thousands of students who will check this forum daily.

For anyone else that wants to help contribute to this forum, both by asking questions and by answering them, 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! :)

Hello Jake,

Thank you so much Jake, I am doing these because when I attended you lectures you very generously gave out very helpful tips for the thousands of students that have attended your lecture and Im just aspired to become a person like you! I owe you a big thanks and I can only repay this gratitude in the form of helping more HSC students around the community.

Personally however, I am a little bit stuck on a proof if you are available to help me jake. It is to use mathematical induction to prove nCr = n!/r!(n-r)!

Thank you very much Jake
Best Regards

Happy Physics Land
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: jakesilove on January 29, 2016, 10:31:21 pm
Hello Jake,

Thank you so much Jake, I am doing these because when I attended you lectures you very generously gave out very helpful tips for the thousands of students that have attended your lecture and Im just aspired to become a person like you! I owe you a big thanks and I can only repay this gratitude in the form of helping more HSC students around the community.

Personally however, I am a little bit stuck on a proof if you are available to help me jake. It is to use mathematical induction to prove nCr = n!/r!(n-r)!

Thank you very much Jake
Best Regards

Happy Physics Land

Hey Happy Physics Land.

Could you post the exact question you're asking about? Like how its phrased in a past paper etc. After having a crack, I did some research and the consensus seems to be (if I got the question right) that the only 'proof by induction' is a by definition proof, which I am still happy to post however is somewhat less satisfying. Maybe I'm misinterpreting the question?

Jake :)
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: Happy Physics Land on January 29, 2016, 10:44:46 pm
Hey Jake

That question wasnt exactly a past paper question but l was just curious about how to use mathematical induction to prove that nCr = n!/r!(n-r)! This way of doing it through mathematical induction is the more tedious way but l honestly had no idea where to start with this proof except for substituting r = 1. If possible would you like to kindly send me a photo of your solution? Thank you so much Jake and sorry for taking up so much of your time because of my curiosity!

Regards

Happy Physics Land
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: jakesilove on January 29, 2016, 11:23:24 pm
Hey Jake

That question wasnt exactly a past paper question but l was just curious about how to use mathematical induction to prove that nCr = n!/r!(n-r)! This way of doing it through mathematical induction is the more tedious way but l honestly had no idea where to start with this proof except for substituting r = 1. If possible would you like to kindly send me a photo of your solution? Thank you so much Jake and sorry for taking up so much of your time because of my curiosity!

Regards

Happy Physics Land

Hey Happy Physics Land!

I've written out, not quite a solution, but my thoughts on the question. From my research, it seems like you really can't solve this question using induction (I've outlined my thoughts below regarding why that is so). Still, I hope this is somewhat helpful, if not just in terms of methodology. If anyone has any other solution or ideas, I'd love to hear them! Keep being enthusiastic!

(http://i.imgur.com/dKjEKFS.png?1)
(http://i.imgur.com/zVDKCBd.png?1)
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: Happy Physics Land on January 30, 2016, 12:03:51 am
HEY Jake!

Yes Jake I see your point now, essentially mathematical induction cannot be applied in this case because it would lead me into a spiral of continuously proving for something that's established as a rule. I was just very interested to see whether it would work but yes I understood everything you said regarding why mathematical induction would be inappropriate in this case. Thank you so much for your time and I AM SO SORRY FOR HAVING BOTHERED YOU WITH A QUESTION LIKE THIS! Thank you for your perseverance with my ridiculous question Jake you are a very responsible teacher!

Best Regards
Happy Physics Land
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: Happy Physics Land on February 01, 2016, 07:53:16 pm
Hey Jake,

A little bit stuck on an inverse function question, may I please get a hand with part C of question 9?

(http://i.imgur.com/bZTPVvr.jpg)
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: jakesilove on February 01, 2016, 08:35:00 pm
Hey Jake,

A little bit stuck on an inverse function question, may I please get a hand with part C of question 9?

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

Hey Happy Physics Land!

I love this question, because it looks really difficult, and then when I explain it to you I promise you'll kick yourself. I seriously think questions like this, which end up having intuitive, as opposed to learnt, answers, are the hardest to get right in an exam situation. Never the less, here is my proof. There are heaps of ways of proving this but I like this one because it's simple, and to be honest, obvious once you've thought about it enough. I don't mean to suggest that this is an easy question: it really isn't. But the proof IS easy, once you know it :)

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

Thanks for your questions! I really encourage everyone to have a go, post something or answer a question, because having a community like this behind you will seriously add to your learning this year.

For anyone else that wants to help contribute to this forum, both by asking questions and by answering them, 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: 98 in 3U Maths: Ask me Anything!
Post by: Happy Physics Land on February 01, 2016, 08:49:35 pm
Hey Happy Physics Land!

I love this question, because it looks really difficult, and then when I explain it to you I promise you'll kick yourself. I seriously think questions like this, which end up having intuitive, as opposed to learnt, answers, are the hardest to get right in an exam situation. Never the less, here is my proof. There are heaps of ways of proving this but I like this one because it's simple, and to be honest, obvious once you've thought about it enough. I don't mean to suggest that this is an easy question: it really isn't. But the proof IS easy, once you know it :)

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

Thanks for your questions! I really encourage everyone to have a go, post something or answer a question, because having a community like this behind you will seriously add to your learning this year.

For anyone else that wants to help contribute to this forum, both by asking questions and by answering them, 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! :)

Oh god .............. I really feel like kicking myself right now. It just looks so much easier once I saw your proof. Those information that I obtained in part a) and part b) made me think that this question must have something to do with the location of stationary points or points of inflection .... but WOW. Thank you so much Jake for this assistance and for unlocking that logical part of my brain. I think now that I have seen this sort of question, I should be able to solve other questions of the same type in the future. All in all, thank you very much Jake you are awesome! :D

Best Regards
Happy Physics Land
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: Phillorsm on February 01, 2016, 10:32:48 pm
Hello Im also a year 12 student doing my maths extension I, and I have encountered similar issues beforehand as you, because binomial is very tedious and the sigma notations always scare me because I am unfamiliar with it. So I began doing questions first from maths in focus which provides easy questions on binomial theorem and then moved on to do more exercises in the cambridge book. It is very challenging however beneficial to do those development and extension questions as well because it is likely that your teacher will confront you with a similar style question.

Here are several tips that I think has helped me a lot with binomial:
1. It is very important to remember that the binomial co-efficients always starts with nC0, not nC1
2. When proving binomial identities, a lot of those can be related back to pascal's triangle. So if you are confused about whats the significance of the proof or what you are trying to prove, write down the first few rows of pascal's triangle to give you a better observation of what exactly you are trying to prove
3. DO NOT BE SCARE OF SIGMA NOTATIONS! A lot of my friends instantly give up as soon as they seen sigma notation because its such a weird representation of a series of numbers. Sometimes we dont think of sigma notation as normal maths, but rather, some "alien language". But its VERY CRUCIAL TO REMEMBER that SIGMA NOTATIONS ARE OUR FRIENDS. It is just A SERIES, nothing more, just A SERIES OF NUMBERS. It is helpful for us because instead of having to tediously look a long, boring chain of numbers, a simple sigma notation essentially summarise it for us in simple expressions. On the bottom of sigma notation there is r= some number or k = some number, this just means that for the expression next to the sigma sign, the initial variable is what r or k represents. E.g. for 3^r, r = 0, that means we start with 3^0. On the top of the sigma sign there is usually "n", which is indicative that the series terminates at r = n, whatever that n value maybe. E.g. for 3^r, we terminate at 3^n.
4. It is beneficial sometimes when solving binomial questions to expand the binomial out. If it is too long an expansion, just write out the first 3-4 terms and the last 3 terms. This helps us to find patterns that can help us to solve the question.
5. In Binomial questions associated with integration or differentiation, we almost always find a value for x (i.e. let x = something) to make our solution look more similar to what the question requires for us to prove/find. A sneaky tip is that HSC examiners would usually write the question in a way that students will let x = 0 or 1.
6. When we are proving an identity in binomial theorem, its not always compulsory to start with the side thats more complicated. This is counter-intuitive to what we have always been learning because we are always used to solving something thats looks more intimidating because there is a higher chance that we can somehow manipulate it to make it look more neat/tidy, and resemble the other side of the equation. In binomial theorem, this is not always the case. For example, consider the proof for "Sum of nCr from r=0 to r=n) = 2^n". Logically, we would begin with the left hand side because it is more complicated and we would hope for a neat result to come out in the end. However, if we begin with the right hand side it will be much easier because RHS = 2^n = (1+1)^n = sum of (nCr x 1^r) from r=0 to r=n. Since 1^r is always 1, we can effectively prove that 2^n = sum of nCr from r=0 to r=n.
7. It almost always helpful that when you are stuck on a binomial proof question to go back to the basics of expanding (1+x)^n, or remembering that (1+x)^n = the sum of (nCr x x^r ) from r= 0 to r=n.
8. When finding the constant term that involves expanding two binomials, expand both and select one term from each binomial expansion that will cancel each other's variable out when multiplied together, leaving us with just a number.
9. Transformations of (1+x)^n will always change the position of the greatest co-efficient in the expansion. (1+x)^n will have its greatest co-efficient at the centre, (1+3x)^n will have its greatest co-efficient shifted to the right and (1+5x)^n will have its greatest co-efficient shifted even further to the right. Adversely, (3+x)^n will have its greatest co-efficient shifted to the left and (5+x)^n will have its greatest co-efficient shifted even more to the left and so on.

These are all just some of my tips that l found very helpful to know. Im not sure how much this will help you but yeah good luck in everything this year!

Best Regards

Happy Physics Land

Thankyou so much both Jake and Happy Physics Land! Your responses have given me a new confidence that binomial isn't as intimidating as I think it is :)
Jake, if you don't mind, could you please show me the solution to question 5b from the 2001 3u paper?
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: IkeaandOfficeworks on February 01, 2016, 10:35:34 pm
Hi, I'm a bit confused with this induction question. I was wondering if you could explain to me the key steps. Thank you!

The sum of consecutive odd positive integers is divisible by 4.
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: jamonwindeyer on February 01, 2016, 11:18:46 pm
Thankyou so much both Jake and Happy Physics Land! Your responses have given me a new confidence that binomial isn't as intimidating as I think it is :)
Jake, if you don't mind, could you please show me the solution to question 5b from the 2001 3u paper?

Hey Phillorsm! Awesome question, and glad to hear that binomial is growing you! I'll be honest and say it is still is one my least favourite parts of teaching the course, though . That being said, I mustered the courage to tag in for Jake and write a quick solution for you!  ;) this question works best by writing out a whole bunch of terms, so try doing that and following on with this quick solution. If you need a bit more detail, please let me know!  :)

(http://i.imgur.com/Suhftxk.png)

Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: jamonwindeyer on February 01, 2016, 11:39:01 pm
Hi, I'm a bit confused with this induction question. I was wondering if you could explain to me the key steps. Thank you!

The sum of consecutive odd positive integers is divisible by 4.

Hi there IkeaandOfficeworks! I bought a wardrobe from you a few weeks back  ;)

So this is an awesome question, thanks so much for posting it.

However, I'm interested, because I don't think you can prove this with induction, because it isn't necessarily true!

The sum of consecutive odd positive integers is divisible by 4. So that means 1+3 should be divisible by 4. And it is. So should 1+3+5? And already we hit an exception the rule.

I am thinking that perhaps there is something missing from your question, some other condition.  Or equally likely, I'm missing something because it is 11:30  ;)

So, please get back to us! However, in the mean time, I'll leave you with a general method for induction questions.

1 - Express the question algebraically. Put it into a formula you can manipulate and change to suit your needs, usually in terms of n (EG - Divisibility induction questions like this would have some expression equal to 4M, where M is some integer, some involve inequalities, etc). This is actually sometimes the hardest part!
2- Prove the result for the lowest value you need to (usually 1, sometimes higher)
3 - Now, assume the result is true for n=k. Then, use this result to prove for  n=k+1. Almost always, this involves rearranging to allow a substitution. The algebra is normally fairly easy, but it is hard to know where to go. The solution? Practice makes perfect!
4 - Conclude that since the result is true for n=1, and n=k+1, it is true for n=k+1=2, n=2+1=3, etc etc! This ending is really important, lots of people forget it, and its worth a mark by itself!

I hope this helps for your induction questions, feel free to keep throwing them at us! Sorry I couldn't be of more help with this one  :)

Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: Happy Physics Land on February 01, 2016, 11:54:50 pm
Hey Phillorsm! Awesome question, and glad to hear that binomial is growing you! I'll be honest and say it is still is one my least favourite parts of teaching the course, though . That being said, I mustered the courage to tag in for Jake and write a quick solution for you!  ;) this question works best by writing out a whole bunch of terms, so try doing that and following on with this quick solution. If you need a bit more detail, please let me know!  :)

(http://i.imgur.com/Suhftxk.png)

Hey Ikeaand Im sorry everything is out of stock (Haha lm funny!):

There are two major issues with this question that you have proposed, and I think this is a slightly more challenging question than the usual mathematical induction questions which simply throw you an equation and tell you to prove the identity.

In this question, however:
1. It mentioned CONSECUTIVE ODD positive integers, however didnt specify how many there are. Initially I thought this would be an extremely challenging problem because the question sounds exactly like to prove "no matter how many consecutive positive odd integers there are, the sum of all these terms will all be divisible by 4". BUT, when you set out a few simple examples, you will discover that this is not the case. Lets start off with 1+3, they are consecutive and they add to 4, which is divisible by 4. Sweet. Now let's consider 1+3+5, they are consecutive however they add to 9, which is not divisible by 4. Furthermore, if we consider 1+3+5+7+9, we will get a sum of 25 which is also not divisible by 4. This creates a discrepancy to the way I initially defined the question. So I gave it a thought and I discovered that if we only do two consecutive odd numbers, e.g. 1+3 = 4, 3+5 = 8, 5+7 = 12, 7+9 = 16, these sums are all divisible by 4. So I think what the question really means is "to prove that the sum of ADJACENT odd positive integers is divisible by 4".
2. Now, another problem is that we are not provided with an equation to prove, which means that unfortunately we will need to construct our own equations, which, is easy enough. If we let n = the lower of the two adjacent odd numbers, and n+2 = the greater of the two adjacent odd numbers,

Here is my worked solution:
(http://i.imgur.com/KaPojqq.jpg)

And at the end of your mathematical induction proof, make sure that you put a concluding statement "Since the it is true for n=1, assumed true for n=k and proved true for n=k+1, then by the principle of mathematical induction, the sum of consecutive odd positive integers is divisible by 4".

It is possible, however, that I have understood the question differently from your teacher, and if this is the case, please inform me and I will upload another solution as soon as possible!

Best Regards
Happy Physics Land
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: Happy Physics Land on February 01, 2016, 11:58:13 pm
Hi there IkeaandOfficeworks! I bought a wardrobe from you a few weeks back  ;)

So this is an awesome question, thanks so much for posting it.

However, I'm interested, because I don't think you can prove this with induction, because it isn't necessarily true!

The sum of consecutive odd positive integers is divisible by 4. So that means 1+3 should be divisible by 4. And it is. So should 1+3+5? And already we hit an exception the rule.

I am thinking that perhaps there is something missing from your question, some other condition.  Or equally likely, I'm missing something because it is 11:30  ;)

So, please get back to us! However, in the mean time, I'll leave you with a general method for induction questions.

1 - Express the question algebraically. Put it into a formula you can manipulate and change to suit your needs, usually in terms of n (EG - Divisibility induction questions like this would have some expression equal to 4M, where M is some integer, some involve inequalities, etc). This is actually sometimes the hardest part!
2- Prove the result for the lowest value you need to (usually 1, sometimes higher)
3 - Now, assume the result is true for n=k. Then, use this result to prove for  n=k+1. Almost always, this involves rearranging to allow a substitution. The algebra is normally fairly easy, but it is hard to know where to go. The solution? Practice makes perfect!
4 - Conclude that since the result is true for n=1, and n=k+1, it is true for n=k+1=2, n=2+1=3, etc etc! This ending is really important, lots of people forget it, and its worth a mark by itself!

I hope this helps for your induction questions, feel free to keep throwing them at us! Sorry I couldn't be of more help with this one  :)

Hey Jamon:

Jamon you pretty much covered what I did but can you check whether we should prove for n=k+2 or n=k+1? Because in this case we are dealing with CONSECUTIVE ODD NUMBERS that differ by 2, do you reckon we should prove for n=k+2?

Best Regards
Happy Physics Land
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: jamonwindeyer on February 02, 2016, 12:19:39 am
Hey Jamon:

Jamon you pretty much covered what I did but can you check whether we should prove for n=k+2 or n=k+1? Because in this case we are dealing with CONSECUTIVE ODD NUMBERS that differ by 2, do you reckon we should prove for n=k+2?

Best Regards
Happy Physics Land

Hey HappyPhysicsLand! I was providing a general method, so it entirely depends on the question. But yes, in this case, it would be n=k+2. If it were consecutive multiples of 5, it would be n=k+5, and so on. Apply to the situation at hand.

Also, I think it is highly likely that two consecutive odd integers is what the question specifies. Any even multiple of odd numbers would satisfy the condition, but two is what was most likely asked, which you've covered excellently.

Also, very impressed with your use of set notation in the solution. Little tip, you can use an N (with the same style typography as your Z) to denote all positive integers (and usually including 0), or more correctly, the set of all natural numbers . Virtually no difference to what you've done, and certainly not a necessity, just FYI!
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: jamonwindeyer on February 02, 2016, 12:24:02 am
Hey Jamon:

Jamon you pretty much covered what I did but can you check whether we should prove for n=k+2 or n=k+1? Because in this case we are dealing with CONSECUTIVE ODD NUMBERS that differ by 2, do you reckon we should prove for n=k+2?

Best Regards
Happy Physics Land

Oh, and I almost forgot, don't forget to conclude your induction proof! You've proved for n=k+2, you need to extrapolate this for further numbers. Just a simple, it is true for n=1, and n=k+2, so it is true for n=1+2=3, n=3+2=5, etc, for any sum of two consecutive odd positive integers, will suffice.
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: Happy Physics Land on February 02, 2016, 10:16:39 am
Oh, and I almost forgot, don't forget to conclude your induction proof! You've proved for n=k+2, you need to extrapolate this for further numbers. Just a simple, it is true for n=1, and n=k+2, so it is true for n=1+2=3, n=3+2=5, etc, for any sum of two consecutive odd positive integers, will suffice.

Hello Jamon:

Thank your for your recommendations and also for giving approval to my induction proof :D !  I think that the use of N notation to represent all the positive integers is quite a clever idea and I really liked the format of the conclusion, very succinct but effective as a concluding statement! But overall thank you for checking my solution!
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: jakesilove on February 02, 2016, 11:03:51 am
Hello Jamon:

Thank your for your recommendations and also for giving approval to my induction proof :D !  I think that the use of N notation to represent all the positive integers is quite a clever idea and I really liked the format of the conclusion, very succinct but effective as a concluding statement! But overall thank you for checking my solution!

Hey all, just thought I'd quickly tag in here.

In regards the the Induction question, Happy Physics Land I think you're right on the money. The question MUST be asking about only two consecutive odd numbers, just because otherwise it isn't true. Which leads me to think that the question itself is phrased badly.
So I think overall a great group effort that eventually got to the right place.

Great job all!

Jake
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: cajama on February 06, 2016, 07:26:52 pm
Hi ive got a question on permutations. This is probably a simple question but my mind is twisted right now so I can't seem to get b) and c) right. Not sure either if 4000 and 5000 is a typo for 40000 and 50000 respectively as the answers are small numbers??

a) How many different arrangements can be made from the numbers 3,4,4,5 and 6? ( which i figured was 5!/2! =60)
b) how many arrangements form numbers greater than 4000? (ans: 48)
c) how many form numbers less than 5000? (ans: 36)
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: nerdgasm on February 06, 2016, 07:52:10 pm
I think there is a typo; 40000 and 50000 should be what is written in the question (given the answers provided), or else every 5-digit arrangement would be larger than both 4000 and 5000.

Assuming that 40000 and 50000 are respectively meant:
For part b), I find it easier to look at which arrangements are not counted, because there are fewer cases to consider. In this case, only those arrangements starting with a 3 will be not counted. So, let's assume the starting digit of the arrangement is 3, so we have 3 _ _ _ _.
We then have to place 4, 4, 5 and 6 into the remaining four slots. This can be done in 4!/2! = 12 ways, by a similar argument to what you did in part a). All other arrangements work, so the total number of successful arrangements is 60 - 12 = 48.

For part c), we can again use the same 'counting the complement' technique. This time, we exclude those arrangements starting with 5 or 6. By the same working out as in part b), we can deduce that if the starting digit is 5, there are 12 arrangements (as you then have 5 _ _ _ _ with 3, 4, 4, 6 to place), and there are also 12 arrangements if the starting digit is 6 (6 _ _ _ _ , with 3, 4, 4, 5 to place). Hence, the total number of arrangements less than 50000 is 60 - 12 -12 = 36.

It is also possible to count the arrangements that are included for part c), but you do have to be a bit careful: if the starting digit is 3, then there are 12 successful arrangements. However, note that if the starting digit is 4, then you have 4 _ _ _ _ with 3, 4, 5, 6 to place. Because the remaining digits are all distinct, there are 4! ways to place them, and hence there are 24 successful arrangements. Therefore, the number of successful arrangements is 12 + 24 = 36, which matches the result we obtained earlier.
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: Happy Physics Land on February 06, 2016, 07:53:02 pm
Hi ive got a question on permutations. This is probably a simple question but my mind is twisted right now so I can't seem to get b) and c) right. Not sure either if 4000 and 5000 is a typo for 40000 and 50000 respectively as the answers are small numbers??

a) How many different arrangements can be made from the numbers 3,4,4,5 and 6? ( which i figured was 5!/2! =60)
b) how many arrangements form numbers greater than 4000? (ans: 48)
c) how many form numbers less than 5000? (ans: 36)

Hey Cajama:

Firstly I just wanna to assure you here that they are not typos, Im fairly certain that the questions are correct despite the answer does seem quite ridiculous. Good job on answering part a), remembering to divide by 2! which shows that you have recognised that there is a repetition.

Now for part B, I would like to invite you to draw just four underline dashes, each representing a number that can be placed in that spot. (i.e. __ __ __ __)

On the first underscore dash, there are only 4 numbers that can be placed there， which are 4, 4, 5, 6 (you cannot select 3 as the first digit because the number has to be greater than 4000). On the second underscore, there are also only 4 numbers that can be placed there (because a number, we dont know which one, has already been selected, leaving us with 4 options to choose from.) On the third underscore, we will only have 3 options to choose from because we have already selected two of them from all the numbers that we are provided with. And finally on the last underscore we will only have 2 options to choose from because we have already selected 3 numbers from all the numbers that the question provides us with.

Hence, the total number of different arrangements will seem to be 4 x 4 x 3 x 2 = 96 ways. BUT, WE HAVE NOT YET ACCOUNTED FOR THE REPETITION OF 4 IN THE NUMBERS THAT WE ARE PROVIDED WITH. Hence similar to what you have done in part a), it is necessary for you to divide 96 by 2! and you will obtain an answer of 48.

Ok, now we are up to part c). This one will be extremely similar to part b). I would recommend you to follow what I did in part b) and attempt part c) yourself to check whether you have truly understood what I have done. I will still post the solution below, but have an attempt on yourself.

So, similar to part b), draw four underscores __ __ __ __, each representing a digit.

On the first digit, we can only have three options of numbers, because from all the numbers we are given, only three of them will suit the criteria of "a number that is less than 5000". These three numbers, evidently, are 3, 4, 4. The second digit will have 4 options for us to choose from, because there is a total of 5 numbers which we are provided with and we have already selected one. The third digit is the same, we will have 3 options to choose from because we have already chosen 2 numbers from the 5 for the first two digits. And for the final digit, we will have 2 options to choose from because we have already chosen 3 numbers for the first 3 digits.

Hence the total amount of numbers that we can make below 5000 will be (3 x 4 x 3 x 2) / 2! = 36 (2! is to account for the repetition of 4 in the numbers that we are provided with)

Anyways great questions, hope you have understood my solution. Don't hesitate to post more questions if you need further assistance!!! :)

Best Regards
Happy Physics Land
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: Happy Physics Land on February 06, 2016, 07:57:26 pm
I think there is a typo; 40000 and 50000 should be what is written in the question (given the answers provided), or else every 5-digit arrangement would be larger than both 4000 and 5000.

Assuming that 40000 and 50000 are respectively meant:
For part b), I find it easier to look at which arrangements are not counted, because there are fewer cases to consider. In this case, only those arrangements starting with a 3 will be not counted. So, let's assume the starting digit of the arrangement is 3, so we have 3 _ _ _ _.
We then have to place 4, 4, 5 and 6 into the remaining four slots. This can be done in 4!/2! = 12 ways, by a similar argument to what you did in part a). All other arrangements work, so the total number of successful arrangements is 60 - 12 = 48.

For part c), we can again use the same 'counting the complement' technique. This time, we exclude those arrangements starting with 5 or 6. By the same working out as in part b), we can deduce that if the starting digit is 5, there are 12 arrangements (as you then have 5 _ _ _ _ with 3, 4, 4, 6 to place), and there are also 12 arrangements if the starting digit is 6 (6 _ _ _ _ , with 3, 4, 4, 5 to place). Hence, the total number of arrangements less than 50000 is 60 - 12 -12 = 36.

It is also possible to count the arrangements that are included for part c), but you do have to be a bit careful: if the starting digit is 3, then there are 12 successful arrangements. However, note that if the starting digit is 4, then you have 4 _ _ _ _ with 3, 4, 5, 6 to place. Because the remaining digits are all distinct, there are 4! ways to place them, and hence there are 24 successful arrangements. Therefore, the number of successful arrangements is 12 + 24 = 36, which matches the result we obtained earlier.

Hmm ok this is quite interesting. We have actually defined the question differently, you saw it as all the numbers that can be made and are greater than 4000 or less than 5000 and I saw the question as all the four-digit numbers than are great than 4000 or less than 5000. Im not too sure about the wording of this question or maybe is it just a typo of 40000 and 50000.
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: jakesilove on February 06, 2016, 09:58:44 pm
Hi ive got a question on permutations. This is probably a simple question but my mind is twisted right now so I can't seem to get b) and c) right. Not sure either if 4000 and 5000 is a typo for 40000 and 50000 respectively as the answers are small numbers??

a) How many different arrangements can be made from the numbers 3,4,4,5 and 6? ( which i figured was 5!/2! =60)
b) how many arrangements form numbers greater than 4000? (ans: 48)
c) how many form numbers less than 5000? (ans: 36)

Great answers all! Glad this forum is developing so quickly, and there are so many people willing to jump in and help others out!!!

Jake
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: nerdgasm on February 07, 2016, 01:34:50 am
Hmm ok this is quite interesting. We have actually defined the question differently, you saw it as all the numbers that can be made and are greater than 4000 or less than 5000 and I saw the question as all the four-digit numbers than are great than 4000 or less than 5000. Im not too sure about the wording of this question or maybe is it just a typo of 40000 and 50000.

You're definitely right. the suggested answers work just as well if the question is considering 4-digit arrangements. I think that your interpretation is more likely to be correct  and that there was no typo. Good job on your explanation too, I thought it was really clear!

*(The following is probably not part of the 3U maths course)*:
I think that the reason why we were able to get the same numerical answers despite our different interpretations is as follows:
Let us define a finite 'pool' of n members (the members are not necessarily distinct). I claim that the number of different n-member arrangements is the same as the number of (n-1)-member arrangements.

My idea is as follows: I propose a mapping from the set of n-member arrangements to the set of (n-1)-member arrangements. This mapping basically chops the last member off the n-member arrangement to get an (n-1)-member arrangement.

To show that this mapping is injective (or one-to-one), I use proof by contradiction: assume we have two distinct n-member arrangements that map to the same (n-1)-member arrangement. Then, it follows the two n-member arrangements must agree in their 1st, 2nd ... (n-2)th and (n-1)th positions. But because the pool is finite, it follows that there is only one member in the pool left to make the nth position. Therefore, the two n-member arrangements must in fact be the same.

To show that this mapping is surjective (or onto), note that for any (n-1)-member arrangement, there must be exactly one member from the original pool that was not used. Stick this member onto the end to form a n-member arrangement, which will map to the (n-1)-member arrangement.

Therefore, as this mapping is both injective and surjective, it follows that there are exactly the same number of n- and (n-1)-member arrangements from a finite pool of n members.

I think this explanation also explains our similarity in answers to parts b) and c) as well: in part b), since the starting digit (to be excluded) is 3 in both of our interpretations, we are essentially trying to place 4, 4, 5 and 6 into a 3-member arrangement or a 4-member arrangement in order to work out how many to exclude, which can be done in exactly the same number of ways. And similarly in part c), we are trying to place 4, 4, 5 and 6 into a 3- or 4-member arrangement if the starting digit is 3, and we are trying to place 3, 4, 5 and 6 into a 3- or 4-member arrangement if the starting digit is 4, both of which can be done in exactly the same number of ways.
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: foodmood16 on February 09, 2016, 09:01:07 pm
Hey I have a test coming up and I am having trouble with ext locus and parabola. The question is
Find the equation of the chord of contact AB of tangents drawn from an external point (x1, y1) to the parabola x2 = 12y

Thankyou  :)
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: Happy Physics Land on February 09, 2016, 09:48:04 pm
Hey I have a test coming up and I am having trouble with ext locus and parabola. The question is
Find the equation of the chord of contact AB of tangents drawn from an external point (x1, y1) to the parabola x2 = 12y

Thankyou  :)

Hey foodmood16,

I did get an answer to your question, it was simply a matter of substituting into the equation of chord of contact. Just double check with your answer just in case I got it wrong (when questions are too easily solved I feel a bit intimidated). But yes great question doe because just about everyone tends to forget about this chord of contact equation.

(http://i.imgur.com/36BiedQ.jpg)
(sorry for the dodgy diagram l drew)

Best Regards
Happy Physics Land
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: Neutron on February 10, 2016, 06:23:32 pm
Hey! I was just practicing some curve sketching and I was wondering for curves that look like these:

y= 3x/(x^2+9)

How would you know the shape it forms? Cause really all I knew was that it intercepts at the origin, is odd and has a horizontal asymptote at 0.. Thanks, sorry if this sounds dumb it's just that it seems a bit dodgy how I'm assuming it looks like a sorta sideways s (sorry I can't really describe and I don't know how to draw/paste it in here).. Thank you!

Neutron
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: Happy Physics Land on February 10, 2016, 07:02:53 pm
Hey! I was just practicing some curve sketching and I was wondering for curves that look like these:

y= 3x/(x^2+9)

How would you know the shape it forms? Cause really all I knew was that it intercepts at the origin, is odd and has a horizontal asymptote at 0.. Thanks, sorry if this sounds dumb it's just that it seems a bit dodgy how I'm assuming it looks like a sorta sideways s (sorry I can't really describe and I don't know how to draw/paste it in here).. Thank you!

Neutron

Hey Neutron,

This is a really good question that can be typically encountered in both 3 unit and 4 unit exams. My solution here is more for the 3unit students that have not done as much graphing as the 4 unit students. If you are a 4 unit students, another alternative of approaching this question that I do recommend is through sketching x^2 + 9 first (which is just a parabola), then sketch 1/(x^2+9) (with x=0 as the asymptote and using special values such as when y = 0, 1 or -1 to help you, these are tricks that you would have learnt in sketching reciprocal functions), and in the end use multiplication of ordinates (again using special points where y = 0, 1 and -1) to sketch the graph.

My approach here is similar to yours. Except I have found the limits as x approaches infinity and I have constructed a table of values to help myself to see what the shape will resemble. In this table of values I only used positive values of x because since this is an odd function, you can simply turn the graph on positive side of x-axis by 180 degrees anticlockwise to find the graph on the negative side of the x-axis. I have also circled when x=3, y =1/2 as the maximum turning point in the positive graph through observing the table of values. Im assuming here that the question does not want you to use calculus, so we can only infer that there is a max tp. at (3,1/2). If you can use calculus, then the maximum and minimum turning points can be easily found and the curve would have been very easy to draw.

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

If you have any concerns or confusions please dont hesitate to ask! :D

Best Regards
Happy Physics Land
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: Neutron on February 10, 2016, 07:16:18 pm
Thank you so much HPL! Yeah yeah it makes sense now.. One more thing though, how do you know whether a point is a point of discontinuity (open circle) or an asymptote?
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: Happy Physics Land on February 10, 2016, 08:18:04 pm
Thank you so much HPL! Yeah yeah it makes sense now.. One more thing though, how do you know whether a point is a point of discontinuity (open circle) or an asymptote?

Another very good question indeed, and this definitely is something that will be hard to distinguish between.

When you are provided with an equation, you should become suspicious of a point of discontinuity under two situations:
1. when a y-value of your curve approaches zero as x approaches the asymptote but the curve looks like as if it's going to intersect the asymptote
2. when you have f(x)g(x) but the domains are not the same (a typical example is xlnx, where domain of y = x is defined for all real x, but the domain for y = lnx is only defined for x>0)

A point of discontinuity is essentially when the function is undefined for both x and y-values of this particular point. For example, lets raise the y = xlnx again. x= 0 would be undefined for lnx because x>0, y = 0 is undefined because it is impossible to get a result of 0 from a log function. Hence the point (0,0) would be a point of discontinuity. When you become suspicious that there is a point of discontinuity, substitute in x and y-values of that point to confirm whether or not the point is discontinuous.

Finding a discontinuous point is not easy and it will take practise to develop an instinct to become suspicious of the existence of such points. In general, whenever you see lnx involved, you should keep an eye out on possible points of discontinuity.
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: foodmood16 on February 10, 2016, 08:20:21 pm
Hey foodmood16,

I did get an answer to your question, it was simply a matter of substituting into the equation of chord of contact. Just double check with your answer just in case I got it wrong (when questions are too easily solved I feel a bit intimidated). But yes great question doe because just about everyone tends to forget about this chord of contact equation.

(http://i.imgur.com/36BiedQ.jpg)
(sorry for the dodgy diagram l drew)

Best Regards
Happy Physics Land

Thankyou Happy Physics Land! That helps a heap as I didn't quite understand it when it was explained in class :)
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: jakesilove on February 10, 2016, 09:17:54 pm
Thankyou Happy Physics Land! That helps a heap as I didn't quite understand it when it was explained in class :)

For those who didn't know: HPL is a legend and should be revered.

Thanks man!

Jake
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: Happy Physics Land on February 10, 2016, 09:24:58 pm
For those who didn't know: HPL is a legend and should be revered.

Thanks man!

Jake

Hahaha thank you jake, I'm just striving to be like you one day! You are my inspiration!!!!
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: katherine123 on February 16, 2016, 06:48:43 pm
in what kind of sequence will you use to sketch y=4sin[3(x-(/6))] 0<=x<=2
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: Happy Physics Land on February 16, 2016, 06:52:41 pm
in what kind of sequence will you use to sketch y=4sin[3(x-(/6))] 0<=x<=2

Is the question y = 4sin[3(x-1/6)] or something else??
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: RuiAce on February 16, 2016, 06:59:33 pm
in what kind of sequence will you use to sketch y=4sin[3(x-(/6))] 0<=x<=2

First, read the comment above as that is important.

However, assuming what he said was true:

Firstly, sketch y=sin(x) as a reference.
Then, apply the period change. y=sin(3x) changes the period of T=2π to T=2π/3. Hence, the graph repeats itself every 2π/3 units instead.
Next, apply the phase shift. The -1/6 (which, I have a feeling should've been -π/6 but regardless) will shift the graph to the RIGHT by 1/6 unit
Finally, add in the amplitude. This makes the maximum value y=4 instead of y=1 and similarly minimum value of y=-4.
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: jakesilove on February 16, 2016, 07:04:19 pm
Is the question y = 4sin[3(x-1/6)] or something else??

Hey all! These all look like great answers, so I thought I'd just chip in with what the graph would actually look like.

(http://i.imgur.com/Pw4CY8f.png?1)

Jake
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: katherine123 on February 16, 2016, 08:48:13 pm
sorry the pie symbol didnt appear

y=4sin[3(x-pie/6)]-2
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: RuiAce on February 16, 2016, 09:08:28 pm
sorry the pie symbol didnt appear

y=4sin[3(x-pie/6)]-2

Then my solution still holds, except replace 1/6 with π/6 instead.
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: Happy Physics Land on February 16, 2016, 09:46:03 pm
sorry the pie symbol didnt appear

y=4sin[3(x-pie/6)]-2

Hey Katherine

If you are stuck you can refer to my solution if you'd wish to

(http://i.imgur.com/O3xSntC.jpg)

Best Regards
Happy Physics Land
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: Neutron on February 24, 2016, 02:41:06 pm
Hey! Okay so i don't seem to be able to find an easy way to do this (like I got an answer but the method was waaay too complicated I reckon but I got -8 degrees and 48 seconds and -24 degrees and 54 seconds??) D: I was wondering whether you guys could help me! thanks :D

Solve the following equation:
2cos2ϴ=1-3sin2ϴ   0≤ϴ≤360

Thank you!

Neutron
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: jakesilove on February 24, 2016, 06:49:37 pm
Hey! Okay so i don't seem to be able to find an easy way to do this (like I got an answer but the method was waaay too complicated I reckon but I got -8 degrees and 48 seconds and -24 degrees and 54 seconds??) D: I was wondering whether you guys could help me! thanks :D

Solve the following equation:
2cos2ϴ=1-3sin2ϴ   0≤ϴ≤360

Thank you!

Neutron

Hey Neutron!

I also can't find an easy way to answer that question! I get the same answers, although make sure that since the range is between 0 and 360, you add 360 degrees to your two answers! If anyone can think of an easy solution, I'd love you to post it!

Jake
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: RuiAce on February 25, 2016, 12:24:09 am
Hey! Okay so i don't seem to be able to find an easy way to do this (like I got an answer but the method was waaay too complicated I reckon but I got -8 degrees and 48 seconds and -24 degrees and 54 seconds??) D: I was wondering whether you guys could help me! thanks :D

Solve the following equation:
2cos2ϴ=1-3sin2ϴ   0≤ϴ≤360

Thank you!

Neutron
The presence of the 2cos(2θ) is not able to eliminate the 1 as no matter what expansion of the double angle formula is used, a 2 will pop out instead of a 1. This makes expanding all double angles, unfortunately, a folly.

The neatest method here would be to attack 3sin(2θ)+2cos(2θ) with the auxiliary angle method. Yes, admittedly this isn't tidy either.
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: Phillorsm on February 29, 2016, 10:15:22 pm
Hey Jake, or whoever answers this...
So In your lectures you mentioned that with Induction you are allowed to skip steps in the proof to save time. I was just wanting to check with you if this is what you meant by that, and if it would get all the marks :)

P.S. Sorry if the image is a bit small, it wouldnt let me upload anything over 512kB.
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: RuiAce on February 29, 2016, 11:11:04 pm
Hey Jake, or whoever answers this...
So In your lectures you mentioned that with Induction you are allowed to skip steps in the proof to save time. I was just wanting to check with you if this is what you meant by that, and if it would get all the marks :)

P.S. Sorry if the image is a bit small, it wouldnt let me upload anything over 512kB.

This is interesting. I would ALWAYS restate the LHS and RHS before rearranging the equation by either substituting in the assumption or just algebra. Also, when you use the assumption you should always state "by assumption" or something along the lines of it e.g. by inductive hypothesis.

And your final statement, whilst it can be short, has to be there. Just say "Hence true by induction"

Edit: Also you skipped heaps of algebra. That's going to confuse the examiner.
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: Happy Physics Land on February 29, 2016, 11:20:19 pm
Hey Jake, or whoever answers this...
So In your lectures you mentioned that with Induction you are allowed to skip steps in the proof to save time. I was just wanting to check with you if this is what you meant by that, and if it would get all the marks :)

P.S. Sorry if the image is a bit small, it wouldnt let me upload anything over 512kB.

Yeah hmm if you recall at the end of Jake's induction lecture he said the best way to write your conclusion would be "since its true for n=1, assumed true for n=k and proven true for n=k+1, then by the principle of mathematical induction the statement is true" This will be the most secure way to write your conclusion, just in case you encounter an old-fashion marker who wants a complete formal style conclusion.
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: jakesilove on February 29, 2016, 11:36:35 pm
Hey Jake, or whoever answers this...
So In your lectures you mentioned that with Induction you are allowed to skip steps in the proof to save time. I was just wanting to check with you if this is what you meant by that, and if it would get all the marks :)

P.S. Sorry if the image is a bit small, it wouldnt let me upload anything over 512kB.

Hey!!

I definitely agree with HPL: You need to write out the full concluding remark. However, in regards to the algebra I would recommend doing a few more steps. Definitely expand brackets/collect like terms etc. Basically, what I was suggesting is that if you can clearly see how the LHS is going to equal the RHS, do some algebra steps and then just pretend you've quickly collected like terms and simplified. Only skip steps if you can see exactly how it's going to work out: as a general rule, try to do each step of algebra!

Jake
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: amandali on March 05, 2016, 09:20:43 am
how to sketch y=inverse cos(x^2)
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: jakesilove on March 05, 2016, 12:06:06 pm
how to sketch y=inverse cos(x^2)

Hey Amandali!

The first thing to remember with inverse trigonometric graphs is their domain and range. We know that inverse cos has a domain of

$-1

and a range of

$0

Then, all I would recommend is plotting points! See what happens when you sub in x=-1, x=-0.5, x=0, x=0.5, x=1. This will give you a general idea of the graph, so that you can sketch the final graph!

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

Jake
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: amandali on March 06, 2016, 03:00:39 pm
how to solve  lim    tan3x/tan2y
x->0
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: jakesilove on March 06, 2016, 06:16:38 pm
how to solve  lim    tan3x/tan2y
x->0

Hey! I assume you actually meant 2x, instead of 2y?

In that case, we know that the limit of tan(a)/a as a approaches zero is equal to one. Therefore, the limit of tan(ax) as x approaches zero is equal to ax!

For tan(3x), as x approaches zero, the value will approach 3x. For tan(2x), as x approaches zero the value will approach 2x. Therefore, we now have 3x/2x=3/2, which is our answer!
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: WLalex on March 15, 2016, 09:12:53 am
Hello Im also a year 12 student doing my maths extension I, and I have encountered similar issues beforehand as you, because binomial is very tedious and the sigma notations always scare me because I am unfamiliar with it. So I began doing questions first from maths in focus which provides easy questions on binomial theorem and then moved on to do more exercises in the cambridge book. It is very challenging however beneficial to do those development and extension questions as well because it is likely that your teacher will confront you with a similar style question.

Here are several tips that I think has helped me a lot with binomial:
1. It is very important to remember that the binomial co-efficients always starts with nC0, not nC1
2. When proving binomial identities, a lot of those can be related back to pascal's triangle. So if you are confused about whats the significance of the proof or what you are trying to prove, write down the first few rows of pascal's triangle to give you a better observation of what exactly you are trying to prove
3. DO NOT BE SCARE OF SIGMA NOTATIONS! A lot of my friends instantly give up as soon as they seen sigma notation because its such a weird representation of a series of numbers. Sometimes we dont think of sigma notation as normal maths, but rather, some "alien language". But its VERY CRUCIAL TO REMEMBER that SIGMA NOTATIONS ARE OUR FRIENDS. It is just A SERIES, nothing more, just A SERIES OF NUMBERS. It is helpful for us because instead of having to tediously look a long, boring chain of numbers, a simple sigma notation essentially summarise it for us in simple expressions. On the bottom of sigma notation there is r= some number or k = some number, this just means that for the expression next to the sigma sign, the initial variable is what r or k represents. E.g. for 3^r, r = 0, that means we start with 3^0. On the top of the sigma sign there is usually "n", which is indicative that the series terminates at r = n, whatever that n value maybe. E.g. for 3^r, we terminate at 3^n.
4. It is beneficial sometimes when solving binomial questions to expand the binomial out. If it is too long an expansion, just write out the first 3-4 terms and the last 3 terms. This helps us to find patterns that can help us to solve the question.
5. In Binomial questions associated with integration or differentiation, we almost always find a value for x (i.e. let x = something) to make our solution look more similar to what the question requires for us to prove/find. A sneaky tip is that HSC examiners would usually write the question in a way that students will let x = 0 or 1.
6. When we are proving an identity in binomial theorem, its not always compulsory to start with the side thats more complicated. This is counter-intuitive to what we have always been learning because we are always used to solving something thats looks more intimidating because there is a higher chance that we can somehow manipulate it to make it look more neat/tidy, and resemble the other side of the equation. In binomial theorem, this is not always the case. For example, consider the proof for "Sum of nCr from r=0 to r=n) = 2^n". Logically, we would begin with the left hand side because it is more complicated and we would hope for a neat result to come out in the end. However, if we begin with the right hand side it will be much easier because RHS = 2^n = (1+1)^n = sum of (nCr x 1^r) from r=0 to r=n. Since 1^r is always 1, we can effectively prove that 2^n = sum of nCr from r=0 to r=n.
7. It almost always helpful that when you are stuck on a binomial proof question to go back to the basics of expanding (1+x)^n, or remembering that (1+x)^n = the sum of (nCr x x^r ) from r= 0 to r=n.
8. When finding the constant term that involves expanding two binomials, expand both and select one term from each binomial expansion that will cancel each other's variable out when multiplied together, leaving us with just a number.
9. Transformations of (1+x)^n will always change the position of the greatest co-efficient in the expansion. (1+x)^n will have its greatest co-efficient at the centre, (1+3x)^n will have its greatest co-efficient shifted to the right and (1+5x)^n will have its greatest co-efficient shifted even further to the right. Adversely, (3+x)^n will have its greatest co-efficient shifted to the left and (5+x)^n will have its greatest co-efficient shifted even more to the left and so on.

These are all just some of my tips that l found very helpful to know. Im not sure how much this will help you but yeah good luck in everything this year!

Best Regards

Happy Physics Land

Thanks so much, this is a great overview and simplification of binomial
Title: Re: 3U Maths Question Thread
Post by: atarz on March 19, 2016, 08:39:11 pm
Hi, just wondering,
we have only just done inverse trigonometry but there was a question in our test that we had not learnt how to attempt. We had to find the integral of the inverse of sin. How would you attempt this question?
Title: Re: 3U Maths Question Thread
Post by: Happy Physics Land on March 19, 2016, 10:16:02 pm
Hi, just wondering,
we have only just done inverse trigonometry but there was a question in our test that we had not learnt how to attempt. We had to find the integral of the inverse of sin. How would you attempt this question?

Hey atarz!

If possible can you please supply me with the actual question? If the question only asks for the integral of sin-1(x), then I have provided a solution below. It is quite infrequent for school exams to ask you such questions, especially when 4 unit candidates can solve it relatively easily using integration by parts. I cant see easier ways of doing this except for using Integration by parts which definitely isnt a 3 unit concept. You can try doing this question also with areas (i.e. the area under sin-1x). I'm suspecting that there may be a part beforehand in the question that may help with integrating inverse of sine using 3u methods?

Anyways, I have posted my solution below through integration by parts:

(http://i.imgur.com/wOiKW9q.jpg)

Best Regards
Happy Physics Land
Title: Re: 3U Maths Question Thread
Post by: jamonwindeyer on March 20, 2016, 03:17:50 am
Hey atarz!

If possible can you please supply me with the actual question? If the question only asks for the integral of sin-1(x), then I have provided a solution below. It is quite infrequent for school exams to ask you such questions, especially when 4 unit candidates can solve it relatively easily using integration by parts. I cant see easier ways of doing this except for using Integration by parts which definitely isnt a 3 unit concept. You can try doing this question also with areas (i.e. the area under sin-1x). I'm suspecting that there may be a part beforehand in the question that may help with integrating inverse of sine using 3u methods?

Anyways, I have posted my solution below through integration by parts:

(http://i.imgur.com/wOiKW9q.jpg)

Best Regards
Happy Physics Land

Nice! Another method for integration by parts would be by a substitution at the start, it simplifies things considerably. But yeah, I don't think this can be done without some application of integration by parts, not that I can see right now anyway! Perhaps I am spoiled by the method  ;)

$x = \sin{u} \\ dx = \cos{u}du \\ u=\sin^{-1}x$

$I = \int u\cos{u}du = u\sin{u}-\int\sin{u} \\ = u\sin{u}+\cos{u} + C \\ x\sin^{-1}{x}+\sqrt{1-x^2}+C$

Title: Re: 3U Maths Question Thread
Post by: RuiAce on March 20, 2016, 07:42:55 am
Hi, just wondering,
we have only just done inverse trigonometry but there was a question in our test that we had not learnt how to attempt. We had to find the integral of the inverse of sin. How would you attempt this question?

As this is a 3U forum, if you were asked such a question in the 3U exam there is a probability of 100% that the integral came with BOUNDARIES and was a DEFINITE integral

You need to draw out the graph of the inverse sine function, and then find
A = area of rectangle - integral of sin(y)dy for certain boundaries.

Integration by parts is NOT in the 3U course, as already mentioned.
Title: Re: 3U Maths Question Thread
Post by: Happy Physics Land on March 20, 2016, 11:30:45 am
As this is a 3U forum, if you were asked such a question in the 3U exam there is a probability of 100% that the integral came with BOUNDARIES and was a DEFINITE integral

You need to draw out the graph of the inverse sine function, and then find
A = area of rectangle - integral of sin(y)dy for certain boundaries.

Integration by parts is NOT in the 3U course, as already mentioned.

OMG yes.... after all inverse sine x comes from the original curve sine y..... of course... this way the question is actually a lot easier. We are essentially just finding the area of sin y within the domain 0 <= x <= pi/2 and then using a rectangle to minus that area. And then because arcsin x is defined from -pi/2 to pi/2, just times the resultant area by 2. Am I right in saying this rui?
Title: Re: 3U Maths Question Thread
Post by: starkidshani on March 20, 2016, 12:13:47 pm
Hi,
I'm having trouble with an Integration By Substitution question:

∫(x-2)/(√x+2) dx
using u^2 =x+2

Can you help me out? Thanks very much :)

Title: Re: 3U Maths Question Thread
Post by: Happy Physics Land on March 20, 2016, 01:40:06 pm
Hi,
I'm having trouble with an Integration By Substitution question:

∫(x-2)/(√x+2) dx
using u^2 =x+2

Can you help me out? Thanks very much :)

Hey shani!

I have posted the solution below, l guess the important thing in this question is to realise that x-2 = u-4 which is kind of tricky to realise I guess (Btw Im just assuming here that you meant sqrt(x+2) not sqrt(x) + 2. If the denominator is sqrt(x) + 2 then the outcome would be different). If you have any further questions please don't hesitate to ask! :)

(http://i.imgur.com/K4JkC88.jpg)

Best Regards
Happy Physics Land
Title: Re: 3U Maths Question Thread
Post by: jamonwindeyer on March 20, 2016, 01:53:24 pm
OMG yes.... after all inverse sine x comes from the original curve sine y..... of course... this way the question is actually a lot easier. We are essentially just finding the area of sin y within the domain 0 <= x <= pi/2 and then using a rectangle to minus that area. And then because arcsin x is defined from -pi/2 to pi/2, just times the resultant area by 2. Am I right in saying this rui?

It would depend on the exact boundaries! The boundaries would determine how much area of the siny curve you need, and then if we have boundaries either side of x=0, then yep we can multiply by two to take advantage of symmetry!  For example, if the boundaries are from x=1 to -1, you would find the area under the sine curve from y=0 to pi/2, subtract from the rectangle, and then multiply by two, if that makes sense :)
Title: Re: 3U Maths Question Thread
Post by: RuiAce on March 20, 2016, 06:47:45 pm
Hi,
I'm having trouble with an Integration By Substitution question:

∫(x-2)/(√x+2) dx
using u^2 =x+2

Can you help me out? Thanks very much :)

Not too sure why HPL used that substitution when you were given that specific one.
{u}^{2}=x+2\Rightarrow2u\,du=dx \\ \left(u>0\right) \\\begin{align*}\int{\frac{x-2}{\sqrt{x+2}}\,dx}&=\int{\frac{{u}^{2}-4}{u}\,\left(2u\,du\right)}\\&=\int{\left(2{u}^{2}-8\right)du}\\&=\frac{2{u}^{3}}{3}-8u+C\\&=\frac{2\sqrt{{\left(x+2\right)}^{3}}}{3}-8\sqrt{x+2}+C\end{align*}
Title: Re: 3U Maths Question Thread
Post by: Happy Physics Land on March 20, 2016, 07:43:11 pm
Not too sure why HPL used that substitution when you were given that specific one.
{u}^{2}=x+2\Rightarrow2u\,du=dx \\ \left(u>0\right) \\\begin{align*}\int{\frac{x-2}{\sqrt{x+2}}\,dx}&=\int{\frac{{u}^{2}-4}{u}\,\left(2u\,du\right)}\\&=\int{\left(2{u}^{2}-8\right)du}\\&=\frac{2{u}^{3}}{3}-8u+C\\&=\frac{2\sqrt{{\left(x+2\right)}^{3}}}{3}-8\sqrt{x+2}\end{align*}

Wait what I thought I did use that substitution? Its just with the numerator I subtracted 4 from u to make it the same as the numerator.
Title: Re: 3U Maths Question Thread
Post by: jamonwindeyer on March 20, 2016, 08:39:28 pm
Wait what I thought I did use that substitution? Its just with the numerator I subtracted 4 from u to make it the same as the numerator.

You used:

$u=x+2$

RuiAce used:

$u^2=x+2$

Both work quite well actually!
Title: Re: 3U Maths Question Thread
Post by: Happy Physics Land on March 20, 2016, 09:25:42 pm
You used:

$u=x+2$

RuiAce used:

$u^2=x+2$

Both work quite well actually!

Yeah I agree Jamon, I realised that only shortly after I posted my comment. I guess at the end of the day it doesnt matter what you substitute it with. After all, substitution is only a technique, and if that technique helps you to solve the question then I guess regardless which mechanism you adopt you would get the same outcome.
Title: Re: 3U Maths Question Thread
Post by: RuiAce on March 20, 2016, 09:31:59 pm
Yeah I agree Jamon, I realised that only shortly after I posted my comment. I guess at the end of the day it doesnt matter what you substitute it with. After all, substitution is only a technique, and if that technique helps you to solve the question then I guess regardless which mechanism you adopt you would get the same outcome.

You mean in 4U and beyond it doesn't matter.

You should, of course, use the substitution given to you in 3U.
Title: Re: 3U Maths Question Thread
Post by: Neutron on March 24, 2016, 10:12:49 pm
Hey guys!

I was wondering whether you could show me how to sketch the function y=tan-1(tanx) over the domain -π≤x≤π <3 thank youu

Neutron
Title: Re: 3U Maths Question Thread
Post by: Happy Physics Land on March 24, 2016, 10:27:48 pm
Hey guys!

I was wondering whether you could show me how to sketch the function y=tan-1(tanx) over the domain -π≤x≤π <3 thank youu

Neutron

Hey Neutron!

Tan-1(tanx) will cancel each other out, leaving us simply with y = x. So all you need to do is plot the points (pi, pi) and
(-pi, -pi), and then connect these two points (essentially just the linear graph y=x).

If you really need me to show you the sketch just tell me, but Im sure you are capable of doing it! :)

Best Regards
Happy Physics Land
Title: Re: 3U Maths Question Thread
Post by: chloe9756 on March 25, 2016, 11:20:41 am
hey, i've been stuck on this question. i keep getting the answer 30.

In a tennis club, there are five married couples available to play a "mixed doubles" match, that is, a match in which a combination of one man and one woman play against a combination of another man and another woman. In haw many ways can a group of four persons be chosen for this match if:

(b) a man and his wife may not play in the match either as partners or as opponents.

Title: Re: 3U Maths Question Thread
Post by: atarz on March 25, 2016, 11:59:49 am
Thanks Happy Physics land, I think the markers wanted us to work it out by finding the area under the graph in relation to the y-axis. Thank you for the integration by parts though, it was very helpful.  :)
Title: Re: 3U Maths Question Thread
Post by: jamonwindeyer on March 25, 2016, 12:24:57 pm
hey, i've been stuck on this question. i keep getting the answer 30.

In a tennis club, there are five married couples available to play a "mixed doubles" match, that is, a match in which a combination of one man and one woman play against a combination of another man and another woman. In haw many ways can a group of four persons be chosen for this match if:

(b) a man and his wife may not play in the match either as partners or as opponents.

Hey Chloe!

Okay, so we have here a combinations question! I find the best way to do these is to separate into any separate categories, in this case, men and women. Doing that, let's proceed!

So let's say we choose the two males first. This would be a combination of 2 males from a possible 5, and so:

$\text{Male Choices} = (^5_2) = 10$

Now, we can pick their partners. Two of the women are not allowed to be chosen, because their husbands are in the game already. Therefore, we are selecting 2 women from 3 possible choices:

$\text{Female Choices} = (^3_2) = 3$

Now at this point, the answer of 30 becomes apparent:

$\text{Male Choices}\times\text{Female Choices}=30$

Now, I was a little confused, and I think the wording of the question is poor here. We haven't considered which team the women choose to join in the second step. That is, when we choose the women, they can join the teams in one way, or they can swap sides and join in another way. So, we multiply our final answer by 2:

$\text{Final Answer} = 30\times2 =60$

Now, the question only said how many ways the groups can be chosen, which I suppose implicitly suggests that we consider different teams as well, but I'd totally forgive someone for not doing it, because technically, changing the teams does not change the group.

But yep, that's where the 60 comes from. Hope this helps Chloe!  ;D

Title: Re: 3U Maths Question Thread
Post by: Neutron on March 25, 2016, 11:33:00 pm
Hey Neutron!

Tan-1(tanx) will cancel each other out, leaving us simply with y = x. So all you need to do is plot the points (pi, pi) and
(-pi, -pi), and then connect these two points (essentially just the linear graph y=x).

If you really need me to show you the sketch just tell me, but Im sure you are capable of doing it! :)

Best Regards
Happy Physics Land

Yeah I know that but the answer gave me something very trippy :'( I'll see if I can upload a pic later D:
Title: Re: 3U Maths Question Thread
Post by: RuiAce on March 25, 2016, 11:48:04 pm
Yeah I know that but the answer gave me something very trippy :'( I'll see if I can upload a pic later D:

$\text{What HPL did wrong was that he considered the graph of }y=\tan{\left(\tan^{-1}{x}\right)}\\\text{Let us first consider the graph of } y=\tan{x} \text{ for the domain } -\frac{\pi}{2} < x < \frac{\pi}{2}\\\text{This just gives us one ordinary section of the tangent curve. Now, take note that over the domain }-\frac{\pi}{2} < x < \frac{\pi}{2}\\ \text{we actually WILL get the immediate result of }\tan^{-1}{\left(\tan{x}\right)}=x$
$\text{So for this domain, we do indeed sketch the line }y=x\text{ taking careful note to mark a discontinuity at the endpoints (hole in graph.)}$

$\text{Regard, now, the domain }\frac{\pi}{2} < x < \pi \\ \text{Obviously, }y=\tan{x} \text{ looks the same for this domain as for} -\frac{\pi}{2} < x < 0$

$\text{And here's the trap. Take careful note that the identity} \\ \tan{\left(\tan^{-1}{x}\right)}=x \\ \text{does hold true for all real }x. \\ \text{This occurs because the inverse tangent function is monotone and has no discontinuity!}$

$\text{On the contrary, }\tan^{-1}{\left(\tan{x}\right)}=x \text{ only holds true for }-\frac{\pi}{2} < x < \frac{\pi}{2}! \\ \text{This occurs because the tangent function, whilst it is monotonic increasing, has one discontinuity every }\pi\text{ intervals! (Periodicity of the tangent function)}\\ \text{The discontinuity takes the special form of an asymptote here, of course.}$

$\text{From here, you need to consider exactly what happens, and WHY the graph actually is just the line }y=x \dots \\ \dots \text{SHIFTED.}$

Your hint is to think about both the domains AND ranges of both the tangent function, and it's inverse.
Title: Re: 3U Maths Question Thread
Post by: Happy Physics Land on March 26, 2016, 12:01:31 am
$\text{What HPL did wrong was that he considered the graph of }y=\tan{\left(\tan^{-1}{x}\right)}\\\text{Let us first consider the graph of } y=\tan{x} \text{ for the domain } -\frac{\pi}{2} < x < \frac{\pi}{2}\\\text{This just gives us one ordinary section of the tangent curve. Now, take note that over the domain }-\frac{\pi}{2} < x < \frac{\pi}{2}\\ \text{we actually WILL get the immediate result of }\tan^{-1}{\left(\tan{x}\right)}=x$
$\text{So for this domain, we do indeed sketch the line }y=x\text{ taking careful note to mark a discontinuity at the endpoints (hole in graph.)}$

$\text{Regard, now, the domain }\frac{\pi}{2} < x < \pi \\ \text{Obviously, }y=\tan{x} \text{ looks the same for this domain as for} -\frac{\pi}{2} < x < 0$

$\text{And here's the trap. Take careful note that the identity} \\ \tan{\left(\tan^{-1}{x}\right)}=x \\ \text{does hold true for all real }x. \\ \text{This occurs because the inverse tangent function is monotone and has no discontinuity!}$

$\text{On the contrary, }\tan^{-1}{\left(\tan{x}\right)}=x \text{ only holds true for }-\frac{\pi}{2} < x < \frac{\pi}{2}! \\ \text{This occurs because the tangent function, whilst it is monotonic increasing, has one discontinuity every }\pi\text{ intervals! (Periodicity of the tangent function)}\\ \text{The discontinuity takes the special form of an asymptote here, of course.}$

$\text{From here, you need to consider exactly what happens, and WHY the graph actually is just the line }y=x \dots \\ \dots \text{SHIFTED.}$

Your hint is to think about both the domains AND ranges of both the tangent function, and it's inverse.

Yeah I know that but I was assuming a -pi <= x <= pi
Title: Re: 3U Maths Question Thread
Post by: RuiAce on March 26, 2016, 12:04:55 am
Yeah I know that but I was assuming a -pi <= x <= pi

Tangent discontinues at pi/2. Not pi.
Title: Re: 3U Maths Question Thread
Post by: Happy Physics Land on March 26, 2016, 12:42:38 am
Tangent discontinues at pi/2. Not pi.

Oh
wait
Yeah
True......
Sorry Neutron bro I failed you!
Title: Re: 3U Maths Question Thread
Post by: amandali on March 26, 2016, 06:47:02 am
How do you do the last part  for the first question ?
Title: Re: 3U Maths Question Thread
Post by: RuiAce on March 26, 2016, 09:08:58 am
How do you do the last part  for the first question ?

Mind posting the solutions to the previous parts? They seem to be useful for the answer to d)

Not sure if the perms/combs question was intentionally placed there but that one isn't too bad.
$\text{Trying to identify what we need is easy. It is only the final result that is a bit awkward to deduce.}\\\text{The amount of ways 6 people can be chosen from 15 is naturally}{15\choose 6}\\\text{and the amount of ways 6 people can be arranged in a circle is }5!$

$\text{Effectively, we want to combine the situations, so we would like to multiply our results to get an unsimplified answer of:} \\ 5!{15\choose 6}\\\text{So we can obviously rewrite the binomial coefficient in terms of it's definition:}\\5!\cdot\frac{15!}{9!6!}$

\text{A sneaky factorial trick comes into play here. Note that}\\\begin{align*}n!&=n(n-1)(n-2)\dots3\cdot2\cdot1\\&=n\left((n-1)(n-2)\dots3\cdot2\cdot1\right)\\&=n(n-1)!\end{align*}\\\text{This effectively allows us to rewrite the 6! into something more convenient in the denominator.}

$5!\cdot\frac{15!}{9!\cdot6\cdot5!} \\ \text{And now the 5! cancels out to leave us with our final answer} \\ \frac{15!}{9!\cdot6}$
Title: Re: 3U Maths Question Thread
Post by: RuiAce on March 26, 2016, 09:22:58 am
How do you do the last part  for the first question ?

$\text{Ok, it seems as though}\\{A}_{n}=5000\left(\left(1+r\right)+{\left(1+r\right)}^{2}+\dots+{\left(1+r\right)}^{n}\right)-M\left(1+\left(1+r\right)+\dots+{\left(1+r\right)}^{n-1}\right)\\{A}_{n}=\frac{5000\left(1+r\right)\left({\left(1+r\right)}^{n}-1\right)}{r}-\frac{M\left(1+r\right)\left({\left(1+r\right)}^{n}-1\right)}{r}$
$\text{So we need}\\0=\frac{5000\left(1+r\right)\left({\left(1+r\right)}^{20}-1\right)}{r}-\frac{8000\left({\left(1+r\right)}^{n}-1\right)}{r}\\\because r\neq0\\0=5000\left(1+r\right)\left({\left(1+r\right)}^{20}-1\right)-8000\left({\left(1+r\right)}^{n}-1\right)\\0=\left({\left(1+r\right)}^{20}-1\right)\left(5000\left(1+r\right)-8000\right)\\\because r > 0\\0=\left(5000\left(1+r\right)-8000\right)$

Except, of course, this is the rigorous approach. I'm not too sure what the question means by to use the calculator.
Title: Re: 3U Maths Question Thread
Post by: xus2016 on March 27, 2016, 02:16:09 pm
How do you integrate cosec3xcot3x?
Title: Re: 3U Maths Question Thread
Post by: Happy Physics Land on March 27, 2016, 04:30:49 pm
How do you integrate cosec3xcot3x?

Hey Xus!

Interesting question! Ok to do this question, we would use 3 unit integration by substitution, I have posted my solution below, if you have any further questions please dont hesitate to ask! :) :)

(Sorry if the image is a bit blurry, still using brick phones hehe :D )

(http://i.imgur.com/2qcrkrq.jpg)

Best Regards
Happy Physics Land
Title: Re: 3U Maths Question Thread
Post by: amandali on March 27, 2016, 04:55:00 pm
Mind posting the solutions to the previous parts? They seem to be useful for the answer to d)

Not sure if the perms/combs question was intentionally placed there but that one isn't too bad.
$\text{Trying to identify what we need is easy. It is only the final result that is a bit awkward to deduce.}\\\text{The amount of ways 6 people can be chosen from 15 is naturally}{15\choose 6}\\\text{and the amount of ways 6 people can be arranged in a circle is }5!$

$\text{Effectively, we want to combine the situations, so we would like to multiply our results to get an unsimplified answer of:} \\ 5!{15\choose 6}\\\text{So we can obviously rewrite the binomial coefficient in terms of it's definition:}\\5!\cdot\frac{15!}{9!6!}$

\text{A sneaky factorial trick comes into play here. Note that}\\\begin{align*}n!&=n(n-1)(n-2)\dots3\cdot2\cdot1\\&=n\left((n-1)(n-2)\dots3\cdot2\cdot1\right)\\&=n(n-1)!\end{align*}\\\text{This effectively allows us to rewrite the 6! into something more convenient in the denominator.}

$5!\cdot\frac{15!}{9!\cdot6\cdot5!} \\ \text{And now the 5! cancels out to leave us with our final answer} \\ \frac{15!}{9!\cdot6}$

these are the solutions for previous parts
Title: Re: 3U Maths Question Thread
Post by: RuiAce on March 27, 2016, 10:49:35 pm
How do you integrate cosec3xcot3x?
Hey Xus!

Interesting question! Ok to do this question, we would use 3 unit integration by substitution, I have posted my solution below, if you have any further questions please dont hesitate to ask! :) :)

(Sorry if the image is a bit blurry, still using brick phones hehe :D )

(http://i.imgur.com/2qcrkrq.jpg)

Best Regards
Happy Physics Land

$\text{Please be advised, that whilst the following integrals are not explicitly stated}\\\text{the first one was on the past HSC "Standard Integrals" sheet, which the reference sheet replaced in 2016.}\\\text{There should be no harm in utilising the following results appropriately.}\\\\\int{\sec{x}\tan{x}\,dx}=\sec{x}+C\\\int{\cot{x}\csc{x}\,dx}=-\csc{x}+C\\\int{\csc^{2}{x}\,dx}=-\cot{x}+C$

Note that the internationally accepted form of cosecant is csc, which is what LaTeX observes. Interpret as cosec always.
Title: Re: 3U Maths Question Thread
Post by: RuiAce on March 27, 2016, 10:58:14 pm

these are the solutions for previous parts
$\text{Ah, so me being ever so careless did misread a bit of the question after all! No worries, let's fix that}$
$\text{Now, I am interested in your answer for (ii).}\\\text{Note that when the rate is given }8\text{ percent we have}\\{A}_{n}=5000{\left(1.08\right)}^{n}-M\left(\frac{{1.08}^{n}-1}{1.08-1}\right)$
$\text{Suppose now, that we don't know the value of the rate. In that case, we have to replace all instances of 0.08 with }r\\{A}_{n}=5000{\left(1+r\right)}^{n}-M\left(\frac{{\left(1+r\right)}^{n}-1}{1+r-1}\right)$
$\text{Then, we are told both: }{A}_{20}=0,\, M=8000\\0=5000{\left(1+r\right)}^{20}-8000\left(\frac{{\left(1+r\right)}^{20}-1}{r}\right)$
$\text{Now, just use the calculator and repeatedly guess and check.}$
Title: Re: 3U Maths Question Thread
Post by: mmadeleine on March 29, 2016, 08:33:59 pm
Hi all,
Really struggling with part C of this 3D trig question, I just can't seem to get it out...
Help would be very much appreciated!

Thankyou!!  :)
Title: Re: 3U Maths Question Thread
Post by: katherine123 on March 29, 2016, 09:23:55 pm
Not sure how to get C
Title: Re: 3U Maths Question Thread
Post by: katherine123 on March 29, 2016, 09:38:08 pm
why is it   10C7(1/4)^7(3/4)3           not just (1/4)^7(3/4)3
C=combination
Title: Re: 3U Maths Question Thread
Post by: ATWalk on March 29, 2016, 10:04:33 pm
Hi all,
Really struggling with part C of this 3D trig question, I just can't seem to get it out...
Help would be very much appreciated!

Thankyou!!  :)

Hey, here's my solution. Sorry I had to upload in two photos because of file size limits. Also, sorry if it's a bit messy, I find 3D trig questions like this hard to do neatly. If you're confused feel free to reply again. Happy to help.
Title: Re: 3U Maths Question Thread
Post by: RuiAce on March 30, 2016, 12:19:41 am
Hi all,
Really struggling with part C of this 3D trig question, I just can't seem to get it out...
Help would be very much appreciated!

Thankyou!!  :)

I will assume the above solution is correct. I am just going to provide tips to doing 3D trigonometry questions.

