Hey Jake:
Hope you are impressed by this first ever 4 unit question of the year. I have little idea how to sketch this graph would you mind giving me a hand on this question? Thank you fam! :)
Essentially the question is asking me to sketch y = ln (1-|x|)
Thank you so much Jake!
(http://i.imgur.com/bxnzaML.jpg)
Best Regards
Happy Physics Land
Hey HPL!
Glad I got a question from the Extension Two course! I really enjoyed this course, but seriously struggled with it, so would encourage anyone who needs help to post up some difficult questions (like the one you have asked!)
I'm sorry that the graphing is sort of dodgy, the software that I usually used has magically disappeared but I'll try find it again soon! Hope the answer helps!
(http://i.imgur.com/Rf6NaF1.png?1)
(http://i.imgur.com/jGEatyN.png?1)
I'm thinking of making a "Graph Sketching" Extension Two resource, but that's probably still a while off.
Jake :)
Hey Jake!
Thank you so much for your help I'm really grateful towards this very elaborative answer. It took me a while to kind of understand the range part and I think it was a really good idea to use domain and range and to check whether the function is even or odd. My initial approach was kind of like trying to start off with the parent function of y = lnx and just translate it and reflect it about the x axis and then take the absolute value of the graph. Well my method didnt quite work out. But yeah I was just ignorant of the first principles of logarithmic functions and this is definitely an effective approach I can adopt in the future. Thank you very much Jake for always helping me in times of need! :D
Best Regards
Happy Physics Land
It was a really difficult question mate, so definitely don't kick yourself for not getting it!
My biggest recommendation, for a graph that you don't have any idea how to sketch, is always to PLOT INTELLIGENT POINTS! Here,you could have just tried x=1,x=2, x=3... x=-1, x=-2, x=-3.... etc. and eventually got a good idea of what the graph should be. Whilst this probably won't get you full marks, at least it will get you something!
Still, having a good approach is always better. Start with Domain/Range, think about Even/Odd, and examine important locations (approaching asymptotes etc.).
Would love you to post any other questions you're struggling with! Love you long time HPL:)
Jake
Hey Jake!
I'm currently doing 3 sciences, 4u maths and 2u english and was planning on dropping to 10 units (I would drop biology) however, I'm slightly hesitant as this would mean every single one of my units would count. Right now, I'm behind on biology which is giving me the incentive to drop it so I wont have to invest time catching up. What are your thoughts? Since you did 12 units? Did you feel it was more beneficial? And also, what would you say is the best way to approach 4U maths and do you think it's doable for everyone? I get stumped on quite a few questions and I feel like I'm more a rote learner than an innovative thinker if that makes sense.. Also, what are your thoughts on dot point summaries for Physics and Chemistry? Necessary or nah? My school puts a lot of emphasis on it but I find that it's extremely time consuming and I don't exactly trust my own notes anyway, but I'm scared I'm disadvantaging myself by not writing them! Thanks Jake!
Hey Jake!
I'm currently doing 3 sciences, 4u maths and 2u english and was planning on dropping to 10 units (I would drop biology) however, I'm slightly hesitant as this would mean every single one of my units would count. Right now, I'm behind on biology which is giving me the incentive to drop it so I wont have to invest time catching up. What are your thoughts? Since you did 12 units? Did you feel it was more beneficial? And also, what would you say is the best way to approach 4U maths and do you think it's doable for everyone? I get stumped on quite a few questions and I feel like I'm more a rote learner than an innovative thinker if that makes sense.. Also, what are your thoughts on dot point summaries for Physics and Chemistry? Necessary or nah? My school puts a lot of emphasis on it but I find that it's extremely time consuming and I don't exactly trust my own notes anyway, but I'm scared I'm disadvantaging myself by not writing them! Thanks Jake!
Thank you both Happy Physics Land and Jake for your amazingly comprehensive responses! They were definitely helpful but I have a question on what Jake said about the summarising another student's notes. Do you mean I highlight them and study them as though they were my own or do I continue to write my own but take theirs into consideration? Sorry for my confusion!
Neutron
Hey Jake! How's it going? Hope everything is good! Anywho, I have a question from polynomials. The question asks me to solve an equation 'given that the roots form an arithmetic series'... But my teacher hasn't really gone through series ( unless it's just saying 'plus or minus k )... Here's the question! Q6!!!!!
(http://images.tapatalk-cdn.com/16/02/10/872438724fa260c62a8b25f0b073ccec.jpg)
Damn Jake, looks like I missed out on some delicious polynomial questions :/
Thank you Jake! I think that's the same method a friend of mine used! But I don't understand it very well... I'll have to keep looking HAHA HOWEVER I have accidentally come across the same solution using a different method and I think this may benefit others too! Here it is! Also, would this be a valid method to use??? (http://images.tapatalk-cdn.com/16/02/11/2f8545333d5348e2fefc5317609e9435.jpg)
Hey xXCandyDXx! Glad Jake could be of help!
In addition, a tip for spotting when to use the "sums of roots" formulae. Pretty much, if a question follows the following template:
Find the roots of ____________________ given that __________________
or even
The roots of ___________________ form an arithmetic series. Find the value of k (some unknown in the polynomial))
Essentially, if you are given a polynomial with something to find, and all you are given is some information about the roots, chances are that these formulae will help you. They are very powerful in surprising ways.
While I'm here, I have an issue with the method you posted above. It's very possible I am misunderstanding, but it seems like the notions of root and factor are being confused. Remember, factors can be multiplied together to get back to the original, roots are just solutions to the polynomial equal to zero.
So, in the first few lines, we multiply the roots together and then form an identity with the original polynomial. This doesn't make sense to me. We could do it with factors, but multiplying the roots of a polynomial together has no relationship with the original, besides the relationships highlighted in the formulae used by Jake's method.
The factor rule makes perfect sense. The first part of the solution however, while the correct answer is obtained, does not make as much sense to me, and I would argue that it is incorrect. I'd love to hear others interpretations on the matter! ;D
Hey Jamon!
Thanks for clarifying when to use the sum and product formulas for Polynomial questions!
Also, I agree that her method is actually incorrect, but as luck would have it gets to the right answer. Here's why
(http://i.imgur.com/WhUlMoD.png?1)
Jake
He really is a god
Jake you are such a god what an elaborative response!
Please send help Almighty Jake when you are available!!! This has been screwing me over!!! :-[
(http://i.imgur.com/aQWttKH.jpg)
(http://i.imgur.com/D9yBL9R.jpg)
Thank you so much Jake! <3 (I dont know why the images are sideways doe...)
He really is a god
Hey HPL.
Typical you giving me incredibly difficult 4U maths questions at Midnight. Couldn't resist though, so this is the solution I came up with! An extremely, extremely difficult graphing question, and I definitely may have missed some components, so if anyone in the community wants to correct me please do.
(http://i.imgur.com/ISRRvZQ.png?1)
(http://i.imgur.com/u4CLi2b.png?1)
(http://i.imgur.com/dMXmX2Z.png?1)
(http://i.imgur.com/QzAvog5.png?1)
(http://i.imgur.com/iIOXMHX.png?1)
Like I said, if anyone has any corrections, suggestions, or anything like that, please feel free to post below. Extension 2 is an incredibly difficult course, one that even University professors often struggle with.
Still, this is a lot of fun (to me, I know I'm a bit of a nerd) so keep posting, and keep answering questions! We're starting to build a real community, and I love watching it grow.
Yours,
Jake :)
Like I said, if anyone has any corrections, suggestions, or anything like that, please feel free to post below. Extension 2 is an incredibly difficult course, one that even University professors often struggle with.
For the poly ques how do u find the leading coefficient? I only know that there are 2 known roots
and for the trig ques i dont know how to simplify it after implicitly differentiating
please help thank you
exsiny + eysinx = 0Hey dii!
After differentiating with respect to x and grouping the dy/dx onto one side we get
dy/dx = -(exsiny + eycosx) / (excosy + eysinx)
Going back to the main equation that they give us (this was the tricky part) Rearranging we get.
exsiny = - eysinx
Then subbing that in to:
dy/dx = -(exsiny + eycosx) / (excosy + eysinx)
We get
dy/dx = -(-eysinx + eycosx) / (excosy - exsiny)
= (eysinx - eycosx) / (excosy - ex siny)
Factorise out the ey from the top and ex from the bottom and you get the answer
This was a pain to type up haha, what program or website do you use jake to post the solutions?
[tex][/tex]
exsiny + eysinx = 0
After differentiating with respect to x and grouping the dy/dx onto one side we get
dy/dx = -(exsiny + eycosx) / (excosy + eysinx)
Going back to the main equation that they give us (this was the tricky part) Rearranging we get.
exsiny = - eysinx
Then subbing that in to:
dy/dx = -(exsiny + eycosx) / (excosy + eysinx)
We get
dy/dx = -(-eysinx + eycosx) / (excosy - exsiny)
= (eysinx - eycosx) / (excosy - ex siny)
Factorise out the ey from the top and ex from the bottom and you get the answer
This was a pain to type up haha, what program or website do you use jake to post the solutions?
hi how do u integrate (x+1)/(x^2+x+1)^1/2
Hello! Me again :P Heh so I was trying to complete some 4U curve sketching when I came across a few silly problems I realised I haven't fully understood. The question is as follows:
Sketch the graph of y=lxl-lx-4l . Use this graph to solve the inequality lxl-lx-4l>2
So I managed to sketch it by plotting points and I was wondering whether there's an easier method? Like how would you know whether the absolute value graph resembles a backwards Z like this one or one that looks more like \_/ ? And secondly, I tried solving
2=|x|-|x-4| to find the x coordinate but I kept getting the answer as -1 when the real answer is 3.. Could you please show me how to solve these equations ^^'' heh sorry this is kinda dumb
Neutron
how to integrate x^2sinx thanks :)
hi! There's this question I don't really understand:
'A cuboid tank is open at the top and the internal dimensions of its base are xm and 2xm. The height is hm. The volume of the tank is Vm^3 and the volume is fixed. Let S m^2 denote the internal surface area of the tank.'
hence find S in terms of V and x
Thanks a lot in advance! :)
Hey Ms She:
To deal with these questions, the most immediate things I would have done are to find the surface area in terms of x and h and the volume in terms of x and h. Because the question asks us to find S in terms of V and x, we know that we need to substitute h with an expression in terms of V and x in order to eliminate the h from the surface area equation. anyways, you can check out what l mean through observing my working out below. If you have any further questions dont hesitate to ask me! :)
(http://i.imgur.com/ixaxogq.jpg)
Best Regards
Happy Physics Land
Hey Ms She:Thankyou!!
To deal with these questions, the most immediate things I would have done are to find the surface area in terms of x and h and the volume in terms of x and h. Because the question asks us to find S in terms of V and x, we know that we need to substitute h with an expression in terms of V and x in order to eliminate the h from the surface area equation. anyways, you can check out what l mean through observing my working out below. If you have any further questions dont hesitate to ask me! :)
Best Regards
Happy Physics Land
Which integer n do you take? would u use n=0,1,2,3 or n=0,+-1, +-2
so when the angle is 4 theta then there will be 4 solutions , 3theta=3 solutions
I have the answer attached below
how do you simplify this?There is no neat simplification to this: this is simplified completely by brute force.
how to do part 2 (ii)
why does z1 , z2 and z1+z2 have to be collinear
Hey Jake,
I'm currently in Year 12 and I'm doing 4U mathematics. I've always done quite well in maths up until around the beginning of year 12, where I've just been struggling with my 4u and 3u maths marks. I'm working quite hard at the moment, although I definitely could be working harder. I usually have a strong grasp on the basic knowledge (component A) parts of maths papers, however I'm never really quite prepared for Component B style questions. In my latest exam, I received 22/25 for the Component A section and a 6/25 for Component B. I'm also having trouble finding a balance between 3u and 4u, as I generally just do 4U maths most days of the week. I was wondering if you had any tips in regards to this kind of issue.
Thanks for your help,
Kezz
for Part (iv) how do you know p is greater than q
the answer is -inverse cos x - root (1-x^2) but mine is -1/2inversecosx -1/2 root (1-x^2)
the answer is -inverse cos x - root (1-x^2) but mine is -1/2inversecosx -1/2 root (1-x^2)
Hey guys ... Having some trouble starting this question... I don't know how to start it correctly...
(http://images.tapatalk-cdn.com/16/03/09/33f5e34bbf1ab29843be0ead1b96e750.jpg)
Oh no wait never mind ... I get it LOL thanks again !!!
I dont how its progressed from last 3rd line to last 2nd line :/
im lost :/
how do you normally start with these questions involving i(n) I(n-1)
How do you do these questions? thanks
Hey guys!
Sort of a dumb question but I was doing some questions and one of the questions was:
"If P(x)=x^4-2x^3-x^2+6x-6 has a zero 1-i, find the zeros of P(x) over C"
Now I know that 1+i is also a root (conjugate root theorem) but how do you know there aren't 4 complex roots? Like how come you can assume it's just two complex roots idk I'm tired and confused so sorry if this is very basic!
Neutron
And I'll add that, in case you aren't sure, we can infer that first quadratic root by multiplying the two complex roots together:
Hey guys I need some help here.. How do I do this question? I know that q, r and s are coefficients but ... How would you relate them ?.. I don't know, I'm lost and confused thanks a heap! @.@(http://images.tapatalk-cdn.com/16/03/21/23b64e7b1d37b614fc81612701e0a552.jpg)
And I threw the pencil here; Somehow I accidentally got a negative stuck in there so unless the question has a typo, I had a mistake in algebra at some point.
Ohhh I got it!!! (http://images.tapatalk-cdn.com/16/03/21/cc2592586b07a55ab58a9c1e8dea7725.jpg) but how about the next part :((((
Ohhh I got it!!! (http://images.tapatalk-cdn.com/16/03/21/cc2592586b07a55ab58a9c1e8dea7725.jpg) but how about the next part :((((
Actually, hang on (in regards to the diagram). Shouldn't the product of roots be negative s?
Is it better if i memorise this or show working out?
Im stucked with this ques
What i did was changing sec^3(x) --> secx(1+tan^2(x)
so 1/cosx(1+sin^2(x)/cos^2(x))
I dont get
-1/n[cosxsin^(n-1)x]
since I(n)= sin^n(x)
Is it better if i memorise this or show working out?
Is it better if i memorise this or show working out?
I'm sorry, but Jamon is right. You have to use partial fractions or a trigonometric substitution for that one.
Anyway side note @Jamon I'm sorry but please take care of any questions HPL can't! I'm not sure when Jake is back but I'm on a vacation at Melbourne right now with grandparents so I might miss the posts. Thanks! :)
I'm sorry, but Jamon is right. You have to use partial fractions or a trigonometric substitution for that one.
Anyway side note @Jamon I'm sorry but please take care of any questions HPL can't! I'm not sure when Jake is back but I'm on a vacation at Melbourne right now with grandparents so I might miss the posts. Thanks! :)
Jake should be back tonight actually
HPL is right! I'm back and scrolling the forums, you guys have all done an incredible job keeping up to date on questions. I seriously, seriously appreciate the hard work you've all put in!
But yeah, back from my break and ready to answer some forum questions!
Thank guys; you are absolutely instrumental in maintaining the growth of this community, and we here at Atar Notes couldn't appreciate you more.
Jake
Damn Jamon and Rui, we can't keep all the questions to ourselves anymore...Oh my goodness.
Jake just rocks up and breaks the equilibrium we have been trying to maintain for the past 2 weeks, and according to Le Chatelier's principle .... hehehe
Damn Jamon and Rui, we can't keep all the questions to ourselves anymore...
Jake just rocks up and breaks the equilibrium we have been trying to maintain for the past 2 weeks, and according to Le Chatelier's principle .... hehehe
Hey come on man I didn't even do Chemistry! Don't confuse me ;)
Yeah I forgot you did legal studies, well in this CASE, I reckon you do have the right to remain silent ... :) :)
Yeah I forgot you did legal studies, well in this CASE, I reckon you do have the right to remain silent ... :) :)
Hey lads Im a bit stuck with this conic question, would anyone mind lending me a hand? Thank you very much!!! :D (I hope rui doesnt hit me for asking this question)
The chord AB of the hyperbola xy=c2 subtends a right angle at a point P on the curve. Prove that AB is parallel to the normal at P.
I couldnt really start this one because Im struggle to even draw the diagram.
Hey lads Im a bit stuck with this conic question, would anyone mind lending me a hand? Thank you very much!!! :D (I hope rui doesnt hit me for asking this question)
The chord AB of the hyperbola xy=c2 subtends a right angle at a point P on the curve. Prove that AB is parallel to the normal at P.
I couldnt really start this one because Im struggle to even draw the diagram.
Hey! Below is my answer. Hope it makes sense!
(http://i.imgur.com/5ZnpIcN.png?1)
Jake
Thank you so much Jake! I understand it a lot better now! Im just wondering, were the coordinates meant to be A = (ca, c/a), B = (cb, c/b) and P = (cp, c/p)? Because I've always learnt that you let x=ct and y = c/t, where t is the variable. If Im the one getting confused here please correct me Jake!!! Thank you very much! I would have loved to contribute more to the forum but because half-yearlies are on currently I'm a bit tight with time, sorry for dropping the burden on you guys! I shall be back to join you guys soon! :))))))
Best Regards
Happy Physics Land
Hey Katherine! That is not the approach I would use, though it might work! I'll show how I would do it with integration by parts:
Through integration by parts:
Now the integration of sec(x) is another job in itself. There is a common trick for figuring out this integral. Transform the integral like so:
Then, use a substitution:
Follow this process to find that:
Taking the formula from above and adding this, and rearranging, will give us the final solution:
Let me know if anything here is unclear!! It is a long and difficult question ;D
is this method the same as :
Let say I= integral (x*lnx)
let u=x v=lnx
I=(Integrating u)*(leave v) - integral of (integrating u*differentiate v)
is this method the same as :
Let say I= integral (x*lnx)
let u=x v=lnx
I=(Integrating u)*(leave v) - integral of (integrating u*differentiate v)
why isnt this B
how do you do this ques?
i know that the angle will be 90
`hey, can someone please help me with this...i have little/to no idea how to do it... :)
find d/dx of:
thankyou so much in advance
okay...thankyou
do you know how to do it without use using logs or In?
i just subbed e^x into (1+√t)^5 into the equation...then subbed 1 in and then just F(e^x) - F(1)...but it's not working :(Facepalm
no they don't...is there a way to integrate the function without using any of those things?
there's another question you might be able to use to explain it (i attached it) ...because we haven't done logs at school yet...so i'm thinking they may not ask those type of questions
Wait I'm just thinking, shouldn't the answer for this one have beenbut how does the chain rule fit...we don't even need to do any differentiation or integration cos h equestion is saying to differentiate then integrate...?
Because I'm inclined to say we missed something important - the chain rule
but how does the chain rule fit...we don't even need to do any differentiation or integration cos h equestion is saying to differentiate then integrate...?
Actually it said to integrate then differentiate. So the fundamental theorem of calculus allows us to work our way around that.how would you do that? :)
how would you do that? :)
not sure why its D
how to solve this?
please help thanks
im not sure how to determine whether point P lies above or below the x axis
for questions like arg(z-1) = arg(z+1) + pie/4
Hello friends!
I was wondering whether you could help me integrate tan^2 (x) sec^2 (x) dx ? Thank you, I seem to be having trouble ughhh
Neutron
Alternatively, if the reverse chain rule is a bit too hard to see, just apply the substitution u=tan(x).
At the Ext 2 level though, you do want to practice identifying the reverse chain rule
Hey Neutron!
This question looks deceptively difficult, but the answer ends up being quite straight forward! The trick is recognising what you need to use (ie. reverse chain rule), and not falling into the trap of using a more complicated method (ie. Integration by parts). Anyway, my answer is below, hope it helps!
(http://i.imgur.com/SpAgXmp.png?1)
Jake
Omg I'm honestly feeling so dumb right now, thank you so much guys :') I kept trying to expand the sec^2 and the tan^2 and ugh that was a mess oops
- make x/[(x-1)^3(x-2)] into partial fractions
why is it A/(x-1) + (BX+C)/(x-1)^2 + D/(x-2) instead of AX+B/(x-1)^2 + C/(x-2)
- Is the cover up method commonly used and when does it not work?
- Will 4x-3/[x^3(x+1)] decompose into A/x+ B/x^2 + C/x^3 + D/(x+1) and why isnt B linear and C a quadratic
1)how to make 1/(x^2+2x-1) into impartial fractions
is it not decomposible since the denominator cant be factorised
2) how to make 1/[x(x^2-1)^2] into impartial fraction
so will it be
A/x + B/(x-1) + C/(x-1)^2 + D/(x+1) + E/(x+1)^2
how to integrate this x^3/ (x^2-x-3) thanks
First one is done by a u-substitution (or just by inspection), whereas the second is done using completing the square.
That was awesome! When I attempted it I had no idea what to do after long division with that 4x - 3 / x2 - x - 3 thing
How to make x^3/ [(x-1)(x^2+2)^2] into impartial fractions:
ans: (8/9-1/9x)/(x^2+2) + (2/3x-4/3)/(x^2+2)^2 +1/9/(x-1)
doesnt the numerator becomes a constant when
you make something like 1/(x+2)^2 to A/(x+2) + B/(x+2)^2
How to make x^3/ [(x-1)(x^2+2)^2] into impartial fractions:
ans: (8/9-1/9x)/(x^2+2) + (2/3x-4/3)/(x^2+2)^2 +1/9/(x-1)
doesnt the numerator becomes a constant when
you make something like 1/(x+2)^2 to A/(x+2) + B/(x+2)^2
I'll also add, on the topic of Partial Fractions, that over half of my lecture on signal transformations this morning was work on Partial fraction decomposition and the Heaviside method. 2nd year university courses teaching very standard 4U content. Another example of why 4 Unit math is an amazing head start for many Math, Science, and Engineering degrees!! ;D
To clarify on Rui's response:
When you need a partial fraction decomposition of something with a fully factorised quadratic term on the bottom, for example:
Then it gets broken down like so:
For an irreducible quadratic factor, the numerator takes the form (Cx + D) ;D
Also remember this technique only works for proper fractions, where the degree of the denominator is larger than the degree of the numerator.
Hope this helps you a tad! ;D
so for eg. 1/(x^3+1)^2
the partial fractions will be (Ax^2+Bx+C)/(x^3+1) + (Dx^2+Ex+F)/(x^3+1)^2
since the denominator has irreducible cubic factor , the numerator will be quadratic
Bad idea.
You gotta admit though... without Wolfram or something some really messy partial fractions like that above one are tedious :P
Wolfram Alpha has definitely lightened my workload numerous times in the last 3 semesters! ;D
Hey,
I'm currently doing the volumes of solids topic and the question is:
By taking slices work out the volume of the area enclosed within the circle (x-1)^2 + y^2 = 1 rotated about the y-axis
I'm not getting to the answer and I don't know where the mistake is. I would appreciate some help
Thank you :)
Fe
Hey,
I'm currently doing the volumes of solids topic and the question is:
By taking slices work out the volume of the area enclosed within the circle (x-1)^2 + y^2 = 1 rotated about the y-axis
I'm not getting to the answer and I don't know where the mistake is. I would appreciate some help
Thank you :)
Fe
To do in terms of volumes is in attachment (just in case jamon's didn't suffice). Like he said though, it's pretty similar :)
Hey guys!
I'm a bit stuck on which substitution to use with integrating x/(x2 + x)3/2. Would anyone mind helping me out a bit? Thank you so much!!!
how to solve
integral of 1/(1+tanx) thanks
how to solve
integral of 1/(1+tanx) thanks
how to solveDoing these questions by hand today cause I'm not at home and typing LaTeX.
integral of sinx /[ 2+cosx - cos^2(x) ] and integral of 1/(e^2x + 4e^x + 9)
The substitution isn't directly obvious for the second one, but the neat thing about exponentials is that you can "introduce" the substitution by multiplying top and bottom by e^x.
The final answer is subject to inaccuracy(http://uploads.tapatalk-cdn.com/20160502/fb43c67499fe5e2c60eb802047daeaeb.jpg)(http://uploads.tapatalk-cdn.com/20160502/9d1eb17d236f7a94c3094a8e75c85d2e.jpg)
Tis a good answer, but I think you forgot to carry the 1/9 for the second two parts of the integral so the coefficients of the log should be -1/18 and the tan inverse is -2/9(5)^1/2
That's correct. That last inverse tangent term should be:
Nasty integral ;D
Tis a good answer, but I think you forgot to carry the 1/9 for the second two parts of the integral so the coefficients of the log should be -1/18 and the tan inverse is -2/9(5)^1/2
integral of 1/(x^2-9)^3/2
I let x=3sec(B)
and im left with
integral of 1/9*cos(B)*cosec^2(B)
how to solve
integral of 1/[x*(x^2+1)^1/2] with limits between 1 and 2?
Hello,
I'm supposed to do this question with both washers and cylinders but I can't get either to work
Find the volume of a torus formed with the circle (x-2)2 + y2 = 1 is rotated about the y axis
Thank you :)
how do i integrate x^4*e^(-x)
how do i integrate x^4*e^(-x)
how to solve these
1.integral of x*inverse sin^-1(x^2)
2.integral of ln[x+(x^2+1)^1/2] after proving d/dx[ln[x+(x^2+a^2)] = 1/[(x^2+a^2)^1/2]
how to solve these
1.integral of x*inverse sin^-1(x^2)
2.integral of ln[x+(x^2+1)^1/2] after proving d/dx[ln[x+(x^2+a^2)] = 1/[(x^2+a^2)^1/2]
1st ques: im asked to find the integral of (x^2+4)^1/2 and i used the substitution x=2tanB which gave me integral of 4*sec^3(B)
2nd ques: integral of e^x*sin(4x)cos(2x) with no limits
i changed the above integral to e^x*[sin(2x)+sin(6x)] using trig property
answer for ques 2 = e^x/370*[37sin(2x)+5sin6x-74cos(2x)-30cos(6x)] +C
These are really just unnecessarily tedious integrals if you ask me(http://uploads.tapatalk-cdn.com/20160512/feee17ed8ba91420d9485044b2499a93.jpg)
(https://cdn.meme.am/instances/500x/68157159.jpg)
Are you ok Jamon
HSC integrals - only partial fractions are tedious :P
(Except for I think 2014 when the last question of the paper was a huge surprise)
As a Class of 2014 graduate, I concur
As a Class of 2014 graduate, I concur
Though I didn't take MX2, I actually heard about that question. It's interesting because in my opinion I actually think the MX1 paper for 2014 was a little easier than the standard, or maybe I just got lucky with the subjects ;D
(http://uploads.tapatalk-cdn.com/20160514/ea8c6a7d4e9b7cbc39be832d47f38c1e.jpg)
im stuck after making sin^n(x) to sin^n-2(x)sin^2(x)
(http://uploads.tapatalk-cdn.com/20160514/ea8c6a7d4e9b7cbc39be832d47f38c1e.jpg)
im stuck after making sin^n(x) to sin^n-2(x)sin^2(x)
I recommend that for the classic examples such as sinn(x) you simply memorise the natural approach, which is to split it up by a power of one and not two in this case. (Method of course demonstrated by Jamon already.)
(http://uploads.tapatalk-cdn.com/20160516/a6eb482deff5136ea9b679311af571bc.jpg)
got stuck after using integration by parts
(http://uploads.tapatalk-cdn.com/20160517/b10d17745781ddd5cdb9a417c3b51240.jpg)
i did part (i) but not sure how it is used for part (ii)
(http://uploads.tapatalk-cdn.com/20160518/5576cae1283ab996cac28240bd41a835.jpg)Hey!
not sure how to do part b) part a is written on top
(http://uploads.tapatalk-cdn.com/20160518/5576cae1283ab996cac28240bd41a835.jpg)
not sure how to do part b) part a is written on top
(http://uploads.tapatalk-cdn.com/20160523/610b5295ae58ef1f1167e411034acdde.jpg)
help with this ques thanks
what r the shortcut to simplify these
what r the shortcut to simplify these
Hey Katherine! Do you mean what steps they used to get to the end of each line?
I'll pull the binomial-style term out of each one to show you the process. I'll just do the outline, I might skip a step here or there, let me know if anything doesn't make sense! ;D
Here are the first two:
I was just posting a reply! Damn you and your superior LaTex skills
Damn sorry man! I think we need a system between Rui, you and myself ;)
how do i find volume generated using cylindrical shell method
when area bounded by y=1-x^2 and y=1-x is rotated About x axis
1. find the volume generated using cylindrical shells when region between y=2x^2 and y=x^4-2x^2 is revolved around y axis
whats the height?
2.find volume generated by using cylindrical shell method when area enclosed by x^2/25+ y^2/16=1 is rotated about x=8
the region bounded by y=e^x, x=1 and y=1 is rotated about line x=1. Find the volume of solid
i used the slicing method
and got
(1-ln(y)) as the radius
v=integral of π(1-ln(y))^2 dy
is this correct or does cylindrical method only apply for this ques
the answer is 2π(e-2) units^3
(http://uploads.tapatalk-cdn.com/20160601/bdbc395aeadb78cf7cc7fd75c95009e4.jpg)(http://i1164.photobucket.com/albums/q566/Rui_Tong/Screen%20Shot%202016-06-02%20at%208.51.42%20AM_zpslaf9bjd6.png)
how to find the radius of smaller and bigger circle using slicing method.
the aim is to find the volume when rotated about y axis
How do you find the restriction in this question
The hyperbola H has equation xy = 16.
(a) Sketch this hyperbola and indicate on your diagram the positions and coordinates
of all points at which the curve intersects the axes of symmetry.
(b) P(4p, ), where p > 0, and Q(4q, ), where q > 0, are two distinct arbitrary
points on H. Find the equation of the chord PQ.
(c) Prove that the equation of the tangent at P is x + p2y = 8p.
(d) The tangents at P and Q intersect at T. Find the coordinates of T.
(e) The chord PQ produced passes through the point N(0, 8).
(i) Find the equation of the locus of T.
(ii) Give a geometrical description of this locus.
I only need part ii) i think the restriction was x can only be between 0 and 4
This is a more general question (with no specifics) - but its basically on how to do well at circle geometry.
Do you have any particular tips that could help in an exam situation when I'm stumped by the question. I've heard in general that there is a lot of cyclic quads, angles standing on the same arc etc. but I was wondering more if you had a particular process you would go through.
Thanks!!
You should be decently skilled (if not already mastered) circle geometry at the Extension 1 level before attempting problems that target Extension 2 students.
At the Extension 1 level, I was already familiar with what my theorems "LOOKED LIKE". What does this mean? Examples:
- Alternate angles look like Z angles on parallel lines
- Angles standing on same arc theorem looked like a nice M shape to me
- Alternate segment theorem involves a tangent and a triangle.
However how you visualise it may differ.
When I tackle an Extension 2 question, I don't label 50 thousand things at once if it's way too irrelevant. I work in one direction.
That is, I look at what I am trying to prove (or find).
Then I try to either work forwards, or backwards. That is, I look at the LHS or the RHS, then I start looking for an angle/side that equals to THAT ONLY. THEN, I keep going.
At the Extension 2 level you should also be prepared to manipulate lots of things. Similar triangles and cyclic quadrilaterals are examples of cliches, however even base angles of isosceles triangle are important. Equal radii is something I always keep at the back of my head regardless of if I need to use it.
Progressively, I build up to the final answer. But in doing so, I only consider RELEVANT information, not EXTRANEOUS information that may be useful for another part but not this present part.
A rare trap is the occurrence of trigonometry in circle geometry. When that happens, always look out for any right angled triangles BEFORE you attempt sine/cosine rule.
Ques: the region enclosed between the curve y=cos^-1(x) , the lines x=-1 and y=pie/2 is made to revolve about the axis y=-1 . find the volume of solid generated(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpsq360p6ps.png)
im not sure what the height is
the ans: 480
I did 3 cases 1 M, 2M and 3M : so 5!+ 6P5/2! + 7P5/5!
but it's not working
need help with this ques
A ring of altitude 2h is generated by revolving about y-axis the area of the segment bounded by the circle x^2+ y^2=a^2 and the chord of length 2h that is parallel to the y-axis. By using cylindrical shell method, show volume is given by v=[4π(h)^3]/3
can you briefly explain the method of pairing cylindrical shell and when is it applicable?
where v=2πsA
s=distance from axis of rotation
A=area
(http://uploads.tapatalk-cdn.com/20160615/e6242109cab4f5b0451622052b73d6cd.jpg)
how do prove this thanks
got stuck on proving this ques(http://uploads.tapatalk-cdn.com/20160616/a93853910c20d07f8e845e6fc493b05d.jpg)
how do i do part (iii)
Hello!
I think I remember somewhere that one moderator had a small checklist that they went through after completing each question (to check for stuff like units, constants, dxs etc.).
Would you be able to remind me what that is?
Thanks!
ive found the equation chord for PQ shown below and Mpo * Mqo =-1If this is one of the further questions in the Cambridge textbook then it might not be so easy to do it. These questions do not reflect the difficulty of the HSC (even though they are valid in reflecting question type) as the question is not broken down into parts for you - you have to discover everything yourself.
and im not sure how to progress
If this is one of the further questions in the Cambridge textbook then it might not be so easy to do it. These questions do not reflect the difficulty of the HSC (even though they are valid in reflecting question type) as the question is not broken down into parts for you - you have to discover everything yourself.
Damn, got about half as far as you did before you posted ;) these ellipse questions can actually be pretty nasty, I'm glad I didn't have to deal with them in 2014 :PThing is, when someone asks you a question and it's clearly out of Cambridge 4U, all you have to do is decipher the worked solutions :P
Thing is, when someone asks you a question and it's clearly out of Cambridge 4U, all you have to do is decipher the worked solutions :P
1. the base of a solid is the circle x^2+y^2=8x and every plane section perpendicular to the x-axis is a rectangle whose height is one third of the distance of the plane of the section from the origin. Show that the volume of the solid is v=64π/3 units^3(http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI21062016_zpsuwx6tey2.jpg)
2.A dome has a circular base of radius=10metres. Cross sections perpendicular to the base and one axis are parabolas who height is the same as the base width. Show that the area of the parabolic cross-sections is 8y^2/3 square metres
1. the base of a solid is the circle x^2+y^2=8x and every plane section perpendicular to the x-axis is a rectangle whose height is one third of the distance of the plane of the section from the origin. Show that the volume of the solid is v=64π/3 units^3
2.A dome has a circular base of radius=10metres. Cross sections perpendicular to the base and one axis are parabolas who height is the same as the base width. Show that the area of the parabolic cross-sections is 8y^2/3 square metres
(http://uploads.tapatalk-cdn.com/20160621/c5b8bebc5e795f6ef77ac064f25b0925.jpg)
(http://uploads.tapatalk-cdn.com/20160621/ab9bbc9e5e76c5a739da26fc95a4627d.jpg)
need help with these ques
(http://uploads.tapatalk-cdn.com/20160621/c5b8bebc5e795f6ef77ac064f25b0925.jpg)
(http://uploads.tapatalk-cdn.com/20160621/ab9bbc9e5e76c5a739da26fc95a4627d.jpg)
need help with these ques
how do i do part (i)
how do i do part iii)
i got
Part i) h=[(root3)*a]/2
part ii) ZY=xa/b
I'll let Rui correct me if this isn't quite 4 Unit friendly, but this is actually very simple if we know the formula for a sphere in three dimensions (x,y,z):Haha don't even need a z-axis
At the cross section, the height above the xy plane (the z-value) is h, so we can substitute:
What we notice here is that this is actually the formula for a circle. We've just found the formula for the outer circle of the cross section, with that term on the right being the square of the radius like normal. So:
Let me know if this makes sense! And Rui might come along and give a better method, I'm not sure if 3-dimensional surface equations are what would be the normal approach ;D
Haha don't even need a z-axis
Cheers Rui, that's much easier, too much 3-dimensional calculus has spoiled me ;)
(http://uploads.tapatalk-cdn.com/20160623/807dca2a1118738f9aa3aceccbbec6d9.jpg)Can't see what areas to compare. I would've just attacked it with an analysis of monotonic behaviour.
does this require comparison of areas
questions for harder ex 1
1. For any real x,y,z,u prove that (x^2+y^2)(z^2+u^2)>and equal to (xz+yu)^2
2.for any real numbers x,y,z,a,b,c, prove that (x^2+y^2+z^2)(a^2+b^2+c^2)>and equal to (ax+by+cz)^2
Here is one massive take out from Jamon's answer:
Where possible, you should feel safe to work BACKWARDS. Sometimes trying to work forwards just doesn't produce an obvious result, and you must start from somewhere else.
@Jamon however, because we're working backwards we don't want to use \implies, because that's for working forwards! We can always use \iff wherever appropriate, however preferably use \impliedby (left sided implication)
help with this ques thanks
1)for a,b>0 show that (a+b)/2 greater than and equal to [(a^2+b^2)/2]^1/2
2) if 0<t<1, show that 1/2<1/(1+t)<1
3) 1/2(x^3+y^3)greater than equal to [(x+y)/2]^3
(http://uploads.tapatalk-cdn.com/20160702/bed5022433ec1c9394cc43c5d79dcb47.jpg)(http://uploads.tapatalk-cdn.com/20160702/a7a5b5d4b26453b72bfdf6813c8a4335.jpg)
need help with these inequality ques thanks
(http://uploads.tapatalk-cdn.com/20160702/bed5022433ec1c9394cc43c5d79dcb47.jpg)(http://uploads.tapatalk-cdn.com/20160702/a7a5b5d4b26453b72bfdf6813c8a4335.jpg)
need help with these inequality ques thanks
(http://uploads.tapatalk-cdn.com/20160703/7333561a2dbd2be6d9cf862b8b57f7e5.jpg)(http://uploads.tapatalk-cdn.com/20160703/a085abe0cc473ed77d56a4a7b42ea586.jpg)I dare say that heaps of information is missing somewhere to make the 3 variable AM-GM inequality easy to prove.
(http://uploads.tapatalk-cdn.com/20160703/9493d5f5a0b89e5fe76172b22b46a9f9.jpg)
is my starting working correct and how do i progress
for resistance force, do i normally include k as a constant
(http://uploads.tapatalk-cdn.com/20160703/0e9e456f8a2dc778045d4cda51b5deca.jpg)
Hey! I don't have time for a full solution (I'm sure Rui will jump in), so I'll just give some quick pointers. The start of the first question looks fine, great job! What you want to do next is integrate between definite limits. So, the right hand side (dv) can be integrated between 30 and v, whilst the left hand side can be integrated between time 0 and T. Once you've done that, you know that the highest point will be when v is zero (ie. the object is stationary). Set v=0 and an answer should pop out neatly! Great job setting up the question.
As for the next question, your problem lies in your integration. I don't know how your school does it, but using indefinite integrals, and finding the +C term each time, is generally more difficult than just using definite integrals, like we did above. Can you see which limits to use?
Hope this helps!
Jake
(http://uploads.tapatalk-cdn.com/20160703/9493d5f5a0b89e5fe76172b22b46a9f9.jpg)In regards to Q21
is my starting working correct and how do i progress
for resistance force, do i normally include k as a constant
(http://uploads.tapatalk-cdn.com/20160703/0e9e456f8a2dc778045d4cda51b5deca.jpg)
(http://uploads.tapatalk-cdn.com/20160706/38e78a8db6661af362df6bf51d890da0.jpg)
how to do part (ii) for ques 4
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(http://uploads.tapatalk-cdn.com/20160706/38e78a8db6661af362df6bf51d890da0.jpg)
how to do part (ii) for ques 4
(http://uploads.tapatalk-cdn.com/20160706/83dcb292ab8221de867e6b4aa4965444.jpg)
why is the boundary between 0 and U/3 for the second integration not between U and U/3 since the initial speed is U
ive done part i but not sure how to do part ii (2nd picture)
(http://uploads.tapatalk-cdn.com/20160708/71756e4a3c2bdd84ebe34469ae7ced09.jpg)
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how to do part v) and vi)
i have problems with these types of questions especially these types.Bringing back memories of CSSA or something. This was a damn tough reduction formula.
OMG that took forever!!!
I just can't explain how much I envy your natural Mathematical ability. Love you matey :)
[tex]\text{Finally, rewrite the expression being clever enough as to use what we have in (v)}\\ \text{that is to say, use what we know!In an exam, if you see this type of question what is the best way to know what to do and should i leave these types of questions last
OMG that took forever!!!
In an exam, if you see this type of question what is the best way to know what to do and should i leave these types of questions last
In an exam, if you see this type of question what is the best way to know what to do and should i leave these types of questions lastPretty sure this question was the last question on the paper that year so that's saying something.
Is it ok if someone could tell me how to recognise and also memorise the difference of integration in 'Volumes' between finding the volume respecting to dy and finding the volume respecting to dx. Like the height and the radius, because i get coinfuse when it gets to dy, Thank you :)What do you mean?
Hi Jake,Yes. You cannot EVER have a resource better than past papers. And you can never do enough of them either.
I couldn't find a general Math Extension 2 forum, so I hope it's fine to post more general questions about the course here.
I was wondering if you could give me some tips on how to study for Extension 2 in the lead up to trials. Are Past Papers the best thing to be focusing on? Or is there something else that I should be focusing on like specifically just the harder areas of the course?
Hi Jake,
I couldn't find a general Math Extension 2 forum, so I hope it's fine to post more general questions about the course here.
I was wondering if you could give me some tips on how to study for Extension 2 in the lead up to trials. Are Past Papers the best thing to be focusing on? Or is there something else that I should be focusing on like specifically just the harder areas of the course?
Yes. You cannot EVER have a resource better than past papers. And you can never do enough of them either.
There are most certainly harder areas of the course. Take the entire harder 3U topic as a basic example. But even if you want to shift your focus to doing, say, lots of textbook revision on it, you still cannot get better than a student who does heaps of papers.
Reason being? Only through past papers, can you be trained into the correct ways of thought required for the exam.
_____________________
That being said, however, for every individual and every subject the best studying methods and techniques will vary. One way is to reconsolidate everything - go through your textbook (unfortunately I am still in the midst of composing the Ext 2 notes to advertise them), and do some review questions whilst you're at it. Then, you can just work through past papers as though you were doing homework, OR you can try doing one open book when you're ready. The further you go, the more confident you should be and not require your book. Towards the end, try to replicate full exam conditions.
And never be scared to use the solutions if you're stuck. if you don't work through the solutions (copy them out for practice), how can you learn?
Hey Mark,Correction: You wrote a list of fucking do this shit
Definitely post any questions you want here! Specific, course related stuff, or general, study related stuff.
I would just be doing past papers, as many as you can get your hands on. Do them over and over, the tough stuff and the stuff you're comfortable with. If you find yourself really failing to understand a specific section, definitely go ahead and try to learn it again, but generally just do past papers.
In Ext 2, I also wrote a list of stupid mistakes I often made, or things that I needed to make sure to remember to do. I would have a list of questions I couldn't figure out, and felt that I should, and every day go through them again until I was totally confident with them. Basically, as I was doing past papers, I identified weaknesses and tried as best I could to fix them.
At the end of the day, past papers are your best friend. Don't focus on the harder areas specifically, because there are the same number of marks allocated to the whole course. Spending 10 hours on harder three unit might get you an additional one mark, but spending 10 hours on complex numbers might give you full marks in that section. Just do past papers, over and over.
Also, post questions that you have here! Get your class into the forums, so you can all help each other out, and learn through teaching (would seriously recommend this).
Catch you around the forums :)
Jake
Correction: You wrote a list of fucking do this shit
Hahahah thanks Rui and Jake, loving the banter btw
Too true. And out of curiosity what school did you go to? Like was it one that normally achieves quite high?
How much time should one ideally spend studying 4u maths per week/day?Varies for person to person.
Just a rough idea
Im in yr 11 currently and doing great in ext 1 maths. Im considering doing ext 2.
Varies for person to person.
I did most of my 4u work during my free periods at school so I can't speak soundly enough from experience. I just did everything that was for homework (until exam time). Jake might have pointers here.
Yes ii is correct.
Since the absolute values are around x + 1 not just x itself.
Therefore we take the positive of x + 1 which is x > -1 and flip that around, disregarding the negative side.
What you are referring to is just ln(|x|)
for lAst ques is it possible to have 2 answers since you dont know whether the angle is obtuse or acute and the location of the vectors?I have no clue why edit post wasn't working.
Help with questions c and d please and thank you ;D
Help with questions c and d please and thank you ;D
(http://uploads.tapatalk-cdn.com/20160719/fcb0de7d958f3fd102564ce662b28824.jpg)
is there always linear relationship for triangles/trapezium only
ie x=mh+c or is using similar triangles better way to find x
You could go for the x=mh+c method but I'm too used to similar triangles myselfBumping this. I figured out the x=mh+c method. Yes, there always is one.
I remember seeing it before but I forgot where I saw it. It looked visibly more appealing
(http://uploads.tapatalk-cdn.com/20160723/2b90b86e42ceb3ccc60f85d487efa7f8.jpg)Nothing. Your answer is the exact same as what WolframAlpha said.
what did i do wrong for part d)
(http://uploads.tapatalk-cdn.com/20160723/07a98e47d05cbddee2bf4ce9b80907f4.jpg)
Nothing. Your answer is the exact same as what WolframAlpha said.(http://uploads.tapatalk-cdn.com/20160724/8e9d77f1623770b5bc8dd09b0623f70b.jpg)
Hey there!This proof uses a tiny bit of intuition, which is mostly trained at the university level. It also really tests if you know that a monotonic function satisfies the horizontal line test: it is one-to-one. It's perfectly understandable to a 4U student however calls on a bit above their skills required.
How can I prove inequalities with composite functions
e.g. How do i prove this: sin(x) - x + (x^2) / 2 > 0
Can i have a quick rundown of the conical pendulum?For these questions you need to analyse where the forces actually are.
Im not sure where normal (reaction force) comes in?
is the upward force(opposite mg) always (N+Tcosα) for ques like 4
Was wondering if anyone had a set of all the proofs we need to know to replicate? I'm fine with most eqns, but when we get those fundamentals questions, you really just have to know... taMost of the time if you know your actual content you will be able to make up the proofs on the spot without roting them. However from memory here are some important ones
Bless your cotton socks. Thank you kindly
Love how polite these forums are
Hey Rui, I have two questionsQ1: Even with reduction formulae, you usually follow the rule of LIATE with integration by parts (assuming that IBP is necessary).
