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AQA A2 MFP2 Further Pure 2 – 24th June [Exam Discussion Thread] watch

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    (Original post by Jm098)
    What's the expression for alpha^3 + beta^3 + gamma^3?
    (a+b+g)^3 - 2(a+b+g)(ab+bg+ga)
    At least I think so...
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    Can anyone provide me with a quick step by step method of how to attack proof by induction questions please? Really struggle with them
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    (Original post by C0balt)
    I will accept my fate as soon as I see 9 questions with hard-core DM question as 9th.
    I'm still holding out a bit of hope that this could be nice because surely it'll just be compensation for how awful FP3 and M3 were.

    That hope will promptly disappear as soon as I can read the first question and AQA suddenly decide that the gamma function is on the syllabus or some other curveball.
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    http://filestore.aqa.org.uk/subjects...W-QP-JUN11.PDF

    On 7b when it says "the three other roots of the equation", isn't there an unlimited amount. Like theta = 6pi/5, 7pi5, 8pi/5, etc.

    ALSO, on part c), why do you reject -root5.
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    I'm having trouble understanding last year's partial fractions question, could someone please show their solution

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    Cheers
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    (Original post by Buntar)
    I'm having trouble understanding last year's partial fractions question, could someone please show their solution

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    Cheers
    I posted an explanation on page 5. I hope it helps!
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    (Original post by TheLifelessRobot)
    I posted an explanation on page 5. I hope it helps!
    Perfect, exactly what I was after!

    Thanks a lot
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    (Original post by 2014_GCSE)
    http://filestore.aqa.org.uk/subjects...W-QP-JUN11.PDF

    On 7b when it says "the three other roots of the equation", isn't there an unlimited amount. Like theta = 6pi/5, 7pi5, 8pi/5, etc.

    ALSO, on part c), why do you reject -root5.
    For part b, you're right there are infinite solutions for theta, but they give the same value as the first 4,  \tan(\pi/5)=\tan(11 \pi/2) , so its theta mod  \pi .
    You reject  -\sqrt 5 for c as you know that  \tan(2\pi/5) and  \tan(\pi/5) are both positive and so cannot multiply to give an negative value.
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    On 6c) how do you get costheta from root(4-sin^2(theta))

    http://filestore.aqa.org.uk/subjects...W-MS-JUN14.PDF
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    (Original post by 2014_GCSE)
    On 6c) how do you get costheta from root(4-sin^2(theta))

    http://filestore.aqa.org.uk/subjects...W-MS-JUN14.PDF
    I think what you mean is how do you get \cos\theta from \sqrt{4-(2\sin\theta)^2} since the substitution is x=2\sin\theta.

    In which case that is worrying to hear right before the FP2 exam, but nonetheless you get it from this:

    \sqrt{4-(2\sin\theta)^2} =\sqrt{4-4\sin^2\theta}=\sqrt{4(1-\sin^2\theta)}=2\sqrt{\cos^2 \theta}.

    Now you can be very hand-wavy (just like the mark scheme is here) and just say this is equal to 2\cos\theta.

    (If you want to be technically correct, then you would actually write 2\sqrt{\cos^2\theta}=2|\cos \theta| and we can actually let \theta \in [-\frac{\pi}{2},\frac{\pi}{2}] since this gives the full range of x=\sin\theta (so the substitution is still valid). Now \cos\theta \geq 0 on this interval and so |\cos\theta|=\cos\theta and we can then carry on as normal.
    Notice that the restriction we placed on \theta also allows us to take arcsines.)

    Hopefully the reason you were confused was that you were actually trying to get \cos\theta from \sqrt{4-\sin^2\theta}, which is impossible.
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    (Original post by mazsaz)
    Can anyone provide me with a quick step by step method of how to attack proof by induction questions please? Really struggle with them
    This is probably too late, but if you have a statement P(n) which you want to prove for all positive integers n:

    Prove P(1) (usually easy).
    Assume that P(k) is true. Use that to prove that P(k+1) is true.
    Then make some statement like 'P(1) is true and P(k)\RightarrowP(k+1), so P(n) holds for all positive integers n by induction.'

    I must warn you that it is much more important to understand how induction works rather than memorising the method, since the method could vary (as an example, you might only have to prove P(n) for all even positive integers in which case you'd prove P(2) and then assume that P(k) is true (where k is an even positive integer) and use that to prove that P(k+2) is true). Usually the way in which it'll vary will be that maybe you have to prove P(0) or P(2) as your base case.
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    (Original post by JackHKeynes)
    (a+b+g)^3 - 2(a+b+g)(ab+bg+ga)
    At least I think so...
    You sure? Haha, thanks
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    (Original post by Jm098)
    You sure? Haha, thanks
    That's right but it's just a 3 instead of a 2 behind the minus sign
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    I wish you all the best with this exam. Let us hope that AQA will be like C4 after that C3
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    Fair paper... I didn't know how to do induction though
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    grade boundary predictions?? I reckon it was slightly harder than last year
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    (Original post by C0balt)
    Fair paper... I didn't know how to do induction though
    Nor the 1+x^2/1-x^2 integration.. Nor the argw question..
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    What did people get for the hyperbolic integration
    • Thread Starter
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    Unofficial MS: http://www.thestudentroom.co.uk/show....php?t=4184065
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    (Original post by C0balt)
    What did people get for the hyperbolic integration
    6sqrt3 + something arcosh2 + 5
 
 
 
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