1. Draw the 3D diagram (or use the given one where applicable), but then draw triangles SEPERATELY
2. Always keep in mind any common sides
3. Label angles where required.
Title: Re: 3U Maths Question Thread
Post by: RuiAce on March 30, 2016, 12:33:02 am
Not sure how to get C

$\text{This was quite a challenging multiple choice question from the 2013 HSC Mathematics Ext 1 final exam.}\\\text{The formal way to attempt this question is to consider cases seperately due to the presence of absolute value.}$
$\text{Recall from the definition of absolute value}\\ \left|x\right|= \begin{cases}x\text{ when }x\ge 0\\-x\text{ when }x<0\end{cases}$
$\text{So we have}\\\left|x+2\right|+\left|x-3\right|=\begin{cases}-(x+2)-(x-3)\text{ when }x<-2\x+2)-(x-3)\text{ when }-2\le x<3\\(x+2)+(x-3)\text{ when }x\ge 3\end{cases}$ \text{Case 1: For }x<-2:\\ \begin{align*}-2x+1&=5\\\Leftrightarrow x&=-2\end{align*}\\\text{Therefore this case yields no solution} $\text{Case 2: For }-2\le x<3:\\ 5=5\\\text{This is trivially true for all values of }x\\\text{Thus, the entire domain }-2\le x<3\text{ belongs to the solution}$ \text{Case 3: For }x\ge 3:\\\begin{align*}2x-1&=5\\\Leftrightarrow x&=3\end{align*}\\\text{Therefore, this case yields the one solution }x=3 $\text{Combining all cases, we have:}\\ -2\le x\le 3\\\text{Which is by default the solution to the inequality}\\(x+2)(x-3)\le 0 \\ \text{i.e. } {x}^{2}-x-6\le 0$ Edited on: Also, take note that for the equation y=|x+2|+|x-3| one is able to sketch the corresponding graph by using either: a) Addition of ordinates b) Considering the cases seperately like as above. Title: Re: 3U Maths Question Thread Post by: RuiAce on March 30, 2016, 12:38:10 am why is it 10C7(1/4)^7(3/4)3 not just (1/4)^7(3/4)3 C=combination $\text{According to the binomial probability distribution, the answer is indeed automatically }\binom{10}{7}{\left(\frac{1}{4}\right)}^{7}{\left(\frac{3}{4}\right)}^{3}$ $\text{This is because the student can get ANY of three questions wrong out of the bundle of ten.} \\ \text{In effect, when the student gets Q1, 5 and 8 wrong, is a seperate case to getting Q2, 5 and 10 wrong.}$ $\text{The whole basis of binomial probability is to address the multitude of cases that arise when dealing with such vague questions.}\\\text{The most simplistic analogy is that tossing HHT on a coin is different to HTH, but marks the same overall outcome of 2 heads 1 tail.}$ Title: Re: 3U Maths Question Thread Post by: katherine123 on March 30, 2016, 11:27:36 am Ive attached the question and the answer to part ii Im not sure why rx=t Title: Re: 3U Maths Question Thread Post by: jakesilove on March 30, 2016, 01:16:01 pm Ive attached the question and the answer to part ii Im not sure why rx=t Hey! Great question, as the explanation isn't great. Basically, it's just an algebraic trick that let's us use the results from part i) in part ii)! We know that a solution to $e^t=\frac{1}{t}$ is approximately 0.56 from part i). This holds true for any value of t: ie. for any formula e to the power of x equals 1 over x, a rough solution is 0.56. We therefore know that a solution to the equation $e^{rx}=\frac{1}{rx}$ also has a solution at $rx=0.56$ however it may be tough for some students to see exactly why (the fact that this is the same equation as the one in part i)). Therefore, the answers let rx = t to make it visually easier to see why the answer translates. I hope this makes sense; it's a little profound, but once you've wrapped your head around it it makes sense. Jake Title: Re: 3U Maths Question Thread Post by: katherine123 on March 31, 2016, 06:16:36 am part iii) please thanks Title: Re: 3U Maths Question Thread Post by: RuiAce on March 31, 2016, 08:57:58 am part iii) please thanks $\text{Recap for anyone else interested:}\\\text{(i) is use of the exterior angle of a triangle, and (ii) is the alternate segment theorem}$ $\text{In the larger circle, according to the alternate segment theorem we have}\\\angle TAC=\angle ABD\\\text{and in the smaller circle, the same theorem also tells us that}\\\angle XDB=\angle BAD$ $\text{Upon equating the expressions in (i) and (ii) we arrive at}\\\angle ABD+\angle XDB=\angle TAC+\angle CAD\\\text{So upon substituting in what we just found out}$ $\angle ABD+\angle BAD=\angle ABD+\angle CAD\\\angle BAD=\angle CAD\\\text{Hence }AD\text{ bisects }\angle BAC$ Title: Re: 3U Maths Question Thread Post by: RussellKindler on April 01, 2016, 02:33:13 pm Hi could i please have help with this question: DIfferentiate: $y=ln(cos^-1(2x)^2$ Thanks a lot in advance. Title: Re: 3U Maths Question Thread Post by: jakesilove on April 01, 2016, 02:59:07 pm Hi could i please have help with this question: DIfferentiate: $y=ln(cos^-1(2x)^2$ Thanks a lot in advance. Hey Russel! I really hope that I interpreted your question correctly (that it was inverse cos squared, not 2x squared). This was a really tough question, because you need to use so many different rules (Chain, ln, inverse trig etc.). However, I've posted my solution below! Hope it helps :) (http://i.imgur.com/dhhsNPS.png?1) Jake Title: Re: 3U Maths Question Thread Post by: imtrying on April 02, 2016, 07:50:47 pm Hey:) Just wondering if someone could give me a hand with the attached question? Thank you so much :D Title: Re: 3U Maths Question Thread Post by: jakesilove on April 02, 2016, 07:59:04 pm Hey:) Just wondering if someone could give me a hand with the attached question? Thank you so much :D Hey! For this question, we just need to know the general domain of the inverse sin graph. To know this, either you can recall the graph, test it on a calculator, or memorise the domain! We know (from one of these methods) that the domain of $y = sin^{-1}(a)$ is $-1< a <1$ Therefore, for $y = 2sin^{-1}(\frac{x}{2})$ We just set $-1<\frac{x}{2}<1$ Therefore $-2 < x < 2$ is the answer! Hope this helps :) Jake Title: Re: 3U Maths Question Thread Post by: ATWalk on April 04, 2016, 11:57:53 am Hey, Could I please have help on part iii of this question? It's really confusing me. Thanks in advance. Title: Re: 3U Maths Question Thread Post by: jakesilove on April 04, 2016, 01:50:17 pm Hey, Could I please have help on part iii of this question? It's really confusing me. Thanks in advance. Hey!! This was actually a crazy projectile motion question: by far the hardest I've ever seen. I don't know if the method I used is the easier, or if it will even make a lot of sense once I've written it out, but it does get to the right answer. I'm sorry I can't be of more help: I'm kind of still reeling from the difficulty of the question! I hope the answer does make sense, but if you'd like me to clarify any specific bit please let me know. (http://i.imgur.com/EdMsnri.png?1) (http://i.imgur.com/tKFp3eF.png?1) Jake Title: Re: 3U Maths Question Thread Post by: ATWalk on April 04, 2016, 03:22:05 pm Thanks Jake! Unfortunately I've also got another question that I'm finding difficult and I'd be grateful for help... Hopefully this one is easier to solve. Could I have help just with part i? Part ii I'll attempt later. Title: Re: 3U Maths Question Thread Post by: jakesilove on April 04, 2016, 06:52:24 pm Thanks Jake! Unfortunately I've also got another question that I'm finding difficult and I'd be grateful for help... Hopefully this one is easier to solve. Could I have help just with part i? Part ii I'll attempt later. Hey! I'm sorry that I don't have time at the moment to figure out where I went wrong: That is the general structure, and I hope it helps! You're asking some crazy hard questions. (http://i.imgur.com/lNUpdMW.png?1) Jake Title: Re: 3U Maths Question Thread Post by: Goodwil on April 07, 2016, 04:55:22 pm Hi there, I'm having trouble with these two questions. Any help would be appreciated. Thanks 1. Show that f(x)= x^3+2x-4 has only 1 root 2. Given log(xy^3)=m and log(x^3y^2)=p, find log(sqrt(xy)) in terms of m and p (all logs are base 'b') Title: Re: 3U Maths Question Thread Post by: jakesilove on April 07, 2016, 05:22:38 pm Hi there, I'm having trouble with these two questions. Any help would be appreciated. Thanks 1. Show that f(x)= x^3+2x-4 has only 1 root 2. Given log(xy^3)=m and log(x^3y^2)=p, find log(sqrt(xy)) in terms of m and p (all logs are base 'b') Hey! These are some seriously tricky questions, but once you've noted the general technique you'll be able to approach similar questions with ease. Take a look at my working! (http://i.imgur.com/6jfEI6b.png?1) Jake Title: Re: 3U Maths Question Thread Post by: katherine123 on April 09, 2016, 10:34:48 am Should i be taken the positive velocity since v>0 when t=0 ans for part ii)1-e^-t Title: Re: 3U Maths Question Thread Post by: jakesilove on April 09, 2016, 10:58:20 am Should i be taken the positive velocity since v>0 when t=0 ans for part ii)1-e^-t Hey! My proof is below :) this question is mainly about applying the relevant velocity/displacement formulas, and then ensuring you integrate correctly. I hope the solution helps! (http://i.imgur.com/brhWLpH.png?1) Jake Title: Re: 3U Maths Question Thread Post by: Happy Physics Land on April 09, 2016, 06:02:42 pm Hey! My proof is below :) this question is mainly about applying the relevant velocity/displacement formulas, and then ensuring you integrate correctly. I hope the solution helps! (http://i.imgur.com/brhWLpH.png?1) Jake Actually I had almost the same question in my 3u exam, the hard part in this question for a lot of my mates was actually to realise the fact that you have to implement natural log in the integral Title: Re: 3U Maths Question Thread Post by: RuiAce on April 10, 2016, 09:07:38 am Actually I had almost the same question in my 3u exam, the hard part in this question for a lot of my mates was actually to realise the fact that you have to implement natural log in the integral There is a very VERY subtle trick placed there that reveals that an exponential might've been needed. If you look closely at the initial equation, recalling that v is just dx/dt, this is actually motion in the form of exponential decay. The final answer was virtually doomed to have an exponential within it. The consequence is that because we find dt/dx before dx/dt, we find t = f(x) + C first. Because f(t) and g(x) are bound to be mutually inverse functions (or even just by looking at the inverse function theorem dx/dt dt/dx = 1) it's essentially unavoidable that the natural log must appear halfway through. You can try explaining this to those kids at your school if you wish Title: Re: 3U Maths Question Thread Post by: katherine123 on April 16, 2016, 01:50:31 am a particle moves in SHM of period 8 hours and amplitude 6 metres. When t=3 hours, x=4m by using x-b=asin(wt) find centre of motion Title: Re: 3U Maths Question Thread Post by: RuiAce on April 16, 2016, 09:18:48 am a particle moves in SHM of period 8 hours and amplitude 6 metres. When t=3 hours, x=4m by using x-b=asin(wt) find centre of motion $\text{To find }w\text{ using the period:}\\T=\frac{2\pi}{w}\Leftrightarrow 8=\frac{2\pi}{w} \Leftrightarrow w=\frac{\pi}{4}\\\text{Then since the amplitude is 6 immediately deduce that }a=6\\\text{So right now we have}\\x-b=6\sin{\left(\frac{\pi}{4}t\right)}$ \text{When }t=3,\, x=4\\\begin{align*}\therefore 4-b&=6\sin{\frac{3\pi}{4}}\\\Leftrightarrow b&=4-6\left(\frac{1}{\sqrt{2}}\right)\\\Leftrightarrow b&=4-3\sqrt{2}\end{align*} Title: Re: 3U Maths Question Thread Post by: lazydreamer on April 17, 2016, 07:02:10 pm Hi, need some help with this differentiating trig func. question :) wasn't sure which thread this question should go but...here~ thanks in advance! Title: Re: 3U Maths Question Thread Post by: jakesilove on April 17, 2016, 07:44:30 pm Hi, need some help with this differentiating trig func. question :) wasn't sure which thread this question should go but...here~ thanks in advance! Hey! This was quite a tough one: I did it wrong twice, working through methods that ended up being way easier, but gave you a different result that became even more difficult to prove equaled the RHS. In the end, brute forcing the entire thing was the easiest way to go! I hope my answer below is helpful: If you're ever unsure, just attack the question using the rules you know! (http://i.imgur.com/eMi5Uy7.png?1) Jake Title: Re: 3U Maths Question Thread Post by: RuiAce on April 17, 2016, 10:58:59 pm Hey! This was quite a tough one: I did it wrong twice, working through methods that ended up being way easier, but gave you a different result that became even more difficult to prove equaled the RHS. In the end, brute forcing the entire thing was the easiest way to go! I hope my answer below is helpful: If you're ever unsure, just attack the question using the rules you know! (http://i.imgur.com/eMi5Uy7.png?1) Jake $\text{Since this is 3U, the shortcut was to bombard the above expressions using auxiliary angle method at the very start.}$ $\frac{d}{dx}\left(\frac{\sin{\left(x+\frac{\pi}{4}\right)}}{\cos{\left(x+\frac{\pi}{4}\right)}}\right)$ Title: Re: 3U Maths Question Thread Post by: jakesilove on April 17, 2016, 11:16:09 pm $\text{Since this is 3U, the shortcut was to bombard the above expressions using auxiliary angle method at the very start.}$ $\frac{d}{dx}\left(\frac{\sin{\left(x+\frac{\pi}{4}\right)}}{\cos{\left(x+\frac{\pi}{4}\right)}}\right)$ Wow I can't believe I missed that: A really simple solution to what is otherwise quite a difficult question. Thanks for pointing this out! Jake Title: Re: 3U Maths Question Thread Post by: lazydreamer on April 18, 2016, 10:10:02 am Hey! This was quite a tough one: I did it wrong twice, working through methods that ended up being way easier, but gave you a different result that became even more difficult to prove equaled the RHS. In the end, brute forcing the entire thing was the easiest way to go! I hope my answer below is helpful: If you're ever unsure, just attack the question using the rules you know! (http://i.imgur.com/eMi5Uy7.png?1) Jake oh i didn't try to approach it from the RHS...my bad haha $\text{Since this is 3U, the shortcut was to bombard the above expressions using auxiliary angle method at the very start.}$ $\frac{d}{dx}\left(\frac{\sin{\left(x+\frac{\pi}{4}\right)}}{\cos{\left(x+\frac{\pi}{4}\right)}}\right)$ Mind=blown xD thank you both for your help! Title: Re: 3U Maths Question Thread Post by: IkeaandOfficeworks on May 02, 2016, 05:15:58 pm Hi!, how do you find the inverse of y=x^{3}+3x. Thank you! :D Title: Re: 3U Maths Question Thread Post by: jakesilove on May 02, 2016, 05:56:37 pm Hi!, how do you find the inverse of y=x^{3}+3x. Thank you! :D Hey! This would require factorisation of cubic equations, which is immensely complicated and way beyond any curriculum dotpoint (you can check it out here http://www.math.vanderbilt.edu/~schectex/courses/cubic/). Is it possible that you have recorded the question wrong? Perhaps I have missed something, but I can't think of a way to solve this. Sorry! Jake Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on May 02, 2016, 10:59:36 pm Hey! This would require factorisation of cubic equations, which is immensely complicated and way beyond any curriculum dotpoint (you can check it out here http://www.math.vanderbilt.edu/~schectex/courses/cubic/). Is it possible that you have recorded the question wrong? Perhaps I have missed something, but I can't think of a way to solve this. Sorry! Jake IkeaandOfficeworks, in case you were still curious, this is courtesy of Wolfram Alpha ;) $f^{-1}{x}=\frac{\sqrt[3]{2}}{\sqrt[3]{\sqrt{x^2+4}-x}}-\frac{\sqrt[3]{\sqrt{x^2+4}-x}}{\sqrt[3]{2}}$ Title: Re: 3U Maths Question Thread Post by: amandali on May 03, 2016, 01:20:53 pm not sure how to do this for eg. (1+x)^3(1+x)^5 why is the coefficient (3C3)(5C0) + (3C2)(5C1) + (3C1)(5C2) + (3C0)(5C3) not (3C0)(5C0) + (3C1(5C1)..... Title: Re: 3U Maths Question Thread Post by: RuiAce on May 03, 2016, 02:30:06 pm not sure how to do this for eg. (1+x)^3(1+x)^5 why is the coefficient (3C3)(5C0) + (3C2)(5C1) + (3C1)(5C2) + (3C0)(5C3) not (3C0)(5C0) + (3C1(5C1)..... $\text{The question you typed up makes it a bit ambiguous as to what term you're interested in. What is that expression equal to?}\\\text{As for the question that you showed in the image}$ ${\left(1+x\right)}^{6}={\left(1+x\right)}^{3}{\left(1+x\right)}^{3}\\ \text{Comparing terms with }x^2:\\ \text{We extract }\binom{6}{2}\text{ from the LHS. From the RHS:}$ $\text{We extract }\binom{3}{0}\text{ from the first bracket, and }\binom{3}{2}x^2\text{ from the second bracket.}\\\text{We extract}\binom{3}{1}x\text{ from the first bracket, and }\binom{3}{1}x\text{ from the second bracket.}\\\text{We extract}\binom{3}{2}x^2 \text{ from the first bracket, and }\binom{3}{0}\text{ from the second bracket.}\\$ $\text{Hence, by equating coefficients on }x^2\\ \binom{6}{0}=\binom{3}{0}\binom{3}{2}+\binom{3}{1}+\binom{3}{1}+\binom{3}{2}\binom{3}{0}\\\text{to which the answer falls out.}$ Title: Re: 3U Maths Question Thread Post by: RuiAce on May 03, 2016, 02:36:09 pm $\text{If I am to assume that your question wants you to find an expression for }\binom{8}{3}\\\text{We then consider }{\left(1+x\right)}^{8}={\left(1+x\right)}^{3}{\left(1+x\right)}^{5}\text{ as you stated.}\text{The term in }x^3\text{ on the LHS is obviously }\binom{8}{3}$ $\text{In the RHS:}\\\text{We extract}\binom{3}{0}\text{ from the first bracket, and }\binom{5}{3}x^3\text{ from the second bracket.}\\\text{We extract}\binom{3}{1}x\text{ from the first bracket, and }\binom{5}{2}x^2\text{ from the second bracket.}\\\text{We extract}\binom{3}{2}x^2\text{ from the first bracket, and }\binom{5}{1}x\text{ from the second bracket.}\\\text{We extract}\binom{3}{3}x^3\text{ from the first bracket, and }\binom{5}{0}\text{ from the second bracket.}$ $\text{Hence} \\ \binom{8}{3}=\binom{3}{0}\binom{5}{3}+\binom{3}{1}\binom{5}{2}+\binom{3}{2}\binom{5}{1}+\binom{3}{3}\binom{5}{0}$ $\textbf{Alternatively}\text{, if we consider } {\left(1+x\right)}^{8}={\left(1+x\right)}^{3}{\left(x+1\right)}^{5}$ $\text{Using a similar approach, we find that }\\\binom{8}{3}=\binom{3}{0}\binom{5}{0}+\dots+\binom{3}{3}\binom{5}{3}$ $\textbf{This is not a coincidence. This is perfectly normal, as do not forget:}\\\binom{n}{k}=\binom{n}{n-k}$ Title: Re: 3U Maths Question Thread Post by: katherine123 on May 05, 2016, 03:18:56 pm how do i find the coefficient of x^(-8) in the expansion of (x-1/x)^6 times (x+1/x)^8 when im asked to find the greatest coefficient eg. (1-2x)^9 am i supposed to exclude the negative when im trying to find k from T(k+1)/T(k) >1 since the greatest coefficient is the number largest in magnitude how do i know if there r no 2 greatest coefficient with different sign eg. -80 and 80 Title: Re: 3U Maths Question Thread Post by: jakesilove on May 05, 2016, 03:49:17 pm how do i find the coefficient of x^(-8) in the expansion of (x-1/x)^6 times (x+1/x)^8 when im asked to find the greatest coefficient eg. (1-2x)^9 am i supposed to exclude the negative when im trying to find k from T(k+1)/T(k) >1 since the greatest coefficient is the number largest in magnitude how do i know if there r no 2 greatest coefficient with different sign eg. -80 and 80 Hey! I've put the answer to your first question below. You can definitely do it more algebraically, but I think it would take approximately the same amount of time. I also think that this method leaves you less likely to make a mistake! (http://i.imgur.com/v1S6jrw.png?1) Jake Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on May 05, 2016, 07:43:02 pm how do i find the coefficient of x^(-8) in the expansion of (x-1/x)^6 times (x+1/x)^8 when im asked to find the greatest coefficient eg. (1-2x)^9 am i supposed to exclude the negative when im trying to find k from T(k+1)/T(k) >1 since the greatest coefficient is the number largest in magnitude how do i know if there r no 2 greatest coefficient with different sign eg. -80 and 80 Hey Katherine! For your second question, I would look at it like this. If I had the question: $\text{What is the largest coefficient in the expansion of }(1-2x)^9\text{?}$ That is extremely ambiguous, and I think it is unlikely to be asked in that form. Much more likely is: $\text{What is the MAGNITUDE of the largest coefficient in the expansion of }(1-2x)^9\text{?}$ As you say, without specifying the magnitude, you really aren't sure whether the question wants the number furthest from zero on the number line (greatest magnitude), or the number furthest to the right (greatest in the algebraic sense). The answer to the question above would, by the way, be 5376! I hope this helps a little, sorry if I misinterpreted the question ;D Title: Re: 3U Maths Question Thread Post by: amandali on May 05, 2016, 09:57:30 pm is there a faster way to do this instead of sin both sides , let α=cos^-1(4/5) and β=cos^-1(3/5) and prove sin(β+α)=1 inverse cos^-1 (4/5) + inverse cos^-1(3/5) = pie/2 Title: Re: 3U Maths Question Thread Post by: jakesilove on May 05, 2016, 11:13:17 pm is there a faster way to do this instead of sin both sides , let α=cos^-1(4/5) and β=cos^-1(3/5) and prove sin(β+α)=1 inverse cos^-1 (4/5) + inverse cos^-1(3/5) = pie/2 Hey hey! I've attached my answer below: For questions like this, you generally want to set up some right angled triangles and figure it out from there. It usually isn't this conveniently set up. Generally, you'll create the triangles, find sine of alpha and sine of beta, cosine of alpha and cosine of beta, expand whatever you're trying to solve (using double angle formulas) and just plug in the values you've found. (http://i.imgur.com/xalJTa0.jpg) Jake Title: Re: 3U Maths Question Thread Post by: RuiAce on May 06, 2016, 09:44:36 am Hey hey! I've attached my answer below: For questions like this, you generally want to set up some right angled triangles and figure it out from there. It usually isn't this conveniently set up. Generally, you'll create the triangles, find sine of alpha and sine of beta, cosine of alpha and cosine of beta, expand whatever you're trying to solve (using double angle formulas) and just plug in the values you've found. (http://i.imgur.com/xalJTa0.jpg) Jake Food for thought $\text{That is always true for all general cases. However for problems such as this}\\\text{If a student can identify a Pythagorean triad then drawing a right angle triangle with BOTH }\alpha, \beta\text{ inside} \\ \text{is considerably faster as then you just use the angle sum of a right-angled triangle.}$ Title: Re: 3U Maths Question Thread Post by: RuiAce on May 06, 2016, 09:47:29 am how do i find the coefficient of x^(-8) in the expansion of (x-1/x)^6 times (x+1/x)^8 when im asked to find the greatest coefficient eg. (1-2x)^9 am i supposed to exclude the negative when im trying to find k from T(k+1)/T(k) >1 since the greatest coefficient is the number largest in magnitude how do i know if there r no 2 greatest coefficient with different sign eg. -80 and 80 $\text{With your second question, so as you and Jamon both affirmed we want the largest magnitude.}\\\text{When writing out the general term, what you should do is ISOLATE the }{(-1)}^{k}\\\text{ term, and then perform the subsequent calculations normally WITHOUT a -1 getting in the way.}$ $\text{For the given example, you will consider it as follows:}\\ T_{k}={(-1)}^k{x}^{k}\binom{9}{k}{2}^{k}\\ \Rightarrow \left| {C}_{k} \right|=\binom{9}{k}{2}^{k}$ $\text{As for if you turn up with both }-80\text{ and }80\text{, you may be forced to state both.}$ Note: Greatest coefficient has not been examined in the HSC since the 1980s or something. You must still study it, but it can be the least of your worries. Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on May 06, 2016, 11:42:22 am $\text{With your second question, so as you and Jamon both affirmed we want the largest magnitude.}\\\text{When writing out the general term, what you should do is ISOLATE the }{(-1)}^{k}\\\text{ term, and then perform the subsequent calculations normally WITHOUT a -1 getting in the way.}$ $\text{For the given example, you will consider it as follows:}\\ T_{k}={(-1)}^k{x}^{k}\binom{9}{k}{2}^{k}\\ \Rightarrow \left| {C}_{k} \right|=\binom{9}{k}{2}^{k}$ $\text{As for if you turn up with both }-80\text{ and }80\text{, you may be forced to state both.}$ Note: Greatest coefficient has not been examined in the HSC since the 1980s or something. You must still study it, but it can be the least of your worries. That's true, it was first asked in 1988 ;D But be careful, it still appears in Extension 1 Trial Papers relatively frequently, and from memory it popped up in one of my school assessments. Certain schools LOVE it, so I wouldn't dismiss it (and indeed, never dismiss anything!), it's a good thing to know. ;D My lecturer for Vector Calculus actually wrote something on this style of question, with some discussion and extension problems. For the curious, you can have a read here, but it is beyond the scope of the course ;D Title: Re: 3U Maths Question Thread Post by: amandali on May 09, 2016, 10:48:41 pm help with this ques thanks let n be a positive even integer expand and simplify (a+b)^n + (a-b)^n what i got was 2(nC0)a^n + 2(nC2)(a^(n-2))b^2 + 2(nC2)(a^(n-4))(b^4) ....+ 2(nCn)(b^n) Title: Re: 3U Maths Question Thread Post by: RuiAce on May 10, 2016, 09:23:59 am help with this ques thanks let n be a positive even integer expand and simplify (a+b)^n + (a-b)^n what i got was 2(nC0)a^n + 2(nC2)(a^(n-2))b^2 + 2(nC2)(a^(n-4))(b^4) ....+ 2(nCn)(b^n) This looks right Title: Re: 3U Maths Question Thread Post by: IkeaandOfficeworks on May 10, 2016, 11:35:04 am Hi guys! I need help with this question: tan^-1 (3/4) = 2tan^-1 (1/3) Thank you! Title: Re: 3U Maths Question Thread Post by: RuiAce on May 10, 2016, 12:13:25 pm Hi guys! I need help with this question: tan^-1 (3/4) = 2tan^-1 (1/3) Thank you! \text{To show that }\tan{(LHS)}=\tan{(RHS)}\\ \begin{align*}\tan{(RHS)}&=\tan{\left(2\tan^{-1}{\frac{1}{3}}\right)}\\&\stackrel{2\angle}{=}\frac{2\tan{\left(\tan^{-1}{\frac{1}{3}}\right)}}{1-\tan^2{\left(\tan^{-1}{\frac{1}{3}}\right)}}\\&=\frac{\frac{2}{3}}{1-\frac{1}{9}} \quad \because \tan{\left(\arctan{x}\right)}=x\\&=\frac{3}{4}\\&=\tan{\left(\tan^{-1}\frac{3}{4}\right)}\\&=\tan{(LHS)}\end{align*} $\text{Hence, by taking tangent inverse of both sides (permissible as arctan is continuous and monotone over }\mathbb R\text{)}\\ LHS=RHS\\ Q.E.D$ Title: Re: 3U Maths Question Thread Post by: katherine123 on May 14, 2016, 08:57:25 pm how to find coefficient of x^8 in the expansion of (x^2-2x+3)^5 Title: Re: 3U Maths Question Thread Post by: RuiAce on May 14, 2016, 09:37:41 pm how to find coefficient of x^8 in the expansion of (x^2-2x+3)^5 \text{By brute force, expand using the binomial theorem however being very careful:}\\ \begin{align*}&{\left(x^2-2x+3\right)}^5\\&=\binom{5}{0}x^{10}+\binom{5}{1}x^{8}{(3-2x)}+\binom{5}{2}x^6{(3-2x)}^2+\binom{5}{3}x^4{(3-2x)}^3+\binom{5}{4}x^2{(3-2x)}^4+\binom{5}{5}{(3-2x)}^5\end{align*} $\text{The purpose of this choice is to clear out the problematic }x^2\text{ term from causing too much chaos.}\\ \text{And now we attempt to extract any terms containing }x^8$ \begin{align*}\text{The term }&\binom{5}{0}x^{10}& \text{ has no term containing }x^8\\ \text{The term }&\binom{5}{1}x^8{(3-2x)}&\text{ contains the term }\binom{5}{1}{x}^{8}\binom{1}{0}3\\ \text{The term }&\binom{5}{2}x^6{(3-2x)}^2&\text{ contains the term }\binom{5}{2}x^6\binom{2}{2}{\left(-2x\right)}^2\\ \text{The term }&\binom{5}{3}x^4{(3-2x)}^3&\text{ has no term containing }x^8\end{align*}\\ \text{And neither does the remainder of the lot} $\text{Hence the coefficient of }x^8\text{ is }\\ \binom{5}{1}\binom{1}{0}3+\binom{5}{2}\binom{2}{2}{(-2)}^2$ Title: Re: 3U Maths Question Thread Post by: amandali on May 17, 2016, 05:45:44 pm (http://uploads.tapatalk-cdn.com/20160517/592faed7af0f73f148da4f3132e865a6.jpg) ques: 7 letter words are formed using the letters of UNUSUAL. One of the words is selected at random. find probability that the word had none of the U together im not sure what i did wrong but ans is 2/7 Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on May 17, 2016, 10:47:55 pm (http://uploads.tapatalk-cdn.com/20160517/592faed7af0f73f148da4f3132e865a6.jpg) ques: 7 letter words are formed using the letters of UNUSUAL. One of the words is selected at random. find probability that the word had none of the U together im not sure what i did wrong but ans is 2/7 Hey amandali! Your approach was definitely on the right track, you are very close!! The issue is that I think there are subtle things at play when considering 2U's together, I want to try considering it a little bit differently: Focusing on two U's together. Let us first consider the possibility that the two u's must be at either end. This can occur two ways. Then, we need another letter next to them (to block the 3rd U). We can choose from 4 remaining non-U's, so we then multiply by 4. Then, we arrange the remaining 4 letters, 4! Now, we actually haven't done anything that would require dividing by 3! here (we've not changed the order of the u's with respect to each other), so the answer is simply: $n_\text{ends}=2\times4\times4! = 192$ Now consider the two U's somewhere in the middle, as you have done. Stick the u's in the middle of the packet, that's where they must be, no probability involved yet. Now, we can pick two letters from four remaining options to stick either side. That is a combination. Then, we can swap their place, multiply by two (you could also just consider a permutation in the previous step). Then, we can order the word in 4! ways. That gives: $n_\text{middle}=\frac{4!}{2!2!}\times2\times4! = 288$ Add these, plus the 120 you got in the last part of your solution, and we end up with 600 total ways the letters can be arranged so that U's are together. Therefore: $P = 1-\frac{600}{840} = \frac{2}{7}$ I hope this helps! That was a tricky one, stumped me for a bit there! ;D Title: Re: 3U Maths Question Thread Post by: jakesilove on May 17, 2016, 11:30:47 pm Hey amandali! Your approach was definitely on the right track, you are very close!! The issue is that I think there are subtle things at play when considering 2U's together, I want to try considering it a little bit differently: Focusing on two U's together. Let us first consider the possibility that the two u's must be at either end. This can occur two ways. Then, we need another letter next to them (to block the 3rd U). We can choose from 4 remaining non-U's, so we then multiply by 4. Then, we arrange the remaining 4 letters, 4! Now, we actually haven't done anything that would require dividing by 3! here (we've not changed the order of the u's with respect to each other), so the answer is simply: $n_\text{ends}=2\times4\times4! = 192$ Now consider the two U's somewhere in the middle, as you have done. Stick the u's in the middle of the packet, that's where they must be, no probability involved yet. Now, we can pick two letters from four remaining options to stick either side. That is a combination. Then, we can swap their place, multiply by two (you could also just consider a permutation in the previous step). Then, we can order the word in 4! ways. That gives: $n_\text{middle}=\frac{4!}{2!2!}\times2\times4! = 288$ Add these, plus the 120 you got in the last part of your solution, and we end up with 600 total ways the letters can be arranged so that U's are together. Therefore: $P = 1-\frac{600}{840} = \frac{2}{7}$ I hope this helps! That was a tricky one, stumped me for a bit there! ;D This is the first time I've had a problem with factorials being donated as '!'. Reading through your (well written) answer, I kept seeing a super enthusiastic Jamon explaining how probability works. 'Now, we just sub this in! Then, divide by this! Isn't maths just so very fun!'. Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on May 18, 2016, 11:36:57 am This is the first time I've had a problem with factorials being donated as '!'. Reading through your (well written) answer, I kept seeing a super enthusiastic Jamon explaining how probability works. 'Now, we just sub this in! Then, divide by this! Isn't maths just so very fun!'. Ahaha, but maths is so very fun, right? :o Title: Re: 3U Maths Question Thread Post by: RuiAce on May 18, 2016, 05:40:42 pm Confession: I purposely didn't answer that question because I don't want to associate with perms and combs again until next semester when I'm doing it in discrete maths. Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on May 18, 2016, 10:24:03 pm Confession: I purposely didn't answer that question because I don't want to associate with perms and combs again until next semester when I'm doing it in discrete maths. Ahaha! Well if you handle the probability questions next semester I'll handle the integrals ;) Title: Re: 3U Maths Question Thread Post by: RuiAce on May 19, 2016, 01:01:09 pm Ahaha! Well if you handle the probability questions next semester I'll handle the integrals ;) I can handle probability :P just want to avoid perms and combs Title: Re: 3U Maths Question Thread Post by: amandali on May 20, 2016, 08:22:34 pm (http://uploads.tapatalk-cdn.com/20160520/80af3be4933c70eec30dd468e0d2ddde.jpg) need help with this ques thanks Title: Re: 3U Maths Question Thread Post by: RuiAce on May 20, 2016, 09:06:25 pm (http://uploads.tapatalk-cdn.com/20160520/80af3be4933c70eec30dd468e0d2ddde.jpg) need help with this ques thanks $\text{This is the somewhat infamous case of the 2n-th power equating terms.}$ $\text{Consider the expression }{(1+x)}^n{(1+x)}^n={(1+x)}^{2n}\\ \text{We do this because the coefficient of }x^n\text{ in the expansion of the RHS is conveniently just }\binom{2n}{n}$ $\text{This immediately implies that we need to consider the coefficient of }x^n\text{ in the LHS. Prepare to extract terms.}$ \begin{align*}\binom{n}{0}\text{ from the first bracket can be paired with }& \binom{n}{n}x^n\text{ from the second bracket}\\ \binom{n}{1}x\text{ from the first bracket can be paired with }& \binom{n}{n-1}x^{n-1}\text{ from the second bracket}\\\binom{n}{2}x^2\text{ from the first bracket can be paired with }& \binom{n}{n-2}x^{n-2}\text{ from the second bracket}\\ &\vdots\\ \binom{n}{n-1}x^{n-1}\text{ from the first bracket can be paired with }& \binom{n}{1}x\text{ from the second bracket}\\ \binom{n}{n}x^n\text{ from the first bracket can be paired with }& \binom{n}{0}\text{ from the second bracket}\\ \end{align*} $\text{Hence the coefficient of }x^n\text{ on the LHS is given}\\ \binom{n}{0}\binom{n}{n}+\binom{n}{1}\binom{n}{n-1}+\binom{n}{2}\binom{n}{n-2}+\dots+\binom{n}{n-1}\binom{n}{1}+\binom{n}{n}\binom{n}{0}\\=\binom{n}{0}^2+\binom{n}{1}^2+\binom{n}{2}^2+\dots+\binom{n}{n-1}+\binom{n}{n}\\ \because \binom{n}{k}=\binom{n}{n-k}$ Worthwhile mention: The harder binomial coefficient proofs always include at least one of the following: Symmetry property of binomial coefficient: nCk=nCn-k Pascal's identity:n+1Ck+1=nCk+nCk+1 Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on May 20, 2016, 11:03:37 pm $\text{This is the somewhat infamous case of the 2n-th power equating terms.}$ $\text{Consider the expression }{(1+x)}^n{(1+x)}^n={(1+x)}^{2n}\\ \text{We do this because the coefficient of }x^n\text{ in the expansion of the RHS is conveniently just }\binom{2n}{n}$ $\text{This immediately implies that we need to consider the coefficient of }x^n\text{ in the LHS. Prepare to extract terms.}$ \begin{align*}\binom{n}{0}\text{ from the first bracket can be paired with }& \binom{n}{n}x^n\text{ from the second bracket}\\ \binom{n}{1}x\text{ from the first bracket can be paired with }& \binom{n}{n-1}x^{n-1}\text{ from the second bracket}\\\binom{n}{2}x^2\text{ from the first bracket can be paired with }& \binom{n}{n-2}x^{n-2}\text{ from the second bracket}\\ &\vdots\\ \binom{n}{n-1}x^{n-1}\text{ from the first bracket can be paired with }& \binom{n}{1}x\text{ from the second bracket}\\ \binom{n}{n}x^n\text{ from the first bracket can be paired with }& \binom{n}{0}\text{ from the second bracket}\\ \end{align*} $\text{Hence the coefficient of }x^n\text{ on the LHS is given}\\ \binom{n}{0}\binom{n}{n}+\binom{n}{1}\binom{n}{n-1}+\binom{n}{2}\binom{n}{n-2}+\dots+\binom{n}{n-1}\binom{n}{1}+\binom{n}{n}\binom{n}{0}\\=\binom{n}{0}^2+\binom{n}{1}^2+\binom{n}{2}^2+\dots+\binom{n}{n-1}+\binom{n}{n}\\ \because \binom{n}{k}=\binom{n}{n-k}$ Worthwhile mention: The harder binomial coefficient proofs always include at least one of the following: Symmetry property of binomial coefficient: nCk=nCn-k Pascal's identity:n+1Ck+1=nCk+nCk+1 I remember the first time I tried this proof in my HSC I tried to do it with the full factorial expansions of each term, ever try it that way? I mean, not that you ever would, my teacher quickly steered me right ;D Title: Re: 3U Maths Question Thread Post by: RuiAce on May 20, 2016, 11:18:12 pm I remember the first time I tried this proof in my HSC I tried to do it with the full factorial expansions of each term, ever try it that way? I mean, not that you ever would, my teacher quickly steered me right ;D Well... $\binom{n}{0}^2+\binom{n}{1}^2+\dots+\binom{n}{n}^2\\=(n!)^2\left[\sum_{k=0}^{n}{\frac{1}{(k!)^2\left((n-k)!\right)^2}}\right]\\=\frac{(n!)^4}{(n!)^2}\left[\sum_{k=0}^{n}{\frac{1}{(k!)^2\left((n-k)!\right)^2}}\right]\stackrel{say}{=}\frac{(2n)!}{n!n!}\\ \Leftrightarrow (n!)^4 \sum_{k=0}^{n}{\frac{1}{(k!)^2\left((n-k)!\right)^2}} = (2n)!\\ \Leftrightarrow (n!)^3\sum_{k=0}^{n}{\frac{1}{(k!)^2\left((n-k)!\right)^2}}=(2n)(2n-1)\dots n$ You'd end up having to prove THAT ugly thing somehow, oh dear Not too sure if symmetry in the sum is required to get there without reverting steps and reintroducing the binomial coefficient but LOL not attempting that. Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on May 21, 2016, 12:47:58 am Well... $\binom{n}{0}^2+\binom{n}{1}^2+\dots+\binom{n}{n}^2\\=(n!)^2\left[\sum_{k=0}^{n}{\frac{1}{(k!)^2\left((n-k)!\right)^2}}\right]\\=\frac{(n!)^4}{(n!)^2}\left[\sum_{k=0}^{n}{\frac{1}{(k!)^2\left((n-k)!\right)^2}}\right]\stackrel{say}{=}\frac{(2n)!}{n!n!}\\ \Leftrightarrow (n!)^4 \sum_{k=0}^{n}{\frac{1}{(k!)^2\left((n-k)!\right)^2}} = (2n)!\\ \Leftrightarrow (n!)^3\sum_{k=0}^{n}{\frac{1}{(k!)^2\left((n-k)!\right)^2}}=(2n)(2n-1)\dots n$ You'd end up having to prove THAT ugly thing somehow, oh dear Not too sure if symmetry in the sum is required to get there without reverting steps and reintroducing the binomial coefficient but LOL not attempting that. Ahaha I think I got roughly to your third line before getting help back in 2014, definitely looks as disgusting as I remember it. If you are at UNSW and do Math 1131/41 next semester, you'll do stuff on series (mostly convergence tests though), so who knows, it might actually be doable :P Title: Re: 3U Maths Question Thread Post by: RuiAce on May 21, 2016, 08:50:38 am Ahaha I think I got roughly to your third line before getting help back in 2014, definitely looks as disgusting as I remember it. If you are at UNSW and do Math 1131/41 next semester, you'll do stuff on series (mostly convergence tests though), so who knows, it might actually be doable :P *51, taking maths with actuarial :P Too bad it's not an infinite sum or it would've been tempting to play around with a Taylor approximation. Finite sums are so different to limiting sums. Title: Re: 3U Maths Question Thread Post by: jakesilove on May 21, 2016, 10:12:58 am *51, taking maths with actuarial :P Too bad it's not an infinite sum or it would've been tempting to play around with a Taylor approximation. Finite sums are so different to limiting sums. I didn't realise you were at UNSW! Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on May 21, 2016, 11:56:48 am *51, taking maths with actuarial :P Too bad it's not an infinite sum or it would've been tempting to play around with a Taylor approximation. Finite sums are so different to limiting sums. Ah that's right I forgot they separated you guys Title: Re: 3U Maths Question Thread Post by: RuiAce on May 21, 2016, 01:46:30 pm I didn't realise you were at UNSW! Aha check signature :P Ah that's right I forgot they separated you guys I miss the days when maths used to be my easiest subject. Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on May 21, 2016, 08:55:41 pm Aha check signature :P I miss the days when maths used to be my easiest subject. Welcome to the dark side! Title: Re: 3U Maths Question Thread Post by: amandali on May 22, 2016, 02:03:03 pm (http://uploads.tapatalk-cdn.com/20160521/08452256328b16bf7ee1ee08f11e3ab0.jpg) i cant do part a) and part b (iii) for part b (i) i got [ (1+x)^(n+1)-1 ] / n Title: Re: 3U Maths Question Thread Post by: jakesilove on May 22, 2016, 02:21:54 pm (http://uploads.tapatalk-cdn.com/20160521/08452256328b16bf7ee1ee08f11e3ab0.jpg) i cant do part a) and part b (iii) for part b (i) i got [ (1+x)^(n+1)-1 ] / n Hey! This is a standard trick that you will learn to utilise as you continue doing these types of questions. Always just look for something that looks like it has been differentiated or integrated, and have a crack! (http://i.imgur.com/2bKjyXl.jpg?1) Jake Title: Re: 3U Maths Question Thread Post by: RuiAce on May 22, 2016, 02:32:07 pm (http://uploads.tapatalk-cdn.com/20160521/08452256328b16bf7ee1ee08f11e3ab0.jpg) i cant do part a) and part b (iii) for part b (i) i got [ (1+x)^(n+1)-1 ] / n $\text{Part ii gives }\\ 1+(1+x)+{(1+x)}^2+\dots+{(1+x)}^n=\frac{1\left({(1+x)}^{n+1}-1\right)}{1+x-1}=\frac{{(1+x)}^{n+1}-1}{x}$ $\text{In part iii, on the RHS we ignore the }-\frac{1}{x}\text{ for obvious reasons}\\ \text{The coefficient of }x^r\text{ in the expansion of }\frac{{(1+x)}^{n+1}}{x}\text{ is the SAME}\\\text{as the coefficient of }x^{r+1}\text{ in the expansion of }{(1+x)}^{n+1}\\ \text{So on the RHS we get }\binom{n+1}{r+1}\text{ as required.}$ Think about why that is true. $\text{In the LHS, take note that the expansion of ALL of these yields NO term with }x^r:\\ 1, (1+x), {(1+x)}^2, {(1+x)}^3, \dots, {(1+x)}^{r-1}$ $\text{The first term to yield }x^r\text{ is }{(1+x)}^r\text{ so we start counting from there.}$ \begin{align*}\text{From }{(1+x)}^r&\text{ we extract }\binom{r}{r}x^r\\\text{From }{(1+x)}^{r+1}&\text{ we extract }\binom{r+1}{r}x^r\\\text{From }{(1+x)}^{r+2}&\text{ we extract }\binom{r+2}{r}x^r\\ &\vdots \\\text{From }{(1+x)}^{n-2}&\text{ we extract }\binom{n-2}{r}x^r\\\text{From }{(1+x)}^{n-1}&\text{ we extract }\binom{n-1}{r}x^r\\\text{From }{(1+x)}^{n}&\text{ we extract }\binom{n}{r}x^r\end{align*} $\text{So the coefficient of }x^r\text{ on the LHS is}\\ \binom{n}{r}+\binom{n-1}{r}+\binom{n-2}{r}+\dots+\binom{r}{r}$ $\text{Then finally equate.}$ Title: Re: 3U Maths Question Thread Post by: katherine123 on May 24, 2016, 10:24:03 pm for part a) i got prob(win1st or 2nd turn)= p+rq and im not sure how to progress from there Title: Re: 3U Maths Question Thread Post by: RuiAce on May 25, 2016, 09:31:56 am for part a) i got prob(win1st or 2nd turn)= p+rq and im not sure how to progress from there $\text{This was the last question of the 2014 MX1 HSC paper. The trick is to draw an infinite tree diagram.}$ \text{To complete part a)}\\ \begin{align*}p+rq&=p+r(1-(p+r))\text{ (complementary probabilities})\\ &=p+r-pr-r^2 \\ &=(1-r)(p+r)\end{align*} $\text{For part b), it can be shown that}\\ \text{P(Win on 3rd or 4th)}=\left(r^2-r^3\right)(p+r)=r^2(1-r)(p+r)\\ \text{And so on. Proof is left as an exercise to the reader...}$ \text{So the limiting probability ends up being}\\ \begin{align*}P(\text{A wins})&=(1-r)(p+r)+r^2(1-r)(p+r)+r^4(1-r)(p+r)+\dots\\ &\stackrel{GP}{=}(1-r)(p+r)\frac{1}{1-r^2}\\ &=\frac{p+r}{1+r}\text{ by factorising the diff. of two square.}\end{align*} Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on May 25, 2016, 09:33:48 am for part a) i got prob(win1st or 2nd turn)= p+rq and im not sure how to progress from there Hey Katherine! I actually sat this HSC Paper, this is the last question, it was a bit of a doozy ;) Okay, so you are at this stage: $P(\text{Win in First Two Turns})=p+rq$ Great! Now, with these sorts of questions it is useful to keep an eye on where we are going. The required form has no 'q', so let's get rid of it. How? Well, we know the sum of all possible outcomes must be 1, therefore: $q=1-p-r \\ \therefore P = p+r(1-p-r) \\ = p+r-rp-r^2 \\ = (p+r)(1-r) \quad\text{ as required}$ Have a go at the next question yourself, or check the solution below if you get stuck! It is a tricky question: Spoiler Okay, so the fact that we are considering the probability that A eventually wins the game, suggests that we are going to need some sort of infinite sum. Now, in the paper this actually annoyed me, because Part A totally throws you off track about how to start Part B, only coming back in near the end of the problem. Tricky! Let's reconsider the probability of winning after the first or second turn: $P = p+rq$ Now the third turn, then the fourth turn: $P = p+rq+r^2p \\ P = p+rq+r^2p+r^3q$ What we have set up is a pair of geometric sequences, which we can manipulate using infinite sum formulae to get the answer. Note that we can use this formula because, in both cases, the common ratio is less than 1 ;D $P = p(1+r^2+r^4...)+q(r+r^3+r^5...)$ $\therefore P_\infty = p(\frac{1}{1-r^2})+q(\frac{r}{1-r^2}) \\ =\frac{p+rq}{1-r^2} \\ =\frac{(p+r)(1-r)}{(1+r)(1-r)} \quad\text{from above} \\ = \frac{p+r}{1+r} \quad\text{as required}$ Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on May 25, 2016, 09:35:20 am $\text{This was the last question of the 2014 MX1 HSC paper. The trick is to draw an infinite tree diagram.}$ \text{To complete part a)}\\ \begin{align*}p+rq&=p+r(1-(p+r))\text{ (complementary probabilities})\\ &=p+r-pr-r^2 \\ &=(1-r)(p+r)\end{align*} $\text{For part b), it can be shown that}\\ \text{P(Win on 3rd or 4th)}=\left(r^2-r^3\right)(p+r)=r^2(1-r)(p+r)\\ \text{And so on. Proof is left as an exercise to the reader...}$ \text{So the limiting probability ends up being}\\ \begin{align*}P(\text{A wins})&=(1-r)(p+r)+r^2(1-r)(p+r)+r^4(1-r)(p+r)+\dots\\ &\stackrel{GP}{=}(1-r)(p+r)\frac{1}{1-r^2}\\ &=\frac{p+r}{1+r}\text{ by factorising the diff. of two square.}\end{align*} Simultaneous solutions it seems ;D Virtually identical for you Katherine so either or ;D Title: Re: 3U Maths Question Thread Post by: RuiAce on May 25, 2016, 06:44:02 pm Simultaneous solutions it seems ;D Virtually identical for you Katherine so either or ;D You know I sat your MX1 paper right :P Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on May 25, 2016, 09:34:31 pm You know I sat your MX1 paper right :P I do, you mentioned it a little why ago I'm pretty sure (during that conversation when I mentioned that it was an easier paper than usual) :o Title: Re: 3U Maths Question Thread Post by: amandali on May 27, 2016, 05:22:34 pm (http://uploads.tapatalk-cdn.com/20160527/dcbb8ed6d88c7b128d76746d4e8c5a9f.jpg)h how to do ques 2 thanks Title: Re: 3U Maths Question Thread Post by: katherine123 on May 27, 2016, 06:04:44 pm question: a die is rolled 100 times. What is the most likely number of sixes that will be thrown? Ans: 16 am i supposed to use the greatest coefficient for these kind of ques cuz the answer i got was wrong using this method nCr*(1/6)^r*(5/6)^(n-r) divided by nC(r-1)*(1/6)^(r-1)*(5/6)^(n-r+1) or am i supposed to do (1/6)*100 Title: Re: 3U Maths Question Thread Post by: jakesilove on May 27, 2016, 11:17:47 pm (http://uploads.tapatalk-cdn.com/20160527/dcbb8ed6d88c7b128d76746d4e8c5a9f.jpg)h how to do ques 2 thanks Hey! This was a bit of a tricky question, and I actually think I formulated my logic wrong (I assumed that you can only form a parcel from the shirts made in a specific day, but then the maximum number of shirts that could be used from the sister is 20, so that doesn't quite work out). However, I did get to the right answer! I hate probability, but I hope this helps you out a bit :) (http://i.imgur.com/wUbNNhy.png?1) Jake Title: Re: 3U Maths Question Thread Post by: jakesilove on May 27, 2016, 11:29:33 pm question: a die is rolled 100 times. What is the most likely number of sixes that will be thrown? Ans: 16 am i supposed to use the greatest coefficient for these kind of ques cuz the answer i got was wrong using this method nCr*(1/6)^r*(5/6)^(n-r) divided by nC(r-1)*(1/6)^(r-1)*(5/6)^(n-r+1) or am i supposed to do (1/6)*100 Hey! This is a really tough one, and you could really do it a number of ways. I would start by saying that, as a matter of probabilities, each value should come up 16 times (16*6=96). This is if the die was perfectly weighted. Then, we can take a quick look at the remaining four rolls. Basically, if it is more likely than not that a six comes up in 4 rolls, then it is more likely than not that a six will be rolled 17 times in 100 rolls. This calculation is just a simple binomial, and I'm not fabulous with LaTex so won't attempt it. Overall, it will be 1 minus the probability of rolling NO sixes in four rolls. The problem is that, when I compute the probability of rolling a 6 in 4 rolls, I get 0.482. Therefore, 1-0.482=0.518, which is greater than 50%. By that calculation, you should be slightly more likely to roll 17 6s than 16 6s. Someone may have to point out a mistake that I made! Jake Title: Re: 3U Maths Question Thread Post by: FallonXay on June 03, 2016, 03:55:59 pm Hii! I'm having trouble with the following question, any help would be much appreciated :) Thanks! Title: Re: 3U Maths Question Thread Post by: katherine123 on June 03, 2016, 04:22:14 pm 1. particle has a=2-2x, initially v=4m/s , x=0 a)find v in terms of x what i got was v=(16+4x-2x^2)^1/2 how do i know whether v is positive or negative b) find the greatest v do i normally just let a=0 for min/max velocity what do i have to do if the ques asks for max/min acceleration particle starts from the origin and has v=cos^2(x) a= -2sin(x)*cos^3(x) x= tan^-1(t) inverse where 0<=x<π Describe the motion of particle from its initial position to its limiting position (2 marks) the graph i get for displacement vs time is an inverse tangent curve with range: 0<=y<π/2 Im not sure what's required for questions that ask me to describe the motion the answer i wrote was: the particle is moving to the right of origin with a speed that is increasing at a decreasing rate Title: Re: 3U Maths Question Thread Post by: jakesilove on June 03, 2016, 04:59:12 pm Hii! I'm having trouble with the following question, any help would be much appreciated :) Thanks! Hey! I've attached my solution before. This is just a straightforward application of the derivative formula for inverse trigonometric functions: you need to be extremely comfortable with these kinds of questions! Hope this helps :) (http://i.imgur.com/bJCnpGQ.png?1) Jake Title: Re: 3U Maths Question Thread Post by: jakesilove on June 03, 2016, 05:06:14 pm 1. particle has a=2-2x, initially v=4m/s , x=0 a)find v in terms of x what i got was v=(16+4x-2x^2)^1/2 how do i know whether v is positive or negative b) find the greatest v do i normally just let a=0 for min/max velocity what do i have to do if the ques asks for max/min acceleration particle starts from the origin and has v=cos^2(x) a= -2sin(x)*cos^3(x) x= tan^-1(t) inverse where 0<=x<π Describe the motion of particle from its initial position to its limiting position (2 marks) the graph i get for displacement vs time is an inverse tangent curve with range: 0<=y<π/2 Im not sure what's required for questions that ask me to describe the motion the answer i wrote was: the particle is moving to the right of origin with a speed that is increasing at a decreasing rate Hey! Can't say I fully understand what you're actually asking, but I'll have a go. I assume you're asking how to tell whether, when you square root v, you should make the equation positive or negative. Essentially, you just sub a point in and see which one works; in this case, you were right to choose the positive route as initially the particle is moving to the right (velocity is positive, so it is assumed to be moving right). In terms of finding the maximum and minimum velocity/acceleration etc. for questions like this, you usually let the particle be at its endpoints or center. Unlike classical motion (ie, 2U stuff), it becomes increasingly difficult to just differentiate for t and find a standard max/min. Instead, we just have to remember that maximum velocity will occur when the particle is at the center of motion, and maximum acceleration will occur when the particle is at its endpoints (ie. either side of its motion). Your second question is absolutely right; all that's being asked is a general description of displacement, velocity and acceleration! Plus, you did it in a succinct and mathematically accurate way :) Jake Title: Re: 3U Maths Question Thread Post by: FallonXay on June 03, 2016, 05:16:08 pm Hey! I've attached my solution before. This is just a straightforward application of the derivative formula for inverse trigonometric functions: you need to be extremely comfortable with these kinds of questions! Hope this helps :) (http://i.imgur.com/bJCnpGQ.png?1) Jake ok, thanks. I understand this much, but i'm having trouble further simplifying it. The final answer provided in the textbook is -1/√ (9-x2). How did the 3 in the denominator cancel out? Title: Re: 3U Maths Question Thread Post by: RuiAce on June 03, 2016, 05:19:35 pm ok, thanks. I understand this much, but i'm having trouble further simplifying it. The final answer provided in the textbook is -1/√ (9-x2). How did the 3 in the denominator cancel out? By simply expanding it in. $-\frac{1}{3\sqrt{1-\frac{x^2}{9}}}=-\frac{1}{\sqrt{9}\sqrt{1-\frac{x^2}{9}}}=-\frac{1}{\sqrt{9\left(1-\frac{x^2}{9}\right)}}$ $\text{In general: }\frac{d}{dx}\arccos{\frac{x}{a}}=\frac{-1}{\sqrt{a^2-x^2}}\\ \text{You should memorise this.}$ Title: Re: 3U Maths Question Thread Post by: FallonXay on June 03, 2016, 05:23:12 pm By simply expanding it in. $-\frac{1}{3\sqrt{1-\frac{x^2}{9}}}=-\frac{1}{\sqrt{9}\sqrt{1-\frac{x^2}{9}}}=-\frac{1}{\sqrt{9\left(1-\frac{x^2}{9}\right)}}$ $\text{In general: }\frac{d}{dx}\arccos{\frac{x}{a}}=\frac{-1}{\sqrt{a^2-x^2}}\\ \text{You should memorise this.}$ ohhh, i see!!! Thank you so much! ;D Title: Re: 3U Maths Question Thread Post by: RuiAce on June 03, 2016, 05:26:31 pm 1. particle has a=2-2x, initially v=4m/s , x=0 a)find v in terms of x what i got was v=(16+4x-2x^2)^1/2 how do i know whether v is positive or negative b) find the greatest v do i normally just let a=0 for min/max velocity what do i have to do if the ques asks for max/min acceleration particle starts from the origin and has v=cos^2(x) a= -2sin(x)*cos^3(x) x= tan^-1(t) inverse where 0<=x<π Describe the motion of particle from its initial position to its limiting position (2 marks) the graph i get for displacement vs time is an inverse tangent curve with range: 0<=y<π/2 Im not sure what's required for questions that ask me to describe the motion the answer i wrote was: the particle is moving to the right of origin with a speed that is increasing at a decreasing rate Your answer to the second question is, as Jake pointed out, correct. To make it clearer for you: $\text{Given that }x=\arctan{t}\\\text{ We first realise that the particle is indeed approaching }x=\frac{\pi}{2}\text{ at an increasing slower rate.}\\ \text{We say that }x\text{ is therefore }\textit{decelerating}$ $\text{To prove that the particle is always moving in the same direction? Well, since }v=\cos^2{x}\\ \text{ We must have have }v>0\text{ for all }x\text{ since something squared cannot ever be negative}$ $\text{Obviously, it's moving to the right, as you already stated.}$ For the first question, this is a classic example of a question that I HATE because it gives you insufficient information to properly derive an expression for the sign of the velocity. In your case, it is safest to just say that = + sqrt(16+4x-2x2) if v≥0 = - sqrt(16+4x-2x2) if v≤0 $\text{Or if I think about it again, you could say }\\ |v|=\sqrt{16+4x-2x^2} \\ \text{But I slightly prefer the former because this here is speed, not velocity.}$ $\text{The second question can technically be done in multiple ways.}\\ \text{One way is to do what you said, that is, let }\ddot{x}=0\\ \text{However, the trick with SHM is as Jake already pointed out, that acceleration is maximised when velocity equals to 0.}$ Title: Re: 3U Maths Question Thread Post by: amandali on June 05, 2016, 08:58:38 am why is the answer 2/5 Title: Re: 3U Maths Question Thread Post by: RuiAce on June 05, 2016, 09:24:37 am why is the answer 2/5 $\text{A question like this is best done by simply counting.}\\ \text{The total number of outcomes is obviously }5^2=25$ $\text{The favourable outcomes are easy to list:}\\ (5,4), (5,3), (5,2), (5,1), (2,1)\\ (4,3), (4,2), (4,1), (3,2), (3,1)\\ \text{So there are ten such outcomes.}$ $\frac{10}{25}=\frac{2}{5}$ $\text{Alternatively, in a similar probabilistic analysis:}\\ Pr(\text{Kim wins}) = \underbrace{\frac{1}{5}\times \frac{4}{5}}_{\text{Kim spins 5}} + \overbrace{\frac{1}{5} \times \frac{3}{5}}^{\text{Kim spins 4}} + \underbrace{\frac{1}{5} \times \frac{2}{5}}_{\text{Kim spins 3}} + \overbrace{\frac{1}{5} \times \frac{1}{5}}^{\text{Kim spins 2}}\\ \text{Note that if Kim gets a 1 then she can't win}$ $\text{The more advanced approach (hard to decipher, but you may understand as you take Ext 2)}\\ \text{is to consider symmetric probabilities.}$ $\text{Observe that the probability of a draw is obviously }\frac{1}{5}\text{ because there's obviously 5 favourable options}.\\ \text{But that means that we have a total probability of }\frac{4}{5}\\ \textit{But note that the probabilities of Kim and Mel winning are evenly split!}\\ \text{This is because for Kim, if you swap who gets which number, then Mel wins}\\ \textit{Hence, as the probabilities are split this way, the probability any specific person wins is }\frac{2}{5}$ Title: Re: 3U Maths Question Thread Post by: imtrying on June 06, 2016, 05:28:48 pm Hey:) I'm struggling a bit with the attached question from the Binomial Theorem topic, wondering if I could have a hand? Thanks so much 8) Title: Re: 3U Maths Question Thread Post by: RuiAce on June 06, 2016, 05:35:21 pm Hey:) I'm struggling a bit with the attached question from the Binomial Theorem topic, wondering if I could have a hand? Thanks so much 8) $\text{Just use the definition of the binomial coefficient }\binom{n}{k}=\frac{n!}{k!(n-k)!}$ (Aka. the definition of the combinatoric operator) $\frac{\binom{8}{3}}{6!}=\frac{\frac{8!}{3!5!}}{6!}=\frac{8!}{3!5!6!}$ $\text{Which can be simplified in a variety of ways, and the most tidiest is generally} \\ \frac{8!}{3!5!6!}=\frac{8\times 7\times 6!}{3!5!6!}=\frac{8 \times 7}{3!5!}$ $\text{You could go further, but then it becomes questionable as to if you're really in factorial notation anymore.}\\ \frac{8 \times 7}{3!5!} = \frac{8 \times 7}{3! \times 5 \times 4 \times 3!} = \frac{2 \times 7}{5(3!)^2}$ Such a question would not be asked in the exam except for as multiple choice as the final answer is too ambiguous Title: Re: 3U Maths Question Thread Post by: katherine123 on June 08, 2016, 03:14:14 pm i dont get part (iii) for question 1 ques2 : the answer is 35/72 Title: Re: 3U Maths Question Thread Post by: jakesilove on June 08, 2016, 04:18:20 pm i dont get part (iii) for question 1 ques2 : the answer is 35/72 Hey! I'll just quickly answer your second question: honestly, I'll leave the first to someone who can use LaTex better than me. To draw a Blue ball from the first bag, there is a $\frac{5}{8}$ chance. To draw a Blue ball from the second bag, there is a $\frac{7}{9}$. So, to get the probability of picking two blue balls, we just multiple these together! The answer will be $\frac{35}{72}$ as you told us! You can answer the other questions in a similar fashion: The probability of getting one yellow and one blue ball will be the probability of getting a yellow THEN a blue plus the probability of getting a blue THEN a red. The probability of getting at least one yellow ball will be the probability of getting a red and a yellow plus the probability of getting two yellows (or, more easily, 1 minus the probability of getting two blues!). Jake Title: Re: 3U Maths Question Thread Post by: RuiAce on June 08, 2016, 06:54:54 pm i dont get part (iii) for question 1 ques2 : the answer is 35/72 $\text{This was one of the questions that contributed to the debate that the 2013 HSC was "too hard". This was one of the Ext 1 questions.}$ $\text{We first combine the first two parts. Note that }{(1+x)}^{4n}={(1+x^2+2x)}^{2n}\text{ by perfect squares.}\\ \text{Hence, we are most likely interested in the coefficient of }x^{2n}\text{ in the messy thing in part ii}$ $\text{The sigma is not easy to tackle. Clearly, we must consider }\textit{EACH INDIVIDUAL term of the sum}\\ \text{to get our desired result. We will utilise the given formula to achieve this.} \\ \binom{2n}{k}x^{2n-k}{(x+2)}^{2n-k} = \binom{2n}{k}\binom{2n-k}{0}2^{2n-k}x^{2n-k}+\binom{2n}{k}\binom{2n-k}{1}2^{2n-k-1}x^{2n-k+1}+\dots + \binom{2n}{k}\binom{2n-k}{2n-k}2^0x^{4n-2k}$ $\text{When }k=0\text{ we have }\binom{2n}{0}x^{2n-0}{(x+2)}^{2n-0}\text{ from the sum.} \\ \text{Looking at the given expression, the term involving }x^{2n}\text{ is the first term }\binom{2n}{0}\binom{2n-k}{0}2^{2n}x^{2n}$ $\text{When }k=1\text{ we have }\binom{2n}{1}x^{2n-1}{(x+2)}^{2n-1}\text{ from the sum.}\\ \text{Looking at the given expression, the term involving }x^{2n}\text{ is the second term }\binom{2n}{1}\binom{2n-1}{1}2^{2n-1-1}{x}^{2n-1+1}$ $\text{When }k=2\text{ we have }\binom{2n}{2}x^{2n-2}{(x+2)}^{2n-2}\text{ from the sum.}\\ \text{Looking at the given expression, the term involving }x^{2n}\text{ is the unrevealed third term }\binom{2n}{2}\binom{2n-2}{2}2^{2n-2-2}x^{2n-2+2}$ $\vdots$ $\text{When }k=n\text{ we have }\binom{2n}{n}x^{2n-n}{(x+2)}^{2n-n}\text{ from the sum.}\\ \text{Looking at the given expression, the term involving }x^{2n}\text{ is the final term }\binom{2n}{n}\binom{2n-n}{n}2^0x^{4n-2n}$ $\textbf{Why did I stop here?}\\ \text{Two reasons: First one is that the final answer stops at }n\text{ and doesn't go further.}\\ \text{Second reason is that if I attempt }k=n+1\text{, I get }\textbf{NO}\text{ terms involving }x^{2n}\text{ anymore! You can try this out}$ $\text{So it would appear that on the RHS, the coefficient of }x^{2n}\text{ is the sum of all of these:}$ $\binom{2n}{0}\binom{2n-0}{0}2^{2n}+\binom{2n}{1}\binom{2n-1}{1}2^{2n-1-1}+\binom{2n}{2}\binom{2n-2}{2}2^{2n-2-2}+\dots + \binom{2n}{n}\binom{2n-n}{n}2^0$ $\text{To which if you carefully look at the sum, this just equals to }\\ \sum_{k=0}^{n}{2^{2n-2k}\binom{2n}{k}\binom{2n-k}{k}2^{2n-2k}}$ $\text{Equate with the coefficient of }x^{2n}\text{ on the LHS, that is, }\binom{4n}{2n}\text{ to get your desired result.}$ $\textbf{Note: This is one of the EXTREMELY hard questions the HSC has ever asked.}\\\textbf{If any of it does not make sense, PLEASE alert me at once!!!}$ Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on June 08, 2016, 10:45:11 pm $\textbf{Note: This is one of the EXTREMELY hard questions the HSC has ever asked.}$ I distinctly remember doing this question in the lead up to my Extension 1 exam, it was nasty, very grateful to have not seen this question under pressure :o Title: Re: 3U Maths Question Thread Post by: RuiAce on June 09, 2016, 07:31:59 am I distinctly remember doing this question in the lead up to my Extension 1 exam, it was nasty, very grateful to have not seen this question under pressure :o Back when I was studying for Ext 1 I couldn't get it out :P Title: Re: 3U Maths Question Thread Post by: katherine123 on June 10, 2016, 06:12:57 pm how to do this question ? the region in the first quadrant bounded by x=y-y^3, x=1, y=1 is rotated about y axis. Find the volume of solid generated Title: Re: 3U Maths Question Thread Post by: RuiAce on June 10, 2016, 07:38:40 pm how to do this question ? the region in the first quadrant bounded by x=y-y^3, x=1, y=1 is rotated about y axis. Find the volume of solid generated $\text{We will not use any of the fancy stuff given this was posted in the 3U section.} \\ \text{The equation }y-y^3=0\text{ has roots at }y=0, y=\pm 1\\ \text{but in the first quadrant, we will disregard }y<0$ (http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpslpz6iaxa.png) $\text{We note that the line }x=1\text{ forms the upper curve}\\ \text{ and the curve } x=y-y^3\text{ forms the lower curve.}$ $\text{Since the volume is just }V=\pi \int_a^b{\left( x_\text{upper}^2-x_\text{lower}^2 \right) dy}\\ \text{or alternatively }V=\pi \int_a^b{\left(R^2-r^2\right)dy}$ \begin{align*} V&= \int_0^1 {\left( 1^2 - \left(y-y^3\right)^2\right) dy} \\ &= \int_0^1{\left(-y^6+2y^4-y^2+1\right) dy} \\ &= \dots = \frac{97}{105}\end{align*} Note: If the curve was what was confusing, notice a trick for this question. (This trick is a bit of a coincidence because of how neat the question is.) Because the region is bounded by x=1 AND y=1, we could just use the inverse relation y=x-x3 and rotate this region about the x-axis instead. We get the same integral: $\int_0^1{\left(1^2-\left(x-x^3\right)^2\right)dx}$ (http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpshktrz4tq.png) Title: Re: 3U Maths Question Thread Post by: amandali on June 10, 2016, 09:58:43 pm (http://uploads.tapatalk-cdn.com/20160610/67eee88c7355af3076a061cbd31a15d9.jpg) need help with part c) Title: Re: 3U Maths Question Thread Post by: RuiAce on June 10, 2016, 10:32:59 pm (http://uploads.tapatalk-cdn.com/20160610/67eee88c7355af3076a061cbd31a15d9.jpg) need help with part c) $\text{Since initially we have }M=49000\\ \text{The initial rate is just }\\ \frac{dM}{dt}=--\frac{1}{2}\ln{\frac{1}{2}}(49000-1000)=24000\ln{\frac{1}{2}}$ $\text{Recall that also, } \frac{dM}{dt} = -kAe^{-kt} = 24000 \ln{\frac{1}{2}} \cdot e^{-kt}$ \text{So since we want }\frac{dM}{dt}\text{ to equal one quarter of its initial value:}\\ \begin{align*}6000 \ln{\frac{1}{2}} &= 24000 \ln{\frac{1}{2}}\cdot e^{-kt}\\ \Rightarrow \frac{1}{4} &= e^{-kt} \\ \Rightarrow \ln{4} &= kt \\ \Rightarrow 2 \ln{2} &= t \cdot \frac{1}{2} \ln{2} \end{align*} I'll leave the rest to you. Basically put this value for t back into the equation for M. This working out is subject to some inaccuracy Title: Re: 3U Maths Question Thread Post by: amandali on June 11, 2016, 11:11:39 am (http://uploads.tapatalk-cdn.com/20160610/fb24f3c84c059b3547637cbc2e071e50.jpg) need help with this ques thanks Title: Re: 3U Maths Question Thread Post by: RuiAce on June 11, 2016, 11:54:11 am (http://uploads.tapatalk-cdn.com/20160610/fb24f3c84c059b3547637cbc2e071e50.jpg) need help with this ques thanks $\text{It just so happens that the coefficient of }x^n\text{ in the expression on the LHS}\\ \text{is that exact sum. To see this:}$ \begin{align*}\text{We extract }\binom{n}{n}x^n\text{ from }{(1+x)}^n & \text{ and match it with }\binom{n}{0}{(-1)}^0 \text{ from }{(1-x)}^n \\ \text{We extract }\binom{n}{n-1}x^{n-1}\text{ from }{(1+x)}^n & \text{ and match it with }\binom{n}{1}{(-1)}^1x \text{ from }{(1-x)}^n \\ \text{We extract }\binom{n}{n-2}x^{n-2}\text{ from }{(1+x)}^n & \text{ and match it with }\binom{n}{2}{(-1)}^2x^2 \text{ from }{(1-x)}^n \\ &\vdots \\ \text{We extract }\binom{n}{1}x\text{ from }{(1+x)}^n & \text{ and match it with }\binom{n}{n-1}{(-1)}^{n-1} \text{ from }{(1-x)}^n \\\text{We extract }\binom{n}{0}\text{ from }{(1+x)}^n & \text{ and match it with }\binom{n}{n}{(-1)}^n \text{ from }{(1-x)}^n \\ \end{align*} $\text{Summing these up, the coefficient of }x^n\text{ on the LHS is given}\\ \binom{n}{n}\binom{n}{0}(-1)^0+\binom{n}{n-1}\binom{n}{1}(-1)^1+\binom{n}{n-2}\binom{n}{2}(-1)^2+\dots+\binom{n}{1}\binom{n}{n-1}(-1)^{n-1}+\binom{n}{0}\binom{n}{n}(-1)^n\\ =\sum_{r=0}^{n}{\binom{n}{n-r}\binom{n}{r}(-1)^r}=\sum_{r=0}^{n}{(-1)^r\binom{n}{r}^2} \\ \text{recalling the symmetry property of the binomial coefficient: }\binom{n}{r}=\binom{n}{n-r}$ $\text{The RHS is also a bit difficult to deal with. Theoretically, the coefficient of }\\ x^n\text{ in the expansion of }{\left(1-x^2\right)}^n\text{ is given }(-1)^\frac{n}{2}\binom{n}{\frac{n}{2}} - \textit{Do you see why?}$ $\text{If }n\text{ is odd, this doesn't even exist, thus for odd }n: \\ \sum_{r=0}^{n}{(-1)^r\binom{n}{r}^2}=0$ $\text{We have no problems if }n\text{ is even though. If }n\text{ is even then we're done.}\\ \sum_{r=0}^{n}{(-1)^r\binom{n}{r}^2}=(-1)^\frac{n}{2}\binom{n}{\frac{n}{2}}$ Additional note: If you test for various odd values of n (that is, physically expanding it), you will find that indeed there IS no term involving xn. Title: Re: 3U Maths Question Thread Post by: katherine123 on June 11, 2016, 09:45:06 pm how to do part b) and c) ans for b = 101/144 and ans for c =0.113) for part c) what i did was 20C12*(5/8)^12*(3/8)^8 + 20C12*(7/9)^12*(2/9)^8 Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on June 11, 2016, 10:42:12 pm how to do part b) and c) ans for b = 101/144 and ans for c =0.113) for part c) what i did was 20C12*(5/8)^12*(3/8)^8 + 20C12*(7/9)^12*(2/9)^8 Hey Katherine!! For Part B, it is just the probability of a blue ball from one plus the probability from the other: $P(\text{Blue})=\frac{1}{2}\times\frac{7}{9}+\frac{1}{2}\times\frac{5}{8}=\frac{101}{144}$ We can use this result for Part C. You had the right idea! But the two probabilities are: $\frac{101}{144} \quad\text{for a blue ball} \\ \frac{43}{144} \quad\text{for a yellow ball}$ These are from Part B. Try Part C with the binomial probability method again with those valuess and the answer should come out!! Let us know if it doesn't and we'll lend a hand ;D Title: Re: 3U Maths Question Thread Post by: amandali on June 12, 2016, 02:49:28 am (http://uploads.tapatalk-cdn.com/20160611/4ad4e133b5469c87d70cb6793c70502b.jpg)(http://uploads.tapatalk-cdn.com/20160611/1c1b825b3e9315d84c60ed406d4ba3d5.jpg) for part c ii) i dont get how the last constant becomes (pCp)(qCp) Title: Re: 3U Maths Question Thread Post by: RuiAce on June 12, 2016, 08:08:01 am (http://uploads.tapatalk-cdn.com/20160611/4ad4e133b5469c87d70cb6793c70502b.jpg)(http://uploads.tapatalk-cdn.com/20160611/1c1b825b3e9315d84c60ed406d4ba3d5.jpg) for part c ii) i dont get how the last constant becomes (pCp)(qCp) This is because you stopped at the wrong term in the last term. You stopped at pCp (xp) * qCq (1/x)q But this is not a constant term. This results in xp-q which doesn't necessarily equal to xp-p! Because p≤q, you must stop at pCp (xp) * qCp (1/x)p = (pCp)(qCp)x0 (as you don't know if p<q or p=q) To view this, consider a simple example where p=2 and q=3. Clearly, when p=2 and q=3, the last term is not a constant term Title: Re: 3U Maths Question Thread Post by: amandali on June 12, 2016, 03:15:00 pm for part b) what is the reason that the velocity is negative i dont understand part c)iii) ive attached the answer it below Title: Re: 3U Maths Question Thread Post by: RuiAce on June 12, 2016, 03:24:11 pm for part b) what is the reason that the velocity is negative i dont understand part c)iii) ive attached the answer it below $\text{When }t=0, x=0\text{ and }v=1$ \begin{align*}\frac{d}{dx}\left(\frac{1}{2}v^2\right)&=x-1 \\ \Rightarrow \frac{1}{2}v^2 &= \frac{x^2}{2}-x+C \\ \Rightarrow v^2 &= x^2 - 2x + C\\ \Rightarrow 1 &= 0 - 0 + C \\ C &= 1 \end{align*} \text{So we have }\\ \begin{align*} v^2&=x^2-2x+1 \\ \Rightarrow v^2 &= (x-1)^2 \\ \Rightarrow v &= \pm(x-1) \end{align*} $\textbf{To satisfy that v=1} \text{ when }x=0\text{, we are forced to take the negative case (check this)}\\ \therefore v=1-x$ That is the reason. To satisfy v=1, not v=-1. Title: Re: 3U Maths Question Thread Post by: RuiAce on June 12, 2016, 03:40:52 pm for part b) what is the reason that the velocity is negative i dont understand part c)iii) ive attached the answer it below $\text{With correct calculations, your answer to part ii) was then:} \\ \binom{4}{4}{\left(\frac{1}{2}\right)}^5 + \binom{5}{4}{\left(\frac{1}{2}\right)}^6 + \binom{6}{4}{\left(\frac{1}{2}\right)}^7\\ \text{If this didn't make sense, please alert me about it.