For reduction integrals, how do you know which one to put as u and which one to put as dv?Is there any general rule to choose? And also when the question usually asks e.g. Show I (n+1) = n/n+1 I(n) when would you recommend subbing in LHS or RHS or is it up to individual intuition.
why is the vertical component Ncosx=mg instead of N+Tcosx=mgFor what question? In Q4 I got the exact same as you got.
(http://uploads.tapatalk-cdn.com/20160729/4967a683358303f5ec34f36f6c6a4267.jpg)I can't make sense out of that (ii) because mixing Z and omega around with z and w is doing my head in. If Z=z and omega=w then the expression would be false because the LHS has a term in Z2, Z3 and Z4 but the RHS does not. So those 1's would also have to be Z's.
how to do part ii)
(http://uploads.tapatalk-cdn.com/20160729/9fb440162049e0890723552c5d27b503.jpg)
how to do part ii) isnt (x^4+1)(x^2+1) irreducible over real fields
for the rectangular hyperbola, am i supposed to memorise the focus (+-root2*c, +-root2*c), directices x+y=-+ root2*c, vertex (c,c) or is there a simple derivation method
For what question? In Q4 I got the exact same as you got.
For Q7 the two tensions are acting horizontally and x = 0o implies cos(x) = 1
forgot to attach the ques
(http://uploads.tapatalk-cdn.com/20160730/2f3156eb9de64ae93a3e59e259b8919e.jpg)
how do to this ques? am i required to know how to prove this?
for ques c is (0,0) excluded?
for ques c is (0,0) excluded?
for ques c is (0,0) excluded?Yeah gonna back up Jake on the second one because
Hi! I should be sleeping but,
With graphs like these x^2y^2 - x^2 + y^2=0 how would you know what the shape is? Like even after finding stationary points etc, thank you!
Hi! I should be sleeping but,You should refrain from thinking about the shape of the graph because for something as ugly as that the shape does not have a name and in fact, it's virtually impossible to predict what it is.
With graphs like these x^2y^2 - x^2 + y^2=0 how would you know what the shape is? Like even after finding stationary points etc, thank you!
Ahh yeah it came up in half yearlies, fingers crossed that it won't come up again :/ But anyhow, with complex loci such as:For the locus arg(z-z1)-arg(z-z2)=α
arg(z+1)-arg(z-1)=pi/3
It looks like an arc of a circle, but how do you know whether to draw this arc above or below the x axis? Thanks ^^
Hey guys for part iii of this question how do you get what b equals to. I've attached the worked solution as well but i don't understand where the third line comes from (how the p^4 just factorises out? :S). Thanks!b is the product of the roots. Note that the general quadratic Ax2+Bx+C has sum of roots α+β=-B/A, αβ=C/A
how do you do this guys?
after proving that lhs=rhs do you use t formula? because it says "hence" :S
WOAAAH that's so smart lol thnx!
how do you get this guys??
how do you do this?
This was just a brute-force question, hope the steps make sense!Woooaaaahhh Jake got a 4U question before me
(http://i.imgur.com/DkMZuK5.png)
Woooaaaahhh Jake got a 4U question before me
Actually I was just studying for my discrete maths test.
guys how do you do this?
How do you get that?
Yo guys how do you do part iii. Rui I tried doing the method you did by taking it to I0 but i ended up with:I subbed in values for n on WolframAlpha. That question has a mistake; it is (n+1)(n+2)/2 as you said.
1/(In) = (n+2)(n+1), am i doing something wrong?
hey can someone get this? i got the answer except with a 1/2 infront :S
Thanks Rui!
Also how do do you do the q attached?
Hello!It is extremely rare that a projectile question should appear in the 4U exam. As far as my memory goes, the only time it was asked was sometime during 2003-2006 where they combined it with resisted motion.
Anyone got some top tips for how to deal with harder projectile motion questions that you don't know how to do?
Any help would be much appreciated!!
Thanks in advance :)
hey guys is ln(x/2) the same as ln( (tanx/2 - 1) / (tanx/2) )? Also if it is true, how? Thanks!!Nope.
oh then how do you do the q attached. my ans was ln( (tanx/2 - 1) / (tanx/2) ) but the answer is ln(x/2). Thanks.
sweeeet thanks Rui!!I think I've seen this question before. It's been tagged as faulty.
Also can i just ask another question. How do you do part ii for the q attached. Obviously you have to use part i because 'hence' but how? What i was thinking was to use by parts but I don't think that's gonna work.
It's been tagged as faulty.
So the questions wrong?
how do you do this man ?
Thanks Jake!
Guys for part ii of the question attached. How do you do get the reduction In ?
Thanks Jake!
Guys for part ii of the question attached. How do you do get the reduction In ?
Thanks Rui! yeah i eventually got it :P(http://uploads.tapatalk-cdn.com/20160815/8fc5c3656b474dce408513c10d76ef78.jpg)
Guys check out the integration q attached. Its pretty interesting never seen anything like it. how does one go about solving something like that. At first I thought partial fractions but i don't think that's going to work. Im guessing substitution, idek. Any thought??
(http://uploads.tapatalk-cdn.com/20160815/8fc5c3656b474dce408513c10d76ef78.jpg)
I briefly looked at it and first deduced that it had gone well beyond the scope of the HSC as any exponentiation on itself is tremendously hard to handle. I.e. x^x
But after spending 4 more minutes on it I realised wait no it's too ridiculous
how do you do part ii guys?Hint (since I'm having dinner right now)
Hint (since I'm having dinner right now)
Let x=tan(theta)
Inspiration: because the derivative of tan is sec squared and because 1+tan^2=sec^2 AND the boundaries were too convenient
far out thanks!!
Also how do you do part ii for this q?
Thank you Rui!!!
I have another question. Part ii - i don't get how you do it, i tried using part i but there's also a B, what do you with that?
Woaah thats so cool! This is probably gonna be a dumb q, but how do you know you're left with sin2nx, like how do you know that that term doesn't get cancelled out?Because the last term has nothing to pair up and vanish with.
Oh yeaah i get it!
Hey i just have one more question, last question for the night (hopefully). How do you do it? :P
Guys when doing volumes, when do you split the area into two sections and then find the volume of each and add the two separate volumes to get the total volume?Don't quite understand this question. I analyse what areas are relevant before I analyse the relevant volume. And the methods for area analysis should be familiar from 2U.
how do you do this q guys?Before I have a go at this (lacking materials), have you drawn out a(n appropriate) diagram?
.Before I have a go at this (lacking materials), have you drawn out a(n appropriate) diagram?
OHHH mate thanks i get it!!This is ridiculous. I'll leave the proof of the volume of a square pyramid as an exercise. Note: The volume of any pyramid is V = 1/3 A h, where A is the area of the base and h is the vertical height.(http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI16082016_zpssm1jjg34.jpg)
Also need help in the one attached. Wth is it even ?? :S
This is ridiculous. I'll leave the proof of the volume of a square pyramid as an exercise. Note: The volume of any pyramid is V = 1/3 A h, where A is the area of the base and h is the vertical height.(http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI16082016_zpssm1jjg34.jpg)
do we have to put this line for hsc (the line with the sigma notation), like do we lose marks if we don't, because in our school our teacher never told us to put this line in :S
guys how do you do this :S
Thanks Rui!!
Guys the question attached is so simple but I can't seem to get it :/
The answer to part i is: 4pi.a.f(x).dx How?? its so weird
Nup it's fine if you go straight to the integral :)Are you sure?
Thanks Rui :DI also got 32
You see for the q attached, i keep getting 32 for my ans, but the actual answer is 24 (part ii) :S
wait part ii is easy, because you just use part a (which is attached)
But i have a new problem, how in the world do you do part a. ??
I tried following the answers but yh this question is something. Thanks in advance :)What's the confusion? Is there a particular part about the answers that makes no sense? (Sorry, but your question is generic and I don't know where to focus my response on.)
Sorry should've been more specific. But yeah it was how to tackle part 2. Thanks, really helped :)If part 2 was the hard part (which is understandable), I recommend you try attempting the example I provided. See if you can use a similar method to prove the 4 variable AM-GM.
Btw did anyone manage to figure out the two volumes questions I posted previously, I still don't get them :( . thanks!
Just because of the volume (lol) of questions you've been asking, do you think you could show us some working out before we tackle the questions you post? Just so we can see where we can help you specifically
Guys how do you even do this question???? :SYou need to draw diagrams.
need help with these quesb) looks like a question that just involves tackling with units.
a)a disc is free to rotate in a horizontal plane about an axis through its centre O. A small object P is placed on disc so that OP=0.2m. Contact between the particle and the disc is rough and limiting friction of the particle can experience is 0.5 times the normal reaction between particle and disc. the disc then begins to rotate. Find angular v. of the disc when particle is about to slip.
b)to start a lawn mower, a green keeper keeper needs to pull a chord wound around a grooves wheel of diameter 12cm at speed 147cm/s. find angular velocity of wheel in revolutions per min
need help with these ques
a)a disc is free to rotate in a horizontal plane about an axis through its centre O. A small object P is placed on disc so that OP=0.2m. Contact between the particle and the disc is rough and limiting friction of the particle can experience is 0.5 times the normal reaction between particle and disc. the disc then begins to rotate. Find angular v. of the disc when particle is about to slip.
b)to start a lawn mower, a green keeper keeper needs to pull a chord wound around a grooves wheel of diameter 12cm at speed 147cm/s. find angular velocity of wheel in revolutions per min
Hi. So i follow this question (with ofc additional workout since the solution skips a lot of steps) right up till this fraction: [ (a+b)^2 - [(a+b)^2]/2 ]/[...]
I get wats going on with the denominator since: (a^1/2 - b^1/2)^2 ≥ 0 ... (a^2 + b^2) ≥ 2ab ... [(a^2 + b^2)^2] / 4 ≥ (ab)^2
therefore [a^2 + b^2] / (ab)^2 ≥ (a^2 + b^2) / [(a^2 + b^2)^2 / 4] since the denominator of the former is smaller than the latter. But I don't get wats going on with the numerator of the first fraction I typed. Thanks :)
Hi,Are we talking about the 1999 HSC exam? Because Q2c) of that does not have Re(z).
I'm stuck on a few questions from paper 1999
Q2c) why is Re(z)= -1 an empty circle?
Q2e) isn't turning anticlockwise x i, and turning clockwise x -i ?
So answer should be -->AP= -Z(1)i? [But ANS is -->AP =Z(1)i]
(I tried to upload the pics of the question, but it didn't work)
Mod edit: Merged posts. Try to keep everything together :)
Hi,Regarding 2e) That is an anticlockwise rotation. Keep in mind the direction of the vector.
I'm stuck on a few questions from paper 1999
Q2c) why is Re(z)= -1 an empty circle?
Q2e) isn't turning anticlockwise x i, and turning clockwise x -i ?
So answer should be -->AP= -Z(1)i? [But ANS is -->AP =Z(1)i]
(I tried to upload the pics of the question, but it didn't work)
Mod edit: Merged posts. Try to keep everything together :)
What about Q2e) ?
Also, for Q6bvi) I followed through i to v) but when I got up to vi) what's U, U(o) and V? and also, T(o) is the total time if air resistance is ignored so this should be smaller than T (which if the total time with air resistance-I think) ? >>because an object falling with no air resistance will fall quicker than an object falling against air resistance. [But the ANS is T(o) > T]
Thanks.
Hi, how do u solve this question?Can't exactly follow your logic how 6!3! eliminates the cases when the white roses are planted together in your working.
In how many ways can 5 different red roses and 3 different white roses be planted in a straight line next to a fence so that no 2 white roses are together?
I personally did: 8! - 6! x 3! but this equates to 36 000 and the answer is 14 400. Can you please also tell me wat is wrong with my working out. Thanks :)
Can't exactly follow your logic how 6!3! eliminates the cases when the white roses are planted together in your working.
Thanks! Well I did 6! x 3! as the white roses being together as you would consider the 3 roses as a group and therefore there would be 6! ways of arranging them with the 5 roses and 3! ways of themselves being arranged in their own group. My wording is not probably the best but that's how I usually tackle these sorts of questions and I'm not sure why this method doesn't get me to the answer like it usually does. Do u know y that is the case?
Hi znaser,Judging by your 6*3!, you probably forgot that all the red roses are different as well.
I might be wrong, but I'm pretty sure that with your method, for the white flowers; it's just 6x3! as if you imagine the 3 flowers as a bunch, you can order them with only 6 choices and not 6! choices.
So instead, the answer should be 8!-3!x6= 40284, unless if I have done something horribly wrong :)
I follow you! That covers cases with all three white roses together, but if you think carefully, it actually doesn't cover two at a time. So, you aren't subtracting enough cases, which is why your answer is higher than it should be ;D
I love the method you use, and often it works really well, just not in this case, you need to be very careful with probability!
Hi znaser,
I might be wrong, but I'm pretty sure that with your method, for the white flowers; it's just 6x3! as if you imagine the 3 flowers as a bunch, you can order them with only 6 choices and not 6! choices.
So instead, the answer should be 8!-3!x6= 40284, unless if I have done something horribly wrong :)
Can't exactly follow your logic how 6!3! eliminates the cases when the white roses are planted together in your working.
I'm no expert in permutations and combinations, but I believe Rui's answer is incorrect. To demonstrate this, we provide the following counterexample:
Suppose the red roses are fixed as Rui imposed. Then observe that another viable arrangement is:
_RRRR_R_
Here, there are only three blank spots to fill. Thus, Rui's calculation does not capture the entire spectrum of cases and I conjecture that to achieve the correct number of permutations using this method, one would have to apply brute force to obtain all possible arrangements of red roses that leave at most one blank spot in between any two that occur. This would prove to be an inefficient process.
I believe one approach that seems right, though I'm not entirely convinced that it is, is to group two white roses together, treating them as a single object, and evaluate the number of permutations this yields, then subtracting this off from the total number of arrangements, namely 8!. The important observation to make is that this will account for the case where all three white roses are together, inclusively. So we need not treat the two cases separately as requiring two independent calculations.
When two white roses are grouped together, we now have 7 distinct objects. These are arranged in 7! ways, however, the two white roses that are coupled can be chosen in 3P2 = 3! ways. So we thus have 7!3! ways this can happen.
The total number of arrangements where no two white roses are together is calculated as 8! - 7!3! = 10 080. However, this is lower than Rui's answer which I alluded to have understated the true answer since he only accounted for one special arrangement of the red roses. Hopefully, I still have the right idea in my method but just need to make a small alteration. Feel free to correct me where I have gone wrong.
I believe that Rui's answer accounts for that; he has all possible places for the white roses to be placed. Taking his example:
We would put white roses in the first, fifth, and last positions to obtain your arrangement :)
Ah so it does involve inclusion-exclusion.
Hi Mei2016,
I think it would be 8! - 6! x 6 x 3! as that would remove the possibility for 1 white rose to be separated but 2 to be together. You get the answer but it may be a fluke 😅. But yh either way rui's method is probably better in this case as it's much more simpler.
Hey znaser,
You mean I got the answer? Do you know what it is? Also, where did this question come from? (as it would normally be like 5 girls and 3 boys...)
I really like Rui's way of doing it, so if you have the answers, what source is if from?
Also, to continue with my method, (because I had previously neglected the white flowers being together case), I've come to:
All possibilities-all white tog-considering the case of 2 white flowers tog and one separate;
8!-6!x6x3!-5!x6x5x2= 7200 (which is exactly half of Rui's answer, and that's so interesting, but I don't know what's wrong with it)
Hi,
This is Q6 iv) from the 1997 HSC paper. I just don't get the N.B part in the answers.
(i.e how the _< turned into a < sign)
Hi,
This is Q6 iv) from the 1997 HSC paper. I just don't get the N.B part in the answers.
(i.e how the _< turned into a < sign)
Hi Rui,If you're going to make an inequality strict you have to be explicit about why.
Thanks for your help. I know what a rational number is, I was just previously confused with the wording of their explanation.
In the test when we do this (taking the equality out) do we need to explain it in words like how the book does?
Hint for the time being:This was the best I could come up with.
The question just means to find the normal reaction the surface exerts on the particle.
Remember: According to Newton's Third Law of Motion, the normal reaction the surface exerts on the particle has the same magnitude as the force the particle exerts onto the bowl.
> > > Please do not double post < < <
guys quick question:This is just asking you to prove the reflection property for the ellipse. It'll be in your textbook.
Find the equation of the normal to the ellipse at P (2cos(theta), sqrt(3)sin(theta)) and prove that it bisects angle SPS' where S and S' are the foci
I can get the normal but i just can't get the second part of the question, thanks for the help!
This is just asking you to prove the reflection property for the ellipse. It'll be in your textbook.
Some questions in past HSC papers (I think there was one in 2009) also made students prove the reflection property. They did it for the hyperbola, though, which tweaks the proof a little.
Check one of the past papers such as 2009. There's a small proof as to how you can then use similar triangles to induce your required result.
The answers prove that PS/PS' = ST/S'T and therefore the normal bisects angle SPS' :S how does that work?? (BTW T is the x intercept of the normal)
need help with these ques:I do not take credit for an answer that's not mine. All credit goes to the person i seek for help. (I honestly could not understand Q2 at all when I first saw it without an appropriate diagram.)
1.a particle of mass M is attached to one end of a light inelastic string of length `2a and the other end is fastened to a fixed point O on a smooth horizontal table. Another particle also of mass M is attached to the midpoint of the string. The system then rotates in a horizontal circle and in contact with the table at an angular velocity of w. Find the ratio of the tensions in each part of the string.
2.a find of mass 0.6kg is attached to a point P on a string AB of length 1.4m, where AP is 0.8m. The ends A and B are attached to 2 points 1.0m apart in a vertical line, A being above B. THe ring is made to travel in a horizontal circle with speed v m/s.
what is the smallest possible values of v if neither portion of string is slack?
Apparently, the scenario to Q1 is similar to this one.
yo guys does anybody know what papers contain circular motion questions, I've been looking everywhere and can't seem to find any. thanks!That's because the HSC is highly unlikely to ask a question that's bluntly related to circular motion. It's virtually always on one of the applications. of it - conical pendulum or banked tracks.
Hey just quickly, do you guys know of any little pointers to remember for mechanics, i think it occurs mostly in conical pendulum. What i mean is for e.g. when N=0 the particle is no longer touching the surface and also if w^2>0 then the particle is moving. Does anybody know anything else to keep an eye out for?0≤θ<π/2 if you do it correctly. (If the angle is 0 then obviously you're at rest though, and that's basically the same thing as saying ω2=0 so you usually don't get that happening.)
how to do last part
(http://uploads.tapatalk-cdn.com/20160907/9f62dd8cc7b37f797025648a47ad2cee.jpg)
need help with this ques thanks
Hey guys how do you do this ??
Hey thankyou so much. This particular question was from the catholic trial 1984...I don't usually do questions that are really old, it's just that i need to get good in maths HSC, theyre the only things pulling my atar up :L
Note that the things here are negative as:
a) The centre of the planet is defined as x=0
b) Gravity brings things BACK to the centre of the planet
And this is where I got lost. I kinda had to use HSC physics here... so I reckon this question is pushing the boundaries of the 4U course.
Can you please stop avoiding this question I've asked many times and tell us what is the source of these questions?
Hey thankyou so much. This particular question was from the catholic trial 1984...I don't usually do questions that are really old, it's just that i need to get good in maths HSC, theyre the only things pulling my atar up :LThat was all you had to say though haha. That explains why these questions are extremely difficult - they're going beyond 1990.
Hello y'all! I come with a complex numbers question (I have no bloody clue what's going on with this one):
Let the points A1,A2,...,An represent the nth roots of unity, w1, w2, ..., wn, and suppose P represents any complex number z such that lzl=1.
a) Prove that w1+w2+...+wn=0
b) Show that PA2i=(z-wi)(z (conjugate) - wi (conjugate) for i=1,2,...,n
c)Prove that:
Σ (from i=1 to n) PA2i=2n
Sorry it's really hard to type it out on this forum because of all the squares and subscripts and conjugate bars and sigmas and ah i hope it still makes sense because it doesn't to me! Thanks guys :)
Neutrons
How do you do this?
If abs(a)>2abs(b), prove that 2abs(a-b)>abs(a)
thanks!
Hi everyone can u please help me with this question, it was from my school's trial:
Here is a question I was asked during the 3U lecture break. I will address it in the next post.Unfortunately, I was unable to generate a better proof than what I offered up on the day for a 2 mark question. I will give a full statement of the proof I had then. It may be worth mentioning that I considered an approach that utilises stationary points, however felt it was a bit overkill for a 2 mark question. The stationary points do not have a tidy form: \( x=q+\frac{q-r}{-1\pm \sqrt{\frac{p-r}{p-q}}} \)
(http://uploads.tapatalk-cdn.com/20160926/0d2e6f8afe0be737393f3ee0a40a7d87.jpg)
Apologies for the poor quality photo
how to do part iii)
(http://uploads.tapatalk-cdn.com/20160926/0ccb89568cb328c0c61d36a0e6e40d42.jpg)I think the division by 3! is because of the issue with ordering here.
for part d ii) do i always need to divide 3! (factorial) because there are same number of people in all the groups
(http://uploads.tapatalk-cdn.com/20160926/04d8ba88c0f9b5da660ba5786b64748a.jpg)
how to do part c)
Hi everyone !!! I know this may sound a bit basic but I am kind of confused ... With relation to resisted motion in a horizontal line and vertically in Mechanics ... When do we use -mkv and just -kv??? I can see in some working out that mass does actually make a difference even though it cancels out in some cases ... I'm just confused as to when I'm supposed to add the m in -kv or -kv^2.... Yeah that's my question thanks !!!They either give it to you, or they will say something along the lines of "force per unit mass".
They either give it to you, or they will say something along the lines of "force per unit mass".
Generally, the HSC is nice enough to just give you it and you can easily guess which to use. But this doesn't happen in textbooks.
In general, when "per unit mass" is mentioned, you are using mkv.
how to do part a i)
Hey!
Could you guys please help me with the last question of the 2011 paper? Thanks, I don't even understand the question haha
Neutron
Would I be able to get some help with this question:
"A stone of mass 6kg is tied at one end of a 3 metre long strong. The other end is fixed to a point O. Find the tension in the string when the mass is rotating at 40 revolutions per minute"
Thank you :)
Hi! I'm a bit confused about this 2011 HSC question:
8(b) A bag contains seven balls numbered from 1 to 7. A ball is chosen at random
and its number is noted. The ball is then returned to the bag. This is done a total
of seven times.
(i) What is the probability that each ball is selected exactly once?
(ii) What is the probability that at least one ball is not selected?
(iii) What is the probability that exactly one of the balls is not selected?
I am fine with parts i and ii, but iii had me a bit confused, and I'm not really sure I quite understand the given answer. My answer was
7/7x6/7x5/7x4/7x3/7x2/7x6/7 -> i.e. probability of six balls being selected exactly once, then one of those six balls being selected again. I don't see what is incorrect with this, if you could explain that would be great!
how to do part iv)Evaluating:
if 1<x<y , is x-y>0 ??
if 1<x<y , is x-y>0 ??1 < x < y
thanks guys, also how do you do the q attached.
the question asks to solve for x
wow that was real neat. Thanks Rui. So whenever we have absolute values and we need to solve do we always use cases like you did?Yeah. Whenever a course of action isn't obvious, split the absolute values into cases.
Back with another question!
I'm super confused about the last part. Doesn't it essentially mean having to prove the existence of the circle in the very question??
Thanks :)
how to do part iii)
Hey, I'm new here so I'm not sure how this works haha
this is probably really simple but how would you graph
i) 1/f(x)
ii) [f(x)]^2
thank you :)
Hey, I'm new here so I'm not sure how this works haha
this is probably really simple but how would you graph
i) 1/f(x)
ii) [f(x)]^2
thank you :)
Hi. I'm not sure of the working out for this question. The answer is A.
A more brute force approach that would work is a case by case analysis.
Hello!
Another out-of-left-field question, but I was wondering how you would order the HSC Extension 2 papers in difficultly (from 2001 to 2015) or in rough equivalents?
Thanks again!
That's hard. I need to look at every single paper to give an answer to this one.And now that I had a quick glance (I'm not spending 45 hours of my life doing all those papers...) this is what I reckon
Jamon went through the effort of making a guide for 4U curve sketching. Consider reading it.
Hey! Welcome to the forums! You've done the exact right thing; post up any questions you've got, and we're here to answer them! You can also answer others' questions if you feel up to it :)
For these kinds of questions, you just need to think about what happens at critical points on the graph. So, let's start with 1/f(x).
We know that at x approaching infinity, y will approach 2. That means that, for 1/f(x), as x approaches infinity, y will approach 1/2. We also know that y equals -1 and x=2, therefore for 1/f(x), when x=2, y=1/(-1)=-1.
Now, we need to recognise that if we divide by zero, there will be an asymptote at that point. We know that at x=0, y=0. Therefore, on our new graph, there will be an asymptote at x=0!
On the left hand side, we know that as x reaches negative infinity, y reaches infinity. As 1/infinity=0, our new graph will reach zero as x reaches negative infinity, and reach infinity as x reaches zero from the left.
This is super hard to explain. I have to run, but hopefully someone can give a better explanation/show you the second graph before I come back!
Thanks so much guys :)Looking good. Well done :)
I had a read of Jamon's guide and it was really helpful!!
Here's my attempt at the questions
Do you lose marks in the HSC for not putting working out in your answers?Yes.
And now that I had a quick glance (I'm not spending 45 hours of my life doing all those papers...) this is what I reckon
From easiest to hardest
2014
2015
2012
2013
2010
2011
2007
2009
2004
2006
2005
2002
2003
2001
Reminder: This is very easily debatable/arguable.
Yes.
That's why some questions are worth 3 marks, not 1 mark each. You're actually AWARDED marks for working out.
If you magically write down an answer, people will think ok where did you get it from
would the working out in the sample answers be a good indication to how much they require?Maths sample answers are somewhat superior to every other course because they answer the question. But they may or may not make sense and you should use Excel Success One or MANSW published books.
How does this work? i've never really understood how do it and when to apply it.
first ques: the answer assumes 0<=t<=1 for part ii) and i dont get why thats the caseYou've already asked the second question and it was answered on the previous page.
2nd ques; dont know how to do part iii)
how would you find the asymptotes for a parametric equation like
Sorry could you please explain why x≠2? Is it because as t --> infinity, x --> 2?
Hey could I have a hand with this question? I'm not sure which formulae to use
Hey could I have a hand with this question? I'm not sure which formulae to usei.e. (So this is just adding to Jake's answer)
Could anyone have a look at this question I'm stuck on? Would you use slices or shells? Or is it a different method that I'm missing? Thanks :)
Hi. I can't seem to understand the working out of part (i).
Note: Consider an alternate source of answers to BOSTES' ones.
__________________________________
Hi guys,
Could you guys help me out with this probability multiple choice question
I always seem to get the harder probability questions wrong, does anyone have any advice for this topic that could help?
Cool thanks for the breakdown RuiYou should see me. I'm dying with my first year uni combinatorics.
I absolutely dread probability :'(
Hey guys how do you part c, the answer is: Incorrect, n=1020 is one counter example
wth does this mean, could someone explain the q and answer please!
Thankss
OHHH that makes more sense now, thankyou. But in an exam situation, how do you come up with a number like 1020 to prove by contradictionFor that particular question, the inspiration was that
Hey could someone help with this question please! :)
hey guys how do you part b, im so close yet so far :/ i keep getting (-1)1-k instead of (-1)k+1
Just curious but is it possible to self learn 4U maths?Things I successfully self learnt:
I've heard mixed rumors from people saying things such as "4U requires critical thinking", "you need a maths tutor for 4U or else you'll do bad", "yes 4U is self learnable" etc but the thing is I don't know what is right and what is wrong
Guys how do you prove the attached question by mathematical induction :S
thankyou <3
Just wondering if anyone has any tips on how to recognise which substitution to use in Integration for 4U.
In 3U integration is the easiest topic for me because majority of the time they give you the substitution to use and it just becomes a matter of algebra but with 4U clearly they don't and I always get stuck in recognising what substitution to use especially when trig is involved. Are there any little markers to recognise or patterns that i am just not seeing?
Guys how do you do this?:Let me say this now, there is a 0% chance of a geometric induction
Prove by mathematical induction that the sum of the exterior angles of a n sided convex polygon is 360 degrees
Also in general is there any technique to do geometry induction proofs?
OHH that makes so much sense, I had not thought of it in that way. Thank you!!Post anytime you want and I'll be there as soon as I'm free (or Jake will) :)
I've given up on maths for the night but I definitely will post up some questions tomorrow (cause there were a few where the worked solutions only left me more confused)
how does this line: e1/(n+1) < 1 + 1/n < e1/n become this line: (1+1/n)n < e < (1+1/n)n+1Hint: Split the inequality up and work on two sides at once. Then recombine them.
Like in the solutions there's no other lines in between, so i was just wondering how they got that second line :S
Hint: Split the inequality up and work on two sides at once. Then recombine them.
how to do part ii and iv
guys how do you do part ii :S
Not sure how to do the first 2 questions. Thanks!
guys what do the thetas in conics represent??The angle it makes with the x-axis at the origin, just like with a circle
guys how do you do the attached q
ans: 8!/(4!x4!)
Hey guys whats the general solution to the following qs: (I put more than one coz they're pretty simple.. I think)The former is wrong. Look again.
1) sin(x-pi/3) = cos2x
2) sin3x + sinx = 0
3) tan2x + cot3x =0
Also is the general solution for cosx = 0 , x= (2n+1)pi OR +/-pi/2 + 2npi ?
Thanks heaps for your help ;D
thanks Rui, one more thing- can you quotes the formula for the chord of contact (in conics) or do you have to prove it?Prove it. You only quote things that they give you in the exam.
Hi, if you take the reciprocal of a function that has an oblique asymptote, what happens to the asymptoe? (HSC 2012 Q14bi) --> an oblique asymptote with equation y=2x-1 turned into an asymptote with equation y=0. How do you know? And is this the case for all oblique asymptotes? Thanks.
Oh my, thank you for your explanation. That makes so much more sense now :))). Could you please give an example of what you meant with y=a and y=1/a? I'm trying to make sense of it but I think Im just confusing myself :')
should we draw graphs and other diagrams in pen or pencil? I was just wondering because of the whole scanning process and them not being able to see your work unless its in black pen.I drew all of my graphs in pen last year
guys how do you do this... i can't seem to get the table of values right :/Show us your working
Show us your workingI don't really have working, this is what i have for the seven ordinates:
I don't really have working, this is what i have for the seven ordinates:7 evenly spaced coordinates just mean that there should be 6 equally-sized 'intervals' between the ordinates.
0, pi/8, 3pi/8, pi/2, 5pi/8, 6pi/8, pi
The only problem is these aren't evenly spaced :/
7 evenly spaced coordinates just mean that there should be 6 equally-sized 'intervals' between the ordinates.
Hence, take x=...
0, π/6, 2π/6, 3π/6, 4π/6, 5π/6, 6π/6
i.e.
0, π/6, π/3, π/2, 2π/3, 5π/6, π
Howdu figure that out :OSince there are 6 intervals, consider the difference in the boundaries and divide by 6
Since there are 6 intervals, consider the difference in the boundaries and divide by 6that's so cool, thanks for sharing!
(π-0)/6 = π/6
Hence, start at 0 and increase by π/6, six times. It's actually a standard trick that 2U kids need to know, although Simpson's rule is, of course, extremely rare in 4U
hey quickly, is the q attached wrong? because cos and sin aren't for hyperbolas, or can this q still be done?It can still be done. It's just extensively unconventional.
(http://uploads.tapatalk-cdn.com/20161019/4cc71cd3339cce81c19e5aca09c15b0f.jpg)Technically yes, the dotted line is correct for c). But I never drew it like that in the HSC. I just kept it blocked because of the ≤.
for answer to part c (below) is it correct to have dotted line for the outer line of circle and the vertical line
(http://uploads.tapatalk-cdn.com/20161019/4dd7f38c176d1491867514836c3de400.jpg)
how to do part b
(http://uploads.tapatalk-cdn.com/20161019/33387b03a6d5b7d94071c5a5d7fd51c3.jpg)
(http://uploads.tapatalk-cdn.com/20161019/4cc71cd3339cce81c19e5aca09c15b0f.jpg)b):
for answer to part c (below) is it correct to have dotted line for the outer line of circle and the vertical line
(http://uploads.tapatalk-cdn.com/20161019/4dd7f38c176d1491867514836c3de400.jpg)
how to do part b
(http://uploads.tapatalk-cdn.com/20161019/33387b03a6d5b7d94071c5a5d7fd51c3.jpg)
(http://uploads.tapatalk-cdn.com/20161019/50301deeb599b5b14a2ecf9661392f0e.jpg)(http://uploads.tapatalk-cdn.com/20161019/c55ed55960bd54bd986dd218bd5af0f1.jpg)(1+3i) and (9+3i) are the foci.
Hi, is there a way to do ii) without finding the equation of the ellipse? Is the locus saying the distance of z from the origin to (1+3i) + distance of z from the origin to (9+3i) = 10? I.e. PS + PS' = 10? For so then wouldn't the origin be a point on the ellipse?
guys is the equation for asymptotes y=+/-b/a x right?, but how come the answer to this q is +/4/3 x not +/-3/4 x :S. In conics does 'a' ALWAYS represent the number under x, i always thought a was the bigger number and b was the smaller no.
wait so is 'a' always the one under x and 'b' the one under y? like when you use b^2=a^2(e^21), B^2 = 16 and a^2=9 right?For x^2/a^2-y^2/b^2=1 yeah
For x^2/a^2-y^2/b^2=1 yeah
is it the same for ellipses as well?
Hey! Looks like you're mixing up your formulas a bit; I would take a look at this seriously comprehensive document HERE and come back if you have any more questions!thanks dude
(1+3i) and (9+3i) are the foci.
S and S' do NOT lie on the x-axis. So the origin doesn't have to play a role here.
And of course, PS+PS' = 10 implies that the length of the major axis is 10 (hence the length of the semi-major axis is 5)
So your interpretation is correct. But your interpretation does not mean that the origin has to lie on the ellipse.
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Oh thank you. But then how do you sketch the locus without finding the equation? As in how do you know it lies strictly in the first quadrant? And after finding 2a, how do you find 2b without the equation?You don't know everything. Some things you have to infer.
You don't know everything. Some things you have to infer.
2a is of course, given.
But the distance between the two foci is 2ae. Recall that SS' = 2ae
You are given the foci in the question.
Since the foci are at (1,3) and (9,3), the focal length is 4.
Hence 4 = ae
So e=4/5
Then you can use b^2=a^2(1-e^2) to find the length of the minor axis.
Note also that the ellipse's centre is also at (5,3)
Given the centre, and the length of the major and minor axes, you should be able to sketch the ellipse.
image 1: how to do part v)
image 2: how to do part ii) and iii)
Haha here comes another question :'). For harder 3U, I don't quite get the logic behind proving f(x)>/= 0 for x>0. Can you please interpret this is more detail? Thanks.Just read those dot points again. Read them one at a time. And use a visual aid if you need it.
image 1: how to do part v)
image 2: how to do part ii) and iii)
(http://uploads.tapatalk-cdn.com/20161020/5e8255e0847472beeba2252a3a473974.jpg)
Can you please walk me through this question
I think it'd be easy to do this part graphically as wellWould they let that go in 4U? Cause people mess up drawing to scale graphs.
Would they let that go in 4U? Cause people mess up drawing to scale graphs.
Also it's not a 1 marker
I'd definitely back a graphical solution to an inequality as full marks, if it's draw to scale. Smart, easy solution; the sort of this 4U markers love. Your solution is kind of a formal 'proof' of the graphical solution; definitely better maths, but like bruh. Cmon. Easy answers = good answers.If you really wanted to be easy then "by inspection" is the best answer please.
guys whats the difference between uniform and non-uniform circular motion??Non-uniform circular motion is not in the HSC 4U course.
Ohhhhh so is it like this: if we didn't show that g(0)=0, then the curve could be below the x axis somewhere between 0<x<infinity. But seeing that is is =0 at 0 then after x=0 the curve must be >0. Which is what you pretty much said-omg took me a while. LMAO wow thank you!!!Lel glad that we figured out the issue :P
image 1: how to do part v)
image 2: how to do part ii) and iii)
(http://uploads.tapatalk-cdn.com/20161020/e3196b8c42fca4b4a1caf9a368362a1d.jpg)Where you said since f"(x) > 0 for all x > 0 you might want to put in brackets (by part (i))
Would this be the correct layout to part iii) of the previous question? :)
(http://uploads.tapatalk-cdn.com/20161020/f34a06fb6272d9c3fc52bc4d93a233bb.jpg)Note that F is NOT perpendicular to N here. Here's some things done to the diagram to make it clearer. Note: Corresponding angles on parallel lines used.
how to do part i) im confused with the angles for normal force
(http://uploads.tapatalk-cdn.com/20161020/d8e0df8773cce450704d080e4194724c.jpg)
Yo! Can someone please help with the last question of 2015? I have no clue what's going on! Thank you
(http://uploads.tapatalk-cdn.com/20161020/d8e0df8773cce450704d080e4194724c.jpg)
Yo! Can someone please help with the last question of 2015? I have no clue what's going on! Thank you
how to do part ii)
how to do part ii)
hello... just wondering if someone could help me... :PYou ALWAYS use radians unless you see that circle o floating in the air. Only when that circle is there do you use degrees.
I always struggle knowing whether to use radians or degrees in certain situations>? how do I distinguish coz i'd HATE to lose a mark just for not typing it into my calculator!!
thnx :D
I don't understand this question.... Help please??Is there a typo here. Equality never holds for what you gave and I know two possible questions you intended to give.
Show that | |z1|-|z2| | < |z1+z2|. State the condition for
equality to hold.
Is there a typo here. Equality never holds for what you gave and I know two possible questions you intended to give.
Edit actually it does hold if z2=0 but that makes me think the question has a typo more now
Is there a typo here. Equality never holds for what you gave and I know two possible questions you intended to give.
Edit actually it does hold if z2=0 but that makes me think the question has a typo more now
its just supposed to be less than or equal to, but otherwise there aren't any typos
Looks like equality holds if z2=-z1Oh right, true.
Oh right, true.
Equality if z2=-z1 or any one of z1, z2 being equal to 0
Oh I see. Sorry, entirely my fault, must've been too hungry because I made a false assumption.
I must be having a really bad day.
I just did it by paper. I knew what I was doing but I articulated it wrong every single time. Terrible impression.
(http://uploads.tapatalk-cdn.com/20161025/bb63fbd2d028a782e4f0ebcba704fc13.jpg)
Seriously so sorry for the hassle.
All good, my teacher explained it for me today :)Awesome (y)
how do you do this 😁If you're lazy with LaTeX just say conj(z)
if a and b are two non-zero complex numbers, show that if a/b = ik,
(conjugate of a)b + (conjugate of b)a = 0
sorry I'm not sure how to type the conjugate sign hahah
If you're lazy with LaTeX just say conj(z)
Is k a constant? If so, is it real or complex?
If ODEF is a rhombus, where O, D and F represent complex numbers 0, 4 +7i, and 7 + 4i respectively, find
a) the number represented by the point E
b) the length of the diagonals
The area of the rhombus
how would you describe the locus of arg(z-i) - arg(1-iz) = pi/4 geometrically?
Prove |z1z2| = |z1||z2|
Find |3-2i|
Hey guys could someone please help me with question 2? Thank you so much! :)Pointers on part a), because you can't do the other two parts without it.
Bit unsure how to do the question. I've expanded the binomial and now I'm a bit lost.
i) Use the binomial theorem to expand (CosX+iSinX)^3
ii) Use DeMoivre's theorem and your result from part (i) to prove that
cos^3(X)=1/4Cos3X+3/4CosX
Where X is theta ( not sure how to type that :( )
Thanks, Wales
http://puu.sh/sCJ9I/789660a744.png
I keep struggling with (ii), I get i but ii gets me nowhere :( It's some drawn out proof...
What is the difference when graphing f(x2)) to [f(x)]2 ?Careful with the typos in your BBC code there; you forgot to [/sup] the first one because you forgot the forward slash
Points P and Q are the endpoints of focal chord of the ellipse x^2/a^2 + y^2/b^2 = 1. if the parameters at P and Q are θ and α, show that the eccentricity is given by e = sin(θ-α)/(sinθ-sinα)
help please :-\
Sketch |w+3i| / |w-4i| = 1
I tried to let w = x + iy and realize the denominator but I don't understand how to do it with the three terms I end up with.
http://puu.sh/sNOEr/240e0f639a.jpg
I get part i but for part ii I don't quite get the arg(z1+z2) expansion. Do I throw in Z1 and Z2 and simplify the expression?
If z^3 = 1, z cannot equal 1, show that (1+z)^5 = - z
I have tried using demoivre's theorem and mathematical induction or just substituting a value for z but I just can't get an answer. Any helpful hints as to how to approach this?
Find a and b.Is that 1+31 or 1+3i there on the top?
2+i=(1+31)/(a+bi)
Is that 1+31 or 1+3i there on the top?I'd assume that it's 1+3i, yet again, too quick for me
Also, this is just elementary complex numbers division. Have you been taught how to do this
I'd assume that it's 1+3i, yet again, too quick for me
Is that 1+31 or 1+3i there on the top?
Also, this is just elementary complex numbers division. Have you been taught how to do this
Prove by induction.When typing in words, feel free to use conj(z1) for the conjugate
a) Conjugate(z1+z2+...zn)=conjugate(z1)+conjugate(z2)+...conjugate(zn)
b) Conjugate(z1*z2*...zn)=conjugate(z1)*conjugate(z2)*...conjugate(zn)
7. A chord AB of a circle makes an angle theta with the diameter passing through A. If the area of the minor segment is one-quarter the area of the circle, show that sin2theta=pi/2 - 2theta. Sovle this equation graphically.
8. A chord AB of a circle subtends an angle theta at the centre of the circl. If the perimeter of the minor segment is one-half the circumference of the circle, show that 2*sin(theta/2)=pi-theta. Solve this equation graphically.
Hey, could I please get some help with this question :)
1) P(3secθ,2tanθ) and Q(3secθ,-2tanθ) are two points on the hyperbola x2/9 - y2/4 = 1. The normal at P meets the line OQ at R. Show that the locus of R is a concentric hyperbola
Thank you!!!
Hey! Let's start by finding the equation for the normal. I'm going to use implicit differentiation here.
We want to find the normal, so
At P,
The equation of the normal will be
At the point P,
Okay, now let's find the line OQ.
So the equation is
Using simultaneous equations, we can find the point of intercept.
Jesus, this is getting complicated. I assume I made a mistake; I'll leave this here, and explain how the rest would work. Find the point of intersection (both x and y value), then figure out the locus (by subbing one into the other). I have to head out or I'd redo this; good luck!
Thanks Jake I worked it out, you got the gradient for OQ the wrong way round haha :)
Hi Could i please get some help on this question
Much appreciated
[(http://uploads.tapatalk-cdn.com/20170126/ba281edde4d2a25281757fd9281ad27e.jpg)
Hi
could please help me out with these questions
thank you so much!
really appreciate it!
Q9 shouldn't be as hard, unless it gets icky. Post up any progress in working.
this really helps!
thank you!
for question 9,
i keep getting the wrong answer
i am not sure what im doing wrong
thanks
could you also help me with this question
i have no idea how to do it,
thank you!
I assume you got to the point where you had from there ignore the 4 obviously and try dividing the numerator and the denominator by cos^2.
Then you're in the position to make another substitution.
Didn't know what was assumed or not in HSC, apologies!
The integral of cosec-squared is known, and is negative cot. Just like how the integral of sec-squared is tan.
could you also help me with this questionIn general, you have to make the substitution t=tan(x/2)
i have no idea how to do it,
thank you!
Didn't know what was assumed or not in HSC, apologies!All g
In general, you have to make the substitution t=tan(x/2)
Because of how that's written, i.e. a + b sin(x) where a and b are either 1 or -1, you can manipulate the Pythagorean trigonometric identity that de used.All g
Didn't know what was assumed or not in HSC, apologies!
Hi! Could you please help me with this question: find the Cartesian equation of the locus of arg(z-4)-arg(z+4) = pi/4?For now, I am going to redirect you to the discussion here.
Thank you! :)
For now, I am going to redirect you to the discussion here.
Use the resources I have linked to.
Hey could please I get some help on these :)
1) P(cp,c/p) and Q(-cp,-c/p) are points on the rectangular hyperbola xy=c2. The normal at P to the hyperbola meets the curve again at R. Find the coordinates of R, and hence show that \(\angle PQR\) is a right angle
2) P(3p,3/p) and Q(3q,3/q) are points on different branches of the hyperbola xy=9. The tangents at P and Q meet at T. Find the locus of T if the positions of P and Q vary so that the chord PQ passes through (0,4).
I managed to find the locus as x=9/2, but the answer says that it is x=9/2, y<0. I was wondering why y has to be less than 0? T is \(( \frac{6pq}{p+q} ,\frac{6}{p+q})\) for reference
Thank you!
Hey!
Let's find the normal at P
So, the normal will be
So,
This meets the hyperbola again where y=c^2/x
Solve this, and you'll get
The first one is obvious (it's the point from which the normal is from!), but the second one is what we're looking for.
The y value will be
Now, you just need to find the equation of the lines from P to Q, and Q to R. Show that the two gradients are inverse negatives of each other, and bam! Q1 done.
I'll work on Q2 now.