}$ $\text{Now, using a similar analogy, we can say that the probability person }A\text{ or person }B\\ \text{wins on or before the NINTH turn is:}\\ \binom{4}{4}{\left(\frac{1}{2}\right)}^5 + \binom{5}{4}{\left(\frac{1}{2}\right)}^6 + \binom{6}{4}{\left(\frac{1}{2}\right)}^7+ \binom{7}{4}{\left(\frac{1}{2}\right)}^8+\binom{8}{4}{\left(\frac{1}{2}\right)}^9 \\ \textit{Why did I do this? To prove a point.}$ $\text{Take note, that in this game, SOMEONE had to win by the NINTH game.} \\ \text{Because five points score a win, we are basically playing a best-5-out-of-9 game}\\ \text{The significance is that there IS NO TENTH game. Thus we end the count here.}$ $\text{In part iii, we generalise it. Instead, now we are playing a}\\ \text{best }(n+1)\text{ out of }(2n)\text{ game.} \\ \text{We realise that using a method similar to part a), the probability }\\ \text{that person A wins the series is:}$ $Pr(\text{A wins }(n+1)) \\ = \binom{n}{n}\left(\frac{1}{2}\right)^{n+1}+\binom{n+1}{n}\left(\frac{1}{2}\right)^{n+2}+\binom{n+2}{n}\left(\frac{1}{2}\right)^{n+2}+\dots+\binom{2n}{n}\left(\frac{1}{2}\right)^{2n+1}$ $\text{But the important thing to remember is that both players have an equal probability of winning!} \\ \text{The probability that A wins is the exact same as B wins}\\ \text{Hence, }Pr(\text{A wins}) = \frac{1}{2}$ $\therefore \binom{n}{n}\left(\frac{1}{2}\right)^{n+1}+\binom{n+1}{n}\left(\frac{1}{2}\right)^{n+2}+\binom{n+2}{n}\left(\frac{1}{2}\right)^{n+2}+\dots+\binom{2n}{n}\left(\frac{1}{2}\right)^{2n+1}=\frac{1}{2}\\ \text{So by simply multiplying both sides by }2^{2n+1}$ $\binom{n}{n}2^n+\binom{n+1}{n}2^{n-1}+\binom{n+2}{n}2^{n-2}+\dots+\binom{2n}{n}=2^{2n}$ This was a nice question. I recalled it from the 2015 paper, because I remember asking my colleagues how they went. This is my first time actually doing the question though. Tell me if something makes no sense. Title: Re: 3U Maths Question Thread Post by: IkeaandOfficeworks on June 13, 2016, 03:44:47 pm Hey guys! can you help me with these motion questions? Thank you! Title: Re: 3U Maths Question Thread Post by: RuiAce on June 13, 2016, 03:54:13 pm Hey guys! can you help me with these motion questions? Thank you! $\ddot{x}=\frac{d}{dx}\left(\frac{1}{2}v^2\right)=-\frac{1}{6}x^{\frac{1}{3}}$ \text{So upon integrating we have} \\ v^2=-\frac{1}{4}x^{\frac{4}{3}}+C \\ \text{And we're told that when }x=0, v=2\\ \begin{align*} \implies 4&=C\\ \therefore v^2&=-\frac{1}{4}x^{\frac{4}{3}}+4\end{align*} \text{We want to find the displacement when it first comes to rest, so that's simply saying }v=0: \\ \begin{align*}0&=-\frac{1}{4}x^{\frac{4}{3}}+4 \\ \implies x^{\frac{4}{3}} &= 16 \\ \implies x&=\pm 8 \end{align*} Title: Re: 3U Maths Question Thread Post by: RuiAce on June 13, 2016, 03:59:24 pm Hey guys! can you help me with these motion questions? Thank you! $\ddot{x}=\frac{d}{dx}\left(\frac{1}{2}v^2\right)=x-3\\ \text{So upon integrating and rearranging we have} \\ v^2 = x^2 - 6x + C$ \text{We are given that when }x=1, v=2\text{ so }\\ \begin{align*} 4 &= 1-6+C \\ \implies C&=9\end{align*} \text{Hence, we have }\\ \begin{align*}v^2&=x^2-6x+9 \\ \implies v&=\pm(x-3)\end{align*}\\ \text{by considering the perfect square} $\text{To satisfy that }v=2\text{ when }x=1\text{ we clearly must take the negative case.} \\ \therefore v = 3-x$ $\text{Then simply substitute }x=0$ Title: Re: 3U Maths Question Thread Post by: amandali on June 13, 2016, 04:37:54 pm i did integration and got stucked Title: Re: 3U Maths Question Thread Post by: jakesilove on June 13, 2016, 05:11:16 pm i did integration and got stucked Hey Amandali! I've attached the solution before. Using a method like this is pretty standard, however usually in the 3U course they lead your through it a bit more. Still, if you ever get a question like this, just think about what you are TRYING to prove and basically brute-force your way there! Sorry that it's hand written; Rui and Jamon still haven't taught me LaTeX. (http://i.imgur.com/DLB1Nq2.jpg?1) Jake Title: Re: 3U Maths Question Thread Post by: RuiAce on June 13, 2016, 05:12:41 pm i did integration and got stucked $\text{Observe that the binomial coefficient steps up in increments of 2.}\\ \text{When this happens, the general trick is that you must consider the expansions of both}\\ {(1+x)}^n\text{, AND }{(1-x)}^n$ $\text{The fact that the question omits the last term, whilst being mathematically incorrect,} \\ \text{is specifically there so that you }\textit{don't worry about it and just focus on the first terms.}$ If you're interested in why it's mathematically incorrect, it's because well an assumption is being made that the sum goes to infinity. \text{We have: }\\ \begin{align*}\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\binom{n}{3}x^3+\binom{n}{4}x^4+\dots+\binom{n}{n}x^n&= {(1+x)}^n \\ \binom{n}{0}-\binom{n}{1}x+\binom{n}{2}x^2-\binom{n}{3}x^3+\binom{n}{4}x^4-\dots+{(-1)}^n\binom{n}{n}x^n&= {(1-x)}^n \end{align*} $\text{Therefore, upon adding these expressions:}\\ 2\left(\binom{n}{0}+\binom{n}{2}x^2+\binom{n}{4}x^4+\dots\right)={(1+x)}^n+{(1-x)}^n$ $\textit{Finding a constant of integration is so boring...}\\ \text{...so let's use a definite integral to help us out:} \\ 2\int_0^1{\left(\binom{n}{0}+\binom{n}{2}x^2+\binom{n}{4}x^4+\dots\right)dx}=\int_0^1{\left({(1+x)}^n+{(1-x)}^n\right)dx}$ Oops: Got a message saying that a reply was already posted. So I'll leave this working out here and leave the remainder of the question for you to try out Title: Re: 3U Maths Question Thread Post by: RuiAce on June 13, 2016, 05:14:07 pm Sorry that it's hand written; Rui and Jamon still haven't taught me LaTeX. Oi. I read this :P Title: Re: 3U Maths Question Thread Post by: jakesilove on June 13, 2016, 05:17:13 pm Oi. I read this :P That was the idea ;) Title: Re: 3U Maths Question Thread Post by: RuiAce on June 13, 2016, 05:17:39 pm That was the idea ;) Well it's not my fault that we all have exams :P Title: Re: 3U Maths Question Thread Post by: RuiAce on June 13, 2016, 06:27:08 pm Hey Amandali! I've attached the solution before. Using a method like this is pretty standard, however usually in the 3U course they lead your through it a bit more. Still, if you ever get a question like this, just think about what you are TRYING to prove and basically brute-force your way there! Sorry that it's hand written; Rui and Jamon still haven't taught me LaTeX. (http://i.imgur.com/DLB1Nq2.jpg?1) Jake Oh wait, slight correction after looking at this again You forgot the (annoying) +C when integrating (The final answer is unaffected though because when you subtract the two resultant equations, the C cancels out) Title: Re: 3U Maths Question Thread Post by: jakesilove on June 13, 2016, 06:29:58 pm Oh wait, slight correction after looking at this again You forgot the (annoying) +C when integrating (The final answer is unaffected though because when you subtract the two resultant equations, the C cancels out) Yeah I thought about including that, but thought it trivial (if one integral equals another, obviously any constant is going to cancel, so generally I just exclude it from the start). Title: Re: 3U Maths Question Thread Post by: RuiAce on June 13, 2016, 06:40:03 pm Yeah I thought about including that, but thought it trivial (if one integral equals another, obviously any constant is going to cancel, so generally I just exclude it from the start). Well in the second line it's kinda important because otherwise, that statement at that step is false Title: Re: 3U Maths Question Thread Post by: jakesilove on June 13, 2016, 06:52:39 pm Well in the second line it's kinda important because otherwise, that statement at that step is false True, true; call it laziness! Title: Re: 3U Maths Question Thread Post by: amandali on June 13, 2016, 11:06:23 pm how to do last part thanks Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on June 14, 2016, 12:05:54 am Well in the second line it's kinda important because otherwise, that statement at that step is false It's funny, at HSC level they stress this sort of thing, my Complex Analysis lecturer is happy for us to omit it in very particular circumstances where it is apparent that the constant is irrelevant. Interesting how the frame of thought changes (though I doubt a marker would take away anything in a HSC exam) ::) Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on June 14, 2016, 12:25:19 am how to do last part thanks Hey hey! For a change in direction you are looking for the velocity being zero. To turn back, an object must first stop completely (if we restrict to one-dimensional travel). So, looking at the previous parts of the question, simply solve for the x-values that cause your expression for velocity squared to be equal to zero!! You probably got the expression in this way. $\frac{1}{2}v^2 = \frac{1}{2}x^2+\frac{3}{2}x +C$ Using initial conditions, we can find the constant and yield: $v^2=x^2+3x-4$ Now we can solve for this equal to zero (usual methods for solving a quadratic), and we obtain two solutions: $x=-4, x=1$ The answer we are looking for is x=1. Think of it this way, the particle starts moving towards the origin from the right. When it reaches x=1, it will turn around and go back the other way. From this point, it will never come back; we can gleam this from equations we have obtained. Acceleration is always positive, it is never going to be accelerated in the negative direction, it is going, going gone ;) Let me know if this doesn't make sense! It is kind of intuitive, which is a little easier than the rigorous mathematical approach (unnecessary for this question) ;D Title: Re: 3U Maths Question Thread Post by: RuiAce on June 14, 2016, 08:35:09 am It's funny, at HSC level they stress this sort of thing, my Complex Analysis lecturer is happy for us to omit it in very particular circumstances where it is apparent that the constant is irrelevant. Interesting how the frame of thought changes (though I doubt a marker would take away anything in a HSC exam) ::) HSC students only know so much so what can you emphasise :P That's sorta why I chose a definite integral approach though. Ever since 4U mechanics, I've tended to using definite integrals wherever possible, if it makes my calculations a bit easier without loss of accuracy. Title: Re: 3U Maths Question Thread Post by: amandali on June 14, 2016, 01:20:28 pm ques: How many different arrangements of the word MAMMOTH can be made if only five letters are used? i dont get the answer is there an alternative way of doing this Title: Re: 3U Maths Question Thread Post by: RuiAce on June 14, 2016, 05:19:17 pm ques: How many different arrangements of the word MAMMOTH can be made if only five letters are used? i dont get the answer is there an alternative way of doing this I'm pretty sure I explained this question recently in the 4U question thread Title: Re: 3U Maths Question Thread Post by: jakesilove on June 14, 2016, 05:28:45 pm ques: How many different arrangements of the word MAMMOTH can be made if only five letters are used? i dont get the answer is there an alternative way of doing this $\text{Case one: 1 M }\\ \text{Your letters are }M,A,O,T,H\\ \text{ which can obviously be arranged in }5!\text{ ways as you said.}$ $\text{Case two: 2 M's} \\ \text{Your letters are }M, M, \_ , \_ , \_ \\ \text{The blanks can be any three letters chosen out of the bundle }A, O, T, H\\ \text{to which there are }\binom{4}{3}\text{ ways of doing so.} \\ \text{So there are }\frac{5! \times \binom{4}{3}}{2!}=240 \text{ such arrangements.}$ $\text{Case three: 3 M's}\\ \text{Your letters are }M, M, M, \_ , \_ \\ \text{Now, the blanks are any two out of the four, i,e, }\binom{4}{2}\text{ choices.} \\ \text{So there are }\frac{5! \times \binom{4}{2}}{3!}=120\text{ ways of doing so.}$ $120+240+120=480$ $\text{Obviously, the division by }2!\text{ and }3!\text{ for case 2 and 3 respectively} \\ \text{arise due to the repeated letter }M$ 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. The above is Rui's method! Hope it helps :) Title: Re: 3U Maths Question Thread Post by: IkeaandOfficeworks on June 14, 2016, 07:16:11 pm Can you help me with this motion question? Thanks guys! :D A ball is projected so that its horizontal range is 45 metres and it passes through a point 22 1/2 metres horizontally from and 11 1/4 metres vertically above the point of projection. Find the angle of projection and the speed of projection. Title: Re: 3U Maths Question Thread Post by: RuiAce on June 14, 2016, 08:59:52 pm Can you help me with this motion question? Thanks guys! :D A ball is projected so that its horizontal range is 45 metres and it passes through a point 22 1/2 metres horizontally from and 11 1/4 metres vertically above the point of projection. Find the angle of projection and the speed of projection. $\text{I will assume the following results for now. Proof will be completed in the next post.}\\\text{The value of gravity is not known.}$ \text{The range of a projectile is given }\\ x_R=\frac{2V^2\sin{\theta}\cos{\theta}}{g} \\ \begin{align*}\therefore 45&=\frac{2V^2\sin{\theta}\cos{\theta}}{g} \\ \iff \frac{g}{2V^2}&=\frac{\sin{\theta}\cos{\theta}}{45}\end{align*} $\text{The Cartesian equation of motion is given} \\ y=-\frac{gx^2}{2V^2}\left(\sec^2{\theta}\right)+x\tan{\theta} \\ \text{So since the particle passes through }\left(22.5,11.25\right)$ \begin{align*}11.25&=-\frac{506.25g}{2V^2}\left(\sec^2{\theta}\right)+22.5\tan{\theta}\\ \text{Then,}&\text{ by substituting in in the earlier equation}\\ 11.25&=-11.25\sin{\theta}\cos{\theta}\left(\sec^2{\theta}\right)+22.5\tan{\theta} \\ \iff 11.25&=-11.25\tan{\theta}+22.5\tan{\theta} \\ \iff \tan{\theta}&=1 \\ \iff \theta&=\frac{\pi}{4} \end{align*} \text{Because the range of the projectile can be rewritten with the sine}\\ \text{double angle formula, we have:}\\ \begin{align*} 45&=\frac{V^2\sin{2\theta}}{g} \\ \therefore 45 &= \frac{V^2}{g} \\ \iff V&=\sqrt{45g} \end{align*} \\ \text{as }\sin{\frac{\pi}{2}}=1 Title: Re: 3U Maths Question Thread Post by: RuiAce on June 14, 2016, 09:10:15 pm \text{To prove all projectile motion formulae, we recall:} \\ \begin{align*}\ddot{x}&=0 \\ \therefore \dot{x}&=V\cos{\theta} \\ \therefore x&=Vt\cos{\theta}\\ \\ \ddot{y}&=-g \\ \therefore \dot{y}&=-gt+V\sin{\theta} \\ \therefore y&=-\frac{gt^2}{2}+Vt\sin{\theta}\end{align*} \\ \text{An assumption was made that the projectile was fired from and lands on the ground.} Note: In the exam, please make sure to evaluate each four of the constants of integration PROPERLY, i.e. when t=0, x'=Vcosθ etc. \text{At the maximum range, the projectile returns to level ground again:} \\ \text{The time of flight is therefore given by }y=0\\ \begin{align*}\implies -\frac{gt^2}{2}+Vt\sin{\theta}&=0 \\ \implies -\frac{gt}{2}&=V\sin{\theta} \\ t&=\frac{2V\sin{\theta}}{g}\end{align*}\\ \text{having safely cancelled out }t\text{ as }t=0\text{ reflects the initial time.} $\text{So for the range, substitute the time of flight into the horizontal displacement formula:}\\ x_R=V\cos{\theta}\left(\frac{2V\sin{\theta}}{g}\right)$ $\text{For the Cartesian equation of motion, rearrange the horizontal displacement formula:} \\ t=\frac{x}{V\cos{\theta}}\\ \text{Substituting this into the vertical displacement formula:}$ \begin{align*}y&=-\frac{g}{2}{\left(\frac{x}{V\cos{\theta}}\right)}^2+V\sin{\theta}\left(\frac{x}{V\cos{\theta}}\right)\\ &= -\frac{gx^2}{2V^2}\left(\sec^2{\theta}\right)+x\tan{\theta}\end{align*} Title: Re: 3U Maths Question Thread Post by: imnotdani on June 20, 2016, 06:58:16 pm Can i please have help with this question? :( Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on June 20, 2016, 07:10:01 pm Can i please have help with this question? :( Hey there! You sure can, the first bit is to handle the integrand: $y^2=(\cos{x}+1)^2=\cos^2{x}+1+2\cos{x}$ Now we complete the integration. We'll use our double angle result to handle the cos squared, which is something you may have seen before, let me know if not! $V = \pi\int^\frac{\pi}{2}_0(\cos^2{x}+1+2\cos{x})dx \\=\pi\int^\frac{\pi}{2}_0(\frac{1}{2}(1+\cos{2x})+1+2\cos{x})dx \\ =\pi\left[\frac{x}{2}+\frac{1}{4}\sin{2x}+x+2\sin{x}\right]^\frac{\pi}{2}_0 \\ =\frac{\pi}{4}+\frac{\pi}{2}+2\pi \\ = \frac{11\pi}{4} \quad u^3$ Let me know if anything here doesn't quite make sense!! ;D Title: Re: 3U Maths Question Thread Post by: imnotdani on June 20, 2016, 07:21:12 pm Hey there! You sure can, the first bit is to handle the integrand: $y^2=(\cos{x}+1)^2=\cos^2{x}+1+2\cos{x}$ Now we complete the integration. We'll use our double angle result to handle the cos squared, which is something you may have seen before, let me know if not! $V = \pi\int^\frac{\pi}{2}_0(\cos^2{x}+1+2\cos{x})dx \\=\pi\int^\frac{\pi}{2}_0(\frac{1}{2}(1+\cos{2x})+1+2\cos{x})dx \\ =\pi\left[\frac{x}{2}+\frac{1}{4}\sin{2x}+x+2\sin{x}\right]^\frac{\pi}{2}_0 \\ =\frac{\pi}{4}+\frac{\pi}{2}+2\pi \\ = \frac{11\pi}{4} \quad u^3$ Let me know if anything here doesn't quite make sense!! ;D Hey! Thanks for the quick reply, but I still don't quite get it. I understand YOUR working method, but the answers say it's pi( 3pi + 8 ) units3 :(? [or the answers could be completely wrong that works too] Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on June 20, 2016, 08:22:31 pm Hey! Thanks for the quick reply, but I still don't quite get it. I understand YOUR working method, but the answers say it's pi( 3pi + 8 ) units3 :(? [or the answers could be completely wrong that works too] Woops! I did make an error, let me redo the last section, I forgot the pi out the front for the first few terms! $=\pi\left(\frac{\pi}{4}+\frac{\pi}{2}+2\right) \\ = \pi\left(\frac{3\pi}{4}+2\right) = \frac{\pi(3\pi+8)}{4}$ It seems as though your answer might be missing the divided by 4 bit on the bottom? I triple checked with Wolfram Alpha and this is definitely correct ;D sorry for that error!! I'm on uni exams at the moment so my heads a bit all over the shop ;) Title: Re: 3U Maths Question Thread Post by: RuiAce on June 20, 2016, 08:28:11 pm I'm getting Jamon's second answer. And definitely so is WolframAlpha http://www.wolframalpha.com/input/?i=integrate+pi(cosx%2B1)%5E2+from+0+to+pi%2F2 Title: Re: 3U Maths Question Thread Post by: imnotdani on June 20, 2016, 08:37:07 pm Thanks heaps to both of you ^_^ Title: Re: 3U Maths Question Thread Post by: conic curve on June 22, 2016, 09:39:45 pm Is this for prelim or HSC? Title: Re: 3U Maths Question Thread Post by: RuiAce on June 22, 2016, 10:53:46 pm Is this for prelim or HSC? As you can see, there are no prelim threads on this forum. Hence, you should assume the answer to your question is both. Title: Re: 3U Maths Question Thread Post by: birdwing341 on June 27, 2016, 04:59:39 pm Having a bit of a struggle over here... Grain is ejected from a chute at the rate of 0.1 metres cubed a minute and is forming a heap on a flat horizontal floor. The heap is in the form of a circular cone of semi-vertical angle 45 degrees. Find the rate, in metres per minute, at which the height of the cone i s increasing at the instant 3 minutes after the opening of the chute. Answer is 0.0732. Any help would be appreciated :) Title: Re: 3U Maths Question Thread Post by: jakesilove on June 27, 2016, 05:50:35 pm Having a bit of a struggle over here... Grain is ejected from a chute at the rate of 0.1 metres cubed a minute and is forming a heap on a flat horizontal floor. The heap is in the form of a circular cone of semi-vertical angle 45 degrees. Find the rate, in metres per minute, at which the height of the cone i s increasing at the instant 3 minutes after the opening of the chute. Answer is 0.0732. Any help would be appreciated :) Hey! This was a bit of an odd question, mainly because you have to figure out the height after 3 minutes yourself. I hope my working out is sufficient: let me know if I can be more explicit! (http://i.imgur.com/pgTDChb.jpg) Also, it's on it's side. Sorry, not sure how to fix that! Jake Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on June 27, 2016, 05:53:26 pm Having a bit of a struggle over here... Grain is ejected from a chute at the rate of 0.1 metres cubed a minute and is forming a heap on a flat horizontal floor. The heap is in the form of a circular cone of semi-vertical angle 45 degrees. Find the rate, in metres per minute, at which the height of the cone i s increasing at the instant 3 minutes after the opening of the chute. Answer is 0.0732. Any help would be appreciated :) Edit: Jake just beat me to a solution, but I finished it, so I'll post it anyway. Maybe it will be equally useful, we used the same method ;) Hey there! This is a chain rule question. These questions can be gnarly, but they all have a very similar approach. First, let's take the question and break it down into some equations. We are told that the cone is being formed from a flow of 0.1 cubic metres per minute. That is, if we take V to be in cubic metres and t to be in minutes: $V=0.1t \\ \frac{dV}{dt}=0.1$ Now, we need a second formula to apply the chain rule. Remember the formula for a volume of a cone: $V = \frac{1}{3}\pi r^2h$ Now this would be great, except we don't want the radius in there. So, to remove it, we use the angle that they gave us. Draw the cone, and you'll be able to see that through some right-angled trigonometry, we can deduce: $r=h$ Or, you might just be able to tell from the info they give you, both are fine! So, that volume formula becomes: $V = \frac{1}{3}\pi h^3 \\ \frac{dV}{dh} = \pi h^2$ Now we apply the chain rule: $\frac{dV}{dt} = \frac{dV}{dh}\times\frac{dh}{dt} \\ 0.01=\pi h^2 \times \frac{dh}{dt} \\\therefore \frac{dh}{dt} = \frac{0.01}{\pi h^2}$ Now to find the final answer, we require the height at 3 minutes. At 3 minutes, we know the volume is 0.03, therefore: $0.3 = \frac{1}{3}\pi h^3 \\\therefore h = \sqrt[3]{\frac{0.9}{\pi}}$ Substituting this into the formula above will give you the answer! ;D Title: Re: 3U Maths Question Thread Post by: RuiAce on June 27, 2016, 05:56:43 pm Edit: Jake just beat me to a solution, but I finished it, so I'll post it anyway. Maybe it will be equally useful, we used the same method ;) It was about time he got a question :P Title: Re: 3U Maths Question Thread Post by: jakesilove on June 27, 2016, 06:00:04 pm It was about time he got a question :P Yeah was pretty happy with myself, sorry matey Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on June 27, 2016, 06:00:47 pm Yeah was pretty happy with myself, sorry matey The race continues ;) or we just need to actually make a system like we always say we are going to aha Title: Re: 3U Maths Question Thread Post by: birdwing341 on June 27, 2016, 06:02:46 pm Many thanks, it seems like a classic case of calculator mistake for me, but that's alright :). Yeah was pretty happy with myself, sorry matey Glad someone could get a self-confidence boost!! Title: Re: 3U Maths Question Thread Post by: jakesilove on June 27, 2016, 06:06:01 pm Many thanks, it seems like a classic case of calculator mistake for me, but that's alright :). Glad someone could get a self-confidence boost!! Calculator mistakes are the worst, especially for a question like this where you've done a solid piece of Mathematics to get to the answer. Rui and Jamon are just too good at the typing-up-maths-on-the-forums-thing, it's difficult to compete. Title: Re: 3U Maths Question Thread Post by: RuiAce on June 27, 2016, 06:34:11 pm The race continues ;) or we just need to actually make a system like we always say we are going to aha But I mean if I see a question unanswered for 3 hours and you guys are MIA... :P Calculator mistakes are the worst, especially for a question like this where you've done a solid piece of Mathematics to get to the answer. Rui and Jamon are just too good at the typing-up-maths-on-the-forums-thing, it's difficult to compete. Actually I would've written this one out just cause of the triangle haha Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on June 27, 2016, 06:35:46 pm But I mean if I see a question unanswered for 3 hours and you guys are MIA... :P Will undoubtedly happen on many more occasions so you are welcome to answer as many or as few as you like ;D Title: Re: 3U Maths Question Thread Post by: jakesilove on June 27, 2016, 06:35:46 pm But I mean if I see a question unanswered for 3 hours and you guys are MIA... :P Actually I would've written this one out just cause of the triangle haha So you're saying I win? Pretty sure that means I win. Take heed kiddos; the HSC is not a competition, but Maths sure is. xoxo Title: Re: 3U Maths Question Thread Post by: 140498 on June 28, 2016, 08:20:48 pm Moderator Action: Added LaTex formatting for easier reading. $f(x) = 2 - \frac{x}{2} \quad (x ≥ 0) \\ g(x) = 2(x-2) \quad (\text{All Real }x)$ Find all values for x for which f[g(x)] = x = g[f(x)]. Title: Re: 3U Maths Question Thread Post by: RuiAce on June 28, 2016, 09:06:44 pm If f(x) = 2 - (x)1/2, x ≥ 0 and g(x) = (x-2)2, for all x. Find all values for x for which f[g(x)] = x = g[f(x)] $\text{Interpreting as }f(x)=2-\frac{x}{2}$ \begin{align*}f(g(x))&=2-\frac{g(x)}{2}\\ &= 2-\frac{\frac{x-2}{2}}{2}\\ &=\frac{5}{2}-\frac{x}{4} \end{align*} \begin{align*}f(g(x))&=x\\ \iff \frac{5}{2}-\frac{x}{4}&=x\\ \iff x&=2\end{align*} \begin{align*}g(f(x))&=\frac{f(x)-2}{2}\\ &=\frac{\left(2-\frac{x}{2}\right)-2}{2}\\ &= -\frac{x}{4}\end{align*} \begin{align*}g(f(x))&=x\\ \iff -\frac{x}{4}&=x\\ \iff x&=0\end{align*} $\text{Therefore no such value for }x\text{ satisfies the three handed equation simultaneously.}$ Moderator Action: Question Fixed. Title: Re: 3U Maths Question Thread Post by: 140498 on June 28, 2016, 10:08:47 pm If f(x) = 2 - (x)^1/2, x ≥ 0 and g(x) = (x-2)^2, for all x. Find all values for x for which f[g(x)] = x = g[f(x)] the (1/2) and 2 are both powers I forgot to put them in sorry Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on June 28, 2016, 10:13:25 pm Moderator Action: Added LaTex formatting for easier reading. $f(x) = 2 - \frac{x}{2} \quad (x ≥ 0) \\ g(x) = 2(x-2) \quad (\text{All Real }x)$ Find all values for x for which f[g(x)] = x = g[f(x)]. Hey, welcome to the forums!! ;D I think I interpret the question differently than Rui, I'd do the following (very brief, the full working would be similar to Rui's) $f\left(g(x)\right) = 4+x$ $g\left(f(x)\right) = -x$ But this gives the same result as Rui anyway. Could you perhaps upload a photo of the original question? ;D PS - Let me know if you need any help finding your way around the forums :) Title: Re: 3U Maths Question Thread Post by: 140498 on June 28, 2016, 10:36:25 pm If f(x) = 2 - (x)^1/2, x ≥ 0 and g(x) = (x-2)^2, for all x. Find all values for x for which f[g(x)] = x = g[f(x)] the (1/2) and 2 are both powers I forgot to put them in sorry Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on June 28, 2016, 10:47:36 pm If f(x) = 2 - (x)^1/2, x ≥ 0 and g(x) = (x-2)^2, for all x. Find all values for x for which f[g(x)] = x = g[f(x)] the (1/2) and 2 are both powers I forgot to put them in sorry Ahhhhh okay cool, let's try again: $f\left(g(x)\right) = 2-\sqrt{(x-2)^2} = 4-x \\ g\left(f(x)\right) = (2-x^{\frac{1}{2}}-2)^2 = x$ $4-x = x = x \\\therefore x=2 \quad\text{is the only solution}$ There we go, I hope that helps! ;D Title: Re: 3U Maths Question Thread Post by: 140498 on June 28, 2016, 10:52:24 pm That's what I got but the questions asks for values (plural) which made me unsure Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on June 28, 2016, 11:35:10 pm That's what I got but the questions asks for values (plural) which made me unsure Even when asking for all values, one is still acceptable!! Let me take another look though... Oh, I did forget to take into account the fact that the square root of the first line can be positive or negative!! This would create a second option: $f\left(g(x)\right) = 2\mp(x-2)=x \text{ or } 4-x$ So actually, taking into account that first option, any value of x greater than or equal to zero will work!! Wow, what a difference a little sign error can make (I can never go more than a few weeks without making one of these) ;) sorry about that! Does this make sense? :) Title: Re: 3U Maths Question Thread Post by: 140498 on June 28, 2016, 11:48:54 pm I understand how you got to where you are, I do not however know how to express the answer that is what is troubling me. Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on June 29, 2016, 12:17:36 am I understand how you got to where you are, I do not however know how to express the answer that is what is troubling me. Sure! So having reached that point, you would likely write something like this: $\therefore \text{the given result is true for } x\ge0$ Does that help? ;D Title: Re: 3U Maths Question Thread Post by: WLalex on June 29, 2016, 07:53:38 am Hey was wondering if someone could please help me out with this strange binomial question for the 2009 HSC. I even had a look at the answers and had absolutely no idea how to do it :P I know its long...sorry and thanks in advance :) Title: Re: 3U Maths Question Thread Post by: RuiAce on June 29, 2016, 10:03:18 am Hey was wondering if someone could please help me out with this strange binomial question for the 2009 HSC. I even had a look at the answers and had absolutely no idea how to do it :P I know its long...sorry and thanks in advance :) Pulling strings here having me do a whole page...! $\textbf{2009 HSC Maths Ext. 1 Q6}$ $\text{b) (i) The series is geometric with first term }(1+x)^r\text{ and common ratio }(1+x)\\ \text{Note how there are }n-r+1\text{ terms}$ \begin{align*}\therefore (1+x)^r + (1+x)^{r+1}+\dots + (1+x)^{n} &= (1+x)^r \left(\frac{(1+x)^{n-r+1}-1}{(1+x)-1}\right)\\ &= (1+x)^r \left(\frac{(1+x)^{n-r+1}-1}{x}\right) \\ &= \frac{(1+x)^{n+1}-(1+x)^r}{x}\end{align*} $\text{Judging by the LHS of the equation we seek to prove, it is clear that we want to equate the coefficient on }x^r\\ \text{On the LHS, the coefficient of }x^r\text{ is automatically }\\ \binom{r}{r}+\binom{r+1}{r}+\dots + \binom{n}{r}$ $\text{The RHS is slightly harder to handle. Notice how we have }\frac{1}{x}\text{ being an obstacle.}\\ \text{To address this, we reanalyse the situation:}$ $\text{The coefficient of }x^r\text{ in } \frac{(1+x)^{n+1}-(1+x)^r}{x}\\ \text{is the same as the coefficient of }x^{r+1}\text{ in }(1+x)^{n+1}-(1+x)^r$ $\text{From }(1+x)^{n+1}\text{ we immediately extract }\binom{n+1}{r+1}\\ \text{However we extract nothing from }(1+x)^r\text{ as the power is too low!}\\ \text{Hence, the coefficient of }x^r\text{ in the RHS is just }\binom{n+1}{r+1}\\ \text{Equate coefficients to finish the proof.}$ ___________________________ $\text{b) (ii) (1) The only way to analyse this question is to realise that on the line }y=x\text{, there are EXACTLY }n\text{ points.}\\ \text{This is simply because we have }n\text{ rows and }n\text{ columns, and we're considering the main diagonal of the square grid.}\\ \text{If we just choose ANY 2 points, we have a total of }\binom{n}{2}\text{ possible outcomes.}$ $\text{b) (ii) (2) Note that we are always choosing just TWO points. This is why the value of }r\\ \text{in }\binom{n}{r}\text{ will always be 2.}$ $\text{Work our way UP from the main diagonal }y=x\\ \text{On the diagonal line right above, we now have }n+1\text{ points to choose 2 from}\\ \text{On the next one, we then have }n+1\text{ points to choose 2 from}\\ \vdots \\ \text{This will continue, so that on the third last diagonal we have 3 points to choose from 2 from}\\ \text{In the second lsat diagonal we have 2 points to choose 2 from}\\ \text{In the last diagonal, we have none. Because how can we choose two from just ONE point?}$ $\text{So if we sum up just these we get:}\\ \binom{2}{2}+\binom{3}{2}+\dots + \binom{n-1}{2}+\binom{n}{2}$ $\text{However we aren't done. We only have a bit more than half of the equation!}\\ \text{This is because we must recall that BELOW the main diagonal }y=x\text{ the exact scenario applies AGAIN.}\\ \text{Hence, by symmetry}\\ S_n = \binom{2}{2}+\binom{3}{2}+\dots + \binom{n-1}{2}+\binom{n}{2}+\binom{n-1}{2}+\dots +\binom{3}{2}+\binom{2}{2}$ _____________________________ $\text{b) (iii) Rewrite the expression into this appropriate form}\\ S_n = 2\left[ \binom{2}{2}+\binom{3}{2}+\dots+\binom{n-1}{2} \right]+\binom{n}{2}$ $\text{Note that we can use the expression we derived in part a) treating }r=2\text{ and swapping }n\text{ with }n-1\\ \therefore S_n=2\binom{n}{3} +\binom{n}{2}$ \text{Finally, do battle with the messy algebra}\\ \begin{align*}S_n&=2\binom{n}{3}+\binom{n}{2}\\ &= \frac{2n!}{3!(n-3)!}+\frac{n!}{2!(n-2)!}\\ &= \frac{2n(n-1)(n-2)(n-3)!}{6(n-3)!}+\frac{n(n-1)(n-2)!}{2(n-2)!}\\ &=\frac{n(n-1)(n-2)}{3}+\frac{n(n-1)}{2} \\ &= n(n-1)\left(\frac{n-2}{3}+\frac{1}{2}\right)\\ &=n(n-1)\left(\frac{2n-1}{6}\right) \end{align*} Title: Re: 3U Maths Question Thread Post by: RuiAce on June 29, 2016, 10:23:39 am Even when asking for all values, one is still acceptable!! Let me take another look though... Oh, I did forget to take into account the fact that the square root of the first line can be positive or negative!! This would create a second option: $f\left(g(x)\right) = 2\mp(x-2)=x \text{ or } 4-x$ So actually, taking into account that first option, any value of x greater than or equal to zero will work!! Wow, what a difference a little sign error can make (I can never go more than a few weeks without making one of these) ;) sorry about that! Does this make sense? :) I understand how you got to where you are, I do not however know how to express the answer that is what is troubling me. Sure! So having reached that point, you would likely write something like this: $\therefore \text{the given result is true for } x\ge0$ Does that help? ;D I disagree (http://i1164.photobucket.com/albums/q566/Rui_Tong/Screen%20Shot%202016-06-29%20at%2010.10.48%20AM_zpsfao3hqma.png) $\text{The composition }p(x)=g(f(x))\text{ defined for }x\ge 0\text{ has the equation }g(f(x))=x\\ \text{as correctly stated }$ \text{The composition }h(x)=f(g(x))\text{ defines a piecewise function }\\ \begin{align*} f(g(x))&=2-\sqrt{(x-2)^2}\\ &=2-|x-2| \end{align*}\\ \text{for all real }x $\therefore f(g(x))=\begin{cases}x & \text{for }x < 2 \\ 4-x &\text{for }x \ge 2\end{cases}$ $\text{Clearly, because }f(g(x))\equiv g(f(x))\text{ for }x<2\text{, we have }0\le x < 2\text{ being a part of the entire solution.}$ $\text{When }x \ge 2\text{ we solve }x=4-x \iff x=2$ $\text{Hence the solutions are }0\le x \le 2$ Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on June 29, 2016, 10:38:01 am \text{The composition }h(x)=f(g(x))\text{ defines a piecewise function }\\ \begin{align*} f(g(x))&=2-\sqrt{(x-2)^2}\\ &=2-|x-2| \end{align*}\\ \text{for all real }x Hmm yep you're right, the subtraction being there implies the sign we should take in this case. No idea what I was thinking above, had a bad streak of mistakes :P Title: Re: 3U Maths Question Thread Post by: 140498 on June 29, 2016, 11:11:16 am If dy/dx = 5x and dx/dt = -2 Find: dy/dt and d^2y/dt^2 Title: Re: 3U Maths Question Thread Post by: jakesilove on June 29, 2016, 11:22:39 am If dy/dx = 5x and dx/dt = -2 Find: dy/dt and d^2y/dt^2 We know that $\frac{dy}{dx}=5x$ $\frac{dx}{dt}=-2$ To find dy/dt, we can use the relationship that we learn in 3U: $\frac{dy}{dt}=\frac{dy}{dx} \frac{dx}{dt}$ Since we have all the information required, we can just sub everything in! $\frac{dy}{dt}=-10x$ Similarly, $\frac{d^2y}{dt^2}=\frac{d^2y}{dx^2} \frac{d^2x}{dt^2}$ $\frac{d^2y}{dx^2}=\frac{d}{dx} 5x= 5$ $\frac{d^2x}{dt^2}=\frac{d}{dt} (-2)= 0$ Therefore, subbing all this in, you get $\frac{d^2y}{dt^2}=5*0=0$ Hope that made sense! Title: Re: 3U Maths Question Thread Post by: RuiAce on June 29, 2016, 11:41:40 am To find dy/dt, we can use the relationship that we learn in 3U: $\frac{dy}{dt}=\frac{dy}{dx} \frac{dx}{dt}$ Lol that's just the chain rule. Also I'm not too sure if it works like that for the second derivative... https://en.wikipedia.org/wiki/Chain_rule#Higher_derivatives Which makes me also question the validity of this problem at the Extension 1 level. \text{From }\frac{dy}{dt}=-10x\\ \begin{align*}\frac{d^2y}{dt^2}&=\frac{d}{dt}(-10x)\\&= \frac{d}{dx}(-10x) \cdot \frac{dx}{dt}\\ &= -10\cdot -2\\ &= 20\end{align*} Title: Re: 3U Maths Question Thread Post by: jakesilove on June 29, 2016, 11:54:13 am Lol that's just the chain rule. Also I'm not too sure if it works like that for the second derivative... https://en.wikipedia.org/wiki/Chain_rule#Higher_derivatives Which makes me also question the validity of this problem at the Extension 1 level. \text{From }\frac{dy}{dt}=-10x\\ \begin{align*}\frac{d^2y}{dt^2}&=\frac{d}{dt}(-10x)\\&= \frac{d}{dx}(-10x) \cdot \frac{dx}{dt}\\ &= -10\cdot -2\\ &= 20\end{align*} To be honest, I couldn't remember whether they call it the Chain rule in High School or they just told you it was a thing that was true. I also assumed they couldn't do the method you wrote out above, so assumed mine had to be true. I was probably/definitely wrong with that second half, sorry! Title: Re: 3U Maths Question Thread Post by: RuiAce on June 29, 2016, 11:59:11 am To be honest, I couldn't remember whether they call it the Chain rule in High School or they just told you it was a thing that was true. I also assumed they couldn't do the method you wrote out above, so assumed mine had to be true. I was probably/definitely wrong with that second half, sorry! Haha they explicitly teach the chain, product and quotient rules in 2U for a reason. Yeah like I said, the validity of the problem at 3U seems dodgy. At the 4U level this is a trivial implicit differentiation question. Whilst the method I used still sticks within the boundaries of 3U, the ability to see the change of position of 'dx' can be hard. Title: Re: 3U Maths Question Thread Post by: 140498 on June 29, 2016, 02:03:10 pm Prove: cos^-1(-x) = pi - cos^-1(x) Title: Re: 3U Maths Question Thread Post by: RuiAce on June 29, 2016, 02:04:20 pm Prove: cos^-1(-x) = pi - cos^-1(x) $\text{There are countless ways to prove this.}\\ \text{Examples include calculus, compound angles}$ You will need to provide context if you want a specific proof. Title: Re: 3U Maths Question Thread Post by: 140498 on June 29, 2016, 02:28:36 pm Compound angles please(it is withing inverse functions), I have proved it but not sure if it is right. Would just like to check Title: Re: 3U Maths Question Thread Post by: RuiAce on June 29, 2016, 02:36:47 pm Compound angles please(it is withing inverse functions), I have proved it but not sure if it is right. Would just like to check Sure thing. (Also, lol I made a mistake you don't really need compound angles; you just need ASTC for this particular one which eases out the proof) Note that arccos is just the international notation for cos-1 \text{For }-1\le x\le 1 \\ \begin{align*}\cos RHS &=\cos (\pi - \arccos x)\\ &= -\cos(\arccos x)\\ &= -x\\ \cos LHS&= \cos (\arccos (-x))\\ &= -x \\ \therefore \cos LHS &= \cos RHS\\ \therefore LHS &= RHS\end{align*}\\ \text{as }\arccos\text{is continuous and monotone} Note that your working does not have to be set out exactly identically. If you want me to provide feedback on your working out then provide it. Title: Re: 3U Maths Question Thread Post by: jakesilove on June 29, 2016, 02:58:30 pm Compound angles please(it is withing inverse functions), I have proved it but not sure if it is right. Would just like to check If it's within inverse functions, I think the easiest (and neatest) way to prove this relationship is to create a function $y=arccos(-x)+arccos(x)$ Differentiate the thing, using the formulas you've learnt in the inverse function section of the course. You'll see it is equal to zero, suggesting that the gradient of this function is zero (first derivative = gradient). If the gradient is zero, then the original function can only be a horizontal line (ie. for any value of x, there is only one value for y). From there, proving it equals Pi (as the question asks) is simple: sub in any value for x. Whether you put x=1, x=3.14 or x=1.6893*10^8, you'll get Pi out as the y value. As the function has a gradient of 0, this solution holds true for any value of x, therefore the original relationship must be true! Jake Title: Re: 3U Maths Question Thread Post by: RuiAce on June 29, 2016, 03:29:58 pm If it's within inverse functions, I think the easiest (and neatest) way to prove this relationship is to create a function $y=arccos(-x)+arccos(x)$ Differentiate the thing, using the formulas you've learnt in the inverse function section of the course. You'll see it is equal to zero, suggesting that the gradient of this function is zero (first derivative = gradient). If the gradient is zero, then the original function can only be a horizontal line (ie. for any value of x, there is only one value for y). From there, proving it equals Pi (as the question asks) is simple: sub in any value for x. Whether you put x=1, x=3.14 or x=1.6893*10^8, you'll get Pi out as the y value. As the function has a gradient of 0, this solution holds true for any value of x, therefore the original relationship must be true! Jake Yeah I agree that calculus is the faster method here. He just wanted a compound angles proof. $\text{Advice on putting a value of }x\\ \text{Just put }x=1\text{ or }x=0\text{ for convenience}\\ \text{But you have to put something between }-1\text{ and }1\text{ or arccos doesn't exist!}$ Title: Re: 3U Maths Question Thread Post by: WLalex on June 29, 2016, 03:38:13 pm Pulling strings here having me do a whole page...! $\textbf{2009 HSC Maths Ext. 1 Q6}$ $\text{b) (i) The series is geometric with first term }(1+x)^r\text{ and common ratio }(1+x)\\ \text{Note how there are }n-r+1\text{ terms}$ \begin{align*}\therefore (1+x)^r + (1+x)^{r+1}+\dots + (1+x)^{n} &= (1+x)^r \left(\frac{(1+x)^{n-r+1}-1}{(1+x)-1}\right)\\ &= (1+x)^r \left(\frac{(1+x)^{n-r+1}-1}{x}\right) \\ &= \frac{(1+x)^{n+1}-(1+x)^r}{x}\end{align*} $\text{Judging by the LHS of the equation we seek to prove, it is clear that we want to equate the coefficient on }x^r\\ \text{On the LHS, the coefficient of }x^r\text{ is automatically }\\ \binom{r}{r}+\binom{r+1}{r}+\dots + \binom{n}{r}$ $\text{The RHS is slightly harder to handle. Notice how we have }\frac{1}{x}\text{ being an obstacle.}\\ \text{To address this, we reanalyse the situation:}$ $\text{The coefficient of }x^r\text{ in } \frac{(1+x)^{n+1}-(1+x)^r}{x}\\ \text{is the same as the coefficient of }x^{r+1}\text{ in }(1+x)^{n+1}-(1+x)^r$ $\text{From }(1+x)^{n+1}\text{ we immediately extract }\binom{n+1}{r+1}\\ \text{However we extract nothing from }(1+x)^r\text{ as the power is too low!}\\ \text{Hence, the coefficient of }x^r\text{ in the RHS is just }\binom{n+1}{r+1}\\ \text{Equate coefficients to finish the proof.}$ ___________________________ $\text{b) (ii) (1) The only way to analyse this question is to realise that on the line }y=x\text{, there are EXACTLY }n\text{ points.}\\ \text{This is simply because we have }n\text{ rows and }n\text{ columns, and we're considering the main diagonal of the square grid.}\\ \text{If we just choose ANY 2 points, we have a total of }\binom{n}{2}\text{ possible outcomes.}$ $\text{b) (ii) (2) Note that we are always choosing just TWO points. This is why the value of }r\\ \text{in }\binom{n}{r}\text{ will always be 2.}$ $\text{Work our way UP from the main diagonal }y=x\\ \text{On the diagonal line right above, we now have }n+1\text{ points to choose 2 from}\\ \text{On the next one, we then have }n+1\text{ points to choose 2 from}\\ \vdots \\ \text{This will continue, so that on the third last diagonal we have 3 points to choose from 2 from}\\ \text{In the second lsat diagonal we have 2 points to choose 2 from}\\ \text{In the last diagonal, we have none. Because how can we choose two from just ONE point?}$ $\text{So if we sum up just these we get:}\\ \binom{2}{2}+\binom{3}{2}+\dots + \binom{n-1}{2}+\binom{n}{2}$ $\text{However we aren't done. We only have a bit more than half of the equation!}\\ \text{This is because we must recall that BELOW the main diagonal }y=x\text{ the exact scenario applies AGAIN.}\\ \text{Hence, by symmetry}\\ S_n = \binom{2}{2}+\binom{3}{2}+\dots + \binom{n-1}{2}+\binom{n}{2}+\binom{n-1}{2}+\dots +\binom{3}{2}+\binom{2}{2}$ _____________________________ $\text{b) (iii) Rewrite the expression into this appropriate form}\\ S_n = 2\left[ \binom{2}{2}+\binom{3}{2}+\dots+\binom{n-1}{2} \right]+\binom{n}{2}$ $\text{Note that we can use the expression we derived in part a) treating }r=2\text{ and swapping }n\text{ with }n-1\\ \therefore S_n=2\binom{n}{3} +\binom{n}{2}$ \text{Finally, do battle with the messy algebra}\\ \begin{align*}S_n&=2\binom{n}{3}+\binom{n}{2}\\ &= \frac{2n!}{3!(n-3)!}+\frac{n!}{2!(n-2)!}\\ &= \frac{2n(n-1)(n-2)(n-3)!}{6(n-3)!}+\frac{n(n-1)(n-2)!}{2(n-2)!}\\ &=\frac{n(n-1)(n-2)}{3}+\frac{n(n-1)}{2} \\ &= n(n-1)\left(\frac{n-2}{3}+\frac{1}{2}\right)\\ &=n(n-1)\left(\frac{2n-1}{6}\right) \end{align*} Thank you so so much! Very quick as well, impressive :P Title: Re: 3U Maths Question Thread Post by: 140498 on June 29, 2016, 04:15:28 pm Thanks so much for the previous answer Thanks for the help, this should be my last one in a while: an inverted cone of radius 20cm and depth is filled with water. The water flows out through the apex of the cone at a constant rate of 10pi cm^3s^-1. Find the rate at which the water is falling when the depth of the water is 20cm. Just like my other questions i have an answer, but it is a take home task and i would just like to check my answer. Title: Re: 3U Maths Question Thread Post by: RuiAce on June 29, 2016, 04:25:39 pm Thanks so much for the previous answer Thanks for the help, this should be my last one in a while: an inverted cone of radius 20cm and depth is filled with water. The water flows out through the apex of the cone at a constant rate of 10pi cm^3s^-1. Find the rate at which the water is falling when the depth of the water is 20cm. Just like my other questions i have an answer, but it is a take home task and i would just like to check my answer. Some clarity needed. Are we saying that the depth of the full cone is also 20cm? (And by consequence we're basically saying the initial rate that water is falling out?) Title: Re: 3U Maths Question Thread Post by: RuiAce on June 29, 2016, 04:39:03 pm If yes (http://i.imgur.com/lmgjqOV.jpg) Title: Re: 3U Maths Question Thread Post by: 140498 on June 29, 2016, 05:07:13 pm My computer is stuffing up it did not copy properly all information is correct except: radius = 20 depth = 30 Really sorry :( Title: Re: 3U Maths Question Thread Post by: RuiAce on June 29, 2016, 05:13:28 pm Just repeat the working out then for 30/20=h/r So r=2h/3 Title: Re: 3U Maths Question Thread Post by: 140498 on June 29, 2016, 05:29:35 pm ok thanks :) Title: Re: 3U Maths Question Thread Post by: 140498 on June 29, 2016, 05:34:44 pm Because the vessel is losing water that means the rate of change is negative right? Title: Re: 3U Maths Question Thread Post by: RuiAce on June 29, 2016, 06:10:57 pm Because the vessel is losing water that means the rate of change is negative right? Sure Or we could define positive as 'losing water' whereas negative is 'gaining water' ;) Title: Re: 3U Maths Question Thread Post by: 140498 on June 29, 2016, 06:18:12 pm Thanks heaps man, Sorry for all the dumb questions (in another setting where it is not an assessment I would not be second guessing myself) Just quickly how was the most effective way you found studying for ext 1, just do past papers or was there something else. Title: Re: 3U Maths Question Thread Post by: RuiAce on June 29, 2016, 06:22:20 pm Thanks heaps man, Sorry for all the dumb questions (in another setting where it is not an assessment I would not be second guessing myself) Just quickly how was the most effective way you found studying for ext 1, just do past papers or was there something else. You're breaking all my own work and HSC policies if I'm doing an assignment for you. Past papers all the way. That's for whatever maths you do. Title: Re: 3U Maths Question Thread Post by: 140498 on June 29, 2016, 06:26:15 pm No no no, i have my own answers I was just checking to see if I had gotten them right. That is why they were relatively easy questions that I was asking, just to make sure. Sorry for not clearing that up Title: Re: 3U Maths Question Thread Post by: katherine123 on July 02, 2016, 10:00:08 am how do i do last part (iii) Title: Re: 3U Maths Question Thread Post by: RuiAce on July 02, 2016, 12:33:53 pm $\textbf{2006 HSC Mathematics Ext 1}$ $\text{To do this question, simply consider the equation of the first particle.}\\ \text{Particle 1 is ascending when }\dot{y}>0$ \begin{align*}\therefore V\sin\theta - gt &> 0 \\ t &< \frac{V\sin \theta}{g} \end{align*} $\text{So by using the result from part (ii) we require:}\\ \frac{a\cos \theta}{2V(1-\sin \theta)}<\frac{V\sin \theta}{g} \\ \text{The rearranging isn't too messy at all.}$ This question just required you to interpret the question and realise "oh, have to go back to where I started". It's more of a deceptively easy question. Title: Re: 3U Maths Question Thread Post by: katherine123 on July 03, 2016, 02:10:16 am first ques: not sure how to do last part ie finding t second ques: not sure how to do last part Title: Re: 3U Maths Question Thread Post by: RuiAce on July 03, 2016, 11:12:28 am first ques: not sure how to do last part ie finding t second ques: not sure how to do last part As always, keep in mind that questions before the year 2001 do not accurately reflect the nature of the current HSC. $\textbf{2000 HSC 3U Additional + 3/4U Common}$ $\text{We are talking about distances but we just have speed. Let }u=\frac{D}{t}$ $\text{Observe that }F\text{ describes fuel used per HOUR. It is a RATE.}\\ \text{Total fuel }k\text{ is therefore }k=Ft$ \begin{align*}D&=ut\\ &= \frac{uk}{F}\\ &= \frac{uk}{Au^3+\frac{B}{u}}\\ &= \frac{ku^2}{Au^4+B} \\ \implies \frac{dD}{du}&=\frac{2ku\left(B-Au^4\right)}{\left(Au^4+B\right)^2}\end{align*}\\ \text{following a rather messy quotient rule.} $\text{For obvious reasons, distance is always going to increase. And we want to maximise distance, so let }\frac{dD}{du}=0\\ \text{Solutions are }u=0\text{ or }u^4=\frac{B}{A}\text{ but obviously the former is the minimum, so we take the latter.}$ $\text{The ratio of the new speed to the old speed is}\\ \frac{\text{New speed}}{\text{Old speed}}=\frac{\left(\frac{B}{A}\right)^{\frac{1}{4}}}{\left(\frac{B}{3A}\right)^{\frac{1}{4}}}=3^{\frac{1}{4}}\approx 132/100\text{ (nearest percent)}$ $\therefore \text{about 32 percent}\text{ faster.}$ Title: Re: 3U Maths Question Thread Post by: RuiAce on July 03, 2016, 11:30:48 am first ques: not sure how to do last part ie finding t second ques: not sure how to do last part $\textbf{2001 HSC Mathematics Ext. 1 Q7}$ $\text{The nasty differentiation could be made simpler through implicit differentiation}\\ \text{but as Ext 1 students may not know about it, use brute force.}$ \begin{align*}\theta&=\cos^{-1} \left( \frac{1}{\sqrt{2}}\sin \alpha + \frac{1}{2\sqrt{2}} \cos \alpha \right) \\ \implies \frac{d\theta}{d\alpha} &= -\frac{\left( \frac{1}{\sqrt{2}}\cos \alpha - \frac{1}{2\sqrt{2}} \sin \alpha \right)}{\sqrt{1-\left( \frac{1}{\sqrt{2}}\sin \alpha + \frac{1}{2\sqrt{2}} \cos \alpha \right)^2}}\end{align*} $\text{Set }\frac{d\theta}{d\alpha} =0 \text{ for the stationary point to have }\tan \alpha =2 \iff \alpha=\tan^{-1}2$ For the sake of graphing, by punching tan-1(2) into your calculator we find that alpha is approximately 1.107 $\text{Draw up a right-angled triangle with opposite side length 2 and adjacent side length 1}\\ \text{Consequently the hypotenuse is 5}\\ \therefore \cos \alpha=\frac{1}{\sqrt{5}}, \sin \alpha=\frac{2}{\sqrt{5}}$ $\text{Subbing back in for }\theta\text{ we have }\theta=\cos^{-1}\frac{\sqrt{10}}{4}$ $\lim_{\alpha \to 0}\theta = \lim_{\alpha \to 0}\cos^{-1} \left( \frac{1}{\sqrt{2}}\sin \alpha + \frac{1}{2\sqrt{2}} \cos \alpha \right) =\cos^{-1}\frac{1}{2\sqrt{2}}\\ \text{So as }\alpha \to 0, \theta \to\cos^{-1}\frac{1}{2\sqrt{2}} \approx 1.209$ $\lim_{\alpha \to \frac{\pi}{2}}\theta = \lim_{\alpha \to \frac{\pi}{2}}\cos^{-1} \left( \frac{1}{\sqrt{2}}\sin \alpha + \frac{1}{2\sqrt{2}} \cos \alpha \right) =\frac{\pi}{4}\\ \text{So as }\alpha \to \frac{\pi}{2}, \theta \to \frac{\pi}{4}$ $\text{Putting together the information gives a graph looking somewhat like the attachment.}$ Note: An alternate approach would've been to use an auxiliary angle transformation at the start. This is good, if you can do it by INSPECTION. Title: Re: 3U Maths Question Thread Post by: conic curve on July 09, 2016, 07:45:20 pm A friend of mine needed help with this question: 1. a. Consider the function f(x)=e^x/(1+e^x). Find f'(x) and deduce that f(x) is increasing for all x b. State the range of f(x) c. Find the inverse function f^-1(x) d. Draw y-f(x) and y=f^-1(x) on the same diagram Title: Re: 3U Maths Question Thread Post by: jakesilove on July 09, 2016, 07:52:11 pm A friend of mine needed help with this question: 1. a. Consider the function f(x)=e^x/(1+e^x). Find f'(x) and deduce that f(x) is increasing for all x b. State the range of f(x) c. Find the inverse function f^-1(x) d. Draw y-f(x) and y=f^-1(x) on the same diagram Hey! My solution is below (http://i.imgur.com/bax8bCK.png?1) (http://i.imgur.com/srOuRoe.png?1) Jake Title: Re: 3U Maths Question Thread Post by: conic curve on July 09, 2016, 08:15:51 pm Hey! My solution is below (http://i.imgur.com/bax8bCK.png?1) (http://i.imgur.com/srOuRoe.png?1) Jake Thanks ;D Title: Re: 3U Maths Question Thread Post by: RuiAce on July 09, 2016, 09:23:09 pm A friend of mine needed help with this question: 1. a. Consider the function f(x)=e^x/(1+e^x). Find f'(x) and deduce that f(x) is increasing for all x b. State the range of f(x) c. Find the inverse function f^-1(x) d. Draw y-f(x) and y=f^-1(x) on the same diagram $\text{For part a) it is possible to immediately apply the quotient rule like Jake did. However why not be clever and do this first:}\\ f(x)=\frac{e^x+1}{1+e^x}-\frac{1}{1+e^x}=1-\frac{1}{1+e^x}\\ \text{Notice all I did was add and take away 1 in the numerator. This effectively means it is not changed.}$ \text{The derivative is now easier to compute by the chain rule.}\\ \begin{align*}f(x)&=1-(1+e^x)^{-1}\\ \implies f^\prime (x)& = e^x(1+e^x)^{-2}\\ &= \frac{e^x}{(1+e^x)^2}\end{align*} ________________________ $\text{For part b), the formal way to analyse the range is to use part a)}\\ \text{BECAUSE }f(x)\text{ is monotonic increasing for all real }x\\ \text{we analyse the behaviour as }x\text{ goes to positive and negative infinity as was shown.}$ $\lim_{x \to -\infty}\frac{e^x}{1+e^x} = \frac{0}{1+0}=0\\ \lim_{x\to \infty}\frac{e^x}{1+e^x} =\lim_{x\to \infty}\frac{1}{1+e^{-x}}=\frac{1}{1+0}=1$ $\text{Because infinity is not a number, combining the fact that }f\text{ is monotonic increasing gives us}\\ 0 < y < 1$ $\text{Two limits were used that should be memorised.}\\ \lim_{x\to \infty}e^{-x}=0 \qquad \lim_{x\to -\infty}e^x=0$ ________________________ I have no further input for part c) - it could not be any simpler :) ________________________ $\text{When sketching the graph of a function and its inverse, the one thing to remember is}\\ \text{The graph of the inverse }y=f^{-1}(x)\text{ is the graph of the original }y=f(x)\textbf{ reflected about the line }y=x$ $\text{It should be obvious that the asymptotes to }y=f(x)\text{ are }y=0 and y=1\text{ by Jake's calculations and the range.}\\ \text{By using our conventional method of }x=0\text{ we find a }y\text{-intercept at 1}.\\ \text{Putting this information together yields the graph of }y=f(x)\text{ shown.}\\ \text{Simply reflect about the line }y=x\text{ to generate the graph of }y=f^{-1}(x)$ Title: Re: 3U Maths Question Thread Post by: IkeaandOfficeworks on July 09, 2016, 09:30:48 pm Hi Guys! Im having problems with this question: The perimeter of a circular sector is 20 cm. The radius is increasing at the rate of 5cm/s . At what rate is the angle of the sector changing when the radius length is 10 cm? Also at what rate is the area changing when the radius length is 10 cm. Thanks guys! Title: Re: 3U Maths Question Thread Post by: RuiAce on July 09, 2016, 09:36:28 pm Hi Guys! Im having problems with this question: The perimeter of a circular sector is 20 cm. The radius is increasing at the rate of 5cm/s . At what rate is the angle of the sector changing when the radius length is 10 cm? Also at what rate is the area changing when the radius length is 10 cm. Thanks guys! $\text{We need to somehow bring it back down to }\frac{d\theta}{dr}$ $\text{Draw an arbitrary sector. The sector is bound by the arc length, and two radii.}\\ \text{But we know that the arc length has formula }l=r\theta$ \begin{align*}\therefore P=20&=2r + r\theta\\ \iff r\theta &= 20-2r\\ \iff \theta &= \frac{20}{r}-2\end{align*} $\therefore \frac{d\theta}{dr}=-\frac{20}{r^2}$ \text{Using the chain rule:}\\ \begin{align*}\frac{d\theta}{dt}&=\frac{d\theta}{dr}\, \frac{dr}{dt}\\ \therefore \frac{d\theta}{dr}&=-\frac{20}{r^2}\times 5\\&=-1 \text{ rad s}^{-1}\end{align*}\\ \text{when }r=10 _________________________ $\text{For the area, we note that }A=\frac{1}{2}r^2\theta\\ \therefore A=\frac{1}{2}r^2\left(\frac{20}{r}-2\right)=10r^2-r\\ \therefore \frac{dA}{dr}=20r-1$ $\text{Repeat the process.}$ Title: Re: 3U Maths Question Thread Post by: jakesilove on July 10, 2016, 10:39:31 am Just in case anyone's interested feel free to try out this question 1. a. SKetch f(x)=(e^x)-4, showing clearly the points of intersection with the axes and the equations of any asymptotes b. On the same diagram, sketch the graph of the inverse function f^-1(x), showing clearly any important features c. Explain why the x-coordinate of any points of the intersection y=f(x) and y=f^-1(x) satisfies e^x - x - 4=0 d. Show the equation e^x - x - 4 =0 between x=0 and x=2 and use the method of 'halving by intervals' to find this root correct to the nearest whole number Also anyone here wishing to try a few challenge circle geometry questions feel free to If you're looking to post challenge questions, just for your own and our edification, you can find the challenge question thread here. If you actually need help with these questions, because you're not sure how to do them, then this is the place to post them; but no point posting them here just because they are 'hard'. Happy to write up a solution, if you want one :) Jake EDIT BY JAMON: Moved these quoted questions to a more appropriate location ;D Title: Re: 3U Maths Question Thread Post by: conic curve on July 10, 2016, 10:42:59 am If you're looking to post challenge questions, just for your own and our edification, you can find the challenge question thread here. If you actually need help with these questions, because you're not sure how to do them, then this is the place to post them; but no point posting them here just because they are 'hard'. Happy to write up a solution, if you want one :) Jake Oh whoops, my bad ahaha but what if they're not "hard" in that sense but you still want to test someone? Title: Re: 3U Maths Question Thread Post by: RuiAce on July 10, 2016, 10:44:12 am Oh whoops, my bad ahaha but what if they're not "hard" in that sense but you still want to test someone? Please reserve this thread for SOS pleas, not for tests. Not sure how obvious it actually was but to me it was very clear that you just took the circle geometry questions straight out of maths in focus. Title: Re: 3U Maths Question Thread Post by: jakesilove on July 10, 2016, 10:45:14 am Oh whoops, my bad ahaha but what if they're not "hard" in that sense but you still want to test someone? If you want us to answer a question that you can't, because you need help, post them here. This thread is to help you feel more comfortable with Extension 1, to be better equipped walking into your exam. We have a lot of questions we need to answer. If there are questions you're just interested in, rather than actually need help with, then that's great! We just need to prioritize questions people are struggling with. So, if you're interested, post it in another thread. If you need help, post it here. Easy as that! Title: Re: 3U Maths Question Thread Post by: conic curve on July 10, 2016, 10:59:04 am Please reserve this thread for SOS pleas, not for tests. Not sure how obvious it actually was but to me it was very clear that you just took the circle geometry questions straight out of maths in focus. What is SOS? SO for tests I put it in the challenge thread Yes it was from MIF but I had to crop the image Title: Re: 3U Maths Question Thread Post by: RuiAce on July 10, 2016, 11:00:26 am What is SOS? SO for tests I put it in the challenge thread Yes it was from MIF but I had to crop the image Save our souls In other words a plea for help. Title: Re: 3U Maths Question Thread Post by: conic curve on July 10, 2016, 11:03:11 am Save our souls In other words a plea for help. Oh, okay :D Title: Re: 3U Maths Question Thread Post by: humble mango on July 13, 2016, 08:01:05 pm Hi! This is a very generic question compared to the others, but I'm doing Prelims next term, and i was wondering for all exams (HSC Trials,Prelims) involving 3D trigonometry, are diagrams always given?? I have done a lot of practice questions involving them but there are certain ones without diagrams and require you to draw them, and i really struggle at drawing the correct diagram whilst being crammed with tons of info on bearings, numbers and letters. Title: Re: 3U Maths Question Thread Post by: RuiAce on July 13, 2016, 08:15:10 pm Hi! This is a very generic question compared to the others, but I'm doing Prelims next term, and i was wondering for all exams (HSC Trials,Prelims) involving 3D trigonometry, are diagrams always given?? I have done a lot of practice questions involving them but there are certain ones without diagrams and require you to draw them, and i really struggle at drawing the correct diagram whilst being crammed with tons of info on bearings, numbers and letters. Generally, yes. You don't know for sure ever, but every paper I've seen has given you the diagram for 3D trig. Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on July 13, 2016, 11:55:50 pm Hi! This is a very generic question compared to the others, but I'm doing Prelims next term, and i was wondering for all exams (HSC Trials,Prelims) involving 3D trigonometry, are diagrams always given?? I have done a lot of practice questions involving them but there are certain ones without diagrams and require you to draw them, and i really struggle at drawing the correct diagram whilst being crammed with tons of info on bearings, numbers and letters. Hey there humble mango!! Aha Rui is right, although I have seen one question with it in a weird Trial, I remember thinking "wtf" when I saw it back in 2014 ;D but you are pretty safe I reckon! Some tips on the diagram in case it does pop up; underline all the angles and data you are given and make sure that all of it goes in, don't miss anything, that's a good start. Besides that, draw the diagram in pieces, adding information piece at a time and crossing off as you go. Draw in pencil and fix mistakes. Or, alternatively, scribble a draft to make sure everything fits together nicely before the final version. Finally, take your time! The diagram is crucial for understanding and interpreting the question, best to spend some time to get it right ;D Title: Re: 3U Maths Question Thread Post by: katherine123 on July 14, 2016, 02:30:54 pm for part iv) is it okay to say: since <TAQ=<QCB (from part iii) therefore converse of equal chord RQ subtend equal angle therefore CAQR is a cyclic quad given <AQT=90 therefore <AQT= <CRA=90 (equal chord subtend equal angle in cyclic quad CAQR) therefore AR is perpendicular to CB Title: Re: 3U Maths Question Thread Post by: RuiAce on July 14, 2016, 02:50:52 pm for part iv) is it okay to say: since <TAQ=<QCB (from part iii) therefore converse of equal chord RQ subtend equal angle therefore CAQR is a cyclic quad given <AQT=90 therefore <AQT= <CRA=90 (equal chord subtend equal angle in cyclic quad CAQR) therefore AR is perpendicular to CB Not written with as much sophistication as my picky self would prefer e.g. <AQT = <AQC = <CRA however enough has been covered to say it is most certainly correct. Title: Re: 3U Maths Question Thread Post by: jakesilove on July 14, 2016, 02:51:31 pm for part iv) is it okay to say: since <TAQ=<QCB (from part iii) therefore converse of equal chord RQ subtend equal angle therefore CAQR is a cyclic quad given <AQT=90 therefore <AQT= <CRA=90 (equal chord subtend equal angle in cyclic quad CAQR) therefore AR is perpendicular to CB Definitely comprehensive enough for full marks! Nice one Title: Re: 3U Maths Question Thread Post by: zoe_rammie on July 15, 2016, 11:25:50 am Hi Jake (and whoever else reads this), Just to get straight to the point, I don't quite understand the answer to this binomial question (Taken from CSSA Trial 2012) (Image attached, because I don't know how to paste it into this textbox LMAO much technology yay) And alsoooo, Isn't it so much easier to just belt out factorial notation, and slay LHS=RHS Title: Re: 3U Maths Question Thread Post by: jakesilove on July 15, 2016, 12:52:20 pm Hi Jake (and whoever else reads this), Just to get straight to the point, I don't quite understand the answer to this binomial question (Taken from CSSA Trial 2012) (Image attached, because I don't know how to paste it into this textbox LMAO much technology yay) And alsoooo, Isn't it so much easier to just belt out factorial notation, and slay LHS=RHS Hey Zoe, First of all, your description of a potential method of answering this question is absolutely gold. Keep at it, love your work. Your method, and the method in the answers, is actually the same. If you do what you were planning on doing, that's great, but you would have to equate coefficients as there isn't another way to compare the RHS to the LHS and prove that they are equal. So, you would equate coefficients, and get exactly the same answer as they did. They're answer probably isn't comprehensive enough anyway; if your method gets you the right relationship, you're absolutely doing the right thing. If you wanted a more comprehensive proof, though, let us know, however as far as I can tell your "do shit get marks" methodology is spot on. Jake Title: Re: 3U Maths Question Thread Post by: RuiAce on July 15, 2016, 04:09:36 pm Hi Jake (and whoever else reads this), Just to get straight to the point, I don't quite understand the answer to this binomial question (Taken from CSSA Trial 2012) (Image attached, because I don't know how to paste it into this textbox LMAO much technology yay) And alsoooo, Isn't it so much easier to just belt out factorial notation, and slay LHS=RHS Firstly, exactly what Jake said. Inspection of coefficients hastes the process (possibly without you realising it because it appears as more writing) because you just look at it and pull things out. $\text{Rather than expanding }(1+x)^n\text{ and }(x+1)^n\text{, the skill of just looking at the expressions}\\ \text{and being able to equate coefficients is important. Without this skill, you would not be able to prove something such as}\\ \binom{8}{3}=\binom{5}{0}\binom{3}{3}+\binom{5}{1}\binom{3}{2}+\binom{5}{2}\binom{3}{1}+\binom{5}{3}\binom{3}{0}$ Your proof using the definition of the binomial coefficient is in no way whatsoever flawed. The method of equating coefficients is just manipulation of an identity, whereas the definition just uses subtraction. Of course, such a question would most likely not be asked in the HSC because it will be too offputting. EVERYONE will think the definition is faster to use. Title: Re: 3U Maths Question Thread Post by: ATWalk on July 15, 2016, 05:26:04 pm Hi, Could someone please attempt this question and tell me which answer they get? I got: $1/a(tan(sin^-1(x/a)))-1/a(sin^-1(x/a)) + C$ I was just wondering whether or not I got hopelessly lost somewhere... It's a super hard question. :-\ Thanks in advance! P.S. Sorry if the way I wrote my solution is confusing. Not that good with the LaTex stuff. Title: Re: 3U Maths Question Thread Post by: Happy Physics Land on July 15, 2016, 06:31:09 pm Hi, Could someone please attempt this question and tell me which answer they get? I got: $1/a(tan(sin^-1(x/a)))-1/a(sin^-1(x/a)) + C$ I was just wondering whether or not I got hopelessly lost somewhere... It's a super hard question. :-\ Thanks in advance! P.S. Sorry if the way I wrote my solution is confusing. Not that good with the LaTex stuff. Hey ATWalk! I did the question and I think our answers are out by a factor of 1/a. It was, after all, a very tedious question. I really could have been wrong. I will stick my solution here so that you can see any mistakes you have made. If you dont see anything wrong with yours, chances are I could have made a mistake. (http://i.imgur.com/KcWd6XX.jpg) Best Regards Happy Physics Land Title: Re: 3U Maths Question Thread Post by: conic curve on July 15, 2016, 06:52:16 pm Hey guys Just curious but how would you tackle 3D triginometry? Title: Re: 3U Maths Question Thread Post by: RuiAce on July 15, 2016, 07:06:31 pm Hey guys Just curious but how would you tackle 3D triginometry? Jamon gave some tips which will be on his lecture slides when they're released. One tip for 3D trig is to draw your 3D diagram, then draw your triangles separately. This reduces the ease of getting lost in your diagram Title: 3U Maths Question Thread Post by: RuiAce on July 15, 2016, 07:10:21 pm (http://uploads.tapatalk-cdn.com/20160715/4f1c105f12d0ea53e5fd02f4eaf123f1.jpg) I get what HPL got. The answers are most likely wrong. Edit: in fact so does WolframAlpha Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on July 15, 2016, 07:32:47 pm Hey guys Just curious but how would you tackle 3D triginometry? As well as Rui's tips, I mainly spoke about working backwards. I need this tangent ratio in my answer, okay, I need this side then, oh, that means I need to use this triangle. Etc, etc. Extension HSC 3D Trig questions will commonly lead you through the process of getting the answer, the trick is just spotting the right triangles to use that give you the correct answer. Always keep your eye on the answer you need to guide your working. I also leave 3D Trig questions till last, just to make sure they don't waste time I need for other questions ;D Title: Re: 3U Maths Question Thread Post by: amandali on July 16, 2016, 10:07:53 am (http://uploads.tapatalk-cdn.com/20160715/93dc3fb158b7ae7f1d4dca6284a5925c.jpg) how to do part iv ) Title: Re: 3U Maths Question Thread Post by: RuiAce on July 16, 2016, 01:05:08 pm (http://uploads.tapatalk-cdn.com/20160715/93dc3fb158b7ae7f1d4dca6284a5925c.jpg) how to do part iv ) Do you have a solution to this question, or the source of it? I requested a solution of this question and it used concepts WELL beyond that of the HSC. Title: Re: 3U Maths Question Thread Post by: jakesilove on July 16, 2016, 01:15:38 pm Do you have a solution to this question, or the source of it? I requested a solution of this question and it used concepts WELL beyond that of the HSC. Totally agree: I can prove that it equals e, but only using university level maths. I'm sure you can prove the relationship somehow using iii) but honestly it doesn't fall out as easily as I would have expected. Title: Re: 3U Maths Question Thread Post by: RuiAce on July 16, 2016, 01:20:15 pm Totally agree: I can prove that it equals e, but only using university level maths. I'm sure you can prove the relationship somehow using iii) but honestly it doesn't fall out as easily as I would have expected. The proof I got given demonstrated how by the monotone sequence theorem the binomial expansion as a sum is greater than 2 upon inspection. It also assumes n!