Okay, so for Q2, I'm actually not sure! Might have to call in the infinite wisdom of RuiAce here
hey jake, or whoever decides to answer this. im having trouble with this question and its kinda put me off the whole of fitzpatrick ahahha. but i was just wondering how we are supposed to approach this? question 6b
(http://16325884_1357187997635286_1495004055_o.jpg)
hello again i have a lot of questions about section 3.2 of 4u cambridge, mostly the parts involving multiple parametric points on ellipses or hyperbola. so the parts with chords between points: P (acosß, bsinß) and Q (acos∂, bsin∂). firstly, in the explanation/ working to show the lines equation and gradient there are a lot of steps missing and i cant understand how they get anywhere, mostly the gradients? secondly, how relevant/do we need to know this at all because my teacher doesnt even know how to do it really i dont think? and if we need to know, is it at a level of deriving, im gonna assume so because it cant hurt but still?
so yea if anyone wants to shoot up some kind of working for the gradients that would be great. thankyouuuuuuuu i hate this whole Ex.
p.s. im probably gonna reply with a couple more questions because there isnt a lot of explanations really, or i dont get them. but thankyou
Hey could please I get some help on these :)
1) P(cp,c/p) and Q(-cp,-c/p) are points on the rectangular hyperbola xy=c2. The normal at P to the hyperbola meets the curve again at R. Find the coordinates of R, and hence show that \(\angle PQR\) is a right angle
2) P(3p,3/p) and Q(3q,3/q) are points on different branches of the hyperbola xy=9. The tangents at P and Q meet at T. Find the locus of T if the positions of P and Q vary so that the chord PQ passes through (0,4).
I managed to find the locus as x=9/2, but the answer says that it is x=9/2, y<0. I was wondering why y has to be less than 0? T is \(( \frac{6pq}{p+q} ,\frac{6}{p+q})\) for reference
Thank you!
They've used the sums to products formulas to simplify the gradients
(http://www.sosmath.com/trig/prodform/img6.gif)
ok so see the photos for q2:
but pretty much y<0 because the point p and q are on separate branches of a standard xy = c^2, so if we say q is on the branch in the 3rd quadrant, then its coordinates are all negative Q (3q, 3/q) and hence q is negative.
and when u rearrange 3(p+q) = 4pq
into p + q = 4pq/3
and sub into y = 6/(p + q)
you get: y = 9/2pq
and since p is positive and q is negative, y cannot be greater than 0, and it cant equal 0 because there isnt a variable on the top of the fraction.
uhhhhh i dont think ive learnt this, how do you get these equations? do we need to know how to derive?
uhhhhh i dont think ive learnt this, how do you get these equations? do we need to know how to derive?
Hi, can someone help me with this question?
I feel like I know how to do it and I just suck at algebra or something. I've tried both realising the denominator and just forming one fraction, cross multiplying then solving the complex and real components separately as real and imaginary components.
Thanks :)
Idk what to do, are you allowed to cross multiply since it isn't a proof?
Idk what to do, are you allowed to cross multiply since it isn't a proof?
WOW that sol is so nice, that's a pretty good trick factorising the i out at the start, i'll keep it in mind. Thanks !
This is the clever way going about it. The standard way is, as mentioned above, realising the denominator.
This is the clever way going about it. The standard way is, as mentioned above, realising the denominator.
this is such a nice proof, but i was just wondering what did you do in the middle step where u got half angles for cis? what exactly were the steps/ method you used
Hi! I've been struggling with complex numbers alot and was wondering if i could get some help with the questions below. I have tried to make sense of them, but just end up confused. Any help on any of them would be greatly appreciated!!With these questions though, you're not going to get anywhere without a diagram. For any question, post up a diagram with any progress, and we can finish it off.
Thanks for your help on 1, it does make sense. But I'm also stuck on 3. This is the diagram I have drawn so far.
Hey guys
I've been doing some revision questions on conics and I came across this one that I'm struggling with.
Any help would be appreciated :D
How do you systematically make sure you can answer any circle geo question or is it not possible. Also, do you have a checklist or something you run through when u see a circle geo question?
Write the equation of the locus at point P that moves such that its distance from (3,0) is 4/5 as its distance from the line x=2.
(http://i.imgur.com/DisGkJL.png)
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Hey guysAre you sure there hasn't been a typo anywhere?
I know I've been asking a lot of questions here but I have a topic test soon and I want to make sure all my understanding is solid :)
Could someone help with this conics question?
I got the locus as x+8y=0 but apprantely there is a restriction to the domain..how do I find that out?
I got the locus as x+8y=0 but apprantely there is a restriction to the domain..how do I find that out?As per this GeoGebra simulation, your locus is a correct one
Answer says -16/sqrt(17)<=x<=16/sqrt(17) for the restriction..have no idea how they got that.
Just adding...Pretty sure I addressed this.
For the chord to exist, the line must join two points at the circumference of ellipse, which means it cannot exist at certain points due to its gradient.
[url=https://ggbm.at/BEZYGzee]
Pretty sure I addressed this.
Also, it is ok for the two points to coincide.
Hey, so I was pretty happy with how much of this question I think I got but couldn't get the final result (this was in the last question of 2000 4u HSC). I got the first part then tried t-method, and tried subbing in the hint and equating imaginary parts but my algebra got super messy and I don't exactly know how to get the second part. Is there any faster way using the properties of Imaginary parts etc.? Also, what are your opinions on people not even bothering with these questions cos they would be very tough to get in the exam? My teacher says if you want a safe B6 get all the easy questions (all other than last/8) 100% and look through the rest if you have time. In 4u there is marginal benefit benefit to your atar going from a 90-92 to a 94-95 even though it requires far more effort. My teacher says your better off maximising your 3u or other subjects as there is a huge difference between 90-92 and 94-95 in english, chem, phys, eco etc. Like these questions are fun for me to do so I will probably try them anyway but I can see their point.
oh dang I didn't apply de Moivres again with the (costheta+isintheta)^2 only the power n so i was getting random half powers and I couldn't get it. Thanks man, its pretty annoying this stuff isnt onlineYou said this was a past HSC paper question?
Hey guys could i have some help with part b please? Thanks :)
Find the equation of the tangents to the ellipse 4x^2+9y^2=36 given that these tangents are perpendicular to the line 3x+2y=5.Judging by what the question asks it's likely you made a little mistake somewhere in your working, especially since you were only off by a factor. Can you please post it up?
I got 2x-3y+6=0
Answer is 2x-3y+6sqrt(2)=0
Judging by what the question asks it's likely you made a little mistake somewhere in your working, especially since you were only off by a factor. Can you please post it up?
Hi!!Which method do you want? Shells is easier unless you haven't been taught how to use it.
could i get some help on this volumes question please,
i have done it but i keep getting the wrong answer
thank you!
The ellipse x^2/a^2 + y^2/b^ =1 meets the y=axis at C and D. Tangents drawn to C and D on the ellipse meet the tangent in (xcos(θ))/a + (ysin(θ))/b=1 at the points E and F respectively. Prove that CE*DF=a^2
HEYYYY Jake, I was struggling to answer the following questions and I was wondering if you could help me with them? thank you SOOOO MUCHH!!!There are several questions here, and we will not do them all without seeing your attempts at them. I will provide the answer to a random choice of two.
HEYYYY Jake, I was struggling to answer the following questions and I was wondering if you could help me with them? thank you SOOOO MUCHH!!!
Hi guys could i have some help with conics question 3 and 5b please? Thank you!
Hi guys could i have some help with conics question 3 and 5b please? Thank you!
any help would be greatly appreciated :) thanks
Hey hey! Here is the working for this question! ;D
You can then manipulate that answer any way that suits you - I hope that helps! ;D I'll get to your next question in a little bit ;D
Hey Jamon!
Thank you so much for this, if i was trying to manipulate it to get the answer 11/10 +3/10i how would you go about getting it?
I also don't understand how to do these questions (3 a & b).
Just came back to do those! Ta Rui ;D
Edit: On the answer above, I wrote the question wrong in the second step: I'll go back and fix the errors in my working and then change Rui's to match, sorry! ;D
Hey hey! Here is the working for this question! ;D
You can then manipulate that answer any way that suits you - I hope that helps! ;D I'll get to your next question in a little bit ;D
Can i just ask- how did you get to this (iz=(3+2i)z-3-2i) step? I don't understand the continuation from the previous step to this one.
Heres the step in between! It's just expansion and then re-factorisation (or partial expansion, if you will) ;D
Hey hey! Here is the working for this question! ;D
You can then manipulate that answer any way that suits you - I hope that helps! ;D I'll get to your next question in a little bit ;D
When you go on to this ((3+i)z=3+2i) step, where did the iz on the left hand side go?
This makes sense! However, I tried doing part (b.) and it didn't work out as well as I thought it would...From your second line:
From your second line:
Solve simultaneously from here :)
please help
How would you approach this question? I tried expanding it out but it got really messy and I just got confused :(
Hey!
Start by writing everything out in cis form.
Then, use DeMoivre's
Finally, use cis rules to get the answer. Remember that when we divide one complex number in mod-arg form by another, we divide their modulus and subtract their argument. I'll leave that bit to you :)
Hey Jake- thank you! That makes sense, however, I don't really understand what to do after this...
Heyyyy I keep forgetting how to do this question someone please help hahaha ty
(http://uploads.tapatalk-cdn.com/20170221/bdbe195a2dd1697db249517b4b54ede9.jpg)
please helpHint: Use trigonometry.
I don't know/forgot how to do this question. Please help. Doing 1part is sufficient (then i can do the other as both questions are similar,,,, probably?). Thank youSomewhat untidy.
Just what I wrote above; when we divide two complex numbers in cis notation, we divide their modulus and subtract their argument. So,
You can use this to get the final answer
Like this?
Looks perfect to me ;D well done!!
The answers say 3π/10 though so I don't know where I went wrong.
Remember; on the Argand diagram, you can add or subtract 2*pi without it changing the answer! This is because the point goes 'full circle'. In fact, the HSC prefers that you write your arguments between -Pi and Pi. In this case, subtract 20*pi/10 (ie. 2*pi) to get the answer you're looking for :)
OHH yea. How does it come so easily to you haha, thank you again Jake!
Is someone able to help me with one of these questions? I don't remember the pattern.I've only shown working out in the first question. The rest are answers (if you were already given answers, this is quite useless). I don't think exam questions require working out, however i might be wrong. I've done all the questions as mostly those questions given are different and there is no specific pattern (not really a pattern).
I don't see how part i) is related to ii). Soooo stressful. Please help, thanks again.
This questions seems easy enough but i just don't trust my proof (it's kinda dodgy). Could someone show me how it's done. Thank you.
Omg... I just realised this morning when I was thinking about it nonchalantly. Sleeep really does help! I used b^2=a^2(e^2-1) and didnt bother checking. Sorry for wasting your time rui.Sleep is a beautiful thing. I had a lot of those moments over the last two years of my maths life.
Hi,
I'm having trouble with this polynomial question. It would be wonderful if I could get some help on it.
1. Find the value of k such that y=x^2-4x+6 and y=k-6x-x^2 are tangential to each other at some point.
Thank you!!
Hi, I have another polynomial question and it would be great if I could get some help on this
1. Find the values of k for which the polynomial p(x)=2x^3-9x^2+12x-k has
a) one distinct real root
b) repeated roots
Thank you!!
Does anyone know how to represent z^6=1 in mod arg form?
Does anyone know how to represent z^6=1 in mod arg form?
Hi, it would be great if I could get some help on this question.
1.
a) If w is a seventh root of 1, w does not equal to 1, show that w^3+w^2+w+1+1/w+1/w^2+1/w^3=0
b) By letting z=w +1/w reduce this equation into a cubic equation in z.
Thank you!!
Thanks Jake! Could you possibly help with this one as well? Its Q. 18 and I've drawn the diagram, but I don't know how to find the diagonal.
Thanks Jake! Could you possibly help with this one as well? Its Q. 18 and I've drawn the diagram, but I don't know how to find the diagonal.
Woah. Rare instance that Jake beat me.
Mate, I got one earlier today as well. Pick up your game son.What did you want me to whilst I was dining out 8)
What did you want me to whilst I was dining out 8)
Ninja answer a 4U question under the table, obviously.
Hey! Weird question; you're not usually asked to find the cartesian equation of something on the argand diagram. Still, let's think about what we know.
Firstly, the line must be perpendicular to the diagonal AC. This is important, because we can find the gradient of the first diagonal, and use that to find the perpendicular gradient. Secondly, the line must bisect the diagonal AC. Now, we have a gradient and a point; all we need to find the equation!
Let's replace A and C with points on the cartesian plane; A(0, 3) and C(4, -5). The gradient of the line between them will be
So, the gradient of the perpendicular line will be 1/2.
Now, the midpoint of AC is (2, -1). So, the cartesian equation of the line perpendicular to the diagonal AC will be
So, we get the equation
Great! That's the first part down. Now, the second point wants us to show that
Let's let z=x+iy
And we're done!
This makes perfect sense. Thank you so much! The only thing that confused me was when you substituted the gradient of the perpendicular line into the (y=mx+b) equation. You stated earlier that the gradient would be 1/2, however, when you substitute it into the (y=mx+b) equation, you've written it as -1/2.Transcription error
Thank you so much otherwise Jake!
Also if anyone could help me with this question.. Should I using De Moivre's theorem?
How would you do this question?
_____
Thanks alot RuiAce :) !!
i get how you get equation 1 and 2
but i dont get how you added equations 1 and 2 and simplified from there , could you please do those steps in more depth :)
Also for your last case is it meant to be pi/4 for all real number and not pi/2?
Same denominator and boundaries , so it's just adding two fractions. The top is then equal to the bottom so it cancels out and equals 1.
Thanks Shadowxo :) :)
just wondering in the end it should be pi/4 for all cases in reference to rui ace's orignal solution ?
Can someone help me with this? I don't know how to make it true for n=1
_____
Hey ruiAce , just a question in regards to your solution?
How did you know and get this trick ? How did you figure it out / get to it?
Thanks :) !
Thanks Rui! You make it so simple haha.
I get really confused on the n=k+1 part, any help would be greatly appreciated! :)
You are so helpful. Thank you so much for taking time to write it all out Rui!!Hint: Part c) is the useful part in the worked example. Commence by \(\text{Let }x=\tan \theta\) then play around with the algebra until you can use the triple-angle identity like they did
If you have time, would possibly be able to give me some help on this question as well? Thank you for all your help so far!!!
Hint: Part c) is the useful part in the worked example. Commence by \(\text{Let }x=\tan \theta\) then play around with the algebra until you can use the triple-angle identity like they did
Feel free to post any progress if you get a bit lost halfway through :)
OK... I've gotten abit lost. HAHA.
Nearly there! :) From your last line:
I have no idea how to do part b. and c. of this question. I've tried watching videos but the examples are all different, and theres no pattern to them so I'm having a hard time working them out...
Hey Rui, could you (or anybody) help me with this question. Thank youI'm trying to look for any shortcuts right now. I'll edit this post later.
Hi,
I'm having trouble with taking the limits of an equation to find where the curve approaches. For example, in the question below, I know I could take points but I would still like to know how to take the limits of this question just for future reference.
Thanks in advanced! :)
Hmm, ok, thank you!Sketch y=sin(x) first.
It seems I'll just have to practise more.
Btw, for equations which use two equations, like log [sin x], how would you recommend sketching them, because I've been considering asymptotes and mainly subbing in values. Surely there is a faster way to approach this?
Sorry I seem to be asking really basic questions.
I've just started the Graphs topic and this question seems completely foreign to me. Could anyone please guide me through it? I've had a look at the solutions but nothing makes too much sense.
Cheers, Wales
Hey guys could I please have some help with this conics question? :-\
how would you do this question?
Evaluate the following using integration by parts
how would you do this question?
Evaluate the following using integration by parts
Using integration by parts
Then, you can use a bunch of trig identities to simply down to
according to Wolfram Alpha. Don't necessarily think you need to do this in an exam situations. Also, I'm sure there are plenty of other ways to solve this question, some of which would have yielded the simplified answer above straight away.
THanks jakesilove :)
but i dont get how you went from
how did you get
Also how did you get the =
in one go ?
Fair enough, I definitely didn't explain myself very well. We want to integrate dv, so
This integrates in the normal way
Thanks so much for
However, we can use trigonometric identities to get
as required.
For the last one, I just used the reverse chain rule. Recall that
Using this, and noting that
we can recognise that
And that's where I got it! If you can't do that in one go, that's no problem at all (I've had a lot of practice). Take your time, set out your working however you want. However, over time, integrations like this will become natural :)
How would you do this question?
Use a substitution and an integration by parts to evaluate each of the following
Rui, could you help me out with part ii). Thank you!Fair warning: My activity is reduced for this month. I won't be able to get to everything and when I do it might take a while to get there.
Fair warning: My activity is reduced for this month. I won't be able to get to everything and when I do it might take a while to get there.Thanks Rui.
Thanks Rui.Ahh I'll still be around, just not as regularly. Besides Jake's floating around too
NOOOOoooo...my life saver. Thankfully, its only for a month. Good luck with whatever your up to!
How many numbers between 1 and 2002 such that the sum of the digits is divisible by 5?
Hey Mahan! Just before we answer these questions, just wondering, are they questions you struggle with? Are they homework questions? Because they all seem far beyond any curriculum I know of.Hey Jakesilove! These questions are not directly out of HSC related textbooks but apparently in order to solve them we don't need any knowledge beyond HSC.I agree that they are tricky but after spending some time they should be doable.All the questions that I have posted are either come from other sources, Not HSC related but can be solved using only HSC knowledge, or I heard it from my friends.Sometimes I post them to share them with others or sometimes I'm curious to see whether someone comes up with a different idea or I might not know the answer to that question. For this particular one, I'm curious to see whether someone comes up with some interesting idea to solve it, because there are always multiple methods to solve counting questions, at the same time share some rather unusual and challenging question for those who like challenge.
Hey, this question relates to the question in posted in the 3U thread. The answer 126deg 52min for part (V). How can we deduce the angle is obtuse?Honestly, I could not see how this is possible unless you drew a diagram that was reasonably to scale. I simulated the scenario on GeoGebra and there was no way that a free-hand diagram would've implied it.
Use a substitution and an integration by parts to evaluate each of the followingIt's not. They're the same answer.
For this question and answer attached where am i going wrong?
Why is my answer different to the supplied answer?
For the error in this question.(attached)Pretty much, +C is the cause of it. Because in indefinite integration, we always have to keep in mind that the statement is true to within a constant.
i said the working out forgot to add an integration constant(ie +c)
when an +c is added the value of c=-1
therefore the equation becomes loge(|x|)=loge(|x|), which is true .
Would this be right/ enough ?
Hi, it would be great if I could get some help on this question
1) Find the equations of the tangent and normal to the ellipse x=4cos theta, y=2sin theta at the point where theta= -pi/4
Thank you!!
Hi, it would be great if I could get some help on this question
1) Find the equations of the tangent and normal to the ellipse x=4cos theta, y=2sin theta at the point where theta= -pi/4
Thank you!!
First, we need an equation. We know thatAha.
Clearly,
We want to find the tangent at the point
Honestly Rui, I thought I'd get this one.
But your way is 100% faster.
Far out.
Aha.
Pro tip: If things are given to you parametrically, try to avoid Cartesian stuff
Pretty much, +C is the cause of it. Because in indefinite integration, we always have to keep in mind that the statement is true to within a constant.
If you ask me though, "true to within a constant" is the key thing that needs to be stated. Of course, you can, by computation, find the value of C (which is -1 or 1, depending on if your C is on the LHS or RHS) to reunite the solutions if you wish.
Could someone please help me with this question
How would you do this question?Note:
Note:
thanks Mahan :)Hmm.
but the question says find cosh(3x) in terms of cosh(x) and sinh(x), your answer only has cosh(x) not sinh(x)?
Hmm.
What answer do they want you to get to? Because the convention is to stop at where he did.
Is sinh and cosh even part of the HSC?Nope.
Hmm.
What answer do they want you to get to? Because the convention is to stop at where he did.
i am not sure myself Rui ;), our teacher just gave us this question as an extension and said we should be able to do it from the info providedBut then again, if you get it in terms of only cosh it's still in terms of both cosh and sinh. This is because you're basically adding \(0\sinh x\) anyhow.
But then again, if you get it in terms of only cosh it's still in terms of both cosh and sinh. This is because you're basically adding \(0\sinh x\) anyhow.
Thanks RuiAce :)Should be, yeah.
basically so mahans answer should be fine right?
is there a way you could get cosh(3x) in terms of cosh(x) and sinh(x), with both sinh(x) and cosh(x) in the final answer?
Should be, yeah.
Hi! Can someone please help me with part 3 of this question :) I've read the board of studies solutions but I still don't get what I'm supposed to do!
_____________________________
For the question: Factorise 4x^4+1 asa product of real polynomials.
I got (2x^2+1)(2x^2-1)
But the answer is (2x^2+2x+1)(2x^2-2x+1), why is it so?
Hey Rathin! It's been a while ;D
I think you've used the difference of two squares shortcut when you shouldn't have, or otherwise misread the question?
I'll leave Rui/Jake to tag in with the actual method if you need it (I'm not sure of the 4U approach to this) ;D
Edit: Nevermind, you'd just complete the square (sort of, not the way you'd normally do) - Pretty sure you'd just do this ;D
I hope that helps ;D
Can someone please help me with the last two parts of this question? Thank you :DPart iv) needs the result of b), which you need to provide.
Part iv) needs the result of b), which you need to provide.
Thank you! :) This is part b)Hint: in 3U, when you were doing Newton's method, you learnt that a root is between -1 and -1/2 if f(-1) and f(-1/2) change signs. The result of part b) is supposed to help make your life easier when you have to deal with f(-1/2).
Suppose that z^7=1, z is not equal to 1.First two parts addressed in post #885 albeit with different pronumerals
i) deduce that z^3+z^2+z+1+1/z+1/z^2+1/z^3=0.
ii) by letting x=z+1/z, reduce the equation in i to a cubic equation in x
iii) Hence deduce that cospi/7cos2pi/7cos3pi/7=1/8
Hint: in 3U, when you were doing Newton's method, you learnt that a root is between -1 and -1/2 if f(-1) and f(-1/2) change signs. The result of part b) is supposed to help make your life easier when you have to deal with f(-1/2).
There are many other solutions better than the BoS ones which you should consider reviewing. However if those don't make sense, post up your interpretation of whichever solutions you looked at and we will fill in the remainder of the puzzle.
HiThis is the equation of a hyperbola, so you can divide each sides by 4 to get it in the general form. Now the equation of the asymptotes are y= b/a or y=-b/a. In this hyperbola, a is 2 and b is 1. So the asymptotes are y=-1/2x and y=1/2x.
i am not sure how to find the asymptote
do you just do lim x approaches 0?
thank you
This is the equation of a hyperbola, so you can divide each sides by 4 to get it in the general form. Now the equation of the asymptotes are y= b/a or y=-b/a. In this hyperbola, a is 2 and b is 1. So the asymptotes are y=-1/2x and y=1/2x.
Alternatively, you can try factorising the LHS
so (x-2y)(x+2y)=4. As the RHS is a non-zero constant, the LHS must not be equal to 0.
Therefore x-2y =/= 0 and x+2y =/= 0
So the asymptotes are y=1/2x and y=-1/2x
a) Show that the roots of y^4+y^3+y^3+y+1=0 are y=cos(2kpi/5)+1sin(2kpi/5), k=1,2,3,4.Your expression is wrong. cos 36deg = -1/2 - cos 72deg.
b) Hence deduce that cos36 deg = 1/2 + cos 72 deg.
^ I need help with part b.
How would you go about doing these questions ? (attached)I'm just gonna give this warning now.
Hi, it would be great if I could get some help on this question
1) Prove that if T (x0, y0) lies on a directrix on the hyperbola x2/a2 - y2/b2=1 then the chord of contact will be a focal chord
Thank you!!
I'm just gonna give this warning now.
Hyperbolic functions are NOT a part of the 4U course. In the previous case it was okay because they defined the functions for you and just made you do algebraic manipulation. I will not do any more questions that are beyond the scope of the course.
First image - Use similar identities to the trig ones.
Second image - Consider sinh for the first and cosh for the others
yea i know they are not rui :), but i think that maths is a very beautiful subject and so it doesn't hurt here and there to do questions outside of the scope of the 4U course :) ;)x=3sinh(u)
next time i will post a disclaimer before posting these questions. :)
Just wondering rui for the second image, what sinh(u) substituition would i have to make, i have tryed x=sinh(u) for the 1st question of the 2nd image , it didnt work ?
Hey there is no solution let alone answer to this question within the past paper. Idk how to get an expression such as the ones in the choices, but i tried subbing each of the MC and i got C (working out may be wrong though). This method seems too long to me for a MC question. Could someone show me a more efficient way? THank you.
Hi. It would be great if I could get some help on this question.This question involves a Riemann sum. (Which is not 'explicitly' in the syllabus.)
Hi, I can't seem to do part (i) by using the given statement. Any help would be great thanks!
Note: The question actually says x>0 and not x≥0. However, I have assumed otherwise because if x>0 then for part i), we should have > as well instead of ≥.
Hi,
It would be great if someone could help me with this question
1) The curve on the Argand diagram for which ||z-3| - |z+5||=4 is a hyperbola.
a) Find the eccentricity of this hyperbola
Thank you!!
Hey,
I'm kinda stuck on questions that require you to find the tangents to the ellipse or hyperbola that are parallel to a line. What's the best method for those types of questions? For example, find the tangents to the ellipse 16x^2 + 25y^2 =400 which are parallel to the line y = x+2. Thanks!!
Hey! First, recall that |PS-PS'|=2a. It seems sort of intuitive that the focus of the hyperbola will be z=3 and z=5, right? If not, try think about some values for z that hold true, and the shape kinda pops out on its own. You can use the 'location' of the focus', plus the knowledge that a=2, to find the eccentricity.
Hey,
I'm kinda stuck on questions that require you to find the tangents to the ellipse or hyperbola that are parallel to a line. What's the best method for those types of questions? For example, find the tangents to the ellipse 16x^2 + 25y^2 =400 which are parallel to the line y = x+2. Thanks!!
PLEASE HELPThree-dimensional curves are not a part of the HSC
Hi, it would be great if I could get some help on this question
1) P(x1,y1) is a variable point on the hyperbola x2/a2 - y2/b2 =1 such that the tangent at P cuts the asymptotes at M and N.
a) Find the equation of the tangent at P (which I've already done)
b) Find the coordinates of M and N
Thank you!!
Hi does anyone know how to do question 2C, because the complementary event is just having none of the windows open, which can only be done in 1 way.
I am not sure how to approach it
Thanks
(http://uploads.tapatalk-cdn.com/20170325/27fa092f784d778ae149c771e97576b7.jpg)
C, because the complementary event is just having none of the windows open, which can only be done in 1 way.
Note obviously that the total number of arrangements is 2^4 = 16, which is where we get 16-1=15 from.
Also this question
It looks simple but i just cant get it
Thank you!
(http://uploads.tapatalk-cdn.com/20170326/7633d4393531b4f770eed32afca6c045.jpg)
How would you do this question? (attached)
Heey
I got part a by subbing z= i.
How do u solve part b?
You guys are awesome!!
Hi! Can someone please help me with part b of this question? Thank you! :)
Hi can someone please help me find the time taken for this resisted motion question. thanks very much!!In the future, please provide any progress.
Hi! Is there a way to recognise when to draw a major/minor arc that's above or below the x-axis for the type of question
arg(z±(...)/z±(...)) = π/(...)?
Thank you :)
I keep forgetting why so I'll link to one of Eddie Woo's videos
can anybody help me out with this question. thank you
In this question i got pi as the answer but i did it through sketching on the argand diagram. Is the answer pi? and is this the simplest method?
Thank you so much ;D
im gonna have alot questions for today :'(. I don't know how to do part b) for part a) however i got A(0,-b/tan(theta)) B(0,(a^2+b^2)tan(theta)/b). Sorry I don't have any answers to this exam paperHint: On scrap paper, start by considering a circle where AB IS the diameter of. Then, since A and B both lie on the y-axis, we know that the center also lies on the y-axis.
let the other root be 'a.'
By doing the sum of roots, a+i=(1+i)/(2-i)
By realising the RHS, we get 1/5+3i/5=a+i.
Now solving for 'a,' you get a=1/5(1-2i)
Hi
Could i please get some help on this mechanics question
what is retardation?
thank youuu
Hi
Could i please get some help on this mechanics question
what is retardation?
thank youuu
I know this isn't what it's supposed to be, but MAKE SURE TO INCLUDE YOUR STATES WHEN YOU WRITE CHEMICAL EQUATIONS!The man has a point
WOOPS sorry
Here it is,
(Question 5)
A particle...
(http://uploads.tapatalk-cdn.com/20170406/5c6c6ae7ba8c46153119ff303099de3f.jpg)
Unique means nothing else like it i.e it's the only solution to the equation
oh ok so I need to show that it is the only real solution (ie all other roots are complex)?
or is sufficient to just implicitly differentiate (you get x+y=0) and sub it into the curve to get that equation (x^5+x^2+4=0)?
I think it should be sufficient to differentiate and sub as long as you justify it (i.e tangent only intersects the ellipse once)
I think it should be sufficient to differentiate and sub as long as you justify it (i.e tangent only intersects the ellipse once)There's no ellipse here. That's a 5th power on y there.
Hi
Could i get some help on question 3b
thank you
Hi,
could i get some help on this question please
Hi,
could i get some help on this question please
Here's an alternate method for part ii
That's a heaps better method. Cheers for posting it up!
Hey! Firstly, since all the coefficients are real, we know that either ONE of the roots is real, or THREE of the roots are real (as imaginary roots will come in complex conjugates).
The easiest way to answer the first half is by differentiating the function, it proving that it is non-decreasing. So,
For a>0, this derivative is ALWAYS positive. Thus, there will only be one real root, as the graph will never 'turn around'. It's helpful to sketch an example of a non-decreasing function (ie. gradient is always positive), to prove to the marker that you know why this fact results in one real root.
Now, if two roots are equal, we can right them as
Using sum and product rules.
and
We know from above that
so
Now, subbing alpha into the original equation, we get
Not sure where we are going with this; let's keep going?
We are trying to prove that
Subbing it what we've found above;
Which looks like it equals zero! Bit of a round about way of getting there, but hey, it works!
(http://uploads.tapatalk-cdn.com/20170412/ae72a154ec188b771b9ad6e2288d5dc5.jpg)
Hey guys could someone help me with 3b and c, I can't quite work it out algebraically
(http://uploads.tapatalk-cdn.com/20170413/1bd49af4c7a5ebef05864e2acdab9efb.jpg)The natural domain of the first is x>0 but for the second it's x>0 AND x<-3/2
I hope this question is better, not sure how to do 8a, the rest is fine but just abit confused on whether its algebraic or word and where to go? Sub in value less than 0 and show it doesn't work because it x^2 +3x/2 has to be greater than 0?
Hi, it would be great if I could get some help on this integration questionIs there a bracket missing around the denominator?
1. Integrate 1/5+4x+x2 dx
Thank you!!
Is there a bracket missing around the denominator?
(If yes, hint: complete the square)
Didn't think of that. Thank you!!
Hi
could i please get some help on part b please
thanks
We know that the net force, on the way down, will be
Also,
Subbing in what we found the first time,
And that should get you to the answer. Possible I messed up a sign/factor somewhere, but that's the general set-up :)
v'=e^x^2Look here. You differentiated your v', instead of integrated it.
v=2xe^x^2
Look here. You differentiated your v', instead of integrated it.
Help with this please
∫sin(x)sin(3x)dx
Hey, wouldn't sin(x)sin(3x)=1/2(cos(2x)-cos(4x))?No idea why I wrote down sin again. Let me edit that
No idea why I wrote down sin again. Let me edit that
By IBP I got I=(sin(3x)cos(x)-3cos(3x)sin(x))/8 +c which is what the textbook has as its answer.
:)
this is a stupid question but RuiAce how did you find the Dv/dhJust differentiate it.
(http://uploads.tapatalk-cdn.com/20170422/fee98dccb1a6ec7f5c1bf8092883d003.jpg)Using the first part,
Please help me question 4c, I feel I've done it before, but a solution would be so helpful
Hi
could i get some help on this question please
i feel like I have done it before but not getting the answer
thank you! :)
Hi
Sorry and this question please
I feel like it is very physicsy (hhaha not a word)
thank you!
HiThe ray is going to alternate through which focus it passes through. If you shine a light from, or directly through one focus, it will always be reflected to the other focus. This is the reflection property of the ellipse.
Sorry and this question please
I feel like it is very physicsy (hhaha not a word)
thank you!
Hey! For this question, simply use the distance formula to find the distance between the pointsIt says Cartesian coordinates in the question. Give me a sec.
Where the eccentricity takes its usual form. You'll find that the distance between the point and the 'positive' foci is
and the distance between the point and the 'negative' foci is
Hi
could i get some help on this question please
i feel like I have done it before but not getting the answer
thank you! :)
hiI just simulated it on GeoGebra and the question is wrong. (Most likely a typo.)
Could i please get some help on question 8 please
thank youuu
Hi! I'm confused about how to prove the reflection property for the hyperbola! Also, would the geometrical properties for the hyperbola also apply to rectangular hyperbolae (especially the type with its asymptotes being the x & y axis)? Thank you :DA rectangular hyperbola is just a hyperbola with eccentricity √2, so any property with the hyperbola will always apply to the rectangular hyperbola.
I just simulated it on GeoGebra and the question is wrong. (Most likely a typo.)Thank you!
Because ellipse ii is literally enclosed INSIDE ellipse i, a tangent to ellipse i can be shown to never touch ellipse ii.
Could I please have the working out and answers to these questions
(http://i67.tinypic.com/ycq49.jpg)
Using the first part,
Also as a reminder,
Note
Similarly
Then
This comes from the facts:
and
and
(http://uploads.tapatalk-cdn.com/20170427/48cf7999750683dbddda5b64f0288bab.jpg)
This is an absolute monster question. I think atleast ahahaha any help is ridiculously appreciated
Hello, i need some help with the following
1) If x,y>0 show that 1/x + 1/y > or equal to 4/(x+y)
2) Using cauchy's inequality, prove that 2(a^3 + b^3 + c^3) greater than or equal to ab(a+b) + ac(a+c) + bc(b+c) greater than or equal to 6abc
Thank you :c
Hello, i need some help with the following
1) If x,y>0 show that 1/x + 1/y > or equal to 4/(x+y)
2) Using cauchy's inequality, prove that 2(a^3 + b^3 + c^3) greater than or equal to ab(a+b) + ac(a+c) + bc(b+c) greater than or equal to 6abc
Thank you :c
I don't see how that can be simplified any further and neither does Wolfram
Omg THANKYOU tho I don't know series. The answer is this if it helps. Sorry for such a random Q and it probably would've helped a lot if I'd shown where u actually need to get to sorrryy(http://uploads.tapatalk-cdn.com/20170429/79b9b702bd22d5b5028501de74152757.jpg)
Maybe this will help? All g if it doesn't just stumped
HEY, could anyone help me out with this integration? THANK YOU. I edited the red minus in(forgot to rewrite it after erasing it). The answer is also attached.The method of computing your integral can be found in post #1040. In your scenario, take R = 6
Sorry Rui, I dont understand how you got from the first to second step and how the integration of the first part of the 2nd step is equal to the first part of the 3rd step. Could you explain? Thanks
Sorry. Before I meant i didn't know how to get from the first line to the second line on the right hand side. Also instead i tried differentiating it to check the integration but i got different.Oh my bad, this whole time there were typos floating around.
That being said, these are 4U integration tricks that you are expected to know.I see. Thanks for the help Rui!
Hi! Could someone please help me with this question? The exercise is about using the substitution t=tanθ/2. Thank you :)What textbook is this? That substitution is a horrible idea; u = sin(x) would be a much better substitution.
What textbook is this? That substitution is a horrible idea; u = sin(x) would be a much better substitution.Yeah that's what I thought as well! It's from the old version of Fitzpatrick but it's in the same exercise in the new one as well. Thanks Rui :)
could i get some help on this question too please
could i get some help plz
thanks
Hey guys can someone help me with part 3? Thanks!Previous post deleted upon seeing part a)
\begin{align*}&\quad \left[(1-x\cos x)-i(x\sin x)\right][(1-x\cos x)+i(x\sin x)]\\ &= (1-x\cos x)^2+(x\sin x)^2\\ &= 1-2x\cos x + x^2(\cos^2x+\sin^2x)\\ &= 1-2x\cos x + x^2\end{align*}
Cis notation is ew.
I might've made an algebra mistake because I don't quite agree with what they got for b.
(http://uploads.tapatalk-cdn.com/20170506/b60a124d5b5956cf9eb1dbeb793dfd9b.jpg)Can't quite follow all of your working. I understand the use of 4a = length of latus rectum, but I don't understand how you found the area of the parabolic cross section to use as the volume of your slice.
4b I feel like I'm Doing something wrong I'm getting 128, the answer is 16?
My working:
(http://uploads.tapatalk-cdn.com/20170506/5ab50ad3287d818f31aad2d40c72f21c.jpg)
Hi! Do we need to know the condition of the focal chord for conics? Or is that only needed in 3u parametrics? Thank you :)Parabola: The tangents at the end of a focal chord intersect at right angles on the directrix
Can't quite follow all of your working. I understand the use of 4a = length of latus rectum, but I don't understand how you found the area of the parabolic cross section to use as the volume of your slice.
So simply put, if the chord is a chord of contact from the directrix, it is focal.Thanks Rui :D
HI,Hints: Use PS=e*PM and PS'=ePM' to take care of the LHS.
just a conics question:
I can do everything else except q2b(gamma).. helppp!!!
Thanks so much :)))
Hints: Use PS=e*PM and PS'=ePM' to take care of the LHS.
If you need the full solution please provide the answers to the previous part.
Hello, I'm having trouble starting off this volumes question, any help is appreciated ;D(http://i.imgur.com/LPMpy9C.png)
....
Just dropping by with a integration question.
Integral of (1+x)/SQRT(1-x-x^2) dx
From Coroneos' 100 Integrals question 15.
Cheers, Wales
Hi! I'm having trouble integrating cot^3x, and cot^5x. Thank you :)This integral works the exact same as \(\int \tan^3 x\, dx\). You should have a go at that one first.
hi
HiA bit all over the place. Feel free to point out any areas of confusion
Could i get some help on this question
A bit all over the place. Feel free to point out any areas of confusion
(http://i.imgur.com/FfE4rrr.jpg)
(http://uploads.tapatalk-cdn.com/20170519/5042b959a17e098c0ceaa0192c2f81d1.jpg)
Please help
heyo, i have a question on the below (i think) inequality question. i got up to c)ii), i kinda fudged the induction process (doesnt rly make much sense) can anyone help me with step by step for that? (assume c)i) uve already done).
thanks :)
hi could i please get some help on this question thanks
heey so for this, do we differentiate 3 times?? I got -9/8 is it right
To finish off the question, the last root is easy to find. Just use the sum of roots formula.
The triple root is indeed 2ah thank you :)
Having trouble using the sum of roots method for this :/You could go for \(ax^2+bx+c\) but if you look carefully it should be beyond obvious that a=1 as the polynomial is monic (leading coefficient 1)
they compare terms in (x+2)^2 (x^2 +ax +b)
but why did they write x^2 +ax +b instead of ax^2 + bx +c
You could go for \(ax^2+bx+c\) but if you look carefully it should be beyond obvious that a=1 as the polynomial is monic (leading coefficient 1)ohhhhhhhhhhhhh
at y=f^2 (x)For the sake of disambiguity, are you referring to \(f(f(x))\) or \([f(x)]^2\)?
why is the turning point squared of f(x)?? i dont get it :/
For the sake of disambiguity, are you referring to \(f(f(x))\) or \([f(x)]^2\)?f(f(x))
oh that makes sense! :)
In general, no rule immediately justifies how to find the values of x satisfying that criteria. These zeroes can potentially be hard to find
oh that makes sense! :)Not too sure what you mean here?
sorry to ask another thing, but does that mean:
if I graph y=(x^2 -1)^3 then I can see that f'(x) is 0 so we have x = + or -1
and then I make this function in terms of f(x) and see that x=0
Not too sure what you mean here?like you know how f(x) should =0 because f cube x 's derivative is f'(x) multiplied by derivate of f'f(x) (and this is made up of f(x) and f'(x))
ah I get you
ohh I get you! Thank you SOO much for explaining all this so well Rui :D
Hi! pls help with the 'describe how the shape of this curve changes as λ increases from 1 towards 2' part
Thanks!
Evaluate the integral between 0 and -1 sqrtx+2/sqrt1-x dxFor the sake of disambiguity, do you mean \( \frac{\sqrt{x+2}}{\sqrt{1-x}}\), \(\frac{\sqrt{x}+2}{\sqrt{1}-x} \) or \(\sqrt{x}+\frac{2}{\sqrt{1}}-x \)?
Thank you!
What's the focus and directrix of a line? (with e=infinity)In essence, we will have a pair of intersecting lines and not just a single line.
In essence, we will have a pair of intersecting lines and not just a single line.
Consider what happens as the eccentricity is increased. As the eccentricity increases, the foci become further distanced and the directrices approaches the y-axis. When the eccentricity becomes infinite, the foci will be infinitely away and the directrices coincide on the y-axis.
This also makes intuitive sense, as PS/PM = e. As e->inf, we have either PS -> inf, or PM -> 0.
It may be worth considering that this case is the opposite of the circle, where the foci coincide at the origin to become the centre, and the directrices are infinitely far away
(http://uploads.tapatalk-cdn.com/20170528/4c8fbaeefbf0c308c274745c18beefd1.jpg)Honestly this question appears nonsensical.
Please help question 3
Hey guys can I grab some help with this whole question I'm seriously struggling (http://uploads.tapatalk-cdn.com/20170530/3cc03a64824ffd60ab8e131028ac200a.jpg)to get itThe significance is in that it means the force acts against the direction of the position/velocity respectively.
Hey, can someone please explain these perms and combs questions, I'm at a loss
1) Four sets of twins go to a party. In how many ways can these children be arranged in pairs so that no child is with his twin?
2) How many ways can three integers be chosen which are in arithmetic progression from the numbers 1 to 101?
I'm not confident with my working out for the first one. Can I please have the answer?Thanks so much omg. The first one is 60
Honestly this question appears nonsensical.
The question says that the particle accelerates, until it reaches the maximum velocity. Then, once it is at the maximum velocity it starts travelling the distance S that it's supposed to.
And then they say \(T_1\) is the time taken for the car travel the full distance at maximum velocity. But it doesn't travel the full distance at maximum velocity?
If the question meant something else then it was worded horribly.
The significance is in that it means the force acts against the direction of the position/velocity respectively.
In the future, where possible please avoid the sideways photos. This request is a bit selfish but it stems from the fact that I personally hate mechanics and get put off from doing it quite easily......(http://uploads.tapatalk-cdn.com/20170601/617ba62b13d717431bb7d9ee56ed6ef7.jpg)(http://uploads.tapatalk-cdn.com/20170601/c3aacaf262b4c38ae2f2384e11f2b964.jpg)(http://uploads.tapatalk-cdn.com/20170601/f7146d1d4807dd50b9d62bf8838cfae3.jpg)
thankyou soo much rui, i worked out the first question if you want me to post it? thankyou soo much.Up to you. Good idea though for the benefit of the others
Hey, can someone please explain these perms and combs questions, I'm at a lossThis Q1 took me ages to do and I had to call in my second year math friends to help me out. It dumbfounded me big time until I realised that I had mucked up the very last piece of my analysis...
1) Four sets of twins go to a party. In how many ways can these children be arranged in pairs so that no child is with his twin?
2) How many ways can three integers be chosen which are in arithmetic progression from the numbers 1 to 101?
This Q1 took me ages to do and I had to call in my second year math friends to help me out. It dumbfounded me big time until I realised that I had mucked up the very last piece of my analysis...
Having a bit of trouble with conics. Part 2. Would be appreciated if you could guide me in it right direction.ST should be easy to find because T is on the x-axis so the distance is horizontal.
Cheers, Wales
Absolute monster question: or atleast idk how to do it.Clarity needed. Do you mean the circled one? The arrowed one was never crossed out
Help greatly appreciated(http://uploads.tapatalk-cdn.com/20170605/5ee6c4d840af4268c551a4a3e6adea96.jpg)
Clarity needed. Do you mean the circled one? The arrowed one was never crossed out
Surely you just sub in V=25.7 and v=10?
Hey! I was just wondering if you have tips on how to solve 4U maths inequalities - I've started with them and needless to say, I don't feel too enthusiastic about them - like where do you even start and how do you show your working? Thanks a bunch! :)You should identify what technique you use to prove those inequalities, be it calculus, something elementary, induction or simple algebra etc. (Not sure what else you might need off the top of your head.)
Hi! Could someone please explain how to prove that x^2<xy<y^2 if 0<x<y? I've tried using the fact that (x-y)^2≥0 and I've multiplied the given inequality by y, but now I'm stuck :P Thank you! :D
Oh dear I can't believe I didn't think of that... Thanks so much Rui! ;D
Similarly for y
Hi, could I please get some help with 7b and 8? Would 7b relate to or use 7a's result? And I have no idea where to even start for 8! Thank you :)
Hi, could I please get some help with 7b and 8? Would 7b relate to or use 7a's result? And I have no idea where to even start for 8! Thank you :)
I leave the formal structure of the proof as your exerciseThank you so much for your help Rui! ;D
Help I don't even know where to begin with this one (or is this level of difficulty outside what the HSC expects of us?)The question looks wrong to me.
The question looks wrong to me.
Could someone please help me with this question.