/(n-k)! < nk.to prove the < 3 component. Title: Re: 3U Maths Question Thread Post by: jakesilove on July 16, 2016, 01:23:15 pm The proof I got given demonstrated how by the monotone sequence theorem the binomial expansion as a sum is greater than 2 upon inspection. It also assumes n!/(n-k)! < nk.to prove the < 3 component. Actually, I think that that's fine. Once you get to the Taylor serious of e, you can easily calculate that the number will be between 2 and 3 by summing up some terms. Maybe that's the best way? Title: Re: 3U Maths Question Thread Post by: RuiAce on July 16, 2016, 01:28:45 pm Actually, I think that that's fine. Once you get to the Taylor serious of e, you can easily calculate that the number will be between 2 and 3 by summing up some terms. Maybe that's the best way? Well I mean, in 3U we can't assume the Taylor series either. But I mean, we can't assume that the sequence will be monotone without proof right? Whilst we're here I want to learn them already. Slight motivation to want to go back to uni to learn them next sem. Title: Re: 3U Maths Question Thread Post by: zoe_rammie on July 16, 2016, 01:38:11 pm Hey Zoe, First of all, your description of a potential method of answering this question is absolutely gold. Keep at it, love your work. Your method, and the method in the answers, is actually the same. If you do what you were planning on doing, that's great, but you would have to equate coefficients as there isn't another way to compare the RHS to the LHS and prove that they are equal. So, you would equate coefficients, and get exactly the same answer as they did. They're answer probably isn't comprehensive enough anyway; if your method gets you the right relationship, you're absolutely doing the right thing. If you wanted a more comprehensive proof, though, let us know, however as far as I can tell your "do shit get marks" methodology is spot on. Jake So would this cut it? Title: Re: 3U Maths Question Thread Post by: jakesilove on July 16, 2016, 01:40:24 pm Well I mean, in 3U we can't assume the Taylor series either. But I mean, we can't assume that the sequence will be monotone without proof right? Whilst we're here I want to learn them already. Slight motivation to want to go back to uni to learn them next sem. Using the proof below, and letting x equal one, we can actually directly find the Taylor series (although obviously a 3U student won't know that; they'll just do the maths). Then, instead of recognising the final step, by typing in the first few terms into a calculator, a student could find that the limit is between 2 and 3! (http://i.imgur.com/pRiOkTQ.png?1) Title: Re: 3U Maths Question Thread Post by: jakesilove on July 16, 2016, 01:42:50 pm So would this cut it? I reckon so! I don't remember if the question asked you to "use" a specific part of the question prior to the proof, and if it did you may not get all the marks using your method. However, I think that your method is the quickest way to get to an answer. Title: Re: 3U Maths Question Thread Post by: RuiAce on July 16, 2016, 01:49:56 pm So would this cut it? I'm with Jake; it's fine. Nice Pokemon Go reference Title: Re: 3U Maths Question Thread Post by: conic curve on July 17, 2016, 10:12:10 am Can someone here please help me with this cos^-1[cos1050]. This is what I did: cos1050=cos330 cos^-1[cos-30] (answer can't be -30 because it is not in the restriction) Thanks Title: Re: 3U Maths Question Thread Post by: jakesilove on July 17, 2016, 10:46:17 am Can someone here please help me with this cos^-1[cos1050]. This is what I did: cos1050=cos330 cos^-1[cos-30] (answer can't be -30 because it is not in the restriction) Thanks You were pretty much there! Type $cos^{-1}({cos{-30}})$ into your calculator, and you'll see the answer is 30. This comes about due to the even/odd nature of the graph etc. but really, if you're ever struggling, see if your calculator can give you the answer! Title: Re: 3U Maths Question Thread Post by: jakesilove on July 17, 2016, 10:48:22 am Can someone here please help me with this cos^-1[cos1050]. This is what I did: cos1050=cos330 cos^-1[cos-30] (answer can't be -30 because it is not in the restriction) Thanks In fact, you could have done this right from the start. To do the maths properly, you needed to recognise that $cos{30}=cos{-30}$ Title: Re: 3U Maths Question Thread Post by: RuiAce on July 17, 2016, 10:55:07 am $\text{The way to do this without a calculator is covered better after you've been taught inverse trig in the HSC}\\ \text{Which, by then you would be using radians.}$ $\text{But the idea is }\cos^{-1}(\cos x)=x\text{ if and only if }0^\circ \le x \le 180^\circ$ $\text{So if you can somehow force your argument to be in that domain (notice that Jake used the fact cos is an even function)}\\ \text{Then you make your way out.}$ Title: Re: 3U Maths Question Thread Post by: conic curve on July 17, 2016, 11:03:48 am $\text{The way to do this without a calculator is covered better after you've been taught inverse trig in the HSC}\\ \text{Which, by then you would be using radians.}$ $\text{But the idea is }\cos^{-1}(\cos x)=x\text{ if and only if }0^\circ \le x \le 180^\circ$ $\text{So if you can somehow force your argument to be in that domain (notice that Jake used the fact cos is an even function)}\\ \text{Then you make your way out.}$ Thanks Jake and Rui How do I do this: if a=tan^-1 (-1/3) find cosa and sina The reason why I am confused is because you can't have negative dimensions Does this mean I'll need o draw it on a coordinate geometry graph? Title: Re: 3U Maths Question Thread Post by: RuiAce on July 17, 2016, 11:14:20 am Thanks Jake and Rui How do I do this: if a=tan^-1 (-1/3) find cosa and sina The reason why I am confused is because you can't have negative dimensions Does this mean I'll need o draw it on a coordinate geometry graph? $\text{Since }a=\tan^{-1}\frac{1}{3}\text{ instead of arbitrarily giving you }\tan a = \frac{1}{3}\\ \text{You know that it is in the first quadrant.}$ $\text{Now you have to draw up a right angled triangle.}$ $\text{Recall that the tangent ratio is just }\frac{\text{opposite side}}{\text{adjacent side}}\\ \therefore opp = 1, \quad adj = 3$ $\text{Using Pythagoras' Theorem you have }hyp=\sqrt{10}$ $\text{So }\sin a = \frac{1}{\sqrt{10}}, \quad \cos a = \frac{3}{\sqrt{10}}$ Title: Re: 3U Maths Question Thread Post by: vamshimadas on July 17, 2016, 01:46:03 pm Quick Question on Parametrics: Find the equation of the tangent to the parabola x=4t, y=2t^2 at the point where t=3. Thank you :) Title: Re: 3U Maths Question Thread Post by: jakesilove on July 17, 2016, 01:54:18 pm Quick Question on Parametrics: Find the equation of the tangent to the parabola x=4t, y=2t^2 at the point where t=3. Thank you :) You can do this a whole bunch of ways. I think it's easiest to find the parabola first, differentiate, and find the tangent the usual way etc. $x=4t$ $y=2t^2$ $y=2(\frac{x}{4})^2$ $y=\frac{x^2}{8}$ Then, we can find the tangent the normal way! First, you will need to x and y coordinates at t=3. We can sub this t value straight into the equations we are given to find them. $x=12$ $y=18$ I'll leave you to find the tangent from there: just use normal calculus! You can also use the chain rule method (ie find the first derivative of x with respect to t, y with respect to t, multiply them etc.). That may or may not be quicker, just depends on how you visualise the maths. Jake Title: Re: 3U Maths Question Thread Post by: jakesilove on July 17, 2016, 01:55:15 pm dx/dt= 4, dy/dt=4t dy/dx= dy/dt * dx/dt = 4t * 1/4 = t --> t=3, dy/dx=3 Thanks High Tide! That's the other method I alluded to at the end of my response; clearly quicker if you are more comfortable with this methodology. Title: Re: 3U Maths Question Thread Post by: RuiAce on July 17, 2016, 02:16:54 pm dx/dt= 4, dy/dt=4t dy/dx= dy/dt * dx/dt = 4t * 1/4 = t --> t=3, dy/dx=3 $\text{Your parametric differentiation should say }\frac{dy}{dx}=\frac{dy}{dt}\div \frac{dx}{dt}$ $\text{If it were multiply it'd be the chain rule: }\frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx}$ Thanks High Tide! That's the other method I alluded to at the end of my response; clearly quicker if you are more comfortable with this methodology. Could've just asked. It does have a name you know :P Title: Re: 3U Maths Question Thread Post by: conic curve on July 17, 2016, 09:16:06 pm Could someone here please help me with this question: I've seen the working out for this question but I'm struggling to understand the theory behind it. It would be lovely if someone could solve it for me and explain to me the theory behind it (because that's what I'm struggling with) Greatly appreciated Thanks ;D Title: Re: 3U Maths Question Thread Post by: RuiAce on July 17, 2016, 09:40:48 pm Could someone here please help me with this question: I've seen the working out for this question but I'm struggling to understand the theory behind it. It would be lovely if someone could solve it for me and explain to me the theory behind it (because that's what I'm struggling with) Greatly appreciated Thanks ;D \text{Proof 1: Proof by Pythagorean identity }\sin^2x+\cos^2x=1\\ \begin{align*}\cos LHS &= \cos (\sin^{-1}x)\\ &= \sqrt{1-\sin^2(\sin^{-1}x )}\\ &=\sqrt{1-x^2}\\ \cos RHS &= \cos (\cos^{-1} \sqrt{1-x^2})\\ &= \sqrt{1-x^2} \end{align*}\\ \therefore \cos LHS = \cos RHS\\ \therefore LHS = RHS\\ \text{taking inverse cosine of both sides as }cos^{-1}\text{ is continuous and monotone} $\text{Proof 2: Proof by ratio definition}\\ \text{Let }\alpha=\sin^{-1} x \implies \sin \alpha = \frac{x}{1}\\ \text{Draw a right-angled triangle with opposite side }x\text{ and hypotenuse }1\\ \text{Then, by Pythagoras' theorem, the adjacent side has length }\sqrt{1-x^2}\\ \therefore \cos \alpha = \frac{\sqrt{1-x^2}}{1} \implies \alpha = \cos^{-1}\sqrt{1-x^2}$ Title: Re: 3U Maths Question Thread Post by: conic curve on July 17, 2016, 09:50:46 pm \text{Proof 1: Proof by Pythagorean identity }\sin^2x+\cos^2x=1\\ \begin{align*}\cos LHS &= \cos (\sin^{-1}x)\\ &= \sqrt{1-\sin^2(\sin^{-1}x )}\\ &=\sqrt{1-x^2}\\ \cos RHS &= \cos (\cos^{-1} \sqrt{1-x^2})\\ &= \sqrt{1-x^2} \end{align*}\\ \therefore \cos LHS = \cos RHS\\ \therefore LHS = RHS\\ \text{taking inverse cosine of both sides as }cos^{-1}\text{ is continuous and monotone} $\text{Proof 2: Proof by ratio definition}\\ \text{Let }\alpha=\sin^{-1} x \implies \sin \alpha = \frac{x}{1}\\ \text{Draw a right-angled triangle with opposite side }x\text{ and hypotenuse }1\\ \text{Then, by Pythagoras' theorem, the adjacent side has length }\sqrt{1-x^2}\\ \therefore \cos \alpha = \frac{\sqrt{1-x^2}}{1} \implies \alpha = \cos^{-1}\sqrt{1-x^2}$ Um, did you take the restriction into account because that's what confused me the most about this question? Title: Re: 3U Maths Question Thread Post by: RuiAce on July 17, 2016, 10:04:32 pm Um, did you take the restriction into account because that's what confused me the most about this question? Mathematicians call these types of proofs "Proofs without loss of generality" This means that the proof most certainly lacks rigour in that some assumptions had to be made to arrive at the conclusion. However, the idea is that because the proof is so similar to the other cases, making such an assumption does not damage the final result. The Pythagorean identity assumes that both sine and cosine are positive simultaneously. The triangle proof produces an even greater flaw in that it assumes we will always be in the first quadrant. Title: Re: 3U Maths Question Thread Post by: amandali on July 19, 2016, 02:08:03 am (http://uploads.tapatalk-cdn.com/20160718/570133fec414d0cb8f9b4840f48dd1b9.jpg) how to do ques 4a) ans is 5:28pm to 9:12 pm Title: Re: 3U Maths Question Thread Post by: jakesilove on July 19, 2016, 10:10:04 am (http://uploads.tapatalk-cdn.com/20160718/570133fec414d0cb8f9b4840f48dd1b9.jpg) how to do ques 4a) ans is 5:28pm to 9:12 pm Hey! So we need to start out by recognising the general form of an equation like this. $x=Asin(nt+b)+c$ Now, we spend some time filling in all the blanks. Firstly, we know that A is the amplitude, which will just be half the distance from high tide to low tide. So, $x=3.2sin(nt+b)+c$ Now, we know that period is calculated by $\frac{2\pi}{n}=P$ so $\frac{2\pi}{n}=6.25*2$ Taking period in hours (and therefore time in hours). We solve this, and plug in what we have $x=3.2sin((\frac{2\pi}{12.5}t+b)+c$ Now, we can see that the center of motion is halfway between high and low tide. This is the value for c, so we can plug that in easily $x=3.2sin((\frac{2\pi}{12.5}t+b)+11.4$ To find b, we have to sub stuff in. It's up to you how you do this part; I think it's easiest to let t=0 at low tide (8.2m) and solve from there. You just need to remember that, at high tide, t doesn't equal 7:20pm, it will equal 6 hours of 15 minutes. $8.2=3.2sin((\frac{2\pi}{12.5}(0)+b)+11.4$ $-3.2=3.2sin(b)$ $b=-\frac{\pi}{2}$ So, the overall equation is $x=3.2sin((\frac{2\pi}{12.5}(t)-\frac{\pi}{2})+11.4$ From there, I'll leave it to you. All you need to do is set the height equal to 13.3 and find times t for which the relationship is true. Make sure to find two times. Those will be two 'hour' values, which you must add to the initial 1:05pm. Once you've done a bunch of questions like this, it becomes pretty easy, because you always answer it in the same way! Jake Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on July 19, 2016, 10:13:27 am (http://uploads.tapatalk-cdn.com/20160718/570133fec414d0cb8f9b4840f48dd1b9.jpg) how to do ques 4a) ans is 5:28pm to 9:12 pm Edit: Jake snuck in, but I did it differently, so I'll post this anyway. Both are correct! Hey amandali! This one is tough because we have to form the function for simple harmonic motion ourselves. Let's do that first, and there are multiple ways. The amplitude is given by halving the difference between low tide and high tide, that is, 3.2 metres. The equilibrium position is given by the halfway point between those, meaning, 11.4 metres. The period can be calculated by considering the fact that it there is 6 hours and 15 minutes (375 minutes) between low and high tide, therefore, the period of the harmonic motion is double that, 750 minutes. Therefore; $T = \frac{2\pi}{n} \\ \therefore n=\frac{2\pi}{T} = \frac{\pi}{375}$ We put all of that data together to get a function for height in terms of time, which as implied, is simple harmonic: $h = 11.4-3.2\cos{\frac{t\pi}{375}}$ For this version, t=0 at 1:05pm. Note that I use the cosine function so that it can start at the extrema instead of the equilibrium as required. There are multiple ways to form this function! This is just my way ;D So we need times when the height is greater than 13.3 metres, so, let's solve the equality first. $11.4-3.2\cos{\frac{t\pi}{375}}=13.3 \\\cos{\frac{t\pi}{375}} = -\frac{1.9}{3.2} \\\therefore \frac{t\pi}{375} = 2.206,4.706 \\\therefore t=263.4,486.6$ We find that those (remember t=0 is 1:05pm) correspond to the answers you gave ;D hope this helps!! The funny thing about this question is that there is no calculus, just testing how well you can form the simple harmonic motion expressions ;D Title: Re: 3U Maths Question Thread Post by: RuiAce on July 19, 2016, 10:13:29 am $\text{Amplitude: }\frac{14.6-8.2}{2}=3.2\\ \text{Equilibrium position: }x_0=8.2+3.2=11.4\\ \textbf{Half }\text{of the period: }7:20pm - 1:05pm = 375min\\ \therefore n=\frac{\pi}{375}$ $\text{For convenience, set }t=0\text{ to be at }1:05pm$ x=3.2 \cos \left(\frac{\pi}{375}t + \alpha \right)+11.4\\ \text{When }t=0, x=8.2\\ \begin{align*}\therefore 8.2&=3.2\cos \alpha + 11.4\\ -1 &= \cos \alpha\\ \alpha&=\pi\\ \\ \therefore x&=3.2 \cos \left(\frac{\pi}{375}t + \pi \right)+11.4\\ x&=-3.2 \cos \frac{\pi t}{375}+11.4\end{align*} \text{We need }x \ge 13.3\\ \begin{align*}-3.2 \cos \frac{\pi t}{375}+11.4 &\ge 13.3 \\ -3.2 \cos \frac{\pi t}{375} &\ge 1.9 \\ \cos \frac{\pi t}{375} &\le \frac{-1.9}{3.2}\end{align*} $\textbf{This last bit is hard, because we're trying to solve an inequality with a trigonometric ratio in it.}\\ \text{Advice: Sketch out a graph of }y=\cos x\text{ at this point and find the }y\text{-coordinate }-\frac{1.9}{3.2}=-0.59375\\ \text{Recall that time starts from }t=0\text{, so your sketch should be from }0\le x \le 2\pi$ $\text{From your graph, }\textbf{infer that:}\\ \text{When }y=-0.59375\\ x=\pi - \arccos 0.59375\\ \text{or} \\ x=\pi+\arccos 0.59375\\ \text{(Not surprising as we needed a second and third quadrant angle.)}$ Don't forget arccos just means cos-1 $\text{Reading off your graph, when }\cos x \le -0.59375\text{, the region is sandwiched between the }x\text{-values.}\\ \text{That is to say:}\\ \pi - \arccos 0.59375 \le x \le \pi + \arccos 0.59375$ $\therefore \pi - \arccos 0.59375 \le \frac{\pi t}{375} \le \pi + \arccos 0.59375\\ \frac{375( \pi - \arccos 0.59375)}{\pi} \le t \le \frac{375( \pi + \arccos 0.59375)}{\pi} \\ 263.382... \le t \le 486.6176...$ So, the overall equation is $x=3.2sin((\frac{2\pi}{12.5}(t)+\frac{\pi}{2})+11.4$ From there, I'll leave it to you. All you need to do is set the height equal to 13.3 and find times t for which the relationship is true. Make sure to find two times. Those will be two 'hour' values, which you must add to the initial 1:05pm. Once you've done a bunch of questions like this, it becomes pretty easy, because you always answer it in the same way! Jake I thought you disappeared so I started doing the question anyway :P it was too late for me to realise that wait this one is actually a toughie. Anyway I worked with minutes so it's slightly different. I also didn't get to check my working out fully yet Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on July 19, 2016, 10:15:29 am $\text{Amplitude: }\frac{14.6-8.2}{2}=3.2\\ \text{Equilibrium position: }x_0=8.2+3.2=11.4\\ \textbf{Half }\text{of the period: }7:20pm - 1:05pm = 375min\\ \therefore n=\frac{\pi}{375}$ $\text{For convenience, set }t=0\text{ to be at }1:05pm$ x=3.2 \cos \left(\frac{\pi}{375}t + \alpha \right)+11.4\\ \text{When }t=0, x=8.2\\ \begin{align*}\therefore 8.2&=3.2\cos \alpha + 11.4\\ -1 &= \cos \alpha\\ \alpha&=\pi\\ \\ \therefore x&=3.2 \cos \left(\frac{\pi}{375}t + \pi \right)+11.4\\ x&=-3.2 \cos \frac{\pi t}{375}+11.4\end{align*} \text{We need }x \ge 13.3\\ \begin{align*}-3.2 \cos \frac{\pi t}{375}+11.4 &\ge 13.3 \\ -3.2 \cos \frac{\pi t}{375} &\ge 1.9 \\ \cos \frac{\pi t}{375} &\le \frac{-1.9}{3.2}\end{align*} $\textbf{This last bit is hard, because we're trying to solve an inequality with a trigonometric ratio in it.}\\ \text{Advice: Sketch out a graph of }y=\cos x\text{ at this point and find the }y\text{-coordinate }-\frac{1.9}{3.2}=-0.59375\\ \text{Recall that time starts from }t=0\text{, so your sketch should be from }0\le x \le 2\pi$ $\text{From your graph, }\textbf{infer that:}\\ \text{When }y=-0.59375\\ x=\pi - \arccos 0.59375\\ \text{or} \\ x=\pi+\arccos 0.59375\\ \text{(Not surprising as we needed a second and third quadrant angle.)}$ Don't forget arccos just means cos-1 $\text{Reading off your graph, when }\cos x \le -0.59375\text{, the region is sandwiched between the }x\text{-values.}\\ \text{That is to say:}\\ \pi - \arccos 0.59375 \le x \le \pi + \arccos 0.59375$ $\therefore \pi - \arccos 0.59375 \le \frac{\pi t}{375} \le \pi + \arccos 0.59375\\ \frac{375( \pi - \arccos 0.59375)}{\pi} \le t \le \frac{375( \pi + \arccos 0.59375)}{\pi} \\ 263.382... \le t \le 486.6176...$I thought you disappeared so I started doing the question anyway :P it was too late for me to realise that wait this one is actually a toughie. Anyway I worked with minutes so it's slightly different. I also didn't get to check my working out fully yet Your working is fine. And we seriously need a system... Title: Re: 3U Maths Question Thread Post by: RuiAce on July 19, 2016, 10:16:58 am Your working is fine. And we seriously need a system... He messaged me about the qn haha I was intending to just step out of the way, but then it was as if he vanished into thin air Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on July 19, 2016, 10:21:04 am He messaged me about the qn haha I was intending to just step out of the way, but then it was as if he vanished into thin air I need to be included in these messages! 8) Title: Re: 3U Maths Question Thread Post by: amandali on July 19, 2016, 01:09:07 pm (http://uploads.tapatalk-cdn.com/20160718/65e4b1f299235eb64ebf2dfe16ecff82.jpg) i dont get ques 4ii) the ans is written below Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on July 19, 2016, 02:08:21 pm (http://uploads.tapatalk-cdn.com/20160718/65e4b1f299235eb64ebf2dfe16ecff82.jpg) i dont get ques 4ii) the ans is written below Hey again! This is a Bernoulli Trial, so we could consider the different ways it can happen using the Binomial distribution. However, we have to consider the fact that the number of trials changes based on the events to that point. That said, for me, it is easier to break it down into the 3 possibilities: 3 sets played, 4 sets played, or 5 sets played. 3 sets played is an easy one, Novak must win in straight sets: $P(\text{Novak-Straight Sets})=\left(\frac{2}{3}\right)^3 =\frac{8}{27}$ For the rest, we need to adjust our thinking. If they are playing until three sets are won, and we are considering Novak winning, then that means Novak must win the last game. So, for when we want to consider the probability of winning in 4 sets, we only need to consider a Bernoulli Trial with 3 sets, because we know the fourth will be won by Novak. $(n+a)^3=n^3+3n^2a+3na^2+a^3$ So we take the appropriate term from here, and multiply by 2/3 to consider the probability of Novak winning the 4th set. So it's binomial distribution for the first 3 sets, then just basic probability in the fourth, because we know Novak needs to win it: $P(\text{Novak in 4 sets})=3n^2a\times\frac{2}{3}=3\times\left(\frac{2}{3}\right)^2\times\frac{1}{3}\times\frac{2}{3}=\frac{8}{27}$ This corresponds to the second term of your answer (I think there is a dictation error in the solution, because it out by a multiple of 2/3. We do the same thing for winning in 5 sets; consider the binomial probability of Novak winning 2 sets out of 4 (since we don't know the order), and then multiplying by 2/3 to consider the probability of him winning in the 5th. This corresponds to the last term of your answer, and if you add everything together, you get the answer ;D I hope that helps!! ;D Title: Re: 3U Maths Question Thread Post by: RuiAce on July 19, 2016, 02:10:05 pm Can you please tell us where you got this question from? (http://uploads.tapatalk-cdn.com/20160715/93dc3fb158b7ae7f1d4dca6284a5925c.jpg) how to do part iv ) Title: Re: 3U Maths Question Thread Post by: amandali on July 19, 2016, 06:31:37 pm Can you please tell us where you got this question from? i got this from 2015 independent trial exam paper Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on July 19, 2016, 06:53:48 pm i got this from 2015 independent trial exam paper Based on that question, I'm very glad I didn't sit it ;) Title: Re: 3U Maths Question Thread Post by: amandali on July 19, 2016, 10:26:27 pm Hey! So we need to start out by recognising the general form of an equation like this. $x=Asin(nt+b)+c$ Now, we spend some time filling in all the blanks. Firstly, we know that A is the amplitude, which will just be half the distance from high tide to low tide. So, $x=3.2sin(nt+b)+c$ Now, we know that period is calculated by $\frac{2\pi}{n}=P$ so $\frac{2\pi}{n}=6.25*2$ Taking period in hours (and therefore time in hours). We solve this, and plug in what we have $x=3.2sin((\frac{2\pi}{12.5}t+b)+c$ Now, we can see that the center of motion is halfway between high and low tide. This is the value for c, so we can plug that in easily $x=3.2sin((\frac{2\pi}{12.5}t+b)+11.4$ To find b, we have to sub stuff in. It's up to you how you do this part; I think it's easiest to let t=0 at low tide (8.2m) and solve from there. You just need to remember that, at high tide, t doesn't equal 7:20pm, it will equal 6 hours of 15 minutes. $8.2=3.2sin((\frac{2\pi}{12.5}(0)+b)+11.4$ $-3.2=3.2sin(b)$ $b=-\frac{\pi}{2}$ So, the overall equation is $x=3.2sin((\frac{2\pi}{12.5}(t)-\frac{\pi}{2})+11.4$ From there, I'll leave it to you. All you need to do is set the height equal to 13.3 and find times t for which the relationship is true. Make sure to find two times. Those will be two 'hour' values, which you must add to the initial 1:05pm. Once you've done a bunch of questions like this, it becomes pretty easy, because you always answer it in the same way! Jake how do we know that it's a sin or cos function i assumed it starts at low tide so i put my function as x=-acos(nt+alpha)+13.3 but it didnt work out Title: Re: 3U Maths Question Thread Post by: RuiAce on July 19, 2016, 10:28:26 pm It really doesn't matter because $\sin \left(x-\frac{\pi}{2}\right)=-\cos x$ You have three sets of solutions to look at for that question so Also it does start at low tide ________________ Merged posts I think it is the 13.3 bit that was incorrect, the actual cosine function itself is fine! Where did the 13.3 come from for you? :D Ohh right wrong equilibrium Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on July 19, 2016, 10:33:16 pm how do we know that it's a sin or cos function i assumed it starts at low tide so i put my function as x=-acos(nt+alpha)+13.3 but it didnt work out I think it is the 13.3 bit that was incorrect, the actual cosine function itself is fine! Where did the 13.3 come from for you? :D Title: Re: 3U Maths Question Thread Post by: timothynguyenn22 on July 21, 2016, 11:01:41 am Hey guys this is my first time on here, I'm struggling with probability right now (mainly permutations and combinations). Could someone please help me with this question? The letters of the word PRINTER are arranged in a row. Find the probability that: (a) the word starts with the letter E, (b) the letters I and P are next to one another, (c) there are three letters between N and T, (d) there are at least three letters between N and T. Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on July 21, 2016, 11:16:00 am Hey guys this is my first time on here, I'm struggling with probability right now (mainly permutations and combinations). Could someone please help me with this question? The letters of the word PRINTER are arranged in a row. Find the probability that: (a) the word starts with the letter E, (b) the letters I and P are next to one another, (c) there are three letters between N and T, (d) there are at least three letters between N and T. EDIT: I didn't spot the duplicate R, that will change things substantially! Hey there Tim, welcome!! ;D super happy to have you around the forums ;D I can definitely help you out! Okay, so for all of these questions, we will need the total number of arrangements of the letters. The number of ways we can arrange unique objects in a line is just n!, where n is the number of objects. However, we have two duplicate letters, so we must divide that by 2! to compensate: $\text{Total Arrangements}=\frac{7!}{2!}$ Okay, for Part A, we just put the E down first, then arrange the rest of the letters in a line behind it. So, we are only arranging 6 elements, meaning, 6! is the number of ways we can do that. We put that over the the total number of arrangements to get the answer. We can ignore the 2! here, because it would be present on both the top and bottom, and would thus cancel. $P(\text{E at Front})=\frac{6!}{7!}=\frac{1}{7}$ For Part B (and this is a common approach), let's consider the P and I as a single letter grouped together (taped together, can't be pulled apart). So, we are now arranging 6 elements again, since the P is stuck to the I. However, we can THEN swap the P and I around, swap the order they are taped. This gives the calculation below, 6! multiplied by 2 for the swap. Again, we can ignore the 2! in this calculation: $P(\text{P and I Together})=\frac{6!\times2}{7!}=\frac{4}{7}$ Three letters between N and T requires a similar approach, but NOW we bring in permutations. First, we need to select the letters to put between the N and the T, and the order counts, so we use a permutation (forgive my terrible version of the permutation notation below). However!! The number of ways we can arrange these letters depends on whether we get 0, 1 or 2 R's (this is a nasty question). If we have 0 R's, there is only we are selecting 3 from the 3 non R letters: $\text{P}^3_3=6$ Then, we swap the N and T as before, then consider the ordering: $P_1=\frac{6\times2\times\frac{3!}{2!}}{\frac{7!}{2!}} = \frac{1}{70}$ If we have 1 R, then we have selected 2 non R letters to go with it. So, we use a combination to make those 2 selections, then multiply by 3! to arrange: $\text{C}^3_2\times3!=18$ We then multiply by two and arrange, no duplicates in the arrangement: $P_2=\frac{18\times2\times3!}{\frac{7!}{2!}} = \frac{6}{70}$ If we have two R's, we've picked one non R letter (3 ways to do this), then we arrange it with 2 R's. Similar to above, this means dividing the final result by 2! because we have two alike elements. $\text{C}^3_1\times\frac{3!}{2!}=9$ Then, we arrange, no duplicate elements in the final arrangement either: $P_3=\frac{9\times2\times3!}{\frac{7!}{2!}} =\frac{3}{70}$ And now, we add these to get the final probability (same answer as before, did I screw up?) $P(\text{3 Letters Between}) = \frac{10}{70}=\frac{1}{7}$ I'm going to leave the final part for you, but basically, we want at least 3, so you'll be applying the principles I used above for 4 and 5 letters, then adding together (there are other methods as well). - Find the number of ways we can arrange the letters between N and T - Double this result because we can swap N and T around - Arrange the grouping as a BIG LETTER using the factorial principle I hope this helps! Let me know if you need anything explained in a little more detail, I moved fast ;D I tried to take my working already and adapt it to the duplicate letter, I'm not 100% confident, if someone spots an error please correct me! Title: Re: 3U Maths Question Thread Post by: RuiAce on July 21, 2016, 01:33:28 pm Yeah I'm in agreement here. I'll type up my answer. $\text{Keeping the total arrangements, for part a)}\\ \text{Fix }E\text{ to be at the front. Then the other six letters still need arranging.}\\ \text{This is done in }\frac{6!}{2!}\text{ways.}\\ \text{Hence a): }\frac{\frac{6!}{2!}}{\frac{7!}{2!}}=\frac{1}{7}$ $\text{For part b), treat the }I\text{ and }P\text{ as one single letter first. Then we have six letters (with repeated R) to arrange.}\\ \text{This is also done in }\frac{6!}{2!}\text{ ways}\\ \text{But the }P\text{ and }I\text{ can go in any order. This is done in 2! ways.}\\ \text{Hence we have }\frac{\frac{6!}{2!} \times 2!}{\frac{7!}{2!}}=\frac{2}{7}$ $\text{For part c), we need to consider the favourable outcomes:}\\ N\, \_\, \_\, \_ \,T\, \_ \, \_\\ \_\, N\, \_\, \_\, \_\, T\, \_\\ \_\, \_\, N\, \_\, \_\, \_\, T\\ \text{The blanks can be filled in }\frac{5!}{2!}\text{ ways (never forget R is repeated).}\\ N\text{ and }T\text{ can have their orders swapped. Total arrangements }=2!\\ \text{So since there are 3 favourable cases:} \\ \frac{3 \times 2! \times \frac{5!}{2!}}{\frac{7!}{2!}}=\frac{1}{7}$ $\text{For part d), make sure you also consider when there are 4 or 5 letters between }N\text{ and }T\\ \text{You will find that when there are 4 letters between, there's only two favourable outcomes.}\\ \text{When there's 5 letters between, there's only one favourable outcome (N and T are the endpoints!)}$ An interesting to note: Even forgetting about the repetition of R, the final answers (probabilities) do not change. Note that there is no conditioning on R EXCEPT for the fact it's repeated in the original word. This means that 1/2! appears in both the numerator AND the denominator! All that has happened was that we limited the amount of total outcomes and favourable outcomes here. This leads to an interesting observation - the probability is not dependent on repetition of letters provided they aren't impacted in any other way whatsoever. Title: Re: 3U Maths Question Thread Post by: amandali on July 22, 2016, 08:25:41 pm how to do ii) Title: Re: 3U Maths Question Thread Post by: RuiAce on July 22, 2016, 08:40:26 pm how to do ii) I don't like this question $\textbf{2005 HSC Mathematics Extension 1}$ $\text{Maintain the fact that }PT=\sqrt{450^2+2000^2}=2050\\ \text{Adjust your analysis for part i) to get that }\tan \frac{\alpha}{2}=\frac{r}{PT} \iff r=2050 \tan \frac{\alpha}{2}\\ \text{If this step was not obvious, consider the top view of the triangle with point }P\text{ and length }2r\\ \text{(Basically, don't let }\alpha=0.1\text{ too early)}$ \begin{align*}\frac{dr}{dt}&=\frac{dr}{d\alpha}\,\frac{d\alpha}{dt}\\ &= 1025\sec^2\frac{\alpha}{2} \times 0.02\end{align*}\\ \text{Substitute }\alpha=0.1\text{ to obtain your answer} Title: Re: 3U Maths Question Thread Post by: conic curve on July 23, 2016, 09:13:00 pm \text{Proof 1: Proof by Pythagorean identity }\sin^2x+\cos^2x=1\\ \begin{align*}\cos LHS &= \cos (\sin^{-1}x)\\ &= \sqrt{1-\sin^2(\sin^{-1}x )}\\ &=\sqrt{1-x^2}\\ \cos RHS &= \cos (\cos^{-1} \sqrt{1-x^2})\\ &= \sqrt{1-x^2} \end{align*}\\ \therefore \cos LHS = \cos RHS\\ \therefore LHS = RHS\\ \text{taking inverse cosine of both sides as }cos^{-1}\text{ is continuous and monotone} $\text{Proof 2: Proof by ratio definition}\\ \text{Let }\alpha=\sin^{-1} x \implies \sin \alpha = \frac{x}{1}\\ \text{Draw a right-angled triangle with opposite side }x\text{ and hypotenuse }1\\ \text{Then, by Pythagoras' theorem, the adjacent side has length }\sqrt{1-x^2}\\ \therefore \cos \alpha = \frac{\sqrt{1-x^2}}{1} \implies \alpha = \cos^{-1}\sqrt{1-x^2}$ What if you did the triangle method for this question? Title: Re: 3U Maths Question Thread Post by: RuiAce on July 23, 2016, 09:14:18 pm What if you did the triangle method for this question? Oh, the ratio definition proof is the triangle proof. I just gave it a different name there. Title: Re: 3U Maths Question Thread Post by: conic curve on July 23, 2016, 09:21:56 pm -tan(alpha)=tan(beta) -Alpha=beta Is this possible since they both represent angles? Title: Re: 3U Maths Question Thread Post by: RuiAce on July 23, 2016, 09:24:44 pm -tan(alpha)=tan(beta) -Alpha=beta Is this possible since they both represent angles? $\text{There's a criteria that }-\frac{\pi}{2} < \alpha < \frac{\pi}{2}\\ \text{and }-\frac{\pi}{2} < \beta < \frac{\pi}{2}\text{ that needs to be met here.}$ $\text{Otherwise it's a bit flawed. }\tan^{-1}(\tan x)=x \text{ if and only if } -\frac{\pi}{2} < x < \frac{\pi}{2}$ $\text{If }\alpha\text{ and }\beta\text{ lie in that domain then yes that is correct.}$ $\text{Also keep in mind that tan being an odd function plays a role here.}\\ - \tan x = \tan (-x)$ Title: Re: 3U Maths Question Thread Post by: conic curve on July 23, 2016, 09:37:10 pm $\text{There's a criteria that }-\frac{\pi}{2} < \alpha < \frac{\pi}{2}\\ \text{and }-\frac{\pi}{2} < \beta < \frac{\pi}{2}\text{ that needs to be met here.}$ $\text{Otherwise it's a bit flawed. }\tan^{-1}(\tan x)=x \text{ if and only if } -\frac{\pi}{2} < x < \frac{\pi}{2}$ $\text{If }\alpha\text{ and }\beta\text{ lie in that domain then yes that is correct.}$ $\text{Also keep in mind that tan being an odd function plays a role here.}\\ - \tan x = \tan (-x)$ This is a completely different question I already solved the previous question This is the question now: Title: Re: 3U Maths Question Thread Post by: RuiAce on July 23, 2016, 09:42:55 pm Bit confused here. What do you mean by that expression on the LHS? $\sqrt{1-x^2}\tan^{-1}(-x)\\ \text{or }\tan^{-1}(-x\sqrt{1-x^2})$ Or was the square root not meant to be there Title: Re: 3U Maths Question Thread Post by: conic curve on July 23, 2016, 09:44:33 pm Bit confused here. What do you mean by that expression on the LHS? $\sqrt{1-x^2}\tan^{-1}(-x)\\ \text{or }\tan^{-1}(-x\sqrt{1-x^2})$ Or was the square root not meant to be there whoops my bad sorry Basically that question but without the squareroot value Thanks Title: Re: 3U Maths Question Thread Post by: RuiAce on July 23, 2016, 09:53:10 pm \text{This is a bit of a weird question but here's one way to tackle it.}\\ \text{Keeping in mind that for this question, forcibly }-\frac{\pi}{2} < \alpha < \frac{\pi}{2} \text{ and } -\frac{\pi}{2} < \beta < \frac{\pi}{2} \\ \text{So } -\pi < \alpha+\beta < \pi \\ \begin{align*}\tan(\alpha+\beta)&=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\\ &= \frac{-x+x}{1+x^2}\\\therefore \tan(\alpha+\beta)&=0\\ \therefore \alpha+\beta&=0 \\ \therefore \beta&=-\alpha\end{align*} \begin{align*}\tan^{-1}x&=\beta\\ \therefore \tan^{-1}x &= -\alpha\\ \alpha&=-\tan^{-1}x\\ \therefore \tan^{-1}(-x)=-\tan^{-1}x\end{align*} Notes: $\tan^{-1}(-x)=\alpha \iff -x=\tan \alpha\\ \tan^{-1}x=\beta \iff x=\tan \beta\\ \text{for the domain restrictions on }\alpha,\beta$ Title: Re: 3U Maths Question Thread Post by: conic curve on July 23, 2016, 09:56:51 pm \text{This is a bit of a weird question but here's one way to tackle it.}\\ \text{Keeping in mind that for this question, forcibly }-\frac{\pi}{2} < \alpha < \frac{\pi}{2} \text{ and } -\frac{\pi}{2} < \beta < \frac{\pi}{2} \\ \text{So } -\pi < \alpha+\beta < \pi \\ \begin{align*}\tan(\alpha+\beta)&=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\\ &= \frac{-x+x}{1+x^2}\\\therefore \tan(\alpha+\beta)&=0\\ \therefore \alpha+\beta&=0 \\ \therefore \beta&=-\alpha\end{align*} \begin{align*}\tan^{-1}x&=\beta\\ \therefore \tan^{-1}x &= -\alpha\\ \alpha&=-\tan^{-1}x\\ \therefore \tan^{-1}(-x)=-\tan^{-1}x\end{align*} Notes: $\tan^{-1}(-x)=\alpha \iff -x=\tan \alpha\\ \tan^{-1}x=\beta \iff x=\tan \beta\\ \text{for the domain restrictions on }\alpha,\beta$ Um, there was no restriction given on the question Whenever I have a question with a restriction, I usually post it up and in this case I didn't hence there is no restriction Next time I'll make it more clearer for you so then you don't do the wrong things Title: Re: 3U Maths Question Thread Post by: RuiAce on July 23, 2016, 10:04:24 pm $\text{I didn't do anything wrong.}\\ \text{Because we are given }\beta = \tan^{-1}x\\ \text{We automatically have }-\frac{\pi}{2} < \beta < \frac{\pi}{2}$ $\text{This is cause the range of the function }f(x)=\tan^{-1}x\\ \text{is }-\frac{\pi}{2} < y < \frac{\pi}{2}$ $\text{The pinpoint is that we were given it as }\beta = \tan^{-1}x\text{, not }\tan x =\beta\\ \text{The latter imposes no massive restriction, but the former imposes an important one} \textbf{ automatically}.$ Title: Re: 3U Maths Question Thread Post by: conic curve on July 23, 2016, 10:14:45 pm I saw a solution to a similar question to this and this was how I ended up doing it: Title: Re: 3U Maths Question Thread Post by: RuiAce on July 23, 2016, 10:19:46 pm Lol fair enough. But that solution lacks rigour in my eyes. Without proper justification that both alpha and beta lie between -pi/2 and pi/2, cancelling out the tan's is not properly justified. (Or it could be I'm too used to uni maths and everything has to be presented with 100% precision) Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on July 24, 2016, 12:24:18 am Lol fair enough. But that solution lacks rigour in my eyes. Without proper justification that both alpha and beta lie between -pi/2 and pi/2, cancelling out the tan's is not properly justified. (Or it could be I'm too used to uni maths and everything has to be presented with 100% precision) I'd be leaning towards saying that the solution would probably be passable at this level of study, but I could be mistaken. The train of thought is solid enough ;D that being said, I recommend getting in the habit of writing all restrictions in the HSC, even those which are seemingly inconsequential or even obvious, to make the marker happy and to make sure your reasoning is clear, catch tricks, etc etc ;D you'd almost definitely get away with it, but better safe than sorry :D Title: Re: 3U Maths Question Thread Post by: RuiAce on July 24, 2016, 01:26:47 am I'd be leaning towards saying that the solution would probably be passable at this level of study, but I could be mistaken. The train of thought is solid enough ;D that being said, I recommend getting in the habit of writing all restrictions in the HSC, even those which are seemingly inconsequential or even obvious, to make the marker happy and to make sure your reasoning is clear, catch tricks, etc etc ;D you'd almost definitely get away with it, but better safe than sorry :D It's as you said in the lecture I'm the type of person who will get you out for anything in maths :P Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on July 24, 2016, 01:50:13 am It's as you said in the lecture I'm the type of person who will get you out for anything in maths :P You keep me on my toes, that's for sure ;) Title: Re: 3U Maths Question Thread Post by: dtinaa on July 24, 2016, 02:48:53 pm Hey everyone, I found this hectic induction question but I couldn't find the solutions, if anyone can do it please reply!!! Also the image was too big to attach so I just put it onto a document... Title: Re: 3U Maths Question Thread Post by: RuiAce on July 24, 2016, 03:37:17 pm Hey everyone, I found this hectic induction question but I couldn't find the solutions, if anyone can do it please reply!!! Also the image was too big to attach so I just put it onto a document... $\text{The base case of }n=1\text{ has}\\ LHS=\frac{1}{1+1}=\frac{1}{2}\\ RHS=1-\frac{1}{2}=\frac{1}{2}\\ \text{So as }LHS=RHS\text{, our base case is true.}$ $\text{Making an assumption that the statement is true for }n=k:\\ \frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}+\dots+\frac{1}{2k}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{2k-1}-\frac{1}{2k}$ $\textbf{Why is our required to prove so hard to generate for the inductive step?}\\ \text{Because of how our terms go.}$ $\text{Observe how we must prove true for }n=k+1.\text{ What is the rule when }n=k+1?\\ \text{On the }LHS\text{notice that our first term when }n=k\text{ is }\frac{1}{k+1}\\ \text{This means that our first term when }n=k+1\textbf{ is actually }\frac{1}{k+2}\\ \text{Indeed, the last term also happens to be }\frac{1}{2(k+1)}=\frac{1}{2k+2}$ \text{On the }RHS\text{, notice how we increase by TWO terms for every value of }n:\\ \begin{align*}n=1 &\rightarrow 1-\frac{1}{2}\\ n=2 &\rightarrow 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\\ n=3 &\rightarrow 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\end{align*}\\ \text{Hence when }n=k+1\text{, instead we have }\\ 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{2k-1}-\frac{1}{2k}+\frac{1}{2k+1}-\frac{1}{2k+2} $RTP: \frac{1}{k+2}+\frac{1}{k+3}+\dots+\frac{1}{2k}+\frac{1}{2k+1}+\frac{1}{2k+2}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{2k-1}-\frac{1}{2k}+\frac{1}{2k+1}-\frac{1}{2k+2}$ \begin{align*}LHS&= \frac{1}{k+2}+\frac{1}{k+3}+\dots+\frac{1}{2k}+\frac{1}{2k+1}+\frac{1}{2k+2}\\ &= \left(\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}+\dots+\frac{1}{2k}\right)+\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{1}{k+1}\\ &=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{2k-1}-\frac{1}{2k}\right)+\frac{1}{2k+1}+\frac{1}{2(k+1)}-\frac{1}{k+1} \tag{by hypothesis}\\ &= 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{2k-1}-\frac{1}{2k}+\frac{1}{2k+1}-\frac{1}{2k+2}\\ &= RHS \end{align*} Hence true by induction Title: Re: 3U Maths Question Thread Post by: alexander.chu1 on July 25, 2016, 06:16:54 pm Hey guys this is my first time on here, I'm struggling with probability right now (mainly permutations and combinations). Could someone please help me with this question? The letters of the word PRINTER are arranged in a row. Find the probability that: (a) the word starts with the letter E, (b) the letters I and P are next to one another, (c) there are three letters between N and T, (d) there are at least three letters between N and T. Hi Tim, I think the answer are 1/7 , 2/7,1/7, and 2/7. Please correct me if i am wrong thanks! Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on July 25, 2016, 07:38:03 pm Hi Tim, I think the answer are 1/7 , 2/7,1/7, and 2/7. Please correct me if i am wrong thanks! Hey Alex, welcome to the forums!! ;D let me know if you need any help finding things! Those answers match Rui and I's, you are correct indeed! ;D Title: Re: 3U Maths Question Thread Post by: katherine123 on July 28, 2016, 10:36:02 am how to find domain of y= inverse sin(1-x^2) Title: Re: 3U Maths Question Thread Post by: RuiAce on July 28, 2016, 10:43:11 am how to find domain of y= inverse sin(1-x^2) $\text{The natural domain of }y=\sin^{-1}x\text{ is }-1 \le x \le 1\\ \text{or alternatively }|x|\le 1$ $\text{Effectively, whatever is enclosed inside the inverse sine (or we can call the 'argument' here)} \\ \text{must lie between }\pm 1$ $\text{Handling this is awkward. }-1 \le 1-x^2 \le 1 \text{ becomes }-2 \le -x^2 \le 0\\ \text{Which becomes }0 \le x^2 \le 2$ $x^2 \ge 0\text{ is a redundancy as this is obviously always true. }\\ x^2 \le 2\text{, however, becomes }-\sqrt{2} \le x \le \sqrt{2}\\ \textit{Do you see why?}$ $\text{So since we only have one condition, the final answer is }-\sqrt{2} \le x \le \sqrt{2}$ Title: Re: 3U Maths Question Thread Post by: conic curve on July 31, 2016, 09:35:45 am Sorry but I've got a mind blank right now How do I differentiate y=2sin^-1 squareroot of x Thanks Title: Re: 3U Maths Question Thread Post by: RuiAce on July 31, 2016, 09:48:25 am $\text{Just chain rule it}\\ \frac{d}{dx}2\sin^{-1}\sqrt{x} = \frac{1}{2\sqrt{x}}\times \frac{2}{\sqrt{1-(\sqrt{x})^2}} =\frac{1}{\sqrt{x}\sqrt{1-x}}$ Title: Re: 3U Maths Question Thread Post by: conic curve on July 31, 2016, 09:53:58 am $\text{Just chain rule it}\\ \frac{d}{dx}2\sin^{-1}\sqrt{x} = \frac{1}{2\sqrt{x}}\times \frac{2}{\sqrt{1-(\sqrt{x})^2}} =\frac{1}{\sqrt{x}\sqrt{1-x}}$ Thanks Do I chain rule everything with a coeffiient in front of the inverse trig? I tend to get confused with coefficients and differentating inverse trig Similar to how I think 2tantheta=tan2theta Title: Re: 3U Maths Question Thread Post by: RuiAce on July 31, 2016, 10:11:18 am Thanks Do I chain rule everything with a coeffiient in front of the inverse trig? I tend to get confused with coefficients and differentating inverse trig Similar to how I think 2tantheta=tan2theta What do you mean? I used the chain rule because of the stupid square root inside the inverse sine, not the coefficient Title: Re: 3U Maths Question Thread Post by: conic curve on July 31, 2016, 10:25:11 am What do you mean? I used the chain rule because of the stupid square root inside the inverse sine, not the coefficient So if you had y=2 squareroot of (sin^-1x) would you still have to use the chain rule? Title: Re: 3U Maths Question Thread Post by: jakesilove on July 31, 2016, 10:34:51 am So if you had y=2 squareroot of (sin^-1x) would you still have to use the chain rule? That's a totally different question. The general form for differentiating inverse sin is $y=asin^{-1}(f(x))$ $\frac{dy}{dx}=\frac{a*f'(x)}{(1-(f(x))^2)^{1/2}}$ Title: Re: 3U Maths Question Thread Post by: massive on July 31, 2016, 11:21:18 am $\text{The natural domain of }y=\sin^{-1}x\text{ is }-1 \le x \le 1\\ \text{or alternatively }|x|\le 1$ $\text{Effectively, whatever is enclosed inside the inverse sine (or we can call the 'argument' here)} \\ \text{must lie between }\pm 1$ $\text{Handling this is awkward. }-1 \le 1-x^2 \le 1 \text{ becomes }-2 \le -x^2 \le 0\\ \text{Which becomes }0 \le x^2 \le 2$ $x^2 \ge 0\text{ is a redundancy as this is obviously always true. }\\ x^2 \le 2\text{, however, becomes }-\sqrt{2} \le x \le \sqrt{2}\\ \textit{Do you see why?}$ $\text{So since we only have one condition, the final answer is }-\sqrt{2} \le x \le \sqrt{2}$ Could they ask us to sketch a graph of that function^ And how would you even sketch that graph?? Title: Re: 3U Maths Question Thread Post by: RuiAce on July 31, 2016, 11:25:14 am Could they ask us to sketch a graph of that function^ And how would you even sketch that graph?? In Extension 2, of course In Extension 1, highly unlikely. The domain can be found because all that you're taught in the 3U course is sufficient to find it. But the graph itself is something you would have to be guided through, i.e. stationary points etc. Title: Re: 3U Maths Question Thread Post by: conic curve on July 31, 2016, 06:27:41 pm For this question (that I have attached below) why do we take 3 outside of the integral rather than -3 (look at 2nd step) Title: Re: 3U Maths Question Thread Post by: jakesilove on July 31, 2016, 06:30:20 pm For this question (that I have attached below) why do we take 3 outside of the integral rather than -3 (look at 2nd step) The reason they take 3 instead of -3 is because integration to inverse cos requires a negative coefficient. (http://image.slidesharecdn.com/lesson17-inversetrigonometricfunctionsslides-090311131526-phpapp01/95/lesson-17-inverse-trigonometric-functions-41-728.jpg?cb=1236777506) They could just have easily taken out -3, and integrated to sin instead of cos. Hope that makes sense! Jake Title: Re: 3U Maths Question Thread Post by: RuiAce on July 31, 2016, 06:58:38 pm $\text{I will admit though, I'd have turned it into }-3 \arcsin \frac{x}{3}+C\text{ myself as well.}\\ \text{Unless it's simple harmonic motion. But that's the only exception.}$ Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on July 31, 2016, 07:29:11 pm The reason they take 3 instead of -3 is because integration to inverse cos requires a negative coefficient. They could just have easily taken out -3, and integrated to sin instead of cos. Hope that makes sense! Jake $\text{I will admit though, I'd have turned it into }-3 \arcsin \frac{x}{3}+C\text{ myself as well.}\\ \text{Unless it's simple harmonic motion. But that's the only exception.}$ For clarity conic, both answers are correct, just depends what you want to do with it ;D Title: Re: 3U Maths Question Thread Post by: conic curve on July 31, 2016, 08:48:36 pm For clarity conic, both answers are correct, just depends what you want to do with it ;D Thanks Jamon Anyways I am confused with the following inverse trig integration questions For the first Integral I attached below, why do we remove 1/2 out to the front? Ignore the second because I believe that this is a similar principle to the first For the third why do we use polynomial long division? Also could someone please help me with the second attached file and explain the theory behind this because I'm not too sure what the underlying theory behind this is (I mean I've seen the working to it but I don't understand what's going on) Thanks Title: Re: 3U Maths Question Thread Post by: RuiAce on July 31, 2016, 08:59:56 pm $\text{With that third one, the fundamental idea is that to integrate some rational function}\\ \text{We cannot have an improper fraction - the degree of the numerator must ALWAYS be less than that of the denominator.}\\ \text{Otherwise, there is no standard form available for us to compute that integral.}$ \text{This polynomial long division is easier to compute.}\\ \begin{align*}\frac{x^4+x^2+1}{x^2+1}&= \frac{x^4+x^2}{x^2+1}+\frac{1}{x^2+1}\\ &=\frac{x^2(x^2+1)}{x^2+1}+\frac{1}{x^2+1}\\ &= x^2+\frac{1}{x^2+1}\end{align*} $\text{So }\int \frac{x^4+x^2+1}{x^2+1}dx=\int \left(x^2+\frac{1}{x^2+1}\right)dx = \frac{x^3}{3}+\arctan x + C$ Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on July 31, 2016, 09:00:47 pm Thanks Jamon Anyways I am confused with the following inverse trig integration questions For the first Integral I attached below, why do we remove 1/2 out to the front? Ignore the second because I believe that this is a similar principle to the first For the third why do we use polynomial long division? Also could someone please help me with the second attached file and explain the theory behind this because I'm not too sure what the underlying theory behind this is (I mean I've seen the working to it but I don't understand what's going on) Thanks Hey! And the first one turns into one inverse trig integral and one logarithm integral with that factorisation applied. Breaking down the expression: $\frac{2+x}{1-x^2}=\frac{1}{2}\quad \frac{1+2x}{(1-x^2)} = \frac{1}{2}\left(\frac{1}{1-x^2}-\frac{2x}{1-x^2}\right)$ So that term on the left is an inverse trig integral, the one on the right is a log. I'll leave a formal answer to you, but the final answer you get should be (assuming I is the initial integral): $I =\frac{ \sin^{-1}{x}}{2}+ \ln{\sqrt{1-x^2}}$ You could adjust how that logarithm is expressed if you wish ;D Edit: Wait, I made a stupid assumption actually, that first one isn't an inverse trig integral. I fell into a pattern, apologies. Rui is right, that first one cannot be solved easily by 3U methods :o Title: Re: 3U Maths Question Thread Post by: RuiAce on July 31, 2016, 09:03:56 pm Thanks Jamon Anyways I am confused with the following inverse trig integration questions For the first Integral I attached below, why do we remove 1/2 out to the front? Ignore the second because I believe that this is a similar principle to the first For the third why do we use polynomial long division? Also could someone please help me with the second attached file and explain the theory behind this because I'm not too sure what the underlying theory behind this is (I mean I've seen the working to it but I don't understand what's going on) Thanks Wait woah woah woah woah woah. That Q1 is not a 3U question. $\text{Whilst }\int \frac{2}{1+x^2}dx \text{ easily becomes }2\arctan x + C\\ \int \frac{2}{1-x^2}dx\textbf{ cannot be computed using Extension 1 methods.}$ Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on July 31, 2016, 09:08:34 pm Wait woah woah woah woah woah. That Q1 is not a 3U question. $\text{Whilst }\int \frac{2}{1+x^2}dx \text{ easily becomes }2\arctan x + C\\ \int \frac{2}{1-x^2}dx\textbf{ cannot be computed using Extension 1 methods.}$ Too true, fell into my pattern there aha woops ;) you'll need partial fractions conic!! That's a 4U concept ;D Title: Re: 3U Maths Question Thread Post by: conic curve on July 31, 2016, 09:10:11 pm Too true, fell into my pattern there aha woops ;) you'll need partial fractions conic!! That's a 4U concept ;D This was the working to it (attached below) Title: Re: 3U Maths Question Thread Post by: RuiAce on July 31, 2016, 09:10:36 pm This was the working to it (attached below) Explain how your minus became a plus Title: Re: 3U Maths Question Thread Post by: conic curve on July 31, 2016, 09:13:50 pm Explain how your minus became a plus Whoops sorry I made an error. The original was supposed to be Integral of (2+x) over (1+x^2) ahaha sorry :-[ The thing is I don't get the underlying theory behind this and why we move 1/2 in front of the integral ??? Title: Re: 3U Maths Question Thread Post by: RuiAce on July 31, 2016, 09:16:38 pm That's what did it then. $\text{For the one that goes to tangent inverse it doesn't matter.}\\ \text{However the idea for the logarithm one is to use the trick }\int \frac{f^\prime(x)}{f(x)}dx = \ln [f(x)]+C$ $\frac{d}{dx}(1+x^2)=2x\\ \text{If you have }x\text{ but not }2x\text{, you're gonna have to cancel out the}2\text{ by multiplying by }\frac{1}{2}\\ \text{so that you don't change what you have.}$ Woop. Jamon beat me Title: Re: 3U Maths Question Thread Post by: conic curve on July 31, 2016, 09:21:04 pm $\text{With that third one, the fundamental idea is that to integrate some rational function}\\ \text{We cannot have an improper fraction - the degree of the numerator must ALWAYS be less than that of the denominator.}\\ \text{Otherwise, there is no standard form available for us to compute that integral.}$ \text{This polynomial long division is easier to compute.}\\ \begin{align*}\frac{x^4+x^2+1}{x^2+1}&= \frac{x^4+x^2}{x^2+1}+\frac{1}{x^2+1}\\ &=\frac{x^2(x^2+1)}{x^2+1}+\frac{1}{x^2+1}\\ &= x^2+\frac{1}{x^2+1}\end{align*} $\text{So }\int \frac{x^4+x^2+1}{x^2+1}dx=\int \left(x^2+\frac{1}{x^2+1}\right)dx = \frac{x^3}{3}+\arctan x + C$ Thanks for the help ;D Anyways the very moment I made a minor mistake, I scared the living daylights out of everyone ahaha Sorry about that guys :-[ :'( Title: Re: 3U Maths Question Thread Post by: massive on August 01, 2016, 12:33:42 am Hey guys, i just have a quick question; many application to the physical world questions have a question where they basically say "describe the movement of the particle". I was just wondering how you interpret the displacement, velocity and acceleration to describe its motion? Like if one is positive and another is negative, or if all are negative or all are positive, and combinations like that. Also what is a limiting position Thanks heaps!! Title: Re: 3U Maths Question Thread Post by: RuiAce on August 01, 2016, 07:49:43 am Hey guys, i just have a quick question; many application to the physical world questions have a question where they basically say "describe the movement of the particle". I was just wondering how you interpret the displacement, velocity and acceleration to describe its motion? Like if one is positive and another is negative, or if all are negative or all are positive, and combinations like that. Also what is a limiting position Thanks heaps!! You need to understand what motion actually is in the context of our course. When we talk about motion, we are talking about motion on a straight line. We define a point called the origin (at x=0) and the particle either moves to the left of it, or to the right of it. We define positive displacement as the particle being to the right of the origin, and negative displacement being to the left of the origin. The value of x tells us basically how much to the right/left it is. What do you mean by combinations? Just analyse each particle separately if you have two or more of them. Don't juggle multiple things at once without reason. You should well know that velocity describes the rate of change in the displacement. But because we defined positive as right and negative as left, if a velocity is positive it is moving towards the right. Conversely, if it is negative, the particle moves towards the left. v=0 is described as "at rest" because the particle doesn't move. You should also well know that acceleration describes the rate of change in the velocity. This tells us whether the particle is speeding up or not. If a=0 then the particle is moving at constant velocity (or constantly at rest, if a=0 and v=0 simultaneously). If a is positive, the particle is trying to accelerate to the right, whereas if a is 0, the particle is trying to accelerate to the left. E.g. x=0, v=2, a=-5: Particle at the origin, moving to the right at 2ms-1 however accelerating back towards the left at 5ms-2 When they ask about a limiting position, however, they are talking about what happens to the particle as time goes to infinity. If you got something like x=e-t+1, then clearly as t approaches infinity, e-t approaches 0. So the limiting position is x=1 You should have seen something similar with limiting velocity in the context of the 4U course Title: Re: 3U Maths Question Thread Post by: massive on August 01, 2016, 06:00:24 pm Hey guys, how do you do part ii and iii. Title: Re: 3U Maths Question Thread Post by: RuiAce on August 01, 2016, 08:20:46 pm Hey guys, how do you do part ii and iii. (http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI01082016_zpsizhyzxyf.jpg)(http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI01082016_0001_zpsdosyzk5b.jpg) Title: Re: 3U Maths Question Thread Post by: conic curve on August 01, 2016, 08:46:15 pm Am I wrong or are the answers wrong? Title: Re: 3U Maths Question Thread Post by: massive on August 01, 2016, 08:47:19 pm (http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI01082016_zpsizhyzxyf.jpg)(http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI01082016_0001_zpsdosyzk5b.jpg) Hey thanks so much for the reply. I was just wondering, is the time of flight 20 seconds for both particles. because the answers say that for particle A its 20s but for particle B its 10s. You think the answers are wrong? Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 01, 2016, 08:50:53 pm Am I wrong or are the answers wrong? You are correct, but the answer is correct too! $\frac{1}{\sqrt{1-9t^2}}=\frac{1}{3\sqrt{\frac{1}{9}-t^2}}=\frac{1}{3}\quad\frac{1}{\sqrt{\frac{1}{9}-t^2}}$ Using a=1/3 in your formula, the answer comes out! That's where the 1/3 out the front comes from ;D Title: Re: 3U Maths Question Thread Post by: conic curve on August 01, 2016, 08:57:22 pm You are correct, but the answer is correct too! $\frac{1}{\sqrt{1-9t^2}}=\frac{1}{3\sqrt{\frac{1}{9}-t^2}}=\frac{1}{3}\quad\frac{1}{\sqrt{\frac{1}{9}-t^2}}$ Using a=1/3 in your formula, the answer comes out! That's where the 1/3 out the front comes from ;D I am still confused ??? Title: Re: 3U Maths Question Thread Post by: RuiAce on August 01, 2016, 09:16:24 pm Hey thanks so much for the reply. I was just wondering, is the time of flight 20 seconds for both particles. because the answers say that for particle A its 20s but for particle B its 10s. You think the answers are wrong? It would be 10s for particle B simply because it started ten seconds later. I am still confused ??? Don't see your point. $\text{It can be shown using that rule, that }\int \frac{dx}{\sqrt{a^2-b^2x^2}}=\frac{1}{b}\sin^{-1}\left(\frac{bx}{a}\right)+C$ _______ \text{What Jamon did, which utilises the special case }\int \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\frac{x}{a}+C\text{ was this.}\\ \begin{align*}\int \frac{dt}{\sqrt{1-9t^2}}&= \int \frac{dt}{\sqrt{9\left(\frac{1}{9}-t^2\right)}}\\ &= \int \frac{dt}{3\sqrt{\frac{1}{9}-t^2}}\\ &= \frac{1}{3}\int \frac{dt}{\sqrt{\left(\frac{1}{3}\right)^2-x^2}}\\ &= \frac{1}{3} \sin^{-1} \frac{x}{\frac{1}{3}}+C\\ &= \frac{1}{3}\sin^{-1}3x+C\end{align*} Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 01, 2016, 09:23:13 pm I am still confused ??? That's alright! Let me do the integral for you. Do you understand the algebra I applied above? Basically when we factorise out the 9 from the square root symbol, moving it outside that symbol means we must instead write 3 (the square root of 9). Let me know if that's the issue for you and I'll do the full working. Assuming it's the integral itself: $\int\frac{dt}{\sqrt{1-9t^2}}=\frac{1}{3}\int\frac{dt}{\sqrt{\frac{1}{9}-t^2}}\\=\frac{1}{3}\sin^{-1}{\frac{x}{\frac{1}{3}}} \quad\text{Using the rule you quoted, with }a=\frac{1}{3}\\=\frac{1}{3}\sin^{-1}{3x}$ Does that help? What specifically is troubling you? Title: Re: 3U Maths Question Thread Post by: conic curve on August 01, 2016, 09:58:18 pm That's alright! Let me do the integral for you. Do you understand the algebra I applied above? Basically when we factorise out the 9 from the square root symbol, moving it outside that symbol means we must instead write 3 (the square root of 9). Let me know if that's the issue for you and I'll do the full working. Assuming it's the integral itself: $\int\frac{dt}{\sqrt{1-9t^2}}=\frac{1}{3}\int\frac{dt}{\sqrt{\frac{1}{9}-t^2}}\\=\frac{1}{3}\sin^{-1}{\frac{x}{\frac{1}{3}}} \quad\text{Using the rule you quoted, with }a=\frac{1}{3}\\=\frac{1}{3}\sin^{-1}{3x}$ Does that help? What specifically is troubling you? Yeah I get it now. Thanks Ruiace and Jamon Title: Re: 3U Maths Question Thread Post by: massive on August 01, 2016, 10:19:58 pm hey guys for part b, the answers say that the max value occurs when t=1, how the heck do you figure that out?? Title: Re: 3U Maths Question Thread Post by: RuiAce on August 01, 2016, 10:23:27 pm hey guys for part b, the answers say that the max value occurs when t=1, how the heck do you figure that out?? \text{By using the conventional next derivative equals 0 method to maximise.}\\ \text{Since }Q(0)=10\\ \begin{align*}A(1+0)e^0&=10\\ A=10\\ \end{align*} $\text{So }Q(t)=10(1+t)e^{-0.5t}\\ Q^\prime(t) = 5(1-t)e^{-0.5t}\text{ by product rule and rearranging.}\\ \text{The maximum for }Q(t)\text{ occurs when }Q^\prime(t)=0\\ \text{Solving gives }t=1$ Title: Re: 3U Maths Question Thread Post by: massive on August 01, 2016, 10:44:48 pm \text{By using the conventional next derivative equals 0 method to maximise.}\\ \text{Since }Q(0)=10\\ \begin{align*}A(1+0)e^0&=10\\ A=10\\ \end{align*} $\text{So }Q(t)=10(1+t)e^{-0.5t}\\ Q^\prime(t) = 5(1-t)e^{-0.5t}\text{ by product rule and rearranging.}\\ \text{The maximum for }Q(t)\text{ occurs when }Q^\prime(t)=0\\ \text{Solving gives }t=1$ lmaoo whatta noob q Title: Re: 3U Maths Question Thread Post by: massive on August 02, 2016, 11:07:17 am how do you show part ii?? thnx! Title: Re: 3U Maths Question Thread Post by: jakesilove on August 02, 2016, 12:06:58 pm how do you show part ii?? thnx! Hey! Without doing the actual maths for this question, I imagine that you have to prove 2 things. a) The particle is initially travelling away from the Origin (at t=0, velocity is positive) b) The velocity of the particle is always positive (for any t, velocity is greater than or equal to zero) This means that the particle will never 'turn around' and head back towards the origin, so it will never hit the Origin! Jake Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 02, 2016, 12:10:57 pm how do you show part ii?? thnx! Hey! So I'll expand on Jake's comment and perhaps tackle it a little differently, a little intuitive: $a = \frac{d^2x}{dt^2} = 2x-3 \\\therefore a>0 \quad \forall x>\frac{3}{2}$ Since the particle starts from rest at x=4, acceleration will always be acting in the positive direction (since there is no negative velocity at the beginning, there can't be later). Therefore, the particle will never return to the origin ;D Title: Re: 3U Maths Question Thread Post by: RuiAce on August 02, 2016, 12:11:07 pm how do you show part ii?? thnx! \text{The particle definitely does not pass through the origin because if }x=0\\ \begin{align*}V^2&=2(0-0-4)\\&=-8\\ &<0\end{align*}\\ \text{But how can the square of the velocity be negative?} $\textbf{Shortcuts}$ Title: Re: 3U Maths Question Thread Post by: humble mango on August 02, 2016, 01:00:12 pm Hi again! Thanks for the help last time :) , um, I'm having an issue with permutations and combinations. I don't really get the some number questions and "alternating" questions at all and was wondering if you guys could help me again. The numbers 1,2,3,4,5,6,7,8,9, are arranged in a straight line. Find the number of possible arrangements if: i) All the odd numbers occur next to each other. ii) All the odd numbers occur in increasing order, reading from left to right. 