Thanks in advance :)
Hi! Could someone please explain the process of making a trig substitution (the kind that looks like the one in this question). Thank you so much! :)Most of it is like an ordinary substitution where you let x = f(u). This one isn't exactly any different so I'm not sure what you want me to explain
Most of it is like an ordinary substitution where you let x = f(u). This one isn't exactly any different so I'm not sure what you want me to explainI was confused about when and how to change the domain! Thanks Rui :D
Ah. Usually when you have x = f(u) instead of u = f(x), you try to put a restriction on the domain of u (or in this case, theta) so that over that domain the function is invertible. (i.e. the function is "locally" invertible.) That gets out out of troubles, e.g. if you had to substitute x = u2 and your original boundaries were 4 and 9, you no longer have to worry about a ± and just use 2 and 3.Thanks again Rui! You've been so helpful ;D
HEY GUYS I WAS WONDERING WHAT YOU THOUGHT (ESP MODERATORS) WHAT WAS THE BEST WAY TO PREPARE FOR TRIALS (OTHER THAN THAT BRILLIANT NEW TOPIC TEST FOR EXT 2 WHICH IVE ORDERED). So what is it? Textbook Qs all day or trial questions all the way? I personally did a tonne of trials before half yearlies and just didn't find them challenging in a good way... It's just a tonne of easy-ish questions with some half hard ones or ridiculously hard - I felt like I didn't cover much of the course either. And half of my study I spent quite uninterested/unmotivated (I realise trials will change that ahahahha and I need to revise- so they are a really good option). So yea what do you legends think. What's the way to go? Smash out textbooks for each topic or focus on trial papers? Or maybe something completely different hahaha just take notes?By the time it was trials for 4U I honestly gave up textbooks. Only past papers can emulate what you're going to get on the day.
I felt like I didn't cover much of the course either.If you do a LOT of past papers, yes you will. No guarantee that you're going to cover every cornerstone of the course, but you're going to have covered quite a fair load.
Well honestly, look, it's not easy to simply 'recommend' an amount because everyone needs to put in amount that works for them. In my opinion, whatever constitutes a 'good' amount is an amount that allows you to get to the level you want to be, without impairing your performance in other subjects. Of course if you want to be like me and put in a tad too much effort into 4U that's possible, but you risk reducing your marks in the other subjects even more so think about if that's worth it.
There's nothing wrong with skipping some easy questions every now and then especially if they're becoming a bore and time-wasting. Just don't neglect them completely because yeah, silly mistakes.
(Them relationship struggles though...)
Unfortunately it's been too long and I don't remember exactly how I planned out my trial study plan, or I probably would've just told you about it
I'm planning on picking up the Atarnotes 4u books and giving them a go. I'm doing that in combination with set sheets from my tutor. I just don't feel confident in doing the work.At uni, I'm always second guessing my work. I would write something down and doubting it because there was probably a silly mistake in there somewhere.
I ALWAYS second guess my work, I'm always afraid to post solutions to 2U questions (even though I've sat the course) in the fear I'll get something wrong and I feel that it's affecting me here as well.
Do you have any advice for this? I've never been particularly talented in Maths but don't find myself struggling with the maths but more interpreting the question and being confident in approaching it.
I'm planning on picking up the Atarnotes 4u books and giving them a go. I'm doing that in combination with set sheets from my tutor. I just don't feel confident in doing the work.
I ALWAYS second guess my work, I'm always afraid to post solutions to 2U questions (even though I've sat the course) in the fear I'll get something wrong and I feel that it's affecting me here as well.
Do you have any advice for this? I've never been particularly talented in Maths but don't find myself struggling with the maths but more interpreting the question and being confident in approaching it.
At uni, I'm always second guessing my work. I would write something down and doubting it because there was probably a silly mistake in there somewhere.
At the start, just roll with it. Literally, don't even bother checking it all and just move on. Do that until you've done every question you can easily do and all that's rest is the toughies that gets you a band E4.
THAT is when you go check. Focus on actually getting the question done, because if all your mistakes are arithmetic errors but your method is right, there's only so much they can deduct.
Of course, if you arrive at a contradiction you should know to backtrack. But if you reach an answer that makes sense and might only be off by a bit, just move on.
It's hard to be perfect at maths and just not make mistakes. Those that don't do that are who get state ranks, and that's only about what, 10 out of 3500 people? You're most going to need to let some mistakes happen.
The important thing is to get into the habit of moving on and accepting it for as long as you can. Go back only when it's the right time to in an exam.
With posting solutions on here, look being honest I make mistakes too. I try my best to make sure I do nothing wrong but more often than expected it just is plain wrong. And I can understand why some people are probably too shy to post solutions in front of me as well - I do have a habit of critiquing where the faults are as well. I get used to it after a bit though - people critique mine as well. That really just takes getting used to, I'd say.
(http://uploads.tapatalk-cdn.com/20170618/ea60f5392396eaace41e327deae8a61d.jpg)I don't get what they have. I get -1/(n+1) not +1/(n+2)
Help pls what even is this??
Thanks Rui. I really don't think I can help u there AHHAHA I'll get back to u if I work anything out or if my teacher gets it. HOW DO U JUST KNOW HOW TO DO IT hahahaTbh it's pretty typical in 3U binomial theorem. Have you covered the whole topic?
(http://uploads.tapatalk-cdn.com/20170620/de50412b3a58aed1191806b585e1b9b4.jpg)If there's no velocity surely that means there's no centripetal force, and thus the only two forces in play are the normal reaction and the gravity?
I seriously can't get question 1a. I think they did something weird I can't see or work out. Thankyou.
But there is still a frictional force or something, so u can't use componentsWhere's the friction if you aren't moving?
Where's the friction if you aren't moving?
heyo!
i was just wondering if there is any smart tips or tricks to do graph transformations for cos^-1(f(x)), sin^-1(f(x)) and tan^-1(f(x))?
so far, I've had to get domain, range, intercepts, then limits for x->0 & infinity, and then finally draw the general transformation, plugging into calculator to make sure each value is correct. this takes an absurd amount of time and i just wanna know some good tricks that i can apply to this transformation thatll make my life easier, like for reciprocal graphs when f(x) ->0+, 1/f(x) -> infinity+
Thanks :D :D
Main qThis question is arbitrary and only makes sense if we assume that the rotation is about the y-axis (the line x=0, as opposed to say x=1). Assuming this:
Shouldn’t limits be -a and a instead of a and 0 when you do a volumes by cylindrical shells method question? (not sure if i've explained it right - i'd post a picture but not sure how to work this)
thanks so much :) your help is appreciated a lot!!
Hi,
In regards to the arg (z) bit of complex numbers, how do you know if the locus of say, arg (z-2) - arg (z+2) = pi/2 is the semicircle above or below the x - axis? And for solving a question say, Re [(z-i)/(z+1)] how can you find arg (z) geometrically instead of algebraically? Thanks!
If you're looking for what a graph would look like, you can do it online :D I recommend Desmos Graphing CalculatorWhich really is not obvious at all.
It would look like this...
(http://i.imgur.com/GUOOj1t.jpg)
If you're looking for what a graph would look like, you can do it online :D I recommend Desmos Graphing Calculator
It would look like this...
(http://i.imgur.com/GUOOj1t.jpg)
Heey how would I graph
y^2=x^4 (4 + x)
Thank you :)
thank you Rui!! :D
By now you should be familiar with techniques on drawing square root curves, e.g. Block out what's below the x-axis
(http://uploads.tapatalk-cdn.com/20170625/2fd1362ad5dfd16f62dba5b476377ca3.jpg)Check that you didn't misplace your theta
Hey guys q2, the answer is:
(g x cot(theta))/4(n x Pi)^2
But I'm getting tan? Just wondering if you guys could check it - can't see where I'm getting it wrong
Heey alsoo
Find the volume of rotation when the region bounded by the x and y axes, x=2 and the curve y= 1/(x^2 - 4x +13) is rotated by the y axis
I attempted it and got an answer of 4pi/3 arctan 2/3 + piln (9/13) but tbh got no idea
Thank you :)
For the circular motion around a circle of x^2+y^2=r^2, with x=rcostheta, y=rsintheta, x*=r(-sintheta)dtheta/dt. y*=rcosthetadtheta/dt. I don't get how the last two equations were derived.
I don't get the second last step to the last step, why does the cos change to a -sintheta and why does the theta appear at the top of the derivative?
I don't get the second last step to the last step, why does the cos change to a -sintheta and why does the theta appear at the top of the derivative?
I don't get the second last step to the last step, why does the cos change to a -sintheta and why does the theta appear at the top of the derivative?Like I said, implicit differentiation. This is a very standard 4U technique that you're expected to know.
Thanks! Can I've some help with this question?What's h in this question? Is there a diagram that goes with the question that should be provided?
A particle P is attached by a light string of length l to a fixed point A, and P describes a horizontal circle with uniform angular velocity w around the vertical AC. Prove that if theta is the inclination of the string to the vertical, then h does not depend on l, and find a expression for h in terms of w.
Find the tension in the string when the particle P is rotating at 1 rev/s, the mass of P is 1kg and the length of the string is 35cm. Take pi^2=10.
Ooh also please explain this thank youuu :
Solve z^2 + (1+3i)z -8 -i =0
ahh i get it now, thank u :)
Hey for complex numbers how do we indicate the locus if it one of the axes? (Eg locus of Im(z) = |z|)You can just call it the y-axis (or imaginary axis) surely
The combined air and road resistance of a car in motion is proportional to v2, where v is its speed. When the engine is disengaged the car moves down an incline making an angle arcsin(1/30) with the horizontal, with a velocity of 30m/s. Find the force required to drive the car up the incline with a steady speed of 24m/s, given that the mass of the car is 1200kg.(http://uploads.tapatalk-cdn.com/20170630/afe8b8a0f064af9ef0b6367c09b2e695.jpg)An assumption had to be made that g=10 to make this work
Answer = 656N
(http://uploads.tapatalk-cdn.com/20170702/5cc9639d16214eed5a3f4625ef72e8f6.jpg)Of course we can't have two yellows because we know that one jellybean has to be black
Ok so the answer is b. But it doesn't consider two yellows being selected - or somethings idk. I got 10/41 so yea I'm way off - help much appreciated
Hello! Could you please explain the solutions to these questions?(Briefly recall that three lines being concurrent means that they intersect at a common point)
all of Q8 and part ii) for c)
Thanks :)
(Briefly recall that three lines being concurrent means that they intersect at a common point)
...
Hi :)
the ansewr to this question is C - i understand that there must be a vertical tangent at x, but how would you choose between a and c (concavity ??)
thanks :)
For a question like that, I'd probably just test each of the quadrants. Quadrant 1, x and y are both positive so dy/dx is positive. Quadrant 2, y is positive and x is negative, so dy/dx is negative etc.
Then you just have to choose between c and d, when x approaches 0 dy/dx approaches positive/negative infinity so it's c
Hope this helps :)
Sketch:These are just random implicit curves that authors throw 4U students when they're bored. Please indicate the source of these questions.
(x^2+y^2)^2=2(x^2-y^2)
and x^2+y^2=3xy
These are just random implicit curves that authors throw 4U students when they're bored. Please indicate the source of these questions.Ohhh thanks, yeah I was wondering why the teacher was setting us so unseenly and obscure questions lol xP Thanks, I feel happier knowing I won't have to cover them at all XD They came from his holiday homework. We don't usually get this much, but since it's the holidays before the last school assessment he decided to give us extra :P
(My honest opinion though - not even worth attempting unless you've done polar curve sketching, which is not in the HSC course. Part b) can possibly be done by using the quadratic formula on \(y\) though.)
a) Find integral from 0 to pi of [dx/(5-4cosx)]
b) Un=integral from 0 to pi of [cosnxdx/(5-4cosx)], show that U(n+1)+U(n-1)-5/2*Un=0.
c) Calculte U0, U1, and hence find U2, U3.
hiii :)
how would you do this question HELPPp
thankkkk
Can anyone solve this Volumes questions? Its Questions 15.a) from the 2015 James Ruse Trial Paper:Assuming a diagram was provided.
The diagram shows a segment of the circle X2 + Y2 = r2 which is rotated about the y axis to from a collar. This collar is thus a sphere was a symmetrical hole through it. Let the hole be of the height 2h as shown.
Use the method of cylidical shells to show that the volume of the material in the collar is given by the integral4pi(integral between r and the square root of (r2-h2) of X multiplied by the square root of (r2-X2)dx)
Evaluate the integral to show that the volume of material in the collar is a function of h only independent of r.
Sidenote: How do I enter maths symbols into my posts?
Assuming a diagram was provided.
It should be clear why I took a positive square root.
Last bit subject to a bit of computational inaccuracy, but the method is still the same.
LaTeX guide
Can anyone explain why (D) is the correct answer?
If In=integral tan^nx dx for n>=0 show that In=[1/(n-1)]*tan^(n-1)x-I(n-2) for n>=2. What's the setting out for proving n>=2.The normal setting out.
The normal setting out.
Note that for the integral of tann[/sup(x) you should not be using by parts, but only a Pythagorean identity.
Thanks!These really standard ones are ones you should memorise the method of.
Show that for In=integral sec^nxdx >=0, In=1/(n-1)tanxsec^n-2x+(n-2)/(n-1)*I(n-2) for n>=2.
For these type questions, how do we know which way to start? Because I can see several ways this question might work. (take out sec squared which gives sec^n-2 which matches the I(n-2) expression a little, or change it into secx*sec^n-1x and differentiate the sec^n-1x to get the I(n-2) (I tried this but got a really long expression that I was stuck unsure of what to do with) etc) but how do I know is the right method?
Hello, please help with d) iv)I had covered a bit of this during my lecture. The first steps are to let z=x+iy to reduce the expressions given.
Thanks :)
If In=integral from 0 to 1 of x(1-x^3)^n for n>=0 show that In=[3n/(3n+2)]*I(n-1) for n>=1. Hence find an expression for In in terms of n for n>=0.
I had covered a bit of this during my lecture. The first steps are to let z=x+iy to reduce the expressions given.
One possible plan for the sketch:
1. Sketch \(x^2-y^2=3\) and \(xy = 2\) separately
2. Identify the regions satisfying \(x^2-y^2<3\) and \(xy<2\)
3. Add in the restriction \( 0 < x^2-y^2\) and \(0 < xy\) using your knowledge of the 4 quadrants. You may wish to do this separately from the rough sketches you had in step 2.
4. Combine the regions required.
Could someone please explain how to solve this polynomial question. Thanks so much.
Thank you! I realised that I wasn't able to get the answer at first because I didn't consider the quadrants, but now I got it :)Ah yep, that was most likely the hardest bit about that question.
Could someone please explain how to solve this polynomial question. Thanks so much.
Here is my solution.Whilst it looks nice I feel as though this is way too overkill when you can just use a simple algorithm and the irrational conjugate theorem for this type of question...
Whilst it looks nice I feel as though this is way too overkill when you can just use a simple algorithm and the irrational conjugate theorem for this type of question...
Whilst it looks nice I feel as though this is way too overkill when you can just use a simple algorithm and the irrational conjugate theorem for this type of question...
Is the irrational conjugate theorem in HSC? Never heard of itYes and it works exactly like the complex conjugate root theorem, which states that if a polynomial has real coefficients but has a complex root then another root must be its complex conjugate.
Hi! Could someone please help me with this question? I don't understand why the answers say that dV=2y(4-x)dx. I've tried to use similar triangles but I'm not sure if that's even relevant here... Thank you :)
Can understand if that didn't make sense but I'm too sleepy now so if something doesn't puzzle up I'll attend to it tomorrowThanks so much Rui, I think I get it now! ;D Is the 4 coming from the dotted line in the diagram?
Thanks so much Rui, I think I get it now! ;D Is the 4 coming from the dotted line in the diagram?Pretty much that's one way of thinking about it
Find V ifThis was difficult to read. In the future, when +'s and -'s are present please use an extra space so that the terms don't look clumped. Same with "thetainvcos" - a space is beneficial. Also, please use sqrt for the square root, and preferably "arccos" for invcos.
v=integral from 0 to h, of invcos(x/h)-(x/h)root(1-(x/h)^2) dx given that integral of invcos(theta)d(theta)=thetainvcos(theta)-root(1-theta^2) and let theta=x/h.
In the diagram the circles XYPS and XYRQ intersect at the point X and Y, and PXQ, PYR, QSY, PST and QTR are straight lines.The diagram they offered was terrible in the fact that it did not make QSY appear to be a straight line.
(i) Explain why ∠STQ=∠YRQ+∠YPS.
(ii) Show that ∠YRQ+∠YPS+∠SXQ=pi.
(iii) Deduce that STQX is a cyclic quadrilateral.
(iv) Let ∠QPY=a and ∠PQY=b. Show that ∠STQ=a+b.
Can I have help with iv?
hiii :))Hint: You will have to assume that the area of an ellipse is \(\text{Area}=\pi AB\), where A and B are the lengths of the major and minor axes.
I'm 100% lost on how to do this question please help
thanks!
Could somebody please help me with this one? I'm trying to get an answer independent of a (their answer is sqrt(3)+pi/3). I've tried the substitution x=asinӨ and integration by parts. Any help would be much appreciated :)Just looking at this I can tell that the answer will not be independent of a, due to the upper bound being unpleasant. Presumably the upper bound on the integral is incorrect and should be \(a\).
Just looking at this I can tell that the answer will not be independent of a, due to the upper bound being unpleasant. Presumably the upper bound on the integral is incorrect and should be \(a\).My teacher and I could only come to that conclusion after changing the limits. It's not often that a mistake exists both in the question and the answer. Thankyou very much!
Edit: According to Wolfram, if the upper bound is \(a\) then the answer is what's given, except with a minus.
Hey Rui,That one isn't easy to explain mathematically. Consider a more intuitive approach
Could you explain why
has the asymptote y=x?
That one isn't easy to explain mathematically. Consider a more intuitive approach
Also, LaTeX doesn't handle it well when you use double spaces.
Thanks!Tbh, I think if you dropped all the spaces in your LaTeX (except for when splitting some functions up) it would've rendered properly.
The latex killed me :( Took me longer trying to get the latex than your explanation (and didn't even get the latex in the end).
why dont we set one person in this and have to reduce the n part by another 1?
usually we say for n around a table it's (n-1)! because we set one person so they aren't the same. why don't we set one here? the answer is (n-4)!4! but what if n=4 then it's 4! and that's not right. is it?
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Compare the two cases. When n=4, what actually happens is that we've chosen everyone. This leaves nobody else left to use as, pretty much, a "basis" for the circular arrangement to handle the (n-1)! issue. So we actually have to cater for it manually.
Provided n>4, we have a way around it because there will always be one leftover guy. So as a result of the proof of the (n-1)! result, we address it in pretty much the way(s) explained above.
As far as 4U goes, it would probably have made sense to assume n>4 here. But pretty good point though.
HISimple reason: Cube rooting (or even square rooting) isn't that simple when you're dealing with the complex numbers, as opposed to the real numbers.
It is true that (cis(2π/3))3=1 but why isn't it true when both sides are cube rooted? (cis2π/3≠1).
Thankss
Heya, could someone please explain how they found the number of terms? I understand the rest of the question :)Since we want to find when Alan eventually picks a black jellybean, we need to consider every case.
Since we want to find when Alan eventually picks a black jellybean, we need to consider every case.Omg thanks so much!!! I get it :D
...
Omg thanks so much!!! I get it :DGlad I could help! ^-^
Hey can someone please help me with these 2 questions. Thanks in advance.
help please, i really don't understand the process to get here in the solution - maybe you have an alternative? or could u explain how u got there? should i add the solution?Why not?
idk ahhaha - just thought it wouldn't be necessary and i was hoping there was another wayThat actually didn't cross my mind during the time I had to think about it. I'll probably explain it later tonight though (unless someone gets there before me) because of a few more classes I have today that I want to focus on.
why isnt the answer B? - if the centre is half way to the vertex Q, shoulds it be half OQ, which gives 1/2 (i+1)z
why isnt the answer B? - if the centre is half way to the vertex Q, shoulds it be half OQ, which gives 1/2 (i+1)zWell what did they say it was?
Well what did they say it was?
(I haven't forgotten about your binomial question; but I just got home and I'm having dinner first.)
sorry they said it was CWait... but isn't that correct? It goes from iz to z so it should be z - iz = (1-i)z
idk ahhaha - just thought it wouldn't be necessary and i was hoping there was another way
Also, I believe they intended to write a2n-r in the place of ar but made a mistake in transcript
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ok so it's kind of a weird comparison/equating of co-efficients for different expansions, i can see what you did but I'm struggling with the reasoning. is it just re-denoting the question so that you can show a new relationship between a with the power? - seems liek they just gave u a question where u just use the answer to find something - i can't see a straightforward process other than changing the powers.Isn't that what you're meant to do in a proof? Look at the answer and find something?
Could i please get some help with part i.
Thanks :)
Hi! Could someone please help me with proving this statement? I'm stuck on how to get the left hand side, but I managed to use (x-y)^2≥0 to get the right hand side (not sure if I'm meant to do that though)! Thank you :DThat question looks wrong to me (if we're to assume, as always, x and y are positive). Are you sure the fraction on the RHS isn't meant to be entirely under the square root?
That question looks wrong to me (if we're to assume, as always, x and y are positive). Are you sure the fraction on the RHS isn't meant to be entirely under the square root?Yes it is! I'm so sorry, I drew it wrong in the picture!
I know you were at my lecture so i know you know what I'm talking about :P I wrote the bottom line first and worked my way up.Thanks so much Rui! ;D The tips at your lecture have helped me like inequalities questions a lot more now knowing that there's different ways to approach them! :D
pls help
(x+iy)^2=7-24i
find x and y
ohhh!
If P(acosθ , bsinθ ) and Q(acos(-θ ), bsin(-θ )] are the eremities of the latus rectum x=ae of the ellipse x^2/a^2y^2/b^2=1.
Show that PQ has length 2b^2/a
P(asecθ , btanθ ) lies on the hyperbola x^2/a^2-y^2/b^2=1 with foci S(ae,0) and S'(-ae,0).
a) Show that PS=a(esecθ -1) and PS'a(esecθ +1)
b) Deduce that |PS-PS'|=2a
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P(asecθ , btanθ ) and Q(asecphi, btanphi) lie on the hyperbola x^2/a^2-y^2/b^2=1. Use the result that the chord PQ has the equation (x/a)*cos[(θ -pi)/2]-(y/b)*sin[(θ +phi)/2]=cos[(θ +phi)/2] to show that if PQ is a focal chord, then tanθ /2tanphi/2 takes one of the values of (1-e)/(1+e) or (1+e)/(1-e).
P(2rt(3), 3rt(3)) is one extremity of a focal chord on the hyperbola x^2/3 - y^2/9=1. Find the coordinates of the other extremity Q.
Show that cos4θ =8(cosθ )^4-8(cosθ )^2+1.
a) Solve the equation 8x^4-8x^2+1=0 and deduce the exact values of cospi/8 and cos5pi/8.
b) Solve the equation 16x^4-16x^2+1=0 and deduce the exact v alues of cospi/12 and cos5pi/12.
Moderator action: Posts merged. At times like these, please resort to the modify at the top right corner of a post, to refrain from multi-posting.
Could i get help with part 3. Thanks!In the future, please provide progress on the previous parts. Here, part i would've been useful.
Thanks :D About the post mergings - I was actually considering doing that before I multi-posted (I think I've been told of for that before and know it's something generally not appreciated in this thread) but I ran into a problem where, if I spent time typing all those questions before, by the time I posted them all at once, there may have already been a solution offered to the first, you get what I mean? (Geezus I feel like I'm doing such a terrible job explaining sorry, but it's just that, if I post 1 problem, in the time it takes for me to type another, someone (like you) can read it and give me a solution - effectively saving time) so that's why I multi-posted. What should I do about this problem in the future?Questions like these are very unlikely to fall under the umbrella of "quick" questions; their solutions require over 10 minutes to be typed up properly, making this unlikely. If you have a series of small questions there's nothing wrong with simply asking them all at once, otherwise preferably wait for one (or if it isn't overly excessive, maybe two) to be answered already before posting more.
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Questions like these are very unlikely to fall under the umbrella of "quick" questions; their solutions require over 10 minutes to be typed up properly, making this unlikely. If you have a series of small questions there's nothing wrong with simply asking them all at once, otherwise preferably wait for one (or if it isn't overly excessive, maybe two) to be answered already before posting more.
Show that cos5θ=15(cosθ)^5-20(cosθ)^3+5cosθ. Hence
a) Solve the equation 16x^5-20x^3+5x-1=0 and deduce the exact values of cos(2pi/5) and cos(4pi/5)
b) Solve the equaion 32x^5-40x^3+10x-1=0 and deduce that
i) cospi/15+cos7pi/15+cos13pi/15+cos19pi/15=-1/2
ii) cospi/15cos7pi/15cos13pi/15cos19pi/15=1/16.
If you have trouble replicating the same procedure you should post up your working instead for guidance/feedback.
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The next question is done via a similar method (with differences being that no exact values make it faster, and the equation to solve is weirder). To be honest, these methods are all in my 4U Notes book; you need to take a look at it.
Thanks! Yeah I dunno why, I'm just not very good at those. For part b i) I got one of the roots as cos(-11pi/15), and the only one that doesn't match up to any of the other roots I got was cos19pi/15. I was wondering how you got from cos(-11pi/15) to cos(19pi/15).
can someone explain this both in words and equations - struggle with dummy variables
Hi!There is no such thing as best textbooks to rely on HSC questions/preparation, especially given that the ultimate focus is always past papers.
I was wondering what are the best textbooks for 4U? I've heard that while cambridge is good for the harder questions, it isn't really realistic for the types of questions in the HSC? (Any recs for 3U as well? Is Fitzpatrick 'better' than Cambridge?)
Thank you
Hi!I'm not sure if you are a year 12 or 11, if you are a year 12 then this isn't as helpful. but if u are going into year 12 i hope this helps u pick and choose.
I was wondering what are the best textbooks for 4U? I've heard that while cambridge is good for the harder questions, it isn't really realistic for the types of questions in the HSC? (Any recs for 3U as well? Is Fitzpatrick 'better' than Cambridge?)
Thank you
Hi -In the future, please provide a link or screenshot to the question.
I need a quick explanation of HSC 2016 q 15 b ii
So I can work to the answer without the limits of the integral (d and b) - I don't understand why they're there. I've read the suggested solutions posted by Rui. But I still don't get it. In other questions where I'm integrating to get an expression for time, I just integrate, solve for any constant and then substitute a point such as x in to get time. Why is it that in this question a definite integral is required?
Many thanks
In the future, please provide a link or screenshot to the question.
Are these my solutions that I posted ages ago last year you're referring to?
Yup - the one here: https://atarnotes.com/forum/index.php?topic=168150.0
Thank you so much Rui!Should be under 6.3.2 Vertical Resisted Motion
Do you know what section it is in, in your MX2 notes, I can't seem to find it.
Could someone help me out with this integration using the t method please?
Integrate tanx/(1+cosx)
cheers
Hi guys, can someone help me on this question please?
Find the number of arrangements that can be made using the letters of the word PARTICULAR without altering the relative positions of vowels and consonants
I don't understand the part where it says the 'relative positions of vowels and consonants.
Cheers Atar team
Hey RuiAce ,What group is this?
Obviously I want to join ur atar notes forum but I'm currently unable to do so.
May u plz guide me and let me join ur Ex2 math science group :) ;)
I'm not sure if you are a year 12 or 11, if you are a year 12 then this isn't as helpful. but if u are going into year 12 i hope this helps u pick and choose.
3U:
Terry Lee - is very good for harder level questions, a lot are HSC esk, some are just ridiculous. overall one of the top recommends personally. Explanations for the exercises are usually not great, but the worked solutions are pretty good.
Cambridge - very good explanations (quite wordy) and a lot of questions to practice with. Some are just not HSC type Questions and are just tedious, strange, or just why? but they aren't irrelevant and all the questions are very good for drawing out some kind of understanding and for testing yourself (i personally think there are too many and some are unnecessary to do)
Fitzpatrick - is very well written and nicely set out, it's not as hard as the TL and Cambridge but it has some really good, tough questions (i don't use it that much but it's good). the explanations are very good and its a pretty book ngl
AtarNotes - is something i recommend for revision (topic specific which helps when u only need to study for a couple of topics for an exam)--> to me there aren't enough questions to practice on or learn stuff. but when it comes time for exams the questions are challenging, they cover almost all types of questions you can be asked and the solutions are so comprehensive that it's super easy to fix any holes.
Papers - are ultimately the best for revision holistically. HSC, or trial you are gonna be challenged and get some solid practice. if you do 30 papers you will undoubtably be exponentially better (but u need to stick with it through hard Qs).
NOTE: YOU WILL NEED TO REVISE CONSISTENTLY TO KEEP ON TOP OF THINGS SO WORK YOU DID DOESN'T GO TO WASTE
4U:
Terry Lee - is again amazing, and very very challenging. (some topics are better than others, so in 4u ull probably have to do a couple textbooks to work out the best)
Cambridge - i honestly don't use it much, not very appealing and the questions are weird (good challenge exercises and diagnostic tests). explanations are good but it's just not as good as others.
Sydney Grammar Notes - ARE EASILY THE BEST RESOURCE overall - explanations are amazing, and as rui said it's the closest to Cambridge 3U. There are lots of Questions, lots of chapters and they go from easy to medium to hard (i forgot what they are called) and if you had to do just one this is the one (in my opinion because it covers all topics pretty well). still i would say use this in conjunction with terry lee. we used these notes from my teacher throughout - but i think if u can do this for the whole course.
Fitzpatrick - is good, I don't use it much because it's a bit repetitive (for what i did), is definitely useful and has good questions. better for starting out than challenging urself, but it asks a lot of questions which is good.
Atarnotes - i haven't started doing, but from reading it they are good challenging questions for revision. if they are as good as the 3u ones then it is very worthwhile
Trials - are again just the best thing for revision and practice but u have to do a lot of them to cover everything since the course is massive. and u need to spend time working things out (dont be afraid to use the solutions just use them well dont copy)
For harder ext 1 I recommend a book by kinny lewis. like just lots of Questions and really good solutions. They are hard, but they cover most things which is amazing. still should do everything under the sun for this topic since its expected to be like 30% of the course by NESA.
yea so there you go. I'm sure there are more out there, but that's my top picks - can't go too wrong with any but it's good to use a couple in my opinion.
HOPE THIS HELPS.
There is no such thing as best textbooks to rely on HSC questions/preparation, especially given that the ultimate focus is always past papers.
For some decently worthwhile mentions, if you're using the small Cambridge book that is only ever so decent. As for question types, they are either too easy or too hard.
On the other hand, things like Terry Lee's textbook are generally recommended for selective schools, due to the difficulty of the questions.
4U Fitzpatrick is decent, provided the newer version is chosen. The old version is quite useless.
What's generally regarded as a standout is the Sydney Grammar textbook, as it is most similar to the 3U Cambridge textbook. This textbook, however, must be purchased from the school. It was never published (presumably due to the new syllabus things), and comes as only a PDF.
Thank you for your advice. Is the Sydney Grammar book available for students who don't go there?Yes.
Hi, Could I get help on this proof:This was one of the 3U questions in 2009.
(n-1)Ck+(n-2)Ck+(n-3)Ck+...+kCk=nC(k+1)
Thankss
Can anyone help with v please?
Thanks
This is because if \(\beta\) or \(\gamma\) were chosen to be the right angle instead, the same argument would work. This falls from just the fact that \(\alpha,\beta,\gamma\) are just the angles of the triangle.
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Remark: Internationally, cosec is denoted as "csc", and is what LaTeX renders. Hence it will be what I type.
How do you find the cot equation?That was what you proved in iv.
halp plsJust looking at this, part b is clearly wrong, as \( w = z(\cos\theta+i\sin\theta)\) so \( w\overline{z} = z^2(\cos\theta-i\sin\theta)\)
Just looking at this, part b is clearly wrong, as \( w = z(\cos\theta+i\sin\theta)\) so \( w\overline{z} = z^2(\cos\theta-i\sin\theta)\)ohh okay thank you
help with this please(https://i.imgur.com/7nqK4RQ.png)
hiii if i wanna sketch locus of arg (z+1/z-i)I've never spent the time to fully understand the locus. I just remember how to draw it
then do i first see where its undefined and put open circles on the graph at those points? I think i should but i only ever heard of it once so I'm not sure.. thank you
edit: please show how to do this? the solution is a semi circle but how
hellooo I don't understand this question or the solutions given pls help :)
thanks in advanced
Hi all,I would always state further constructions made, e.g produce AB to P. Else the reader might be confused.
Just a general question if anyone knows.
If we copy down a diagram (e.g. circle geometry) and add a new element (e.g. a new point, or line), do we still need to write down what was added?
Thanks :)
I would always state further constructions made, e.g produce AB to P. Else the reader might be confused.
But if you're in a rush for time you might get away without doing it.
How would I do this question?
A funfair game has the following set up: a player may toss two balls into any of k chutes and if both balls are returned via a single chute, the player wins. What is the probability that both balls enter via different chutes, but are returned via the same chute, where that chute is neither of the chutes by which any ball entered?
The answer is (k^2 - 3k +2)/ k^3
Thank you!
Hey,
This implicit differentiation is simple but I didn't know that dy/dx being a constant meant that the original relation represents two parallel lines.
So I was wondering if this is just something you have to know or if there is some explanation for how they concluded that.
Can someone please help me with this probability q. Thanksa) Total no. of outcomes is given by 9P3 = 504
b) Again, total probability is 9P3.Remark: Total outcomes is different to total probability. The sum of all probabilities is 1.
Now, if we were to randomly choose three numbers, there's only one way to do so.
So, the total number of arrangements we'll get is: 1 + 1 + 1 + 1 + 1 + ... + 1, since in each arrangement, there's only one way in arranging them in descending order.
The fact that there are 9C3 ways of arranging them is revealed when we count the fact that (1, 2, 3) arranged would be the same as (1, 3, 2) arranged, which is the same as (3, 2, 1) arranged.
This brings us to conclude that there is a probability of 9C3/9P3 = 1/6
Hey Atar team,
Could someone give me a hand on 2016 ext 2 multi choice for 7,9, and 10 please?
Cheers
Hi,At some point I had reached the limit for third party hosting with photobucket. I'll extract the images and put them on another image hosting site instead.
I was just view the 2016 solutions that RuiAce post and Q16 says I need to update 3rd party hosting. How do I do that
Cheers
i dont get the jump from the equals to the inequality sign. 2011 HSC q8ci btw. but i only need the working explained not the question. it is basically the triangle inequality thing - if so could u explain both how it works and how the hell we were meant to see that
ok so i get the inequality. but wth is m. like is it the coefficient of every single term??? that seems so sus and impossible. how can u factorise it out if its the coeffiecient of each??
hiii i need help with this questionAlready addressed in the compilation.
thank you so much!!! :)
Hello, sorry I have another question -
what is the effect on a general hyperbola as e approaches infinity ???? i know that it approaches a 'pair of vertical lines' (directrices) but i don't understand why that is :))
thank you
hey! this is my very first post so please excuse me if this doesn't work out. anywho I wanted to ask about 4u. how hard is it? I reallyyy want to try it and I know I'm capable of doing it if I work for it but Ive been hearing things like this subject is killer and when you do it you'll be looking like a zombie from the lack of sleep, which has made me reconsider taking it but what do you recommend?I may have forgotten to address these exact rumours, but they were basically what I cleared up at the start of the lecture.
Hey atar team, how would you solve this question?
Cheers
Hi Atar team, Could someone help me with this question please?This is just 1/2, because the other 1/2 is when the number 1 lies somewhere to the right of the number 2.
Thanks
heey im not sure if I'm getting the right answer for this,That looks fine (although you should be using absolute values for ln|2x+3| and ln|2x-3|).
basically I need to now integrate -1/(6 (2x+3)) and 1/(6 (2x+3)) for partial fractions,
is the answer -1/12 ln (2x+3) + 1/12ln (2x-3)
yo Rui can u tell the ext 2 methods? XD
find the values of a and b such that x^4 -x^3 +x^2 +ax +b is divisible by (x^2+4)
()
hey not sure if this question's been asked already
but I don't understand how to do this q and the solutions don't make sense
thanks in advance!
Hey,I don't know of any better strategy (until either the night before the exam, where you should be resting, and the morning of the exam, where you should just reconsolidate all the methods/formulas/... you know).
What are the best strategies for the next 3 days to maximise marks before the Ext 2 paper apart from past HSC papers?
Thankyou
hiiiI’m about to drive home but for reference, did you mean w^3=1?
so if we know z^3=1
how do we evaluate (1+2w+3w^2)(1+2w^2+3w)
whats the way we do this?
I’m about to drive home but for reference, did you mean w^3=1?ohh yeah I mean w^3 sorry
ohh yeah I mean w^3 sorry
ii and iii, it says "recorded", why don't we care about the arrangements in the answer, specifically ii?
also why can't we do iii, by going 1-probability of a draw and divide by 2?
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If you mean [1 - Pr(whole match is drawn) ] / 2 then yes you can. But if you do that, just make sure you've covered every single case for a draw.
Terry Lee's answers just chose not to do it.
then don't we need to consider the arrangements in the "total number of arrangements"? or something - otherwise isnt it just the probability of a win, law, draw, draw in any order?Why does the total number of arrangements matter?
Hey, how would I do this question?A and D are integrals of odd functions so they are exactly equal to 0.
Cheers
Hi! Could someone please show me how to prove the triangle inequality? I'm confused about both the geometric and algebraic approaches! Thank you so much :DI'm fairly sure there is no algebraic proof that you're expected to know of in 4U. The first time I proved it algebraically was in first year uni after doing some linear algebra in advance. The only thing that can be proven algebraically is the generalised triangle inequality, and even then you have to assume that it's true for 2 complex numbers before you can prove it for n complex numbers.
I'm fairly sure there is no algebraic proof that you're expected to know of in 4U. The first time I proved it algebraically was in first year uni after doing some linear algebra in advance. The only thing that can be proven algebraically is the generalised triangle inequality, and even then you have to assume that it's true for 2 complex numbers before you can prove it for n complex numbers.Thanks so much Rui! ;D
Hi! Could someone please show me how to prove the triangle inequality? I'm confused about both the geometric and algebraic approaches! Thank you so much :D
2001 HSC 5
(c) A class of 22 students is to be divided into four groups consisting of 4, 5, 6 and
7 students.
(i) In how many ways can this be done? Leave your answer in unsimplified
form.
The answer my book gives is 22C4x18C5x13C16
What I don't understand is why this isn't divided by 4! on account of the ways that the same groups can be selected in different orders, but still essentially be the same.
Many Thanks
Hi,I always memorised it as this (because textbooks can use multiple conventions and it gets annoying)
When finding the eccentricity for a conjugate hyperbola (y^2/b^2-x^2/a^2=1), is it e^2=1+a^2/b^2 or is it e^2=1+b^2/a^2
Thankyou
Hey Rui(https://i.imgur.com/ZWXA6GU.png)
For the 2016 question 9, I dont understand how you did h/4=x/6 with similar triangle method (I saw the 6 from finding the difference of the two lengths but how did you use similar triangles and got 6)?
Thanks
Algebraic proof of the triangle inequality can be done in two ways.Thank you so much! :D
Hey,Retardation is negative acceleration. Resistance is negative force.
In the mechanics section, more specifically F=ma, how do you know when to include the 'm' when integrating? Is there a difference between 'Resistance' and 'Retardation' where you leave the m or take the m out of the equation?
Thanks
HELPPP, i can't work out this question. and i am struggling with my reasoning. i have all the principles for Probability, but i can't apply them and im struggling, like in the last question how i didn't see how you didnt arrange them, well it was because it was an arrangement and was set. but now im sturggling with a lot of probability and was wondering how i could brush up or revise - like just arranging things properly or knowing when not to, or when it's just repeats or basic stuff i am over complicating.hey! thought i'd give Rui a bit of a break and have a go aha :)
i cant do the first part, i would like to attempt the second after you guys give me hints, thankyou.
hey! thought i'd give Rui a bit of a break and have a go aha :)OI TRUUUUUUUU.
So this isn't too bad if you break it down a bit. So let's just put one of that pair in the left-most slot in the first rung (call them Player A). Let's look at the possibilities they verse Player B, depending on where Player B is.
If Player B is next to them in the same bracket (a 1 in 3 chance), the condition is already satisfied. So, 1/3 there. If Player B is in one of the other slots (a 2 in 3 chance), two things need to happen. Player A needs to win their game, and Player B does too. So multiply 2/3 by 1/2, and then 1/2 again, and you get 1/6.
Add these together!
1/3+1/6=1/2
(sorry I don't know how to use the fancy math text like Rui but hopefully this makes sense aha)
and where can i find harder binomial theorem Qs, harder sequences and series? - not really major parts but i have major holes.Your best bet is to try papers from before 2000 (or even 1990) if you want some of these ones. (Or some of the extension level questions in the Cambridge textbook; like they're not really examinable but they definitely are hard.)
Hey,
How do you find the minor and major axis of a locus of an ellipse using complex numbers?
Ex: Iz-3I + Iz+3I = 12
Thanks
in finding the nth root of -1, all textbooks and solutions say it is cis(((2k+1)pi)/n). However isn't the general solution cis((2kpi +/- pi)/n)? Do we ignore the plus/minus for some reason?
Thanks. So for any question, I can safely use only + and sub in k=1?
You can check this by listing out a few cases.
For the ± case, you would have cis(π/n), cis(3π/n), cis(5π/n) and so on, and also the other way.
For the + case, you would also have cis(π/n), cis(3π/n), cis(5π/n) and the rest.
This happens more or less because
1. They both go both ways: you can just sub k=-1 instead of k=1.
2. The trig functions are periodic: \( \sin (2\pi + x ) = \sin x\) and \( \cos (2\pi+x) = \cos x\).
Hey Rui!
Just wondering if you had any general tips for graphing functions when they're part of tan inverse ((tan^-1) f(x) )?
Thanks. So for any question, I can safely use only + and sub in k=1?I've never found any problem with only using +
Now, hints for the second part:
Thanks Rui, that really cleared it up for me! Also, would you be able to share any general tips on attacking 4u probability questions? (like how to answer certain types of questions - whether to use binomial probability, combinations, the box method or something else)Not too sure what the box method is.
Hi RuiAce, for the first round, what is the probability the players will meet? I understand that the number of ways to pick the 8 players is 105, but what is the number of ways that they will meet? I have the book, it says 1/7 but I'm not sure how to arrive at that using your method (I originally was attempting to do it this way)
Thanks
Hi,Similarly is allowed so long as if you replicate the exact same STEPS albeit with different angles/sides you will reach a very similar conclusion.
For part (ii), are we allowed to use 'Similarly to argument in part (i)'? Or would we have to prove it separately? The sample answers allow the similarly argument but not sure if we should follow this.
Thanks.
Ok scrap ALMOST ALL of that - I think I've worked out a way to figure out the initial starting positions properly.That method doesn't get to the correct final answer so I need to check somethingEDIT I'M SORRY ALL OF THIS IS WRONG
Note that the identical groups will only occur for the other 4 people. The pairs that A and C must be with are immune from this because A and C must be in different groups, and that effectively forces all possible groups to be distinct.
The only difference is that we don't have (A,B)(C,D) || (E,F)(G,H) but rather (A,B)(D,E) || (C,F)(G,H). This doesn't actually change the number of arrangements though.
Sanity check: If we regroup the cases, we get 15+45+45 = 105 as we expected.
Edit #2: I've found out what the theoretical values are by working backwards. I'm gonna try again and work out how we can get to them.
Hey for 2015, question 12 (d) I am very confused as to whether to use x or (3-x) for volumes by cylindrical shells. How do we know which one to use?(https://i.imgur.com/roBd70E.png)
Please Help!
https://www.boardofstudies.nsw.edu.au/hsc_exams/2015/exams/2015-hsc-maths-ext-2.pdf
is it ok if you could explain this a bit more? I am still a bit confusedWhat part is the confusing part?
And what was 16ciii? I got 2(3)^5 -3
Yewww I can say I posted in 4u maths 8)Perhaps the easiest way to put it
Helloooo, this is such a dumb question that has no relevance to any mathematical problems whatsoever, but I am so confused about the structure of 4u maths so I thought I might as well ask here since the exam is over ;D Sooo I know you guys sit the exam the same time as Mathematics, meaning you don't do that exam, but do you do the Mathematics course still? Im 10/10 confused and always wanted to know the answer to this
- from your fellow 4 unit (English) student ;)
Perhaps the easiest way to put it
The 4U student is taught the 2U content as well, but then it doesn't get examined. It just becomes assumed knowledge for the 4U student
solve x^2cos^2 theta + x sin2theta +1 =0Assuming that theta is not a typo, what are we trying to solve for here? Because if we're trying to solve for x, the easiest way out is really just the quadratic formula
Would someone please do this. Solving quadratic equations with complex coefficients: finding the complex number equations.\begin{align*}0&=4x^2-4(1+2i)x-(3-4i)\\ x&= \frac{4(1+2i)\pm \sqrt{16(1+2i)^2 +16(3-4i)}}{8}\tag{quadratic formula}\\ &= \frac{4+8i\pm \sqrt{16(1+4i-4)+16(3-4i)}}{8}\\ &= \frac{4+8i \pm \sqrt{0}}{8}\\ &= \frac{1}{2}+ i\end{align*}
4x^2-4(1+2i)x-(3-4i)=0
Hi, guys, do you use De Moivre's Thereom to solve:Handwriting as "cis" is ok
z^5= -32
Thank you
lil questionThat usually isn't true. Do they give more information?
say i have 2 complex roots which are conjugates of eachother. One is alpha and the other is beta.
how do we prove alpha^2=beta and beta^2=alpha
thank you :)
That usually isn't tru6e. Do they give more information?
well alpha=(-1+root 3 i)/2Lol.
beta=(-1-root3 i)/2
verify that a^2=b, b^2=a, a^3=b^3=1, 1+a+b=0
maybe it's related to roots of unity? idk :/
Lol.oh lmao how do I verify all that stuff tho
Yes, I bet you those are the cube roots of unity
Tbh, whilst there might be a crafty solution, at a time like that I would just be lazy and actually do the multiplication by handooh okay thank you Rui
Heya, I tried doing this by subbing in x1+iy1 for w1 and x2+iy2 for w2 but im still not getting it :/
solve for w1, w2
2w1 + 3iw2 =0
(1-i)w1 + 2w2 = i-7
(when i write w1, i mean w subscript 1)
Thank you so much Rui :)\begin{align*}\frac{2}{1+z}&=\frac{2}{1+\cos\theta + i\sin \theta}\\ &= \frac{1}{\frac{1}{2}(1+\cos \theta) + \frac{1}2 i\sin \theta }\\ &= \frac{1}{\cos^2\frac\theta2 + i\cos\frac\theta2\sin\frac\theta2}\\ &= \frac{1}{\cos \frac\theta2} \times \frac{1}{\cos \frac\theta2 + i\sin \frac\theta2}\end{align*}
Alsoo, kinda not getting anywhere with this despite using trig formulas and whatnot:
if z is cistheta
prove that
2/(1+z) = (1+costheta+isintheta)/(1+costheta) = (2cos^2theta - 2isin(theta/2) cos(theta/2))/ 2cos^2(theta/2) = 1-itan(theta/2)
or if you can point me in the general direction? thank you!!!
\begin{align*}\frac{2}{1+z}&=\frac{2}{1+\cos\theta + i\sin \theta}\\ &= \frac{1}{\frac{1}{2}(1+\cos \theta) + \frac{1}2 i\sin \theta }\\ &= \frac{1}{\cos^2\frac\theta2 + i\cos\frac\theta2\sin\frac\theta2}\\ &= \frac{1}{\cos \frac\theta2} \times \frac{1}{\cos \frac\theta2 + i\sin \frac\theta2}\end{align*}ohh yes I got the final part!! Thank you!!