10 people (5 boys and 5 girls) have a birthday party. In how many ways can they be seated if the birthday boy is to be seated between two particular girls. Sorry for the trouble. Title: Re: 3U Maths Question Thread Post by: RuiAce on August 02, 2016, 01:26:48 pm Hi again! Thanks for the help last time :) , um, I'm having an issue with permutations and combinations. I don't really get the some number questions and "alternating" questions at all and was wondering if you guys could help me again. The numbers 1,2,3,4,5,6,7,8,9, are arranged in a straight line. Find the number of possible arrangements if: i) All the odd numbers occur next to each other. ii) All the odd numbers occur in increasing order, reading from left to right. 10 people (5 boys and 5 girls) have a birthday party. In how many ways can they be seated if the birthday boy is to be seated between two particular girls. Sorry for the trouble. $\text{If all the odd numbers are next to each other, then arrange them as one entity, say }'O'\\ \text{Effectively, we want to arrange }2,4,6,8,O\text{ which can be done in }5!\text{ ways}\\ \text{Because }O\text{ contains }1,3,5,7,9\text{, there are }5!\text{ ways of arranging the odd numbers.}\\ \text{So combining we have }5!5! = 14400$ $\text{The next question is }\textbf{appreciably}\text{ harder and I cannot guarantee my answer is accurate.}\\ \text{Notice how the odd numbers are forcibly in the order }1,3,5,7,9\text{. This is deadlocked.}\\ \text{But the even numbers are freed up. They can go before the 1, after the 9 or between any two odd numbers.}\\ \text{There are thus }6\text{ possible positions for any of the }4\text{ even numbers to be at.}\\ \text{Thus we have a total of }4^6\text{ favourable outcomes.}$ The reason I doubt the accuracy of this answer is because I feel it's invalid - I don't think I've catered for the ordering of the even numbers. _____________________ $\text{Because your question says that the boy is to be seated between two 'particular' girls}\\ \text{I am assuming that there can only be two girls to choose from. Not any two of the 10 girls.}\\ \text{Now, firstly we arrange the boy and two girls.}\\ \text{Because the boy is forcibly between them, there's only }2!\text{ ways this can be done.}$ $\text{We now treat the trio as a single entity. So we are basically arranging 8 people in a line.}\\ \text{This can be done in }8!\text{ ways}\\ \text{So the total arrangements is }2!8!$ $\text{If we had to choose the girls, however, we choose two out of 5 so that's done in }\binom{5}{2}\text{ ways}\\ \text{Our final answer would then be }\binom{5}{2}2!8!$ Title: Re: 3U Maths Question Thread Post by: massive on August 02, 2016, 01:38:15 pm How do you do part iii, iv, and v ??? Title: Re: 3U Maths Question Thread Post by: RuiAce on August 02, 2016, 01:43:52 pm This is another attempt at Q2. $\text{We keep the idea that the odd numbers are fixed.}\\ \text{Now, to cater for the order of the even numbers, we firstly take care of the order.}\\ \text{There are 4 even numbers. So we can arrange them in }4!\text{ ways as for the order.}$ $\text{As for where we position the odd and even numbers though, we now treat odd and even numbers}\\ \text{as the same thing. I.e. we are trying to arrange }O,O,O,O,O,E,E,E,E\\ \text{This can be done in }\frac{9!}{4!5!}\text{ ways}$ $\text{Giving us a final answer of }\frac{9!}{4!5!}\times 4!=\frac{9!}{5!}$ Edit: Posts merged How do you do part iii, iv, and v ??? I've noticed that a lot of your questions are projectile motion based. Is there anything about the topic that troubles you? Most of the questions just require knowing what means what (e.g. time of flight means y=0) and braving heavy algebra. Title: Re: 3U Maths Question Thread Post by: massive on August 02, 2016, 02:10:29 pm I've noticed that a lot of your questions are projectile motion based. Is there anything about the topic that troubles you? Most of the questions just require knowing what means what (e.g. time of flight means y=0) and braving heavy algebra. No I understand the main concepts. its just that rn im doing application of calculus to the physical world questions, and the ones im posting (like the one i just posted) are not the ordinary kind. For example, in the one i posted, i got the first two parts that were about projectile motion, but the last three parts are about the target moving, i just dont know how to do that Title: Re: 3U Maths Question Thread Post by: dreamdog10 on August 02, 2016, 02:26:44 pm Hi! So i was wondering when using Newton's method of approximation, whether i need my calculator in radians or degrees such as in the question attached :) Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 02, 2016, 02:29:52 pm Hi! So i was wondering when using Newton's method of approximation, whether i need my calculator in radians or degrees such as in the question attached :) Hey dream dog, radians!! Always radians unless you see the degree symbol (and you'll never see it when doing things with Calculus) ;D Title: Re: 3U Maths Question Thread Post by: jakesilove on August 02, 2016, 02:30:32 pm Hi! So i was wondering when using Newton's method of approximation, whether i need my calculator in radians or degrees such as in the question attached :) Hey! Unless you're working with angles which are specifically in degrees, you should ALWAYS be using radians. Title: Re: 3U Maths Question Thread Post by: dreamdog10 on August 02, 2016, 02:34:57 pm Hey! Unless you're working with angles which are specifically in degrees, you should ALWAYS be using radians. Hey dream dog, radians!! Always radians unless you see the degree symbol (and you'll never see it when doing things with Calculus) ;D Brilliant :) Title: Re: 3U Maths Question Thread Post by: humble mango on August 02, 2016, 04:58:39 pm This is another attempt at Q2. $\text{We keep the idea that the odd numbers are fixed.}\\ \text{Now, to cater for the order of the even numbers, we firstly take care of the order.}\\ \text{There are 4 even numbers. So we can arrange them in }4!\text{ ways as for the order.}$ $\text{As for where we position the odd and even numbers though, we now treat odd and even numbers}\\ \text{as the same thing. I.e. we are trying to arrange }O,O,O,O,O,E,E,E,E\\ \text{This can be done in }\frac{9!}{4!5!}\text{ ways}$ $\text{Giving us a final answer of }\frac{9!}{4!5!}\times 4!=\frac{9!}{5!}$ Thank you so much!! Our teacher has given us practice sheets to do, but didnt really print out the answers or provide a walkthrough on how to complete them ( which i think weakens the point of doing them in the first place :P) Again, thank you. Title: Re: 3U Maths Question Thread Post by: RuiAce on August 02, 2016, 06:28:07 pm How do you do part iii, iv, and v ??? $\text{Anyway, I was unable to offer a solution because I had classes from 2pm to 6pm}$ \text{The fact that the target is moving does not matter. You just need to analyse the moving thing as if it were a projectile.}\\ \text{For the target}\\ \begin{align*}\ddot{x}&=0\\ \dot{x}&=u\\ x&=ut+d\end{align*} \text{Because the particle hits the target WHEN }t=\frac{2V\sin \alpha}{g}\\ \text{AT }x=\frac{2V^2\cos\alpha\sin\alpha}{g}\\ \begin{align*} \frac{2V^2\cos\alpha\sin\alpha}{g}&=\frac{2uV\sin\alpha}{g}+d\\ u&=V\cos \alpha-\frac{gd}{2V\sin \alpha}\\ \end{align*}\\ \text{as required.} $\text{For the remainder of the question we immediately substitute our given information:}\\ u=V\cos\alpha - \frac{V}{4\sqrt{3}\sin\alpha}$ \text{Now, if }u > \frac{V}{\sqrt{3}} \\ \begin{align*}V\cos \alpha - \frac{V}{4\sqrt{3}\sin \alpha}&> \frac{V}{\sqrt{3}}\\ V\left(\cos \alpha - \frac{1}{4\sqrt{3}\sin \alpha}-\frac{1}{\sqrt{3}}\right)&>0\end{align*} \text{Obviously }V>0\text{ so we just need }\\ \begin{align*} \cos \alpha - \frac{1}{4\sqrt{3}\sin \alpha}-\frac{1}{\sqrt{3}} &>0 \\ 4\sqrt{3}\cos \alpha\sin \alpha -1 - 4\sin \alpha &> 0 \end{align*} $\text{What's peculiar is that we have to show that somehow, NO values of }\alpha\text{ will satisfy this equation.}\\ 2\sqrt{3}\sin 2\alpha - 1 - 4\sin \alpha > 0\\ \text{I have verified that this result is true via a graph on DESMOS however as I need to get home first...}\\ \text{...I will post this here for now and continue with the question later.}$ Edit: Let's continue with this question. $\text{This is a method I wanted to avoid but I couldn't. So let's use it.}\\ \text{Let }f(\alpha)=4\sqrt{3}\cos \alpha\sin \alpha -1 - 4\sin \alpha\\ \text{ over the domain }0< \alpha < \frac{\pi}{2}\\ \text{Then }f^\prime(\alpha)=4\sqrt{3}\cos 2\alpha - 4 \cos \alpha$ \text{To find any local maxima/minima let }f^\prime(\alpha)=0\\ \begin{align*}4\sqrt{3}(2\cos^2\alpha-1)-4\cos \alpha&=0 \\ 2\sqrt{3}\cos^2\alpha - \cos \alpha - \sqrt{3}&=0\\ \text{By quadratic }&\text{formula}\\ \cos \alpha&=\frac{\sqrt{3}}{2},-\frac{1}{\sqrt{3}}\end{align*}\\ \text{But because of the domain of }\alpha\text{ our only solution is }\alpha=\frac{\pi}{3} \text{Further note that }\\ \begin{align*}f^{\prime\prime}(x)&=-8\sqrt{3}\sin 2\alpha+4\sin \alpha \\&= -16\sqrt{3}\sin \alpha \cos \alpha + 4\sin \alpha\\ &=4\sin \alpha(-4\sqrt{3}\cos \alpha+1)\\ &<0\end{align*}\\ \text{regardless of what values of }\alpha\text{ we have.}\\ \text{So since }f(\alpha)\text{ is always concave down, that stationary point describes the global maxima.} $f\left(\frac{\pi}{3}\right)=0\\ \text{So plugging everything back in we have }... < 0$ $\textit{Note how when }\theta=\frac{\pi}{3}\text{ we have }u=\frac{V}{3}<\frac{V}{\sqrt{3}}\text{ which is not ok}$ Blech what a long post. By the way where do you get these questions - I feel as though most of them are noticeably harder than the current HSC. $\text{For part (v), well if it hits when it's two distances from }O\text{ then we are saying }\\ 2d=\frac{2V^2\sin \alpha \cos \alpha}{g}\iff gd=V^2\sin \alpha \cos \alpha$ \begin{align*}\therefore V^2\sin \alpha \cos \alpha &= \frac{V^2}{2\sqrt{3}}\\ \sin 2\alpha &= \frac{1}{4\sqrt{3}}\\ \alpha &= \frac{1}{2}\sin^{-1}\frac{1}{4\sqrt{3}}\end{align*} \text{When this happens, from the relation in part (iii)}\\ \begin{align*}\end{align*}u=\left(\cos \left(\frac{1}{2}\sin^{-1}\frac{1}{4\sqrt{3}}\right)-\frac{1}{4\sqrt{3}\sin \left(\frac{1}{2}\sin^{-1}\frac{1}{4\sqrt{3}}\right)}\right)\\ \text{which should satisfy the condition }u<\frac{V}{\sqrt{3}}. \text{I don't know anymore hurry up and help me Jamon/Jake} Title: Re: 3U Maths Question Thread Post by: massive on August 02, 2016, 08:14:27 pm Hey how do you do part iii, that's the only part i don't get. (btw i'm sorry if im annoying you with all these projectile questions, i just have a few that I dont get :-\ ) Title: Re: 3U Maths Question Thread Post by: RuiAce on August 02, 2016, 08:45:26 pm Hey how do you do part iii, that's the only part i don't get. (btw i'm sorry if im annoying you with all these projectile questions, i just have a few that I dont get :-\ ) $\text{Ok this time I'm going to use a method that is extremely peculiar and unprecedented.}\\ \text{This was me trying my luck with quadratics. Part (ii) was done using the discriminant}\\ \text{So I tried approaching part (iii) with quadratics as well.}$ $\text{Note how }\theta_1<\frac{\pi}{4}\text{ and }\theta_2<\frac{\pi}{4}\text{ simultaneously, supposedly.}\\ \text{Because tan is monotonic increasing for }0\le x<\frac{\pi}{2}\\ \text{we have }\tan \theta_1<1\text{ and }\tan\theta_2<1\\ \text{Thus, multiplying and preserving the inequality sign, }\tan \theta_1 \tan \theta_2 < 1$ $\text{Rearranging the Cartesian equation of motion:}\\ x^2\tan^2\theta - 4hx\tan \theta + (x^2+4hy)=0\\ \text{Taking the }\textbf{product of roots}\\ \frac{x^2+4hy}{x^2}=\tan \theta_1 \tan \theta_2\\ \therefore 1+\frac{4hy}{x^2} < 1\\ \therefore \frac{4hy}{x^2}<0 \implies y < 0\\ \text{So since }y\text{ has to be negative (lol) no point above the }x\text{-axis can be hit.}$ I haven't forgetten your other question yet. I will get to it shortly. Title: Re: 3U Maths Question Thread Post by: conic curve on August 02, 2016, 09:09:14 pm Am I doing something wrong? Title: Re: 3U Maths Question Thread Post by: RuiAce on August 02, 2016, 09:21:12 pm Am I doing something wrong? From line 3 to line 4 do not take out the negative and complete the square in the form A - (Bt+C)2 Line 4 to line 5 made no sense. How did -3 and 6 become -1/2 and 1 Title: Re: 3U Maths Question Thread Post by: conic curve on August 02, 2016, 09:23:03 pm From line 3 to line 4 do not take out the negative and complete the square in the form A - (t+C)2 Line 4 to line 5 made no sense. How did -3 and 6 become -1/2 and 1 I factorised the whole thing by 6 Title: Re: 3U Maths Question Thread Post by: RuiAce on August 02, 2016, 09:23:57 pm I factorised the whole thing by 6 If you factorised the whole thing by 6 that should be 18 there not 1/2. Anyway, regardless of if you factorise the 6 out, put the negative back in there and complete the square. \text{So this is how I would do that question}\\ \begin{align*}\int \frac{dt}{\sqrt{3-12t-18t^2}}&=\frac{1}{\sqrt{18}}\int \frac{dt}{\sqrt{\frac{1}{6}-\frac{2}{3}t-t^2}}\\ &= \frac{1}{\sqrt{18}}\int \frac{dt}{\sqrt{\frac{5}{18}-\left(t+\frac{1}{3}\right)^2}}\end{align*}\\ \text{which is now your ordinary sine inverse integral.} Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 02, 2016, 11:48:05 pm $\text{For part (v), well if it hits when it's two distances from }O\text{ then we are saying }\\ 2d=\frac{2V^2\sin \alpha \cos \alpha}{g}\iff gd=V^2\sin \alpha \cos \alpha$ \begin{align*}\therefore V^2\sin \alpha \cos \alpha &= \frac{V^2}{2\sqrt{3}}\\ \sin 2\alpha &= \frac{1}{4\sqrt{3}}\\ \alpha &= \frac{1}{2}\sin^{-1}\frac{1}{4\sqrt{3}}\end{align*} \text{When this happens, from the relation in part (iii)}\\ \begin{align*}\end{align*}u=\left(\cos \left(\frac{1}{2}\sin^{-1}\frac{1}{4\sqrt{3}}\right)-\frac{1}{4\sqrt{3}\sin \left(\frac{1}{2}\sin^{-1}\frac{1}{4\sqrt{3}}\right)}\right)\\ \text{which should satisfy the condition }u<\frac{V}{\sqrt{3}}. \text{I don't know anymore hurry up and help me Jamon/Jake} I think I'd swing at Part (v) a little differently, I interpret the question differently. When it says precisely two distances from O, doesn't that mean it has two solutions? The wording is really weird, but I don't think it relates to the distance d given earlier in the question, nothing suggests that. So my interpretation would be that we just need to prove that, provided that the condition on U holds, then there are two values of d which satisfy the other conditions given. This would make sense, provided the target is moving slow enough, it can hit it on the way up (one distance), or on the way down (another distance). At least I think. I just don't think it relates to the distance d. What do you reckon Rui? ::) Title: Re: 3U Maths Question Thread Post by: RuiAce on August 03, 2016, 08:24:29 am I think I'd swing at Part (v) a little differently, I interpret the question differently. When it says precisely two distances from O, doesn't that mean it has two solutions? The wording is really weird, but I don't think it relates to the distance d given earlier in the question, nothing suggests that. So my interpretation would be that we just need to prove that, provided that the condition on U holds, then there are two values of d which satisfy the other conditions given. This would make sense, provided the target is moving slow enough, it can hit it on the way up (one distance), or on the way down (another distance). At least I think. I just don't think it relates to the distance d. What do you reckon Rui? ::) Yeah seems more legit that way aha I just got completely lost out of the rigour of that question in my fatigue Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 03, 2016, 10:42:40 am Yeah seems more legit that way aha I just got completely lost out of the rigour of that question in my fatigue I don't blame you, that is probably one of the worst projectile questions I've ever seen :P Title: Re: 3U Maths Question Thread Post by: massive on August 03, 2016, 04:55:13 pm Guys i don't get part d and e. btw i really appreciate the help ;D Title: Re: 3U Maths Question Thread Post by: RuiAce on August 03, 2016, 07:09:01 pm Guys i don't get part d and e. btw i really appreciate the help ;D I have to ask this again. What is the source of these questions $\text{With part d), notice how }V^2\textbf{ is gone.}\\ \text{We need to aim to get rid of it.}$ $\text{Using the new piece of information, we know that AT }y=0\text{, specifically }x=R+r$ \begin{align*}0&=h+(R+r)\tan \alpha - \frac{g(R+r)^2}{2V^2\cos^2 \alpha}\\ \frac{g(R+r)^2}{2V^2\cos^2\alpha}&=h+(R+r)\tan \alpha\\ V^2&=\frac{g(R+r)^2}{2\cos^2\alpha(h+(R+r)\tan \alpha)}\end{align*} \text{The inequality stays, so sub this into the expression in part iii)}\\ \begin{align*}\frac{g(R+r)^2}{2\cos^2\alpha (h+(R+r)\tan \alpha)}& \ge \frac{gR}{2\sin\alpha \cos \alpha}\\ (R+r)^2\frac{\sin \alpha}{\cos \alpha}&\ge R(h+(R+r)\tan \alpha)\\ r\tan \alpha(R+r)&\ge Rh\\ \tan \alpha&\ge \frac{Rh}{(R+r)r}\end{align*} Give me a bit, still working on the other one Title: Re: 3U Maths Question Thread Post by: RuiAce on August 04, 2016, 12:16:12 pm Note that MX1 questions from Pre-2001 are tough. The difficulty/rigour of these questions is WELL beyond that of the current HSC. Post them if have them! But we might not give you a full worked solution, rather, guide you through one for yourself. In saying that, here are solutions. THSC. Always feel free to comment further if any part of the provided solutions don't make sense though. Title: Re: 3U Maths Question Thread Post by: conic curve on August 04, 2016, 01:09:14 pm Can we please refrain from too many past HSC questions before the year 2001? The difficulty/rigour of these questions is seriously getting WELL beyond that of the current HSC. In saying that, here are solutions. THSC. Always feel free to comment further if any part of the provided solutions don't make sense though. Isn't it good that you're doing it because you're exposing yourself to questions harder than the HSC? Title: Re: 3U Maths Question Thread Post by: RuiAce on August 04, 2016, 01:24:32 pm Isn't it good that you're doing it because you're exposing yourself to questions harder than the HSC? Sure, if you want to aim that high. I never went before 2001 and I still did reasonably well anyway. It's becoming really hard to provide concise solutions, or sometimes you just get lost lol. Sometimes it can be nearing impossible to provide solutions so easily because it's reaching a point where the lateral thinking required is not that prevalent anymore, and demanded by the students today. Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 04, 2016, 01:34:05 pm Sure, if you want to aim that high. I never went before 2001 and I still did reasonably well anyway. It's becoming really hard to provide concise solutions, or sometimes you just get lost lol. Sometimes it can be nearing impossible to provide solutions so easily because it's reaching a point where the lateral thinking required is not that prevalent anymore, and demanded by the students today. Just to give an indication, I did the 1980's-1990's papers for study back in 2014, my average mark in those decades was a good 10 points lower than my average mark for the more recent exams. There is a definite difference in difficulty ;D That said, there is definitely benefit to doing them. I'd recommend them to anyone aiming for a Band E4 in Extension 1 ;) Keep posting questions if you have them! Just understand that Rui, Jake and myself might not be able to provide proper solutions to the really hard stuff on occasion, if it really stumps us 8) Title: Re: 3U Maths Question Thread Post by: Goodwil on August 04, 2016, 04:20:37 pm Hi, I'm having trouble finding the asymptotes, turning points, inflection points etc to draw these graphs Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 04, 2016, 04:43:51 pm Hi, I'm having trouble finding the asymptotes, turning points, inflection points etc to draw these graphs Hey Goodwil! I'm happy to help, I'll help you find the asymptotes for both, then I'll go through the full process for the second one, hopefully you can apply it to the first ;D Okay, so vertical asymptotes exist where the function is undefined. In both of these cases, that means when the denominator is equal to zero. So, we can do it by inspection if we want, or alternatively: $x^2-4=0 \\ (x-2)(x+2)=0 \\\therefore x=\pm2$ $x^2-2x-3=(x-3)(x+1)=0 \\\therefore x=3, -1$ These are the vertical asymptotes! Let's do the turning points and such for the second function. First, we differentiate using the quotient rule: $y=\frac{2x^3}{x^2-4} \\ \frac{dy}{dx}=\frac{(x^2-4)6x^2-2x^3(2x)}{(x^2-4)^2} \\=\frac{2x^4-24x^2}{(x^2-4)^2}$ To find the turning points, we put this equal to zero (note in the second step we discard the denominator because it can't be equal to zero): $\frac{2x^4-24x^2}{(x^2-4)^2}=0 \\\therefore 2x^4-24x^2=0 \\ 2x^2(x^2-12) = 0 \\\therefore x=0, \pm\sqrt{12}$ Now we'll need our second derivative for the next step, but I'll be honest, that derivative looks sort of gross. It would be another application of chain rule, but I'm going to use magic (Wolfram Alpha) to get the answer: $\frac{d^2y}{dx^2}=\frac{16x(x^2+12)}{(x^2+4)^3}$ Okay, so we do a few things with this. First, we substitute in our x values from earlier to find the nature of the turning points. Doing so, we find that: $x=0\text{ is a horizontal inflexion since at this point } y''=0 \\x=\sqrt{12} \text{ is a minimum since y''>0 at this point} \\x=-\sqrt{12}\text{ is a maximum since y''<0 at this point}$ We also need to put the second derivative equal to zero to find any points of inflexion: $16x(x^2+12)=0 \\\therefore x=0$ We already know about this point, but at least we know there aren't any more ;D Putting all of that information together, and perhaps plotting a few points and considering some limiting cases near the asymptotes, leads us to this graph here. Notice the asymptotes in the correct spots, the inflexion in the middle, and the two max/min points ;D So that's the process, it will be an exact replica for the first question! Asymptotes, first and second derivatives, turning points, inflexions, and then sketch ;D This sort of question, besides the nasty second derivative, is quite standard in MX1 exams so it is handy to know this process really really well ;D does this help? :) Title: Re: 3U Maths Question Thread Post by: massive on August 05, 2016, 10:04:00 am Yeah they're pretty hard, but they're pretty fun at the same time! :P Anyway, how do you find the amplitude for the attached equation (its Simple Harmonic Motion) Title: Re: 3U Maths Question Thread Post by: RuiAce on August 05, 2016, 10:26:46 am Yeah they're pretty hard, but they're pretty fun at the same time! :P Anyway, how do you find the amplitude for the attached equation (its Simple Harmonic Motion) \text{Expand out the compound-angle to get}\\ \begin{align*}y&=6\sin \left(2t+\frac{\pi}{4}\right)+\sin 2t\\ &= 6\left(\frac{1}{\sqrt{2}}\sin 2t+\frac{1}{\sqrt{2}}\cos 2t\right)+\sin 2t\\ &= \left(3\sqrt{2}+1\right)\sin 2t + 3\sqrt{2} \cos 2t \end{align*} $\text{Using the auxiliary angle transformation we can rewrite this as}\\ y=\sqrt{(3\sqrt{2}+1)^2+(3\sqrt{2})^2}\sin (2t+\alpha)\\ \text{where }0<\alpha < \frac{\pi}{2}, \, \tan \alpha=\frac{3\sqrt{2}}{3\sqrt{2}+1}$ $\text{So the amplitude is just }\sqrt{(3\sqrt{2}+1)^2+(3\sqrt{2})^2}=\sqrt{37+6\sqrt{2}}$ Title: Re: 3U Maths Question Thread Post by: jakesilove on August 05, 2016, 10:47:43 am $\text{So the amplitude is just }\sqrt{(3\sqrt{2}+1)^2+(3\sqrt{2})^2}=\sqrt{37+6\sqrt{2}}$ 'Just'. Toughest Amplitude question I've seen Title: Re: 3U Maths Question Thread Post by: RuiAce on August 05, 2016, 10:52:10 am 'Just'. Toughest Amplitude question I've seen I did the exact same question around a month ago Title: Re: 3U Maths Question Thread Post by: jakesilove on August 05, 2016, 10:53:05 am I did the exact same question around a month ago Plus you're hard to impress x Title: Re: 3U Maths Question Thread Post by: RuiAce on August 05, 2016, 10:54:48 am Plus you're hard to impress x Wow ok Title: Re: 3U Maths Question Thread Post by: massive on August 05, 2016, 11:08:31 am Thanks Rui!!! Where would I be without you. Also for the attached equation (SHM) i have to find the speed of the particle when x=2. Do i just sub x=2 into the displacement eqn to get time and then sub that time into the velocity eqn, or is there a better way of doing it? thanks! Title: Re: 3U Maths Question Thread Post by: RuiAce on August 05, 2016, 11:37:59 am I'm on my phone so ceebs with LaTeX right now. If you sub x=2 you will get an infinite number of solutions for t. But then differentiate it to find the velocity and your speed will still be the same. The more clever way is to differentiate FIRST, then use a Pythagorean identity to write v^2 as a function of x instead. Then sub x=2. Also, keep in mind speed is the MAGNITUDE of the velocity. The sign should always be positive for speed. I.e. You want |v|, not v Title: 3U Maths Question Thread Post by: jamonwindeyer on August 05, 2016, 11:44:00 am Thanks Rui!!! Where would I be without you. Also for the attached equation (SHM) i have to find the speed of the particle when x=2. Do i just sub x=2 into the displacement eqn to get time and then sub that time into the velocity eqn, or is there a better way of doing it? thanks! That's probably the easiest way in terms of accessibility, but let me show you Rui's way for something different ;D $x=3\sin{(2t+5)} \\\therefore \sin{(2t+5)}=\frac{x}{3} \\v=6\cos{(2t+5)} \\\therefore \cos{(2t+5)}=\frac{v}{6}$ Now using the Pythagorean identity: $\sin^2{(2t+5)}+\cos^2{(2t+5)}=1 \\ \frac{x^2}{9}+\frac{v^2}{36} = 1 \\\therefore |v|=\sqrt{36-4x^2} = 4.47\text{ms}^{-1}\text{ when x=2}$ Skipped a few steps, but that should give you the idea! ;D -Rui: Popped absolute value brackets around the v. Without the absolute value brackets you need a plus or minus! But thanks for doing the hard stuff Jamon :) - Title: Re: 3U Maths Question Thread Post by: Neutron on August 05, 2016, 12:12:56 pm :O Neutron has something other than a physics question (surprise, i finished my trials :P) Okay, so for some reason I can't get part ii of this, I feel like I'm just tired and not processing properly, but if one of you kind souls could help that would be great!! b) A rescue plane is travelling at an altitude of 120 metres above sea level and a constant speed of 216 km/h towards a stranded sailor. A canister containing a life raft is dropped from the plane to the sailor. i) How long will it take for the canister to hit the water? (Take g=10ms^-2) A current is causing the sailor to drift at a speed of 3.6 km/h in the same direction as the plane is travelling. The canister is dropped when the horizontal distance from the plane to the sailor is D metres. ii) What values can D take if canister lands at most 50 metres away from the sailor? And also, one more! I attached a screenshot :D Thank you guys!! Title: Re: 3U Maths Question Thread Post by: RuiAce on August 05, 2016, 12:32:33 pm :O Neutron has something other than a physics question (surprise, i finished my trials :P) Okay, so for some reason I can't get part ii of this, I feel like I'm just tired and not processing properly, but if one of you kind souls could help that would be great!! b) A rescue plane is travelling at an altitude of 120 metres above sea level and a constant speed of 216 km/h towards a stranded sailor. A canister containing a life raft is dropped from the plane to the sailor. i) How long will it take for the canister to hit the water? (Take g=10ms^-2) A current is causing the sailor to drift at a speed of 3.6 km/h in the same direction as the plane is travelling. The canister is dropped when the horizontal distance from the plane to the sailor is D metres. ii) What values can D take if canister lands at most 50 metres away from the sailor? And also, one more! I attached a screenshot :D Thank you guys!! I'm in a lecture right now so I'll just address the binomial theorem question. $\text{Cleverly factor out 2 from the factorised binomial to get something more useful:}\\ \left(2+\frac{x}{2}\right)^n = 2^n \left(1+\frac{x}{2^2}\right)^n$ $\text{Now, expanding out with the binomial theorem:}\\ \left(1+\frac{x}{2^2}\right)^n=\binom{n}{0}+\binom{n}{1}\frac{x}{2^2}+\binom{n}{2}\frac{x}{2^4}+\dots+\binom{n}{n}\frac{x}{2^{2n}}$ $\text{Substitute in }x=1\text{ because the }x\text{ itself must disappear:}\\ \left(1+\frac{1}{2^2}\right)^n=\left(\frac{5}{2^2}\right)^n=\binom{n}{0}+\binom{n}{1}\frac{1}{2^2}+\binom{n}{2}\frac{1}{2^4}+\dots+\binom{n}{n}\frac{1}{2^{2n}}$ $\text{Then just take the }\binom{n}{0}=1\text{ to the LHS and we are done.}$ $\textit{Note, you can directly expand out what was given and let }x=1\textit{ as well}\\ \textbf{However, you have to divide through by }2^n\textbf{ at a later point.}$ Title: Re: 3U Maths Question Thread Post by: jakesilove on August 05, 2016, 12:34:45 pm :O Neutron has something other than a physics question (surprise, i finished my trials :P) Okay, so for some reason I can't get part ii of this, I feel like I'm just tired and not processing properly, but if one of you kind souls could help that would be great!! b) A rescue plane is travelling at an altitude of 120 metres above sea level and a constant speed of 216 km/h towards a stranded sailor. A canister containing a life raft is dropped from the plane to the sailor. i) How long will it take for the canister to hit the water? (Take g=10ms^-2) A current is causing the sailor to drift at a speed of 3.6 km/h in the same direction as the plane is travelling. The canister is dropped when the horizontal distance from the plane to the sailor is D metres. ii) What values can D take if canister lands at most 50 metres away from the sailor? And also, one more! I attached a screenshot :D Thank you guys!! Sorry that it's a bit messy (and handwritten), I just couldn't be bothered typing it up. (http://i.imgur.com/51RCZVg.jpg?1) You needed to think pretty carefully about what was actually happening, and I think a diagram helped with that. Still, a tricky question, hope my working out makes sense! Jake Title: Re: 3U Maths Question Thread Post by: massive on August 05, 2016, 07:20:11 pm Hey guys how do you determine whether a particle will return to the origin or not?? Title: Re: 3U Maths Question Thread Post by: RuiAce on August 05, 2016, 07:38:27 pm Hey guys how do you determine whether a particle will return to the origin or not?? Does setting x=0 yield a solution? Or if you analyse it using physics, (as an example) if initially x=0 but then v is monotonic then the particle is of course always going to travel away from the origin. It depends on the question. The first method is fool-proof but every question is different. Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 05, 2016, 07:41:45 pm Hey guys how do you determine whether a particle will return to the origin or not?? Hey! There are normally two ways: First, if any expression for position, velocity or acceleration is undefined at the origin, then it will not ever reach there. This doesn't help if it was there once though. More commonly we just need a bit of intuition. If, for example, velocity and acceleration are both positive for all positive t, then the particle will never return to the origin, it will zoom off to infinity. Unfortunately, there are no concrete sets of rules for this sort of thing, because there is always some weird exception or counter-argument. It is best to try and practice getting a good look at the whole situation and interpreting your info (we can help with this!) ;D Title: Re: 3U Maths Question Thread Post by: massive on August 05, 2016, 07:47:01 pm what about something like this? Title: 3U Maths Question Thread Post by: Klexos on August 05, 2016, 10:22:52 pm (http://uploads.tapatalk-cdn.com/20160805/4a76b9eb0f24662e0c40e86bd778796d.jpg) (http://uploads.tapatalk-cdn.com/20160805/a84a1ded04b84c317970903fb0ee63ea.jpg) I have attached my pathetic attempt as well, just for the lols. And my flipped photos are...wow... Is t formulae question merely just subbing stuff in...? Title: Re: 3U Maths Question Thread Post by: RuiAce on August 05, 2016, 10:27:21 pm (http://uploads.tapatalk-cdn.com/20160805/4a76b9eb0f24662e0c40e86bd778796d.jpg) (http://uploads.tapatalk-cdn.com/20160805/a84a1ded04b84c317970903fb0ee63ea.jpg) I have attached my pathetic attempt as well, just for the lols. And my flipped photos are...wow... Is t formulae question merely just subbing stuff in...? Yes it is. Keep going. You end up with (2+2t)/(2t+2t2) which is just 1/t Title: Re: 3U Maths Question Thread Post by: Klexos on August 05, 2016, 10:30:08 pm Yeah I had that result but I don't know what I'm supposed to do with it to make it cot x/2 Title: 3U Maths Question Thread Post by: Klexos on August 05, 2016, 10:32:12 pm Correct?(http://uploads.tapatalk-cdn.com/20160805/0c1433fc26ba8d8300c01321709c9987.jpg) Title: Re: 3U Maths Question Thread Post by: RuiAce on August 05, 2016, 10:36:55 pm Correct?(http://uploads.tapatalk-cdn.com/20160805/0c1433fc26ba8d8300c01321709c9987.jpg) Yes. Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 05, 2016, 11:54:13 pm what about something like this? This one was a little tricky actually, but again, let's use some intuition (you'd want something a little more rigorous in an exam). The particle starts going to the right at 4ms^-1. The acceleration is negative after x=1, and this will eventually turn it around. When it returns to x=1 it is travelling backwards: $v^2=16+6x-2x^3 = 20 \\\therefore v=-\sqrt{20}\text{ms}^{-1}$ The acceleration is zero at this point, and although between x=0 and x=1 it will be positive acceleration, it won't be enough to turn it back around, the thing is going to zoom off to negative infinity. This can also be gleamed from the fact that the expression for velocity only has a single root at about x=2.5, the point where it turned around initially. Besides that, it ain't stopping, so it does return to the origin and then zooms off to negative infinity ;) Title: Re: 3U Maths Question Thread Post by: massive on August 06, 2016, 11:45:54 am Thanks Jamon! Guys for the question attached, in part iii, I know that we have to differentiate the eqn of part ii, but do you differentiate with respect to theta or respect to alpha? Title: Re: 3U Maths Question Thread Post by: RuiAce on August 06, 2016, 12:26:54 pm Thanks Jamon! Guys for the question attached, in part iii, I know that we have to differentiate the eqn of part ii, but do you differentiate with respect to theta or respect to alpha? $\alpha\text{ relates to the angle of the inclined plane, not the nature of the particle.}\\ \text{Because }\theta\text{ is the angle of projection, that is the variable here. However we don't actually need differentiation.}$ \text{Using the product-to-sum formula they gave:}\\ \begin{align*}r&=\frac{\sin (\theta-\alpha)\cos \theta}{\cos^2 \alpha}\\&= \frac{\sin (2\theta-\alpha)+\sin(- \alpha)}{2\cos^2\alpha}\end{align*} $\text{The trick with }\textbf{maximum range}\text{ is that}\\ \text{Because the range of }\sin \theta\text{ is }-1 \le \sin \theta \le 1\\ \text{The maximum range is essentially when }\sin \theta=1$ \sin (-\alpha)\text{ is a constant, but }\sin(2\theta-\alpha)\text{ depends on }\theta\text{ so we have}\\ \begin{align*}r_{max}=R&=\frac{1+\sin (-\alpha)}{2\cos^2\alpha}\\&=\frac{1-\sin \alpha}{2(1-\sin \alpha)(1+\sin \alpha)}\\ &= \frac{1}{2(1+\sin \alpha)}\end{align*} Title: Re: 3U Maths Question Thread Post by: conic curve on August 06, 2016, 06:07:55 pm In how many ways can five writers and five artists be arranged in a circle so that the writers are separated? In how many ways can this be done if two particular artists must not sit next to a particular writer? It's from the Fitzpatrick book. I got the first part by doing 1 * 4! * 5! = 2880 but I can't do the second part. I was thinking 2880 - the number of arrangements with 2 artists sitting next to the write but I can't figure it out. Any help appreciated. Title: Re: 3U Maths Question Thread Post by: anotherworld2b on August 07, 2016, 01:56:25 am Hi i was wondering if i could get help with these two questions Title: Re: 3U Maths Question Thread Post by: RuiAce on August 07, 2016, 07:43:43 am Hi i was wondering if i could get help with these two questions $\text{The presence of the }\frac{\pi}{4}\text{does not actually create that much of a difference}\\ \text{in the method for solving the second one.}$ \begin{align*}\sin \left(\frac{\pi}{4}(3x-1)\right)&=\frac{1}{4}\\ \frac{\pi}{4}(3x-1)&=n\pi + (-1)^n \sin^{-1}\frac{1}{4} \qquad (n \in \mathbb{Z})\\ 3x-1 &= 4n\pi + \frac{4(-1)^n}{\pi}\sin^{-1}\frac{1}{4}\\ x&=\frac{1}{3}\left( 4n\pi +1+ \frac{4(-1)^n}{\pi}\sin^{-1}\frac{1}{4}\right)\end{align*} $\text{An alternate way of writing the general solution in line 2 is }\\ \frac{\pi}{4}(3x-1)=2n\pi + \sin^{-1}\frac{1}{4}\\ \text{or }\frac{\pi}{4}(3x-1)=2n \pi + \pi - \sin^{-1}\frac{1}{4}\\ \text{Then, rearrange and finally round to 2 d.p. where appropriate.}$ _____________________________ $\textbf{The first question goes outside the scope of MX1 as the product to sum formulae are not taught at this level.}$ \text{By converting the expression using the product to sum formulae we have }\\ \begin{align*}2\sin 3x \sin x + \cos 4x &= \frac{1}{2}\\ (\cos 2x - \cos 4x) + \cos 4x &= \frac{1}{2}\\ \cos 2x &= \frac{1}{2}\\ 2x&= 2n \pi \pm \frac{\pi}{3}\qquad (n \in Z)\\ x &=n \pi \pm \frac{\pi}{6}\end{align*}\\ \text{Round as appropriate} Title: Re: 3U Maths Question Thread Post by: conic curve on August 07, 2016, 08:56:42 am If you factorised the whole thing by 6 that should be 18 there not 1/2. Anyway, regardless of if you factorise the 6 out, put the negative back in there and complete the square. \text{So this is how I would do that question}\\ \begin{align*}\int \frac{dt}{\sqrt{3-12t-18t^2}}&=\frac{1}{\sqrt{18}}\int \frac{dt}{\sqrt{\frac{1}{6}-\frac{2}{3}t-t^2}}\\ &= \frac{1}{\sqrt{18}}\int \frac{dt}{\sqrt{\frac{5}{18}-\left(t+\frac{1}{3}\right)^2}}\end{align*}\\ \text{which is now your ordinary sine inverse integral.} Just curious but why specifically take out 1/squareroot of 18 Title: Re: 3U Maths Question Thread Post by: RuiAce on August 07, 2016, 09:04:46 am Just curious but why specifically take out 1/squareroot of 18 To use the standard form of the integral, the coefficient on t must be 1. Otherwise we'd probably need integration by substitution. Title: Re: 3U Maths Question Thread Post by: conic curve on August 07, 2016, 09:08:12 am What am I supposed to do after? Title: Re: 3U Maths Question Thread Post by: RuiAce on August 07, 2016, 09:10:57 am What am I supposed to do after? \text{Likewise, you want the coefficient on }x\text{ to be }\textbf{1}.\\ \begin{align*}\int \frac{1}{1+4x^2}dx&=\int \frac{1}{4\left(\frac{1}{4}+x^2\right)}dx\\ &= \frac{1}{4} \int \frac{1}{\left(\frac{1}{2}\right)^2+x^2}dx\\ &= \frac{1}{4} \left[2 \tan^{-1}(2x) \right]+C\\ &= \frac{1}{2}\tan^{-1}(2x)+C\end{align*} Title: Re: 3U Maths Question Thread Post by: conic curve on August 07, 2016, 09:15:10 am \text{Likewise, you want the coefficient on }x\text{ to be }\textbf{1}.\\ \begin{align*}\int \frac{1}{1+4x^2}dx&=\int \frac{1}{4\left(\frac{1}{4}+x^2\right)}dx\\ &= \frac{1}{4} \int \frac{1}{\left(\frac{1}{2}\right)^2+x^2}dx\\ &= \frac{1}{4} \left[2 \tan^{-1}(2x) \right]+C\\ &= \frac{1}{2}\tan^{-1}(2x)+C\end{align*} I don't get the bit where you went to tan? (from how the integral was converted into tan) Title: Re: 3U Maths Question Thread Post by: RuiAce on August 07, 2016, 09:16:28 am I don't get the bit where you went to tan? (from how the integral was converted into tan) $\text{That's just the formula}\\ \int \frac{1}{a^2+x^2}dx=\frac{1}{a} \tan^{-1} \frac{x}{a}+C$ Title: Re: 3U Maths Question Thread Post by: conic curve on August 07, 2016, 09:33:12 am What did I do wrong here? Title: Re: 3U Maths Question Thread Post by: RuiAce on August 07, 2016, 09:38:21 am What did I do wrong here? Two mistakes in the last line. 1. You dropped off the 1/16 2. The x went missing inside the inverse tangent Title: Re: 3U Maths Question Thread Post by: conic curve on August 07, 2016, 10:10:15 am Two mistakes in the last line. 1. You dropped off the 1/16 2. The x went missing inside the inverse tangent Thanks How do you do this (I think there's no inverse trig integration rule for it) Title: Re: 3U Maths Question Thread Post by: RuiAce on August 07, 2016, 10:29:43 am Thanks How do you do this (I think there's no inverse trig integration rule for it) Requires partial fractions. Not 3U material. Title: Re: 3U Maths Question Thread Post by: conic curve on August 07, 2016, 10:31:50 am Requires partial fractions. Not 3U material. Thanks What am I doing wrong Title: Re: 3U Maths Question Thread Post by: RuiAce on August 07, 2016, 10:36:00 am Thanks What am I doing wrong \text{Wrong completing the square}\\ \begin{align*}u^2+\frac{u}{2}+\frac{1}{2}&=\left(u^2+2u\left(\frac{1}{4}\right)+\frac{1^2}{4^2}\right)-\frac{1}{16}+\frac{1}{2}\\&=\left(u+\frac{1}{4}\right)^2+\frac{7}{6}\end{align*} Title: Re: 3U Maths Question Thread Post by: conic curve on August 07, 2016, 10:52:59 am \text{Wrong completing the square}\\ \begin{align*}u^2+\frac{u}{2}+\frac{1}{2}&=\left(u^2+2u\left(\frac{1}{4}\right)+\frac{1^2}{4^2}\right)-\frac{1}{16}+\frac{1}{2}\\&=\left(u+\frac{1}{4}\right)^2+\frac{7}{6}\end{align*} Thanks How would I do this? Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 07, 2016, 10:56:27 am Thanks How would I do this? Factorise! $\frac{1}{\sqrt{4-3x^2}}=\frac{1}{\sqrt{3}\sqrt{\frac{4}{3}-x^2}}=\frac{1}{\sqrt{3}}\frac{1}{\sqrt{\left(\frac{2}{\sqrt{3}}\right)^2-x^2}}$ That's an inverse sine integral ;D you pull the 1 on root 3 out the front and then use your regular rule ;D Title: Re: 3U Maths Question Thread Post by: conic curve on August 07, 2016, 11:05:00 am Factorise! $\frac{1}{\sqrt{4-3x^2}}=\frac{1}{\sqrt{3}\sqrt{\frac{4}{3}-x^2}}=\frac{1}{\sqrt{3}}\frac{1}{\sqrt{\left(\frac{2}{\sqrt{3}}\right)^2-x^2}}$ That's an inverse sine integral ;D you pull the 1 on root 3 out the front and then use your regular rule ;D What happened to the dx? Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 07, 2016, 11:21:10 am What happened to the dx? Oh, it was just showing you the factorisation to apply, in the context of the integral: $\int^1_{-1}\frac{dx}{\sqrt{4-3x^2}}=\frac{1}{\sqrt{3}}\int^1_{-1}\frac{dx}{\sqrt{\left(\frac{2}{\sqrt{3}}\right)^2-x^2}}$ Then apply the regular rule ;D Title: Re: 3U Maths Question Thread Post by: Spencerr on August 07, 2016, 12:48:20 pm Hey there, could some help explain to me how to graph these functions. 1. Cos (cos inverse x) 2. Cos (sin inverse x) 3.. Cos inverse (cos x) 4. Cos inverse (sinx) 5.. Tan inverse (tan x) 6. Tan (tan inverse x) Sorry these are alot of graphs but I get really confused and I find it hard to wrap my head around these trigonometric inverse questions especially with all the restrictions. Any other general tips in dealing with these questions will also be appreciated :) Thanks Title: Re: 3U Maths Question Thread Post by: RuiAce on August 07, 2016, 01:04:43 pm Hey there, could some help explain to me how to graph these functions. 1. Cos (cos inverse x) 2. Cos (sin inverse x) 3.. Cos inverse (cos x) 4. Cos inverse (sinx) 5.. Tan inverse (tan x) 6. Tan (tan inverse x) Sorry these are alot of graphs but I get really confused and I find it hard to wrap my head around these trigonometric inverse questions especially with all the restrictions. Any other general tips in dealing with these questions will also be appreciated :) Thanks $\text{Keep in mind that }g(y)=\cos^{-1}y, \, -1 \le y \le 1 \\\text{ is defined as the inverse of }f(x)=\sin x\text{ for }0 \le x \le \pi \\ \text{The consequence is that when }-1 \le x \le 1, \text{ we have }\cos (\cos^{-1}x)=x\\ \text{That is to say, they will cancel each other out.}$ $\text{Thus, }y=\cos (\cos^{-1}x)\text{ is just the sketch of }y=x\\ \text{based off the natural domain of inverse cosine, that is }-1 \le x \le 1.$ $\textit{Note: You should label your endpoints.}$ (http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpssryziayr.png) $\text{On the contrary, }\cos x\text{ keeps repeating itself every }2\pi\text{ intervals; it has no restricted domain.}\\ \text{Much more care is needed here.}$ $\text{Notice how when }0 \le x \le \pi\text{, indeed }\cos^{-1}(\cos x)=x.\\ \text{So for }0\le x \le \pi\text{ we ARE sketching }y=x$ $\text{But that is just about it. ONLY when }0 \le x \le \pi, \cos^{-1}(\cos x)=x$ $\text{From here on, we can take two analytical approaches.}\\ \text{The more algebraic method is hard.}\\ \text{The graphical analysis is that keep in mind }\cos^{-1}(-1)=\pi \\ \text{and }\cos^{-1} 1 = 0$ $\text{So when }y=\cos x = 1\text{, we have }y=\cos^{-1}(\cos x) = 0\\ \text{Similarly }y=\cos x = -1\text{ means }y=\cos^{-1}(\cos x)=\pi$ $\text{Thus we can easily compare the two graphs.}\\ \textit{Note that from the restricted domain version, we know that the relationships are always linear.}$(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpsignrt2gh.png) $\text{All you should be doing is comparing the turning points on }y=\cos x\\ \text{to the corners on }y=\cos^{-1}(\cos x)$ More to come. Title: Re: 3U Maths Question Thread Post by: RuiAce on August 07, 2016, 01:15:12 pm $\text{On the other hand, }y=\tan^{-1}x\text{ has no natural domain restriction.}\\ \text{It's maximal domain is ALL real }x\\ \text{So if we try undoing }\tan^{-1}\text{ by applying }\tan\\ \text{We actually get }y=x\text{ in its entirety.}$(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpslawz9eyp.png) $\text{Once again, we need to repeat our analysis for }y=\tan^{-1}(\tan x)\\ \text{In a similar way, because the natural range of }\tan^{-1}(x)\text{ is }-\frac{\pi}{2} < \tan^{-1}x < \frac{\pi}{2}\\ \text{WHEN }-\frac{\pi}{2} < \tan^{-1}x < \frac{\pi}{2},\, \tan^{-1}(\tan x)=x$ $\text{So over this domain we can once again, sketch }y=x\\ \textit{And here's where things get a bit trickier.}$ $\text{Is }\tan \frac{\pi}{2}\text{ defined? The answer is }\textbf{NO}\text{, it is UNDEFINED.}\\ \text{If }\tan \frac{\pi}{2}\text{ is undefined, how can }\tan^{-1} \left(\tan \frac{\pi}{2}\right)\text{ be?}$ $\text{This is due to the asymptotic behaviour of dividing by 0.}\\ \text{Taking tan inverse is supposed to undo it but you can never undo divide by 0.}\\ \text{Whilst as }x\to \frac{\pi}{2}, y \to \frac{\pi}{2},\, x\text{ is undefined at }\frac{\pi}{2}\\ \text{So on top of this Desmos generated graph, you MUST insert a DISCONTINUITY (hole in graph) at the endpoint }\left(\frac{\pi}{2},\frac{\pi}{2}\right)$(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zps7mn8abq7.png) $\text{Similar reasoning show that at }\left(-\frac{\pi}{2},-\frac{\pi}{2}\right)\text{ there is ALSO a discontinuity.}$ $\text{One thing you will have noticed is that whereas inverse cosine becomes full of uncertainty,}\\ \text{this one ALWAYS repeats the }y=x\text{ shape.}$ $\text{The actual reason for this is a bit hard, and has to do with how }\tan^{-1}(\tan x )= x\\ \text{when }x\text{ is between }\pm \frac{\pi}{2}\text{ instead of the yucky }0, \pi$ $\textit{However, I will let you make sense of this yourself. Come back if there's confusion.}$ $\text{Note: Yes, at every endpoint there's a discontinuity.}$ Title: Re: 3U Maths Question Thread Post by: RuiAce on August 07, 2016, 01:29:43 pm $\text{Now, obviously inverse sine can never cancel out with cosine! At least, not directly.}\\ \text{We need to do some kind of manipulation to make the form friendlier.}$ $\text{One thing that you must always be able to recall is the Pythagorean identity. We will make use of it here.}\\ \text{From }\sin^2 \theta + \cos^2\theta = 1\text{ we have }\cos \theta = \pm \sqrt{1-\sin^2\theta} \\ \textit{Note that the plus-minus is not strictly defined and depends on the quadrant.}$ \begin{align*}y&= \cos (\arcsin{x})\\ &=\pm \sqrt{1- \sin^2 (\arcsin x)}\\ &= \pm \sqrt{1-x^2}\end{align*} $\text{How do we decide on whether we take the positive or negative case? We consider the range of }y=\arcsin x\\ \text{Note that }-\frac{\pi}{2} \le y \le \frac{\pi}{2}\\ \text{Over this domain, }\cos y\text{ is }\textbf{ALWAYS}\text{ positive. (Do you realise why?)}$ $\text{Hence, we are just sketching the semi-circle }y=\sqrt{1-x^2}$(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zps5wyvw5hw.png) $\text{For the last one, we can analyse the question similarly.} \\ \text{The only difference is that we are now using sine as our argument.}$ $\text{If instead }\sin x=1\text{ here, we will have }\cos^{-1}(\sin x)=0\\ \text{If instead }\sin x = -1\text{ here, we will have }\cos^{-1}(\sin x)=\pi$ $\text{Thus we will arrive at this graph.}$(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zps5zcvmw48.png) $\text{Why is it that earlier we didn't have something linear going on, but here we do?}\\ \text{This is because the relationship between }\sin x\text{ and }\cos x\text{ is just a shift to the left/right by }\frac{\pi}{2}\\ \text{That is, }\sin\left(x+\frac{\pi}{2}\right)=\cos x$ $\textbf{The shape of the curve doesn't change.}\\ \text{Whereas with inverse sine and cosine, the shape does change and one is the other FLIPPED as well as being translated upwards.}$ $\text{So if }\sin x\text{ or }\cos x\text{ is on the inside, we have less problems}\\ \text{as opposed to throwing }\sin^{-1}x\text{ or }\cos^{-1}x \text{ inside.}$ Title: Re: 3U Maths Question Thread Post by: Spencerr on August 07, 2016, 03:22:04 pm Thanks Rui, This is very well explained, clear and heaps helpful :) Title: Re: 3U Maths Question Thread Post by: Neutron on August 07, 2016, 06:39:35 pm Hey hey! Thank you so much for the help with the other question ^^ I have returned with yet another one: In a given lottery the probability that the jackpot prize is won is 0.013. Successive lottery draws are independent. The jackpot prize is initially 1, 000, 000 and increases by 250 000 each time the prize is not won. Find, correct to 5 decimal places, the probability that the jackpot prize will exceed 5 000 000 when it is finally won. The answer just found the probability for it being won on the 17th draw (First draw when the prize is above 5 000 000) but don't you have to take into account the fact that the prize could be on on the 18th draw, 19th draw, 20th draw etc all the way till infinity? I don't understand how just finding the probability of it being won on the 17th draw is equal to finding the probability that the jackpot prize will exceed 5000 000 when it is finally won D: Thank you legends!! Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 07, 2016, 07:02:29 pm Hey hey! Thank you so much for the help with the other question ^^ I have returned with yet another one: In a given lottery the probability that the jackpot prize is won is 0.013. Successive lottery draws are independent. The jackpot prize is initially 1, 000, 000 and increases by 250 000 each time the prize is not won. Find, correct to 5 decimal places, the probability that the jackpot prize will exceed 5 000 000 when it is finally won. The answer just found the probability for it being won on the 17th draw (First draw when the prize is above 5 000 000) but don't you have to take into account the fact that the prize could be on on the 18th draw, 19th draw, 20th draw etc all the way till infinity? I don't understand how just finding the probability of it being won on the 17th draw is equal to finding the probability that the jackpot prize will exceed 5000 000 when it is finally won D: Thank you legends!! I agree with you Neutron! I'd be taking the sum of an infinite geometric series or something similar, where is the question from? ;D Title: Re: 3U Maths Question Thread Post by: Neutron on August 07, 2016, 07:12:08 pm It's from one of my school's past papers :/ Title: Re: 3U Maths Question Thread Post by: RuiAce on August 07, 2016, 07:15:08 pm Hey hey! Thank you so much for the help with the other question ^^ I have returned with yet another one: In a given lottery the probability that the jackpot prize is won is 0.013. Successive lottery draws are independent. The jackpot prize is initially 1, 000, 000 and increases by 250 000 each time the prize is not won. Find, correct to 5 decimal places, the probability that the jackpot prize will exceed 5 000 000 when it is finally won. The answer just found the probability for it being won on the 17th draw (First draw when the prize is above 5 000 000) but don't you have to take into account the fact that the prize could be on on the 18th draw, 19th draw, 20th draw etc all the way till infinity? I don't understand how just finding the probability of it being won on the 17th draw is equal to finding the probability that the jackpot prize will exceed 5000 000 when it is finally won D: Thank you legends!! \text{So using what you just said.}\\ \text{We first consider the AP going on. We require }T_n=5000000\\ \begin{align*}5000000=1000000+(16)(250000)\end{align*}\\ \text{This is true so }n=17\text{ as you said.} $\text{Yes. I believe that the probability the jackpot prize will reach 5000000 is different to exceeding.}\\ \text{Unless the prize has to be won at 5000000, we can treat }n\text{ as going to infinity.}$ \begin{align*}Pr(n\ge 17) &= (0.987)^{16}(0.013)+(0.987)^{17}(0.013)+(0.987)^{18}(0.013)+\dots\\ &= \frac{(0.013)(0.987)^{16}}{1-0.987}\\ &=0.987^{16}\end{align*} I was about to ask where the source of the question was. Can you ask your school for some clarification? Extra reading: Statisticians refer to these independent events as following a "geometric" distribution. A geometric distribution measures the probability of the first success after having (n-1) fails. It follows the "negative-binomial" distribution for n=1. The probability mass function for the geometric distribution is (1-p)n-1p, where p is the probability of success. Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 07, 2016, 07:41:57 pm It's from one of my school's past papers :/ Yep, consenting with Rui above, his method is how I would solve it. They've solved for the probability that the lottery is won as soon as the prize exceeds 5 million, but as you correctly say, it could be won when the prize is 100 million ;D Title: Re: 3U Maths Question Thread Post by: massive on August 07, 2016, 08:38:02 pm yoo guys, howdu do part ii? Title: Re: 3U Maths Question Thread Post by: RuiAce on August 07, 2016, 09:25:51 pm yoo guys, howdu do part ii? $\frac{AP}{BS}=\frac{5}{3}=\frac{PX}{XS}=\frac{AX}{BX}\\ \text{But }AX+BX=16$ \begin{align*}\frac{5}{3}=\frac{16-BX}{BX}&=\frac{16}{BX}-1\\ \frac{8}{3}&=\frac{16}{BX}\\ BX&=6\\ \therefore AX&=10 \end{align*} $\text{Then since both triangles are right-angled (radius perp. tangent at point of contact)}\\ \text{Use two applications of Pythagoras' Theorem to determine }PX\text{ and }XS\\ \text{Add them up, and don't forget to double because there's two strings.}$ Title: Re: 3U Maths Question Thread Post by: Neutron on August 07, 2016, 10:39:17 pm Ahh thank you so so much!! One more thing from me (sorry) :o This is the question: The point P divides the interval AB in the ratio 3:7. In what external ratio does the point A divide the interval PB? How do you know whether it's 3:10 or 10:3? :O Thank you! Title: Re: 3U Maths Question Thread Post by: RuiAce on August 07, 2016, 10:41:34 pm Ahh thank you so so much!! One more thing from me (sorry) :o This is the question: The point P divides the interval AB in the ratio 3:7. In what external ratio does the point A divide the interval PB? How do you know whether it's 3:10 or 10:3? :O Thank you! Match it up: A---P-------B 3 : 7 |-----10------| So A is closer to P than it is to B. So since P comes first: So 3:10 Title: Re: 3U Maths Question Thread Post by: Jakeybaby on August 07, 2016, 10:49:12 pm Ahh thank you so so much!! One more thing from me (sorry) :o This is the question: The point P divides the interval AB in the ratio 3:7. In what external ratio does the point A divide the interval PB? How do you know whether it's 3:10 or 10:3? :O Thank you! Spoiler (https://i.gyazo.com/63f005f5febd7ff346fe025249c231f0.png) (see spoiler) Therefore, following the 2 steps, the external ratio which A divides PB is 3:10 Edit: Added Spoiler Title: Re: 3U Maths Question Thread Post by: Neutron on August 07, 2016, 11:07:16 pm Match it up: A---P-------B 3 : 7 |-----10------| So A is closer to P than it is to B. So since P comes first: So 3:10 But when you look at questions where you divide externally, normally you go from B to A and then back to P? Ahh another thing (sorry i keep doing more questions and running into more problems) With this question: Warehouse A has 100 computers and the probability that one of these computers is defective is 0.02. Warehouse B has 100 computers, two of which are defective. Joe buys three computers from Warehouse A and three computers from Warehouse B. What is the probability that exactly one of the computers he has bought is defective? Why aren't you allowed to just use binomial probability for both warehouse A and B? The answer used binomial probability for Warehouse A and normal probability for Warehouse B? This is what the solution said: P(exactly one computer defective) =P( 1 defective from A, 0 from B) + P(0 from A, 1 from B) 3C1 (0.02) ^1 (0.98)^2 x 98/100 x 97/99 x 96/98 + 3C0 (0.98)^3 x 2/11 x 98/99 x 3= 0.1095 Why do you have to times by 3 for the second part of the addition ( for the P(0 from A, 1 from B) ) sorry, I think my brain's fried, but tysm Rui!! (Sorry for having to type it up, my school website's down again :/ ) Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 07, 2016, 11:19:50 pm But when you look at questions where you divide externally, normally you go from B to A and then back to P? Ahh another thing (sorry i keep doing more questions and running into more problems) With this question: Warehouse A has 100 computers and the probability that one of these computers is defective is 0.02. Warehouse B has 100 computers, two of which are defective. Joe buys three computers from Warehouse A and three computers from Warehouse B. What is the probability that exactly one of the computers he has bought is defective? Why aren't you allowed to just use binomial probability for both warehouse A and B? The answer used binomial probability for Warehouse A and normal probability for Warehouse B? This is what the solution said: P(exactly one computer defective) =P( 1 defective from A, 0 from B) + P(0 from A, 1 from B) 3C1 (0.02) ^1 (0.98)^2 x 98/100 x 97/99 x 96/98 + 3C0 (0.98)^3 x 2/11 x 98/99 x 3= 0.1095 Why do you have to times by 3 for the second part of the addition ( for the P(0 from A, 1 from B) ) sorry, I think my brain's fried, but tysm Rui!! (Sorry for having to type it up, my school website's down again :/ ) Did this in my 3U lecture ;) So there is a slight difference here. Warehouse A has a probability of 2% that it is defective for every computer. Warehouse B has two defective computers. So, the probability is dependent on how many computers are left in the warehouse! As you buy computers, the probability turns from 2/100, to 2/99, 2/98, etc. The binomial probability idea can only be applied when the probability is identical with every repetition of the experiment, called a Bernoulli Trial. The multiplying by three? That comes from the fact that the defective computer can be the first computer bought from B, OR the second, OR the third. We add the probability of all three possibilities, but that ends up being the same as multiplying by three ;D Title: Re: 3U Maths Question Thread Post by: RuiAce on August 07, 2016, 11:31:10 pm But when you look at questions where you divide externally, normally you go from B to A and then back to P? Ahh another thing (sorry i keep doing more questions and running into more problems) With this question: Warehouse A has 100 computers and the probability that one of these computers is defective is 0.02. Warehouse B has 100 computers, two of which are defective. Joe buys three computers from Warehouse A and three computers from Warehouse B. What is the probability that exactly one of the computers he has bought is defective? Why aren't you allowed to just use binomial probability for both warehouse A and B? The answer used binomial probability for Warehouse A and normal probability for Warehouse B? This is what the solution said: P(exactly one computer defective) =P( 1 defective from A, 0 from B) + P(0 from A, 1 from B) 3C1 (0.02) ^1 (0.98)^2 x 98/100 x 97/99 x 96/98 + 3C0 (0.98)^3 x 2/11 x 98/99 x 3= 0.1095 Why do you have to times by 3 for the second part of the addition ( for the P(0 from A, 1 from B) ) sorry, I think my brain's fried, but tysm Rui!! (Sorry for having to type it up, my school website's down again :/ ) You do? I always just matched up the ordering. If it were division of BP then I would've gotten 10:3 Did this in my 3U lecture ;) I thought the question looked familiar.. Title: Re: 3U Maths Question Thread Post by: conic curve on August 08, 2016, 10:56:57 am Find domain and range of the following functions and sketch their graphs: 1. y=-3sin^-1 x For domain I got -1≤x≤1 For range I was not too sure??? 2. y=cos^-1 2x For domain, I got -1/2≤x≤1/2 Range I'm not too sure. Is it 0≤cos^-1 2x≤π? 3. y=3tan^-1 (x/2) Domain: all x Range: -Ω/2≤3tan^-1 (π/2) ≤pi/2 (I think I got this wrong) Point of inflection: let x/2=0 x=0 Sub in LHS point (and I'm not sure of what I need to do) Could someone here please help me Thanks Title: Re: 3U Maths Question Thread Post by: massive on August 08, 2016, 11:11:55 am hey, how do you do this guys? Title: Re: 3U Maths Question Thread Post by: massive on August 08, 2016, 11:20:36 am Find domain and range of the following functions and sketch their graphs: 1. y=-3sin^-1 x For domain I got -1≤x≤1 For range I was not too sure??? The range for this is -3pi≤y≤0 Find domain and range of the following functions and sketch their graphs: 2. y=cos^-1 2x For domain, I got -1/2≤x≤1/2 Range I'm not too sure. Is it 0≤cos^-1 2x≤π? The range for this is just the normal 0≤y≤pi Find domain and range of the following functions and sketch their graphs: 3. y=3tan^-1 (x/2) Domain: all x Range: -Ω/2≤3tan^-1 (π/2) ≤pi/2 (I think I got this wrong) Point of inflection: let x/2=0 x=0 Sub in LHS point (and I'm not sure of what I need to do) The range for this is -3pi/2<y<3pi/2 im not really sure how you get point of inflexion (sorry) but im pretty sure its still at the origin for this. Title: Re: 3U Maths Question Thread Post by: RuiAce on August 08, 2016, 11:44:40 am Find domain and range of the following functions and sketch their graphs: 1. y=-3sin^-1 x For domain I got -1≤x≤1 For range I was not too sure??? 2. y=cos^-1 2x For domain, I got -1/2≤x≤1/2 Range I'm not too sure. Is it 0≤cos^-1 2x≤π? 3. y=3tan^-1 (x/2) Domain: all x Range: -Ω/2≤3tan^-1 (π/2) ≤pi/2 (I think I got this wrong) Point of inflection: let x/2=0 x=0 Sub in LHS point (and I'm not sure of what I need to do) Could someone here please help me Thanks $\text{Understand what the effect of what adding a coefficient in front of the }\textbf{whole thing actually does}.\\ \text{Open up Desmos and graph }y=\sin^{-1}x\text{ and }y=3\sin^{-1}x\text{ simultaneously. The effect is entirely on the }y\text{-coordinate.}$ $\text{Also note that the range of }\tan^{-1}x\text{ is }-\frac{\pi}{2} < x <\frac{\pi}{2}\\ \text{The inequality is strict. }<\text{, not }\le$ \text{For points of inflexion, let }\frac{d^2y}{dx^2}=0\\ \text{i.e. Use the systematic method you were taught in 2U.} \\ \begin{align*}y&=3\tan^{-1}\frac{x}{2}\\ \implies \frac{dy}{dx}&=\frac{6}{4+x^2}\\ \implies \frac{d^2y}{dx^2} &= -\frac{12x}{(4+x^2)^2}\end{align*} Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 08, 2016, 11:45:53 am 1. y=-3sin^-1 x For domain I got -1≤x≤1 For range I was not too sure??? Quick correction for above as well, the range of this function: $-\frac{3\pi}{2}\le3\sin^{-1}x\le\frac{3\pi}{2}$ We take the normal range for inverse sine and multiply everything by three ;D You also won't need to find the point of inflexion for that last one to do the question, the domain and range should be enough to do it properly ;D (though Rui has shown you how to do it if you want to) ;D Title: Re: 3U Maths Question Thread Post by: RuiAce on August 08, 2016, 12:02:41 pm hey, how do you do this guys? $\text{The normal to the parabola }x^2=4ay\text{ at the point }P(2ap,ap^2)\text{ is }\\ x+py=2ap+ap^3\\ \text{When }a=1\text{ we have }x+py=2p+p^3\\ \text{at }P(2p,p^2) \\ \text{The proof is left as an exercise to the reader.}$(http://i1164.photobucket.com/albums/q566/Rui_Tong/Screen%20Shot%202016-08-08%20at%2011.57.17%20AM_zps1rqgkwn9.png) $\text{Given }2PS=QS\text{, by the converse of Pythagoras' Theorem we wish to prove}\\ (2PS)^2+QS^2 = 5PS^2 = PQ^2$ $S\text{ is the point }(0,1)\\ \text{The distance }PS^2\text{ is equal to }(p^2-1)^2 - (2p)^2$ I have a lecture now so I will resume this later. I have set most of part (i) up in the meantime. _____________ $\text{Whilst I have the time, a proposed solution to find }p:\\ \text{Substitute the equation }x^2=4y\\ \text{into the equation of the tangent. There will be one and only one point of intersection.}\\ \text{This means that the resulting equation should have quadratic discriminant }0\\ \text{This should allow you to show that }p=2\\ \text{Note: The equation of the tangent is }y=px-ap^2=px-p^2$ $\text{If you know the value for }p\text{ then everything else will be easy.}$ I'll hand it over to you from here. Tell me if everything puzzles out eventually. Title: Re: 3U Maths Question Thread Post by: conic curve on August 08, 2016, 01:14:17 pm How would you do this for i and ii would you have to use your circle geometry theorems Title: Re: 3U Maths Question Thread Post by: RuiAce on August 08, 2016, 01:20:21 pm How would you do this for i and ii would you have to use your circle geometry theorems Theorems to use: i) Alternate segment theorem ii) Vertically opposite iii) Alternate segment theorem (in the OTHER circle) Base angles + matching sides on isosceles triangle iv) <ACT = <ART (angle standing on minor arc AT) <ACT = <TBC (base angles on isos ∆) Therefore <ART = <TBC Therefore <ART = ABR (vertically opposite) so you have a new isosceles triangle. Title: Re: 3U Maths Question Thread Post by: massive on August 08, 2016, 03:45:19 pm _____________ $\text{Whilst I have the time, a proposed solution to find }p:\\ \text{Substitute the equation }x^2=4y\\ \text{into the equation of the tangent. There will be one and only one point of intersection.}\\ \text{This means that the resulting equation should have quadratic discriminant }0\\ \text{This should allow you to show that }p=2\\ \text{Note: The equation of the tangent is }y=px-ap^2=px-p^2$ $\text{If you know the value for }p\text{ then everything else will be easy.}$ What is this magic :O?? Can/should you do this for all parametrics qs? wait i just realised, howdu get p=2 :S, the discriminant equals 0, then what?? Also howdu find the coordinates for q without using p=2? Title: Re: 3U Maths Question Thread Post by: RuiAce on August 08, 2016, 07:15:23 pm Yeah I had a play around with that and it did nothing but show 0=0 lol. Ok so: $\text{Solving }x^2=4y\text{ with }x+py=2p+p^3\\ \text{to find the point of intersection between the normal and the parabola:}$ \begin{align*}4x+px^2&=4p(2+p^2)\\ px^2+4x-4p(2+p^2)&=0 \\ x&=\frac{-4\pm \sqrt{16+16p^2(2+p^2}}{2p}\\ &= \frac{4\pm\sqrt{16(1+p^2)^2}}{2p}\\ &=\frac{4\pm 4(1+p^2)}{2p}\\ &=2p \text{ or }-\frac{2(p^2+2)}{p}\end{align*} $x=2p\text{ obviously corresponds to }y=p^2\\ \text{The other one corresponds to the point }Q\\ x=-\frac{2(p^2+2)}{p} \implies y=\frac{(p^2+2)^2}{p^2}\\ \text{upon substituting back into }x^2=4y$ $\text{From earlier }PS^2=p^4-6p^2+1$ $\text{Now, from }QS=2PS \implies QS^2=4PS^2\\ \left( \frac{(p^2+2)^2}{p^2} - 1\right)^2 + \left(-\frac{2(p^2+2)}{p}\right)^2 = 4\left((p^2-1)^2+4p^2 \right)\\ \text{According to WolframAlpha this should give you }p=\pm 2\\ \text{Then take the positive case since }P\text{ is in the first quadrant and carry through from there.}$ Title: Re: 3U Maths Question Thread Post by: massive on August 08, 2016, 08:54:05 pm hey guys howdu do part c, ive been trying for ages idk why i cant get it >:( Title: Re: 3U Maths Question Thread Post by: RuiAce on August 08, 2016, 09:06:38 pm hey guys howdu do part c, ive been trying for ages idk why i cant get it >:( $\text{For some reason I don't get exactly what they got}\\ \text{Using }pq=-1 \iff q=-\frac{1}{p}$ \begin{align*}PQ &=\sqrt{\left(2Ap-2Aq\right)^2+\left(Ap^2-Aq^2\right)^2}\\ &= A\sqrt{4(p-q)^2+\left(p^2-q^2\right)^2}\\ &=A\sqrt{(p-q)^2\left[4+(p+q)^2\right]}\\ &= A\sqrt{(p-q)^2(p^2+2pq+q^2+4)}\\ &= A\sqrt{(p-q)^2(p^2+2+q^2)}\\ &=A\sqrt{\left(p-\frac{1}{p}\right)^2\left(p^2+2+\frac{1}{p^2}\right)}\\ &=A\sqrt{\left(p-\frac{1}{p}\right)^2\left(p+\frac{1}{p}\right)^2}\\ &= A\sqrt{\left(p^2-\frac{1}{p}^2\right)^2}\\ &=A\left(p^2-\frac{1}{p^2}\right) \end{align*} Tell me if I did something wrong. Title: Re: 3U Maths Question Thread Post by: massive on August 09, 2016, 12:40:53 pm hey guys how do you do part b? Also, in general, how do you prove that a coefficient(s) is/are the greatest in an expansion. do you just have to proves the the term in front divided by that term have a value of less than one, thus it is the largest? Title: Re: 3U Maths Question Thread Post by: massive on August 09, 2016, 01:55:16 pm Also how do you do this?? (part ii) Title: Re: 3U Maths Question Thread Post by: RuiAce on August 09, 2016, 02:05:02 pm Also how do you do this?? (part ii) This is what I think: You can either include the digit in the group, or you cannot. There are 10 digits. You can say yes to that digit, or no to that digit. (And note there can't be repetition.) So that's a total of 210 outcomes. (But, if you have to have at LEAST one digit, you can't have no no no no no no no no no no) So in total 210 - 1 ______________ If you were to use part (i), the trick is that you want to find 10C1 + 10C2 + 10C3 + ... + 10C10 I.e. 210 MINUS 10C0 Cheers Jamon :) Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 09, 2016, 02:24:00 pm This is what I think: You can either include the digit in the group, or you cannot. There are 9 digits. You can say yes to that digit, or no to that digit. (And note there can't be repetition.) So that's a total of 29 outcomes. (But, if you have to have at LEAST one digit, you can't have no no no no no no no no no) So in total 29 - 1 ______________ If you were to use part (i), the trick is that you want to find 9C1 + 9C2 + 9C3 + ... + 9C1 I.e. 29 MINUS 9C0 Perfect, except there are 10 digits including 0, so that would be: $2^{10}-1$ hey guys how do you do part b? Also, in general, how do you prove that a coefficient(s) is/are the greatest in an expansion. do you just have to proves the the term in front divided by that term have a value of less than one, thus it is the largest? In general, that method is correct! The usual idea is to take the general form of the k-th coefficient, and the general form of the (k+1)th coefficient, and take the ratio (Tk)/T_(k+1)). As soon as this is greater than one you have the k-th coefficient as greater (remember k can only be a positive integer, no decimals) ;D Don't quite have time to punch out a full solution for (ii), however, I think it would be the exact same thing, except instead of considering the general form of coefficients, you'd consider the general form of TERMS! So: $T_k=C^n_k\times\left(\frac{1}{7}\right)^{n-k}\times\left(\frac{4}{7}\right)^k$ Do the same thing and find a ratio. There should be a value of k which makes it EQUAL to 1, and that will give your equal terms ;D Title: Re: 3U Maths Question Thread Post by: massive on August 09, 2016, 02:53:27 pm Thanks Jamon and Rui!!!! ;D Title: Re: 3U Maths Question Thread Post by: massive on August 09, 2016, 03:36:22 pm guys for this question how do you do it? i just cant figure out the last lines. Title: Re: 3U Maths Question Thread Post by: massive on August 09, 2016, 03:56:37 pm guys for this i got 124, but the answer is 9420??!?! Title: Re: 3U Maths Question Thread Post by: RuiAce on August 09, 2016, 04:06:23 pm guys for this i got 124, but the answer is 9420??!?! $\left(a^2+3a+2\right)^6=\left[(a+2)(a+1)\right]^6\\ \text{and we want the coefficient of }a^4\text{ in that.} \\ \text{Begin pairing terms.}$ \begin{align*}\text{We extract }\binom{6}{0}2^6\text{ from }(a+2)^6&\text{ and pair it with }\binom{6}{4}a^4\text{ from }(a+1)^6\\\text{We extract }\binom{6}{1}a(2^5)\text{ from }(a+2)^6&\text{ and pair it with }\binom{6}{3}a^3\text{ from }(a+1)^6\\\text{We extract }\binom{6}{2}a(2^4)\text{ from }(a+2)^6&\text{ and pair it with }\binom{6}{2}a^2\text{ from }(a+1)^6\\\text{We extract }\binom{6}{3}a^3(2^3)\text{ from }(a+2)^6&\text{ and pair it with }\binom{6}{1}a\text{ from }(a+1)^6 \\ \text{We extract }\binom{6}{2}a^2(2^2)\text{ from }(a+2)^6&\text{ and pair it with }\binom{6}{0}\text{ from }(a+1)^6\end{align*} $\text{So summing up the coefficients}:\\ \binom{6}{0}2^6\binom{6}{4}+\binom{6}{1}2^5\binom{6}{3}+\binom{6}{2}2^4\binom{6}{2}+\binom{6}{3}2^3\binom{6}{1}+\binom{6}{4}2^2\binom{6}{0}=9420$ Title: Re: 3U Maths Question Thread Post by: RuiAce on August 09, 2016, 04:15:44 pm guys for this question how do you do it? i just cant figure out the last lines. $\text{Note how our binomial coefficients go up by 2. This implies we need to address }(1+x)^n\text{ and }(1-x)^n\\ \text{You should remember this trick.}$ \begin{align*}(1+x)^{10}&=1+\binom{10}{1}x+\binom{10}{2}x^2+\binom{10}{3}x^3+\binom{10}{4}x^4+\dots+x^{10}\\(1-x)^{10}&=1-\binom{10}{1}x+\binom{10}{2}x^2-\binom{10}{3}x^3+\binom{10}{4}x^4-\dots+x^{10}\\ \therefore \frac{1}{2}\left[(1+x)^{10}+(1-x)^{10}\right]&=1+\binom{10}{2}x^2+\binom{10}{4}x^4+\dots+x^{10}\end{align*} \text{It should be more obvious now that we just have to put }x=3\\ \text{The RHS falls out immediately. As for the LHS}\\ \begin{align*}\frac{1}{2}\left[(1+3)^{10}+(1-3)^{10}\right]&= \frac{1}{2}(4^{10}+(-2)^{10})\\ &= \frac{1}{2}(2^{10}2^{10}+2^{10})\\ &= 2^{9}\left(2^{10}+1\right)\end{align*} Title: Re: 3U Maths Question Thread Post by: conic curve on August 09, 2016, 09:12:30 pm How would you do this question? This is from the cambridge extension questions part of the book so it might be out of syllabus Title: Re: 3U Maths Question Thread Post by: massive on August 09, 2016, 10:04:19 pm guys,how do you do part ii?? wait i think i got it; do you just expand and equate constants? Title: Re: 3U Maths Question Thread Post by: RuiAce on August 09, 2016, 10:19:54 pm guys,how do you do part ii?? wait i think i got it; do you just expand and equate constants? Yes. (Technically equating coefficients) I have a feeling you may need the symmetry property nCk = nCn-k though. Title: Re: 3U Maths Question Thread Post by: massive on August 09, 2016, 10:25:51 pm I have a feeling you may need the symmetry property nCk = nCn-k though. Whats this :O. Can you teach me what it is? (please) Title: Re: 3U Maths Question Thread Post by: massive on August 09, 2016, 10:31:36 pm Guys for part ii, i can see that u have to integrate, but why are there the negative and positive signs. If u let x= -1 you'll get that but opposite to what you have to prove :S Title: Re: 3U Maths Question Thread Post by: RuiAce on August 09, 2016, 10:34:32 pm Whats this :O. Can you teach me what it is? (please) $\text{This property is crucial. Haven't you seen it?}\\ \binom{n}{k}=\binom{n}{n-k}$ $\text{Proof:} RHS=\binom{n}{n-k}=\frac{n!}{(n-k)!(n-(n-k))!}=\frac{n!}{k!(n-k)!}=\binom{n}{k}$ $\textit{Also for that question it's much more convenient to just put the }x\textit{ back in there}\\ \text{and consider the coefficient of }x^n\text{ in }(1+x)^n(1+x)^n$ $\text{Or alternatively maybe }(1+x)^n(x+1)^n$ Title: Re: 3U Maths Question Thread Post by: RuiAce on August 09, 2016, 10:37:38 pm Guys for part ii, i can see that u have to integrate, but why are there the negative and positive signs. If u let x= -1 you'll get that but opposite to what you have to prove :S $\text{Nope. Let }x=-1\text{ in the expression you're given and the result in (i) falls out immediately.}$ $\text{Then for part ii, try the definite integral}\\ \int_{-1}^{0}\left(1+\binom{n}{1}x+\binom{n}{2}x^2+\dots + \binom{n}{n}x^n\right)=\int_{-1}^0 (1+x)^ndx$ Note that I use a definite integral to save me from the need of finding a constant of integration, and then just subbing in a new value for x Title: Re: 3U Maths Question Thread Post by: massive on August 09, 2016, 10:37:55 pm $\text{This property is crucial. Haven't you seen it?}\\ \binom{n}{k}=\binom{n}{n-k}$ $\text{Proof:} RHS=\binom{n}{n-k}=\frac{n!}{(n-k)!(n-(n-k))!}=\frac{n!}{k!(n-k)!}=\binom{n}{k}$ I think we just touched on it in class, but i'll keep it in mind from now on lol thanks! Title: Re: 3U Maths Question Thread Post by: massive on August 09, 2016, 10:45:38 pm $\text{Nope. Let }x=-1\text{ in the expression you're given and the result in (i) falls out immediately.}$ $\text{Then for part ii, try the definite integral}\\ \int_{-1}^{0}\left(1+\binom{n}{1}x+\binom{n}{2}x^2+\dots + \binom{n}{n}x^n\right)=\int_{-1}^0 (1+x)^ndx$ Note that I use a definite integral to save me from the need of finding a constant of integration, and then just subbing in a new value for x but how do you get the alternating positive and negative signs? for e.g. when u integrate, the second term is x^2, how can that be negative :S Title: Re: 3U Maths Question Thread Post by: RuiAce on August 09, 2016, 10:49:10 pm but how do you get the alternating positive and negative signs? for e.g. when u integrate, the second term is x^2, how can that be negative :S What do you mean? By putting -1 on the bottom you're doing F(0) MINUS F(-1), where F(-1) is the antiderivative of that expansion Title: Re: 3U Maths Question Thread Post by: massive on August 09, 2016, 10:51:42 pm What do you mean? By putting -1 on the bottom you're doing F(0) MINUS F(-1), where F(-1) is the antiderivative of that expansion Yeah but aren't all the signs positive after that? Title: Re: 3U Maths Question Thread Post by: RuiAce on August 09, 2016, 10:57:22 pm \begin{align*}&\quad\int_{-1}^0\left(1+\binom{n}{1}x+\binom{n}{2}x^2+\binom{n}{3}x^3+\dots+\binom{n}{n}x^n\right)dx\\ &= \left[x+\binom{n}{1}\frac{x^2}{2}+\binom{n}{2}\frac{x^3}{3}+\binom{n}{3}\frac{x^4}{4}+\dots+\binom{n}{n}\frac{x^{n+1}}{n+1}\right]_{-1}^0\\ &= 0 - \left[ (-1)+\binom{n}{1}\frac{(-1)^2}{2}+\binom{n}{2}\frac{(-1)^3}{3}+\binom{n}{3}\frac{(-1)^4}{4}+\dots+\binom{n}{n}\frac{(-1)^{n+1}}{n+1}\right] \\&= -\left[ -1+\binom{n}{1}\frac{1}{2}+\binom{n}{2}\frac{-1}{3}+\binom{n}{3}\frac{1}{4}+\dots+\binom{n}{n}\frac{(-1)^{n+1}}{n+1}\right]\\ &= 1-\binom{n}{1}\frac{1}{2}+\binom{n}{2}\frac{1}{3}-\binom{n}{3}\frac{1}{4}+\dots+\binom{n}{n}\frac{(-1)^n}{n+1}\end{align*} What do you mean they're "all" positive? By substituting in -1 and not +1 we force the signs to alternate. Title: Re: 3U Maths Question Thread Post by: massive on August 10, 2016, 12:40:25 am guys how do you do part ii? Title: Re: 3U Maths Question Thread Post by: massive on August 10, 2016, 01:03:13 am I don't know how to do this either :/ Title: Re: 3U Maths Question Thread Post by: jakesilove on August 10, 2016, 01:28:23 am guys how do you do part ii? I would genuinely just do it by guess and check. Here's my method for the whole thing! Sorry about the lack of maths, but it's late. (http://i.imgur.com/iKczRAM.png) Jake Title: Re: 3U Maths Question Thread Post by: massive on August 10, 2016, 01:40:26 am Thanks Jake! Guys I just have one more binomial question (hopefully) I don't know how to do both parts Thanks! Title: Re: 3U Maths Question Thread Post by: jakesilove on August 10, 2016, 01:52:24 am I don't know how to do this either :/ It's pretty goddamned late, so I really hope this is correct. (http://i.imgur.com/hPsYdey.png) Jake Title: Re: 3U Maths Question Thread Post by: jakesilove on August 10, 2016, 01:55:44 am Thanks Jake! Guys I just have one more binomial question (hopefully) I don't know how to do both parts Thanks! I'm not going to be able to have a crack at this one tonight, but by the looks of it you need to expand the binomial, set it equal to it's un-expanded form, then differentiate it. You'll either have to differentiate again, or just set x=1, I'm not entirely sure by the looks of it. Give it a go, show us your working out, and we can have another go in to morning! Title: Re: 3U Maths Question Thread Post by: RuiAce on August 10, 2016, 08:21:57 am Thanks Jake! Guys I just have one more binomial question (hopefully) I don't know how to do both parts Thanks! From memory this was in the 2004 HSC. Like previously said, please consult the answers before you post these up. THSC I'm not going to be able to have a crack at this one tonight, but by the looks of it you need to expand the binomial, set it equal to it's un-expanded form, then differentiate it. You'll either have to differentiate again, or just set x=1, I'm not entirely sure by the looks of it. Give it a go, show us your working out, and we can have another go in to morning! First part's differentiating. Second part is integrating though I'm not going to be able to have a crack at this one tonight, but by the looks of it you need to expand the binomial, set it equal to it's un-expanded form, then differentiate it. You'll either have to differentiate again, or just set x=1, I'm not entirely sure by the looks of it. Give it a go, show us your working out, and we can have another go in to morning! Yeah it's correct. Title: Re: 3U Maths Question Thread Post by: jamgoesbam on August 10, 2016, 05:04:24 pm Hi! Could someone please explain how to work out this multiple choice question? Thanks! (Answer: C) Title: Re: 3U Maths Question Thread Post by: RuiAce on August 10, 2016, 05:16:04 pm Hi! Could someone please explain how to work out this multiple choice question? Thanks! (Answer: C) $\text{We utilise the fact that the }\textbf{angle of projection}\text{ is exactly }0^\circ\\ \text{Note that }\cos 0^\circ = 1$ \text{For projectile A}\\\begin{align*} \ddot{x}&=0\\ \dot{x}&=10\\ x&=10t\end{align*} \text{For projectile B}\\\begin{align*} \ddot{x}=0\\ \dot{x}=20\\ x=20t\end{align*} $\text{Just by looking at that, regardless of the time of flight it is clear that projectile }B\\ext{travels exactly double the distance as projectile }A$ Title: Re: 3U Maths Question Thread Post by: massive on August 10, 2016, 05:57:23 pm This question looks simple but its pretty deadly, no one in my class got it, nor my teacher; help please! Title: Re: 3U Maths Question Thread Post by: RuiAce on August 10, 2016, 06:26:05 pm This question looks simple but its pretty deadly, no one in my class got it, nor my teacher; help please! I only know how to do this one cause I've seen it before, \text{FORCE the 2 and the 5 to bring out a 10:}\\ \begin{align*}2^x=10^z &\implies 2^{xy}=10^{yz} \tag{1}\\ 5^y = 10^z &\implies 5^{xy} = 10^{xz} \tag{2}\end{align*} \text{Multiply 1 and 2:}\\ \begin{align*}2^{xy}5^{xy}&=10^{yz}10^{xz}\\ 10^{xy}&=10^{xz+yz}\\ xy &= xz+yz\\ xy&=z(x+y)\\ \frac{1}{z}&=\frac{x+y}{xy}\\&=\frac{1}{y}+\frac{1}{x}\end{align*} Title: Re: 3U Maths Question Thread Post by: levendibigd on August 10, 2016, 06:44:38 pm Hey! Could I please have an explanation for solution for (a)(ii). I'm a bit confused as to why k=-6 Title: Re: 3U Maths Question Thread Post by: massive on August 10, 2016, 06:46:12 pm guys what is this question even? Title: Re: 3U Maths Question Thread Post by: RuiAce on August 10, 2016, 06:55:16 pm Hey! Could I please have an explanation for solution for (a)(ii). I'm a bit confused as to why k=-6 $\text{Take note that if you square anything, it's positive or equal to 0.}\\ \text{The reason why this method works is simply because if you don't have everything squared,}\\ \text{Something negative will definitely pop up.}\\ \text{This is because }Q(x)\text{ is in the form }(x-\alpha)(x-\beta)+C\text{ where }C = 0$ $\text{Fortunately, }(x-2)\text{ has been squared for us.}\\ \text{We need only introduce }(x+3)\text{ to be squared.}$ $\text{This can only be done if and only if the }\left(x-\frac{k}{2}\right)\text{ term is IDENTICAL to }(x+3)$ $\text{From }\left(x-\frac{k}{2}\right) \equiv (x+3)\\ \text{Equating constant terms: } -\frac{k}{2} = 3 \iff k = -6\\ \text{Note that we can only equate coefficients/constant terms IF we have an IDENTITY.}$ $\textit{Appendix: An identity is something that holds true for ALL }x\\ \text{An example is }2x+5 \equiv 2x+5 \equiv 2(x+1)+3$ Title: Re: 3U Maths Question Thread Post by: RuiAce on August 10, 2016, 07:09:49 pm guys what is this question even? $\text{Let the intermediate side have length }a\\ \text{Then the shorter side has length }a-d\\ \text{and the longer side has length }a+d$ $\text{Where }d\text{ is a common difference to satisfy the A.P.}$(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zps9e7ovz0j.png) $\text{Observe how the }\textbf{larger}\text{ angle is bound by the two }\textbf{shorter}\text{ sides.}\\ \text{In fact, the longest side is ALWAYS opposite to the largest angle, and similarly for the shortest.} \\ \text{This gives us a basis to use trigonometry.}$ $\text{Note that the length of the intermediate side does not matter. Instead}\\ \text{We must use the angle of }\frac{2\pi}{3}\text{ to somehow deduce an expression for the common difference.}$ \text{If we want to ensure we introduce no more than ONE angle, we use the cosine rule:}\\ \begin{align*}(a+d)^2&=a^2+(a-d)^2-2a(a-d)\cos \frac{2\pi}{3}\\ a^2+2ad+d^2&=a^2+a^2-2ad+d^2+a^2-ad\\ 5ad&=2a^2\\ d&=\frac{2a}{5}\end{align*}\\ \text{having safely cancelled out }a\text{ as }a\neq 0\text{, or we would have no triangle.} \text{Now that we have an expression for }d\text{, we can just throw it all into the sine rule.}\\ \begin{align*}\frac{\sin x}{a-\frac{2a}{5}}&=\frac{\sin 120^\circ}{a+\frac{2a}{5}}\\ x&=\sin^{-1}\left(\frac{3}{7}\sin 120^\circ\right)\\ &= 21.7867...^\circ \\ &\approx 22^\circ \end{align*} Title: Re: 3U Maths Question Thread Post by: conic curve on August 10, 2016, 07:52:25 pm The range for this is -3pi≤y≤0 The range for this is just the normal 0≤y≤pi The range for this is -3pi/2<y<3pi/2 im not really sure how you get point of inflexion (sorry) but im pretty sure its still at the origin for this. could you please explain how/why they are the range(s)? Title: Re: 3U Maths Question Thread Post by: RuiAce on August 10, 2016, 07:55:49 pm could you please explain how/why they are the range(s)? $-1\le \sin^{-1}x \le 1\\ -3 \le 3\sin^{-1}x\le 3\\ -3 \le y \le 3$ $\text{ and etc.}$ Title: Re: 3U Maths Question Thread Post by: levendibigd on August 10, 2016, 08:49:59 pm Could I please have some help with this mathematical induction Q Title: Re: 3U Maths Question Thread Post by: RuiAce on August 10, 2016, 08:58:39 pm Could I please have some help with this mathematical induction Q $\text{The base case }n=1\text{ is pretty straightforward as }2^3+(-1)^4=9=3(3)\\ \text{which is divisible by 3}$ $\text{We hypothesise that the statement holds when }n=k\\ 2^k+(-1)^{k+1}=3M\text{ for some integer }M$ $\text{And prove that it holds when }n=k+1\\ RTP: 2^{k+1}+(-1)^{k+2}\text{ is divisible by }3\\ \text{As always, we force out a means of using our assumption. Let's try:}\\ 2^k = 3M - (-1)^{k+1}$ \begin{align*}2^{k+1}+(-1)^{k+2}&=2(2^k)+(-1)^{k+2}\\ &= 2\left(3M-(-1)^{k+1}\right)+(-1)^{k+2}\tag{*}\\ &= 6M - 2(-1)^{k+1}+(-1)^{k+2} \end{align*} $\text{Utilise the fact that }(-1)^2 = 1=-1 \times -1\text{ to get }$ \begin{align*}2^{k+1}+(-1)^{k+2} &=6M + 2(-1)^k + (-1)^k \\ &= 6M+3(-1)^k\\ &= 3\left(2M+(-1)^k\right) \end{align*}\\ \text{which is divisible by 3.} $\text{Hence when it holds for }n=k\text{ it holds for }n=k+1\text{ hence true by induction.}$ Title: Re: 3U Maths Question Thread Post by: levendibigd on August 10, 2016, 09:59:22 pm Thank you, I really appreciate it! Title: Re: 3U Maths Question Thread Post by: massive on August 11, 2016, 12:24:45 am Guys, for part i- how do i show that a limiting sum exists (i got s=cos^2x) For part ii- how do you do it? Thanks! Title: Re: 3U Maths Question Thread Post by: massive on August 11, 2016, 01:44:44 am Guys i just don't get the last part, where is says "find the equations of the tangent lines to the curve..." Title: Re: 3U Maths Question Thread Post by: massive on August 11, 2016, 02:06:10 am Guys again for this I don't get the last part where it says to find the rate at which theta is changing. What i did was find x when theta=pi/4 and sub it into the eqn of part ii and i got 0.2 but the answer is 11 degrees/sec :S Title: Re: 3U Maths Question Thread Post by: RuiAce on August 11, 2016, 06:50:52 am Guys, for part i- how do i show that a limiting sum exists (i got s=cos^2x) For part ii- how do you do it? Thanks! $\text{A limiting sum exists provided that the common ratio }r\\ \text{satisfies }|r| < 1$ $\text{So we need }\tan^2 x < 1\\ \text{which is true for all }-45^\circ \le x \le 45^\circ$ $\text{Our domain is clearly in the answer above so we have a limiting sum.}$ Title: Re: 3U Maths Question Thread Post by: RuiAce on August 11, 2016, 06:57:01 am Guys i just don't get the last part, where is says "find the equations of the tangent lines to the curve..." $\text{Note that any line that passes through the origin will have form }y=mx$ $\text{The hyperbola is an example of a conic section.}\\ \text{We know that for a conic section, the tangent will touch the graph and then never meet it again.}\\ \text{So there can be at most one solution to the equation above. Hence, set the discriminant to equal 0.}$ \begin{align*}\Delta &= 0\\ [-(m+3)]^2-4(4)(m)&=0\\ m^2+6m+9-16m&=0\\ m^2-10m+9&=0\\ (m-9)(m-1)&=0\\ m&=1,9\end{align*} Title: Re: 3U Maths Question Thread Post by: RuiAce on August 11, 2016, 07:06:49 am Guys again for this I don't get the last part where it says to find the rate at which theta is changing. What i did was find x when theta=pi/4 and sub it into the eqn of part ii and i got 0.2 but the answer is 11 degrees/sec :S The answer wants it in degrees/sec. Just convert 0.2 rads into degrees by typing in 0.2 * 180/π Title: Re: 3U Maths Question Thread Post by: jamgoesbam on August 11, 2016, 11:33:00 am Hi again, I didn't quite understand part ii) of this question... how did the answer get ln? I thought when integrating (x^2+3)^-0.5 you could use the reverse chain rule so t = (x^2+3)^0.5/(0.5 x 2x). Could someone please explain? Thank you!! :) Title: Re: 3U Maths Question Thread Post by: jakesilove on August 11, 2016, 11:55:50 am Hi again, I didn't quite understand part ii) of this question... how did the answer get ln? I thought when integrating (x^2+3)^-0.5 you could use the reverse chain rule so t = (x^2+3)^0.5/(0.5 x 2x). Could someone please explain? Thank you!! :) Hey! The thing to note, when doing the reverse chain rule, is that you cannot do it when, inside the brackets, there is a non-linear function. So, you can integrate $y=(2+x)^3$ But you cannot integrate $y=(2+x^2)^3$ using the reverse chain rule. You have to use alternate means, such as the table of standard integrals (as is the case in this question). So, knowing that you can't actually integrate using the reverse chain rule in this case, does the answer described make sense? Like you understand their application of the table of standard integrals? Great question though! Jake Title: Re: 3U Maths Question Thread Post by: jamgoesbam on August 11, 2016, 12:03:28 pm Hey! The thing to note, when doing the reverse chain rule, is that you cannot do it when, inside the brackets, there is a non-linear function. So, you can integrate $y=(2+x)^3$ But you cannot integrate $y=(2+x^2)^3$ using the reverse chain rule. You have to use alternate means, such as the table of standard integrals (as is the case in this question). So, knowing that you can't actually integrate using the reverse chain rule in this case, does the answer described make sense? Like you understand their application of the table of standard integrals? Great question though! Jake Ohhhh, yes that's why they used ln!! I didn't know to do this as they don't provide us with the standard table of integrals anymore (since this was from a 2014 paper, before the formula sheet got introduced). But how would you do it without using the table of integrals? Title: Re: 3U Maths Question Thread Post by: jakesilove on August 11, 2016, 12:31:34 pm Ohhhh, yes that's why they used ln!! I didn't know to do this as they don't provide us with the standard table of integrals anymore (since this was from a 2014 paper, before the formula sheet got introduced). But how would you do it without using the table of integrals? Ahh okay yeah true, I forgot that you don't get that table anymore! In that case, you can't answer the question, so really don't worry about it :) They can't ask you to integrate that. Title: Re: 3U Maths Question Thread Post by: RuiAce on August 11, 2016, 12:40:25 pm Ohhhh, yes that's why they used ln!! I didn't know to do this as they don't provide us with the standard table of integrals anymore (since this was from a 2014 paper, before the formula sheet got introduced). But how would you do it without using the table of integrals? $\text{Can confirm. }\int \frac{dx}{\sqrt{x^2+a^2}}=\ln (x+\sqrt{x^2+a^2})\dots \text{ is no longer examinable.}$ Title: Re: 3U Maths Question Thread Post by: massive on August 11, 2016, 01:07:33 pm how do you do this q guys?? Title: Re: 3U Maths Question Thread Post by: massive on August 11, 2016, 01:45:59 pm guys how do you do part ii? Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 11, 2016, 05:55:12 pm guys how do you do part ii? Hey! I'm about to head to a tutorial, but have a go at just doing a straight differentiation and finding a maxima!! Remember, d is given, so L is what varies, so you are finding dV/dL and putting it equal to zero! That will give you a length of L in terms of d that maximises the volume (remember to check that it is a maxima) ;D if the answer doesn't come out, show me your working and I'll see if I can help you figure out what went wrong ;D how do you do this q guys?? This is tougher. Consider the girth first, it is the shortest distance around the parcel. And then L is the longest side. Based on the dimensions, this gives us two sets of possible equations. The first, if x is the longest side, gives us this: $G=2x+2y\\L+G=3x+2y=100$ Or, if y is the longest: $G=4x\\L+G=y+4x=100$ Whoops, have to go, see what you make of this ;) Title: Re: 3U Maths Question Thread Post by: jamgoesbam on August 11, 2016, 07:49:13 pm Hi again, Another question I couldn't quite grasp - part iv). Could someone please explain? Thank you!! Title: Re: 3U Maths Question Thread Post by: RuiAce on August 11, 2016, 07:54:55 pm Hi again, Another question I couldn't quite grasp - part iv). Could someone please explain? Thank you!! See post #429 here. I have already done this question. Title: Re: 3U Maths Question Thread Post by: massive on August 11, 2016, 10:23:04 pm guys how do you part ii :S? Title: Re: 3U Maths Question Thread Post by: RuiAce on August 11, 2016, 10:36:40 pm guys how do you part ii :S? \begin{align*}\frac{d\theta}{dt}&=\frac{d\theta}{dx}\,\frac{dx}{dt}\\ &= -\frac{\sin^2\frac{\pi}{4}}{0.9}\text{rad km}^{-1} \times 240\text{km hr}^{-1}\\ &= \frac{400}{3}\text{ rad hr}^{-1}\\ &= \frac{400}{3}\div 3600\text{ rad s}^{-1}\\ &= \frac{1}{27}\text{ rad s}^{-1}\end{align*} $\text{A bit mean for a maths question but you must ensure consistency in your units.}$ Title: Re: 3U Maths Question Thread Post by: jakesilove on August 11, 2016, 10:39:11 pm guys how do you part ii :S? These questions are all quite similar, so it is worth noting the steps I take so you can use them in an exam. $\frac{dx}{d\theta}=-\frac{0.9}{sin^2\theta}$ $\frac{dx}{d\theta}=\frac{dx}{dt}*\frac{dt}{d\theta}$ $\frac{dx}{dt}=\frac{1}{15}$ That last piece of information is given in the question, but as it is in kilometers per HOUR, not per SECOND, I divided by 3600. So, $-\frac{0.9}{sin^2\theta}=\frac{1}{15}*\frac{dt}{d\theta}$ $\frac{dt}{d\theta}=-\frac{0.9}{\frac{1}{15}*sin^2\theta}=-\frac{0.9}{\frac{1}{15}*sin^2(\pi/4)}=27$ Therefore $\frac{d\theta}{dt}=\frac{1}{27}$ Title: Re: 3U Maths Question Thread Post by: jakesilove on August 11, 2016, 10:40:08 pm \begin{align*}\frac{d\theta}{dt}&=\frac{d\theta}{dx}\,\frac{dx}{dt}\\ &= -\frac{\sin^2\frac{\pi}{4}}{0.9}\text{rad km}^{-1} \times 240\text{km hr}^{-1}\\ &= \frac{400}{3}\text{ rad hr}^{-1}\\ &= \frac{400}{3}\div 3600\text{ rad s}^{-1}\\ &= \frac{1}{27}\text{ rad s}^{-1}\end{align*} $\text{A bit mean for a maths question but you must ensure consistency in your units.}$ Dammit Title: Re: 3U Maths Question Thread Post by: massive on August 11, 2016, 10:43:53 pm woah i completely missed the km/HOUR, Thanks Rui and Jake!! Just another question, I can't get part iii, i keep getting 24 but the answer is 37 :/ Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 11, 2016, 11:49:08 pm Consider the girth first, it is the shortest distance around the parcel. And then L is the longest side. Based on the dimensions, this gives us two sets of possible equations. The first, if x is the longest side, gives us this: $G=2x+2y\\L+G=3x+2y=100$ Or, if y is the longest: $G=4x\\L+G=y+4x=100$ Alright, I left off here! Let me know if you need a hand with the other question, it is just differentiation and finding a maxima, give it a go! ;D So here, we have two potential equations to start with, let's see how they effect the formulae for volume (in both cases, we'll eliminate y): $y+4x=100 \implies V=x^2y=100x^2-4x^3 \\ 3x+2y=100 \implies V=x^2y=50x^2-\frac{3}{2}x^3$ When we do this, it becomes a little clearer that the top option gives a larger volume for any value of x. I'll leave you to form a mathematical proof with all this information, but that's the basic setup! To show you the idea of finding the maxima/minima, I'll work through this one for you: $V=100x^2-4x^3\\\frac{dV}{dx}=200x-12x^2\\\frac{d^2V}{dx^2}=200-24x\\\frac{dV}{dx}=0\implies x(200-12x)=0 \\\therefore x=0,\frac{50}{3} \\ x\neq0\quad \therefore x=\frac{50}{3} \\ x=\frac{50}{3} \implies \frac{d^2V}{dx^2}=200-400<0 \\\therefore\text{ volume is maximised at }x=\frac{50}{3}$ So the dimensions are (using the formulas above), 50/3 by 50/3 by 100/3 ;D Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 12, 2016, 12:22:31 am woah i completely missed the km/HOUR, Thanks Rui and Jake!! Just another question, I can't get part iii, i keep getting 24 but the answer is 37 :/ Hmm, let's have a look. The absolute best thing to do after that extra payment (unless you know EXACTLY what you are doing) is to hit reset and recalculate absolutely everything. By using the formula proved in Part 1, and then subtracting the additional 5000: $\text{Amount Owing}=20025.11$ From here, we need to re-apply the formula from above, and set the amount owing equal to zero. Do a quick rearrangement: $AR^n=M\left(\frac{R^n-1}{R-1}\right) \\R^n(A-\frac{M}{R-1})=\frac{-M}{R-1}\\R^n=\frac{-M}{(R-1)(A-\frac{M}{R-1})}=1.3178$ You can use logarithms to get the exact answer here, but it needs to be an integer so I will use guess and check, and indeed: $n=37\text{ Months}$ Hope that helps! ;D Title: Re: 3U Maths Question Thread Post by: massive on August 12, 2016, 12:42:21 am thnx Jamon! Title: Re: 3U Maths Question Thread Post by: conic curve on August 15, 2016, 08:10:39 pm How would you do this: 1. Let f(x)=sin^-1 x a. State the domain and rage of f(x) I got "all x" for this and for the range I got -pi/2 =<y=<pi/2 b. Find f'(x) and f''(x) f'(x)=(sin^-1 x)' Then I'm not sure of whatelse to do :( c. Is f'(x) undefined at x=1 and x=-1. Deduce the tangents to the curve at these points d. Show that the line y=x is a tangent to the curve y=f(x) at (0,0) e. Sketch y=f(x) and y=x f. For what value of m does the line y=mx cut y=f(x) at three points g. Investigate the concavity of the curve and find the coordinates of the point of inflextion Thanks guys ;D Title: 3U Maths Question Thread Post by: RuiAce on August 15, 2016, 08:19:33 pm How would you do this: 1. Let f(x)=sin^-1 x a. State the domain and rage of f(x) I got "all x" for this and for the range I got -pi/2 =<y=<pi/2 At this point, you need to read the textbook and make sure you understand what inverse sine is. I'm sorry to say this, but this is because your answer is wrong for a foundation level question. (In particular, the domain) Also, the derivative is also in the textbook Title: Re: 3U Maths Question Thread Post by: MysteryMarker on August 16, 2016, 05:56:42 pm Just came across a geometry question and was wondering if, just as with similar triangles their sides are in corresponding ratios, can this be applied to a parallelogram? Just curious. Cheers. Title: Re: 3U Maths Question Thread Post by: RuiAce on August 16, 2016, 06:00:34 pm Just came across a geometry question and was wondering if, just as with similar triangles their sides are in corresponding ratios, can this be applied to a parallelogram? Just curious. Cheers. Similar polygons with number of sides greater than 3 are not in the course. Hence, no. Also, keep in mind that you need more than all 4 sides to be in corresponding ratios. You also need a few ANGLES to be equal for a pair of quadrilaterals. (Of course, if you know it's a parallelogram you would only need two adjacent sides and one angle) (After all, if you take two parallelograms with sides in matching ratios, but then just "squash" one, well...) Title: Re: 3U Maths Question Thread Post by: MysteryMarker on August 16, 2016, 06:43:16 pm When a polynomial Px is divided by (x-2) and (x+1) the respective remainders are 5 and 8. What is the remainder when px is divided by (x-2)(x+1) Always been confused with these types of questions. Cheers. EDIT: Another question, find the exact value sin(cos-12/3 + tan-1-3/4) Title: Re: 3U Maths Question Thread Post by: Jakeybaby on August 16, 2016, 07:01:47 pm When a polynomial Px is divided by (x-2) and (x+1) the respective remainders are 5 and 8. What is the remainder when px is divided by (x-2)(x+1) Always been confused with these types of questions. Cheers. From the given information, we know that upon division of the polynomial by (x-2), the remainder is 5, therefore P(2) = 5. Also, division with (x+1) leaves 8, therefore P(-1) = 8. Using our general formula: P(x) = (x-1)(x+1)q(x)+Ax+B Substituing in the particular values for x, in this case, 2 and -1. P(-1) = 8, therefore: 8 = -A+B P(2) = 5 5 = 2A+B Solving the simultaneous equations, we get: A = -1 & B = 7 Therefore the remainder present upon the division by (x-2)(x+1) is -x + 7 Title: Re: 3U Maths Question Thread Post by: RuiAce on August 16, 2016, 07:14:57 pm When a polynomial Px is divided by (x-2) and (x+1) the respective remainders are 5 and 8. What is the remainder when px is divided by (x-2)(x+1) Always been confused with these types of questions. Cheers. EDIT: Another question, find the exact value sin(cos-12/3 + tan-1-3/4) $\text{We first expand out the compound angle and also convert anything negative into being positive}\\ \text{To convert from negative to positive, use the fact that tan inverse is an odd function and etc. }(\tan^{-1}(-x)=-\tan^{-1}x)$ \begin{align*} \sin \left(\cos^{-1}\frac{2}{3}+\tan^{-1}\frac{-3}{4}\right)&=\sin \left(\cos^{-1}\frac{2}{3}\right)\cos \left(\tan^{-1}\frac{-3}{4}\right)+\cos\left(\cos^{-1}\frac{2}{3}\right)\sin \left(\tan^{-1}\frac{-3}{4}\right)\\ &= \sin \left(\cos^{-1}\frac{2}{3}\right)\cos \left(\tan^{-1}\frac{3}{4}\right)-\frac{2}{3}\sin \left(\tan^{-1}\frac{3}{4}\right)\end{align*} $\text{This question is disgusting in that you have to draw out }\textbf{TWO}\text{ triangles to assist you.}$ $\text{Let }\theta = \tan^{-1}\frac{3}{4}\text{ so that }\tan \theta = \frac{3}{4}=\frac{\textit{opposite}}{\textit{adjacent}}\\ \text{Sketch out the relevant right angled triangle. The hypotenuse will have length }5\\ \text{So we know that }\sin \left(\tan^{-1}\frac{3}{5}\right) =\frac{2}{\sqrt{13}}\text{ and }\cos \left(\tan^{-1}\frac{4}{5}\right) = \frac{3}{\sqrt{13}}$ $\text{In a similar way, we can show that }\sin\left(\cos^{-1}\frac{2}{3} \right)=\sqrt{5}{3}$ $\text{Finally substitute everything back in.}$ Using our general formula: P(x) = (x-1)(x+1)q(x)+Ax+B Substituing in the particular values for x, in this case, 2 and -1. P(-1) = 8, therefore: 8 = -A+B Comment: Would like to mention that the "formula" is known as the "division transformation". :) Title: Re: 3U Maths Question Thread Post by: MysteryMarker on August 16, 2016, 07:41:44 pm From the given information, we know that upon division of the polynomial by (x-2), the remainder is 5, therefore P(2) = 5. Also, division with (x+1) leaves 8, therefore P(-1) = 8. Using our general formula: P(x) = (x-1)(x+1)q(x)+Ax+B Substituing in the particular values for x, in this case, 2 and -1. P(-1) = 8, therefore: 8 = -A+B P(2) = 5 5 = 2A+B Solving the simultaneous equations, we get: A = -1 & B = 7 Therefore the remainder present upon the division by (x-2)(x+1) is -x + 7 Ah, okay. But how do you know that the remainder for (x+1)(x-2) is in the form Ax + B Cheers. Title: Re: 3U Maths Question Thread Post by: jakesilove on August 16, 2016, 07:45:31 pm Ah, okay. But how do you know that the remainder for (x+1)(x-2) is in the form Ax + B Cheers. The remainder will always be in the form that has a power ONE LESS than the divisor So, if you divide by ax, the remainder will be a constant. If you divide by ax^2+bx+c, the remainder will be Ax+C (ie. x is one power lower than the divisor!). If you divide by x^n, the remainder will have a factor including x^(n-1) Title: Re: 3U Maths Question Thread Post by: RuiAce on August 16, 2016, 07:53:00 pm Ah, okay. But how do you know that the remainder for (x+1)(x-2) is in the form Ax + B Cheers. Furthering onto what Jake said now. $\text{The whole point of the polynomial division algorithm is that we want to reduce what's known as an }\\ \text{improper fraction, into what's a proper fraction. The definitions are similar to primary school definitions for integers.}$ $\text{An IMPROPER fraction is that of which the degree of the numerator is GREATER than or EQUAL to the denominator.}\\ \text{A PROPER fraction is that of which the degree of the numerator is LESS than the denominator's.}$ $\text{When we perform polynomial long division on a polynomial }P(x)\text{ with divisor }A(x)\text{ we get:}\\ \frac{P(x)}{A(x)}=Q(x)+\frac{R(x)}{A(x)}\\ \text{Where }Q(x)\text{ is the quotient and }R(x)\text{ is the remainder.}$ $\text{The remainder is of particular interest to mathematicians. People are indeed interested in how many times something goes in,}\\ \text{but the leftover has its own special properties.}\\ \textit{Theorem: The degree of the remainder is ALWAYS at MOST 1 LESS than that of the divisor.}$ $\text{Because the theorem decrees that it is at MOST, we need to cater for all cases.}\\ \text{Thus, our ARBITRARY remainder (without evaluating the coefficients) must be EXACTLY one less.}\\ \text{(Of course, sometimes the leading coefficient on the remainder is 0 - this is not impossible.)}$ $\text{An informal proof: Perform a division on a polynomial by }P(x)\text{ to a divisor of }A(x)\\ \text{You will keep working your way down, until at some point you have say }x^2\text{ in the divisor but you have to deal with }bx+c\\ \text{No value will go in anymore.}\\ \text{So the remaining }R(x)=bx+c\text{ DOES become your remainder.}\\ \text{Indeed, if your divisor has }x^3\text{ then when you're down to }ax^2+bx+c\text{ you're also screwed. And so on forth.}$ $\text{Naturally, if we take over the }A(x)\text{ we get}\\ P(x)=A(x)Q(x)+R(x)$ Title: Re: 3U Maths Question Thread Post by: Loki98 on August 16, 2016, 09:22:35 pm How would you do the following questions attached? Title: Re: 3U Maths Question Thread Post by: RuiAce on August 16, 2016, 10:01:24 pm How would you do the following questions attached? See post #429 here for the first one. The method for (iii) is given and alluded to for part (ii). I don't have the time to do lengthy questions right now. Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 16, 2016, 11:14:17 pm See post #429 here for the first one. The method for (iii) is given and alluded to for part (ii). I don't have the time to do lengthy questions right now. I do ;D For the first part of the question, we can equate the x and y displacements: $x=Ut\cos{\alpha}\quad y=Ut\sin{\alpha}-\frac{1}{2}gt^2 \\\therefore UT_1\cos{\alpha}=UT_1\sin{\alpha}-\frac{1}{2}gT_1^2\\U\cos{\alpha}=U\sin{\alpha}-\frac{1}{2}gT_1 \quad (T_1\neq0)\\\therefore T_1=\frac{2U}{g}(\sin{\alpha}-\cos{\alpha})$ Skipped a few steps of algebra at the end there, but that should give you the idea ;D Second bit, we set y=0: $0=UT_2\sin{\alpha}-\frac{1}{2}gT_2^2 \\T_2=\frac{2U}{g}\sin{\alpha}\quad(T_2\neq0) \\ T_2=\frac{T_1\sin{\alpha}}{\sin{\alpha}-\cos{\alpha}}\\ T_2=\frac{T_1}{1-\cot{\alpha}}$ That last line was dividing by sine alpha ;D The last one is a little tricky to spot, but consider a rearranged form of part (b): $\frac{T_2}{T_1}=\frac{1}{1-\cot{\alpha}}$ Now remember that T2 happens after T1! This means that the ratio of T2/T1 must be greater than 1, therefore: $\frac{1}{1-\cot{\alpha}}>1\quad(\alpha\neq\frac{\pi}{4})\\ 1-\cot{\alpha}>1+\cot^2{\alpha}-2\cot{\alpha} \\ \cot{\alpha}\left(\cot{\alpha}-1\right)<0$ Solving that quadratic on the open domain of 0 to pi/2, will give you the answer you need ;D Title: Re: 3U Maths Question Thread Post by: RuiAce on August 16, 2016, 11:21:06 pm Aha thanks Jamon :) $\text{Continuing on:}\\ \text{Note how }V\cos \beta\text{ and }U\cos \alpha\text{ both represent}\\ \text{the horizontal component of the velocity, just at }\textbf{different times}: t=0, t=T_1$ $\text{Because in projectile motion we know that the horizontal component of velocity is ALWAYS constant}\\ \text{the above expressions are just different ways of writing the same thing.}\\ \therefore V\cos \beta = U\cos \alpha$ $\text{Using }\alpha=2\beta\text{ we can get a double angle formula. This will be perfect to use the quadratic formula with.}$ \begin{align*}V\cos \beta &= U\cos 2\beta\\ &=U(2\cos^2\beta - 1)\\ 2U\cos^2\beta - V \cos \beta - U &=0 \\ \cos \beta &= \frac{V\pm \sqrt{V^2-4\times 2U \times -U}}{4U}\\ \beta &= \cos^{-1}\left(\frac{V+ \sqrt{V^2+8U^2}}{4U}\right)\end{align*}\\ \text{taking the positive case as the negative one gives an obtuse angle.} Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 16, 2016, 11:25:16 pm Aha thanks Jamon :) $\text{Continuing on:}\\ \text{Note how }V\cos \beta\text{ and }U\cos \alpha\text{ both represent}\\ \text{the horizontal component of the velocity, just at }\textbf{different times}: t=0, t=T_1$ $\text{Because in projectile motion we know that the horizontal component of velocity is ALWAYS constant}\\ \text{the above expressions are just different ways of writing the same thing.}\\ \therefore V\cos \beta = U\cos \alpha$ $\text{Using }\alpha=2\beta\text{ we can get a double angle formula. This will be perfect to use the quadratic formula with.}$ \begin{align*}V\cos \beta &= U\cos 2\beta\\ &=U(2\cos^2\beta - 1)\\ 2U\cos^2\beta - V \cos \beta - U &=0 \\ \cos \beta &= \frac{V\pm \sqrt{V^2-4\times 2U \times -U}}{4U}\\ \beta &= \cos^{-1}\left(\frac{V+ \sqrt{V^2+8U^2}}{4U}\right)\end{align*}\\ \text{taking the positive case as the negative one gives an obtuse angle.} Teamwork! Lucky I checked back in, I was about to write the quadratic formula bit ;) Title: Re: 3U Maths Question Thread Post by: MysteryMarker on August 17, 2016, 07:40:28 am If cosx = -3/5 for 0<x<pi then tan(x/2) = ? Title: Re: 3U Maths Question Thread Post by: RuiAce on August 17, 2016, 08:12:27 am If cosx = -3/5 for 0<x<pi then tan(x/2) = ? $\text{Let }t=\tan\frac{x}{2}\text{ so that }\cos x = \frac{1-t^2}{1+t^2}$ \text{Then it's literally just algebra.}\\ \begin{align*}\frac{1-t^2}{1+t^2}&=-\frac{3}{5}\\ 5-5t^2&=-3-3t^2\\ 8&=2t^2\\ t&=\pm 2\end{align*} $\text{Since }0 < x < \pi\text{ we have }0<\frac{x}{2}<\frac{\pi}{2}\\ \text{So }\frac{x}{2}\text{ belongs in the first quadrant.}\\ \text{Since tan is positive in the first quadrant we thus have }t=\tan \frac{x}{2}=2$ Title: Re: 3U Maths Question Thread Post by: imtrying on August 19, 2016, 09:01:45 pm Hey! I need a hand with this binomial theorem question (its 11f from the 2012 Ext1 HSC: I have done part i) with no issues, the answer is -1760, its just part ii) I'm really stuck on. Thanks, this website is saving my HSC :) Title: Re: 3U Maths Question Thread Post by: jakesilove on August 19, 2016, 09:23:30 pm Hey! I need a hand with this binomial theorem question (its 11f from the 2012 Ext1 HSC: I have done part i) with no issues, the answer is -1760, its just part ii) I'm really stuck on. Thanks, this website is saving my HSC :) Hey! I'm glad you're finding Atar Notes useful! I certainly wish I had it in my HSC. For a question like this, you sort of just brute force your way through until something nice pops out. See my working below! (http://i.imgur.com/anF2dn8.png) Title: Re: 3U Maths Question Thread Post by: imtrying on August 19, 2016, 09:28:33 pm Thank you! Title: Re: 3U Maths Question Thread Post by: conic curve on August 20, 2016, 09:46:42 pm At this point, you need to read the textbook and make sure you understand what inverse sine is. I'm sorry to say this, but this is because your answer is wrong for a foundation level question. (In particular, the domain) Also, the derivative is also in the textbook I got it now Anyways the question was let f(x)=sin^-1 x Following up on that question, the questions were c. Is f'(x) undefined at x=1 and x=-1. Deduce the tangents to the curve at these points d. Show that the line y=x is a tangent to the curve y=f(x) at (0,0) Could someone here please give me hints Thanks Title: Re: 3U Maths Question Thread Post by: RuiAce on August 20, 2016, 09:50:40 pm I got it now Anyways the question was let f(x)=sin^-1 x Following up on that question, the questions were c. Is f'(x) undefined at x=1 and x=-1. Deduce the tangents to the curve at these points d. Show that the line y=x is a tangent to the curve y=f(x) at (0,0) Could someone here please give me hints Thanks Two ways to analyse c), both of which lead to the answer "no". $\text{Method 1: Since }f^\prime(x)=\frac{1}{\sqrt{1-x^2}}\\\text{if you sub in }x=1\text{, you divide by }0\text{. Boom. So it can't exist}$ $\text{Method 2: }f(x)\text{ was never defined at an endpoint to begin with.}$ d) is just your ordinary 2U find the equation of the tangent to y=... at (point, point) just with sine inverse Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 21, 2016, 12:23:12 am $\text{Method 1: Since }f^\prime(x)=\frac{1}{\sqrt{1-x^2}}\\\text{if you sub in }x=1\text{, you divide by }0\text{. }\textbf{BOOM.} \text{So it can't exist}$ Lol'ed ;D Title: Re: 3U Maths Question Thread Post by: RuiAce on August 21, 2016, 02:46:36 pm Lol'ed ;D The ∞th commandment: Thou shall not divide by zero. Title: Re: 3U Maths Question Thread Post by: anotherworld2b on August 21, 2016, 11:42:41 pm I was wondering is there a way to always be able to graph an exponential function and finding the exponential equation from the graph? I am really struggling with exponential graphs in terms of drawing them gievn the equation and being unable to find the equation given the graph. The tranformations that can be possible are also being quite an obstacle for me. Please help :'( Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 22, 2016, 09:52:51 am I was wondering is there a way to always be able to graph an exponential function and finding the exponential equation from the graph? I am really struggling with exponential graphs in terms of drawing them gievn the equation and being unable to find the equation given the graph. The tranformations that can be possible are also being quite an obstacle for me. Please help :'( Perhaps a set of rough rules will help!! Consider the general form of an exponential function: $y=Ae^{Bx+C}+D$ (Note that it doesn't have to be the with an 'e', any number works fine, though if you have a negative number there you'll need to adjust this a little the base won't be - this requires complex analysis) - Higher values of A stretches the curve in the vertical direction. Really, the noticeable impact is that A is the y-intercept, but only if there is no constant C added to the exponent - Higher values of B means the exponent will increase at a faster rate, the right hand side of the exponential will increase its slope noticeably - C causes a horizontal shift to the left (for positive values) or right (for negative values) - D causes a vertical shift upwards (positive values) or downwards (negative values) These rules might seem very wish-washy, and that's because they are, they should only be a rough guide. You aim should be to gradually develop an intuition for what the different parts of the exponential do. Have a play with sketching different functions on Desmos or something similar, and try to notice the patterns in what you see, that is the best way to develop that intuition ;D Title: Re: 3U Maths Question Thread Post by: anotherworld2b on August 22, 2016, 09:39:45 pm Thank you I was also wondering how would you evalulate it as a single power and evaulate. Also does evaluate simply mean solve? Perhaps a set of rough rules will help!! Consider the general form of an exponential function: $y=Ae^{Bx+C}+D$ (Note that it doesn't have to be the with an 'e', any number works fine, though if you have a negative number there you'll need to adjust this a little the base won't be - this requires complex analysis) - Higher values of A stretches the curve in the vertical direction. Really, the noticeable impact is that A is the y-intercept, but only if there is no constant C added to the exponent - Higher values of B means the exponent will increase at a faster rate, the right hand side of the exponential will increase its slope noticeably - C causes a horizontal shift to the left (for positive values) or right (for negative values) - D causes a vertical shift upwards (positive values) or downwards (negative values) These rules might seem very wish-washy, and that's because they are, they should only be a rough guide. You aim should be to gradually develop an intuition for what the different parts of the exponential do. Have a play with sketching different functions on Desmos or something similar, and try to notice the patterns in what you see, that is the best way to develop that intuition ;D Title: Re: 3U Maths Question Thread Post by: RuiAce on August 22, 2016, 09:45:55 pm Thank you I was also wondering how would you evalulate it as a single power and evaulate. Also does evaluate simply mean solve? $\textbf{Evaluate }\text{means to compute a calculation and thus find its value.}\\ \textbf{Solve }\text{means to determine a value for an unknown. That is, solve is used in relation to equations.}\\ \text{E.g. Evaluate }2+2\text{ as opposed to Solve }x=2+2$ \text{For that question you just use your index laws over and over again.}\\ \begin{align*}\left(2^{-\frac{1}{2}}\right)^{-3}\div \left(2^{\frac{3}{4}}\right)^6&=2^{\frac{3}{2}}\div 2^{\frac{9}{2}}\\ &= 2^{-3}\\ &= \frac{1}{8}\end{align*} Note: This question belongs to the 2U section as nothing more is demanded on top of the index laws. Title: Re: 3U Maths Question Thread Post by: anotherworld2b on August 22, 2016, 10:01:57 pm Thank you rui ace I also tried these questions but i am not getting the right answers. I got 2+25 for a Title: Re: 3U Maths Question Thread Post by: jakesilove on August 22, 2016, 10:08:26 pm Thank you rui ace I also tried these questions but i am not getting the right answers. I got 2+25 for a For a) you should get 2 + 125. You've correctly square rooted the 4, but after you square root the 25 (to get 5) you need to put it to the power of 3 (resulting in 125). For b) you can use indices laws to multiply the indices in brackets, then convert to 9 to 3^2. Then, you can add powers when multiplying with the same base Title: Re: 3U Maths Question Thread Post by: anotherworld2b on August 22, 2016, 10:16:11 pm For the answer for a the sheet says its 2/125? Im still a bit confused for b i got (5^2 -3^4)÷ 3^1 For a) you should get 2 + 125. You've correctly square rooted the 4, but after you square root the 25 (to get 5) you need to put it to the power of 3 (resulting in 125). For b) you can use indices laws to multiply the indices in brackets, then convert to 9 to 3^2. Then, you can add powers when multiplying with the same base Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 22, 2016, 10:37:20 pm For the answer for a the sheet says its 2/125? Im still a bit confused for b i got (5^2 -3^4)÷ 3^1 The answer is definitely wrong there, unless the question is ;D For B, you are close! Just that 3^1 should only be dividing the 6^4, the 5^2 is unaffected (remember your order of operations) ;D Title: Re: 3U Maths Question Thread Post by: anotherworld2b on August 22, 2016, 10:49:38 pm I got 5^2 -3^2? But the answer is -2? I also tried this question but im not sure to do next Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 22, 2016, 11:12:57 pm I got 5^2 -3^2? But the answer is -2? I also tried this question but im not sure to do next Take a look at the quick map of how I do it, maybe you'll spot your mistake? $5^2-\left(3^\frac{2}{3}\right)^6\times9^\frac{-1}{2}=5^2-\frac{3^4}{3}=5^2-3^3=25-27=-2$ That second question is broken, as Rui pointed out above (and something I should have noticed given I did complex analysis last semester), a negative base to a fractional exponent is well beyond your scope ;D $(-1)^\frac{3}{5} = -0.31+0.95i \quad\textbf{(Requires Complex Analysis)}$ Title: Re: 3U Maths Question Thread Post by: RuiAce on August 23, 2016, 08:02:19 am @anotherworld2b Just to clear things up - Your textbook is full of typos. I think the answers they provide can be reached if all those TYPOs were amended. Title: Re: 3U Maths Question Thread Post by: anotherworld2b on August 23, 2016, 09:15:45 am I had a feeling some of the answers were wrong but I wasn't completely sure Thank you RuiAce, jamonwindeyer and jakesilove for your help :) @anotherworld2b Just to clear things up - Your textbook is full of typos. I think the answers they provide can be reached if all those TYPOs were amended. Title: Re: 3U Maths Question Thread Post by: FallonXay on August 25, 2016, 04:12:58 pm Hi, I'm having trouble with part b of this question. I've tried finding the cartesian equation of each projectile and equating them, but I keep getting a weird answer. (the textbook answer is x=11.5m and y=8.2m). I'm not sure what I'm doing wrong, help would be appreciated! Thanks! :) Title: Re: 3U Maths Question Thread Post by: RuiAce on August 25, 2016, 05:13:39 pm Hi, I'm having trouble with part b of this question. I've tried finding the cartesian equation of each projectile and equating them, but I keep getting a weird answer. (the textbook answer is x=11.5m and y=8.2m). I'm not sure what I'm doing wrong, help would be appreciated! Thanks! :) Note: Part a) looked fine. When I simulated the relevant Cartesian equations of motion on GeoGebra, I found the diagram that they provided was highly deceptive. Unless my equation was wrong, particle B literally soars right over particle A. When I plugged the values for the answer in, I found that particle A travels through the given coordinates, but not particle B. These are the equations I had for particle B. Tell me what you think. $x=30-10\sqrt{2}t\\ y=-5t^2+10\sqrt{2}t\\ y=-\frac{1}{40}(30-x)^2+(30-x)$ Title: Re: 3U Maths Question Thread Post by: FallonXay on August 25, 2016, 06:24:36 pm Note: Part a) looked fine. When I simulated the relevant Cartesian equations of motion on GeoGebra, I found the diagram that they provided was highly deceptive. Unless my equation was wrong, particle B literally soars right over particle A. When I plugged the values for the answer in, I found that particle A travels through the given coordinates, but not particle B. These are the equations I had for particle B. Tell me what you think. $x=30-10\sqrt{2}t\\ y=-5t^2+10\sqrt{2}t\\ y=-\frac{1}{40}(30-x)^2+(30-x)$ Yeah, I had similar equations but since the velocity is in the opposite direction [relative to the other projectile], acceleration is -10, velocity is -10 - 20sin(45) and therefore displacement is -5t^2 - 20 sin(45) t. (Just a sign difference). Title: Re: 3U Maths Question Thread Post by: RuiAce on August 25, 2016, 06:32:35 pm Yeah, I had similar equations but since the velocity is in the opposite direction [relative to the other projectile], acceleration is -10, velocity is -10 - 20sin(45) and therefore displacement is -5t^2 - 20 sin(45) t. (Just a sign difference). The direction of the velocity is flipped for the horizontal component but not the vertical. If when t=0, y'=-20sin(45), then we'd be implying that the projectile was being fired into the ground, not upwards. Title: Re: 3U Maths Question Thread Post by: FallonXay on August 25, 2016, 06:47:48 pm The direction of the velocity is flipped for the horizontal component but not the vertical. If when t=0, y'=-20sin(45), then we'd be implying that the projectile was being fired into the ground, not upwards. ohhh right, my bad Title: Re: 3U Maths Question Thread Post by: RuiAce on August 25, 2016, 07:09:02 pm Bear in mind maths in focus is full of mistakes. I might be wrong but I'm pretty sure I didn't do anything stupid here (if anyone would like to check, please do). Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 25, 2016, 10:48:43 pm Bear in mind maths in focus is full of mistakes. I might be wrong but I'm pretty sure I didn't do anything stupid here (if anyone would like to check, please do). Cast my eye, see no mistakes ;D Title: Re: 3U Maths Question Thread Post by: Mei2016 on August 27, 2016, 11:49:15 pm Hi, I've only just recently started doing some binomial questions. In a textbook, I got through 8.1 fine, but got stuck on a few 8.2 questions. (The first two questions are from a textbook). Also, I found that I could do some of the past paper questions on binomial theorem, but the last two pics are 2 questions which I didn't how to approach. Thanks. Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 28, 2016, 07:42:07 am Hi, I've only just recently started doing some binomial questions. In a textbook, I got through 8.1 fine, but got stuck on a few 8.2 questions. (The first two questions are from a textbook). Also, I found that I could do some of the past paper questions on binomial theorem, but the last two pics are 2 questions which I didn't how to approach. Thanks. Hey there! I'll give you a bit of a hand! ;D I'm a tad short on time so I'll leave your first question, it has a bit of messy algebra, if no-one has jumped in by this evening I'll come back! Quite a few of your other questions are very similar, and just require equating coefficients for binomial identities. Let me use the 1996 question as an example. For the result we need to prove, the RHS is the coefficient of x^4 on the right hand side of the result we are given (did you spot this?) ;D So, let's see what the coefficient of x^4 is on the LHS: $LHS = \left(\binom{9}{0}+\binom{9}{1}x+...+\binom{9}{9}x^9\right)\left(\binom{4}{0}+\binom{4}{1}x+...+\binom{4}{4}x^4\right)$ Notice that we need pairs of term that give a total power of x^4, yielding: $\binom{9}{0}\binom{4}{4}+\binom{9}{1}\binom{4}{3}...$ I think you see where I'm heading, it gives the LHS of the result, and we are done! ;D Many questions require this method of thinking, it is what is required in your second textbook question. Oops, gotta go (train), however, for your 1992 question: i) x=-1 might work ;) ii) Hmmm, I wonder what integrating the given expression might do... Title: Re: 3U Maths Question Thread Post by: Spencerr on August 28, 2016, 01:38:03 pm For the last one ii) the binomial identity, i've tried integrating, factoring out the x then subbing in x equals -1 and it gives you the lhs but not the rhs :/ Will work on this further But wait! After integrating and subbing in x=0 you get the entire lhs equals to 0 and the rhs equalling 1/n +1 Summing the two would give the final identity. I think I might have broken a maths law haha could someone check? Title: 3U Maths Question Thread Post by: RuiAce on August 28, 2016, 01:48:43 pm For the last one ii) the binomial identity, i've tried integrating, factoring out the x then subbing in x equals -1 and it gives you the lhs but not the rhs :/ Will work on this further But wait! After integrating and subbing in x=0 you get the entire lhs equals to 0 and the rhs equalling 1/n +1 Summing the two would give the final identity. I think I might have broken a maths law haha could someone check? $\text{You can be clever and do it using definite integrals.}\\ \int_{-1}^0 \left[\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\dots+\binom{n}{n}x^n\right]dx=\int_{-1}^0 (1+x)^ndx$ Not too sure how your method broke. I need to test it out. Edit it works so not sure what you meant (http://uploads.tapatalk-cdn.com/20160828/70437dd8f80917746ea00b5c6d2bc605.jpg) Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 28, 2016, 02:36:01 pm For the last one ii) the binomial identity, i've tried integrating, factoring out the x then subbing in x equals -1 and it gives you the lhs but not the rhs :/ Will work on this further But wait! After integrating and subbing in x=0 you get the entire lhs equals to 0 and the rhs equalling 1/n +1 Summing the two would give the final identity. I think I might have broken a maths law haha could someone check? Yeah Rui is right, no math laws broken, you are all sweet ;) Title: Re: 3U Maths Question Thread Post by: Spencerr on August 28, 2016, 03:16:20 pm Ah yes definite integrals gives the answer. I didn't consider the constant term when I indefinitely integrated it. All good :) Title: Re: 3U Maths Question Thread Post by: Mei2016 on August 28, 2016, 10:09:02 pm Thanks Jamon, Spencerr and Rui for you help! For the 1992 q, for the definite integral, how would you know to use 0 and -1 or does any combo of 1/-1/0 still work fine? For the 1996 q, is the pic below the kind of working out required for the 2 marks of the questions? Thanks. Title: Re: 3U Maths Question Thread Post by: RuiAce on August 28, 2016, 10:28:39 pm Thanks Jamon, Spencerr and Rui for you help! For the 1992 q, for the definite integral, how would you know to use 0 and -1 or does any combo of 1/-1/0 still work fine? For the 1996 q, is the pic below the kind of working out required for the 2 marks of the questions? Thanks. Just quickly regarding the definite integral question (I'm busy right now) You can use an indefinite integral. But using anything that's not 0 to evaluate your +C is never a good idea. -1 is then chosen because the signs alternate! Title: Re: 3U Maths Question Thread Post by: Mei2016 on August 28, 2016, 10:35:44 pm So from the textbook questions, Q9 can be solved in a similar way to the 1996 question, but how would you solve Q7? Also, again in the pic is the working to Q9. Is this how you would set out the working to 'show' the given equation? Title: Re: 3U Maths Question Thread Post by: Mei2016 on August 28, 2016, 10:45:21 pm Hi Rui, thanks for answering. I understand why we use x=0 and x=-1 for the indefinite method, but previously, I couldn't see why you had chosen these from a quick glance at the question with the definite method, because I'm used to using to definite method with resisted motion questions where the limits of the integration are given with the initial and final conditions. I really like this method because it's really quick :D Thanks for your time. Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on August 28, 2016, 11:24:09 pm So from the textbook questions, Q9 can be solved in a similar way to the 1996 question, but how would you solve Q7? Also, again in the pic is the working to Q9. Is this how you would set out the working to 'show' the given equation? Hey! That working is definitely sufficient for the marks, a little messy in the middle, you can just not worry about that box if you don't want to, it's obvious what you are up to ;D That question you posted has me a bit stumped. We consider two successive coefficients (the two and three's come from the terms inside the bracket): $C_k=C_{k+1} \\\binom{n}{k}2^{n-k}3^k=\binom{n}{k+1}2^{n-k-1}3^{k+1}\\\frac{n!2^{n-k}3^k}{(n-k)!k!} = \frac{n!2^{n-k-1}3^{k+1}}{(n-k-1)!(k+1)!} \\\frac{2}{n-k}=\frac{3}{k+1}$ Perhaps I'm a bit tired, I can't see how to proceed from here! Do we even have enough info to answer? Happy for someone to tag in ;) Title: Re: 3U Maths Question Thread Post by: RuiAce on August 29, 2016, 12:02:50 am Ok I'm back in action Thanks Jamon, Spencerr and Rui for you help! For the 1992 q, for the definite integral, how would you know to use 0 and -1 or does any combo of 1/-1/0 still work fine? For the 1996 q, is the pic below the kind of working out required for the 2 marks of the questions? Thanks. For your equating coefficients thing, you can just say that the coefficient of xsomething in the expansion of ... is ... You don't have to explicitly expand (1+x)4 or (1+x)13 Title: Re: 3U Maths Question Thread Post by: RuiAce on August 29, 2016, 12:03:46 am Hi Rui, thanks for answering. I understand why we use x=0 and x=-1 for the indefinite method, but previously, I couldn't see why you had chosen these from a quick glance at the question with the definite method, because I'm used to using to definite method with resisted motion questions where the limits of the integration are given with the initial and final conditions. I really like this method because it's really quick :D Thanks for your time. Just pick 0. Usually if your thing is tidy then 0 is involved. But yeah -1 was like I said, because the signs alternated. Sorry but can you mention what year and exactly which question which part for anything else you wish for help with? Title: Re: 3U Maths Question Thread Post by: amandali on August 31, 2016, 03:17:13 pm (http://uploads.tapatalk-cdn.com/20160830/f7ab2bd6fd855e696d342cbb6152427e.jpg) how to do part b ii) Title: Re: 3U Maths Question Thread Post by: jakesilove on August 31, 2016, 03:26:24 pm (http://uploads.tapatalk-cdn.com/20160830/f7ab2bd6fd855e696d342cbb6152427e.jpg) how to do part b ii) Hey! My solution is below :) (http://i.imgur.com/K2pjoxE.png) Jake Title: Re: 3U Maths Question Thread Post by: RuiAce on August 31, 2016, 03:42:27 pm Hey! My solution is below :) (http://i.imgur.com/K2pjoxE.png) Jake Just check your working out again. Your answer gets A > (3√3) / 2 The correct reasoning is that because (2A+3√3)/(3√3) is strictly positive, the other must be positive as well. @amandali if this question is hard to understand I can offer up a GeoGebra simulation if you need it later on. I'm in a lecture now to give a comprehensive solution. Title: Re: 3U Maths Question Thread Post by: RuiAce on August 31, 2016, 07:13:51 pm $\text{Here's the full analysis for 2005 MX1 Q7 b)}\\ \text{Firstly note that the leading coefficient is }A\text{. Since }A\text{ is strictly positive}\\ \text{we know that the parabola ultimately goes from negative infinity, to positive infinity.}$ $\text{i.e. }\lim_{x\to -\infty}Ax^3-Ax+1=-\infty\text{ and }\lim_{x\to \infty}Ax^3-Ax+1=\infty\\ \text{Graphically, it just shows you which way the cubic goes.}$(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpsco1hzpaf.png) $\text{We know that the cubic }\textbf{must have two stationary points}.\\ \text{If a cubic has two stationary points, one must be a local max and the other a local min.}$ $\text{But because the leading coefficient is positive, the left one must be the min and the right one the max! (As shown.)}$ $\text{The point now, is that the stationary point makes the curve around.}\\ \textbf{Effectively, we don't want the curve to turn around and cross the }x\textbf{-axis again!}\\ \text{Otherwise, we would have more than one zero.}$ $\text{As Jake showed, the local maximum }\left(\frac{1}{\sqrt{3}},\frac{2A+3\sqrt{3}}{3\sqrt{3}}\right)\text{ is always above the }x\text{-axis.}\\ \text{So we just have to ensure that the local minimum is also above the }x\text{-axis.}$ $\text{Solving }\frac{-2A+3\sqrt{3}}{3\sqrt{3}}>0 \text{ gives }A<\frac{3\sqrt{3}}{2}$ If you play around with GeoGebra you'll see what we're trying to do. Notice how as A gets larger, the stationary points seemingly poke outwards more. We are basically telling it to not poke over the x-axis. Title: Re: 3U Maths Question Thread Post by: jakesilove on September 01, 2016, 10:22:07 am $\text{Here's the full analysis for 2005 MX1 Q7 b)}\\ \text{Firstly note that the leading coefficient is }A\text{. Since }A\text{ is strictly positive}\\ \text{we know that the parabola ultimately goes from negative infinity, to positive infinity.}$ $\text{i.e. }\lim_{x\to -\infty}Ax^3-Ax+1=-\infty\text{ and }\lim_{x\to \infty}Ax^3-Ax+1=\infty\\ \text{Graphically, it just shows you which way the cubic goes.}$(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpsco1hzpaf.png) $\text{We know that the cubic }\textbf{must have two stationary points}.\\ \text{If a cubic has two stationary points, one must be a local max and the other a local min.}$ $\text{But because the leading coefficient is positive, the left one must be the min and the right one the max! (As shown.)}$ $\text{The point now, is that the stationary point makes the curve around.}\\ \textbf{Effectively, we don't want the curve to turn around and cross the }x\textbf{-axis again!}\\ \text{Otherwise, we would have more than one zero.}$ $\text{As Jake showed, the local maximum }\left(\frac{1}{\sqrt{3}},\frac{2A+3\sqrt{3}}{3\sqrt{3}}\right)\text{ is always above the }x\text{-axis.}\\ \text{So we just have to ensure that the local minimum is also above the }x\text{-axis.}$ $\text{Solving }\frac{-2A+3\sqrt{3}}{3\sqrt{3}}>0 \text{ gives }A<\frac{3\sqrt{3}}{2}$ If you play around with GeoGebra you'll see what we're trying to do. Notice how as A gets larger, the stationary points seemingly poke outwards more. We are basically telling it to not poke over the x-axis. Truuuueeee I didn't actually look at the question (noting that it was a cubic), and then just basically worked backwards from what I expected. Thanks Rui! Title: Re: 3U Maths Question Thread Post by: anotherworld2b on September 02, 2016, 07:06:38 pm I was wondering how do you do these two questions? Title: Re: 3U Maths Question Thread Post by: jakesilove on September 02, 2016, 07:10:16 pm I was wondering how do you do these two questions? Hey! Just for everyone's else sake, THIS IS NOT HSC! Whilst I have done matrices, I don't really understand the terminology used, so won't be able to help you out here. Sorry! Jake Title: Re: 3U Maths Question Thread Post by: anotherworld2b on September 02, 2016, 10:48:48 pm oh okay :) thanks anyway Hey! Just for everyone's else sake, THIS IS NOT HSC! Whilst I have done matrices, I don't really understand the terminology used, so won't be able to help you out here. Sorry! Jake Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on September 02, 2016, 11:32:12 pm I was wondering how do you do these two questions? I can handle this one ;D First, a definition a shear, in case you need it. A shear parallel to a given line through the origin is a transformation in which the component of a vector perpendicular to the line is unchanged, and the component parallel to the line is increased by an amount proportional to the perpendicular component. If the proportionality constant is k, we call it a k–shear. Okay, so what you are looking at here is a linear transformation that applies both a shear, and a rotation, on the cartesian plane. Recall that we can represent any linear transformation with a matrix multiplication, we just need to find a matrix A such that: $T(\vec{x})=A\vec{x} \quad\forall \vec{x}\in\mathbb{R}^2$ How do we do this? Fortunately, we can actually get each column of the matrix A by considering the transformation of the standard basis vectors in the domain, \(\binom{1}{0}, \binom{0}{1}. The results in the co-domain will give us the columns of matrix A.