NOW try "realising" the denominator
hey for this, do i sub in x+iy later??
find x,y if
2z/(1+i) - 2z/i = 5/(2+i)
heey if my complex number has no real parts (e.g. it's in where a=0 and b=z) then how do I know my argument?If you draw the Argand diagram, you'll see any purely imaginary number has argument either \(\frac\pi2\) or \(-\frac\pi2\)
thank you so much Rui omlThose letters E and n look ugly so I'm gonna replace them with Greek letters........ :P
also can't figure this-
E+ni is a root for ax2+bx+c=0
where a and b and c are real
show that an^2 =aE^2 + bE +c
so I figured that the other root is E-ni but I've tried using product of roots, but can't seem to prove this :/
prove by Induction that z1+z2+zn = z1 +z2 +zn
where the LHS has a big conjugate line on the top and RHS has small conjugate lines (lmao)
so I proved true for n=1 and assumed for n=k but im stuck on proving n=k+1
thank you!!! :D
Modify message
z^2-2(1+i)z+8i=0
when we solve we should get (1 + root 3) +i(1-root3) and this other root, but i cant seem to get this :/
thank you rui :)
i dont get this part, moduli of complex numbers removes the i?
i dont get this part, moduli of complex numbers removes the i?\begin{align*}|(x+iy)|^2 &= |-6i|\\ |x+iy|^2 &= 6\\ x^2+y^2&=6\end{align*}
Hello!Yeah for HSC MX2 this is fine
I would like to know if using cis is possible for complex numbers.
Hi Rui,
I read somewhere that apparently it causes issues later on at university using the cis notation? Is this correct?
Thanks
Hey guys, would anybody be keen to help me with some subset graphing?
the question says to illustrate the following subsets of C on the z-plane and one of them is:
Arg[z-(1-iroot3)]=2pi/3
I've been shown a couple ways of how to do these types of questions but I was just after an easy method that will work everytime
So you'd have the ray drawn from \( 1-\sqrt3 i \) making an angle of \( \frac{2\pi}3 \) with the positive real axis.
for any complex numbers z1,z2, show that |z1+z2|^2 + |z1-z2|^2 = 2(|z1|^2 + |z2|^2)
thank you :)
thank u Rui :D
for any complex numbers z1,z2, show that |z1+z2|^2 + |z1-z2|^2 = 2(|z1|^2 + |z2|^2)
thank you :)
2i(y-1)>0Inequalities don't exist with complex numbers (because they don't make sense). You can do inequalities with the real/imag parts, the mod and the arg, but not the complex number itself. The fact that you got there means you most likely made a mistake along the way.
Can you divide by 2i and keep the inequality the same?
To give y>1
Or is there something I'm missng?
An alternative method, without resorting to the modulus-argument form is;Lol right, that's what I failed to think up last night. I knew the linear algebra way but entirely forgot the complex analysis version.
Hey is there a site where is can plug in things including arg and it will sketch it on complex plane? My textbook docent have answers to some of the questions.If you want to use technology, consider feeding GeoGebra the following input:
I've used the complex plane on desmos but I can't or don't know how to type in arg.
If not then can you sketch for me:
-π < arg < π
arg (z - i) = 0
arg ( z + 2) = 3π/4
arg(x+i*y+2)=3pi/4
If you want to use technology, consider feeding GeoGebra the following input:Code: [Select]arg(x+i*y+2)=3pi/4
Note that excluding -2 will be necessary. GeoGebra won't do this, but you will need to.
Thanks Rui,Well yeah -π < arg < π is essentially the whole plane.
Could you please sketch for me -π < arg < π because I don't think geogebra can do it?
Or if not, would you sketch the whole plane excluding the real axis ( i = 0)?
I just don't know where to start when attacking this question? Is sum of roots the right way to go? Would love if I could get a worked solution :)Sum of roots will indeed be necessary but you would certainly not be told to prove this in an exam without any guidance. Please provide the source of the question.
cos(pi/7) = cos(2pi/7) + cos(4pi/7) + 1/2
Hey how would you go about factorising z^5 + 3z^4 - z - 3?I would do it by grouping terms
I got the first 3 factors by finding factors of the constant term that sub into P(z) to get zero.
So far I got P(z) = (z-1)(z+1)(z+3)Q(z)
But I don't know the best way to go about finding Q(x).
I would do it by grouping terms
Edit: I don't know if this is complex numbers or whatever so I'll just leave it like this
If (z + 2 -i) is a factor then (z + 2 +i) is also a factor via the conjugate root theorem (probably a typo on your part) :)
Thankyou! Also,
How do I factorise this?
z^4 + 4z^3 + 3z^2 - 8z - 10 if I am given (z +2 - i) as one linear factor.
I know that (z -2 + i) would be another factor due to a theorem but I don't know what to do to get the other two factors.
Hey :)
I have a 4U assessment coming up and I have this question:
If given the graph of f(x), how would I sketch f^2(x)?
Is this the same as f(f(x))?
Thanks in advance :)
Hi, Could someone please explain how to factorise 2z^2 - 2z +1 over C?Yeah. The quadratic formula gives you the roots under the assumption that the polynomial is monic. If you wish to factorise by quadratic formula, you must always pull out the leading coefficient first (and hence the 2 in front).
I used quadratic formula so that z = 1/2 + or - i/2
So I thought P(z) = (z - 1/2 + i/2)(z - 1/2 - i/2)
But the answer is P(z) = 2(z - 1/2 + i/2)(z - 1/2 - i/2) (s0 there is a 2 out the front, I thought this may have something to do with the polynomial being non-monic but I'm not sure)
Hi Rui, could you please help with this question?The \(\cos 5\theta\) variant was covered in my lecture so I won't go in depth here - just provide the working out.
Use De Moivre's Theorem to express cos3θ in terms of powers of cosθ
Thanks in advance :)
The letters from the word FIFTY are selected at random. Find the probability that the T is chosen somehwere between the 2 F's.
I don't even know where to start with this one, a step by step of the whole process would be greatly appreciated, cheers!
Can someone explain how to do part b and c?Using part a), we know that:
Can someone explain how to do part b and c?
Hey guys,
I'm not sure how to fnd the roots of this or find x and y. Am i supposed to sub z=1, because that only got me a "2i+ix-y-1=0" and I'm not sure how to simplify it. Also is that the traditional way of dealing with questions about roots
How would I factorise x^4 + x^2 +1 over the real numbers. It has no real roots. The roots are 1/2 + or - √3i/2 and -1/2 + or - √3i/2
Thanks in advance.
Hi. How do I know when my curve is able to cross the horizontal asymptote and when it isnt able ?
Also how do I know which part crosses ? On the diagram the left side crosses but the right side doesnt :(
pls help
Cheers.
Hello
I am having trouble with seeing how (-2 plus/minus root 8 x i) over 2,
can be written as -1 plus/minus root 2 x i
Hi, there is this inequality question I am having trouble with. Could someone please assist?What is the source of this question? I know I'm quite braindead but I genuinely don't see how the first inequality is supposed to be of any use right now
Using a^2 + b^2 + c^2 >or= ab + bc + ca,
Show that a^2 + b^2 + c^2 >or= 3*cube root(a^2*b^2*c^2)
Show that if a, b, c, d > 0, then:
a/b + b/c + c/a + d/a >or= 4
Just started doing advanced 3 unit inequalities and I have no idea what to do with this question, any help would be appreciated,
cheers.
Show that if a, b, c, d > 0, then:This is actually one of the questions covered in my 4U notes book. So all I will do is put the working out here.
a/b + b/c + c/a + d/a >or= 4
Just started doing advanced 3 unit inequalities and I have no idea what to do with this question, any help would be appreciated,
cheers.
With that question you asked, the only logical thing to do is to somehow prove that \( ab+ac+bc\ge 3\sqrt[3]{a^2b^2c^2} \), because otherwise you aren't using the hence method. But to do that, you would have to make a very weird substitution to force certain terms to collapse (in your given inequality), and right now I can't see how that works. I won't say that the question is nonsensical yet, but it seems extremely peculiar. (They actually make you prove something using a HORRIBLE method.)
The question I asked was from Fitzpatrick. The question above is also from this textbook. Do you know how to solve either of them. If so could you please show working because I am stumped.
Thanks
Recall that for a rectangular prism, given a fixed volume, the sum of dimensions is minimised when it is a cube.This is really just bringing back memories of Lagrange multipliers...
Consider a rectangular prism with sides a2,b2 and c2 and a cube with sides . It is not hard to check they have the same volume.
Therefore for any a,b,c: as required.
Mod edit: Was authorised to fix up the LaTeX
Hi, is there a way to find the zeros of polynomials with complex coefficients that have a degree higher than 2? Because I am having trouble solving theses types of polynomials. I'm not sure really where to start with them. Thanks!There is no concrete rule of thumb. It always depends on a question-by-question basis.
There is no concrete rule of thumb. It always depends on a question-by-question basis.
Hint: The fact that \(k\) is real means that you can do some rearranging, and then equate the real and imaginary parts. The real parts should give you a quadratic...
But like its not like with polynomials with real coefficients wear you can check if factors of the constant term are zeros right? Like there is nothing like that right?
I have this question,
Find the real numbers k such that z=ki is a root of the equation z^3 + (2+i)Z^2 + (2+2i)z +4 = 0. Hence or otherwise, find the three roots of the equation.
I subbed in ki where there was a z and ended up getting -ik^3 + (-2-i)k^2 + (2i-2)k +4 =0 and I am not sure how to find k from here.
Hint: The fact that \(k\) is real means that you can do some rearranging, and then equate the real and imaginary parts. The real parts should give you a quadratic...
Oh, I wouldn't have.
Alright thanks heaps Rui!!! I got it an everything now from doing what you said but now Im wondering how would you go about solving that polynomial I had before with complex coefficients where I just subbed in ki where there was a z? Could you tell me how you would solve it because I don't think I have ever tried one with complex coefficients that wasn't a quadratic or quartic that could be reduced to a quadratic.
hi i have a question ;D\begin{align*}\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}&= \frac{1+i\text{ cis }(-\theta)}{1-i\text{ cis }\theta}\\ &= \frac{\text{ cis }\left(-\frac{\theta}{2}\right)}{\text{ cis }\frac{\theta}{2}}\times \frac{\text{ cis }\frac{\theta}{2}+i\text{ cis }\left(-\frac\theta2\right)}{\text{ cis }\left(-\frac\theta2\right)-i\text{ cis }\frac\theta2}\\ &= \text{cis }(-\theta)\times \frac{\cos \frac\theta2+i\sin \frac\theta2 + i \left(\cos \frac\theta2-i\sin \frac\theta2\right)}{\cos \frac\theta2-i\sin \frac\theta2-i\left(\cos\frac\theta2+i\sin \frac\theta2\right)}\\ &= \text{cis }(-\theta) \times \frac{(1+i)\left(\cos \frac\theta2+\sin \frac\theta2\right)}{(1-i)\left(\cos \frac\theta2+\sin \frac\theta2\right)}\\ &= (\cos\theta - i\sin \theta) \times i\\ &= \sin \theta + i \cos \theta\end{align*}
how can I prove that (1+sintheta +icostheta)/(1+sintheta-icostheta) = sintheta + icostheta
thank you
Find constants a,b,c such that the polynomial p(x) = x3-6x2+11x-13 is expressible as X3+aX+b where X=x-c. Hence show that the equation p(x)=0 has only one root.
I found a, b, and c quite easily but unsure of how to show that the equation has only one root?
You should be well aware of the fact that this is not 4U material as it requires you to pick the values for \(a, b\) and \(c\) yourself, without any guidance whatsoever.
________________________________________________________
How do we know that p(x-c) is simply a horizontal translation of p(x)?
Hi, I've done part a) and the answer is c = 1/27 a(9b -2a^2) but I really don't understand how to go about b). The answer is C - k = A/27 (9(B-m) -2A^2 ) fixed point has x-coordinate -A/3. Thanks in advanced!!!I'll start you off. This certainly involves using part a) somehow.
I'll start you off. This certainly involves using part a) somehow.
Ok thanks Rui, that makes sense, any further hints?So that gives you the weird expression involving \(C-k\) and according to your answers we can stop there. Which is reasonable, because we now have something that links \(m\) and \(k\) together.
Use De Moivre’s theorem to express cos5theta and sin5theta in powers of costheta and sintheta. Hence, express tan5theta as a rational function of t where t=tantheta and deduce that tan(Pi/5)tan(2pi/5)tan(3pi/5)tan(4pi/5)=5.
I’ve gotten up to having
Tan5theta= (t5-10t3+5t)/(5t4-10t2+1)
But I don’t really know how I’m supposed to use that to deduce the last line in the question. Any help would be appreciated, thanks.
b) How do I identify which roots belong to z^6 + z^3 + 1 = 0
Cheers :D
So that gives you the weird expression involving \(C-k\) and according to your answers we can stop there. Which is reasonable, because we now have something that links \(m\) and \(k\) together.
The second point is actually the fixed point. If you let the first, second and third points have \(x\)-coordinate \(\alpha-d, \alpha, \alpha+d\), you'll see that \(\alpha = -\frac{A}{3} \)
Sorry I'm looking at this question too and I don't understand how to find that c-k = A/27 (9(B-m) - 2A^2)That part was addressed a few posts back when she first posted this question
That part was addressed a few posts back when she first posted this question
Yeah I read this but I am still not sure.Please elaborate further on the confusion.
how would you go about graphing x^3 + y^3 =1
as in what process would you follow
how would you go about graphing x^3 + y^3 =1
as in what process would you follow
Cheers :)
Hey Jake and all 4U lovers. Could I please get some help with some 4U graphing.
I'm quite confused on approaching the absolute value one (| |y|-|x| |=1) and I'd prefer to learn how to do the second one (y=x^3/(x^2-9)) without using calculus.
Hey Jake! Not sure how to prove that Re(alpha)=1.
Hi rui!
I couldnt find the roots for this question
Use the fact that a2+b2>=2ab to prove that:
i) (ab+xy)(ax+by)>=4abxy
ii) ax+by<=1 if a2+b2=1 and x2+y2=1
Just starting out with this stuff still and I think our teacher is just as confused as us, any help would be appreciated, thanks :)
May someone explain what has happened?They just used the compound angle identities.
I've got this question from Fitzpatrick that I can't solve.
Given that x2 + y2 + z2 >= xy + yz + zx
Show that:
1) a2b2 + b2c2 + c2a2 >= abc(a +b +c)
2) (x + y + z)2 >= 3(xy + yz + zx)
Thanks in advanced!!!
Cheeers :'(This question lacks information - it is not stated whether \(a\) and \(b\) are real or not.
This question lacks information - it is not stated whether \(a\) and \(b\) are real or not.
Hello how would i go about finding the integral of the last question?This technically is not 4U material as improper integrals are not in the course. But having said that, many bits and pieces will be provided as it's quite easily doable regardless.
Oh. The answer from the textbook for a and b is real.
a = -2 and b =-2
I cant get the ans though :(
Dear Rui, i dont get why why it is 1- the original curve inside the integral.Like mentioned, area between two curves.
This question is weird:I’m at the gym right now but here’s some thoughts upon first glance.
solve the equation t^4+8t^3-6t^2-8t=1 correct to 3 decimal places.
This question is weird:
solve the equation t^4+8t^3-6t^2-8t=1 correct to 3 decimal places.
God that's a typo, it is t^4+8t^3-6t^2-8t+1=0
______________________
Note that the intuition behind the substitution \( u = t - \frac{1}{t} \) was essentially dependent on the symmetric nature of the coefficients (of the original equation). This kind of intuition isn't expected for a 4U student.
I tried doing this again on paper and for some reason it didn't work..., the u sub I tried on the equation, I get the u^2 but I did not get 8u...
How should I do this?Sketch the locus given and then deduce the max and min values of \( \arg(z) \). Recall that \( \arg(z) \) measures the angle made at 0+0i with respect to the positive real axis.
Cheers
:'(
Show that if x>0, y>0 then
1/x2+1/y2>=8/(x+y)2
Please help, fully worked solution would be appreciated.
Show that if x>0, y>0 then
1/x2+1/y2>=8/(x+y)2
Please help, fully worked solution would be appreciated.
Using expansion (x+ y + z)3, show that (x + y + z)3 >= 27xyz
I've written out expansion but cannot solve question. Any help appreciated.
Prove that, if a>0, b>0, a4 + b4 > a3b + ab3 by writing expansion (a-b)4That second one is another famous way of proving the three-variable AM-GM inequality.
Using x3+y3>=(x/z+y/z)xyz and similar expressions for y3+z3 and z3+x3 to deduce that x3+y3+z3>=3xyz.
These are two questions I am having trouble with. Can anyone help?
Ok thanks heaps for all that Rui and all your help in the past! I have another question. We didn't really cover cube roots of unity or roots of unity in general so I'm unsure about this stuff.
If 1, w1 and w2 are the cube roots of unity, prove that:
w1 = conjugate of w2 = conjugate of (w2)2 - Also, this bit is a mistake. It is equal to just \(w_2\,^2\); no conjugate.
w1 + w2 = -1
w1w2=1
Hi i need help with this q,
for a real number r, the polynomial 8x^3-4x^2-42x+45 is divisible by (x-r)^2. Find the value of r.
thanks
Hi, what are the techniques used to graph y = f(x2)?When what's inside the \(f\) gets altered there usually isn't any strategy. I just rely on common sense.
Thankyou!!
Cheers :)Hint: Just quote \( PS = ePM\), where in your case \(e = \frac35 \)
Show that cos(2kpi/7), where k=1,2,3,4,5,6 are roots of the equation 64x^6-64x^5-48x^4-48x^3+8x^2+8x+1=0I just had a brief at this and they're not. \( \cos \frac{2\pi}7 \) does not go to 0 when I plug it into that thing.
I just had a brief at this and they're not. \( \cos \frac{2\pi}7 \) does not go to 0 when I plug it into that thing.there is a k after pi
there is a k after piYeah.
Trying to help a friend. This question(2a) is ridiculous tho. I've broken it up into two integral of negative inverse cot and inverse tan but I can't work out wat to use as the separating boundary. (1 works but I don't know why)
Being trying this question for the last two days, always getting it wrong
"Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region enclosed within the circle (x-1)2 + y2 = 1 about the y axis."
from Cambridge 4u
Being trying this question for the last two days, always getting it wrong(https://i.imgur.com/qd0VaxHh.jpg)
"Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region enclosed within the circle (x-1)2 + y2 = 1 about the y axis."
from Cambridge 4u
Ahh i got most of RuiAce's solution but couldn't get the second last line of working. Thanks :)If you meant the harder-to-see red marker, basically the first integral represents the integral of an odd function and the second represents the area of a semi-circle with radius 1. (Line above is just splitting the integral up in case I misinterpreted). Let me know if anything else was hard to see or etc
Hey Rui and all maths lovers i saw this integral online and im really trying to crack it but its just too complicated. Please give it a try if you can.
Intergrate [ (ln (x) + 1/x) e^x ] dx
I need help with this question..
Of course, this also means that \( g(x) \) is concave up, but we don't really need this.
Yeah there was a part before which I weirdly proved but I think your method might be better if I can get my head around the first bit.Typo (I'll fix it now), sin x is definitely less than x.
Proved:
sinx > x
but as soon as you start rearranging...
sinx - x > 0
-(x-sinx) > 0
x-sinx < 0 (sign flips because you divide both sides by -1?)
Hey Rui (and other math students),\( y\) increasing means \( y^\prime > 0\). \(y^{\prime\prime} > 0\) means \(y^\prime\) is increasing and \(y\) is concave up. So what you had to solve was \(\frac12 x^{-1/2} > 0 \).
Can you check if I did (a - first box) correctly and help me with (b - second box)?
Consider the graph of the function y=sqrtx
Spelling typo in picture: Curve not Cruve...
Thank you so much Rui (you're very likely going to be saving my 4U life throughout the year).
I just wanted to ask you about the last section of your working where you start off as:
RHS = (given)
= (how did you get this... the pro-numerals went from n to x a bit sudden)
= MHS (given in the question)"
Not sure how to continue here...
Normally I'd try to factorise out sqrt{k+1} but it seems to complicate it even more...
Hey,That's literally just going to be \( y = f(x) \) in both cases.
any general types on drawing y=ln(e^f(x)) where a graph of f(x) is provided for you?
also drawing e^ln(f(x))
Hi!!! I'm struggling with a conics q about ellipses -
M is the midpoint of PQ where P and Q lie on x2/a2 + y2/b2 = 1. O is the centre of the ellipse. Show that the tangents at P and Q intersect on OM produced.
Thank you to anyone that helps!!
Hey Rui please help with another conic question if you're free.I'll come back to this with a proper answer in a few hours time. Basically the idea is that you start by just finding the equation of the chord of contact from (0,4), then locating the points of intersection between the chord of contact and the hyperbola
Two tangents to the hyperbola 4x^2-y^2=36 intersect at (0,4). Find the coordinates for the point(s) on the hyperbola for this to occur.
Hey Rui please help with another conic question if you're free.Let's say that the solution points are at \(\left(\alpha,\beta\right)\ \left(1\right)\). From the equation of the curve, we get:
Two tangents to the hyperbola 4x^2-y^2=36 intersect at (0,4). Find the coordinates for the point(s) on the hyperbola for this to occur.
Holy crap me, Jassycab, you broke it down so wonderfully, my only fear is that in an exam i may use the chord of contact method because its a bit simpler but forming a tangent and subbing in the values is pretty darn insightful.
Hey maths lovers another conics question if you guys have time.
Oh sweet the arg part i can see clearly it's angle made by the x and y axis to the +ve x-axis, the diagram makes it much clear. I'd like to request help for 2 more questions though, again conics.Just from memory I'm pretty sure you arrived at the best method for Q12. (I remember seeing that \(a^2m^2+b^2=k^2\) thing, or something very similar to it quite a lot.)
If un = 2n+2 + 32n+1, show that un+1 = 2un + 7*32n+1. Hence, show that un is divisible by 7 for n≥1.
How do you graph y=ef(x) when given the graph of y=f(x). I think it is simply dealing what is below x-axis and then everything else is kind of higher up than original f(x) but thats just by looking at desmos and looking at some graphs
A bit of help with complex polynomial division please.
Hi, could you plese help me out with this question?
Given that arg(z+1)=pi/6 and arg(z-1)=2pi/3, write z in the form x+iy where x and y are real numbers
Thank you!
Hi I came across this in the 2001 HSC paper and I don't know what to do and can't find solutions. Can anyone assist?Please find the solution for the latter linked in the compilation. (The first one will be added shortly.)
Hi, I'm also unsure about this question. Any help appreciated. Thankyou! :)
Mod edit: Merged. Please use the "Modify" button at times like this to refrain from chain posting.
How do I do this question?
There are 11 people and 2 circular tables, one has 5 seats and the other has 6 seats. How many arrangements are possible?
Hi, can you please explain how to do this question?
Prove that (x + y)^2 >= 4xy
Hence prove that 1/x^2 + 1/y^2 >= 4/(x^2 + y^2)
(https://i.imgur.com/CDuZWhQ.png)
\begin{align*}\therefore z &= 1 \left( \cos \frac\pi3 + i \sin \frac\pi3\right)\\ &= \frac12 + \frac{\sqrt3}{2}i\end{align*}
how would we do the question if the args didn't add up to pi/2 and we couldn't use the circle methodThis is actually a lot harder in general. Whenever you encounter questions like this, usually there's a trick or two behind it all. In general, however, you may need to brute force it and actually find the equation of the ray, before going back to 2U methods for the point of intersection.
This is actually a lot harder in general. Whenever you encounter questions like this, usually there's a trick or two behind it all. In general, however, you may need to brute force it and actually find the equation of the ray, before going back to 2U methods for the point of intersection.
:/ That sounds tricky haha. If you don't mind, could you walk me through finding the equation of the ray? e.g. if the question was like arg(z-1)=pi/4, arg(z+1)=pi/3 find arg(z)The method itself is actually not too bad. Take \( \arg (z-1) = \frac\pi4 \). The ray will just be one part of the line, through the point \( (1,0) \), with gradient \( m = \tan \frac\pi4 = 1 \). So using the point gradient form, you just have \( y = 1(x-1) \).
Also, what are some of the really common complex number qs that one must now. From looking at past trials, finding the roots of unity and factorising over complex/real fields, finding square roots and then solving a quadratic are often the easy marks at the beginning of the test. Can you think of any others, especially to do with vectors (e.g. finding locus)
Thank you!
Hi Rui and other users, I'm not too sure how to do (b) and (c).. I feel like (b) might be related to cyclic quad through constructing another circle but I haven't had much luck doing it.Also with part a), you really should invoke that RASD is a cyclic quadrilateral first (reason: the angles you specified are supplementary), before you draw the circle around the points.
Also with part a), you really should invoke that RASD is a cyclic quadrilateral first (reason: the angles you specified are supplementary), before you draw the circle around the points.
(In this case, to the right of \(DC\)).
Note that the theorem used here is essentially the converse of the "angles subtended to the circumference on the same arc are equal" theorem. It is also one of the reasons why I hate maths in focus, because they don't teach this theorem.
______________________________________________________________________
\begin{align*}\pi - angle DST &= \angle DST \tag{part ii}\\ &= \angle DAR \tag{ext. angle of cyclic quad}\\ &= \angle DSR \tag{part i}\end{align*}
Thank you for your reply. I wanted to ask about the third-to-second-last line for the solution to part (c). I don't understand how Angle DST = Angle DAR through the external angle of cyclic quad. DAR is the external angle to the cyclic quad, but DST is not one of the four corners in cyclic quad ABCD, it's kinda just floating in the middle... I think the external angle of cyclic quad means Angle DAR = Angle DCB?Typo's from doing it before I went to sleep. Fixed
Hi Rui (and other users), so this one is a bit of a challenge.. I'm having some trouble actually visualising what is descriptively told to us so if someone can include a diagram, that'd be amazing.This is what the image looks like:
Hey there guys need some help with polynomials, in the first que i get that 2 roots are complex conugates as all coefficient are real and to have 2 turning points there must be a negative gradient (thus c<0) but the rest is tough to get.What are the questions?
Ohhhh so sorry i completely forgot the ques my bad. Here they are.PDF attached. As stated, will get back to part a) of Q15 later. (I wasn't really able to use the internet at the time of writing up a response.)
And please a bit more help with another hard conics question im having trouble with.
NP is the ordinate of a point P(x1,y1) on the hyperbola x^2/a^2 - y^2/b^2=1. The tangent at p meets the x axis at T. Prove thata (OT)(ON)=a^2, where O is the origin.
oh i see what the k does it shifts the graph up and that restricts the the cubic to intersect the x axis more than once. Thank you heaps Rui deeply appreciate it even if you had no internet. The other conics que i asked is from sk patel (6E q8) textbook, take your time and thank you for helping us out.(No worries :) - Pretty much PDFs are a go-to option if I'm trying to answer it without access to a computer with internet, because I can still use the full LaTeX package. The other option is I pull out a whiteboard, which I didn't have at the time either)
I'm not sure how I'd find the gradient for this question:
Find the equation of the tangent to the curve sqrt(x)+sqrt(y)=sqrt(c) at the point P(a,b) on the curve.
Thanks Sine. I managed to successfully figure out the rest.You proved something was symmetrical about y=0 (even function) by just showing that the replacement of x with -x didn't change the given equation. You do the same thing here by showing that if you replace x with y (and also y with x), it's symmetrical about the line y=x.
Anyway, bumped into another question I wasn't sure how to do...
Prove that x^3+y^3=3xy is symmetrical about the line y=x
How do you do this?If that's not an absolute value then I actually don't know what that means. You can't have \(i\) appearing in an equality by itself.
|2iy|≤4 (Not absolute value)
The question is sketch |z-w|≤4 (w is conjugate of z)That is the absolute value then. (As mentioned above, the modulus and the absolute value mean the same thing.)
But apparently, it is -2≤y≤2
But apparently, it is -2≤y≤2Oh my bad. Technically speaking I should have left it as |y|≤2 Which solves to give -2≤y≤2 (it’s an absolute value inequality).
Find the general solutions of the 1/cos(3x)=1/sin(2x). I got up to the line attached, but I'm not sure if I've lost solutions on the way. (Still haven't figured out how to input fractions and stuff despite Rui's guide)
[tex]\frac{a}{b}[/tex]
At no point anywhere else in the proof did you attempt to 'cancel things out', so that was the only instance that was risky.
I wasn't entirely sure how to continue from that last line in order to get to the answer they were expecting.Hint: Double/Triple angles were overkill. Just use the identity \( \cos \alpha = \sin \left( \frac\pi2 - \alpha \right) \)
The answers had it as:
but technically, I only have the first two solutions...
Hey guys need some help with confusing trig questions.Hints: the second one is a quadratic in tan(x) after expanding and dividing everything by cos^2(x). The last one is just 4sin^2(2x)=1.
The question is to solve for x where 0< x < 2pi
Question 13
I need help with (c). I've put what I've found in (a) and (b) since the question says "Hence".
Question 13The extension question have multiple parts, the sin(2n+1) is part a, part b with the cot^2 is easy, however I don't know how to do part c and d which is the last two questions
Question 24 is insaneNor is it useful to do and not worth the time and effort.
Nor is it useful to do and not worth the time and effort.
A small remark on part b) though - the radius will approach infinity, and the locus ends up being a straight line. Uncoincidentally, that straight line is the perpendicular bisector of the interval joining \(z_1\) and \(z_2\).
If you're interested in the basic ideas behind the geometric method, refer to the Wikipedia page linked.
Is this question beyond the course?No.
I'm really struggling with this, but how can I express 2cis(5pi/12) - 2cis(3pi/4) in mod-arg form? I don't know how to simplify this further. Please help.It's easily doable once we decide to use compound angles but I'm very sceptic about how you had a \( \frac{5\pi}{12} \) in there. Was this the entire question? (If not, send the previous parts.)
It's easily doable once we decide to use compound angles but I'm very sceptic about how you had a \( \frac{5\pi}{12} \) in there. Was this the entire question? (If not, send the previous parts.)This is the entire question. I was able to cruise through the first two parts. For the third part I tried to simplify cis5pi/12 to normal x + iy form and do the same for the other complex number, then add them and work backwards to get the total mod-arg form but that led me down an extremely complicated path. Must have done something wrong on the way.
This is the entire question. I was able to cruise through the first two parts. For the third part I tried to simplify cis5pi/12 to normal x + iy form and do the same for the other complex number, then add them and work backwards to get the total mod-arg form but that led me down an extremely complicated path. Must have done something wrong on the way.
OHHHH. Okay that makes a lot more sense. I just
Carefully note the negative. This is because we're doing a clockwise rotation. If the rotation were anticlockwise, then we'd still use \( +\frac\pi3 \).
\begin{align*}z_2 - z_1 &= z_2 \left( \cos \frac\pi3 + i \sin \frac\pi3 \right)\\ &= 2\left( \cos \frac{5\pi}{12} + i\sin \frac{5\pi}{12} \right)\left( \cos \left(-\frac\pi3\right) + i\sin \left(-\frac\pi3\right) \right)\\ &= 2 \left( \cos \frac\pi{12} + i\sin \frac\pi{12} \right)\end{align*}
Note that there is a slight subtlety in that \(z_2-z_1\) actually points from \(z_1\) to \(z_2\). This contributes to why the rotation is clockwise.
Ohhhhh. Okay that makes a lot more sense. I guess I never saw it in terms of the rotations but kept thinking about the expression algebraically. Thank you so much!
Carefully note the negative. This is because we're doing a clockwise rotation. If the rotation were anticlockwise, then we'd still use \( +\frac\pi3 \).
\begin{align*}z_2 - z_1 &= z_2 \left( \cos \frac\pi3 + i \sin \frac\pi3 \right)\\ &= 2\left( \cos \frac{5\pi}{12} + i\sin \frac{5\pi}{12} \right)\left( \cos \left(-\frac\pi3\right) + i\sin \left(-\frac\pi3\right) \right)\\ &= 2 \left( \cos \frac\pi{12} + i\sin \frac\pi{12} \right)\end{align*}
Note that there is a slight subtlety in that \(z_2-z_1\) actually points from \(z_1\) to \(z_2\). This contributes to why the rotation is clockwise.
This is not so much a maths Q as it is an advice Q.
I have been studying mx2 for an exam on wednesday. The last 4 days i have spent 8 hours a day on time set aside to study the four topics i get tested on (complex #'s, polynomials, conic and graphs).
Nothing is clicking and i cant do even the esiest of questions (apart from in complex #'s) . I dont want to drop the subject but i know im going to fail, and i hate failing.
Does anyone have any tips on a new way to apprach study for this subject? (My test is on tuesday so i only have one more day to study for it, its also worth 30% of my grade)
Hey!
If you're doing 4U maths, that means you're good at maths. If you get the complex numbers section, then that's a really good start to understanding the rest of the 4U maths curriculum.
The fact is that 4U is totally different to every other HSC course. Usually, you'll be walking into exams knowing close to 100% of the content you'll be assessed on. Sure, you'll need to work out how exactly to answer particular questions, but for the most part you're just regurgitating.
4U is nothing like that. When you study 4U maths, you resign yourself to never being quite sure if you'll be able to answer ANY questions, or even understand what the question is getting at. It's a hard bloody subject, no doubt about it. Everyone finds it freakin' difficult, so just know that you're not alone.
Past papers, and looking at the answers to questions and trying to establish patterns for particular classes of questions, is really the best way to study in my opinion. Once you have a basic grasp of the content itself, just do a billion past papers. It sounds like you're doing that at the moment, and it doesn't feel like it's sinking in. However, I promise you that it is helping, even if only a little bit per question.
My advice at this point is to think about the TYPES of questions you'll be getting in your exam, and writing out a brief 'structure' to answer such a question. This is possible for topics like conics and graphs. Failing that, you could even just write out a list of tips and techniques you pick up from the answers to past questions. It's like developing a 'cheat sheet', which you can use to do past papers and make your life a whole lot easier.
Everyone struggles with 4U, even those ridiculously good at Maths. It sucks to feel the way you're feeling, but honestly I'm 100% sure you're better than you think you are. Just keep slugging away, and do your best to answer every question in your exam. Seriously good luck in your study!
Hey!JAKE IS ALIVE.
Hey!
If you're doing 4U maths, that means you're good at maths. If you get the complex numbers section, then that's a really good start to understanding the rest of the 4U maths curriculum.
The fact is that 4U is totally different to every other HSC course. Usually, you'll be walking into exams knowing close to 100% of the content you'll be assessed on. Sure, you'll need to work out how exactly to answer particular questions, but for the most part you're just regurgitating.
4U is nothing like that. When you study 4U maths, you resign yourself to never being quite sure if you'll be able to answer ANY questions, or even understand what the question is getting at. It's a hard bloody subject, no doubt about it. Everyone finds it freakin' difficult, so just know that you're not alone.
Past papers, and looking at the answers to questions and trying to establish patterns for particular classes of questions, is really the best way to study in my opinion. Once you have a basic grasp of the content itself, just do a billion past papers. It sounds like you're doing that at the moment, and it doesn't feel like it's sinking in. However, I promise you that it is helping, even if only a little bit per question.
My advice at this point is to think about the TYPES of questions you'll be getting in your exam, and writing out a brief 'structure' to answer such a question. This is possible for topics like conics and graphs. Failing that, you could even just write out a list of tips and techniques you pick up from the answers to past questions. It's like developing a 'cheat sheet', which you can use to do past papers and make your life a whole lot easier.
Everyone struggles with 4U, even those ridiculously good at Maths. It sucks to feel the way you're feeling, but honestly I'm 100% sure you're better than you think you are. Just keep slugging away, and do your best to answer every question in your exam. Seriously good luck in your study!
I don't understand what part b is asking forSuggestions regarding what its asking: Draw the line through the point P and the focus S. Let P' be where that line meets the ellipse again.
Hi, I'm stuck on this rates of change question:
A ferris wheel has radius 50m and the carriages move at a constant speed of 0.3m/s without stopping. Suppose a carriage is C, the base of the ferris wheel is B and the centre is O, and z = angle <BOC. The position of C can be represented by: [50sin(z),50cos(z)] i.e. coordinates x=50sin(z) and y=50cos(z). Also, the speed of C is given by the formula = sqrt((dx/dt)^2 + (dy/dt)^2).
a) Find dz/dt in radians per second.
b) How long does it take for carriage C to make one complete revolution? (in minutes)
I am having trouble visualising this question.This is just proving the reflection property.
Could I get some help with this question please?
Found another question I wasn't too sure about... Am I mean't to use double sigma? But then I wouldn't know what values to set it equal to since that's normally used when it says something like find the coefficient of x...This was in one of the older 3U papers. It has already been addressed in the corresponding compilation.
The same exact question was in some 3U paper? Hmm, not sure if I see it.. Some links keep redirecting me to solutions I wrote up for others when its under the binomial heading..From memory it was in 2001
The same exact question was in some 3U paper? Hmm, not sure if I see it.. Some links keep redirecting me to solutions I wrote up for others when its under the binomial heading..The explanation can be found here. :)
The explanation can be found here. :)
Are we required to know the proofs for the intersection of tangents and normals of a rectangular hyperbola. My teacher said yes but noted that it had never come in an exam before but my tutor said no since it was too tedious to ask as a question. Whose right? :-\That should be a proof that you can wing on the spot, not one that you should memorise.
I'm trying to teach myself a bit of Volumes and in the first exercise I've kind of stumbled upon this question (highlighted in image), I took multiple routes in order to get to the answer but I think I went wrong somewhere each time in terms of my interpretation of the solid. There are no worked solutions/examples, hence I require your help. Thank You.(https://i.imgur.com/GORo5QPl.jpg)
For conics, is it worth trying to remember the many equations? Or should they all be derived on the spot?The exam will either tell you to derive it on the spot (including the chord of contact, yes), or they'll just give you the formula.
Also, do people derive the chord of contact?
hey, need help for this mechanics questions. very close to the solution they provided but not exactly it. just need help with part d)Your answer is definitely correct.
A submarine traavels underwater with a constant driving force of mF (where m is the mass of the submarine). Water pressure exerts a resistance to its motion of v^2 per unit mass ( v is velocity and v > 0)
a) Show that the equation of motion is d^2x/dt^2 = F - v^2 --------------------------Done
b) Show that the terminal velocity is sqrt(F)-------------------------------Done
c) The cruising speed is half the terminal velocity. How far does the submarine need to travel to reach its cruising speed? (Ans: 1/2ln(4/3))------Done
d) How long does it take to reach its cruising speed? (So close!!!!!!)
answer is t = 1/(2sqrt(F)ln3) -------------------------I'm getting ln3 on numerator!!!!!!!!!!!! so triggered
Would appreciate if someone able to show their method if they get the solution, so I can compare and see where I might have gone wrong. Ty!!!
Hey, had an issue with a mechanics question. I'm probably missing a very obvious part, but I've tried many times and I keep leading myself to these weeeird integrals. Has to be much easier. Someone please help.
A bit of help with an integration que please.
hey rui, another mechanics question,(https://i.imgur.com/GDfbWZN.png)
A particle of unit mass moves in a straight line against a resistance kv^alpha (0<alpha<1), where v is the speed in m/s. It starts at a speed of 3000 m/s and this drops to 1500 m/s in one second. After a further 2 seconds the particle comes to rest. Find alpha, to 2 d.p (Answer: 0.42)
I attached where i got up to, giving about 0.86. I got 3 = (3000)^(1-alpha).Idk where i went wrong :(
OMG, thankyou so much for helping with the +C common denominator error I had. I wasn't going to sleep if I couldn't get that solution.You introduce a new constant, that depends on the old constant.
Still a little fuzzy on the +C idea though, a little embarrassing as a 4u student, but was the problem here that I was trying to multiply +C by a variable/unknown term? coz its okay to do if its just a constant like "2", or "-e^0.1" since C represents any constant anyway?
Hey rui need some help. I've been given this volumes question and also the answer but am really confused how it works. I can do the second part but i'm not sure what to do for the first.
I swear I see this equation all the time with implicit questions, but can't do it T_TThis is a lot that you're throwing in all at once. What progress have you made on every part in question. Or alternatively would you like us to focus on certain bits.
This is a lot that you're throwing in all at once. What progress have you made on every part in question. Or alternatively would you like us to focus on certain bits.
Only the first part of c I could do.. Without knowing the intercepts and all that I cannot progress with it (I need to know how to get it), once I can recognize critical points or intercepts than perhaps I can work out there rest (I wasn't being specific there sorry)
There's no point in spoon-feeding every single bit.
Harder 3U topic. Appeared in HSC 4-unit 1989 past paper.The first few parts only require 2U concepts.
I had a bit of trouble doing the questions from (c) onwards.
Hey Rui, mechanics question - vertical motion1/2k is interpreted to mean \( \frac{1}{2} k\). Did you mean 1/(2k)?
essentially, x = 1/2k (ln(g/(g-kv^2))
Terminal velocity, vT = sqrt(g/k)
I have to show distance particle must fall to reach velocity 1/2vT is x = 1/2k(ln(2/3))............I'm getting x = 1/2k(ln(4/3)) am I wrong?
1/2k is interpreted to mean \( \frac{1}{2} k\). Did you mean 1/(2k)?
Although regardless, I am also getting 4/3 assuming you do need \(v = \frac12 \sqrt{\frac{g}{k}} \). In fact \( \ln \frac23 \) would be weird, because that'd imply displacement was allowed to be negative for a 4U resisted motion question.
Why is the domain of cis(theta) restricted to between -pi and pi?It's not?
Hi,It will look like:
If this was the graph of derivative function f'(x), what would the graph of f(x) look like?
are allowed to use euler's law to prove de moivre's theorem instead of induction?No. Euler's formula is not a part of the course and therefore must never be used for a final answer.
Hey, I just have a polynomial question that I'm a bit confused about :/
I've done part (i) but can't figure out the second part
It can be proven that cos6x =32cos6^(theta) - 48cos^4(theta) + 19cos^2(Theta) - 1
(i) Find the roots of P(x) = 32x^6 -48x^4 + 18x^2 - 1SpoilerMy roots were x = cos(pi/12), cos(5pi/12), cos(7pi/12), cos (11pi/12), and positive and negative 1/root 2
(ii) Show that cos^2(pi/12)+cos^2(5pi/12)=1
HeyyHint: This is just a 3U integral, but of course in 4U you're not given the substitution. Try \( u = 1-x^2\).
Im stuck on integrating x / the root of 1- x^2, with limits at 1/2 and - 1/2
:)
Hint: This is just a 3U integral, but of course in 4U you're not given the substitution. Try \( u = 1-x^2\).
Hey Rui,\[ \int \sqrt{\frac{a+x}{b-x}}dx \text{ and similar forms are usually a little nasty.}\]
what would be the most efficient way of integrating I = (integral) sqrt((4+x)/(3-x)) dx ?
HiLooks legit.
I’ve got an integration with trig substitution question and I’ve got an answer for it but I’m not entirely confident in it and was wondering if you could check it for me please?
The question wanted the integral of the square root of 6x-x2-5 and I got
2sin-1((x-3)/2)+(((x-3)•sqrt(4-(x-3)2))/2) +C
Sorry for all the messy typing I tried uploading a photo but the photo was too big :///
HeyyThe second one should be screaming trig sub. Which I get is unpleasant, but still it should be obvious because of the \( \sqrt{a^2-x^2} \) form..
Which methods do I use to integrate (×+2) / (the root of: ×^2 - 1)
And also the root of: 9 - x^2
Hey Rui, got this question from a past paper. Not sure how to do. (attached to post)
ty for any help
Hey Rui,
Im really stuck on this recurrence formulae question. Any help would be appreciated
"If In = integral (from pi/2 to 0) of sinnxcos2x dx for n >0, show that In = [(n-1)/(n+2)] In-2"
thanks
Hey Rui,A bit of backtracking: I realised at an early onset that \( I_n = \int_0^{\pi/2} \sin^n x\,dx - \int_0^{\pi/2} \sin^{n+2} x \, dx\). But then I ignored that observation and kept trying to find smart ways to do this problem and surely enough, I was getting nowhere. So I started going back to my first idea and thought about how to handle the reduction formula for \(J_n = \int_0^{\pi/2} \sin^n x\,dx \). I recalled that for that one, I always had to split it into \( \sin x \sin^{n-1} x\) regardless. So (albeit with some reluctance) I decided to do that for this one as well. Thankfully it worked.
thanks for that - didn't realise how hard that question was
also, how did you know to change sinnx to sinn-1sinx ?
Hey Rui, back again. 4u curve sketching question here and I'd like to know how exactly they got the integral graph to approach 2 in the final part, as x approaches negative infinity. Also, I had my own kind of way of doing f(e^x), was wondering if there is method to it coz mine is kind of dodgy. Thankyou very much.I don't see at all why the integral graph should approach 2. In fact, I don't know what it should approach at all - the only thing I know is that it must approach something greater than 0.
ruii heyIt’s a bit hidden because the post is old but I actually have talked about it here.
would you happen to have a really simple proof that explains the rotation of a rectangular hyperbola? also if the hyperbola is in the form x^2-y^2=a^2, then do we just use the regular x=a/e for directrix right?