I'll do the first one slow. First, apply the shear to $$\binom{1}{0}$$:

$\binom{1}{0}\implies\binom{1}{3}$

Now, apply the rotation:

$\binom{1}{3}\implies\binom{3}{-1}$

Let me know if you need help with those individual parts of this question! So, the two vectors we need are:

$T\binom{1}{0}=\binom{3}{-1}\\T\binom{0}{1}=\binom{1}{0}$

So the matrix is formed with these as its columns:

$A=\begin{bmatrix}1 & 0\\-3 & 1\end{bmatrix}$

There is a fair bit of decently complex theory at play here, let me know if you need anything clarified! :)

I'll leave you to tackle the second one, same principle, it is just applying the rotation first!
Title: Re: 3U Maths Question Thread
Post by: anotherworld2b on September 03, 2016, 01:13:54 am
For this question im not sure how to start it
Title: Re: 3U Maths Question Thread
Post by: jamonwindeyer on September 03, 2016, 02:57:06 am
For this question im not sure how to start it

Hey! You'll form two sets of simultaneous, do the matrix multiplication on the LHS:

$\begin{bmatrix}a+2b & c+2d\\-3a+b&-3c+d\end{bmatrix}=\begin{bmatrix}12&-1\\7&0 \end{bmatrix}$

Equate corresponding matrix elements to obtain two sets of simultaneous equations, see how that treats you ;)

Disclaimer: Again, non HSC related question

In fact, another world, post your math questions in the relevant board here from now on, I'll be sure to watch it, just to keep this thread for HSC stuff to prevent confusion ;D
Title: Re: 3U Maths Question Thread
Post by: RuiAce on September 03, 2016, 06:31:59 am
For this question im not sure how to start it
Matrices are in no way a part of the HSC course.

I don't mind questions from students outside of NSW however please refer to the syllabus for the HSC before posting questions here.
I can handle this one ;D

First, a definition a shear, in case you need it. A shear parallel to a given line through the origin is a transformation in which the component of a vector perpendicular to the line is unchanged, and the component parallel to the line is increased by an amount proportional to the perpendicular component. If the proportionality constant is k, we call it a k–shear.

How did you even know this anyway  :o

For this question im not sure how to start it

You guys were taught matrix multiplication right? Or does Jamon's answer need clarity
Title: Re: 3U Maths Question Thread
Post by: anotherworld2b on September 03, 2016, 08:16:28 am
Sorry for the inconvenience caused.i wasnt wure where to post my questions
Title: Re: 3U Maths Question Thread
Post by: RuiAce on September 03, 2016, 09:33:23 am
Sorry for the inconvenience caused.i wasnt wure where to post my questions
All good

There are threads for every state to post their questions. But I understand perfectly if they feel isolated or hard to find
Title: Re: 3U Maths Question Thread
Post by: anotherworld2b on September 03, 2016, 10:09:46 am
I tried to make a new topic for a while but all my photos of questions keep failing security checks so i cant post it?
Title: Re: 3U Maths Question Thread
Post by: jamonwindeyer on September 03, 2016, 10:31:41 am
How did you even know this anyway  :o

Actually did an example on shears in a lecture last week, it was timely, I'm doing linear transformations right now ;) (all revision from Math 1B)
Title: Re: 3U Maths Question Thread
Post by: jakesilove on September 04, 2016, 12:01:38 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: 3U Maths Question Thread
Post by: amandali on September 08, 2016, 10:45:18 am

how to do part b ii) a and b
Title: Re: 3U Maths Question Thread
Post by: jakesilove on September 08, 2016, 11:18:29 am

how to do part b ii) a and b

For a), we know that the amount of oil being poured into the hemisphere is

$\frac{\pi}{3} m^3/min$

Therefore, after 8 minutes, there will be a volume of

$8*\frac{\pi}{3} m^3$

Setting this as V,

$8\frac{\pi}{3}=\frac{\pi}{3}(9h^2-h^3)$
$h^3-9h^2+8=0$

Just looking at the equation, it's pretty clear that the solution is h=1m

For b), we need to start by finding the rate of change equation.

$\frac{dh}{dt}=\frac{dh}{dV}*\frac{dV}{dt}$

This should be intuitive by now; set what you want to find on the left hand side, then use what you can find out to produce a right hand side. We know the change in volume over time; we just need to change h over V.

$\frac{dV}{dh}=\frac{\pi}{3}(18h-3h^2)$

$\frac{dh}{dt}=(\frac{\pi}{3}*(18h-3h^2))^{-1}*\frac{\pi}{3}$

I'll leave you to sub the relevant numbers in
Title: Re: 3U Maths Question Thread
Post by: Ali_Abbas on September 08, 2016, 11:31:24 am
For a), we know that the amount of oil being poured into the hemisphere is

$\frac{\pi}{3} m^3/min$

Therefore, after 8 minutes, there will be a volume of

$8*\frac{\pi}{3} m^3$

Setting this as V,

$8\frac{\pi}{3}=\frac{\pi}{3}(9h^2-h^3)$
$h^3-9h^2+8=0$

Just looking at the equation, it's pretty clear that the solution is h=1m

For b), we need to start by finding the rate of change equation.

$\frac{dh}{dt}=\frac{dh}{dV}*\frac{dV}{dt}$

This should be intuitive by now; set what you want to find on the left hand side, then use what you can find out to produce a right hand side. We know the change in volume over time; we just need to change h over V.

$\frac{dV}{dh}=\frac{\pi}{3}(18h-3h^2)$

$\frac{dh}{dt}=\frac{dh}{dV}*\frac{dV}{dt}=(\frac{\pi}{3}(18h-3h^2))^{-1)*\frac{\pi}{3}$

I'll leave you to sub the relevant numbers int

Remark: The resultant degree-3 polynomial in h will, by a consequence of the Fundamental Theorem of Algebra, yield exactly three roots. Although h = 1 was obtained as a solution by inspection, there was no justification for why the remaining two roots are to be rejected. Of course, the simple answer is that intuitively they must not be fitting to the physical context of our scenario. However, it is my opinion that this should be shown. We demonstrate it as follows:

$h^{3} - 9h^{2} + 8 = 0$ $\Longrightarrow h^{3} - 1 - 9h^{2} + 9 = 0$ $(h-1)(h^{2}+h+1) - 9(h-1)(h+1) = 0$ $\Longrightarrow (h-1)(h^{2}-8h-8) = 0$ $\Longrightarrow h = 1 \space or \space h = \frac{8 \pm \sqrt{96}}{2}$ Clearly, one root is negative and the other is greater than three (which is the radius of the semi-sphere). Thus, h = 1 is the only answer which is physically possible.

QED

Note: The Fundamental Theorem of Algebra belongs exclusively to the 4U course. But yes, justification of why certain roots are to be rejected is important.

(Also if the factorisation method is confusing for a 3U student reading this, you can do it by long division.)
Title: Re: 3U Maths Question Thread
Post by: amandali on September 09, 2016, 07:02:43 am

how to do part a  iii)
Title: Re: 3U Maths Question Thread
Post by: RuiAce on September 09, 2016, 06:01:29 pm

how to do part a  iii)
That question doesn't make sense to any of us. Mind telling us where you got it from?
Title: Re: 3U Maths Question Thread
Post by: katherine123 on September 13, 2016, 01:31:12 pm
In each game of chess that Bobby plays against Boris there is a probability of 1/3 that Bobby wins, a prob. of 1/6 that Boris wins, and prob. of 1/2 that the game is drawn. They play 4 games.
i) find prob. that Bobby wins 2 games and Boris wins 2 games
ii) find prob. that Bobby wins 1 game, Boris wins 1 game and the other 2 games are drawn.

i) 4C2(1/3)^2(1/6)^2
ii)4C2(1/2)^2(1/3)(1/6) *2

i get the answer for i) but i dont understand why there is a need to multiply 2 for part ii)

ques2 :
Consider x^2=4ay with focus S. The normal at P(2ap, ap^2) meets y axis at R and triangle SPR is equilateral
i) normal at P is x+py=2ap+ap^3
ii) R(0, a(2+p^2)
iii) prove SP is equal in length to latus rectum, that is 4a units

not sure how to do part iii)
Title: Re: 3U Maths Question Thread
Post by: jamonwindeyer on September 13, 2016, 02:27:21 pm
In each game of chess that Bobby plays against Boris there is a probability of 1/3 that Bobby wins, a prob. of 1/6 that Boris wins, and prob. of 1/2 that the game is drawn. They play 4 games.
i) find prob. that Bobby wins 2 games and Boris wins 2 games
ii) find prob. that Bobby wins 1 game, Boris wins 1 game and the other 2 games are drawn.

i) 4C2(1/3)^2(1/6)^2
ii)4C2(1/2)^2(1/3)(1/6) *2

i get the answer for i) but i dont understand why there is a need to multiply 2 for part ii)

Hey! So the answer without that extra multiplication accounts for selecting which games are drawn/won, then the probability of these occurring. We multiply by two to account for the fact that the two games won by Boris and Bobby can be swapped, as in, once we've chosen which games they'll win, there are then two different allocations to those games :)

ques2 :
Consider x^2=4ay with focus S. The normal at P(2ap, ap^2) meets y axis at R and triangle SPR is equilateral
i) normal at P is x+py=2ap+ap^3
ii) R(0, a(2+p^2)
iii) prove SP is equal in length to latus rectum, that is 4a units

not sure how to do part iii)

For Part (iii), we need to use the fact that the triangle SPR is equilateral, meaning SP=PR=RS. So, if we can prove any of those sides is equal to 4a units, our job is done. Let's just get expressions for all three (if you spot what to do straight away, that's awesome too):

$d_{PR}=\sqrt{(2ap-0)^2+(ap^2-a(2+p^2))^2}=\sqrt{4a^2p^2+4a^2}=2a\sqrt{p^2+1}\\d_{SR}=a(2+p^2)-a=a(1+p^2)\\d_{SP}=\sqrt{(2ap-0)^2+(ap^2-a)^2}=a(p^2+1)$

All of these must be equal by definition, therefore we can say that:

$2\sqrt{p^2+1}=p^2+1\\4p^2+4=p^4+2p^2+1\\p^4-2p^2-3=0\p^2-3)(p^2+1)=0\\p=\pm\sqrt{3}\quad(p^2+1=0\text{ yields no solutions})$ Then we go back and pop that in our distance equation for SP: $d_{SP}=a(1+p^2)=a(1+3)=4a \text{ as required}$ Hope this helps! ;D Title: Re: 3U Maths Question Thread Post by: Ali_Abbas on September 13, 2016, 09:13:40 pm ques2 : Consider x^2=4ay with focus S. The normal at P(2ap, ap^2) meets y axis at R and triangle SPR is equilateral i) normal at P is x+py=2ap+ap^3 ii) R(0, a(2+p^2) iii) prove SP is equal in length to latus rectum, that is 4a units not sure how to do part iii) $\textrm{I have an alternative method (just for fun :D).}$ $\textrm{Observe that we can generate two distinct expressions for the area of} \bigtriangleup SRP \textrm{ by first using the base and perpendicular height formula, and secondly by using the two side and included angle formula (since all interior angles equal to 60 degrees). The inputs for the former approach can be obtained by noting that the horizontal from P to the y-axis is perpendicular to SR (the base). Thus:}$ $A = \frac{1}{2}SR(2ap) = \frac{1}{2}SR*SP*\sin{60} \\ \Longrightarrow 2ap = \frac{\sqrt{3}SP}{2} \\ \Longrightarrow SP = \frac{4ap}{\sqrt{3}}$ $\textrm{Clearly, } p = \sqrt{3} \textrm{ will yield the desired property. To prove this, we can exploit the fact that the gradient of the tangent to the parabola at the point P is equal to p. So p equals to tan of whatever angle the tangent makes with the positive x-axis. But this will evaluate to the angle the tangent at P makes with the horizontal from P to the y-axis (being alternate angles).}$ $\textrm{Now, since RP = SP and the horizontal from P is perpendicular to SR, it follows that this horizontal is the angle bisector of } \angle RPS = 60 \textrm{ degrees. We also have that RP is perpendicular to the tangent at P given that it is a component of the normal. So the angle we seek is equal to 90 degrees minus a half of 60 which equals to 90 minus 30 = 60. Thus, }$ $p = \tan{60} = \sqrt{3}.$ $\textrm{Substituting this into our equation for SP and the result is proved. QED}$ Title: Re: 3U Maths Question Thread Post by: lha on September 17, 2016, 08:14:27 pm Can anyone give me a rundown on parametrics or tell me where i can find something that will explain it easily? Title: Re: 3U Maths Question Thread Post by: conic curve on September 17, 2016, 08:15:44 pm Can anyone give me a rundown on parametrics or tell me where i can find something that will explain it easily? Watch eddie woo videos on Youtube Title: Re: 3U Maths Question Thread Post by: RuiAce on September 17, 2016, 08:30:28 pm Can anyone give me a rundown on parametrics or tell me where i can find something that will explain it easily? Look for where it says parametrics Title: 3U Maths Question Thread Post by: lha on September 19, 2016, 06:52:06 pm What should you do in the 5 mins reading time? Look for where it says parametrics Watch eddie woo videos on Youtube Thank you!!! Edit: Posts merged, avoid double posting, use the Modify and Insert Quote functions instead! Title: Re: 3U Maths Question Thread Post by: jakesilove on September 19, 2016, 07:04:07 pm What should you do in the 5 mins reading time? Thank you!!! Edit: Posts merged, avoid double posting, use the Modify and Insert Quote functions instead! In the five minutes, I usually try to answer as many multiple choice as I can after I quickly flick through the paper. In Ext 1, you're not going to really get far on any major question, so I think multis is the best way to go! Title: Re: 3U Maths Question Thread Post by: RuiAce on September 19, 2016, 08:06:21 pm What should you do in the 5 mins reading time? Thank you!!! Edit: Posts merged, avoid double posting, use the Modify and Insert Quote functions instead! In the five minutes, I usually try to answer as many multiple choice as I can after I quickly flick through the paper. In Ext 1, you're not going to really get far on any major question, so I think multis is the best way to go! On the other hand (but still overlapping) I only do about 1-3 multiple choice questions. I take enough time to look at the paper and actually know what's going on. I look out for questions that will throw me off and brace myself for those. Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on September 19, 2016, 08:10:51 pm On the other hand (but still overlapping) I only do about 1-3 multiple choice questions. I take enough time to look at the paper and actually know what's going on. I look out for questions that will throw me off and brace myself for those. To throw in my 2 cents as well, I'd probably be more similar to Rui. The first thing I do is look through the paper, try and spot any doozy questions, get my brain chugging on them in the background. That could take all of the reading time and that was okay with me ;D if I did have leftover reading time, I did the MC ;D Title: Re: 3U Maths Question Thread Post by: jakesilove on September 19, 2016, 08:12:57 pm On the other hand (but still overlapping) I only do about 1-3 multiple choice questions. I take enough time to look at the paper and actually know what's going on. I look out for questions that will throw me off and brace myself for those. And finally, 5 minutes is enough time to say 'I think I can' about 150 times. At the end of the day, no matter what you do in the 5 minutes reading time, the important thing is to be in the right mindset. If finding tricky questions, and feeling safe in knowing you can figure out how to do them, is the way to make you feel comfortable, go for that. If reading the formula sheet, or listing pneumonic devices, is your thing; do that. Reading time isn't a huge factor in exam success (although, if you use it in a way best suited to you, it can be very effective) Title: Re: 3U Maths Question Thread Post by: massive on September 23, 2016, 01:47:41 am Hey guys how do you do part ii and iii for the q attached?? Title: Re: 3U Maths Question Thread Post by: RuiAce on September 23, 2016, 09:25:14 am Hey guys how do you do part ii and iii for the q attached?? Part (iii) is easy. Clearly r=6 and n=12 so just sub it into the formula for (ii). ________________________________ $\text{There's a bit of symmetry that's going on in part (ii).}\\ \text{One way to do it is to expand the entire thing out and compare/contrast.}\\ \text{I'll leave that method as your (or Jamon's...) exercise.}$ $\text{I will use something akin to integration by substitution.}\\ \text{Just with summation.}\\ \textit{It's similar to the proof of the 4u result }\int_0^af(x)dx=\int_0^af(a-x)dx$ ________________________________ \text{Commence by splitting up the summation.}\\ \begin{align*}LHS&=\sum_{k=0}^{2r}(-1)^k\binom{n}{k}\binom{n}{2r-k}\\ &= \sum_{k=0}^{r}(-1)^k\binom{n}{k}\binom{n}{2r-k}+\sum_{k=r+1}^{2r}(-1)^k\binom{n}{k}\binom{n}{2r-k}\end{align*} $\text{Observe that the first term is something we want to stay there.}\\ \text{We need to do something about the second term.}\\ \text{Firstly, I will add a fudge term.}\\ \textit{This is basically because we want the second sum to match the first.}$ \begin{align*}LHS&= \sum_{k=0}^{r}(-1)^k\binom{n}{k}\binom{n}{2r-k}+\sum_{k=r+1}^{2r}(-1)^k\binom{n}{k}\binom{n}{2r-k}\\ &= \sum_{k=0}^{r}(-1)^k\binom{n}{k}\binom{n}{2r-k}+\left[(-1)^r\binom{n}{r}\binom{n}{2r-r}+\sum_{k=r+1}^{2r}(-1)^k\binom{n}{k}\binom{n}{2r-k}\right]-(-1)^r\binom{n}{r}\binom{n}{2r-r}\\ &= \sum_{k=0}^{r}(-1)^k\binom{n}{k}\binom{n}{2r-k}+\sum_{k=r}^{2r}(-1)^k\binom{n}{k}\binom{n}{2r-k}-(-1)^r\binom{n}{r}\binom{n}{r}\end{align*} \text{Now for the second sum, apply the substitution }u=2r-k\\ \begin{align*}LHS&= \sum_{k=0}^{r}(-1)^k\binom{n}{k}\binom{n}{2r-k}+\sum_{u=0}^{r}(-1)^{2r-u}\binom{n}{2r-u}\binom{n}{2r-(2r-u)}-(-1)^r\binom{n}{r}^2\tag{*}\\ &=\sum_{k=0}^{r}(-1)^k\binom{n}{k}\binom{n}{2r-k}+\sum_{u=0}^{r}(-1)^{-u}\binom{n}{u}\binom{n}{2r-u}-(-1)^r\binom{n}{r}^2\\ &=\sum_{k=0}^{r}(-1)^k\binom{n}{k}\binom{n}{2r-k}+\sum_{u=0}^{r}(-1)^{u}\binom{n}{u}\binom{n}{2r-u}-(-1)^r\binom{n}{r}^2 \end{align*}\\ \text{Notes: }(-1)^{-u}=(-1)^u\text{ and }(-1)^{2r}=1 $\text{But now the first sum is the same as the second!}\\ \text{This is due to the idea of dummy variables.}\\ \text{Just like integratino, here }u\text{ and }k\text{ are dummy variables in summation.}$ $\text{Hence we can rewrite the }LHS\text{ as simply }\\ LHS=2\sum_{k=0}^{r}(-1)^k\binom{n}{k}\binom{n}{2r-k}-(-1)^r\binom{n}{r}^2$ $\text{Sub this back into part a), rearrange and we're done.}$ _____________________________________________________________ $\textit{Note regarding line (*)}\\ \text{This is a bit awkward. Technically, after the change of variable, it should be...}\\ \sum_{u=2r}^{0}(-1)^{2r-u}\binom{n}{2r-u}\binom{n}{2r-(2r-u)}$ $\text{But that makes no sense. Why are we summing backwards?}\\ \text{And why was I allowed to just swap it around?}$ $\text{The answer lies in how if you expand the sum, and just rewrite the terms in the REVERSE order}\\ \text{You get what I wrote.}\\ \text{Note that unlike integration, we actually DON'T have an extra negative appearing for this reason.}$ $\textit{It is for this reason why expanding out the sigma and doing rearranging may be a better approach.}$ Title: Re: 3U Maths Question Thread Post by: Neutron on September 23, 2016, 01:49:59 pm Hi! This is a silly question but I never understood how to do these types of polynomials :/ P(x)=(x+1)(x-3)Q(x)+(-2x+6) Where Q(x) is a polynomial. Find the remainder when P(x) is divided by (x+1)(x-3) Sorry if this is dumb D: Thank you! Neutron Title: Re: 3U Maths Question Thread Post by: RuiAce on September 23, 2016, 02:55:58 pm Hi! This is a silly question but I never understood how to do these types of polynomials :/ P(x)=(x+1)(x-3)Q(x)+(-2x+6) Where Q(x) is a polynomial. Find the remainder when P(x) is divided by (x+1)(x-3) Sorry if this is dumb D: Thank you! Neutron The answer is just (-2x+6) Because this is just the division transformation P(x) = A(x)Q(x) + R(x) Where: P(x) is your original polynomial A(x) is your divisor (x+1)(x-3) Q(x) is your quotient which doesn't matter R(x) is your remainder. If instead you wanted the remainder upon division by x-3 then you would need to use the remainder theorem and find P(3) Title: Re: 3U Maths Question Thread Post by: jamonwindeyer on September 23, 2016, 03:04:21 pm Hi! This is a silly question but I never understood how to do these types of polynomials :/ P(x)=(x+1)(x-3)Q(x)+(-2x+6) Where Q(x) is a polynomial. Find the remainder when P(x) is divided by (x+1)(x-3) Sorry if this is dumb D: Thank you! Neutron Don't worry, I still hate these, always makes me do a double take, not sure why ;) Just to add a bit more to Rui's explanation. Go back to basics (and make it a little more colloquial than normal). Remember that when we divide something, we are really determining how many times the divisor goes (fully) into the dividend, that number being the quotient. The number leftover is the remainder. So really, what we are doing is separating the dividend into [quotient] groups of the divisor, plus the remainder. That's the division transformation: $P(x)=Q(x)D(x)+R(x)$ As Rui said, in your case the question is easy. Our divisor is \((x+1)(x-3), so our remainder is actually in the question itself!!

$R(x)=-2x+6$
Title: Re: 3U Maths Question Thread
Post by: lha on September 25, 2016, 03:53:26 pm
And finally, 5 minutes is enough time to say 'I think I can' about 150 times. At the end of the day, no matter what you do in the 5 minutes reading time, the important thing is to be in the right mindset. If finding tricky questions, and feeling safe in knowing you can figure out how to do them, is the way to make you feel comfortable, go for that. If reading the formula sheet, or listing pneumonic devices, is your thing; do that. Reading time isn't a huge factor in exam success (although, if you use it in a way best suited to you, it can be very effective)

This helps a lot, thank you!
Title: Re: 3U Maths Question Thread
Post by: nimasha.w on September 25, 2016, 04:31:42 pm
hi! can someone please explain this question for me :)
Title: Re: 3U Maths Question Thread
Post by: RuiAce on September 25, 2016, 04:39:02 pm
hi! can someone please explain this question for me :)
$\text{We use the division transformation mentioned not many posts ago.}\\ \text{Observe that our divisor is }x^2-1\\ \text{and the quotient is }3x-1$
$\text{Hence, we can express the polynomial as }\\ P(x)=(x^2-1)Q(x)+(3x-1)$
$x-1\text{ is a linear term (no squared or anything)}\\ \text{So we just use the remainder theorem. By this theorem}\\ \text{The remainder equals }P(1)$
\text{So we just evaluate }P(1)\\ \begin{align*}P(1)=0+3-1=2\end{align*}
Title: Re: 3U Maths Question Thread
Post by: massive on September 26, 2016, 12:20:57 am
Guys how on earth would you do the q attached in an exam?? The answer is C but I don't get how you can get without using something like wolfram alpha. Any tips would be greeeatly appreciated.
Title: Re: 3U Maths Question Thread
Post by: massive on September 26, 2016, 07:58:14 am
another one, how do you do part ii guysss??
Title: Re: 3U Maths Question Thread
Post by: RuiAce on September 26, 2016, 08:39:34 am
Guys how on earth would you do the q attached in an exam?? The answer is C but I don't get how you can get without using something like wolfram alpha. Any tips would be greeeatly appreciated.
That question is kinda mean.

Draw a graph, but draw it to SCALE.

The only solution you can guarantee is x=0. The rest you cannot solve via algebra; you can only do so by graph.
Title: Re: 3U Maths Question Thread
Post by: RuiAce on September 26, 2016, 08:41:28 am
another one, how do you do part ii guysss??
$\text{The particle starts off at rest, however if you look at the acceleration}\\ \text{That thing is always positive.}\\ \ddot{x}>0\text{ as }\frac{1}{4+x^2}>0\text{ for all }x$
$\text{So the particle always ACCELERATES in the positive direction.}\\ \text{If it starts from rest, and always accelerates like that, by consequence }v>0\text{ as well.}$
Title: Re: 3U Maths Question Thread
Post by: jamonwindeyer on September 26, 2016, 03:55:19 pm
Guys how on earth would you do the q attached in an exam?? The answer is C but I don't get how you can get without using something like wolfram alpha. Any tips would be greeeatly appreciated.

To clarify above, we are drawing two separate graphs:

$y=-2x\\y=3\pi\sin{(x)}$

The points of intersection of this graph will be solutions to the equation (can you see why?) ;D

To emphasise something even further from Rui's explanation, definitely draw to scale. Like, crazy to scale. You'll get this graph:

(https://www4c.wolframalpha.com/Calculate/MSP/MSP54761ff1f1ebbh398f4400002icigif5g85d96dg?MSPStoreType=image/gif&s=62)

It is very easy to miss those two solutions on the right there. There are 3 solutions in the domain ;D
Title: Re: 3U Maths Question Thread
Post by: lha on September 26, 2016, 07:39:46 pm
Hi can anyone help me with question 12b)i) and ii) in the 2015 hsc 3u exam? Its not letting me attach a pic of the question so if someone could look it up that would be great! I have no idea how to do these types of questions so any help is extremely appreciated!
Title: Re: 3U Maths Question Thread
Post by: RuiAce on September 26, 2016, 08:05:52 pm
Hi can anyone help me with question 12b)i) and ii) in the 2015 hsc 3u exam? Its not letting me attach a pic of the question so if someone could look it up that would be great! I have no idea how to do these types of questions so any help is extremely appreciated!
These are one mark circle geometry questions. Tips:

(i) <ACD is right next to it and labelled 30 degrees. That's something I immediately keep in mind. And then just read the question. They explicitly tell you BD is a diameter, so what does that say about <BCD?

(ii) I see the angle formed by a chord (AD) and a tangent (XD). I immediately think about alternate segment theorem. There are two triangles to use alternate segment theorem on, so just look at both of them and find a useful once.

Remember - You're only looking for like one thing for a one marker. Don't overanalyse in these scenarios.
(Exclusion: That 1 marker is the last question of the paper. Maybe think a bit there)
Title: Re: 3U Maths Question Thread
Post by: lha on September 26, 2016, 08:46:56 pm
These are one mark circle geometry questions. Tips:

(i) <ACD is right next to it and labelled 30 degrees. That's something I immediately keep in mind. And then just read the question. They explicitly tell you BD is a diameter, so what does that say about <BCD?

(ii) I see the angle formed by a chord (AD) and a tangent (XD). I immediately think about alternate segment theorem. There are two triangles to use alternate segment theorem on, so just look at both of them and find a useful once.

Remember - You're only looking for like one thing for a one marker. Don't overanalyse in these scenarios.
(Exclusion: That 1 marker is the last question of the paper. Maybe think a bit there)

Thank you that helps but i think you got the questions mixed up, I need help on 12b) not 12a). Sorry!
Title: Re: 3U Maths Question Thread
Post by: RuiAce on September 26, 2016, 09:24:27 pm
Damn it lol thought something was weird!

For the correct question

(i) This is one of the more fundamental proofs that are included in the textbook. The derivation isn't hard - a chord being "focal" means it runs through the focus. For that parabola, the focus is (0,a). So all you do is just sub that in.

(ii) This now requires you to call in a bit of knowledge. Look closely at the coordinates of P. The coordinates of P are (2ap, ap^2) and (8a, 16a) simultaneously! That means all you have to do is equate.

Regardless of whether you equate x or y coordinates, you're going to get p=4.

So if PQ is a focal chord, you can safely use part (i) and sub back in.
Title: Re: 3U Maths Question Thread
Post by: lha on September 26, 2016, 09:27:39 pm
I just realised how to edit my posts hahaha

Well I have another question. How do you do q14a)ii) from the 2015 hsc paper?
Title: Re: 3U Maths Question Thread
Post by: RuiAce on September 26, 2016, 09:52:26 pm
That makes sense, thank you!!

Also, could you please explain how to do 13a)iii) from the 2015 hsc paper?

Mod edit: You're most welcome, but please stick to editing prior posts in these situations over unnecessarily posting
$\text{That was perhaps one of the more 'interesting' questions last year.}$
$\text{The given equation shouldn't be overly unfamiliar but may be forgotten.}\\ \text{It is important that you are well aware of this formula just in case:}\\ v^2=n^2(a^2-x^2)$
$\textit{For SHM, when }v=0\text{ we are at the }\textit{endpoints}\\ \text{Hence the endpoints of motion are }3\text{ and }7$
$\text{From this, we deduce that the centre of motion is }5\\ \text{and the amplitude is }2$
$\text{So }c=5\text{ and }a=2\text{. You need to spend time realising why this is true.}$
$\text{It remains to deduce }n\\ \text{But we have one extra piece of information:}\\ \text{When }x=5, v^2=11\textit{ (do you see why?)}$
Title: Re: 3U Maths Question Thread
Post by: lha on September 26, 2016, 10:00:49 pm
How do you do question 14c)iii)?

Btw sorry for all the questions!
Title: Re: 3U Maths Question Thread
Post by: RuiAce on September 26, 2016, 10:04:14 pm
I only understand how you got c=5 but not the rest. How did you find a=2 and what n equals to?

Also im not sure how to edit my previous posts
There's a modify button in the top right corner of every post you make. Just use that and you're all good.

If the endpoints are 3 and 7 then the distance between the endpoints is 4. But the amplitude is the distance between an endpoint and the centre of motion. So a = 2

Just sub in x=5, v2=11 and you can find n.
i.e. you find n LAST.
Title: Re: 3U Maths Question Thread
Post by: IkeaandOfficeworks on September 30, 2016, 04:14:16 pm
Hi guys! Can you help me with this question? Thanks!
Title: Re: 3U Maths Question Thread
Post by: jakesilove on September 30, 2016, 04:55:41 pm
Hi guys! Can you help me with this question? Thanks!

Hey! Check out what I did below; essentially, I knew I needed to pop out the sinh/h, and the rest works out easily!

(http://i.imgur.com/kq1gP8X.png)

Jake
Title: Re: 3U Maths Question Thread
Post by: IkeaandOfficeworks on September 30, 2016, 05:18:35 pm
Thanks! I thought I needed to expand the cos(x+h) so I got confused.    :-X
Title: Re: 3U Maths Question Thread
Post by: jakesilove on September 30, 2016, 05:31:58 pm
Thanks! I thought I needed to expand the cos(x+h) so I got confused.    :-X

Totally fair enough. Basically, if you CAN let h equal zero, then you just do. We needed to get rid of the h in the denominator, so we don't divide by zero, but after that we have no problems!
Title: Re: 3U Maths Question Thread
Post by: IkeaandOfficeworks on September 30, 2016, 06:02:38 pm
Post by: RuiAce on September 30, 2016, 06:06:24 pm
$\text{Use a double angle formula for that one}\\ 1-\cos \theta=2\sin^2\frac{\theta}{2}$
$\text{So then you need to play with}\\ 2\lim_{x\to 0}\left(\frac{\sin\frac{\theta}{2}}{\theta}\right)^2$
It's actually like the question above it. Just remember to be careful with the fraction
Title: Re: 3U Maths Question Thread
Post by: samuels1999 on September 30, 2016, 08:59:41 pm
Hi Jake,

I am doing Extension 1 mathematics next year, as I did really well in prelim Ext 1. The thing is, I have heard many rumours from students in the year above me that it is really hard to do Extension 1 by itself (without Ext 2). However they say, it's really easy for people doing Extension 2. I just wanted to know what your take on that is (how true is the statement?) and whether extension 1 is really worth doing as it only counts for a single unit in my case.

Thanks heaps,
Samuel
Title: Re: 3U Maths Question Thread
Post by: jakesilove on September 30, 2016, 09:07:37 pm
Hi Jake,

I am doing Extension 1 mathematics next year, as I did really well in prelim Ext 1. The thing is, I have heard many rumours from students in the year above me that it is really hard to do Extension 1 by itself (without Ext 2). However they say, it's really easy for people doing Extension 2. I just wanted to know what your take on that is (how true is the statement?) and whether extension 1 is really worth doing as it only counts for a single unit in my case.

Thanks heaps,
Samuel

Hey Samuel!

I would definitely disagree with the rumours that you've heard. I'll admit that there are (some) special circumstances in which doing Ext 2 is beneficial for your study of Ext 1 (mainly, just that extra time spent thinking about Maths), however in general I think you can do just as well, and understand just as much, as any student doing Ext 2. MAYBE Ext 1 seems easy to an Ext 2 student, but that's just because they're often very, very good at maths (and thus doing Ext 2; if they were doing only Ext 1, they would still find it just as easy!).

A better question is asking whether Ext 1 is worth it, since it's only worth one unit. I would still say that it does. The content is definitely less than 2U, which makes sense since you have less class time dedicated to it. You'll do better in 2U if you're studying Ext 1 (lots of techniques translate, and you just spend more time doing maths and thinking about the beauty of numbers), and if you find yourself struggling at any point you can drop the subject. My overall recommendation is that, seeing as you did well in prelim, you should continue your study and decide further down the track (and use this forum if you have any questions!)

Let me know if you have any other specific questions; you've raised some really good points that a lot of students worry about, so I'm glad you asked it here!

Jake
Title: Re: 3U Maths Question Thread
Post by: samuels1999 on September 30, 2016, 09:24:57 pm
Thanks very much Jake.

I also have another question regarding Permutations and Combinations. I personally found it the hardest maths topics I have ever done in the sense that, 1. at times I didn't really know what the question was actually asking and 2. there was no way to check my method/ answer.
I just wanted some advice on how to study the topic: Tonnes of Practice? What textbook maybe (Cambridge/ Maths in Focus)? or Maybe just past papers?

Thanks
Samuel

Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 01, 2016, 01:30:10 am
Hi Jake,

I am doing Extension 1 mathematics next year, as I did really well in prelim Ext 1. The thing is, I have heard many rumours from students in the year above me that it is really hard to do Extension 1 by itself (without Ext 2). However they say, it's really easy for people doing Extension 2. I just wanted to know what your take on that is (how true is the statement?) and whether extension 1 is really worth doing as it only counts for a single unit in my case.

Thanks heaps,
Samuel
Those rumours are ridiculous. Seconding everything Jake said.
Thanks very much Jake.

I also have another question regarding Permutations and Combinations. I personally found it the hardest maths topics I have ever done in the sense that, 1. at times I didn't really know what the question was actually asking and 2. there was no way to check my method/ answer.
I just wanted some advice on how to study the topic: Tonnes of Practice? What textbook maybe (Cambridge/ Maths in Focus)? or Maybe just past papers?

Thanks
Samuel
You do practice questions for every single topic. Singling out perms and combs doesn't make a difference.

But it is certainly a more dodgier topic. It's very easy to misinterpret the question and head in the wrong direction. You need to use many fundamental counting principles.
- And implies multiply
- Difference between selections and arrangements
- Difference between ordered and unordered

Whilst it is possible to categorise different combinatorics questions, it's too easy to go into the wrong category. It really takes heaps of practice. Hence, simply do as many questions as possible. In all honesty, I had minimal confidence with that topic until uni, and even now I'm not that confident anyhow.

I've never used Cambridge for perms and combs. Maths in focus makes it smoother because it's all categorised into topics but the questions in maths in focus aren't hard altogether.
Title: Re: 3U Maths Question Thread
Post by: massive on October 01, 2016, 03:19:17 pm
Hey guys, this question looks pretty simple but i'm kind of perplexed because if both A and B have zero velocity doesn't that mean they're both turning points :S and if that's the case how is b>a. I don't get part i because of this lack of understanding of the question. Thanks for and tips!
Post by: RuiAce on October 01, 2016, 03:44:27 pm
(http://uploads.tapatalk-cdn.com/20161001/ec185216686fdc6cb969583e461bf6a7.jpg)Read the question again. b and a are actually y-coordinates, not x-coordinates

(Or technically I should say they are x-coordinates and not t-coordinates)

They can definitely both be stationary points because SHM is described by a sine curve, which has infintely many stationary points
Title: Re: 3U Maths Question Thread
Post by: massive on October 01, 2016, 04:33:05 pm
(http://uploads.tapatalk-cdn.com/20161001/ec185216686fdc6cb969583e461bf6a7.jpg)Read the question again. b and a are actually y-coordinates, not x-coordinates

(Or technically I should say they are x-coordinates and not t-coordinates)

They can definitely both be stationary points because SHM is described by a sine curve, which has infintely many stationary points

OHH but when it says "whose distances from the same side of the origin..." isn't the origin referring to the centre of motion?
Post by: RuiAce on October 01, 2016, 05:26:06 pm
OHH but when it says "whose distances from the same side of the origin..." isn't the origin referring to the centre of motion?
Recall: The origin is x=0 for our purposes of analysing motion

It just means that additionally 0<a<b or a<b<0
Title: Re: 3U Maths Question Thread
Post by: Goodwil on October 03, 2016, 10:15:28 am
Hi, could someone please explain how to work through this question (both parts) ? I've looked through the solutions and it still doesn't make much sense
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 03, 2016, 10:35:24 am
Hi, could someone please explain how to work through this question (both parts) ? I've looked through the solutions and it still doesn't make much sense
$\text{This question from the 2014 HSC was covered in Jamon's trial HSC lectures.}\\ \text{The question is best approached with the aid of an infinite tree diagram.}$
$\text{Observe that if the spinner lands on }P\text{ or }Q\text{, it ends.}\\ \text{Only if the spinner lands on }R\text{ will there be a next go.}$(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpsn5utqgm4.png)
$\text{The probability that player A wins on the first go is obviously }\boxed{p}$$\text{For the second go to }\textbf{actually even happen}\text{, player A must've rolled }R\\ \text{There is a probability of }r\text{ of this happening.}$
$\text{Then, if A were to win, B must've rolled a }Q.\\ \text{Hence, the probability of winning on the second go is }\boxed{rq}$
\textbf{Using the rule of sum of probabilities: }\boxed{p+q+r=1}\\ \text{So we have}\\ \begin{align*}Pr(\text{A wins on 1st or 2nd go})&=p+rq\\ &=p+r(1-p-r)\\ &= p+r-r(p+r)\\ &=(1-r)(p+r)\end{align*}
________________________________________
$\textit{We are now interested in the probability that he will ULTIMATELY win.}\\ \text{We must consider ALL possible cases.}$
$\text{For a }\textbf{third}\text{ turn to even be possible, player B must've also spun }R\text{ on the second turn.}\\ \text{Then, player A must've spun }P\\ \text{So this has a probability of }r^2p$
$\text{For a }\textbf{fourth}\text{ turn to even be possible, player A must've rolled }R\text{ on the third turn.}\\ \text{Then, player B must've spun }Q\\ \text{So this has a probability of }r^3q$
\text{If we }\textbf{add}\text{ the above results we get}\\ \begin{align*}Pr(\text{A wins on 3rd or 4th turn})&=r^2p+r^3q\\ &=r^2(p+rq)\\ &=r^2(1-r)(p+r)\end{align*}
$\text{Similarly, the 5th and 6th turn follow }r^4p, \, r^5q\text{ respectively.}\\ \text{So we can show that the probability he wins on the 5th or 6th turn is }r^4(1-r)(p+r)$
\textbf{Summing all of these up, we arrive at a geometric progression}\\ \begin{align*}Pr(\text{A ultimately wins})&=(1-r)(p+r)+r^2(1-r)(p+r)+r^4(1-r)(p+r)+\dots\\ &= (1-r)(p+r)(1+r^2+r^4+\dots)\\ &= (1-r)(p+r)\frac{1}{1-r^2}\\ &=\frac{p+r}{1+r}\end{align*}\\ \text{via difference of two squares.}
Title: Re: 3U Maths Question Thread
Post by: yaboiaderler on October 04, 2016, 10:29:24 pm
hiya folk! This is from the 2012 HSC exam (Question 11). I understand how to do part i of the question, but part ii is like...waht?
I'm mainly confused about what a 'Non-zero constant term' is.

Thanks for the help!
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 04, 2016, 10:43:16 pm
hiya folk! This is from the 2012 HSC exam (Question 11). I understand how to do part i of the question, but part ii is like...waht?
I'm mainly confused about what a 'Non-zero constant term' is.

Thanks for the help!
$\text{That just means that the constant term is there, however it can't be }0.\\ \text{It can be }1, 2, -1, 48986203954123\text{ or whatever. Just not }0$
$\textit{An alternate way of thinking about it is that the constant term does not vanish, it stays there.}$
$\text{Return to part (i). From your working, you showed that}\\ \text{The constant term is when }4k-12=0$
$\text{If you used the }\textbf{exact same}\text{ computations, however}\\ \text{using }n\text{ and not the number }12\text{, you would get}\\ 4k-n = 0$
$\text{That rearranges into }n=4k\\ \text{implying }n\text{ must be a multiple of }4$
As an exercise, you may wish to expand (2x3-1/x)n on WolframAlpha. Use n=1,2,3,5,6,7,9,10,11 and you will find there is no constant term (or in other words, the constant term is 0). When n=4,8,12,... however, you'll find you get a constant

Title: Re: 3U Maths Question Thread
Post by: fizzy.123 on October 05, 2016, 01:32:45 am
how do we answer iii? (2014 MX1)
Title: Re: 3U Maths Question Thread
Post by: lha on October 05, 2016, 08:51:09 am
Title: Re: 3U Maths Question Thread
Post by: jakesilove on October 05, 2016, 09:29:40 am
how do we answer iii? (2014 MX1)

From part ii) we find that the slope of the line is t. We also know from part i) that

$x=\frac{2at}{t^2+1} y=\frac{2at^2}{t^2+1}$

The important thing to note here is that the gradient t is equal to RISE OVER RUN, and thus

$t=\frac{y}{x}$

From there, just sub all t values in for y/x (for either x or y equations above, although I think x is easier), solve the equation, and you'll get a circle out. A tough question for sure!

Jake
Title: Re: 3U Maths Question Thread
Post by: jakesilove on October 05, 2016, 09:32:57 am

If you could post the actual question in the future that'd be awesome.

I think the Board of Studies does a good job with this answer, so I've just posted it below. Let me know if you need any clarification on any particular line or working; I feel like it is probably clear, and to be honest I'm not good enough at LaTeX to type out a proper answer.

(http://i.imgur.com/RrSouq1.png)

Jake
Title: Re: 3U Maths Question Thread
Post by: lha on October 05, 2016, 10:02:53 am
If you could post the actual question in the future that'd be awesome.

I think the Board of Studies does a good job with this answer, so I've just posted it below. Let me know if you need any clarification on any particular line or working; I feel like it is probably clear, and to be honest I'm not good enough at LaTeX to type out a proper answer.

(http://i.imgur.com/RrSouq1.png)

Jake

No this is fine, thank you!
Title: Re: 3U Maths Question Thread
Post by: jakesilove on October 05, 2016, 11:11:42 am
No this is fine, thank you!

No problem! I always hated these kinds of questions (still do), so if you find that you struggle with them, don't worry! In my opinion, they are the most difficult maths you do in the 3U course (ie, really hard binom questions) just because of the terminology, logic and guesswork that is required.
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 05, 2016, 02:39:50 pm
No problem! I always hated these kinds of questions (still do), so if you find that you struggle with them, don't worry! In my opinion, they are the most difficult maths you do in the 3U course (ie, really hard binom questions) just because of the terminology, logic and guesswork that is required.
2013 binomial theorem question was insane though to be fair.
Title: Re: 3U Maths Question Thread
Post by: jamonwindeyer on October 05, 2016, 03:43:43 pm
2013 binomial theorem question was insane though to be fair.

Doing it for the MX1 Revision lecture brought back nightmares ;)
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 05, 2016, 03:45:22 pm
Doing it for the MX1 Revision lecture brought back nightmares ;)
You just purposely chose to bring back many baaaad memories
Title: Re: 3U Maths Question Thread
Post by: Goodwil on October 05, 2016, 04:19:01 pm
Hi, can someone please explain how to do this question?
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 05, 2016, 04:30:02 pm
Hi, can someone please explain how to do this question?
$\text{The question must be approached more cautiously}\\ \text{In schools, they teach you to just consider signs.}\\ \text{This approach is inferior. We must use the definition of absolute value:}\\ |x|=\begin{cases}x&\text{ if }x\ge 0\\ -x&\text{ if }x < 0\end{cases}$
$\text{Or by extension}\\ |x-a|=\begin{cases}x&\text{ if }x-a\ge 0\\ -(x-a)&\text{ if }x < 0\end{cases}$
\text{So for our question, we have 3 cases.}\\ \text{Case 1: }x \ge 3\\ \begin{align*}|x+2|+|x-3|&=5\\ (x+2)+(x-3)&=5\\ 2x&=6\\ x&=3\end{align*}\\ \text{So from this case, we }\textit{extract the case } x=3\\ \text{for our solution.}
\text{Case 2: }-2 \le x < 3\\\begin{align*}|x+2|+|x-3|&=5\\ (x+2)-(x-3)&=5\\ 5&=5\end{align*} \\ \text{ALWAYS true}\\ \text{Hence, the entire }-2\le x < 3\text{ belongs in our solution.}
\text{Case 3: }x < -2\\ \begin{align*}|x+2|+|x-3|&=5\\ -(x+2)-(x-3)&=5\\ -2x&=4\\ x&=-2\end{align*}\\ x=-2\text{ does not belong in }x< -2\\ \text{so we extract }\textit{nothing}\text{ out of this case.}
$\text{Combining ALL of the above cases gives us the solution}\\ -2 \le x \le 3$
__________
$\text{It falls to identify which case has the solution }-2\le x \le 3\\ \text{It shouldn't take you too long to realise that }\\ x^2-x-6\le 0\text{ also solves to give }-2 \le x \le 3$
Title: Re: 3U Maths Question Thread
Post by: massive on October 05, 2016, 04:53:59 pm
how do you do this guys?
Title: Re: 3U Maths Question Thread
Post by: massive on October 05, 2016, 04:55:51 pm
sorry i have another question. How do you do part iv, it's one mark but wth :S
Title: Re: 3U Maths Question Thread
Post by: kevin217 on October 05, 2016, 05:29:54 pm
$\text{That just means that the constant term is there, however it can't be }0.\\ \text{It can be }1, 2, -1, 48986203954123\text{ or whatever. Just not }0$
$\textit{An alternate way of thinking about it is that the constant term does not vanish, it stays there.}$
$\text{Return to part (i). From your working, you showed that}\\ \text{The constant term is when }4k-12=0$
$\text{If you used the }\textbf{exact same}\text{ computations, however}\\ \text{using }n\text{ and not the number }12\text{, you would get}\\ 4k-n = 0$
$\text{That rearranges into }n=4k\\ \text{implying }n\text{ must be a multiple of }4$
As an exercise, you may wish to expand (2x3-1/x)n on WolframAlpha. Use n=1,2,3,5,6,7,9,10,11 and you will find there is no constant term (or in other words, the constant term is 0). When n=4,8,12,... however, you'll find you get a constant

On the reference sheet there is two general expansions for binomial theorem. However, if you used the other expansion you get a different answer? I get n = (4/3)k
Title: Re: 3U Maths Question Thread
Post by: MightyBeh on October 05, 2016, 05:44:45 pm
how do you do this guys?
Not 100% confident on this one but if the answer is A I can share my working. ::)

sorry i have another question. How do you do part iv, it's one mark but wth :S

i) for an inverse to exist, f(x) must be a one to one function (there must be a unique y-value for every x-value). By restricting the domain to one side of the turning point at x =2 (after which y-values will begin to repeat) we turn f(x) into a one-to-one function, thus an inverse exists for values including and on one side of the point (2,0), in this case x >= 2.

ii) Domain and range of the inverse are the same as the original function, but swapped. This is because (x,y) → (y,x). The range of the inverse becomes [2,∞). The domain will be [0,∞) because that's the range of the original function.

iii) f(x)=f-1(x) implies that f(x)=x because any intersections between the function and its inverse will occur on the line y=x.
$
(x-2)^2 = x\\
\implies x^2 - 4x + 4 = x\\
\implies x^2 - 5x + 4 = 0\\
\implies (x-4)(x-1) = 0\\
\therefore x = 4, 1\\
\text{f(x) will be equal to }f^{-1}(x) \text{ at x=4 and x=1 }
$

Edit: Make a mistake on this one, fixed it up. ;)
Edit Edit: Make a different mistake that I didn't fix up the first time.
iv) as the inverse function in this case is 2-sqrt(x) because k < 2.
$
f^{-1}(f(k)) = 2 - \sqrt{(k-2)^2}\\
= 2 - (k - 2) = 4 - k
$
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 05, 2016, 06:18:15 pm
On the reference sheet there is two general expansions for binomial theorem. However, if you used the other expansion you get a different answer? I get n = (4/3)k
$\text{That may be right and still works.}\\ \text{Calling upon my skills from discrete maths now.}$
$3n=4k\\ \text{So 4 must divide into 3, or it must divide into }n\\ \text{But 4 obviously does NOT divide into 3.}\\ \text{Hence 4 divides into }n$
$\text{Reminder: }k\text{ and }n\text{ are integers.}$
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 05, 2016, 06:21:31 pm
how do you do this guys?
$\text{Let }O\text{ be the centre of the circle.}$
$\textit{Required theorems:}\\ \text{The radius is perpendicular to the tangent drawn to point of contact}\\ \text{The angle subtended to the centre is double the angle subtended to the circumferecence.}\\ \text{The angle sum of a quadrilateral is }360^\circ$
Inspiration: The tangent is there but we're only interested in the angles. Alternate segment theorem could work but it seems too dodgy here.
$\textit{Required constructions: }\text{Join }OB\text{ and }OD$
$\angle BOD = 2\alpha$
$2\alpha + 90^\circ + 90^\circ + \beta = 360^\circ$
$\text{The correct answer is A.}$
Edit: Actually, if you join BD the alternate segment theorem takes care of it more quickly.
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 05, 2016, 06:24:46 pm
i) for an inverse to exist, f(x) must be a one to one function (there must be a unique y-value for every x-value). By restricting the domain to one side of the turning point at x =2 (after which y-values will begin to repeat) we turn f(x) into a one-to-one function, thus an inverse exists for values including and on one side of the point (2,0), in this case x >= 2.

1. They don't teach interval notation in the HSC. They only use x>=2 to denote [2, inf)
2. There's no definition of 'one-to-one' and 'onto' functions in the HSC either. If a function is one-to-one, it is just called 'invertible'.
No need for bijective functions in the HSC. This is because the HSC defines the range but absolutely ignores the concept of a 'codomain'
Title: Re: 3U Maths Question Thread
Post by: massive on October 05, 2016, 06:31:42 pm
Not 100% confident on this one but if the answer is A I can share my working. ::)

Yhep the answers A, Howdu get it :O

----

And for part iv, the answer is actually 4-k :S
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 05, 2016, 06:32:39 pm
Yhep the answers A, Howdu get it :O

----

And for part iv, the answer is actually 4-k :S
It is A. I just made a typo in my working.
Title: Re: 3U Maths Question Thread
Post by: massive on October 05, 2016, 06:35:09 pm
It is A. I just made a typo in my working.

Oh yeaah thanks mate!  ;D

Also how do you do part iv for the second q attached (ans: 4-k)
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 05, 2016, 06:39:55 pm
Oh yeaah thanks mate!  ;D

Also how do you do part iv for the second q attached (ans: 4-k)
$2+\sqrt{(k-2)^2}\text{ is actually }2+(2-k)\text{, not }2+(k-2)\\ \text{Recall that }\sqrt{x^2}=|x|=\begin{cases}x& x\ge 0\\ -x & x< 0\end{cases}$
$\text{Because }k-2 < 0\\ \sqrt{(k-2)^2}=|k-2|=2-k$
Title: Re: 3U Maths Question Thread
Post by: MightyBeh on October 05, 2016, 06:41:09 pm

1. They don't teach interval notation in the HSC. They only use x>=2 to denote [2, inf)
2. There's no definition of 'one-to-one' and 'onto' functions in the HSC either. If a function is one-to-one, it is just called 'invertible'.
No need for bijective functions in the HSC. This is because the HSC defines the range but absolutely ignores the concept of a 'codomain'

Thanks for that. :)

Oh yeaah thanks mate!  ;D

Also how do you do part iv for the second q attached (ans: 4-k)
I fixed it up, my bad. Because k < 2 the inverse is actually 2 - sqrt(x), not 2 + sqrt(x) like I had before.
$
f^{-1}(f(k)) = 2 - \sqrt{(k-2)^2}\\
= 2 - (k - 2) = 4 - k
$

Edit: Rui beat me to it but yolo  8)
Title: Re: 3U Maths Question Thread
Post by: massive on October 05, 2016, 09:01:49 pm

I fixed it up, my bad. Because k < 2 the inverse is actually 2 - sqrt(x), not 2 + sqrt(x) like I had before.
$
f^{-1}(f(k)) = 2 - \sqrt{(k-2)^2}\\
= 2 - (k - 2) = 4 - k
$

Edit: Rui beat me to it but yolo  8)

wait i still don't get why you take the negative root, even tho k<2
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 05, 2016, 09:07:14 pm
wait i still don't get why you take the negative root, even tho k<2
Uhh I think that's wrong. You don't redefine the inverse function just like that. >_<

Because (k-2)2 > 0 even though k-2 < 0

Just use my working out.
Title: Re: 3U Maths Question Thread
Post by: massive on October 05, 2016, 09:15:54 pm
Uhh I think that's wrong. You don't redefine the inverse function just like that. >_<

Because (k-2)2 > 0 even though k-2 < 0

Just use my working out.

i understand all your working except; since k-2<0 why do we take the negative branch of the abs(k-2) i.e. 2-k ?
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 05, 2016, 09:43:15 pm
i understand all your working except; since k-2<0 why do we take the negative branch of the abs(k-2) i.e. 2-k ?
$\text{Recall that }\sqrt{x^2}=|x|=\begin{cases}x& x\ge 0\\ -x & x< 0\end{cases}$
Title: Re: 3U Maths Question Thread
Post by: massive on October 05, 2016, 09:57:54 pm
Title: Re: 3U Maths Question Thread
Post by: lha on October 06, 2016, 09:19:09 pm
Could someone please help me with question 14b)iii) from the 2012 hsc paper? I have the solutions, i just dont understand it...
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 06, 2016, 09:37:40 pm
Could someone please help me with question 14b)iii) from the 2012 hsc paper? I have the solutions, i just dont understand it...
Please send an image of the question for convenience in the future. Instructions on how to do so may be found here.
$\text{Note that }0 < \theta < \frac{\pi}{2}$
$\text{To satisfy the conditions given, we must solve the following equations.}\\ 250\sin^2\theta \ge 150 \iff \sin^2\theta \ge \frac{3}{5}\\ 125 \le 250\sin 2\theta \le 180 \iff \frac{1}{2}\le \sin 2\theta \le \frac{18}{25}$
$\text{For }\sin^2\theta \ge \frac{3}{5}\text{, since }0 < \theta < \frac{\pi}{2}\\ \text{We are really just solving }\sin \theta \ge \sqrt{\frac{3}{5}}$
$\text{Sketch }y=\sin \theta\text{ for }0 < \theta < \frac{\pi}{2}$
(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpshnjrwhfj.png)
$\text{Computing }\sin^{-1}\sqrt{\frac35}\text{ on the calculator, from the graph we deduce that this case spits out }\boxed{\theta \ge 50.768...^\circ}$
$\text{Now sketch }y=\sin 2\theta\text{ for }0 < \theta < \frac{\pi}{2}$(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zps1aclsfju.png)
$\text{To find the points of intersection, solve }\sin 2\theta = \frac{1}{2}\text{ and }\sin 2\theta = \frac{18}{25}\text{ over }0 < \theta < \frac{\pi}{2}\\ \text{i.e. }0 < 2\theta < \pi$
$\sin 2\theta = \frac{1}{2} \implies \theta = 15^\circ, 75^\circ\\ \sin 2\theta = \frac{18}{25}\implies \theta = 23.027...^\circ, 66.972...^\circ$
$\text{So looking at the graph, we deduce that this case spits out}\\ \boxed{15^\circ \le \theta \le 23.027...^\circ\\ 66.972...^\circ \le \theta \le 75^\circ}$
$\textbf{The final answer needs both equations to hold true simultaneously.}\\ \text{Hence, combining cases, our answer is (to 2 d.p.)}\\ \boxed{66.97^\circ \le \theta \le 75^\circ}$
Title: Re: 3U Maths Question Thread
Post by: lha on October 06, 2016, 10:02:37 pm
Please send an image of the question for convenience in the future. Instructions on how to do so may be found here.
$\text{Note that }0 < \theta < \frac{\pi}{2}$
$\text{To satisfy the conditions given, we must solve the following equations.}\\ 250\sin^2\theta \ge 150 \iff \sin^2\theta \ge \frac{3}{5}\\ 125 \le 250\sin 2\theta \le 180 \iff \frac{1}{2}\le \sin 2\theta \le \frac{18}{25}$
$\text{For }\sin^2\theta \ge \frac{3}{5}\text{, since }0 < \theta < \frac{\pi}{2}\\ \text{We are really just solving }\sin \theta \ge \sqrt{\frac{3}{5}}$
$\text{Sketch }y=\sin \theta\text{ for }0 < \theta < \frac{\pi}{2}$
(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpshnjrwhfj.png)
$\text{Computing }\sin^{-1}\sqrt{\frac35}\text{ on the calculator, from the graph we deduce that this case spits out }\boxed{\theta \ge 50.768...^\circ}$
$\text{Now sketch }y=\sin 2\theta\text{ for }0 < \theta < \frac{\pi}{2}$(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zps1aclsfju.png)
$\text{To find the points of intersection, solve }\sin 2\theta = \frac{1}{2}\text{ and }\sin 2\theta = \frac{18}{25}\text{ over }0 < \theta < \frac{\pi}{2}\\ \text{i.e. }0 < 2\theta < \pi$
$\sin 2\theta = \frac{1}{2} \implies \theta = 15^\circ, 75^\circ\\ \sin 2\theta = \frac{18}{25}\implies \theta = 23.027...^\circ, 66.972...^\circ$
$\text{So looking at the graph, we deduce that this case spits out}\\ \boxed{15^\circ \le \theta \le 23.027...^\circ\\ 66.972...^\circ \le \theta \le 75^\circ}$
$\textbf{The final answer needs both equations to hold true simultaneously.}\\ \text{Hence, combining cases, our answer is (to 2 d.p.)}\\ \boxed{66.98^\circ \le \theta \le 75^\circ}$

Okay thanks! And yes ill make sure to attach a pic next time
Title: Re: 3U Maths Question Thread
Post by: 16rf on October 07, 2016, 10:17:11 am
Could anyone explain the answer to question 13)d)ii) of the 2012 HSC extension 1 exam.
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 07, 2016, 10:35:46 am
Could anyone explain the answer to question 13)d)ii) of the 2012 HSC extension 1 exam.
$\text{We want to know when }C(t)=0.3\\ \text{i.e. }0.3=1.4te^{-0.2t} \iff \boxed{0.3-1.4te^{-0.2t}=0}$
$\text{Let }f(t)=0.3-1.4te^{-0.2t}\\ \text{So using the same product rule as in part (i)}\\ f^\prime(t) = e^{-0.2t}(0.28t-1.4)$
\text{So by Newton's method}\\ \begin{align*}t_1&=t_0-\frac{f(t_0)}{f^\prime(t_0)}\\ &= 20 - \frac{f(20)}{f^\prime(20)}\end{align*}
$\text{Then plug everything in.}$
Title: Re: 3U Maths Question Thread
Post by: lha on October 07, 2016, 01:04:08 pm
Is there a set list of proofs that we need to know for circle geo?
Title: Re: 3U Maths Question Thread
Post by: jakesilove on October 07, 2016, 01:09:00 pm
Is there a set list of proofs that we need to know for circle geo?