Its a bit hidden because the post is old but I actually have talked about it here.ah thank you!:)
And yeah. In particular, \(e=\sqrt2\) in this case
Hey, I'm a bit confused on how you integrate this with t-formula:This is really just a pointless grind and involves nothing clever, only heavily brute forced computations.
1dx/(3+4sin(2x))
Did you get this question off the meme? (It's final answer is actually quite disgusting.)
Did you get this question off the meme? (It's final answer is actually quite disgusting.)I did check the answer and it is disgusting, how do you guess that? (I mean you are right)- how do I approach this whole mess to get to the final answer?
Hello,The concept of Riemann sums are not involved in the MX2 course. Where is this question coming from?
Could I please get some help with part b of this question?
The concept of Riemann sums are not involved in the MX2 course. Where is this question coming from?
Help please :p(https://i.imgur.com/1Egcx0B.png)
Cheers
Hey Rui, this is a different kind of problem but I don't know where else to ask, I'm doing a software major and have encountered a math problem. Using C# on Visual Studio, and for some reason, the square root is coming as square root of -23. It seems to be doing 4 - (3 * 9), instead of 3 * 9/169 which is 27/169 (what I'm meant to get since 3 - that, would give me a value I can square root.Fairly sure programming languages aren't in the HSC altogether, so if this is something more related to uni stuff then please make a new thread in the uni section of the forum.
With inputs:
a = 8
b = 5
Formula:
perimeter = (Math.PI) * (a + b) * (((1 + (3 * ((a - b) * (a - b) / (a + b) * (a + b)) / (10 + Math.Sqrt(4 - (3 * ((a - b) * (a - b) / (a + b) * (a + b)))))))));
the square root im dealing with is on 2nd half of that mess
I know, brackets at end are crazy but there seems to be that many since no syntactical error. During runtime, I placed a breakpoint and perimeter value returned 'NaN', or '0' when variable was hovered over. what is wrong? Also, you've probably realised that there is an approximation for the perimeter of the ellipse, assuming ive copied it right...
Any help would be much appreciated, ty
Hey guys,Not sure what you mean by "tangents" at asymptotes.
how do I recognise horizontal or vertical tangents at a stat point or an asymptote, thanks
Hey, I'm having trouble recognising how to draw a guide graph or how it helps when drawing a graph. This is a question that I didnt quite understand: By investigating the behaviour of the curve y^2=x^2(x-1) at the neighbourhood of x=1 and fr large values of x sketch the curve. And is it possible to tackle this question without calculus??
I am not sure how to integrate (x +2) / (x^2 -1)You need to complete the square for the second one: \( \int \frac{1}{3x^2+8x+10}dx = \int \frac{1}{3(x+1)^2+7}dx \)
And
1/ (3x^2 + 6x + 10)
Can I get the integral of 1/3x and times 8t by the integral of 1/x and the integeral of 1/6x+10?
You need to complete the square for the second one: \( \int \frac{1}{3x^2+8x+10}dx = \int \frac{1}{3(x+1)^2+7}dx \)
\[ \int \frac{x+2}{x^2-1} dx = \int \frac{x}{x^2-1}dx + \int \frac{2}{x^2-1}dx\\ \text{The first integral goes to }\frac12 \ln |x^2-1|\\ \text{The second requires partial fractions.} \]
I don't understand what that third one is supposed to be.
What method do I use to integrate1. That is screaming partial fractions right there.
1/(x(3-x))
(1-x^2)/(1+x^2)
And (sec^2x)/(9 - tan^2 x)
I am having trouble showing that 1/9-10sin^2 = sec^2 X / 9-tan^2 XHint: After dividing top and bottom by \(\cos^2 x\), you need to use a Pythagorean identity involving \(\sec^2 x\) on the bottom
Hi there, I was wondering if there's a chance my version has been misprinted or all of them are like that, but both my HSC Mathematics Extension 1 and 2 TOPIC TESTS books from ATARNotes have significant mistakes in them. The issue is, that there are so many of them. Many of the answers answer a variant of the actual question, with different values etc. Some of them, like Q3 in the Volumes part of the MX2 tests book, have printed the question differently to the answer, such as the question deals with a function with (e^3 . x) whereas the answer answers a question that uses (e^(3x)).Can you please compile the list of mistakes you have found and send them through to my inbox?
There are several spelling errors in my other atarnotes books also.
I love these books, and the topic tests are helpful, but with this many mistakes, its embarrasing to my self and my teacher (who bought them), and stops me from reccommending these products.
\begin{align*}\int_3^8 \frac{x}{x+1-\sqrt{x+1}}dx &= \int_2^3 \frac{u^2-1}{u^2 - u} 2u\,du\\ &= \int_2^3 \frac{2u(u-1)(u+1)}{u(u-1)}\,du\\ &= \int_2^3 (2u+2)\,du\\ &= [u^2+2u]_2^3\\ &= 7\end{align*}
Hey Rui, I'm a bit confused about something in the ATAR Math ext. 2 topic notes...My bad there, went against my own rule. I need to go fix that later.
With your LIATE rule, you said when we're integrating e^xsinx dx that we would differentiate sinx and integrate e^x.
But then in your solution, you integrated sinx and then differentiate e^x.
I feel like I'm missing something, but I don't know what... Pls help!
Hei Rui, conics has been a topic I really struggle as i cant grasp what the question actually asks and the usefulness of sum of roots such as in this question: y=mx+1 is a chord of the hyperbola x^2 - 4y^2 = 16 , that passes through (0,1) for all real values of m. Find the locus of the midpoint of the chord as m varies. (This is a textbook question btw)These questions aren't hard because of the fact they're conics. What makes them hard is trying to find the shortcut that makes the problem a lot simpler, and actually tying the concepts together. This one you have here is one of the classic ones. Note that drawing a diagram is highly recommended.
Hey, I am doing a volumes question where the equation y=x^3+1 between -1 to 0 is rotated about the x-axis and I have to use the slicing method.
The volume is evaluated as:
The volume is evaluated as:ah ok so I don't even need slicing.. make sense...
\[ V = \pi \int_a^b y^2\,dx \]
The slicing formula is \( \delta V = \pi r^2 \times \text{thickness}\). The square is supposed to be there for the radius, which in your case is just \(r = x^3+1\). You should go back and revise the formula.
The volume is evaluated as:They actually need to do a bit more work in MX2. I just skipped all the details in my earlier reply.
\[ V = \pi \int_a^b y^2\,dx \]
Hey, Rui, for complex locus, I don't get when I'm meant to be putting in open circles for intersections and dotted lines...coz I thought dotted lines were only for when there is no =, hence it cannot exist there. So this question I've attached..Part i) shouldn't have any open circles for that region..? It doesn't look like there's any open circles there which should be fine, since nothing is dotted here.
Thankyou for help
Hei Rui,can you help with this induction q, its a cambridge 4u textbook question. I'm not quite sure what exactly im supposed to assume. Thank You in advance!!!Fairly sure you can continue with this question once I give you what to assume.
Hey Rui,Length, height, perpendicular height and radius are 1-dimensional concepts.
for another subject, I don't really know what area, perimeter and volume actually are categorised as. What I mean is, length, height, perpendicular height, radius are dimensions. Therefore, using these, u can get area, perimeter, volume, circumference.etc. But what are these things actually called? These results?
Thankyou in advance
Fairly sure you can continue with this question once I give you what to assume.
[tex]\text{You know that }u_{k+1} = 4u_k - 3u_{k-1}. Is this statement just subbing in k+1 into the 'fact' that shows how each term in the relation is created?, and do you choose to sub in k+1 because you know you are going to use that to prove in the final step??. Thank You!!
Yeah well you're meant to prove it holds when \(n=k+1\) so you're trying to prove that \(u_{k+1} = 2+3^{k+1} \).Fairly sure you can continue with this question once I give you what to assume.Is this statement just subbing in k+1 into the 'fact' that shows how each term in the relation is created?, and do you choose to sub in k+1 because you know you are going to use that to prove in the final step??. Thank You!!
[tex]\text{You know that }u_{k+1} = 4u_k - 3u_{k-1}.
hey Rui a bit of help please with some harder inequalities.What are the conditions on a, b and c?
Hi Rui, is relying on Cauchy's inequalitity AM>= GM too much to prove inequalities a bad habit or should i considering using proofs previously done and expanding on it to proof another result, idk if you understand by what i meant. Plus, when I state i have used it is just saying AM>= GM fine??. Thanks in advance!!Your questions at Terry Lee coaching are kinda rigged so that you have no choice but to use them. In the actual HSC you'll only need it if you prove it beforehand. Preferably, yeah, always use what you've already proven to prove new stuff.
Hey rui please some help with polynomials\begin{align*}4x^5+x&=0\\ x(4x^4+1)&=0\\ x(2x^2+i)(2x^2-i)&=0\end{align*}
solve 4x^5+x =0 over the complex field
Hey Rui,
Can you please help me with question 3 and 4?
Hey Rui,Sketch solution for Q3: That region isn't even gonna be a region. It's literally just gonna be a tiny segment of the ray. First, verify that the ray \( \arg z = \frac\pi3\) and the circle \( |z|=2\) intersect at \(z = 2 \text{cis} \frac\pi3\). (Recall from my lecture that we only need to consider the boundary, which is why we don't care too much about the interior of the circle; \(|z|<2\).)
Can you please help me with question 3 and 4?
Hi Rui, for the inequality proof question posted as an image below, I've been looking at it for a while, just cant see the actually logical reason to why the part highlighted in the box is true. Plus, when i do some questions, is it 'normal' just to sorta guess how some of your facts are bigger or smaller than another fact such that when you combine them you can get easily get to the desired result. I dont know whether that makes sense, also is it necessary to write equality iff - bla bla bla?? Thank You in advance, sorry for it being a mouthful to take in\begin{align*}a^2+2ab+b^2 &\leq 2a^2+2b^2\\ \frac{a^2+2ab+b^2}{4}&\leq \frac{a^2+b^2}{2}\\ \left( \frac{a+b}{2} \right)^2 &\leq \frac{a^2+b^2}{2}\end{align*}
Hello, having trouble with this one integration question.
The solution is:
But I am getting:
Wow thanks, I did it a stupid way that is why I was getting a constant.Whilst the question is doable by considering \(x+iy = \frac{t^2-1}{t^2+1} - \frac{2ti}{t^2+1}\), I find that it's only really viable for the equation of the locus, and becomes kinda useless when dealing with the restriction on \(t\). More or less because optimising \( f(t) = \frac{t^2-1}{t^2+1} \) and \(g(t) = \frac{-2t}{t^2+1}\) over \(0\leq t \leq 1\) is really handwavy unless we use calculus, but that's not just one, but two disgusting quotient rule bashes.
Another question:
Another question:
Thanks for the reply, I was doing this a little differently with less working out:I'll take a closer look at this soon but in the meantime is there a diagram attached to this? My hangry brain cannot visualise it right now
- Find |z| which is 1 so the locus lies on the unit circle.
- Rationalise z and split it up so it is in the form z = x + iy
- Use 0<=t<=1 with x and y to determine domain and arg(z) and then draw the locus. However when trying to find the arg I got 0<=arg(z)<= arctan(-2/0). Here was where I was stuck.
Your response is very good, really helps explain everything!
Another question if you don't mind:
(https://scontent-syd2-1.xx.fbcdn.net/v/t1.15752-9/37820053_408870752968181_1265485472084262912_n.jpg?_nc_cat=0&oh=f69107dcbd32c7ad0860122c19e1c6f6&oe=5BC59D8C)
Removed
Oh yes, sorry I should have shown the first part too, my bad.This is just a brainstorm here:
(https://imgur.com/sFy4bDZ.png)
Hi I need help with this question below on how to approach it and the logic behind the steps. Thank youThis question is simply screaming "multiplication of \(y\)-coordinates" and that's all there is to it.
Sketch y=f(x)=(x^n)e^-x for x>0, n>1
Thank you
Find the volume bounded by y = 10 and y = x + 16/x about x = −2Do you want to use cylindrical shells or washers (slicing)?
hey guys,Firstly, similar to \(y = f(|x|) \) you block out the part of the curve to the left of the \(y\)-axis. This is because regardless of what \(x\) is, \(x^2\) will always be positive, so you're always taking \(f\) of a positive number.
could someone please explain how to graph the last one? Nut quite sure how to do those types of questions!
lol figuring out how to paste an image in was such a struggle
(https://i.imgur.com/sIWXj2V.png)
Hey Rui, can you please help me with Q15a) i and iii? -I don't understand the solutions providedBecause \(P\) is the point \( \left(cp, \frac{c}{p} \right) \), the parameter at \(P\) is \(p\). That's literally it - the parameter at a point is just the parameter that represents the coordinates of that point. The reason why \(Q\) must therefore be \(-p\) is because clearly the ellipse and the hyperbola intersect at only two distinct points. So because \( \left( cp, \frac{c}{p} \right) \) corresponds to \(P\), this leaves us with \( \left( -cp, -\frac{c}{p} \right) \) corresponding to \(Q\). So clearly \(-p\) must be the parameter representing the point \(Q\).
For i) why is p a double root if the hyperbola touches the ellipse at P?
For iii) what do they mean by the parameter at the point Q? Why is it -p? And how do they know that O is the midpoint of PQ?
Thank you in advance(:
Because \(P\) is the point \( \left(cp, \frac{c}{p} \right) \), the parameter at \(P\) is \(p\). That's literally it - the parameter at a point is just the parameter that represents the coordinates of that point. The reason why \(Q\) must therefore be \(-p\) is because clearly the ellipse and the hyperbola intersect at only two distinct points. So because \( \left( cp, \frac{c}{p} \right) \) corresponds to \(P\), this leaves us with \( \left( -cp, -\frac{c}{p} \right) \) corresponding to \(Q\). So clearly \(-p\) must be the parameter representing the point \(Q\).
(It's the same as the parabola \(x^2 = 4ay\). If a point is marked \( P(2ap, ap^2)\), then the parameter at \(P\) is just \(p\).)
It is then easy to show by the midpoint formula that the midpoint of \(P\) and \(Q\) is the origin.
For example, the quadratic equation \( x^2 - 2x + 1 = 0\) has only one unique solution. (In particular, for that one it will be \(x = 1\).) That solution is also a double root of the equation.
It's quite abstract to argue formally, but essentially the idea is to generalise the intersection between a conic and a tangent line at the point of contact, to the intersection between two "touching" conics at their point of contact instead.
ahh i understand, thank youu!!Hint for iv): You want \(n!\) to appear under the power. But you know that \( n! = 1\times 2\times \dots \times n\)....................
While you're at it, can you please help me with this question as well?
bi, ii and iv)
For i) i found f'(x) and let f'(x)=0 to find the stat point, but I cant really get an answer
Hint for iv): You want \(n!\) to appear under the power. But you know that \( n! = 1\times 2\times \dots \times n\)....................
If you've correctly found the stationary point \( x = \frac{c}{n} \), you should then proceed to prove that it is a local minimum and argue that it is a global minimum. You're essentially doing the same things I did in my trial survival lecture regarding the global minimum.
ohhh thanks rui i get it :D
Can you also help me with the 2nd part of di)?
hey can someone please help me with q10I’ve attempted the question but because of its very vague wording there was very little that I could salvage from it. Please provide the final answer for me to confirm with before I post any solution.
thank you
I’ve attempted the question but because of its very vague wording there was very little that I could salvage from it. Please provide the final answer for me to confirm with before I post any solution.
there is no solutions :(
hello, can someone please help me with d) - mathematical induction question and 16a)? i cant seem to get a solution and there are no solutions providedThe induction question is already addressed in the compilation as it is a part of the 2016 paper.
That's cool, but like was looking for just the final **answer**. Anyway here is a sketch solution.(https://i.imgur.com/ZQhzONh.png)
Because the radius of the circular motion isn’t necessarily the same as the radius of the actual hemisphere that makes up the bowl.
why did you introduce another R?
Hey people, so just as I thought mechanics was going alright, good ol' fitzpatrick threw me a bit of a curveball.Your final answer should be in terms of \(U\) and \(V\). You should consider the cases where the particle is going up and going down separately. In the case of the particle going up you will have \( \ddot{x} = -g - kv^2\) and the boundary cases:
A body, projected vertically upwards with speed U, returns with a speed V. Assuming constant gravity and air resistance proportional to the square of the speed, find the total time taken.
I just find it very vague and I'm not really sure how I'm meant to go about doing this, any help would be appreciated, thanks.
Your final answer should be in terms of \(U\) and \(V\). You should consider the cases where the particle is going up and going down separately. In the case of the particle going up you will have \( \ddot{x} = -g - kv^2\) and the boundary cases:Aaaaahhh yes that makes a lot of sense, thanks for that Rui. Just out of curiosity though, I have two different formulas for integrating something with 1/(a2-x2) and the way to tell which one to use is that for one |x|<a and the other you use when |x|>a,but in this question a ended up being the square root of g/k and x was the velocity, which was just v, so how would I go about figuring out if |v| is larger than or less than the square root of g/k??
- Initially, \(t = 0\) and \(v = U\)
- At the max height, \(t = \text{what you want to find}\) and \( v = 0\).
Then when the particle is going down you will have \( \ddot{x} = g - kv^2\).
Aaaaahhh yes that makes a lot of sense, thanks for that Rui. Just out of curiosity though, I have two different formulas for integrating something with 1/(a2-x2) and the way to tell which one to use is that for one |x|<a and the other you use when |x|>a,but in this question a ended up being the square root of g/k and x was the velocity, which was just v, so how would I go about figuring out if |v| is larger than or less than the square root of g/k??You really should be using partial fractions to integrate that expression instead of jump to \( \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right| \). But to compare \(v\) to something like \(\sqrt{\frac{g}{k}} \), you can exploit the fact that \( \ddot{x} > 0\) after you change the orientation to be the more convenient one.
Are we allowed to use the standard integral formulas in the HSC exam or not because of its omission from the paper? For example the ln(x^2 + sqrroot(x^2 +/- a^2) formula.Not anymore in your actual exam. But the odds that they will examine that now are also extremely low.
hey can someone please help me with this question?
-----
let a, b be positive real numbers so that a+b=1. prove: a/(1+a) + b/(1+b) is less that or equal to 2/3
Hi Rui, just a question on volumes of solids of revolution by slices.Fairly sure if you're using slicing the big radius is just \(1\). Not \(1 - \sin y\). (Which will give you \(\delta V = \pi (1 - \sin^2 y)\delta y \))
Thank You!!!
thank you so much. but shouldn't the second equation you boxed be 1>=4ab not < ?Yeah I'll fix the typo shortly
is it ok if you can try part ii of this question? I can't seem to get the volume equation they have.
(https://scontent-syd2-1.xx.fbcdn.net/v/t1.15752-9/40622720_2119421968377060_115003800959320064_n.png?_nc_cat=0&oh=89517562bb0af26a4084b519943bc708&oe=5BFA3EDF)
I don't get how u got x =10 + (10-h)See the edit. It's the exact same method I use in the book
See the edit. It's the exact same method I use in the book
could we be asked hard simple harmonic questions and projectile questions in a 4u exam?They combined resisted motion with projectile motion back in 2003 so I wouldn't rule it out completely
100 post!
They combined resisted motion with projectile motion back in 2003 so I wouldn't rule it out completely
so its possible, but highly unlikely that they will?Yeah pretty much
Hey had this conics question I was stuck on:This is just the usual reflection property of the ellipse. You should work through Q4a of the 2009 paper to attempt this.
Hey Rui, got a conical pendulum question,It looks like I get \(\tan \alpha\) in the denominator as well.
A smooth hollow cone with semi-vertex angle 'alpha' is placed with its axis vertical and vertex down.. A particle moves in a horizontal circle on its inner surface, making n revolutions per second. Find the radius of the circle of motion.
I got r = gtan(alpha)/(4n^2pi^2) but the answer has tan(alpha) in the denominator idk how.
Thanks for help
hey Rui need some help with ext2 combinations que
Heya:)Did you mean the second part? (If so, note that the RHS of your expression has a mistake in the denominator)
from 2012 HSC paper, 16 c) iii
how do we simplify this expression
Hi, there this question is not attached but it is the very last question of the 2005 hsc..... I can do every part but I am struggling to understand the solutions for the last part of the question..... If anyone could somehow explain?See this recent add to the compilation (you will need to scroll down for the last part). The main reason I didn't address this sooner was because I wanted to cover the entire question as it is somewhat demanding, but I was also on vacation until today so I wasn't feeling bothered enough to do it all.
Hi,careful dude, for cube roots of unity w is a complex number, so w-1 cannot be one as it is not purely real, although w^3 is a real number that is one... this type of proof you need to be familiar with, and 1+w+w^2 has to be zero....
For all cube root of unity questions do I do:
w^3=1
w^3-1=0
(w-1)(w^2+w+1)=0
w-1=0 and w^2+w+1=0 equations and use these to find solutions?
Hi,
For all cube root of unity questions do I do:
w^3=1
w^3-1=0
(w-1)(w^2+w+1)=0
w-1=0 and w^2+w+1=0 equations and use these to find solutions?
Hi Rui, just need help with this polynomial question.
Ok so im a little worried doing the 4u maths past papers. Im getting around 65's at this point.Around there typically scrapes the E4 as you say.
What kind of raw marks do I need for a band 6? About 70?
Thanks.
Hey, how do you solve this integral:I’m reading this as \( \int xe^x(1+x)\,dx \) which is just \( \int (x^2+x) e^x\, dx \) and can be handled by two applications of integration by parts. Where are you getting that substitution from?
(xe^x) (1+x)
I tried doing by parts using u=x/1+x but didn't really get anywhere
I’m reading this as \( \int xe^x(1+x)\,dx \) which is just \( \int (x^2+x) e^x\, dx \) and can be handled by two applications of integration by parts. Where are you getting that substitution from?
Sorry that was completely my fault!I tried doing it with let u=1+x and it is a lot nicer than your substitution.. However integrals like e^x/x is a special property not in the syllabus... I end up integrating e^(u-1)-e^(u-1)/u by using substitution and the final answer is a property that we don't even learn in 4U
It was meant to be xe^x/(1+x)
Thanks guys! It was a question on one of the trial papers our school bought (our actual trial was made up of questions from 3~ independent papers) and our teacher briefly mentioned that the question (which was originally xe^x/(1+e^x) ) was wrong and that xe^x/(1+x) was meant to be the correct q. He might have just stuffed it up, but thank you guys for all your help!Both of them are non-elementary actually. I do vaguely recall this of this debuckle actually, but I’m fairly sure whatever the integral was supposed to be should’ve had some boundaries attached to it. Check with your teacher what the integration boundaries were because I think your integrand may have been correct...?
Both of them are non-elementary actually. I do vaguely recall this of this debuckle actually, but I’m fairly sure whatever the integral was supposed to be should’ve had some boundaries attached to it. Check with your teacher what the integration boundaries were because I think your integrand may have been correct...?
The boundaries were 1 to 0 if that helps :)Had to consult with a friend just in case and even then it still looks not doable (damn). If you ever get to see the solutions on the independents paper please keep us updated haha
Hi Rui, can you help me with part (c) of the question. Thanks in advanceUse the equaion from part (i)
Hi Rui, can you help me with part (c) of the question. Thanks in advanceYeah pretty much see above. All that it really wanted you to interpret is that when \(v = V_1\), \(F = F_1\), and also when \(v = V_2\), \(F = F_2\). Then we just do the best thing we can (i.e. cancellation) to get rid of the \(m\), \(r\) and \(\cos \alpha\).
Use the equaion from part (i)Then the friction is reversed and starts going upwards instead. Which makes sense because if the velocity is too small, the particle will start to slide down the banked track, so you wanna push it back up there.
Also I don't know how to do part (ii) so if anyone can help there that would be great..
Hi could someone pls help me w this question thankswhich parts do you need help with specifically?
which parts do you need help with specifically?
Anyway, I'll start you off with part (i), expand the expression (cisx)^5 manually
Hi could someone pls help me w this question thanks
Hey Rui can i get help with these 2 questions. One is circle geo and one is complex. Would need some help on the working out.(https://i.imgur.com/fK6kTCk.png)
Hi Rui, can I have help with this induction q: I was think assume n=k and n=k-1, but the answer's did n=k and n=k+1 just not sure.If it's just the issue of {assume n=k, n=k-1 and prove n=k+1} v.s. {assume n=k+1, n=k and prove n=k+2} either works. Perhaps the reason why they did the latter is because by doing the latter you let \(k\geq 1\), in accordance with what the question provides. Assuming the former makes the working out heaps tidier, just that you need to let \(k \geq 2\) instead.
If it's just the issue of {assume n=k, n=k-1 and prove n=k+1} v.s. {assume n=k+1, n=k and prove n=k+2} either works. Perhaps the reason why they did the latter is because by doing the latter you let \(k\geq 1\), in accordance with what the question provides. Assuming the former makes the working out heaps tidier, just that you need to let \(k \geq 2\) instead.Thats all I needed to know that you!!.
Did you need help with the computations?
(https://i.imgur.com/fK6kTCk.png)
There is a geometric way around this question as well, and it's quite interesting. But I found I had to do a tiny bit more work than this algebraic method.
hello\begin{align*} \int \frac{1}{1-\sin x}\,dx &= \int \frac{1 + \sin x}{1 - \sin^2 x}\,dx\\ &= \int \frac{1 + \sin x}{\cos^2 x}\,dx\\ &= \int \frac{1}{\cos^2 x} + \frac{\sin x}{\cos^2 x}\,dx\\ &= \int \sec^2 x + \sec x \tan x \,dx\\ &= \sec x +\tan x + C \end{align*}
can someone help pleaseee because Im not getting the right answer.
thank you
\begin{align*} \int \frac{1}{1-\sin x}\,dx &= \int \frac{1 + \sin x}{1 - \sin^2 x}\,dx\\ &= \int \frac{1 + \sin x}{\cos^2 x}\,dx\\ &= \int \frac{1}{\cos^2 x} + \frac{\sin x}{\cos^2 x}\,dx\\ &= \int \sec^2 x + \sec x \tan x \,dx\\ &= \sec x +\tan x + C \end{align*}
hello people, need some help please, Q10 from 2014 hsc
helloThey do get harder the further back you go, however anything from 2001 onwards is still relevant, so as far as you can go back to there (factoring in that you also have other subjects to study for).
I just wanted to ask, up to what year/how far back should I go with the hsc past papers since the syllabus changed and the older papers are more difficult?
thank you
sorry can someone help me with these questions please. with the volume question I'm confused because they did not specify which axis its rotated by so how am I supposed to know.Actually that is a fair call with the volumes question. That question is in theory not doable because they haven't specified the line about which the rotation is taken with respect to.
thank you
The angular velocity \( \omega\) is defined by \( \omega = \frac{\d\theta}{dt} \), i.e. the rate at which the angle the particle makes, at the origin in the positive \(x\)-axis, changes with respect to time. That derivation was examined in the 1981 paper (and from memory it should've been stated in my 4U notes book), but it essentially relies on you starting with \( x = r\cos \theta\), \(y = r\sin \theta\) and then using implicit differentiation to obtain results like \( \dot{x} = \frac{d\theta}{dt} \times -r\sin \theta = -r\omega \sin \theta\), etc.the answers for the volumes question did it around the y axis.
That locus essentially describes the arc, with endpoints at \(z_1\) and \(z_2\), going in an anticlockwise direction from \(z_1\) to \(z_2\) and not including the points \(z_1\) and \(z_2\) themselves.
the answers for the volumes question did it around the y axis.Yeah ignore that volumes question. They didn't hint at all that you were meant to do that.
1981 😵😵 does that mean its highly unlikely that they will ask this question, ever? and I don't understand what is the point of that question. how I sone supposed to know wtf they are talking about. I did not do that in school (coz I have a crappy teacher) so when I first saw it I was like wtffffff
for that locus that is what I did. but the answers had something weird
Their diagram looks like it basically gave the proof for the locus. The reason why that locus gives the arc is because if you look at the diagram, using the exterior angle of a triangle we have \( \arg(z-z_1) = \beta + \arg(z-z_2)\). Using \( \arg z - \arg w = \arg \frac{z}{w}\), this rearranges to give the equation you started off with. The arc gets traced out because you can move \(z\) anywhere along that arc, and the same proof is still valid.I don't understand what you mean :(
I don't understand what you mean :((Yep, the angle would be \(\beta\).)
why is the "circle" midway of the quadrant and not touching the x axis like the normal ones like this (but ofc the angle wouldn't be 90, it would be beta)?
(Yep, the angle would be \(\beta\).)
In that particular example, you didn't just take an arbitrary choice of \(z_1\) and \(z_2\). For your original question, \(z_1\) and \(z_2\) could've been anywhere you wanted it to. But for your newer question, you had specifically let \(z_1 = 2\), which represents the point \( (2,0)\) on the Argand plane, and you've also specifically let \(z_2 = -2\), which represents the point \( (-2,0)\) on the Argand plane. Both of these points specifically lie on the \(x\)-axis.
So pretty much, that newer example isn't a "normal" one. It's just a special case.
does that mean if I drew the original question like the answer I just sent, I would get it right (but again not 90, beta)?In theory, I would give you the marks for it.
Hey, I'm new here
If what is
and why?
In theory, I would give you the marks for it.
Although in practice, usually when you aren't told anything about what \(z_1\) and \(z_2\) are, I like to keep my answer 'as arbitrary as possible' in a sense. Placing \(z_1\) and \(z_2\) in places that almost look like you're just chucking them at random places in thin air feels less 'restricted' in a sense, as opposed to doing what that other one did. Copying what they did can unintentionally give the impression that \(z_1\) and \(z_2\) have to lie on the \(x\)-axis no matter what, when really they don't. So whilst it's technically not incorrect to do it that way, it probably isn't the best habit to go towards.
Especially since, potentially in the exam they could give you something like \( \arg \left( \frac{z-3+4i}{z+2-i} \right) = \frac\pi3\) instead.
Let w be a non real cube of unity. Show the other non-real cube root is w^2.For cube root of unity, we have \(z^3 - 1 = 0\), giving us \(z^3 = 1\). So clearly \(z = 1\) is a root. Note that \(z^3 - 1\) can be factorised in the form: \((z - 1)(z^2 + z + 1)\). Finding the roots of the irreducible quadratic is fairly straight forward. We have:
What would be the best method for this?
Hi Rui, with the following question attached, can you please explain why you do the part highlighted and its significance. Thank You!!!With multiple zeroes, we know that at least \(P(x) = P'(x) = 0\). To show that there are no multiple zeroes, we essentially want to show that for any \(x\), \(P(x) = 0 \neq P'(x)\). Notice that
For cube root of unity, we have \(z^3 - 1 = 0\), giving us \(z^3 = 1\). So clearly \(z = 1\) is a root. Note that \(z^3 - 1\) can be factorised in the form: \((z - 1)(z^2 + z + 1)\). Finding the roots of the irreducible quadratic is fairly straight forward. We have:
\[ \begin{align*}z &= \frac{-1 \pm \sqrt{3}i}{2}\end{align*}\]
Taking \(\omega\) to be \(\frac{-1 + \sqrt{3}i}{2}\) and squaring it gives us:
\[ \begin{align*}\omega^2 &= \left(\frac{-1 + \sqrt{3}i}{2}\right)^2 \\ &= \frac{1 - 2\sqrt{3}i - 3}{4} \\ &= \frac{-1 - \sqrt{3}i}{2}\end{align*}\]
which is the other cube root of unity.
Hey, I forgot to mention that the next part of the question was that using the above, prove w^2 + w + 1=0
So would I still be able to use this method and then requote it for part ii?
Thanks!
Hey so uuuhh I need a lil help with question 7b)ii) from 2011 paper.With the screenshots being too large, when that happens you'll need to upload it to an image hosting website like imgur and extract the link instead.
I got part i) easily, it was just a subsitution but then when I was looking at the solutions at the end of part i) they “relabelled” u as x so they could use it in their solution for part ii). Why is it that they can just relabel it as x straight after going through that whole process of substiting u=4-x?? Btw I tried to post screenshots of the question I was talking about but it said it was too large or something?
hey rui plz need some help, a friend gave me and some other mates this 2u que but I find it really hard.Yeah I know the source of this question. I've tried to avoid their own solution here though:
When i first graphed x^2+xy+y^2=0, on desmos, it was just a dot on the origin.
So i just tried a blodge method by letting y=1 and used quadratic formula to get x=(1+sqrt[3] i)/2 and plugged in both values of x&y in the other eq.
After that i got it simplified to 2cos(2015pi/3)=1
But im not sure if this is a full proof method. Plz give it a try.
Hi Rui, this was a question posted on here a few days ago. I'm not sure how to do part c of it, i know its got to do with product and sum of roots but just cant grasp wats happening. Thank You!!
Hi guys, can anyone explain how to sketch im(z)=|z|. Thanks
What are some good time management strategies for the exam? (e.g. how long should I spend on each question, at what point should I move on from a question if it takes too long, what happens if I can't crack Q16)For me, 2 minutes blankly staring at a question is enough for me to say screw it for now. Or occasionally even 1 minute is too much. On the other hand, if it's taken me more than 2 minutes but it's because I've been constantly writing, I'd probably say 5 minutes of writing for one part is a healthy cut-off point.
hey guys some help here plzNote that questions before 2001 do not reflect well the difficulty of the current HSC exams. Here are sketch solutions.
what are "dividers" in probability?Glad it did :)
also thank you for the good luck story on ig. it made me feel calmer :)
Glad it did :)
I've rarely seen the word "divider" come up in probability. Which question did you get that from?
the question was "there are 10 coins to be placed in 4 boxes with max of 4 in each. find the number of ways this can happen"Oh that. Typically that type of question won't appear in the HSC courses, but I have seen it happen. The technique is known as "stars and bars", and some people rename it to "dots and dividers" which basically means the same thing.
and when I asked my friend she used "dividers" so what she did was there are 10 coins and 3 dividers. so 13 objects in total. so it'll be 13!, but since the dividers are identical you do 13!/3! but also the coins are identical so your final answer will be 13!/3!10!
and I don't understand what she did
thank you
However, and this is I'd say a huge however and why I don't anticipate this question to be in the exam. The fact that a max of 4 coins can be placed into each box doubles the difficulty of the question. The problem with the \( \frac{13!}{3!10!} \) is that this doesn't cater for that restriction. For a counterexample, an arrangement that would have been counted could've been this: X | X | X | X X X X X X X (1 in the first three boxes, but 7 in the last).
The usual approach to such a problem is to apply the inclusion exclusion principle on the complement. This gives \( \frac{13!}{3! 10!}\) minus number of arrangements when at least one box has 5 in there plus number of arrangements when at least two boxes have 5 in there. But of course, at this point we've gone beyond the scope of the HSC.
Remark: It turns out that if at least 5 are in any one box, then this becomes \( \frac{8!}{3!5!} \) and if there's at least 5 in any two boxes, there's only \( \frac{3!}{3!0!}=1\) case.
Also, I'd recommend not thinking too hard about this the morning of the exam either.I know I shouldn't be doing math, but yesterday after bio exam I came home sooo exhausted and couldn't do anything because I was so tired which is making me stressed because I feel like I haven't done enough questions, considering the difficulty of the exams !!!!!!!!!
Hi,there is no question attached
Apparently this is a De Moivre’s theorem question but I can’t seem to get this. Please just give me a hint so I can try it myself. I have tried everything. Also, can you tell me how to approach such questions where it says From this, deduce this. I.e am I allowed to work with the second line, the one we are trying to prove?
Thank you!!!
I don't get what u meant by doing the 7 in one box and the "inclusion exclusion principle"The other one - simply put:
also, there was a similar question that said
"10 people arrive at the airport and there are only 4 counters opened. how many ways can 10 people line up in 4 lane queue?"
and my other friend did since there are 10 people so 10!. and there are 13 spots to put 3 dividers which divides them people into the 4 counters. so she just did 13c3 * 10!
and since its practically the same question as the coins, they both give very different answers. why?
I know I shouldn't be doing math, but yesterday after bio exam I came home sooo exhausted and couldn't do anything because I was so tired which is making me stressed because I feel like I haven't done enough questions, considering the difficulty of the exams !!!!!!!!!
Essentially, the inclusion exclusion principle is the reason why I would not worry about this question, because it is not an HSC level technique.thank god!!!!
thank god!!!!For the total outcomes, we essentially throw out all the restrictions.
also, I was looking at your trial handout (thank you for that lecture btw it was an absolute lifesaver!!!! and haha thanks for the chocolates :) ) and I know this is dumb but I can't figure out how you got 4^6 I understood it at the time but now I can't remember what you said
For the total outcomes, we essentially throw out all the restrictions.
Without restriction, each of the 6 people can choose any of the 4 rooms to stay in. So we have \(4\times4\times4\times4\times4\times4\) (one for each of the six people), which is just \(4^6\)
(Haha glad you enjoyed the chocolate, it's become a known thing at my lectures ;))
also should I take a math set, like protractor, compass or nahLike, sure if you want. It helps to draw circles for circle geometry problems where you need to copy the diagram I guess
From,You should only assume stuff from the first equation. You can work with one side at a time for what you wish to prove, but that probably was not the intended approach of the question since it says "deduce".
Z^6 +1= (z^2 -2zcospi/6 +1)(z^2 -2zcospi/2 +1)(z^2 -2zcos5pi/6 +1)
Deduce:
Cos3theta= 4(Costheta-cospi/6)(Costheta-cospi/2)(Costheta-cos5pi/6)
Hi, I wasn’t too sure how to do these sort of questions where it says from this deduce this. Like what are my boundaries. For example, am I allowed to work with the second equation (the one we have to deduce) or not?
Also apparently this is a De Moivres theorem question but I can’t seem to figure out why. If you could please give me a hint on where to start that would be great.
Thank you!!!!!
Thank you so much!!! Just got the answer by letting z=cisthtea. I just had some other questions. How did you know to divide by z^3, did it have soething to do with the cos3theta? Also, is there a general approach to these questions or does it just take experience?Yeah. I saw that \( \cos \frac\pi6\) and the other three were already there so I wanted to leave them untouched. My initial idea was to somehow kill off the \(z^2+1\) with \(z=-i\sin\theta\) or something, but then I decided nope there's gonna be too many sine's floating about.
Hey! can someone pls help me with this question
thank you!!
Hello, could someone help me clarify something.Tbh, that is such a weird question. Normally in the HSC they'd let you assume it.
This is the question:
Show that the following equations represent hyperbolas and find their centre, focu and the equation of directrices and asymptotes.
1) (x-3)^2/64 - (y+1)^2/36 = 1
For the first part of the question, in ‘showing that the equation is a hyperbola’ do I show that e > 1 or something else like |PS-PS’|=2a
Thanks :)
Tbh, that is such a weird question. Normally in the HSC they'd let you assume it.
But yeah, you would either prove that \( \frac{PS}{PM} > 1 \) for a suitable choice of \(S\) and \(M\), or alternatively \( |PS - PS^\prime| = 2a\) for a suitable choice of \(S\) and \(S^\prime\) (well, and \(a\)).
If I were the examiner, I would've been happy with the fact that it is a translation of the curve \( \frac{x^2}{64} - \frac{y^2}{36} = 1\), which is clearly a hyperbola as it is a course assumption. But those methods you hinted are the more formal approaches. Where did this question come from?
Tbh, that is such a weird question. Normally in the HSC they'd let you assume it.
But yeah, you would either prove that \( \frac{PS}{PM} > 1 \) for a suitable choice of \(S\) and \(M\), or alternatively \( |PS - PS^\prime| = 2a\) for a suitable choice of \(S\) and \(S^\prime\) (well, and \(a\)).
If I were the examiner, I would've been happy with the fact that it is a translation of the curve \( \frac{x^2}{64} - \frac{y^2}{36} = 1\), which is clearly a hyperbola as it is a course assumption. But those methods you hinted are the more formal approaches. Where did this question come from?
wouldn't you show that it is in the formIt's not of that form because it's not centred at the origin.
xx yy
--- - ---- = 1
aa bb
because that is how my teacher told me to do it
Hi i need help with this q, if z is a complex number, z/z-i then show that it is imaginary. I let z=x+iy and tried doing it from there but i get stuck when i let re(z) =0 , thanks
Help please :-\
P and Q are variable points on the rectangular hyperbola xy = c2. The tangents at P and Q meet at R. If PQ passes through the point (a,0), find the equation of the locus of R.
Okay,
So I've got the equation of the tangent at P:
x=2cp-p2y
and the tangent at Q
x=2cq-q2y
Simultaneously solve them and R is at:
x= 2cp/(p+q)
y=2c/(p+q)
And the equation of the chord PQ is:
x+pqy=c(p+q), but it passes through (a,0), so we get this:
a=c(p+q)
Now what do I do. I would usually try to form an equation by getting rid of the parameters (p+q), but then pq is left behind so what do I do?
I understand how how to get the roots but not sure how to get to the area and perimeter.
Hello, just a question on 4U Cambridge, can anyone please give me a detailed answer for question 7 and 8 in the link? Thanks :)
Ah, these questions. I remember when I first saw them; it was not a pleasant discovery. I hope the solutions help.Thanks for the help! :)
Hi,
Could I please get some help with these two maths induction questions?
For all positive intergers n, prove by maths induction that:
1) |z1z2...zn| = |z1| |z2| ... |zn|
2) arg(z1z2...zn) = arg(z1) + arg(z2) + ... + arg(zn)
Thank you!
Hey! Let's roll with (1), I'll assume you could prove it for \(n=1\) (it is self apparent), but we'll also want to prove it for \(n=2\) to use later.
We can prove this by letting \(z_1=x_1+iy_1\) and \(z_2=x_2+iy_2\), and substituting.
Find each of those moduli (expand out the LHS) and they will be the same. So you've proved the case for \(n=2\).
Now, the induction assumption for (1) is:
Now let's go with \(n=k+1\) and see what we get:
Let's break this into two pieces, \(a=z_1z_2...z_k\) and \(b=z_{k+1}\). We do this because we've already proven that \(|z_1z_2|=|z_1||z_2|\) above! So by splitting it in two, we can automatically therefore say that:
Now we use our induction assumption!!
And we are done!! Conclude as usual, and you're all set :) the second question is the exact same process, use the result for \(n=2\) to help you generalise it for \(n=k+1\) ;D
Thank you so much! I've done question no. 2 now using the method, and I just wanted to check if my working (esp. for S1) was the most efficient one?
Thanks!
Thank you so much! I've done question no. 2 now using the method, and I just wanted to check if my working (esp. for S1) was the most efficient one?
Thanks!
Hey! Nicely done - Definitely looks cumbersome, but nothing immediately jumps to me as easier. If anyone reading this spots anything, give a shout ;DThe quicker way was to let \(z = r \operatorname{cis} \theta\) instead of go for the Cartesian approach. That reduces the working out by a tiny bit for the modulus, but makes the argument just as fast as the modulus to deal with
Hello,I don't see any attachment?
Can anyone please give me a detailed answer for 4 d) and 4 f) ASAP? Thanks 😁
I don't see any attachment?Sorry, my bad. Here's the attachment of the questions
Thanks for the solutions :)
\begin{align*} \overline{5z} &= \overline{5(x+iy)}\\ &= \overline{5x+5iy}\\ &= 5x-5iy\\ &= 5(x-iy)\\ &= 5\overline{z} \end{align*}
_________________________________________________
\begin{align*} \overline{\left(\frac{1}{z}\right)}&= \overline{\left( \frac{1}{x+iy} \right)}\\ &= \overline{\frac{x-iy}{x^2+y^2}}\\ &= \frac{x+iy}{x^2+y^2}\\ &= \frac{1}{x-iy}\\ &= \frac{1}{\overline{z}} \end{align*}
The steps used are only common techniques in the MX2 course. If you require a step to be expanded on, please state that step.
Hello,I will only do the first one mentioned as they both follow the exact same procedure.
Can anyone please give me a detailed answer for 9 d) and 9 c) ASAP? Thanks 😁
Heyyy. Currently stuck on a question and thought I'd ask some geniuses. Also does one even know the first step to take in questions like these?This particular question is more of a trick. It relies on whether you've seen ratios of lengths before. If you haven't, then the intuition to use a geometric approach becomes far less obvious.
haha thanks heaps guys!
If:
Prove that:
Hello,Hey, don't usually go through the MX2 thread as I didn't do MX2 but considering no one's answered it, I'll give them a crack!
Can anyone please give me a detailed answer for Q4 to Q9 ASAP (I have included some of my working out, Q8 and Q9 I cannot do at all)?. Thanks :)
Hey, don't usually go through the MX2 thread as I didn't do MX2 but considering no one's answered it, I'll give them a crack!
I'll provide hints for you and then a full solution to some of the questions in the spoiler! :)
Q4
Hint 1: If you want to prove that a triangle is right angled, then you may like to use Pythagoras' Theorem! Perhaps, try showing that:
\[ \left|OQ\right|^2 = \left|OP\right|^2 + \left|PQ\right|^2.\] Thank you for the help kind sir!Q4 - full solutionAs hinted, we'll try to employ Pythagoras' Theorem.
First note that given: \(z = x + iy\), we have \(\left|OP\right| = \sqrt{x^2 + y^2}\) and:
\[ \boxed{\left|OP\right|^2 = x^2 + y^2}\]
We also have:
\[\begin{align*}z + iz &= (x + iy) + i(x + iy) \\ &= (x - y) + i(x + y).\end{align*}\]
So we get:
\[ \boxed{\left|OQ\right|^2 = (x - y)^2 + (x + y)^2}\]
Finally, the line segment \(PQ\) is just \(Q - P = ((x - y) - x) + i((x + y) - y) = -y + ix.\)
So we get:
\[ \boxed{\left|PQ\right|^2 = y^2 + x^2}\]
Now, we know that:
\[ \begin{align*}\left|OQ\right|^2 &= (x - y)^2 + (x + y)^2 \\ &= x^2 - 2xy + y^2 + x^2 + 2xy + y^2 \\ &= (x^2 + y^2) + (x^2 + y^2) \\ &= \left|OP\right|^2 + \left|PQ\right|^2.\end{align*}\]
Q5Q5 - full solutionLet:
\[z_1 = r\text{cis}\theta\] and \[z_2 = r\text{cis}\left(\theta + \frac{\pi}{3}\right)\]
We then have the following (using De Moivre's Theorem):
\[ z_1^2 = r^2\text{cis}(2\theta), \quad z_2^2 = r^2\text{cis}\left(2\theta + \frac{2\pi}{3}\right)\]
Adding these two gives us:
\[ \begin{align*}z_1^2 + z_2^2 &= r^2\text{cis}(2\theta) + r^2\text{cis}\left(2\theta + \frac{2\pi}{3}\right) \\ &= r^2\left(\text{cis}(2\theta) + \text{cis}\left(2\theta + \frac{2\pi}{3}\right)\right) \\ &= r^2\text{cis}\left(2\theta + \frac{\pi}{3}\right) \\ &= r^2\text{cis}\left(\theta + \left(\theta + \frac{\pi}{3}\right)\right) \\ &= r\text{cis}\theta \times r\text{cis}\left(\theta + \frac{\pi}{3}\right) \\ &= z_1z_2\end{align*}\]
Q6
Hint 1: You may like to take a geometric approach :) Try drawing up \(z_1\) on an Argand diagram and from the nose of \(z_1\), draw \(z_2\). The result gives us: \(z_1 + z_2\).
Hint 2: The triangle inequality (see Q8) states that:
\[ \left|x+y\right| \leq \left|x\right| + \left|y\right|\]Q6 - full solutionWe note that:
\[ \left|z_1\right| = \left|(z_1 - z_2) + z_2\right|\]
And by the triangle inequality, we see that:
\[ \left|z_1\right| \leq \left|z_1 - z_2\right| + \left|z_2\right|\]
By a similar argument, the same can be said for \(z_2\). That is:
\[ \left|z_2\right| \leq \left|z_1 - z_2\right| + \left|z_1\right|\]
We see that:
\[ \left|z_1\right| - \left|z_2\right| \leq \left|z_1 - z_2\right|, \quad \left|z_2\right| - \left|z_1\right| \leq \left|z_1 - z_2\right|\]
which implies that:
\[ \left|\left|z_1\right|-\left|z_2\right|\right| \leq \left|z_1 - z_2\right|.\]
Q8
Hint: Best bet would be to prove it by induction.Q8 - full solutionBase case is just the simple triangle inequality.
When \(n = k\), we have:
\[ \left|z_1 + z_2 + z_3 + \dots + z_k\right| \leq \left|z_1\right| + \left|z_2\right| + \left|z_3\right| + \dots + \left|z_k\right|.\]
When \(n = k+1\), we have:
\[ \begin{align*}\left|z_1 + z_2 + z_3 + \dots + z_k + z_{k+1}\right| &\leq \left|z_1 + z_2 + z_3 + \dots + z_k\right| + \left|z_{k+1}\right| \tag{by our triangle inequality} \\ &\leq \left|z_1\right| + \left|z_2\right| + \left|z_3\right| + \dots + \left|z_{k+1}\right|\end{align*}\]
which completes the proof.
Q9Q9 - full solutionIf the triangles \(OBA\) and \(OID\) are similar, then it follows that:
\[ \frac{OD}{OA} = \frac{OI}{OB}\]
and thus, we have:
\[ \frac{OD}{OI} = \frac{OA}{OB} = \frac{z_1}{z_2} = OD.\]
Hi there!I'm not a fan of tip-to-tail for half of the complex numbers proofs problems I see. The parallelogram makes it more clear that the main diagonal of the parallelogram is \(z+w\), but then illustrates how the other diagonal represents \(z-w\). It is also the more formal way of doing vector addition.
I was reading through Rui's notes when it's mentioned that the parallelogram method of adding complex number vectors is better than the 'tip to tail' method (which should be "avoided where possible in the MX2 course"). I was wondering what was the reason for this was, given my teacher's strong advocacy for the latter method.
Thanks a bunch!
Thanks Rui! Definitely clears things up. As a follow-up question, I was wondering how you'd go about approaching/ setting out an answer for a question like the one in 2017 - Q13(e)This question doesn't use vector addition directly but rather draws on a key consequence (in fact namely the \(z-w\) issue). Recognising that \(\overrightarrow{DC}\) represents \(c-d\) and etc. originate from the parallelogram, but it can be something that people just memorise.
(https://i.ibb.co/GV2YZ0m/Capture.png)
"By using vectors" - does that mean demonstrate geometrically?
I feel like we'll get to know each other a lot on this forum haha.It is quite tedious.
I'm unsure if I'm way too tired or if this is hard:Find the locus of z ifa) is purely real
b) is purely imaginary
Hello,Please be mindful when using comments like this. It creates unnecessary pressure to rush people who are taking their time to help out.
Can anyone please give me a detailed answer for Q8 ASAP (I have included some of my working out)? Thanks :)
Please be mindful when using comments like this. It creates unnecessary pressure to rush people who are taking their time to help out.Sorry about the ASAP part Rui. Here is question 8 (although I have already written it down on the paper).
There's no question attached.
Sorry about the ASAP part Rui. Here is question 8 (although I have already written it down on the paper).
Thank you for the help Rui! Much appreciated :)
Hi, I was wondering how one would go about proving that:This particular one was addressed in this thread
Sinx+sin2x+sin3x+......+sin(nx)=sin(x(n+1)/2)*sin(nx/2) / sin(x/2)
Knowing that De Moivre's theorem must be used.
Hi :)The simple fix to make it rigorous is to impose the condition that \( -\frac\pi2 \leq \theta \leq \frac\pi2 \). This covers all values in the range of \(y=\sin \theta\), that is \(-1\leq y\leq 1\).
I have a problem that I can't seem to resolve in my head. Perhaps someone knows the answer. I want to compute the integral Of course, to an extension 2 student this is not too difficult, you just use the substitution However, when you eventually get to the line you encounter a problem, because However, everywhere I look I find the solution continue as Is it correct to continue like this, ignoring the absolute values? If yes, why? If not, how can we make it rigorous?
Edit: You made a typo I think, but it was clear enough to me that you substituted in \(\boxed{x=\sin\theta}\)
Hello, I have a few problems that are bugging me. I have attempted some of my working out, but for a couple of the questions I don't know where to go. Can anyone please help me with Q3, Q4, Q8 and Q10? It would be greatly appreciated :)Starting points:
Starting points:Thanks for the help kind sir! :)
Q3 and Q4 both represent standard loci that you are expected to memorise or identify on the spot.
- \( |z-1| = |z-i| \) is the perpendicular bisector of the points representing \(1\) and \(i\) (i.e. the perpendicular bisector of \((0,1) \) and \((1,0)\) Factoring in the inequality, we want the region cut off by this line that includes the point \((1,0)\).
- \( |z-2-2i| \leq 1 \) is the circle centred at \( (2,2)\) with radius \(1\), and the inequality suggests that we want its interior.
- \( \arg(z+3) = \frac\pi3 \) is the ray starting at \(z=-3\) (i.e. the point \((-3,0)\), inclined at an angle of \( \frac\pi3\) and excluding the point \((-3,0)\) itself. The equation of the ray itself is not of interest.
Note however that Q8 is a special locus. You should watch Eddie Woo's video to understand this locus if you were not introduced to it at school to get started on this question.
I will only do Q10 for now as this is still a bombardment of questions, however that particular one is not something taught at school and expected to be memorised.
As usual, keep in mind that we aim to eliminate any parameters when handling locus problems this way. Here, the parameter will be \(\theta\) for part a), and \(r\) for part b).
Hi,If your final result for 2U and 3U were literally 100 and 50 (100 if you do end up continuing 4U) then it becomes pretty hard to say. Because of course something like a 95 in 4U would be the minimum to have a chance of topping off 100 in 2U.
I have a question about doing 4U Maths for my HSC.
I will graduate in 2019, so I've already started by Year 12 year. I also accelerated Maths 2U and 3U, in which I received full marks for both subjects. I also state ranked in 2U Maths (in the top 3). I love maths and am currently doing 4U maths, but I don't see myself getting better than a mid-top-band for 4U maths. So many people have been telling me that it's crazy for me to do 4U maths, because it will make my 2U marks redundant. So what should I do? Will state ranking in 2U help increase my ATAR, in comparison than, say, getting a 95 in 4U maths? What should I do to get the highest ATAR - should I drop 4U maths to keep my 2U marks, or keep 4U maths and ditch my 2U marks? To be honest, my aim is to get the highest ATAR, not to gain skills and learn more (sorry!).
I am doing 14 units right now, including my 3 units of maths, which I have already completed.
Any advice is greatly appreciated.
Thanks
Question from Terry Lee's textbook:Well, most likely in the exam they'd give you a diagram so that you only have to consider one of the endpoints of one particular latus rectum.
The normal at an end P(x1,y1) of the latus rectum of the ellipse x^2/a^2+y^2/b^2=1 meets the y-axis in M, and PN is the abscissa of P (i.e. PN is perpendicular to the y-axis). Prove that MN=a.
I know how to do this but I'm getting confused whether to use the ± sign. For example when finding the x coordinate of P it would be ±ae right? But the answers only use ae. Same when finding the y coordinate of P, which I substitute ae into x^2/a^2+y^2/b^2=1 to find the y value which would be ±b2/a. But again, the answers only use b2/a. Please help :) sorry to disturb your holidays
I was so happy to be done with 4u maths, now starting at UNSW I found out just some time ago I need to do math1131 or math1141 which covers 4u in apparently like a month. Ah RIP to meMaybe post this under the UNSW section next time.
Just in relation to the ATAR Notes MX2 Topic Tests, I came across this question in Complex Numbers:That's interesting; I was surprised to see that that method is not there. Your method is certainly also valid.
"The complex number z satisfies |z-i| = 1. What is the greatest distance that z can be from the point (1,0) in the Argand diagram?"
Intuitively, I thought that the line from (1,0) to the circle |z-i| = 1 with greatest length must pass through the centre of the circle, since the diameter is the longest 'chord' in a circle.
I got the right answer, but I was just wondering whether this was a valid conclusion or not? Would this kind of logic occur in every case or was I just lucky in getting the answer right? The worked solutions have two alternative methods, none of which use this reasoning.
Hi\[ \text{Seeing as though the quadratic in the denominator is irreducible}\\ \text{i.e. it cannot be factored any further without complex numbers}\\ \text{some kind of splitting may be necessary.} \]
I'm fairly new to integration and got a bit stuck with this question type: ∫(x-1)/(x^2+1) dx. Should I be integrating these question with partial fractions or can I do this with inverse trig?
PS: This is my first post on the forum!
Thank you.In the future, please use brackets to indicate what's going on. When reading your gradients, the gradient of PS1 is interpreted as \( m_{PS_1}= \frac{y_0}{x_0} - 4 \), not the intended \( m_{PS_1} = \frac{y_0}{x_0 - 4} \)
I'm going through a topic test and got this question wrong:
The line PN is the normal to the ellipse x2/25 + y2/9 = 1 at P(x0, y0) and S1 and S2 are the foci of the ellipse. Angle NPS1 = alpha and Angle NPS2 = beta. Show that alpha = beta.
(https://i.imgur.com/TJYMg5T.jpg)
I've been doing this question for so long and can't come up with an answer. My teacher said to use the angle between two lines formula. I found that mPN= 25y0/9xo and mPS1=y0/x0 - 4 and mPS2=y0/x0 + 4. But it doesn't work out.
Hello,This question came from the 2016 HSC. Fairly sure it was Q16. Should be in the compilation
I'm having trouble with this question in the ink below. I tried to use an argument approach, but I was stuck midway. Can anyone please help me with this question? Thanks :)
Hi, could someone help me solve part (iii) of this question:Hint: The whole point of the previous part was to show that \(N\) actually lies on the circle. Does this diagram seem to evoke the alternate segment theorem?
(https://i.gyazo.com/d211669d58aa2d3139e74bf1e0574792.png)
I know how to show the angle of inclination between the normal and semi-minor axis of the ellipse is equal, through creating an isosceles triangle. But I don't know how to show the tangents to the circle and the ellipse at the point of intersection is equal to the angle of inclination between the normal.
Hey! My teacher and I are completely stuck on this Polynomials question. We can complete part i), but part ii) & iiI) is a mystery.\begin{align*} x^4+x^3+x^2+x+1&= x^2(x^2+x+1+x^{-1}+x^{-2})\\ &= x^2 (\omega^2-2 + \omega + 1)\\ &= x^2(\omega^2 + \omega - 1)\end{align*}
If w=x+x^(-1)
i)
ii)
iii) Show that the roots of are the four complex roots of and deduce that
\[ \text{Hence }x=\operatorname{cis} \left( \pm \frac{2\pi}{5} \right)\text{ must be the two roots of}\\ x^2 + \frac12 x (1-\sqrt5) + 1 = 0\\ \text{and further by the sum of roots,}\\ 2\cos \frac{2\pi}{5} = \frac{\sqrt5 - 1}{2} \implies \boxed{\cos \frac{2\pi}{5} = \frac{\sqrt5 - 1}{4}}\\ \text{as required.}\]
\[ \text{Similarly }2\cos \frac{4\pi}{5} = -\frac{ \sqrt5 + 1}{2}\\ \text{However since }\cos (\pi - x) = -\cos x\text{, we then have}\\ -2 \cos \frac\pi5 = \frac{\sqrt5+1}{2} \implies \boxed{\cos \frac\pi5 = \frac{\sqrt5 + 1}{2}}\\ \text{as required.}\]
Can someone please help with this question?\[ \text{Because of }\triangle OPQ\text{ being isosceles}\\ \text{it follows that }\overrightarrow{OQ}\text{ must be some }\frac\pi3\text{-rotation of }\overrightarrow{OP}. \]
On an Argand diagram the points P and Q represent the numbers z1 and z2 respectively. OPQ is an equilateral triangle. Show that z12 + z22 = z1z2
Thank you!
That one doesn't use by parts. (Any integral that can be converted to \(I_n = \int \tan^n u \,du\) through substitution doesn't use by parts either.)
\begin{align*} I_n&= \int \tan^n x\,dx\\ &= \int \tan^{n-2}x\tan^2x\,dx\\ &= \int \tan^{n-2}x\sec^2 x\,dx - \int \tan^{n-2}x\,dx \tag{Pythagorean identity}\\ &= \frac{\tan^{n-1}x}{n-1} - I_{n-2} \end{align*}
Oh that makes sense. Would you please be able to explain how you got from the third to the fourth row? Thank you so much!That \(I_{n-2}\) bit should be clear. As for the first integral I basically just reversed the chain rule. However if you're uncomfortable with this, observe that the substitution \(u=\tan x \implies du = \sec^2 x\,dx\) gives
That \(I_{n-2}\) bit should be clear. As for the first integral I basically just reversed the chain rule. However if you're uncomfortable with this, observe that the substitution \(u=\tan x \implies du = \sec^2 x\,dx\) gives
\[ \int \tan^{n-2}x\sec^2 x\,dx = \int u^{n-2}\,du = \frac{u^{n-1}}{n-1}+C = \frac{\tan^{n-1}x}{n-1}+C \]
Hey rui please help with these two questions, I have no absolute idea on how to attempt it except making the denominator 2 for the first oneWhere did this come from? It looks like competition level maths instead of 4U. But what vexes me more is that a quick check on Wolfram shows that \( \sqrt{\frac{2009\times2010\times2011\times2012+1}{4}} \) is actually a rational number, and the 50th digit after the decimal is (not coincidentally) zero.
Hey, I'm not sure if this is part of the syllabus but my teacher gave us some extension questions to do (apparently our school doesn't teach harder 3u saying we ought to have already have learned it...) but could someone help me with this:
By considering the cases x ≥ 0, −1 < x < 0 and x ≤ −1 (or otherwise) prove that 1+x+x^2+x^3+x^4 is always positive.
Can someone please help with this question?If you've managed to find both the circle and the ray, the restrictions are that the circle only excludes \( (0,0) \), but the ray is essentially \( \arg (z-2) = 0 \)
"Find the Cartesian equation of the locus of z such that arg(z-2) = arg(z2). Describe the locus geometrically, noting any restrictions"
I was okay with finding the locus, but I'm stuck on how to determine the restrictions.
Thanks!
Hey, I'm not sure if this is part of the syllabus but my teacher gave us some extension questions to do (apparently our school doesn't teach harder 3u saying we ought to have already have learned it...) but could someone help me with this:
By considering the cases x ≥ 0, −1 < x < 0 and x ≤ −1 (or otherwise) prove that 1+x+x^2+x^3+x^4 is always positive.
Hey, could someone explain to me how to do this:Is this supposed to be under complex numbers? No other use of vectors appears in the current MX2 course.
Prove using vector methods that the midpoints of the sides of a convex quadrilateral form a parallelogram.
I'm not getting the same results for both the sides to prove it's parallel and I don't really get the internet explanations, I think I might need it to simplified a lot!
Is this supposed to be under complex numbers? No other use of vectors appears in the current MX2 course.
I'm not quite sure, my substitute teacher gave us a bunch of maths questions to do, saying her son said was appropriate for us (I think some of it is uni maths since there were some she marked wasn't in our syllabus). I thought it was just vector addition like in physics, but I'm not sure how to go about itI'm not convinced that they understand the current 4U syllabus. It's actually doable within the boundaries of the new MX1 syllabus but not with any of the current syllabuses. Although if you're really interested in a solution, you can post it in the first year uni maths question thread and I'm happy to address it there.
Hello,Follow the usual partial fractions approach. Sub \(x=1\), then \(x=2\) and then \(x=3\) to find your coefficients \(A\), \(B\) and \(C\).
I have partially done the working out, but I don't know what to do after. Can anyone please help me with this question? Thanks :)
Follow the usual partial fractions approach. Sub \(x=1\), then \(x=2\) and then \(x=3\) to find your coefficients \(A\), \(B\) and \(C\).Thank you!
Does anyone know how to attempt the answer to q19 and 20?Considering you've posted two potentially long questions, what attempts have you made so far? Please provide any understanding relevant.
Thank you so much !
Hi, I have an induction question that I'm a bit stuck onYou can actually use induction for both. However, the key thing to remember for both parts is that if \(n \geq 1\), then by definition, \(\boxed{A_{n+1} = \sqrt{1 + A_n}} \). Note - You wrote \(n >1 \), however because of how you defined it I'm actually convinced that you should've written \(n \geq 1\). Because otherwise we cannot compute what \(A_2\) is.
Let A1=1 and for n>1, let A(n+1) (the n+1 is a subscript sorry if it's not clear) = sqrt(1+A(n)). let G=(1+sqrt(5))/2
a) Prove that if A(n) < G, then A(n+1) <G
b) Prove that for all n, A(n+1) > A(n)
I'm a bit confused on whether I should use induction for the first part or for the second part, or whether there's an easier way of doing the question?
Thanks for your help!
You can actually use induction for both. However, the key thing to remember for both parts is that if \(n \geq 1\), then by definition, \(\boxed{A_{n+1} = \sqrt{1 + A_n}} \). Note - You wrote \(n >1 \), however because of how you defined it I'm actually convinced that you should've written \(n \geq 1\). Because otherwise we cannot compute what \(A_2\) is.
\[ \text{For the second one, we can check that when }n=1,\\ A_2 = \sqrt{1+A_1} = \sqrt{1+1} = \sqrt{2} > 1 = A_1\\ \text{as required.} \]
\[ \text{Now assume that for any integer }k\geq 1,\\ \boxed{A_k > A_{k-1}} \]
\[ \text{Then we have}\\ \begin{align*} A_{k+1} &= \sqrt{1+A_k} \tag{by definition}\\ &>\sqrt{1+A_{k-1}} \tag{inductive assumption}\\ &= A_k \tag{by definition} \end{align*}\\ \text{and hence the statement holds when }n=k+1\text{ as well.}\]
Note that we had used the definition twice.
________________________________________________________
\[ \text{For the first, we can check that when }n=1,\\ A_1 = 1 = \frac{1+\sqrt1}{2} < \frac{1+\sqrt5}{2} = G\\ \text{as required.}\]
\[ \text{Now assume that for any integer }k\geq 1,\\ \boxed{A_k < G}\]
\[ \text{Then we have}\\ \begin{align*} A_{k+1} &= \sqrt{1+A_k}\tag{by definition}\\ &< \sqrt{1+G}\tag{inductive assumption}\\ &= \sqrt{1+\frac{1+\sqrt5}{2}} \\ &= \sqrt{\frac{3+\sqrt5}{2}}\\ &= \sqrt{\frac{6+2\sqrt5}{4}}\\ &= \sqrt{\frac{1+2\sqrt5+5}{4}} \tag{completing the square}\\ &= \sqrt{\frac{(1+\sqrt5)^2}4} \\ &= \frac{1+\sqrt5}{2}\\ &= G\end{align*}\\ \text{as required, having taken positive root}\\ \text{noting that }\frac{1+\sqrt5}{2}\text{ is positive.}\]
I'll ask the teacher about the greater than and equal to sign, since the greater than sign was what was written on our worksheet. But if it's any comfort, I completely agree ;)Like yeah it's just harder 3U. Since you posted it on the 4U thread that's fine. I'm gonna swap out all the \(n\)'s for \(k\)'s, except for in the last part.
I have another from a tutoring centre and I'm not sure whether it's just a bit advanced for the course or where it fits in exactly into harder 3u (i think it's applications of inequalities)
Considering 1/x (take x>0), explain why 1/n+1 < the integral from (n+1) to n of dx/x < 1/n
As the integral can be simplified to ln(1 +1/x), do we just graph the intervals to prove that it works (I checked on desmos and it fits nicely)
b) hence or otherwise, deduce (1+1/n)^n < e < (1 + 1/n)^(n+1)
Do we separate the two sides from part a) and then raise it to e (e.g. from 1+1/k to e^ (1/k+1)) and multiply by the two sides and somehow manipulate the result? I'm not sure on how to get the lone e
c) this one is an extension question but it goes like this:
((1+n)^n)/(e^n) <n! < ((1+n)^(1+n))/(e^n)
Would that be relevant for this course or not?
Hey!Your integral should look something like \(V = 2\pi \int_0^2 2x\sqrt{1-(x-1)^2}\,dx\). For this integral, first sub \(u=x-1\). Then, split the integral up, and use the area of a semi-circle and properties of odd functions to help out.
The question I have attached is finding volumes using cylindrical/shell method. I am able to get the diagram and what I think is the right volume integration. However, I am struggling to progress with it. (6.2 Question 7 Cambridge)
I don't understand how you got from the second step to the third step.
Skipped a few steps, but happy to explain further :) Hope this helps
Hi, I'm having trouble with this inequality question:When you write out your answer, you must start from the beginning, which is either the basic fact that \( (a-b)^2 \geq 0\), or some equation they've given to you. Of course, if there's a part i) or something, you can assume the result of that for part ii).
Prove that, if a,b,c and d are any four positive numbers, then ab+cd <= sqr((a^2+c^2)(b^2+d^2))
I worked backwards and got that (a^2+c^2)(b^2+d^2) >=4abcd, and I need to prove that (ab+cd)^2<=4abcd. I don't know how to manipulate the LHS to do that, and also is this the right method?
Also, just generally: what's the best way to approach these kinds of inequality questions?
What are some useful inequality identities to know eg. a^2+b^2>=2ab or are you expected to derive them from the beginning?
I really appreciate you answering my questions!
Hello,The question's really ambiguous with it's wording (in particular when it said "ordinate"). Can you please provide the value of the answer so I know what I'm aiming for?
I haven't got a clue how to do this question in the link below. However, I have attached a separate attachment based on my thoughts of doing this question. Can anyone please help me with this question? Thanks :)
help thanksElaborate on what you need help with please?
The question's really ambiguous with it's wording (in particular when it said "ordinate"). Can you please provide the value of the answer so I know what I'm aiming for?Elaborate on what you need help with please?
heres a better picWould this diagram be helpful?
Hey rui need some of your maths experties plzAfter tracking down the second question I realise that as it was stated, \(x\) and \(y\) were intended to be real. Of course, this wasn't specified in the question, but apparently it was assumed by the author for some unknown reason.
The question's really ambiguous with it's wording (in particular when it said "ordinate"). Can you please provide the value of the answer so I know what I'm aiming for?Elaborate on what you need help with please?Hi Rui, the answer is 7√2➗16 + (3➗16) Ln(1+√2).
Hi Rui, the answer is 7√2➗16 + (3➗16) Ln(1+√2).The thing that completely horrifies me with this question is how they expect you to know how to sketch \(x^{2/3} = y^{2/3} + 1\), and the awkwardness of the wording. Sketching that curve should, in theory, be a completely separate question on its own, guided using stuff like implicit differentiation and identifying the domain of the curve.
Hi, please could I have some help with proving this absolute value inequality:Were you trying to use \( \leq \) there?
l (a+b+c) l = lal + lbl + lcl
The thing that completely horrifies me with this question is how they expect you to know how to sketch \(x^{2/3} = y^{2/3} + 1\), and the awkwardness of the wording. Sketching that curve should, in theory, be a completely separate question on its own, guided using stuff like implicit differentiation and identifying the domain of the curve.Thank you Rui! But surely this question won't be in an Extension 2 exam?
The wording of the question is also ambiguous, as it could represent the region below. They should have specified that they specifically want the region in the first quadrant.
Nevertheless, the question becomes doable with those assumptions put into place (albeit with some awkwardness). The idea is that they want you to find the green area shown below, but with the aid of the blue area. I will start it off.
(https://i.imgur.com/RZVb6De.png)
\[ \text{Note that when }x=2\sqrt{2}, \, y = 1.\\ \text{The combined area of the blue and green region is therefore just }2\sqrt{2}.\]
\[ \text{Note that the equation given rearranges to }x = \left( y^{2/3}+1\right)^{3/2}.\\ \text{The area of the blue region is therefore equal to}\\ \int_0^1 \left( y^{2/3}+1\right)^{3/2}\,dy\]
\[ \text{Let }y = tan^3\theta\implies dy = 3\tan^2\sec^2\theta.\text{ Then}\\ \begin{align*}\int_0^1 \left( y^{2/3} +1\right)^{3/2}\,dy &= \int_0^{\pi/4} \left( \tan^2\theta + 1 \right)^{3/2} (3\tan^3\theta \sec^2\theta)\,d\theta\\ &= 3\int_0^{\pi/4} \sec^2\theta (\sec^2\theta-1)\sec^2\theta\,d\theta\\ &= 3\left( \int_0^{\pi/4} \sec^6\theta\,d\theta + \int_0^{\pi/4} \sec^4\theta\,d\theta \right) \\ &= 3\left( \left[I_6\right]_0^{\pi/4} + \left[I_4\right]_0^{\pi/4} \right)\\ &= 3\left( \frac{\sec^4\frac\pi4 \tan \frac\pi4}{5} - \frac{\sec^4 0 \tan 0}5 - \frac45 \left[I_4\right]_0^{\pi/4} + \left[ I_4 \right]_0^{\frac\pi4}\right)\\ &= 3\left( \frac45 + \frac15 \left[ I_4\right]_0^{\pi/4}\right)\end{align*}\]
The remaining use of the reduction formulae and the final subtraction at the end is left as your exercise.
Subject to minor computational error.
Thank you Rui! But surely this question won't be in an Extension 2 exam?If it were, it would be chopped up into multiple parts.
Show that sinx<x for x>0.Firstly, for the sake of calculus, never use degrees.
Here's my working so far:
Let f(x)=sinx-x
f'(x)=cosx-1 =0 for stat pts
cosx=1
x=0,360
x>0 so x=360 degrees
f''(x)=-sinx
f''(360)=0
so apparently there's a horizontal point of inflexion at x=360
but don't you have to show that it's the absolute max turning point?
Hello,\begin{align*}\int_{-1}^1 \frac{x^2}{e^x+1}dx&=\int_0^1\left(\frac{x^2}{e^x+1}+\frac{x^2}{\frac{1}{e^x}+1}\right)dx\\
I am stuck on an integration question. The question and my working out are attached below. Can anyone please help me out? Thanks :)
If f(x) = x/sqr(1+x^2), prove by mathematical induction for n>=2 that f(f(...(f (x))...)) = x/sqr(1+nx^2), if there are n number of letter f's on the LHS.To help you get started, say you have \(h(x) = x-1\). Then, \(h(h(x)) = h(x-1)\). Of course, clearly this simplifies to \( h(h(x)) = x-2\), but it illustrates the idea - you sub the expression the function is equal to, back into itself.
Please help! I don't understand what f of f(x) means. What exactly am I subbing in when I am trying to prove true for n=2 and assuming true for n=k?
Thank you.
To help you get started, say you have \(h(x) = x-1\). Then, \(h(h(x)) = h(x-1)\). Of course, clearly this simplifies to \( h(h(x)) = x-2\), but it illustrates the idea - you sub the expression the function is equal to, back into itself.
The notation is standard - it is just composition of functions \(g(f(x))\), except \(g(x)\) and \(f(x)\) are taken to be the same thing here. Note we also learnt to use the chain rule in 2U differentiation for composite functions - this let us establish that \( \frac{d}{dx} g(f(x)) = f^\prime(x) \cdot g^\prime(f(x)) \).
\[ \text{So for your question, just for the sake of starting you off,}\\ \text{If }f(x) = \frac{x}{\sqrt{1+x^2}} \]
\[\text{Then}\\ \begin{align*} f(f(x)) &= f\left( \frac{x}{\sqrt{1+x^2}} \right)\\ &= \frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1 + \left( \frac{x}{\sqrt{1+x^2}}\right)^2}}\end{align*} \]
For the base case, you wish to prove that that expression equals to \(\frac{x}{\sqrt{1+2x^2}} \).
Thank you so much you actual genius! I figured it out and I'm so proud.I can look at this one
I also have another question:
If the sums of the first three terms of an AP and a GP are equal and non-zero, the common difference of the AP and the common ratio of the GP are equal and the ratio of the AM to the GM is 1:-2, find the common ratio of the GP. Also, find the relationship between their first terms.
So far I've just defined all the terms and apparently you need to solve them simultaneously but not sure how to do that.
I can look at this onetomorrowlater today if you want but mind if I grab the final answer off you so I know what to aim for?
Hello,
I am stuck on an integration question which I have partially done. Can anyone please help me out, especially with the second part? Thanks :)
Hi,It is not true to assume things like \( \infty\times 0 = 0\), or that \( \frac{\infty}{\infty} = 0\).
Could you help me with a graphing question attached?
I just don't understand why when 1/x is multiplied by f(x)approaching y= infinity, when x approaches infinity on the new graph, the new graph doesn't approach y=0?
Thanks
Hey guyssss, im stuck on these integration questions, could someone please help me?In both cases, you need to factor a 3 from the denominator. For example, for the second, you will obtain \( \int_0^{1/3} \frac{dx}{3\sqrt{\frac49-x^2}}\). Did you try this?
In both cases, you need to factor a 3 from the denominator. For example, for the second, you will obtain \( \int_0^{1/3} \frac{dx}{3\sqrt{\frac49-x^2}}\). Did you try this?
Don't worry; already did it with 1 page of working. Thanks for the input though :)
Hope this helps :)
hey how do you do this one:Hint: You can use a sketch to verify the following inequality.
h/(x+h) < ln(x+h) - ln(x) <h/x . (it's meant to be greater than and equal to btw)
Hello,Two things:
I am stuck on Example 7 part a) in the link below. I have also attached my working out. Can anyone please help me out? Thanks :)
Two things:I realised where I went wrong; thanks :)
\[ \text{Firstly, you've correctly noted that }A = \frac12 x^2 \tan \alpha.\\ \text{So your volume of a cross section should just be }\boxed{\delta V = \frac12 x^2 \tan \alpha\, \delta y}\\ \text{i.e. }\boxed{\delta V = \frac12 (r^2-y^2)\tan \alpha\, \delta y} \]
Not sure why you multiplied \(x^2\) to another \(r^2-y^2\) term in your expression. By doing so, because of the fact that \(x^2+y^2=r^2\), you're basically trying to deal with \(x^4\), or equivalently \((r^2-y^2)^2\)
\[ \text{Secondly, what is }a?\\ \text{The boundaries of integration (and also the sum) are just }-r\text{ and }r \]
Hello,Sorry about the delay with this one. Been ages since I used the linear equation method for the volume of a frustrum so I wanted to be more alert when revising it. I just attached the full solution here.
I am stuck on a volumes question in the attachment below (especially the proving part). Can anyone please help me out? Thanks :)
Sorry about the delay with this one. Been ages since I used the linear equation method for the volume of a frustrum so I wanted to be more alert when revising it. I just attached the full solution here.Wow, this seems top-notch. Thank you! :)
It's essentially based off the exact same method you used. Just with some arbitrary variables, instead of given constants.
(https://i.imgur.com/NPmtuYtl.png)
(https://i.imgur.com/WU2u3B5l.png)
Hey Rui! Following the MX2 lecture today at UTS (was awesome btw!), I've decided to try this thread for help. I think my worst area is Applications of Polynomials (locuses and all that). Here's my question from that topic (from textbook):Glad you enjoyed it! :)
a) Show that the equation of the normal to the parabola y = x^2 at the point P(t, t^2) may be written as
b) Hence, if there exist three normals passing through P(x_0, y_0) then using the result of part (b) in question 7 above show that .
Question a) is easy, just added it for context. The real head-scratcher is part b), and I don't even know where to start. (side note; is there any way to do LaTeX inline?)
I just noticed that I did not include the result mentioned in the question! It's; if f(x) is in the form:Ah cool, the question is definitely doable now aha! That above result is quite important because it holds only when \(f(x)\) has 3 distinct \(x\)-intercepts.
then,
(I'm not actually sure how much information from the previous question is needed, but this is the result obtained)
\( stuff goes here \)
Okay, I have another question (from the same exercise):Your equations are essentially a bit too restrictive.
6 Assume \(cos3\theta = 4cos^3\theta - 3cos\theta \), find in terms of \(\pi\) the exact values of the solutions of
a) \(x^3-3x+1=0\)
The answer in the book begins quite simply, with \(x=kcos\theta\), \(k^3cos^3\theta - 3kcos\theta + 1 = 0\). They then mention that:
"We want \(\frac{k^3}{3k} = \frac{4}{3}\), ∴ \(k^2=4\), ∴ \(k=2\)."
From this, they attain the equation:
... and being able to solve for theta. What I'm confused at is why they wanted that original equation that gave them k. What I originally tried to do was something like:
Why did they divide the two equations? I see that there is no solution for my two equations, which is where I got stuck. Help!
Hello everyone,\[ \text{Since this is a second order recurrence relation}\\ \text{we require a stronger version of induction that requires two base cases.} \]
Just stuck on this question, which has a weird sequence I can't really figure out. Any help with part (iii) would be very much appreciated.
It's from the 2003 HSC, question 6
Hello,\[ \text{Your displacement bit looks fudgy.}\\ \text{I'm with you up to and including this step:}\\ \frac{1}{2k} \left[ \ln (c^2+v^2)\right]_c^0 = -x .\]
I am stuck on this mechanics question in the link below. Can anyone please help me out? Thanks :)
\[ \text{Your displacement bit looks fudgy.}\\ \text{I'm with you up to and including this step:}\\ \frac{1}{2k} \left[ \ln (c^2+v^2)\right]_c^0 = -x .\]After a moment of refining my working out, I got the answer for the first part. I have trouble understanding and doing the second part, can you help me out please? Thanks :)
I can't see why \(c^2\) should equal 1 here - not sure where your explanation comes from. The next step here is to do what you would normally do in 2U, i.e. sub in the boundaries.
\begin{align*} \frac1{2k} \left[ \ln (c^2) - \ln (c^2+c^2) \right] &= -x\\ \frac{1}{2k} \ln \left(\frac{c^2}{c^2+c^2}\right)&=-x \tag{log laws}\\ \frac{1}{2k}\ln \frac12 &=-x\\ \therefore x &= -\frac1{2k}\ln \frac12\\ &= \frac1{2k}\ln 2. \end{align*}
\[ \text{Also I think there's a mistake with the velocity.}\\ \text{From }\int -dt = \frac1k\int \frac{dv}{c^2+v^2}\text{ you should've obtained }\boxed{-x = \frac{1}{k} \left( \frac{1}{c}\tan^{-1} \frac{v}{c}\right)+C} .\]
Note that whilst it is true that \( \int \frac{x}{a^2+x^2}dx = \frac12 \ln (a^2+x^2)+C \), when the numerator is instead a constant you have \( \int \frac{1}{a^2+x^2}\,dx = \frac1a \tan^{-1} \frac{x}a \). The first is a \( \int \frac{f^\prime(x)}{f(x)}\,dx \) pattern, but the second is a 3U standard integral from the inverse trig topic.
After a moment of refining my working out, I got the answer for the first part. I have trouble understanding and doing the second part, can you help me out please? Thanks :)(There's a slight typo in that I wrote \(x\) instead of \(t\) for that last boxed expression. Fixed that now.)
(There's a slight typo in that I wrote \(x\) instead of \(t\) for that last boxed expression. Fixed that now.)The part which says 'Find its velocity when it reaches a point C, having travelled for a time pi/12ck from A. Find the distance AC'. The very bottom part.
When \(t=0\), \(v=c\), so \( C = -\frac{1}{ck}\tan^{-1}\frac{c}{c} = -\frac{1}{ck}\times\frac\pi4\). Then just sub \(v=0\). I'm not sure what else you want elaboration with?
The part which says 'Find its velocity when it reaches a point C, having travelled for a time pi/12ck from A. Find the distance AC'. The very bottom part.Completely missed that bit oops.
Completely missed that bit oops.Ok, thanks for the help :)
\[ \text{After showing that}\\ \boxed{t = \frac1{ck} \left( \frac\pi4 - \tan^{-1} \frac{v}{c} \right)}\\ \text{if we sub }t = \frac\pi{12ck}\text{ we have:} \]
\begin{align*}\frac\pi{12ck} &= \frac1{ck} \left( \frac\pi4 - \tan^{-1} \frac{v}{c} \right)\\ \frac\pi{12} &= \frac\pi4 - \tan^{-1} \frac{v}{c}\\ \tan^{-1} \frac{v}{c} &= \frac\pi4 - \frac\pi{12}\\ &= \frac\pi6\\ \therefore \frac{v}{c} &= \frac1{\sqrt3}\\ v &= \frac{c}{\sqrt3} \end{align*}
From here, we should be able to just sub straight into \( x= -\frac1{2k} \ln (c^2+v^2) + \frac1{2k} \ln (2c^2) \). The computations look a bit nasty but I don't see any curveball in this part either.
Hello,Sorry must've missed this. Here's the integration bit done for you:
I am stuck on both parts of this question in the attachment below. I have done part of the question but I can't seem to integrate the function. Can anyone please help me out? Thanks :)
Sorry must've missed this. Here's the integration bit done for you:Ok, thanks :)
\begin{align*} \int_U^0 \frac{v}{V+v}\,dv &= \int_U^0 \frac{V+v}{V+v} - \frac{V}{V+v}\,dv\\ &= \int_U^0 1 - \frac{V}{V+v}\,dv\\ &= \left[v - V \ln(V+v) \right]_U^0 \end{align*}
(Basically the classic 'add something and subtract the same thing' trick)
Hello,The point \(P\) depends entirely on its parameter \(t\). Since your expression for the area does not have this parameter appearing, i.e. it is independent of whatever \(t\) is, equivalently speaking it is independent of the point \(P\).
I am confused with this question in the attachment below. I know how to find the area of the triangle (stated in my working out), but I don't know why Area (OLM) is independent of the point P. Can anyone please explain this thoroughly? Thanks :)
The point \(P\) depends entirely on its parameter \(t\). Since your expression for the area does not have this parameter appearing, i.e. it is independent of whatever \(t\) is, equivalently speaking it is independent of the point \(P\).Ahhh, I get it now. I didn't quite get this concept since the book does not clearly mention the parameter stuff with that question. Thanks Rui!
(Note that \(c\) is assumed to be a constant, not some parameter that is allowed to vary.)
Currently stuck on these recurrence relation questions.Remember to also use the add/subtract or multiply/divide same thing trick for recurrence relations, not just integration by parts. The first question is relatively similar to what I did in the April lecture.
Any ideas?
Cheers! :)
Remember to also use the add/subtract or multiply/divide same thing trick for recurrence relations, not just integration by parts. The first question is relatively similar to what I did in the April lecture.
\begin{align*}\therefore (nr+1) I_n &= nr I_{n-1}\end{align*}
The [x(1-x^r)^n] bit equates to zero when subbing in the boundaries, so it basically just disappears :)
Can someone please help with this question? Thanks!What attempts have you made so far at this problem? And what is the answer we are aiming for?
Heyy, back with a volume question. Still trying to get my head around cylindrical shell since we haven't done it in class yet.You should've arrived at \(\delta V =2\pi x\left(e^{x^2} - 1 \right)\delta x\) and \(V = 2\pi \int_0^1 xe^{x^2} - x\,dx\). (If you got that then you probably made an accident with the actual integration)
The area bounded by the curve y = e^(x^(2)), the lines x=1 and y=1 is rotated about the y-axis. Use the method of cylindrical shells to calculate the volume of the solid of revolution formed.
I got the answer pi(0.5e^2 -e-1) but I am pretty sure that isn't right since it returns a negative value...
Hey, I'm a bit stuck over this integration question.What textbook is this? This is usually proven via a double-recurrence formula which is outside of the scope of the MX2 course.
Prove the integral of 0 to 1 (x^m)(1-x)^n dx = (m!n!)/(m+n+1)! Hint: think of binomial theorem.
I tried using integration by parts but got stuck. I'm not sure how to use binomial theorem since we haven't really covered that in 3U yet, even though it looks like the binomial theorem formula from my textbook.
Hey everyone, I'm quite stuck on this Harder 3U inequalities question:For this question I took on a bit of an unorthodox method. I'm not too sure at this stage how I'd approach it in a sense similar to AM-GM. For now I'll post what I've seen, but it may not be the intended approach.
a²+b²≥ab+a+b-1
I know that a²+b²≥2ab from the AM≥GM inequality, so I've attempted to prove that 2ab≥ab+a+b-1 hence, ab-a-b+1≥0, but I can't prove that either.
There are no restrictions on the values of a and b. Any help is much appreciated :)
Hello,It's way easier to use the product-to-sum identities (which should be derived) for integrals of that form. Here, we derive:
I have another question on 4u, this time on integration. The question is evaluate the integral ∫(cos6x*cos2x)dx. My working out is shown, but I can't seem to get the answer using substitution. Can anyone please help me out? Thanks :)
Hello,Just taking a look at this now:
I am stuck on Q 23 in the link below. I got the first part and attempted the second part, but I don't know where to go. Can anyone please help me out? Thanks :)
Just taking a look at this now:Thank you Rui! :)
With b) (i), it's basically relating back to where banked tracks originate from. The idea is that the velocity you're travelling at is faster than the velocity that gives you the optimal banking angle. The train therefore has a tendency to skid up the track.
To ensure this does not happen, the lateral force must bring the train somehow downwards. The lateral force must therefore be directed inwards to the centre of the circle of motion.
This can only be done by the rail that is further outwards. The outer rail effectively 'pushes' the train back inwards.
With b) (iii), if you draw your force diagram correctly you should now arrive at \(N\cos \theta -F\sin \theta = mg\) and \(N\sin\theta + F\cos\theta = \frac{mv^2}{500} \). Hint: Multiply both sides of the first equation by \(\sin\theta\), and both sides of the second by \(\cos\theta\)...
What textbook is this? This is usually proven via a double-recurrence formula which is outside of the scope of the MX2 course.
This was from tutoring but I asked both my tutor and my class teacher and they both say it's outside the scope, so I shouldn't worry about it!If they have an answer I'm genuinely curious to see tbh, because I can't see how the coefficients work nicely when the binomial theorem is used.
I have some other questions though (from past assessments from my school).
1. Explain why multiplying a complex number z by cistheta rotates z anticlockwise about the origin, through an angle of theta.
I get the theory behind this, but I'm confused on how to set this out. Would it be best to let z= rcis(delta) and then use the multiplication of polar numbers:
so you get that: z*cistheta= rcis(delta+theta)? But I'm still a bit confused on how to prove it shifts anticlockwise
2. Assume w is a nth root of infinity. Using geometric sums, prove that w+w^2+w^3+...w^n=0
When I tried doing this, I got this: w(1-w^n)/(1-w)= w(1-w^n)/w(1/w-1)
= (1-w^n)/(1/w-1)
was I meant to use smth to do with the property 1+w+w^2=0??
3. If the mod of z=1, prove |a+bz|= |az+b|. You may assume a and b are elements of the Real numbers,.
What I did was:
square both sides: |a+bz|^2= |az+b|^2 . [[I'm just going to let the conjugates = capitalised version of the letter to make it easier to read]
(a+bz)(conjugate(a)+conjugate(bz))= (az+b)conjugate(az)+conjugate(b))
(a+bz)(A+BZ)= (az+b)(AZ+B)
aA+Abz+aBZ+BbZz=aAzZ+AZb+aBz+Bb (but Zz=1)
LHS= |a| +Abz+ aBZ+|b|
RHS= |a|+AZb+aBz +|b|
but these aren't the same? if Zz=1, isn't Z=1/z? So I'm not sure how to get the answer.
Thank you!!
Took me ages bc Word crashed but here's my "latexed" version of their solutionThat's basically the double recurrence relation method actually. It's not the binomial theorem.
That's basically the double recurrence relation method actually. It's not the binomial theorem.
(Double recurrences are outside the scope of MX2 but as seen in that example, they're handled the same way as usual recurrence formulas. i.e. with integration by parts. But I feel as though the question is misleading to mention the binomial theorem in that case.)
I'm struggling with a volumes question help would be great thanks.
A hemisphere has radius r. By considering cross-sections parallel to the base of the hemisphere, show the volume is given by V=2/3(pi)r^3.
Yeah, I think it's like some of those recurrence formulas were, which is probably why they got away with it. would it be something that we could expect?Nah, double recurrence relations are actually not examinable in the HSC. Can't remember where I was first told of this but it is an official statement.