For sure; you can find a great summary HERE!
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 07, 2016, 01:09:56 pm
Is there a set list of proofs that we need to know for circle geo?
You may or may not be surprised, but check  the syllabus. Scroll to around page 28.
Title: Re: 3U Maths Question Thread
Post by: lha on October 07, 2016, 01:18:56 pm
So much to remember  :'( but thank you RuiAce and jakesilove
Title: Re: 3U Maths Question Thread
Post by: jakesilove on October 07, 2016, 01:22:38 pm
So much to remember  :'( but thank you RuiAce and jakesilove

No worries! There is a lot to remember, this is true, but there's also plenty of time before the exam :)
Title: Re: 3U Maths Question Thread
Post by: Blissfulmelodii on October 08, 2016, 03:02:53 pm
Hey,
I know this is probably really really simple but i wasn't in class when this was covered so i was just wondering if someone could explain how to find the value of this question to me?

Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 08, 2016, 03:13:01 pm
Hey,
I know this is probably really really simple but i wasn't in class when this was covered so i was just wondering if someone could explain how to find the value of this question to me?
$\text{That one's undefined.}\\ \text{Did you mean }\lim_{x \to 3}\frac{\sin(x-3)}{(x-3)(x+2)}$
Title: Re: 3U Maths Question Thread
Post by: Blissfulmelodii on October 08, 2016, 03:16:08 pm
$\text{That one's undefined.}\\ \text{Did you mean }\lim_{x \to 3}\frac{\sin(x-3)}{(x-3)(x+2)}$

ohh yes it's that! Sorry I typed it wrong.
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 08, 2016, 03:22:49 pm
$\text{First, note that limits can be split}\\ \lim_{x\to a}f(x)g(x) = \lim_{x\to a}f(x) \times \lim_{x\to a}g(x) = f(a)g(a)\\ \text{(of course, assuming the limit exists.)}$
$\text{We just use the identity }\lim_{x\to 0}\frac{\sin x}{x}=1$
\begin{align*}\lim_{x\to 3}\frac{\sin (x-3)}{(x-3)(x+2)}&= \left(\lim_{x\to 3}\frac{\sin (x-3)}{x-3}\right)\left(\lim_{x\to 3}\frac{1}{x+2}\right)\\ &= 1 \times \frac{1}{5}\\ &= \frac{1}{5}\end{align*}
Title: Re: 3U Maths Question Thread
Post by: Blissfulmelodii on October 08, 2016, 03:26:02 pm
$\text{First, note that limits can be split}\\ \lim_{x\to a}f(x)g(x) = \lim_{x\to a}f(x) \times \lim_{x\to a}g(x) = f(a)g(a)\\ \text{(of course, assuming the limit exists.)}$
$\text{We just use the identity }\lim_{x\to 0}\frac{\sin x}{x}=1$
\begin{align*}\lim_{x\to 3}\frac{\sin (x-3)}{(x-3)(x+2)}&= \left(\lim_{x\to 3}\frac{\sin (x-3)}{x-3}\right)\left(\lim_{x\to 3}\frac{1}{x+2}\right)\\ &= 1 \times \frac{1}{5}\\ &= \frac{1}{5}\end{align*}

ahh okay, that makes sense.
Is there a proof for that identity or is it just something to memorise?
Thank you!
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 08, 2016, 03:28:08 pm

ahh okay, that makes sense.
Is there a proof for that identity or is it just something to memorise?
Thank you!
There is most certainly a proof for it but memorising it is well beyond 3U.
Title: Re: 3U Maths Question Thread
Post by: Blissfulmelodii on October 08, 2016, 04:04:11 pm
14 past papers in and I currently have a stack of questions that I am stumped on, so i apologise in advance for the spam that will most likely occur.
Can someone please explain how to solve this question from the 2008 HSC paper that I have attached below.
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 08, 2016, 04:27:04 pm
14 past papers in and I currently have a stack of questions that I am stumped on, so i apologise in advance for the spam that will most likely occur.
Can someone please explain how to solve this question from the 2008 HSC paper that I have attached below.
If they're all past HSC papers, look at the answers first in the future. You're more than free to address any confusion with the answers though.
$\text{In part (i), observe that the derivative with respect to }t\text{ is written in terms of }x\\ \text{This suggests that we need to use the classic chain rule trick for related rates of change.}$
$\text{Also note the peculiar }\ell^2+x^2\text{ in the denominator.}\\ \text{This looks like a tangent inverse derivative.}\\ \text{Indeed, easily enough, }\tan \theta=\frac{x}{\ell} \iff \theta = \tan^{-1}\frac{x}{\ell}\\ \text{This is unsurprising, as obviously we want to differentiate }\theta\text{ to something.}$
$\text{Differentiating, we get }\frac{d\theta}{dx}=\frac{\ell}{\ell^2+x^2}$
$\text{Read the question again. All that text says about }v\text{ is that }\frac{dx}{dt}=v\\ \text{So by the chain rule, }\frac{d\theta}{dt}=\frac{d\theta}{dx}\,\frac{dx}{dt}=\frac{v\ell}{\ell^2+x^2}$
________________________________________
$\text{Note that }v\text{ and }\ell\text{ are constants. The only thing that changes is }x\\ \text{So we want to maximise }\frac{d\theta}{dt}=\frac{v\ell}{\ell^2+x^2}\\ \text{But this is 1 mark, because you don't need to differentiate. It's easier than what meets the eye.}$
$\frac{v\ell}{\ell^2+x^2}\text{ is a fraction.}\\ \textbf{The larger the denominator, the smaller the fraction.}\\ \text{So we maximise it by making the denominator smaller.}$$\text{The denominator is obviously at its smallest when }x=0\\ \text{Hence the maximum value is }\frac{d\theta}{dt}=m=\frac{v\ell}{\ell^2+0^2}=\frac{v}{\ell}$
________________________________________
$\frac{m}{4}=\frac{v}{4\ell}\\ \text{So we're just solving }\frac{d\theta}{dt}=\frac{v}{\ell}\\ \text{I'll leave this part to you. Just remember to solve for }\theta\text{, not stop at }x$
Title: Re: 3U Maths Question Thread
Post by: Blissfulmelodii on October 08, 2016, 04:49:20 pm
If they're all past HSC papers, look at the answers first in the future. You're more than free to address any confusion with the answers though.
$\text{In part (i), observe that the derivative with respect to }t\text{ is written in terms of }x\\ \text{This suggests that we need to use the classic chain rule trick for related rates of change.}$
$\text{Also note the peculiar }\ell^2+x^2\text{ in the denominator.}\\ \text{This looks like a tangent inverse derivative.}\\ \text{Indeed, easily enough, }\tan \theta=\frac{x}{\ell} \iff \theta = \tan^{-1}\frac{x}{\ell}\\ \text{This is unsurprising, as obviously we want to differentiate }\theta\text{ to something.}$
$\text{Differentiating, we get }\frac{d\theta}{dx}=\frac{\ell}{\ell^2+x^2}$
$\text{Read the question again. All that text says about }v\text{ is that }\frac{dx}{dt}=v\\ \text{So by the chain rule, }\frac{d\theta}{dt}=\frac{d\theta}{dx}\,\frac{dx}{dt}=\frac{v\ell}{\ell^2+x^2}$
________________________________________
$\text{Note that }v\text{ and }\ell\text{ are constants. The only thing that changes is }x\\ \text{So we want to maximise }\frac{d\theta}{dt}=\frac{v\ell}{\ell^2+x^2}\\ \text{But this is 1 mark, because you don't need to differentiate. It's easier than what meets the eye.}$
$\frac{v\ell}{\ell^2+x^2}\text{ is a fraction.}\\ \textbf{The larger the denominator, the smaller the fraction.}\\ \text{So we maximise it by making the fraction smaller.}$$\text{The fraction is obviously at its smallest when }x=0\\ \text{Hence the maximum value is }\frac{d\theta}{dt}=m=\frac{v\ell}{\ell^2+0^2}=\frac{v}{\ell}$
________________________________________
$\frac{m}{4}=\frac{v}{4\ell}\\ \text{So we're just solving }\frac{d\theta}{dt}=\frac{v}{\ell}\\ \text{I'll leave this part to you. Just remember to solve for }\theta\text{, not stop at }x$

When I said stumped I mean not even the worked solutions on BOSTES makes sense to me lol.
Thank you for the explanation, i definitely needed that! It took me a while to understand the second part but I think i get it now. Thanks again  :)
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 08, 2016, 04:52:35 pm
When I said stumped I mean not even the worked solutions on BOSTES makes sense to me lol.
Thank you for the explanation, i definitely needed that! It took me a while to understand the second part but I think i get it now. Thanks again  :)
BOSTES solutions don't always make sense because they just spit it out. Hence the link - solutions written by other sources are usually clearer.

The best solutions come from the MANSW published books but those aren't for free

But as stated above, feel free to come for help with clarifying the answers anytime
Title: Re: 3U Maths Question Thread
Post by: WLalex on October 11, 2016, 01:41:45 pm

Thanks
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 11, 2016, 01:54:20 pm

Thanks
\text{We need to solve the equation }x=0\text{, i.e.}\\ \begin{align*}5+6\cos 2t+8\sin 2t&=0\\ 8\sin 2t + 6\cos 2t &= -5\end{align*}
$\text{The form of the LHS is the standard form which suggests}\\ \text{that you need to apply an auxiliary angle transformation.}\\ \text{Here, we have}\\ R=\sqrt{8^2+6^2}=10\\ \tan \theta = \frac{6}{8} \iff \theta = \tan^{-1}\frac{3}{4}\\ \textit{Alternatively, use the full derivation.}$
$\text{Hence the equation can be rewritten as}\\ 10\sin \left(2t + \tan^{-1}\frac{3}{4}\right)=-5$
$\textit{At this point, it's not really applying SHM knowledge anymore}\\ \textit{It is just solving a trig equation.}$
$\text{Despite the fact that }t> 0\text{, there will be an infinite number of solutions.}\\ \text{We are interested in the }\textbf{first}\text{ possible value of }t.\\ \text{We solve the trigonometric equation as normal:}$
\begin{align*}10\sin \left(2t + \tan^{-1}\frac{3}{4}\right)&=-5\\\sin \left(2t + \tan^{-1}\frac{3}{4}\right)&=-\frac12\\ 2t+\tan^{-1}\frac34&=-\frac{\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}, \dots\\ t&=\frac{1}{2}\left(-\frac{\pi}{6}-\tan^{-1}\frac{3}{4}\right),\frac{1}{2}\left(\frac{7\pi}{6}-\tan^{-1}\frac{3}{4}\right),\frac{1}{2}\left(\frac{9\pi}{6}-\tan^{-1}\frac{3}{4}\right), \dots\end{align*}
$\text{So recalling that time is positive, the }\textbf{first}\text{ possible solution is}\\ t=\frac{1}{2}\left(\frac{7\pi}{6}-\tan^{-1}\frac{3}{4}\right)\\ =\textit{whatever the calculator says}$
Title: Re: 3U Maths Question Thread
Post by: WLalex on October 11, 2016, 02:55:04 pm

ah, perfect. Thanks Rui
Title: Re: 3U Maths Question Thread
Post by: anotherworld2b on October 13, 2016, 08:28:49 pm
Can i Get help with this question please?
Title: Re: 3U Maths Question Thread
Post by: jakesilove on October 13, 2016, 09:11:19 pm
Can i Get help with this question please?

Hey! So, the question is basically asking us to find the stationary point (ie. the minimum value), and then the x value for which this will occur. Naturally, we employ calculus for this!

Our expanded equation is

$y=\frac{-x^3}{50000}+\frac{7x^2}{5000}-\frac{x}{50}$

So, we start by finding the derivative

$\frac{dy}{dx}=\frac{-3x^2}{50000}+\frac{7x}{2500}-\frac{1}{50}$

If we set this equal to zero, and multiply out our denominators, we can use the quadratic formula to find solutions. We find that the first value for x is

$x=\frac{70-10(19)^{0.5}}{3}$

This is where the maximum sag occurs; sub it back into the original equation to get the maximum sag!

Let me know if I can clarify anything

Jake
Title: Re: 3U Maths Question Thread
Post by: anotherworld2b on October 13, 2016, 10:14:26 pm
thank you for you help
I got confused on how to multiply out the denominators and use the quadratic formula to find solutions. How do you do this?  :o

Hey! So, the question is basically asking us to find the stationary point (ie. the minimum value), and then the x value for which this will occur. Naturally, we employ calculus for this!

Our expanded equation is

$y=\frac{-x^3}{50000}+\frac{7x^2}{5000}-\frac{x}{50}$

So, we start by finding the derivative

$\frac{dy}{dx}=\frac{-3x^2}{50000}+\frac{7x}{2500}-\frac{1}{50}$

If we set this equal to zero, and multiply out our denominators, we can use the quadratic formula to find solutions. We find that the first value for x is

$x=\frac{70-10(19)^{0.5}}{3}$

This is where the maximum sag occurs; sub it back into the original equation to get the maximum sag!

Let me know if I can clarify anything

Jake
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 13, 2016, 10:19:07 pm
thank you for you help
I got confused on how to multiply out the denominators and use the quadratic formula to find solutions. How do you do this?  :o
$\frac{dy}{dx}=0\text{ means }\frac{-3x^2}{50000}+\frac{7x}{5000}-\frac{1}{50}=0$
$\text{What's the denominator? 50000. So if we multiply it by 50000 we get}\\ -3x^2+70x-100=0$
$\text{If you don't know the quadratic formula }x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\text{ you need to check your textbook.}$
Title: Re: 3U Maths Question Thread
Post by: anotherworld2b on October 13, 2016, 10:54:25 pm
I was wondering why is it 7x/5000? I got 14x/5000 for some reason?
$\frac{dy}{dx}=0\text{ means }\frac{-3x^2}{50000}+\frac{7x}{5000}-\frac{1}{50}=0$
$\text{What's the denominator? 50000. So if we multiply it by 50000 we get}\\ -3x^2+70x-100=0$
$\text{If you don't know the quadratic formula }x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\text{ you need to check your textbook.}$
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 13, 2016, 11:01:31 pm
I copied Jake's working wrong. That was meant to be 7x/2500 which is equivalent to 14x/5000
Title: Re: 3U Maths Question Thread
Post by: anotherworld2b on October 13, 2016, 11:11:35 pm
Is it supposed to be like this?  ???
Title: Re: 3U Maths Question Thread
Post by: jakesilove on October 14, 2016, 10:35:50 am
Is it supposed to be like this?  ???

Looks right!
Title: Re: 3U Maths Question Thread
Post by: kiwiberry on October 15, 2016, 01:18:08 pm
how do you integrate this :-\
$\int_{0}^{3}(\sqrt{x(x-3)^2}\, dx$

Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 15, 2016, 01:20:56 pm
\text{Note that the square root can be distributed here.}\\ \begin{align*}\int_0^3\sqrt{x(x-3)^2}dx&=\int_0^3[x(x-3)^2]^\frac{1}{2}dx\\ &=\int_0^3x^\frac{1}{2}(x-3)dx\\&=\int_0^3\left(x^\frac{3}{2}-3x^\frac{1}{2}\right)dx\end{align*}
Title: Re: 3U Maths Question Thread
Post by: kiwiberry on October 15, 2016, 01:30:38 pm
ohh thank you!! could you please do this one as well :)
$\int_{0}^{1}\frac{1}{\sqrt{(2-x)(x+1)}}\, dx$
Title: Re: 3U Maths Question Thread
Post by: jakesilove on October 15, 2016, 01:36:28 pm
ohh thank you!! could you please do this one as well :)
$\int_{0}^{1}\frac{1}{\sqrt{(2-x)(x-1)}}\, dx$

Hey! I would expand the the denominator out, and then use completing the square to get something that looks like a^2-(x-c)^2 in the square root. From there, you can easily let u=x-c so that you can integrate using inverse sin/cos! Show us some working out, and if you still can't get it we'd be happy to write up a solution.
Title: Re: 3U Maths Question Thread
Post by: kiwiberry on October 15, 2016, 09:39:43 pm
Hey! I would expand the the denominator out, and then use completing the square to get something that looks like a^2-(x-c)^2 in the square root. From there, you can easily let u=x-c so that you can integrate using inverse sin/cos! Show us some working out, and if you still can't get it we'd be happy to write up a solution.

i think i got it! :D thanks
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 15, 2016, 09:46:26 pm
i think i got it! :D thanks
Looking good
Title: Re: 3U Maths Question Thread
Post by: jamgoesbam on October 17, 2016, 08:39:53 am
Hi there,
I was just wondering how did they get the answer for this question?? Could someone please explain? Thanks! :)
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 17, 2016, 09:33:33 am

Hi there,
I was just wondering how did they get the answer for this question?? Could someone please explain? Thanks! :)
The first two parts are elementary formula work
_________________
$\text{For part iii, they had a mistake in line 1 despite lines 2 and 3 being correct.}\\ \text{First note that }\\\alpha+\beta+\gamma=4\\\alpha\beta+\alpha\gamma+\beta\gamma=5\\\alpha\beta\gamma=1$
$\text{The polynomial given has roots }\alpha, \beta, \gamma\\ \text{Hence, if you sub back in the results above, we can rewrite the polynomial as}//x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\alpha\gamma+\beta\gamma)x-\alpha\beta\gamma$
$\text{In fact, this is always true.}\\ \text{IF a polynomial has roots }\alpha,\beta,\gamma\\ \text{THEN it can always be written as }x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\alpha\gamma+\beta\gamma)x-\alpha\beta\gamma$
___________
$\text{We now use this fact above.}\\ \text{We want a new polynomial, except with THESE roots}\\ 2\alpha,2\beta,2\gamma$
$\text{Hence, in our above formula, we REPLACE }\alpha, \beta, \gamma\\ \text{with }2\alpha,2\beta,2\gamma$
$\text{So our NEW polynomial takes the form}\\ x^3-(2\alpha+2\beta+2\gamma)x^2+(4\alpha\beta+4\alpha\gamma+4\beta\gamma)x+8\alpha\beta\gamma$
$\text{Sub everything back in and our answer will match their final answer}$
Title: Re: 3U Maths Question Thread
Post by: bethjomay on October 18, 2016, 09:18:30 pm
If our working is a little dodgy (I.e made some leaps in my head) but the answer is correct can you still get full marks?
Title: Re: 3U Maths Question Thread
Post by: jakesilove on October 18, 2016, 09:26:08 pm
If our working is a little dodgy (I.e made some leaps in my head) but the answer is correct can you still get full marks?

You certainly can! The only exception to this is in a 'show that' question, because they'll think you're bullshitting since you know the answer, or if you get the question incorrect
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 18, 2016, 09:34:52 pm
If our working is a little dodgy (I.e made some leaps in my head) but the answer is correct can you still get full marks?
Yep leaps are fine unless it's one of those show that questions. If it says show that (or prove that etc.) you should only skip the trivial steps.

The HSC can't penalise you if you're showing an advanced level of algebraic skills and manipulation, until it becomes overboard, which isn't easy.
Title: Re: 3U Maths Question Thread
Post by: jamgoesbam on October 18, 2016, 10:05:17 pm
Hi again,

In a game, two dice are rolled and the score given is the maximum of the two numbers on the uppermost faces. For example, if the dice shows a three and five, the score is a five.
d) Given that one of the dice shows a three, what is the probability of getting a score greater than five?

Title: Re: 3U Maths Question Thread
Post by: jamonwindeyer on October 18, 2016, 11:16:26 pm
Hi again,

In a game, two dice are rolled and the score given is the maximum of the two numbers on the uppermost faces. For example, if the dice shows a three and five, the score is a five.
d) Given that one of the dice shows a three, what is the probability of getting a score greater than five?

Hey! Okay, so for the first question, honestly the least confusing way to think about it is to draw a table of possibilities. Like, with all the stuff we know i  Extension, sometimes we go back to basics and wonder why the hell we were so confused. Here is the table with the possibilities on the outside, and the score on the inside.

(http://i.imgur.com/IihzrAk.png)

The bits in green represent the throws where a '3' appeared; that's our restricted sample space! There are 11 elements in that restricted sample space, and two of those have a score greater than 5 (that's six). So, $$P=\frac{2}{11}$$ ;D
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 18, 2016, 11:20:35 pm
Hi Neutron,
I think, I found an easy answer for your question, 2cos(2x) = 1-3sin(2x).Before I prove it, I need to point out that cos(x) = 0 is not a solution of the equation since that forces LHS=-2(because cos(2x) =2cos^2(x) -1 ) and RHS = 1(since sin(2x) = 2 sin(x) cos(x))
Solution:
2cos(2x) = 1-3sin(2x)
3(cos(2x) + sin(2x)) = 1+cos(2x) = 2cos^2(x)
3(cos^2(x) -sin^2(x) +2sin(x)cos(x) ) = 2cos^2(x) (since cos(2x) = cos^2(x) -sin^2(x) )
collecting cos^2(x) terms yields
cos^2(x) = sin(x) (3sin(x) - 6cos(x))
1 = (sin(x) / cos(x)) ( 3sin(x) -6cos(x) )/cos(x) (dividing by cos^2(x), it is possible since cos(x) is not zero )
1 = tan(x)(3tan(x) - 6)
3tan^2(x) -6tan(x) -1 = 0
tan(x) = 1+2sqrt(3)/3 or 1-2sqrt(3)/3
From there you can find the angles.I do agree with -8 degree but I don't think -24 degree is correct, I put it into the calculator and it gave me
RHS= 1.33 , LHS =3.23
A general tip to tackle to these type of questions where you get ' sin(x)cos(x)' somewhere when you solve it is to represented in terms of tan(x), (If you get lucky! )
I hope that was helpful. (Y)

Please quote the post you are referring to. This is definitely not a recent post you are alluding to, and potentially very outdated.
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: Mahan on October 18, 2016, 11:28:06 pm
Hey! Okay so i don't seem to be able to find an easy way to do this (like I got an answer but the method was waaay too complicated I reckon but I got -8 degrees and 48 seconds and -24 degrees and 54 seconds??) D: I was wondering whether you guys could help me! thanks :D

Solve the following equation:
2cos2ϴ=1-3sin2ϴ   0≤ϴ≤360

Thank you!

Neutron

Hi Neutron,
I think, I found an easy answer for your question, 2cos(2x) = 1-3sin(2x).Before I prove it, I need to point out that cos(x) = 0 is not a solution of the equation since that forces LHS=-2(because cos(2x) =2cos^2(x) -1 ) and RHS = 1(since sin(2x) = 2 sin(x) cos(x))
Solution:
2cos(2x) = 1-3sin(2x)
3(cos(2x) + sin(2x)) = 1+cos(2x) = 2cos^2(x)
3(cos^2(x) -sin^2(x) +2sin(x)cos(x) ) = 2cos^2(x) (since cos(2x) = cos^2(x) -sin^2(x) )
collecting cos^2(x) terms yields
cos^2(x) = sin(x) (3sin(x) - 6cos(x))
1 = (sin(x) / cos(x)) ( 3sin(x) -6cos(x) )/cos(x) (dividing by cos^2(x), it is possible since cos(x) is not zero )
1 = tan(x)(3tan(x) - 6)
3tan^2(x) -6tan(x) -1 = 0
tan(x) = 1+2sqrt(3)/3 or 1-2sqrt(3)/3
From there you can find the angles.I do agree with -8 degree but I don't think -24 degree is correct, I put it into the calculator and it gave me
RHS= 1.33 , LHS =3.23
A general tip to tackle to these type of questions where you get ' sin(x)cos(x)' somewhere when you solve it is to represented in terms of tan(x), (If you get lucky! )
I hope that was helpful. (Y)
Title: Re: 3U Maths Question Thread
Post by: jamonwindeyer on October 18, 2016, 11:30:22 pm
Hi again,

In a game, two dice are rolled and the score given is the maximum of the two numbers on the uppermost faces. For example, if the dice shows a three and five, the score is a five.
d) Given that one of the dice shows a three, what is the probability of getting a score greater than five?

And now for your Calculus questions!

So the first one starts with some calculus:

$g(\theta)=\sin{\theta}-\theta\cos{\theta}\\g'(\theta)=\cos{\theta}+\theta\sin{\theta}-\cos{\theta}\\=\theta\sin{\theta}$

Now in the range given, $$\sin{\theta}$$ is always positive (and so is $$\theta$$, clearly), so our condition is fulfilled!

Unfortunately for the rest of your questions I don't have the expression! Mind uploading what you have so far? I'll take it from there (or maybe this well help enough) ;D
Title: Re: 98 in 3U Maths: Ask me Anything!
Post by: jamonwindeyer on October 18, 2016, 11:32:00 pm
Hi Neutron,
I think, I found an easy answer for your question, 2cos(2x) = 1-3sin(2x).Before I prove it, I need to point out that cos(x) = 0 is not a solution of the equation since that forces LHS=-2(because cos(2x) =2cos^2(x) -1 ) and RHS = 1(since sin(2x) = 2 sin(x) cos(x))
Solution:
2cos(2x) = 1-3sin(2x)
3(cos(2x) + sin(2x)) = 1+cos(2x) = 2cos^2(x)
3(cos^2(x) -sin^2(x) +2sin(x)cos(x) ) = 2cos^2(x) (since cos(2x) = cos^2(x) -sin^2(x) )
collecting cos^2(x) terms yields
cos^2(x) = sin(x) (3sin(x) - 6cos(x))
1 = (sin(x) / cos(x)) ( 3sin(x) -6cos(x) )/cos(x) (dividing by cos^2(x), it is possible since cos(x) is not zero )
1 = tan(x)(3tan(x) - 6)
3tan^2(x) -6tan(x) -1 = 0
tan(x) = 1+2sqrt(3)/3 or 1-2sqrt(3)/3
From there you can find the angles.I do agree with -8 degree but I don't think -24 degree is correct, I put it into the calculator and it gave me
RHS= 1.33 , LHS =3.23
A general tip to tackle to these type of questions where you get ' sin(x)cos(x)' somewhere when you solve it is to represented in terms of tan(x), (If you get lucky! )
I hope that was helpful. (Y)

Welcome to the forums! But that question is nearly 9 months old, love that you are keen to help out but be sure to click to the more recent pages to see the unanswered questions! :)
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 18, 2016, 11:35:12 pm
$\text{Whilst it's been brought up anyway}\\ \text{The easiest way is to bombard it with an auxiliary angle transformation.}$
\begin{align*}3\sin (2x) + 2\cos (2x) = 1\\ \sqrt{13}\sin\left(2x+\tan^{-1}\frac{2}{3}\right)&=1\\ \sin\left(2x+\tan^{-1}\frac{2}{3}\right)&=\frac{1}{\sqrt{13}} \\ 2x+\tan^{-1}\frac{2}{3}&=\sin^{-1}\frac{1}{\sqrt{13}}, 180^\circ -\sin^{-1}\frac{1}{\sqrt{13}} \end{align*}
$\text{Leave it all exact and round at the end}$
Title: Re: 3U Maths Question Thread
Post by: jamgoesbam on October 19, 2016, 12:51:26 pm
And now for your Calculus questions!

So the first one starts with some calculus:

$g(\theta)=\sin{\theta}-\theta\cos{\theta}\\g'(\theta)=\cos{\theta}+\theta\sin{\theta}-\cos{\theta}\\=\theta\sin{\theta}$

Now in the range given, $$\sin{\theta}$$ is always positive (and so is $$\theta$$, clearly), so our condition is fulfilled!

Unfortunately for the rest of your questions I don't have the expression! Mind uploading what you have so far? I'll take it from there (or maybe this well help enough) ;D

Hi Jamon,
Thanks for the help so far! This was the first part of the question I forgot to upload. I couldn't get Part D, and I checked some online answers which I still didn't understand (also attached below). Could you please explain what it all means? :'D Thanks heaps!! :)
Title: Re: 3U Maths Question Thread
Post by: jamonwindeyer on October 19, 2016, 10:17:06 pm
Hi Jamon,
Thanks for the help so far! This was the first part of the question I forgot to upload. I couldn't get Part D, and I checked some online answers which I still didn't understand (also attached below). Could you please explain what it all means? :'D Thanks heaps!! :)

Awesome!! Okay, so if we want $$\frac{dA}{d\theta}=0$$, we really just want this (the denominator can't be zero and neither can w):

$\cos{\theta}\left(\sin{\theta}-\theta\cos{\theta}\right)=0$

Let's try and use what we proved above. If we go back to the function $$g(\theta)$$, $$g(0)=0$$. But the function is always increasing, that's what we proved, so that means that for any other value of $$\theta>0$$, $$g(\theta)>0$$.

So that means above, we resort to:

$\cos{\theta}=0\\\theta=\frac{\pi}{2}\quad(0<\theta<\pi)$

That's what the answer is getting at. Does it help? ;D
Title: Re: 3U Maths Question Thread
Post by: nimasha.w on October 21, 2016, 09:37:00 pm
hi! can i please have some help on finding the latest time in part ii?
Title: Re: 3U Maths Question Thread
Post by: teapancakes08 on October 21, 2016, 10:37:31 pm
Have any tips into showing the working for this question?

2. b) "Given that PQ passes through (0, 6a) find the equation of locus of T"

I worked out the first part which was (a) Find out the point of intersection of T of the tangents (2ap, ap^2) and Q (2aq, aq^2) on the parabola x^2 = 4ay" in which the answer was T ( a(p+q), apq) – found by finding the normal and using simultaneous equations. Would I use T to help solve question 2b?
Title: Re: 3U Maths Question Thread
Post by: jamonwindeyer on October 22, 2016, 12:28:45 am
hi! can i please have some help on finding the latest time in part ii?

Howdy! Sure thing!

Just as a side note, if you need help with a latter part of the question, try and attach your working so far! That way we can just pick up from where you left off :)

So the max height and time you obtained in A:

$h\approx7041m\\t\approx20.41s\\$

Now for Part B, let's get an expression for the angle of travel:

$\tan{\theta}=\frac{\dot{y}}{\dot{x}}\\=\frac{200-9.8t}{200}=1-\frac{9.8}{200}t$

Now we assume that we are working with negative acute angles here; doesn't make sense otherwise. So we are looking at values of $$\theta$$ between -45 and -60 degrees, meaning $$\tan{\theta}$$ ranges from -1 to $$-\sqrt{3}$$.

$\therefore -\sqrt{3}\le1-\frac{9.8}{200}t\le-1\\2\le\frac{9.8}{200}t\le\sqrt{3}+1\\40.8\le t\le55.76$

Title: Re: 3U Maths Question Thread
Post by: jamonwindeyer on October 22, 2016, 12:59:00 am
Have any tips into showing the working for this question?

2. b) "Given that PQ passes through (0, 6a) find the equation of locus of T"

I worked out the first part which was (a) Find out the point of intersection of T of the tangents (2ap, ap^2) and Q (2aq, aq^2) on the parabola x^2 = 4ay" in which the answer was T ( a(p+q), apq) – found by finding the normal and using simultaneous equations. Would I use T to help solve question 2b?

Let me lend a hand! So you are correct, T is essential, since it has the coordinates we need. But we want more information to help eliminate our parameters.

Let's try and use the new info we were given. Let's look at the gradient of PQ:

$m_\text{PQ}=\frac{p+q}{2}$

But it passes through (0,6a), so let's get another expression for that:

$m_\text{PQ}'=\frac{ap^2-6a}{2ap}=\frac{p^2-6}{2p}$

Equating these:

$\frac{p^2-6}{2p}=\frac{p+q}{2}\\\frac{-2}{p}=\frac{q}{2}\\pq=-4$

So this is a new bit of information for us to use to find our locus! Remember, the idea is to tie x and y back in to the T coordinates. Let me show you; start by considering the y-coordinate:

$y_T=apq=-4a$

And actually, believe it or not, that's it. We have an equation that defines a set of points that obey the given conditions! Any point on the line $$y=-4a$$ will satisfy our requirements, and so THAT is the locus of the Point T :)

Title: Re: 3U Maths Question Thread
Post by: fizzy.123 on October 22, 2016, 01:15:24 am
Hi, this isnt an actual maths question, but i was wondering what mark out of 70 you will need for Maths ext 1 for it to be scaled to a 85+?
Title: Re: 3U Maths Question Thread
Post by: FallonXay on October 22, 2016, 09:02:43 am
Hi for this question part ii, how do you solve for the constant B?

Thanks!
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 22, 2016, 09:20:14 am
Hi for this question part ii, how do you solve for the constant B?

Thanks!
One way of approaching this question is to work with negative time.

Let t=0 be when the object was found, which was at 30oC.

A = 22 because that's the ambient temperature (of the park).
Then using t=0, T=30 we can get B=8.

Then, try to work backwards in time to find when T=37.
Title: Re: 3U Maths Question Thread
Post by: FallonXay on October 22, 2016, 09:30:55 am
One way of approaching this question is to work with negative time.

Let t=0 be when the object was found, which was at 30oC.

A = 22 because that's the ambient temperature (of the park).
Then using t=0, T=30 we can get B=8.

Then, try to work backwards in time to find when T=37.

Yeah, I got B=8, although the answers say B=15?>
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 22, 2016, 09:37:36 am
Yeah, I got B=8, although the answers say B=15?>
They didn't do negative time. They did positive time.

If we use positive time, initially the object's temperature is 37, not 30.
Title: Re: 3U Maths Question Thread
Post by: FallonXay on October 22, 2016, 09:41:13 am
They didn't do negative time. They did positive time.

If we use positive time, initially the object's temperature is 37, not 30.

Ahh, I see. Cheers  :)
Title: Re: 3U Maths Question Thread
Post by: FallonXay on October 22, 2016, 11:32:34 am
With this question part (iii), how come you can't solve by mathematical induction? (or can you?)
Title: Re: 3U Maths Question Thread
Post by: atar27 on October 22, 2016, 11:50:22 am
Hey, this may be a stupid question but when they tell us to simplify in a binomial question what do they mean?
I have attached a photo below:

Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 22, 2016, 12:21:46 pm

With this question part (iii), how come you can't solve by mathematical induction? (or can you?)
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 22, 2016, 12:27:51 pm
Hey, this may be a stupid question but when they tell us to simplify in a binomial question what do they mean?
I have attached a photo below:

$\textit{Simplify}\text{ in binomial theorem implies to USE}\\\text{the binomial theorem to achieve a much neater result.}$
$\text{Here, it is clear that we are just considering }(1+x)^{n-1}$
\begin{align*}(1+x)^{n-1}&=\binom{n-1}{0}+\binom{n-1}{1}x+\binom{n-1}{2}x^2+\dots+\binom{n-1}{n-2}x^{n-2}+\binom{n-1}{n-1}x^{n-1}\\ \text{Put }x=1&\text{ as }x\text{ vanishes with no trace.}\\ 2^{n-1}&=\binom{n-1}{0}+\binom{n-1}{1}+\binom{n-1}{2}+\dots+\binom{n-1}{n-2}+\binom{n-1}{n-1}\\ 2^{n-1}-2&=\binom{n-1}{1}+\binom{n-1}{2}+\dots+\binom{n-1}{n-2}\\ 2n(2^{n-2}-1)&=n\binom{n-1}{1}+n\binom{n-1}{2}+\dots+n\binom{n-1}{n-2}\end{align*}
Title: Re: 3U Maths Question Thread
Post by: FallonXay on October 22, 2016, 01:50:18 pm
Hiya, how do you do d (ii) of this question?  :P
Title: Re: 3U Maths Question Thread
Post by: FallonXay on October 22, 2016, 03:30:17 pm
and how would you solve Q6/10?
Title: Re: 3U Maths Question Thread
Post by: jakesilove on October 22, 2016, 03:43:07 pm
Hiya, how do you do d (ii) of this question?  :P

Hey! For Newton's method, we need to set up an equation such that 0.3 is a zero. Recall that Newton's method is a way of finding x-intercepts! Therefore, we create the equation

$f(t)=C(t)-0.3$

You can then use Newton's method in the regular way, starting from t=20s! Let me know if you want an actual worked answer for this
Title: Re: 3U Maths Question Thread
Post by: FallonXay on October 22, 2016, 03:47:57 pm
Hey! For Newton's method, we need to set up an equation such that 0.3 is a zero. Recall that Newton's method is a way of finding x-intercepts! Therefore, we create the equation

$f(t)=C(t)-0.3$

You can then use Newton's method in the regular way, starting from t=20s! Let me know if you want an actual worked answer for this

How does that equation make 0.3 a zero/root?
Title: Re: 3U Maths Question Thread
Post by: jakesilove on October 22, 2016, 03:48:50 pm
and how would you solve Q6/10?

For 6, you can go straight to your reference sheet. Instead of a theta, you have 2x, so just let 2x equal the general solution for arcsin(a) (again, this is on your formula sheet). Divide through by two, and you're done!

10 is an interesting one. I can't really think of a better method than guess and check to be honest. Nah, scratch that. We can see clearly that x=3 is a solution to our given equation. However, x=3 is NOT a solution to a) or b), and if it were we would be dividing by zero. So, c) and d) are left. Unfortunately, x=3 is a solution to both of them, so that's not very helpful. x=2 is also a solution to our given equation, but NOT for d), leaving us with c) as the answer :)

Jake
Title: Re: 3U Maths Question Thread
Post by: jakesilove on October 22, 2016, 03:50:51 pm
How does that equation make 0.3 a zero/root?

We want to find where C(t)=0.3, right? Like, for what value of t will C(t)=0.3. If I SHIFT the graph down 0.3, then what used to be 0.3 is now going to be zero. Thus, it will be an x-intercept. But, just logically, if I want C(t) to equal 0.3, then in that case C(t)-0.3=0.3-0.3=0. Does that make sense? Definitely a difficult concept, but once you've seen it once you'll be able to answer any similar question :)
Title: Re: 3U Maths Question Thread
Post by: FallonXay on October 22, 2016, 03:54:05 pm
We want to find where C(t)=0.3, right? Like, for what value of t will C(t)=0.3. If I SHIFT the graph down 0.3, then what used to be 0.3 is now going to be zero. Thus, it will be an x-intercept. But, just logically, if I want C(t) to equal 0.3, then in that case C(t)-0.3=0.3-0.3=0. Does that make sense? Definitely a difficult concept, but once you've seen it once you'll be able to answer any similar question :)

Ahh, rightyo. Got it - thanks!!!  :)
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 22, 2016, 04:05:09 pm
and how would you solve Q6/10?
For 6, you can go straight to your reference sheet. Instead of a theta, you have 2x, so just let 2x equal the general solution for arcsin(a) (again, this is on your formula sheet). Divide through by two, and you're done!

10 is an interesting one. I can't really think of a better method than guess and check to be honest. Nah, scratch that. We can see clearly that x=3 is a solution to our given equation. However, x=3 is NOT a solution to a) or b), and if it were we would be dividing by zero. So, c) and d) are left. Unfortunately, x=3 is a solution to both of them, so that's not very helpful. x=2 is also a solution to our given equation, but NOT for d), leaving us with c) as the answer :)

Jake
Question 10 was addressed in post #721
Title: Re: 3U Maths Question Thread
Post by: FallonXay on October 22, 2016, 04:29:09 pm
I've got another 3 questions  ;D

Question f [attachment 1]:

Question f ii [attachment 2]:

Question 12 d i [attachment 3]:

P.s sorry for the spam of questions; and thank you for answering all of my questions - much appreciated!!!  :)
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 22, 2016, 04:46:28 pm
I've got another 3 questions  ;D

Question f [attachment 1]:

Question f ii [attachment 2]:

Question 12 d i [attachment 3]:

P.s sorry for the spam of questions; and thank you for answering all of my questions - much appreciated!!!  :)
$\text{First note, in your head, that the domain of }\ln(x^2+e)\text{ is all real }x\\ \text{Now draw a freehand sketch of }y=\ln x$
$y=\ln x\text{ is monotonic increasing.}\\ \text{The higher the value of }x\text{, the higher the value of }y.$
$\text{Now consider our function }f(x)=\ln (x^2+e)\\ \text{What is the lowest (minimum) value of }x^2+e?\\ \text{The answer is }e\\ \text{Hence the minimum of }\ln(x^2+e)\text{ is }\ln e = 1$
$\text{What is the maximum? Well there isn't any.}\\ \text{And note that }\ln(x^2+e)\text{ is obviously continuous.}\\ \text{Hence the range is }y\ge 1$
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 22, 2016, 04:50:41 pm
I've got another 3 questions  ;D

Question f [attachment 1]:

Question f ii [attachment 2]:

Question 12 d i [attachment 3]:

P.s sorry for the spam of questions; and thank you for answering all of my questions - much appreciated!!!  :)
Next one addressed in post #710
Title: Re: 3U Maths Question Thread
Post by: FallonXay on October 22, 2016, 04:58:33 pm
Also, for this question. How come the answers don't adjust the domain so that x+pi/6 = 7pi/3?
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 22, 2016, 05:00:23 pm
Also, for this question. How come the answers don't adjust the domain so that x+pi/6 = 7pi/3?
$\text{That falls outside.}\\ x+\frac{\pi}{6}=\frac{7\pi}{3} \implies x = \frac{13\pi}{6} > 2\pi$
Title: Re: 3U Maths Question Thread
Post by: FallonXay on October 22, 2016, 05:03:14 pm
$\text{That falls outside.}\\ x+\frac{\pi}{6}=\frac{7\pi}{3} \implies x = \frac{13\pi}{6} > 2\pi$

Oh wait oops, my bad - thanks  :)
Title: Re: 3U Maths Question Thread
Post by: jamonwindeyer on October 22, 2016, 05:13:02 pm
I've got another 3 questions  ;D

Question f [attachment 1]:

Question f ii [attachment 2]:

Question 12 d i [attachment 3]:

P.s sorry for the spam of questions; and thank you for answering all of my questions - much appreciated!!!  :)

And I'll tag in for the last one. Notice that when we have a right angle, it is almost definitely related to the product of the gradients being equal to -1 (since they are perpendicular). So:

$m_\text{AC}=\frac{k}{t}\\m_\text{BC}=\frac{-y}{t}$

Therefore we have the relationship:

$\frac{-ky}{t^2}=-1\\y=\frac{t^2}{k}$

Since P is in line with the points B and C, we know that we can simply say that P has coordinates $$\left(t,\frac{t^2}{k}\right)$$. That's the parametric relationship specified in the question, we're done (it is also a parabola, if you wanted to prove that you could!) ;D

Key Point: If theres a right angle in a locus question, find the gradients, put their product equal to -1, and you'll almost always have what you need. :)
Title: Re: 3U Maths Question Thread
Post by: jamgoesbam on October 22, 2016, 05:14:12 pm
Hi there!! I've seen multiple answers to this long binomial question and... nope still don't really get it :'D Could you pleaseee explain this question ('cause you guys have really good explanations)! Thanks heaps!! :)
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 22, 2016, 05:15:54 pm
Hi there!! I've seen multiple answers to this long binomial question and... nope still don't really get it :'D Could you pleaseee explain this question ('cause you guys have really good explanations)! Thanks heaps!! :)
Title: Re: 3U Maths Question Thread
Post by: WLalex on October 22, 2016, 05:35:58 pm
Hey! wondering if someone can help me out with part (iv)

Tried to find the cartesian equation but finding a value for t and substituting that into y and as you can see (I've attached my working), it feels like I'm on a road to nowhere ahaha. So is this the right way to attack this question or am i missing something?

This is usually how I would do these but seems not to be working
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 22, 2016, 05:43:20 pm
Hey! wondering if someone can help me out with part (iv)

Tried to find the cartesian equation but finding a value for t and substituting that into y and as you can see (I've attached my working), it feels like I'm on a road to nowhere ahaha. So is this the right way to attack this question or am i missing something?

This is usually how I would do these but seems not to be working
\text{By comparison, squaring seems appropriate.}\\ \text{Observe that}\\ \begin{align*}x^2&=a^2\left(t^2-2+\frac{1}{t^2}\right)\\ &= a\left[a\left(t^2+1+\frac{1}{t^2}\right)-3a \right]\\ &=a(y-3a)\end{align*}
Title: Re: 3U Maths Question Thread
Post by: WLalex on October 22, 2016, 05:45:26 pm
\text{By comparison, squaring seems appropriate.}\\ \text{Observe that}\\ \begin{align*}x^2&=a^2\left(t^2-2+\frac{1}{t^2}\right)\\ &= a\left[a\left(t^2+1+\frac{1}{t^2}\right)-3a \right]\\ &=a(y-3a)\end{align*}

yeah sorry, it was the only size that fit. Woah that is so much easier...how do we know what method to use?

Thanks also :)
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 22, 2016, 05:47:11 pm
yeah sorry, it was the only size that fit. Woah that is so much easier...how do we know what method to use?

Thanks also :)
t and 1/t, compared to t2 and 1/t2, is immediately suggestive of squaring.

But you need to be careful of what happens AFTER squaring.
Title: Re: 3U Maths Question Thread
Post by: teapancakes08 on October 22, 2016, 08:13:34 pm
Let me lend a hand! So you are correct, T is essential, since it has the coordinates we need. But we want more information to help eliminate our parameters.

Let's try and use the new info we were given. Let's look at the gradient of PQ:

$m_\text{PQ}=\frac{p+q}{2}$

But it passes through (0,6a), so let's get another expression for that:

$m_\text{PQ}'=\frac{ap^2-6a}{2ap}=\frac{p^2-6}{2p}$

Equating these:

$\frac{p^2-6}{2p}=\frac{p+q}{2}\\\frac{-2}{p}=\frac{q}{2}\\pq=-4$

So this is a new bit of information for us to use to find our locus! Remember, the idea is to tie x and y back in to the T coordinates. Let me show you; start by considering the y-coordinate:

$y_T=apq=-4a$

And actually, believe it or not, that's it. We have an equation that defines a set of points that obey the given conditions! Any point on the line $$y=-4a$$ will satisfy our requirements, and so THAT is the locus of the Point T :)

Ah, I see – so you find the gradients and then simplify your way through. Although...I'm little confused as how (p^2 – 6)/2p = (p+q)/2 simplified into -2/p = q/2 . But other than that I understand how to do it now. Thanks so much for the explanation ;D
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 22, 2016, 08:18:27 pm
Ah, I see – so you find the gradients and then simplify your way through. Although...I'm little confused as how (p^2 – 6)/2p = (p+q)/2 simplified into -2/p = q/2 . But other than that I understand how to do it now. Thanks so much for the explanation ;D
I think that should be -3/p there, not -2/p
Title: Re: 3U Maths Question Thread
Post by: jamgoesbam on October 22, 2016, 09:09:36 pm
Hi there! Could someone please explain part e)? Would be much appreciated! :)
Title: Re: 3U Maths Question Thread
Post by: teapancakes08 on October 22, 2016, 09:36:27 pm
I think that should be -3/p there, not -2/p

That...makes more sense, actually. Although I'm still not sure how it simplifies into -3/p = q/2...
I did, however, get the answer:

$y_T=apq=-6a$

By the messy working out below. Would it be okay to write it like that?
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 22, 2016, 09:41:36 pm
That...makes more sense, actually. Although I'm still not sure how it simplifies into -3/p = q/2...
I did, however, get the answer:

$y_T=apq=-6a$

By the messy working out below. Would it be okay to write it like that?
\text{It's nothing more than algebra.}\\ \text{What you did is correct, but this was all Jamon did.}\\ \begin{align*}\frac{p^2-6}{2p}&=\frac{p+q}{2}\\ \frac{p^2}{2p}-\frac{6}{2p}&=\frac{p}{2}+\frac{q}{2}\\ \frac{p}{2}-\frac{3}{p}&=\frac{p}{2}+\frac{q}{2}\\ -\frac{3}{p}&=\frac{q}{2}\end{align*}\\ \text{Whilst I don't expect you to just be like oh that's obvious}\\ \text{You should be able to just take the time and break down into steps what Jamon did.}
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 22, 2016, 09:44:43 pm
Hi there! Could someone please explain part e)? Would be much appreciated! :)
Hints have been offered up by Jamon in post #434. Try those first.

Also consider solutions which can be found here
Title: Re: 3U Maths Question Thread
Post by: teapancakes08 on October 22, 2016, 10:09:20 pm
\text{It's nothing more than algebra.}\\ \text{What you did is correct, but this was all Jamon did.}\\ \begin{align*}\frac{p^2-6}{2p}&=\frac{p+q}{2}\\ \frac{p^2}{2p}-\frac{6}{2p}&=\frac{p}{2}+\frac{q}{2}\\ \frac{p}{2}-\frac{3}{p}&=\frac{p}{2}+\frac{q}{2}\\ -\frac{3}{p}&=\frac{q}{2}\end{align*}\\ \text{Whilst I don't expect you to just be like oh that's obvious}\\ \text{You should be able to just take the time and break down into steps what Jamon did.}

I see. To be honest, I tried figuring out how to unpack Jamon's method but ended up getting frustrated because some figures wouldn't cancel out (cumulated in a page of crossed out working out). It genuinely slipped my mind that you could separate them into separate terms...something that tends to happens a lot. I probably should go do more practice then, huh...

Thanks for the help  ^^
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 22, 2016, 10:14:13 pm
I see. To be honest, I tried figuring out how to unpack Jamon's method but ended up getting frustrated because some figures wouldn't cancel out (cumulated in a page of crossed out working out). It genuinely slipped my mind that you could separate them into separate terms...something that tends to happens a lot. I probably should go do more practice then, huh...

Thanks for the help  ^^
All good. If you tend to run into little mistakes/mishaps though, you should make a list of them. And then look at those before you walk into the exam room.
Title: Re: 3U Maths Question Thread
Post by: chloe9756 on October 22, 2016, 10:37:47 pm
question im stuck on:

1. Consider the series 1+ 2/2 + 3/4 +4/8 +5/16 + 6/32 + ...

a) write out the terms of Sn and 2Sn.

b) subtract to get an expression for Sn.

c) find the limit as n approaches infinity and hence find limiting sum.
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 22, 2016, 10:52:03 pm
\begin{align*}\text{If }S_n&=1+\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\frac{6}{32}+\dots\\ \text{Then }2S_n &= 2 + 2 + \frac{3}{2} + \frac{4}{4} + \frac{5}{8} + \frac{6}{16} + \frac{7}{32} + \dots\\ &\text{So upon subtracting:}\\ S_n &= 2 + (2-1) + \left(\frac{3}{2} - \frac{2}{2}\right) + \left(\frac{4}{4} - \frac{3}{4}\right) + \left(\frac{5}{8}-\frac{4}{8}\right) + \left(\frac{6}{16}-\frac{5}{16}\right)+ \left(\frac{7}{32}-\frac{6}{32}\right)+\dots\\ &= 2+1+\frac12+\frac14+\frac18+\frac{1}{16}+\frac{1}{32}+\dots\end{align*}
$\text{i.e. the trick was to group common denominators.}\\ \text{This is now a geometric series which you should be able to find the limiting sum of.}$
Title: Re: 3U Maths Question Thread
Post by: MagmaMeerkat on October 22, 2016, 11:01:36 pm
Multiple choice question:
Which expression is equal to cos(x) - sin(x)?
A) (root2) cos(x+pi/4)
B) (root2) cos(x-pi/4)
C) (2) cos(x+pi/4)
D) (2) cos(x-pi/4)

And please explain how you can work it out/derive it?
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 22, 2016, 11:03:31 pm
Multiple choice question:
Which expression is equal to cos(x) - sin(x)?
A) (root2) cos(x+pi/4)
B) (root2) cos(x-pi/4)
C) (2) cos(x+pi/4)
D) (2) cos(x-pi/4)

And please explain how you can work it out/derive it?
$\text{This is just smacking it into the auxiliary angle transformation}\\ A\cos x - B\sin x = R\cos (x+\alpha)\\ \text{where }R=\sqrt{A^2+B^2}, \tan \alpha=\frac{B}{A}$
Title: Re: 3U Maths Question Thread
Post by: Sanaz on October 22, 2016, 11:37:40 pm
Can anyone help me with this?

Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 22, 2016, 11:41:40 pm
Can anyone help me with this?
Addressed in post #372 a while back
Title: Re: 3U Maths Question Thread
Post by: WLalex on October 23, 2016, 11:27:29 am
Help with part (iii) please..I got an expression in terms of n, n+1, n+2 and (1/2) but can't figure out how to manipulate it into what the answers wants
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 23, 2016, 11:46:24 am
Help with part (iii) please..I got an expression in terms of n, n+1, n+2 and (1/2) but can't figure out how to manipulate it into what the answers wants
Title: Re: 3U Maths Question Thread
Post by: WLalex on October 23, 2016, 12:36:38 pm
Thanks for that Rui!!

Another one that makes no sense... I got an answers but it was wrong and I don't quite understand what they did (part ii)

Thanks again
Title: Re: 3U Maths Question Thread
Post by: FallonXay on October 23, 2016, 12:54:22 pm
Heyyo again, How do you do both parts of this question? Thanks!  :)
Title: Re: 3U Maths Question Thread
Post by: FallonXay on October 23, 2016, 12:59:33 pm
and part iii of this q  ~ thanks :)
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 23, 2016, 01:00:49 pm
and part iii of this q  ~ thanks :)
Answered further back in post #184
Title: Re: 3U Maths Question Thread
Post by: jamonwindeyer on October 23, 2016, 01:06:42 pm
Thanks for that Rui!!

Another one that makes no sense... I got an answers but it was wrong and I don't quite understand what they did (part ii)

Thanks again

Sure! So let's start by doing what we always do to find tangents to curves, differentiate. I'll call the two graphs $$y_1$$ and $$y_2$$:

$y_1'=re^{rx}\\y_2'=\frac{1}{x}$

So we'll come back to that later; we also need the point of intersection, and at that point of intersection:

$e^{rx}=\log_e{x}\quad (*)$

But remember the gradients from above will need to be equal too (common tangent), so we also have $$re^{rx}=\frac{1}{x}$$.

What we have here are two equations linking $$r$$ and $$x$$; we're going to solve these simultaneously! The goal will be to use the result we were just given somewhere too. With that in mind, let's rearrange one of those equations:

$re^{rx}=\frac{1}{x}\implies e^{rx}=\frac{1}{rx}$

Aha, theres the result from Part (i); we can then say that $$rx\approx0.56$$!

Therefore back in $$*$$, we have: $$\log_e{x}=e^{0.56}$$, and that's an equation you can solve for $$x$$, and then substitute back to find $$r$$; does that make sense? :)
Title: Re: 3U Maths Question Thread
Post by: WLalex on October 23, 2016, 01:14:31 pm
Sure! So let's start by doing what we always do to find tangents to curves, differentiate. I'll call the two graphs $$y_1$$ and $$y_2$$:

$y_1'=re^{rx}\\y_2'=\frac{1}{x}$

So we'll come back to that later; we also need the point of intersection, and at that point of intersection:

$e^{rx}=\log_e{x}\quad (*)$

But remember the gradients from above will need to be equal too (common tangent), so we also have $$re^{rx}=\frac{1}{x}$$.

What we have here are two equations linking $$r$$ and $$x$$; we're going to solve these simultaneously! The goal will be to use the result we were just given somewhere too. With that in mind, let's rearrange one of those equations:

$re^{rx}=\frac{1}{x}\implies e^{rx}=\frac{1}{rx}$

Aha, theres the result from Part (i); we can then say that $$rx\approx0.56$$!

Therefore back in $$*$$, we have: $$\log_e{x}=e^{0.56}$$, and that's an equation you can solve for $$x$$, and then substitute back to find $$r$$; does that make sense? :)

Yes that makes sense thank you! Except one part, how can we assume that rx = 0.56 (I know we found it as an approximate about but howdy you know that it equals rx as there was not mention of that above??)
Title: Re: 3U Maths Question Thread
Post by: jamonwindeyer on October 23, 2016, 01:20:20 pm
Heyyo again, How do you do both parts of this question? Thanks!  :)

Hey! First bit:

$\frac{1}{(k+1)^2}-\frac{1}{k}+\frac{1}{k+1}=\frac{1}{(k+1)^2}-\frac{k+1}{k(k+1)}+\frac{k}{k(k+1)}=\frac{1}{(k+1)^2}-\frac{1}{k(k+1)}=\frac{1}{k+1}\left[\frac{1}{k+1}-\frac{1}{k}\right]$

Clearly this last part is less than zero; what is inside the bracket is always negative as long as $$k$$ is a positive integer (which it is) ;D

For your next question, I'll refer you to the HSC solutions and give you a bit of an explanation, since I'd type the same thing:

(http://i.imgur.com/Ns5Y5CT.png)

So we start with the proof for the lowest case and an assumption, as usual. Basically what they do then is consider one additional term on the LHS, and then they tie in the induction assumption where indicated! Remember, if we've assumed that the previous LHS is less than zero, it's kind of like just adding the $$\frac{1}{(k+1)^2}$$ term onto both sides; the inequality is maintained. Then some algebra using Part (i) is applied ;D
Title: Re: 3U Maths Question Thread
Post by: jamonwindeyer on October 23, 2016, 01:21:23 pm
Yes that makes sense thank you! Except one part, how can we assume that rx = 0.56 (I know we found it as an approximate about but howdy you know that it equals rx as there was not mention of that above??)

Part (i) said:

$e^t=\frac{1}{t}\implies t\approx0.56$

Now we have:

$e^{rx}=\frac{1}{rx}\implies rx\approx0.56$

Essentially, put $$t=rx$$ ;D
Title: Re: 3U Maths Question Thread
Post by: WLalex on October 23, 2016, 01:24:12 pm
ooooooo completely missed that! makes much more sense now thank you greatly
Title: Re: 3U Maths Question Thread
Post by: FallonXay on October 23, 2016, 01:29:48 pm
Hey! First bit:

$\frac{1}{(k+1)^2}-\frac{1}{k}+\frac{1}{k+1}=\frac{1}{(k+1)^2}-\frac{k+1}{k(k+1)}+\frac{k}{k(k+1)}=\frac{1}{(k+1)^2}-\frac{1}{k(k+1)}=\frac{1}{k+1}\left[\frac{1}{k+1}-\frac{1}{k}\right]$

Clearly this last part is less than zero; what is inside the bracket is always negative as long as $$k$$ is a positive integer (which it is) ;D

For your next question, I'll refer you to the HSC solutions and give you a bit of an explanation, since I'd type the same thing:

(http://i.imgur.com/Ns5Y5CT.png)

So we start with the proof for the lowest case and an assumption, as usual. Basically what they do then is consider one additional term on the LHS, and then they tie in the induction assumption where indicated! Remember, if we've assumed that the previous LHS is less than zero, it's kind of like just adding the $$\frac{1}{(k+1)^2}$$ term onto both sides; the inequality is maintained. Then some algebra using Part (i) is applied ;D

Sorry, would you be able to elaborate on the second last line? - I don't really understand the application of part i in the induction.
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 23, 2016, 01:32:32 pm
Sorry, would you be able to elaborate on the second last line? - I don't really understand the application of part i in the induction.
$\text{Part (i) implies that }\frac{1}{(k+1)^2} < \frac{1}{k}-\frac{1}{k+1}$
$\text{So now if we just add }2-\frac{1}{k}\text{ to both sides}\\ 2-\frac{1}{k}< 2-\frac1k+\frac1k-\frac1{k+1}$
$\text{Note that the LHS is just the LHS of what we trying to prove.}\\ \text{So rather than substituting an equation, say, }a=b, b=c\therefore a=b\\ \text{we are subbing in an }\textbf{inequality},\text{ say }a=b, b< c\therefore a < c$
Title: Re: 3U Maths Question Thread
Post by: FallonXay on October 23, 2016, 01:37:33 pm
$\text{Part (i) implies that }\frac{1}{(k+1)^2} < \frac{1}{k}-\frac{1}{k+1}$
$\text{So now if we just add }2-\frac{1}{k}\text{ to both sides}\\ 2-\frac{1}{k}< 2-\frac1k+\frac1k-\frac1{k+1}$
$\text{Note that the LHS is just the LHS of what we trying to prove.}\\ \text{So rather than substituting an inequation, say, }a=b, b=c\therefore a=b\\ \text{we are subbing in an }\textbf{inequality},\text{ say }a=b, b< c\therefore a < c$

Ahh, I see. Cheers  :)
Title: Re: 3U Maths Question Thread
Post by: FallonXay on October 23, 2016, 01:59:08 pm
Hiya, how do you solve these 2 questions? (I'm assuming there's a more efficient way of solving Q9 than expanding it all out haha  :P)
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 23, 2016, 02:03:23 pm
Hiya, how do you solve these 2 questions? (I'm assuming there's a more efficient way of solving Q9 than expanding it all out haha  :P)
Question 5 is best done by guess and check.

The product of roots is clearly -42
Hence A and D are out.

The sum of 2 roots at a time is -41
This may be used to deduce that B is the correct answer.

Note that the presence of the a just means to ignore the sum of 1 root a time.
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 23, 2016, 02:05:43 pm
Hiya, how do you solve these 2 questions? (I'm assuming there's a more efficient way of solving Q9 than expanding it all out haha  :P)
\text{By the division transformation, Q9 implies}\\ \begin{align*}P(x)&=x^4-8x^3-7x^2+3\\&=(x^2+x)Q(x) + ax+3\end{align*}
$\text{As we don't know the quotient }Q(x)\\ \text{ we want it to vanish.}\\ \text{Clearly, the candidate choices to make it vanish are }x=0, x=-1\\ \text{But obviously }x=0\text{ is useless as then }a\text{ poofs.}$
$\text{Hence put }x=-1$
Title: Re: 3U Maths Question Thread
Post by: atar27 on October 23, 2016, 03:06:52 pm

I can't solve for k+1
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 23, 2016, 03:13:27 pm

I can't solve for k+1
$\text{Assumed: }(1\times 2)+(2\times 3)+\dots+((k-1)k) = \frac{1}{3}(k-1)k(k+1)\\ \text{RTP: }(1\times 2)+(2\times 3)+\dots+(k-1)k+k(k+1)=\frac{1}{3}k(k+1)(k+2)$
\text{So by our assumption and factorising:}\\ \begin{align*}LHS&=\frac{1}{3}(k-1)k(k+1)+k(k+1)\\ &= \frac{1}{3}k(k+1)(k-1)+\frac{3}{3}k(k+1)\\ &= \frac{1}{3}k(k+1)[(k-1)+3]\\ &= \frac13k(k+1)(k+2)\\ &= RHS\end{align*}
Title: Re: 3U Maths Question Thread
Post by: MysteryMarker on October 23, 2016, 03:29:52 pm
Hey guys, just got a few questions which I'm not too confident with.

1. |x+1| +x = |x-1|

2. When you are finding the volume of a curve that has been rotated about lets say the x axis, and the question specifically asks to use simpsons rule. Do you just find y2, input the values into the simpsons formula and then multiply it by pi?

3. What is the value of integral of [cos-1xdx]a-a where -1<=a<=1

4. How many times must a die be rolled so that the probability of rolling at least one size is greater than 95%?
Title: Re: 3U Maths Question Thread
Post by: jamonwindeyer on October 23, 2016, 04:06:53 pm
Hey guys, just got a few questions which I'm not too confident with.

1. |x+1| +x = |x-1|

2. When you are finding the volume of a curve that has been rotated about lets say the x axis, and the question specifically asks to use simpsons rule. Do you just find y2, input the values into the simpsons formula and then multiply it by pi?

3. What is the value of integral of [cos-1xdx]a-a where -1<=a<=1

4. How many times must a die be rolled so that the probability of rolling at least one size is greater than 95%?

For your first one, doing that algebraically would probably be gross. I'd take a graphical approach. If you draw the two sides of that equation on the same graph, just pick where they intersect and that's your answer! Drawing the one on the right is easy; the one on the left might be tricky; have a think! Here is what you should get :)

For your second one, yes :)

For your third one, do you mean you need to integrate $$\cos^{-1}{x}$$? Or just evaluate that anti-derivative in the brackets? Snap a pic of the original question? ;D

Your fourth one, we could use binomial theorem and other such things; but let's just think of it simply. Every time we roll a dice, our probability of rolling NOT A SIX is $$\frac{5}{6}$$. But as soon as we fail to do that, we have rolled a six. So, instead of considering how many times we need to roll to have a 95% chance of getting A six, let's just consider how many times we need to roll such that the probability of throwing NO SIXES becomes less than 5%! It's a complementary event ;D

So we seek the lowest integer value of n such that:

$\left(\frac{5}{6}\right)^n<0.05$

By trial and error, the answer is 17 ;D
Title: Re: 3U Maths Question Thread
Post by: MysteryMarker on October 23, 2016, 04:28:00 pm
Oh thanks, and yeah for the third one it asks to integrate cos-1x dx and evaluate it with the bounds given. Its a multiple choice question where the options are

a) 0
b) a x (pi)/2
c) a x (pi)
d) 2 x a x (pi)
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 23, 2016, 05:59:50 pm
Oh thanks, and yeah for the third one it asks to integrate cos-1x dx and evaluate it with the bounds given. Its a multiple choice question where the options are

a) 0
b) a x (pi)/2
c) a x (pi)
d) 2 x a x (pi)

$\text{You do not know of any approach to integrate }\cos^{-1}x\\ \text{Hence, sketch }y=\cos^{-1}x\text{ and colour in the area you require.}$(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpseh5a39zy.png)
$\text{Due to the symmetry on the cos inverse curve}\\ \text{It appears as though the shaded region above }y=\frac{\pi}{2}\\ \text{can literally be }\textit{shifted}\text{ so that you end up with a rectangle.}$
$\text{This triangle will always have height }\frac{\pi}{2}\\ \text{and length }2a$
$\text{Hence, from the area of a rectangle, the answer is }a\pi$
Title: Re: 3U Maths Question Thread
Post by: MysteryMarker on October 23, 2016, 07:01:51 pm
Cheers guys, I'm gonna be spamming this thread with a lot of questions :P

Is there a shorter method to finding the general solution to cos2x = cosx rather than using double angle formula and solving for the quadratic?

Title: Re: 3U Maths Question Thread
Post by: jakesilove on October 23, 2016, 07:11:45 pm
Cheers guys, I'm gonna be spamming this thread with a lot of questions :P

Is there a shorter method to finding the general solution to cos2x = cosx rather than using double angle formula and solving for the quadratic?

Potentially, you can sketch the two graphs and see if it's obvious where the intercepts lie. However, I would always suggest using double angle formula etc. etc. for a question like that :)
Title: Re: 3U Maths Question Thread
Post by: RuiAce on October 23, 2016, 07:13:54 pm
Cheers guys, I'm gonna be spamming this thread with a lot of questions :P

Is there a shorter method to finding the general solution to cos2x = cosx rather than using double angle formula and solving for the quadratic?
Not really. Other methods risk losing solutions. Better off just to pull it safe with the quadratic here

Potentially, you can sketch the two graphs and see if it's obvious where the intercepts lie. However, I would always suggest using double angle formula etc. etc. for a question like that :)
But sketching is tedious pls.
Title: Re: 3U Maths Question Thread
Post by: Cindy2k16 on October 23, 2016, 07:29:14 pm
hi for Q5 a) ii) of 1997 HSC Maths ext http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/97MAT3U.PDF
how are you supposed to know that the fastest speed occurs at a=0 if the particle isnt moving in SHM?