I'm stuck on some other problems (Is it just me or is complex numbers a more complicated area that everyone seems to treat it as? I've heard a lot of people saying it's super easy, which makes me feel awful)
1. Let z= (i+ix)/(1-ix). Show that:
if x is less than or equal to 1, arg(z)= pi/2+2arctan(x)
if x is greater than 1, arg(z)= -3pi/2+2arctan(x)
Challenge q: Are there interesting patterns you can spot as x approaches plus and minus infinity
2. Find the real and imaginary part of (1+cos2theta+isin2theta)/(1+cos2theta-isin2theta). I tried to use the 2cos^2theta-1 for it but it didn't come out nicely at all
Nah, double recurrence relations are actually not examinable in the HSC. Can't remember where I was first told of this but it is an official statement.
________________________________________________________________________
Just before I jump into 1, I want to confirm that you meant \( z = \frac{i+ix}{1-ix} \) as opposed to \( z = \frac{1+ix}{1-ix} \)? Haven't tried it yet, just wanted to double check first.
\[ \text{Double angle formulas should be fine for the second one.}\\ \text{It's just that even more work is necessary after you use it.} \\ \begin{align*} \frac{1+\cos2\theta+i\sin2\theta}{1+\cos2\theta-i\sin2\theta} &= \frac{2\cos^2\theta+2i\sin\theta\cos\theta}{2\cos^2\theta-2i\sin\theta\cos\theta}\\ &= \frac{\cos\theta+i\sin\theta}{\cos\theta-i\sin\theta}\\ &= \frac{\cos\theta+i\sin\theta}{\cos(-\theta) + i\sin(-\theta)}\\ &= \frac{\operatorname{cis}\theta}{\operatorname{cis}(-\theta)}\\ &= \operatorname{cis}2\theta\\ &= \cos2\theta+i\sin2\theta \end{align*} \]
Note that the complex numbers questions you have given are on the upper end of the spectrum - they tend to rock up in Q14-16 instead of Q11-13 of the exam. But at the E4 range, recognising things like \( \operatorname{cis}(-\theta) = \cos\theta - i\sin\theta \) becomes quite common, as you're expected to have a full mastery of trigonometric identities. And of course, the fact that division of complex numbers in mod-arg form involves subtracting their arguments.
Whoops, I made a typo!: the original statement was i(1+ix)/(1-ix) so I just multiplied it in. So it should equal (i-x)/(1-ix). Sorry about that!With the double angle formula, technically speaking you should, and then check the solutions for \(\cos x = 0\), i.e. \( x=2n\pi \pm \frac\pi2\) separately. There is another method that completely avoids this issue, but it's a bit harder to see. You can request to see that method if you wish in a later post.
Also for the previous question, when you divided the fraction by costheta, do we need to say smth like costheta can't equal 0? or is that not required?
Four letters are chosen from the letters of the word TELEGRAPH.Hint: You will need to consider three separate cases:
These four letters are then placed alongside one another to form a four letter arrangement. Find the number of distinct four letter arrangements which are possible, considering all choices. (2 marks)
Help would be much appreciated
Hello! I have a question from class when doing the harder 3u section.
Given that d/dx (lnx) = 1/x,
Prove that:
Thanks a lot!
Hello! I have a question from class when doing the harder 3u section.With this question, I will say that I heavily dislike this question as it has been provided without any context whatsoever, especially since the supposed result to be proved is commonly taken as a definition. Without the source of the problem, or any extra parts that were provided, I fail to see how this is appropriate even at the MX2 level.
Given that d/dx (lnx) = 1/x,
Prove that:
Thanks a lot!
Hey guys, I'm struggling with part iv. (see photo) help would be greatly appreciatedHint: You can try to show that the interval \(QR\) subtends angles of equal size at \( P\) and \(S\), which are both on the same side of \(QR\). That'll help you prove that the points are concyclic.
Thanks!
Hint: You can try to show that the interval \(QR\) subtends angles of equal size at \( P\) and \(S\), which are both on the same side of \(QR\). That'll help you prove that the points are concyclic.
To do so, you may need \(b^2=a^2(1-e^2)\), but it's likely you'll also need the angle between two lines formula to find \( \tan \angle QPR \) and \(\tan \angle QSR\). I haven't tried the problem yet, but I wouldn't be surprised by some nasty algebra.
Damn, im still struggling through the algebra of this question. Maybe i haven't gotten the right y coordinate for R, -sinΘ(a^2-b^2)/b?\[ \angle QPR = 90^\circ\\ \text{because the normal is perpendicular to the tangent by definition.} \\ \text{So we just have to show that }\angle QSR = 90^\circ. \]
Hey! Our class is currently freaking out over this Complex Number - Vector question.\[ \text{As }M\text{ is the midpoint of }AB,\\ \overrightarrow{OM} = \frac{z_1+z_2}{2}.\]
Does anyone have a definitive answer? ;D
Cheers!!
It's very easy to get lost in the vectors so try approaching it somewhat systematically. Use tip-to-tail addition purely for mental purposes as much as you can. Also, I compare MQ to MA as opposed to QM to AM, because things are more obvious to me when they start from the same point.
Hey Guys, can anyone help with the last part of this question? It's from Sydney Grammar 2014 Trials and I don't really understand their solutions.The first one is simply applying \( A = \frac{1}{2}ab\sin \theta \) three times, once for each of the smaller triangles. Note that the angles are all made at the centre of the circle, hence the lengths \(a\) and \(b\) in the area formula equal the length of the radius \(r\). So we just get \( \frac12 r^2 \sin \alpha +\frac12r^2\sin \beta + \frac12 r^2\sin\gamma\).
The answer to i) is (r^2)/2 * (sin alpha + sin beta + sin gamma).
The answer to ii) a) is 1/3 (alpha + beta + gamma), 1/3 ( sin alpha + sin beta + sin gamma).
Hope this helps :)
\[ \text{Intuition suggests that the values }\alpha,\,\beta\text{ and }\gamma\text{ should somehow}\\ \text{correspond to the same ones given in the previous part.}\\ \text{Else it does not make sense.} \]
\[ \text{How the answers start off is basically relying on part i) for some simplifications.}\\ \text{In part i), it is clear that the angles all meet at a point, which gives the}\\ \alpha+\beta+\gamma = 2\pi\text{ bit.} \]
Which is just 3 times the expression for the \(x\)-coordinate of the centroid. So the centroid must have \(x\)-coordinate \( \frac{2\pi}{3} \).
\[ \text{But on one hand, the area of }\triangle TUV\text{ was }\frac12 r^2 (\sin\alpha+\sin\beta+\sin\gamma)\\ \text{whilst the }y\text{-coordinate of the centroid is }\frac{\sin\alpha+\sin\beta+\sin\gamma}{3}. \]
\[ \text{So pretty much a bit of rearranging is what's needed here to show that}\\ \text{the centroid has }y\text{-coordinate }\frac{2\times \operatorname{Area}_{\triangle TUV}}{3r^2}. \]
The answers write the area as \(|\triangle TUV|\) instead. This isn't really common in the HSC but it does get used at times in the real world, so I'll use the absolute value brackets too.
Of course, that was the easy mark to get. By considering the link between the two parts we've more or less reinterpreted the problem. What we wished to do is to maximise \( |\triangle TUV| \), but we see that the \(y\)-coordinate of the centroid of \(\triangle ABC\) happens to be a scalar multiple of it. (Note of course that \(r\) is constant.)
So if we maximise the \(y\)-coordinate of said centroid, we achieve the same goal. This is a much harder mark to get. Naively, we'd be tempted to go back to maximising \(\sin\alpha+\sin\beta+\sin\gamma\), but this problem is still not viable because we only have the one constraint \(\alpha+\beta+\gamma = 2\pi\). This is where we get the intuition that something about the curve \(y=\sin x\) must be useful - we want to do this maximisation with a completely different argument, not via calculus in optimisation.
To my knowledge, concavity hasn't been examined much at all in the HSC - in fact, I only know of one question in 2018 or 2017 that did. This is the idea.
____________________________________________________________________________
\[ \text{Given two points }x_1\text{ and }x_2,\text{ we say that a curve is concave down for all }x_1 \leq x \leq x_2\\ \text{if for every }0\leq t \leq 1,\text{ it is true that}\\ f(tx_1 + (1-t)x_2) \geq tf(x_1) + (1-t)f(x_2).\]
\[ \text{This definition is awkward to work with, so we clarify what is going on.}\\ \text{Essentially, if your curve is concave down over a region, what you can do is draw a straight line segment}\\ \text{from one endpoint of the region, to the other.} \]
\[ \text{A curve is concave }\textbf{down}\text{ if that line segment}\\ \text{is guaranteed to lie }\textbf{below}\text{ the curve about that region.} \]
Note similarly that for a curve concave up over a certain interval, the line segment drawn between the endpoints should lie above the curve.
\[ \text{This is illustrated in the diagram. For example, between the points }A\text{ and }C\\ \text{the curve is concave down.}\\ \text{The interval }AC\text{ is also, in fact, under the curve.} \]
Similarly, between the points \(A\) and \(B\) the curve is concave down, and the line segment \(AB\) is below the curve. Same goes for \(BC\).
The answers claim that this will always be the case, regardless of what \(A\), \(B\) and \(C\) are. This is because the \(x\)-coordinates of these points are respectively \(\alpha\), \(\beta\) and \(\gamma\), which in part (i) we're told are always between \(0\) and \(\pi\). For a MX2 Q16 problem, it should be safe to just assume without proof that \(y=\sin x\) is always concave down for \(0 < x< \pi\) - this can easily be proven using the second derivative, if a graph is not adequately convincing.
(Aside: The second derivative is used as a theorem to prove concavity. The definition, however, is the clunky thing I mentioned earlier.)
\[ \text{So because we know that }AB, \, AC\textbf{ and }BC\text{ lie under the curve }y=\sin x,\\ \text{it follows that }\triangle ABC\text{ also lies under it too.} \]
And after all, since the centroid is just the intersection of the triangles' medians, that is consequently forced under the curve as well.
____________________________________________________________________________
\[ \text{However, the crucial thing to note now is that }\textbf{only}\text{ the }y\text{-coordinate}\\ \text{influences }|\triangle TUV|. \text{ Recall that the }x\text{-coordinate}\\ \text{of the centroid is }\textbf{constant}.\text{ It is always equal to }\frac{2\pi}{3}. \]
This is where advanced intuition becomes necessary. The idea is that we know the centroid can never float above \(y=\sin x\). Rather, the \(y\)-coordinate of the centroid is bounded above by the curve \(y=\sin x\).
So the natural question to ask her is whether or not this bound is attainable. That is, is there a configuration of points \(A\), \(B\) and \(C\) that would ensure that the centroid must lie on the curve as well. This is useful, because if we know it can lie on the curve, and also know that it can't go above it, then we've deduced that the maximum possible \(y\)-coordinate IS when it is on the curve.
\[ \text{The answer to these questions is usually yes,}\\ \text{when we allow for a thing called }\textbf{degeneration}. \]
\[ \text{The idea is to }\textit{collapse }\triangle ABC\text{ somehow.}\\ \text{Without loss of generality, let's start by collapsing }BC. \]
\[ \text{What we are going to do is let }B\text{ and }C\text{ become the same point.}\\ \text{That is, }C=B.\\ \text{Then the concept of }\triangle ABC\text{ no longer exists.} \]
\[ \text{However what happens is that if we gradually bring }C\text{ closer to }B,\\ \text{what happens geometrically is that }\triangle ABC\\ \text{will gradually turn into just another line segment, namely the interval }AB! \]
Once we're left with the interval \(AB\), the "centroid" of \(\triangle ABC\) will be nothing more than a point on said interval. (To be somewhat specific, it will be the point that divides \(AB\) internally in the ratio 1:2, I believe. It could be 2:1 though.)
But the point is this is not enough. We know that \(AB\) is still under the curve \(y=\sin x\), so if the centroid lies on \(AB\), it is also under the curve.
\[ \text{This is why we need to think about the most }\textit{borderline}\text{ case.}\\ \text{This is when we make all }\textbf{three}\text{ points coincide.}\\ \text{That is, set }A = B = C\text{, as the answers have stated.} \]
The intuition behind this is a general rule of mathematics - borderline cases tend to influence optimality (be it minimality or maximality). This can prove useful at times, but trying to recall this in an exam is hard.
____________________________________________________________________________
\[ \text{And the thing is, we see that this is good!}\\ \text{When }A=B=C\text{, we're saying that all 3 points are now the same point}\\ \text{and hence }\triangle ABC\text{ has also collapsed down into a single point.} \]
\[ \text{Consequently, the centroid has also been collapsed to a single point.}\\ \text{But the key idea is, the centroid now has nowhere to run.}\\ \text{Since }A,\, B\text{ and }C\text{ are the same point, the centroid is forced to being the same point as well.} \]
\[ \text{The reason why this is optimal is because we know that }A,\, B\text{ and }C\text{ are on the curve.}\\ \text{Hence the centroid is }\textbf{also}\text{ on the curve }y=\sin x\text{ now!} \]
Thus, we've shown that the centroid both cannot lie above the curve, but can also lie ON the curve. This must therefore give the maximum \(y\)-coordinate.
Basically, the argument was pretty much fully geometric. But in a specific way.
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\[ \text{Once this bit of the puzzle is cracked, everything is easy.}\\ \text{We know that }A = B = C\text{ automatically imposes the condition}\\ \text{that their corresponding }x\text{-coordinates are equal.} \]
\[ \text{Therefore }\alpha = \beta = \gamma.\\ \text{So by subbing back into }\alpha+\beta+\gamma = 2\pi\\ \text{we can conclude that in fact, }\\ \alpha = \beta = \gamma = \frac{2\pi}{3}.\\ \text{Which, of course, can now be plugged back into the area formula.} \]
Ahhhh, it was only worth one mark so I thought maybe there was a quick way but this makes sense, thanks a lot!Note: There is if you have previous parts to help you out.
The first 3 sections were fine and i havnt tried the second part of iv, but for finding SP i used the locus definition of a hyperbola and got the result but i ended up with a y still hanging around. Help would be much appreciated
Hi!!
Can somebody work out how to do part (a)?
Thanks :)
Hi!!In the future, please provide any relevant understanding/working you've already attempted, or a final result to give a direction to head towards.
Can somebody work out how to do part (a)?
Thanks :)
Why is F ignored; is it because of the phrase "no tendency to slip sideways"?
Pretty much answered your own question :)
Basically the force F acting down the slope is going to be some form of motion resistance (basically friction every time). Since you're told that there's no tendency for the particle to slip up and down the slope this implies there is no such force F that exists ie. the only forces you consider are the weight force, and the normal force.
Hope this helps :)
So then the F in the diagram is essentially a red herring?For that question, or at least that specific part, yes.
For that question, or at least that specific part, yes.
In other questions, the \(F\) may be important.
Weird math question my teacher brought up today.
A numbers, 1, 2, 3, and 4 are rearranged to create a 4-digit number.
a) How many total numbers can you create if no number are repeated.
b) What is the sum of all possible 4-digit numbers?
I guess for (b) you can just list them all, but I was wondering if there was a quicker, more "mathematical" way of solving this.
All possible 4-digit numbers would range from 1111 to 4444. As such there 256 numbers, and there is 1/4 chance any single digit appears in a certain place value. ie the answer will be 64 x (4000+3000+2000+1000+400+300+200+100+40+30+20+10+4+3+2+1).
Hope this helps :)
Can someone please help with this question?
P(2p,2/p) and Q(2q, 2/q) are two points on the rectangular hyperbola xy=4. M is the midpoint of PQ.
Find the equation of the locus of M if PQ is a tangent to the parabola y2=4x
Essentially, I determined the locus to be y2=-x/4 (not sure if this is right as I don't have solutions). However, I'm conscious of the fact that there may be restrictions, but I can't seem to find what these may be. How do I determine the restrictions, if any?
Thank you.
Can someone please help with this question?
P(2p,2/p) and Q(2q, 2/q) are two points on the rectangular hyperbola xy=4. M is the midpoint of PQ.
Find the equation of the locus of M if PQ is a tangent to the parabola y2=4x
Essentially, I determined the locus to be y2=-x/4 (not sure if this is right as I don't have solutions). However, I'm conscious of the fact that there may be restrictions, but I can't seem to find what these may be. How do I determine the restrictions, if any?
Thank you.
That locus is correct, I think i did that question in a cssa paper. Does the question ask to specify restrictions? I don't see what the restrictions would be as you've eliminated the parametersMy working out and GeoGebra are claiming \(y^2 = -\frac{x}{2}\). Note that the line(s) demonstrating that \(PQ\) is a tangent to \(y^2=4x\) is hidden, but you can put it back in there.
Hello!Considering they said acceleration acting against the ball instead of the "force", \(a=-g-kv\) would be the correct answer.
For this question:
"A ball of mass m is projected vertically upwards with speed u. The acceleration acting against the ball is proportional to it speed. Find the time (t) taken to reach the greatest height."
I'm simply confused on how to resolve forces. I resolved them like this: "ma = -mg - kv", but the answers resolved them as "a = -g - kv"
Why is that the correct resolution? Thanks in advance!
Hello!
For this question:
"A ball of mass m is projected vertically upwards with speed u. The acceleration acting against the ball is proportional to it speed. Find the time (t) taken to reach the greatest height."
I'm simply confused on how to resolve forces. I resolved them like this: "ma = -mg - kv", but the answers resolved them as "a = -g - kv"
Why is that the correct resolution? Thanks in advance!
They're both correct. However, the force balance equation should always be the first line in any of these types of problems. The equations of motion are always derived from the force balance equation. In this case, "ma = -mg - kv", which means that a = -g-(k/m)*v. Note that k/m is still just a proportionality constant, so you could replace k/m with another constant, called k, for simplicity.Although this makes perfect sense in the real world, from what I have seen they are far more specific about things in the HSC (in what they're looking for). The questions are worded so that students will always go along the exact same line of thinking, with the exact same constant \(k\) in mind. (As opposed to swapping it out for another.)
Although this makes perfect sense in the real world, from chat I have seen they are far more specific about things in the HSC (in what they're looking for). The questions are worded so that students will always go along the exact same line of thinking, with the exact same constant \(k\) in mind. (As opposed to swapping it out for another.)
If we want to follow the rule that the force resolution should be the start, if anything they would like you to turn the resistance into \(mkv\). And then write \(ma = -mg-mkv\).
Yeh I'm aware of that, but in the question it doesn't state what the propotionality constant is, so it should be perfectly valid to use k/m and then combine that into one final constant. They shouldn't lose marks for it - I remember doing this back in the day and it was fine, according to my teacher as well as my friends teachers.What concerns me about this is that whilst it may have been fine once upon a time, marking guidelines may fluctuate in how strict they are. (For example, stuff like L’Hopitals rule is instantly penalised now.)
What concerns me about this is that whilst it may have been fine once upon a time, marking guidelines may fluctuate in how strict they are. (For example, stuff like L’Hopitals rule is instantly penalised now.)Fair enough. I wasn't even aware that there was ever a time where L'Hopitals rule was acceptable lol.
It feels too much like gambling on chance to rely on these handwaves. Although I can respect what teachers have told you, and that they probably still are examiners, this would be something I want to hear from this year’s examiners first-hand as well. I’ve never been comfortable recommending tricks that could potentially be taken as a stretch unless I have first-hand permission granted with it.
Fair enough. I wasn't even aware that there was ever a time where L'Hopitals rule was acceptable lol.Oddly enough, once upon a time it was ok. But the stance now is strictly against it and I think it's a good call by NESA here.
Hey there!Thank you so much! Sorry for the late reply; I had trials.
Firstly, I think the reason you're not getting to the correct solution is because your dv is incorrect!! Remember that adding one to the power, dividing by the new power and the derivative doesn't actually work unless you have a linear function (which is what I think you did there.)
An alternative solution is below:
If there's anything you don't understand, please ask further :)
Hope this helps :)
Can someone please help?The mistake was in how you wrote it: xx5xx7xx9xx.
The numbers 5,6,7,8 and 9 are chosen without replacement and arranged to form a five-digit number. Find the probability that the number formed has the three odd digits in increasing numerical order.
This is what I've attempted to reason (though it's wrong).
Since the 5, 7, and 9 are increasing numerical order, let them be fixed. We're essentially arranging the 6 and 8 around them. Let crosses denote possible positions for 6 and 8.
I ended up with: XX5XX7XX9XX
I then said that the number of arrangements was 8P2 as we're choosing two X's then arranging them around the 5, 7 and 9 - then divide by the total arrangements (5!) to get the answer. I've used similar logic in other questions and have gotten them right, however in this case this is wrong.
Just wondering, why am I wrong/ how do I go about getting the right answer?
Thank you.
The mistake was in how you wrote it: xx5xx7xx9xx.
Consider the following. On one hand, we can do 6x58x7xx9xx. This would give us the number 65879.
On another hand, we can do x65x87xx9xx. This is a different arrangement in how you did 8P2, but observe that it also gives 65879! Hence the number 65879 has been double counted.
The flaw was that you focused too much on the possibility of something like 68579, which actually escapes any double counting. But it only does so because the even numbers got grouped between two odd numbers. It doesn’t work when the even numbers are split.
Instead, a better option would be the following. First arrange the 6 by picking one of the 4 crosses in: x5x7x9x.
Then arrange the 8. Note that now that the 6 has been arranged, we’d have something like x6x5x7x9x. Therefore we pick one out of 5 crosses for the 8.
In total, we therefore have \(4\times5=20\) favourable arrangements, giving a probability of \(\frac{20}{120}=6\).
Note: there is also a clever approach here. Recall the problem of arranging the letters ABCDEFG, if say the E must come before the F. But you can match each arrangement with E before F with another one with F before E. (For example, match BCDEFGA with BCDFEGA.) So exactly half of the arrangements have E before F, I.e. \(\frac{7!}{2}\) arrangements.
Here, the idea is that we can group any configuration of the odd numbers among each other. Take for example 65879. We can match it with:
- 65897
- 67859
- 67895
- 69875
- 69857
So each number with 5,7,9 in order can be matched with exactly five other numbers, without changing the position of the 6 and the 8, but worn 5,7,9 not in order. Hence only one sixth of arrangements have 5,7,9 in order.
Nevertheless, that was a nice idea with the stars and bars like idea. Just executed with a flaw!
Thank you so much! This really helps.That is perfectly mathematically sound. It’s just the more cleaner way of expressing my quicker way :)
Just a thought, and I'm not sure whether this is purely coincidental, but could we potentially say that all of these 5 digit numbers have some sort of arrangement of 5,7, and 9. Now, 5,7 and 9 can be arranged in 3! = 6 ways. Only one such arrangement is them in ascending order (5,7,9). So the probability is 1/6.
Is this a mathematically sound justification or a coincidence?
Will you guys make a new Maths Extension 2 course notes for the new syllabus???Yep.
Hello, I have trouble working out this integral. My working out is attached, the answer is 4𝜋/27 - √3 /9. Can anyone please help me out? Thanks :)Your mistake was pretty deep in actually. It was in the transition from line 7 to line 8, next to the blue text.
Your mistake was pretty deep in actually. It was in the transition from line 7 to line 8, next to the blue text.Ahhh I see, thank you :)
\[ \text{Notice how you wrote}\\ \int_{-\frac1{\sqrt3}}^{\frac1{\sqrt3}} \frac{dx}{1+9x^2} = \int_{-\frac1{\sqrt3}}^{\frac1{\sqrt3}} \frac{dx}{\frac19 + x^2}\\ \text{without appropriately adjusting in the numerator as well!} \]
\[ \text{The correct change is}\\ \int_{-\frac1{\sqrt3}}^{\frac1{\sqrt3}} \frac{dx}{1+9x^2} = \int_{-\frac1{\sqrt3}}^{\frac1{\sqrt3}} \frac{\frac19\, dx}{\frac19 + x^2}\]
Can someone please help with part (iii) for this question? Thank you.Edit: At later glance I realised I slightly messed up the base.
\[ \text{When }n=1,\\ \text{the only configurations for }(r,s)\text{ are }(1,0)\text{ and }(0,1).\\ \text{In both cases, we know that }P(r,s) = 1\text{ from the question.} \]
\[\text{And also, }\\ \frac{1}{2^1}\left[ \binom10 + \binom11\right] = \frac22 = 1.\\ \text{Hence the statement holds for }n=1. \]
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\[ \text{Now assume the statement holds }\textbf{for all}\text{ configurations }(r,s)\\ \text{such that }r+s-1=k. \\ \text{We then need to prove the statement holds for all configurations }(r,s)\\ \text{such that } r+s-1=k+1. \]
That was perhaps the most confusing part of the induction - figuring out the goal of the inductive step.
\[ \text{Firstly, from the definition of the recurrence,}\\ P(r,s) = \frac12 P(r-1, s) + \frac12 P(r,s-1). \]
\[ \text{Now if }r+s-1 = k+1\text{, we know that}\\ (r-1)+s-1 = k\text{ and } r + (s-1) - 1= k.\\ \text{Hence we may use the inductive assumption on }\textit{both}\text{ terms:} \]
\begin{align*}
P(r,s) &= \frac12 P(r-1, s) + \frac12 P(r,s-1)\\
&= \frac12 \frac1{2^n} \left[ \binom{k}0 + \binom{k}1 + \dots + \binom{k}{s-2}+ \binom{k}{s-1} \right] + \frac12 \frac1{2^n} \left[ \binom{k}0 + \binom{k}1 + \dots + \binom{k}{s-2} \right] \\
&= \frac1{2^{n+1}} \left[ \binom{k}0 + \binom{k}1 + \dots + \binom{k}{s-2}+ \binom{k}{s-1} \right] +\frac1{2^{n+1}} \left[ \binom{k}0 + \binom{k}1 + \dots + \binom{k}{s-2} \right]
\end{align*}
Ensure that you understand why the first sum terminates at \(s-1\), whilst the second terminates at \(s-2\).
\[ \text{Now to finish, the tricks here are similar to the induction proof of the binomial theorem.}\\ \text{We need to recall that firstly, }\binom{k}{0} = \binom{k+1}{0} = 1.\\ \text{Also, we need Pascal's identity }\boxed{\binom{N+1}{K+1} = \binom{N}{K} + \binom{N}{K+1}}.\\ \text{With }\textit{both}\text{ properties in mind, we proceed to cleverly rearrange the terms.} \]
\begin{align*}
P(r,s) &= \frac1{2^{n+1}} \left[ \binom{k}{0} + \left[ \binom{k}0 + \binom{k}1\right] + \left[ \binom{k}1+ \binom{k}2 \right]+ \dots + \left[\binom{k}{s-2}+\binom{k}{s-1} \right] \right]\\
&= \frac1{2^{n+1}} \left[ \binom{k+1}{0} + \binom{k+1}{1}+\binom{k+1}{2} + \dots + \binom{k+1}{s-1} \right].
\end{align*}
And this is exactly what we wished to prove, so we are done.
*I did this in a bit of a rush so I may have given too little clarification. Let me know if there's any bit you want me to elaborate on
Hey, would somebody be able to explain why the answer to this question is C? I feel myself going insane and hope somebody can assist me in my time of lunacy. Thanks heaps!
Hey, would somebody be able to explain why the answer to this question is C? I feel myself going insane and hope somebody can assist me in my time of lunacy. Thanks heaps!Above explanation should sum it up pretty much. Here's a diagram to go with it.
Above explanation should sum it up pretty much. Here's a diagram to go with it.
hi, I am not sure the answers for this trial paper for part ii)This requires a somewhat combinatorial argument to justify.
went from the 4th last line to the last line
(or from the 2nd tick to the third tick)
thanks. 8)
.wow, would of never thought of this
Hey there!
You're right in that there is a quicker way; usually, if you're listing all these possibilities and the exhaustive list gets a bit long, using the complement is a better idea.
What the restriction actually does is ensure you can't have 5 or 6 people in a single room ie. the combinations 5/1/0/0 and 6/0/0/0 are invalid. Noting that the total number of combinations is just 46, we subtract (4C1x6C5x3C1x1C1 + 4C1x6C6) (which are just the total number of ways of the above combinations respectively) to find the total with the restriction.
Hope this helps :)
Hi!The main thing is that from 2001 onwards, mark allocation for each part of a question was promised.
Can someone please help with part (v) of this question (from the 1993 MX2 HSC paper)?
Also, just a question, I've heard (but I'm not entirely sure) that the question types in the MX2 HSC changed after 2001 - what exactly does this mean? I know they introduced MC in 2012 but I heard something about a change in 2001 as well.
Thanks :)
Just a question that I came across that even stumped my teacher, would appreciate the help. Its pt. iv) of the first and ii) of the secondFor the first part, we just use a double angle formula on \(\cos \frac\pi5 \).
Hi can someone help me with Question 14???
Thanks :)
Hi Thanks so much for your reply!!! I just had one question, how did you deduce that a = 0 from a^2 + ai / a^2 + b^2 + 1 ???
Hi Thanks so much for your reply!!! I just had one question, how did you deduce that a = 0 from a^2 + ai / a^2 + b^2 + 1 ???
Hi!I'm so sorry, but I still don't understand how we can say a =0??? I'm sorry for troubling you guys :(
So just following up from DrDusk's solution - we are able to deduce that a = 0 since the question specifies that the given expression, z/(z-i) is purely real.
This essentially implies that the imaginary part of the expression is zero, that is, after realising the denominator, the coefficient of i. Since the denominator cannot equal zero, this means the coefficient of i on the numerator, in this case 'a' = 0.
Hi!Wait I think i understand it. So basically we're solving a / a^2 + b^2 +1 = 0, since the imaginary part is 0. from this we get a = 0
So just following up from DrDusk's solution - we are able to deduce that a = 0 since the question specifies that the given expression, z/(z-i) is purely real.
This essentially implies that the imaginary part of the expression is zero. After realising the denominator, we look for the coefficient of i and let this equal zero. In this case, the coefficient of i (the imaginary part) is; ai / (a2 +b2 +1). Since the denominator cannot equal zero, this means that a = 0.
Wait I think i understand it. So basically we're solving a / a^2 + b^2 +1 = 0, since the imaginary part is 0. from this we get a = 0
I'm sorry for bothering you guys again, but can someone help me with this question?\begin{align*}
THANK YOU SO MUCH FOR YOUR HELP!!!
How would I solve this question? I know that if you multiply a complex number by i, you get a rotation of pi/2. I also know that arg(zw)=arg(z)+arg(w), but how would I use these to show the working out??
Any help is appreciated :)
Hi Mani.s_Thank you so much!!!
Speaking in general terms, when you multiply a complex number by (cosθ + isinθ), you're really rotating it anticlockwise about the origin by θ radians. Indeed, a rotation by i will produce an anticlockwise rotation of π/2 since i in mod/arg form is really (cos(π/2) + isin(π/2)).
For this question, you are essentially given a value of θ. What you must do is convert cosθ + isinθ (the mod/arg form) into the Cartesian form - more generally expressed as x + iy. To do so, you use exact trigonometric values, e.g. for b, cos(π/3) + isin(π/3) becomes 1/2 + i√3/2.
I hope this helps!
Hi Mani.s_Thank you so much!!!
Speaking in general terms, when you multiply a complex number by (cosθ + isinθ), you're really rotating it anticlockwise about the origin by θ radians. Indeed, a rotation by i will produce an anticlockwise rotation of π/2 since i in mod/arg form is really (cos(π/2) + isin(π/2)).
For this question, you are essentially given a value of θ. What you must do is convert cosθ + isinθ (the mod/arg form) into the Cartesian form - more generally expressed as x + iy. To do so, you use exact trigonometric values, e.g. for b, cos(π/3) + isin(π/3) becomes 1/2 + i√3/2.
I hope this helps!
I need help in this question, and at the end the messed up letter is meant be be alpha conjucate
Did you justify why the bar vanishes off the coefficients \(a\), \(b\) and \(c\)?
Did you justify why the bar vanishes off the coefficients \(a\), \(b\) and \(c\)?I knew this was coming hahaha ;)
Oh, I see. But how do we know that we need to sub alpha at the start. ThanksFrom the fact that \(\alpha\) is by definition a root of \(ax^2+bx+c=0\).
From the fact that \(\alpha\) is by definition a root of \(ax^2+bx+c=0\).
You're not being asked to show that \(\alpha\) is a root; you're given it. (What has to be shown is that the same result holds when \(\overline{\alpha}\) is subbed in.)
I have this question, and my attempted answer is below, I am not completely sure if it is the right approach.You can't do that because the conjugate doesn't apply to a, b and c IF they are real. We haven't proved this so you can't just use it in your proof like that. Instead you should do this:
Any assistance would be great.
Hey there!
Could you possibly explain your line of thinking when writing 'since the conjugate roots only applies to imaginary components we can say a, b and c are real?'. I'm a little confused - we could properly correct/understand/explain better if you clarified this :)
Perhaps a more clear approach (alternative to the above) would be to let \(\alpha = x+iy\) and \(\bar{\alpha} = x-iy\), thus leading to the fact that both
\(a(x+iy)^2+b(x+iy)+c=0\) and \(a(x-iy)^2+b(x-iy)+c=0\) are true. Adding these up, we see that we have \(2a(x^2-y^2)+2bx+c=0\) - given a, x and y are real, b and c must also be real.
Hope this helps :)
Hi there, I was trying to just say that the conjugate only affects the imaginary part for example, x= a+ib then x bar = a- ib.
But yes, I see your way of solving the question. Many thanks for the help :)
In the Vector question, I am getting a different answer to the book's. My answer and book's answer is attached. Would you please help me where I went wrong or if the book answer is wrong.
This makes things clearer - but as with your original statement 'since the conjugate roots only applies to imaginary components we can say a, b and c are real' - the jump from the first part to the second part is very awkward; the logic that takes you from A to B is very flawed. I can't quite see how or why that was assumed?Perhaps explaining your line of thinking could help us better your understanding :) Though it's true in this case, you can't assume this every time, and you certainly shouldn't be assuming things in the first place - it's much safer to use one of the methods described above :)
The book's answer is wrong. In future, a more helpful check (when given the value of each of the complex numbers!) is to actually compute what z-w is: in this case it's \(\left(1-\frac{1}{\sqrt{2}}\right) + \left(\sqrt{3}+\frac{1}{\sqrt{2}}\right)i\) - which is clearly in the first quadrant, not on the imaginary axis as the book's answer suggests. In actuality, the book's answer and your answer is identical - the book likely hasn't taken care in drawing a diagram properly, as well as positioning the origin of the ray at the actual origin.
Hope this helps :)
Hey there!
Note that while \(\vec{AB}\) represents the same complex number as \(\vec{OC}\), you can't just use that to find \(\vec{OB}\), since \(\vec{OB}\) clearly does not equal \(\vec{OC}\) (one is the diagonal of the square, the other is a side!). What you've actually done is rotate \(\vec{OA}\) clockwise 90 degrees to give \(\vec{OC}\).
There are two ways to do this:
a) Consider that \(\vec{OB}=\vec{OA}+\vec{AB}\) - you've got both values; the first given in the question and the second you've just calculated as equal to \(\vec{OC}\). Add them up and you'll have \(\vec{OB}\).
b) Consider that \(\vec{AB}\) is a rotation anti-clockwise of \(\vec{AO}\) by 90 degrees. Recalling that \(\vec{AO}\) = \(-\vec{OA}\), we find that \(\vec{AO}=-2-i\). Rotating to find \(\vec{AB}\), the computation then becomes identical to the one above via addition of vectors.
There technically is also a 'cheat' way of doing it by inspection; it's a relatively simple square and some people might pick up that B is 1+3i straightaway, but there's no working out involved; for the most part this is only helpful for checking your answer.
Hope this helps :)
Hi there. Yes I see the way of doing it, I think a) is much easier for me to follow, but I don't get how OB=OA + AB, and the fact that we added OA after rotating is not quite clear. Would you please explain me why that is. Thanks
I just need some clarification here about why the following cannot be one of the possible positions.
Thanks
\(\vec{OB}=\vec{OA}+\vec{AB}\) because in general, for any complex numbers X, Y and Z, \(\vec{XZ}=\vec{XY}+\vec{XZ}\) - consider the following diagram:Also, consider the square we have as well, as per the diagram below:Click for the diagram!(https://i.imgur.com/cOHW9hF.png)When we rotate \(\vec{OA}\), we have \(\vec{OC}\), which you'll note is identical to \(\vec{AB}\). From the above, we must then add to have \(\vec{OB}\). Note that here \(\vec{OB}\) is the diagonal of the square; you'll need to add the two sides to get the value of the complex number.Click for the next diagram!(https://i.imgur.com/LfLxdc9.png)
This is clear now. I understand it. Thanks
Hi,
I was wondering what resources I could use to study for the new 3U and 4U courses? Are there certain websites I can find past HSC papers? I'm a bit worried about studying for mathematics with the new syllabuses having come in so any advice/recommendations would be greatly appreciated!
Hi, I am not sure how to do this question, any hints to start off will be much appreciated .For \(y^2 = f(x)\) curves, the curve is always symmetric about the \(x\)-axis. Hence the area under the \(x\)-axis should equal the area above it.
Thanks
In this question we need to find the area of the curve bounded by the abscissas. So we need to make x the subject. I have many operations but I can't find a way of making x the subject.\begin{align*}y &= \frac{x}{\sqrt{x^2+5}}\\ y^2 &= \frac{x^2}{x^2+5}\\ y^2(x^2+5) &= x^2\\ 5y^2 &= x^2 - x^2y^2\\ 5y^2 &= x^2(1-y^2)\\ x^2 &= \frac{5y^2}{1-y^2} \\ x &= \frac{\sqrt{5}y}{\sqrt{1-y^2}}\end{align*}
\begin{align*}y &= \frac{x}{\sqrt{x^2+5}}\\ y^2 &= \frac{x^2}{x^2+5}\\ y^2(x^2+5) &= x^2\\ 5y^2 &= x^2 - x^2y^2\\ 5y^2 &= x^2(1-y^2)\\ x^2 &= \frac{5y^2}{1-y^2} \\ x &= \frac{\sqrt{5}y}{\sqrt{1-y^2}}\end{align*}
Note that the positive square root was taken here because we are working in the first quadrant. In the first quadrant, \(x \geq 0\).
(Note also that this question could be done without remaking \(x\) the subject. One way is to consider the area of a rectangle, minus the area of some region by the \(x\)-axis. But both methods would've relied on the same integration technique.)
Hi, I have a question. Would both these answers be considered valid.They're both correct, but ideally try to use the principle argument where possible. Here, \( \operatorname{Arg}(z) = -\frac\pi2\).
Convert -4i into polar form
4 cis (3pi/2)
4 cis (-pi/2)
They're both correct, but ideally try to use the principle argument where possible. Here, \( \operatorname{Arg}(z) = -\frac\pi2\).Ok. Thanks
Hi!Your working makes perfect sense to me. But just take note that \( \frac{\lambda}{\frac12 \lambda - \frac32} \) reads like a fraction: "lambda over (half lambda - 3 halves)". Try to keep to vector notation: \( \begin{pmatrix} \lambda\\ \frac12 \lambda - \frac32\end{pmatrix} \).
(https://i.imgur.com/ICZD0f3.jpg)
Is my working out right for this? it says answers may vary, but how did they get the answer below?:
(https://i.imgur.com/kLfoqSQ.jpg)
Also, I'm really struggling with these questions:
(https://i.imgur.com/PzSmz03.jpg)
(https://i.imgur.com/gNT8YEr.jpg)
Thanks!
Hi. I am not quite sure why we integrate with the following limits.
Thanks.
Hey there!
One of the ideas that was stressed to me as a student was 'look at what you're working towards'. It's become a really handy hint - note that in the result, the logarithm does not use 'n', rather, it uses '1/n'. Since we know the integral of a linear function is a logarithm, and we can directly see the connection between the given result and the result we're working towards (particularly in the middle term), it makes sense to try out 1/n as an upper limit. As a result of this, this means the right hand side will become 1/n, meaning everything will be multiplied by n, leaving a power on the logarithm. This kind of foresight often reduces the amount of work that needs to be done trialling different limits amongst other things - attempt to use it wherever you can.
Hope this helps :)
Hi!This is actually an extremely demanding question given the scope of MX2. The shortest distance between two lines usually requires knowledge of planes in 3D space or the cross product, both of which aren’t in the syllabus. What is the source of this question?
Another vector question I'm struggling with:
"Find the shortest distance between the line through the points (1, 3, 1) and (1, 5, -1) and the line through the points (0, 2, 1) and (1, 2, -3)"
how do you find the equations using vector methods? what would the subsequent steps be?
Thanks!!!
This is actually an extremely demanding question given the scope of MX2. The shortest distance between two lines usually requires knowledge of planes in 3D space or the cross product, both of which aren’t in the syllabus. What is the source of this question?
I'll leave the vector question for a more competent user while I continue getting accustomed to them myself, but I'll answer your second question :)
There are a few ways to prove to yourself this is true:
- You can prove to yourself that they are one and the same by doing the sum of roots and product of roots of both those quadratics
- You can also consider the roots as per the quadratic equation for each quadratic, inverting them and rationalising the denominator
- Also, consider what happens when you have \(f(\alpha)\) in the first quadratic, and factor out \(\alpha ^2\), and vice versa
Hopefully this will start to make sense! :)
I think my coaching taught us some out of syllabus content for vectors- they said that for completeness of this topic, learning the cross product is necessary. I'd still love to know how it would be done though, but I'm not too fussed about it since it won't be tested :)Yeah, their perspective is understandable, but definitely out of the syllabus. Regardless, with the cross product:
Also, could someone explain this proofs question: Q10d from the cambridge textbook? I've forgotten how roots work- am I meant to use the sum and product of roots?
(https://i.imgur.com/Pkpxczz.png)
thanks so much :D
Yeah, their perspective is understandable, but definitely out of the syllabus. Regardless, with the cross product:Thanks heaps again Rui!
...
Hi, back again :DPart i:
Thanks heaps again Rui!
This proofs topic is so confusing for me (it's such a new concept!). Hopefully it will get a lot easier with practice :'(
(https://i.imgur.com/xKFU7xX.png)
Last one for the day- I asked a few of my friends this and we are all quite mind boggled by this question (cambridge enrichment question). Does anyone know how to do this?
Thanks :)
Hi!!
I have reached the dreaded topic of proving ~inequalities~ :')
I have a general question about working out layout:
e.g. RTP \( (a+b)( \frac{1}{a} + \frac{1}{b} ) \ge 4\)
I like to work backwards by assuming that the proof is correct, shuffling it around until it gives me a property that I know is correct (e.g. in this case I ended up with \( (a-b)^2 \ge 0 \)) and then writing the proof out backwards in my final answer- if I do this, where should I put my original assumption? I know we're not meant to write it as an inequalities since that means that you're assuming that it's true
I was thinking about doing LHS= and RHS= but then that doesn't let me move numerals from either side
orrr is there a better method to solving these that doesn't involve me working backwards?
Thanks so much ;D
Thanks a lot for your help!!! I get it now. Also stuck with part (c) of another question. For these types of questions, I know how to solve the LHS but for the RHS I'm not sure. I tried changing the roots from the equation in part (b) into mod-arg form first and then dividing by z^3 but it didn't work.Dividing by \(z^3\) before subbing in \(z = \cos\theta+i\sin\theta\) seems to be the required approach here.
Hey there!
Consider what happens if you take \((\omega + \omega ^8)(\omega ^2 + \omega ^7)(\omega ^4+ \omega ^5)\), where \(\omega = \cos \frac{2\pi}{9} +i\sin \frac{2\pi}{9}\).
We have that \((\omega + \omega ^8)(\omega ^2 + \omega ^7)(\omega ^4+ \omega ^5) = (2\cos \frac{2\pi}{9})(2\cos \frac{4\pi}{9})(-2\cos \frac{\pi}{9})\).
We also have that by expanding the brackets and simplifying using part (b), \((\omega + \omega ^8)(\omega ^2 + \omega ^7)(\omega ^4+ \omega ^5) = -1\).
Hence, we have that \((2\cos \frac{2\pi}{9})(2\cos \frac{4\pi}{9})(-2\cos \frac{\pi}{9}) = -1 \implies \cos \frac{2\pi}{9} \cos \frac{4\pi}{9} \cos \frac{\pi}{9} = 1\).
What's evident in this question is that you occasionally need to look to manipulate the question into a nicer form - here, the solution only really became obvious for me after manipulating the result to \((2\cos \frac{2\pi}{9})(2\cos \frac{4\pi}{9})(-2\cos \frac{\pi}{9}) = -1\). From this sort of position, it's a lot simpler and much less time consuming to explore possible solutions, especially given a number of results from previous parts of the question. Try this line of thinking with similar questions!
Hope this helps :)
Hey there :)
Imagine you have a number line from 0 to 1 (including 0, not including 1 - the question is kinda dodgy here, since adding one doesn't modify \({\alpha}\)) divvied up into N intervals, each of length 1/N (draw it out!). Counting the numbers they give you, there are N+1 objects and you've drawn up N intervals. You can assign them one by one and there will be an interval that has two numbers in it - since we know that the interval length is 1/N, we can deduce that the difference is less than the length of the interval. Note also that if we put a number at each endpoint of every interval, this would still be the case (as 1, or N/N is excluded - this is why this was key in the first place!).
If you're struggling to visualise it still, I recommend a few things:
- draw it out
- try simpler PHP questions then apply the above to that, see if you can replicate the same ideas here
- come back to it later
- don't worry about it too much, it's only going to stress you out this close to the exam
Hi,I don't think it's too late at all. A more important question would be if your teachers would let you. If they deny you entry, that's a barrier out of your control.
I needed some advice on what to do for my subjects in Year 12. Currently, I have started year 12 term 1 at my school and my subjects are as follows:
4U English
3U Math
Chemistry
Business Studies
I've been content with these subjects and am doing relatively well in them at school. We're starting week 4 at school and I've been thinking; is it too late to consider studying math ext 2? I love doing math but I don't know if I have the right to make that decision yet as Year 12 has already started and so has term classes for math ext 2. My friends in 4U math have already had 2 classes so far. Could anyone give me some advice on what to do about this?