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AQA June 2016 Further Pure 2 Unofficial Mark Scheme Watch

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    (Original post by tanyapotter)
    thank god! i panicked so much

    what would 61/75 be? A?
    I haven't really done enough papers to know if that was hard or easy, perhaps someone else can predict some boundaries?


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    (Original post by jtebbbs)
    Did anyone else get this for 7c? Haven't seen anyone else with this answer yet
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    Angle was 5pi/6, then when cube rooting you get 5pi/18.
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    Did anyone else get this for 7c? Haven't seen anyone else with this answer yet
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    (Original post by jtebbbs)
    Did anyone else get this for 7c? Haven't seen anyone else with this answer yet
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    No the argument was \dfrac{5\pi}{6}.
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    Adding this to the unofficial ms thread too, since it seems like it'd be useful here:

    If anyone wants to know how the induction went:(1+p)^1 \geq 1+p obviously, so base case established.Assume (1+p)^k \geq 1+kp. Clearly we need to multiply by 1+p and we can do so without changing the sign of the inequality because p\geq -1\Rightarrow 1+p \geq 0.

    Therefore (1+p)^{k+1} \geq (1+kp)(1+p)=1+p+kp+kp^2=1+(k+1)p  +kp^2 \geq 1+(k+1)p, since p^2 \geq 0, k \geq 1 \Rightarrow kp^2 \geq 0.

    Therefore result holds for n=k+1 and by induction we have result.
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    (Original post by jjsnyder)
    Angle was 5pi/6
    Not too bad compared to some past papers, De Moivre's wasn't too bad but think a lot of people will have tripped up withs signs on the last 8 marks maybe

    Proof by induction was hardest I've seen

    Most questions could be checked with a calculator so that was nice

    EDIT: replied to the wrong post but that was what I thought of the paper, and yeah just realised that
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    (Original post by IrrationalRoot)
    No the argument was \dfrac{5\pi}{6}.
    Ah you're right, I used the argument from the last part, damn
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    I stupidly didn't divide by the coefficient of 3z^3 to find gamma but the continued to p and q with the right method, would they ecf?
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    (Original post by jtebbbs)
    Not too bad compared to some past papers, De Moivre's wasn't too bad but think a lot of people will have tripped up withs signs on the last 8 marks maybe

    Proof by induction was hardest I've seen

    Most questions could be checked with a calculator so that was nice
    Literally used 5 different functions on my calculator to check my answers, it's great.

    I used the series evaluator, cubic equation solver, definite integration button, derivative (at a point) button and complex number mode.

    So I was pretty convinced I got everything right by the end! XD
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    Yeah, all these answers are right. I'm pretty sure Q7 was induction though and Q5 the argand diagram.
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    (Original post by IrrationalRoot)
    Literally used 5 different functions on my calculator to check my answers, it's great.

    I used the series evaluator, cubic equation solver, definite integration button, derivative (at a point) button and complex number mode.

    So I was pretty convinced I got everything right by the end! XD
    Feel sorry for anyone with a standard calculator lol, I got the wrong argument for 7c and struggled with proof by induction but think I got the rest of the marks
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    (Original post by jtebbbs)
    Feel sorry for anyone with a standard calculator lol, I got the wrong argument for 7c and struggled with proof by induction but think I got the rest of the marks
    Yeah it's pretty broken compared to the standard one haha.

    I've posted a solution for the proof by induction if you haven't seen it and would like to .

    If you've got the rest of the marks you should be fine .
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    (Original post by jjsnyder)
    Angle was 5pi/6, then when cube rooting you get 5pi/18.
    Sorry if this is a little irrelevant, but how has STEP gone for you? Did well in I and II?
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    (Original post by IrrationalRoot)
    Adding this to the unofficial ms thread too, since it seems like it'd be useful here:

    If anyone wants to know how the induction went:(1+p)^1 \geq 1+p obviously, so base case established.Assume (1+p)^k \geq 1+kp. Clearly we need to multiply by 1+p and we can do so without changing the sign of the inequality because p\geq -1\Rightarrow 1+p \geq 0.

    Therefore (1+p)^{k+1} \geq (1+kp)(1+p)=1+p+kp+kp^2=1+(k+1)p  +kp^2 \geq 1+(k+1)p, since p^2 \geq 0, k \geq 1 \Rightarrow kp^2 \geq 0.

    Therefore result holds for n=k+1 and by induction we have result.
    I got up to the expanded brackets but didn't know what to do after that so just wrote the finishing proof by induction statement then left it, didn't have time to go back but oh well
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    Completely failed, absolute ****. Goodbye any chance of getting an A or B
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    (Original post by Jm098)
    I stupidly didn't divide by the coefficient of 3z^3 to find gamma but the continued to p and q with the right method, would they ecf?
    ??
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    (Original post by IrrationalRoot)
    Sorry if this is a little irrelevant, but how has STEP gone for you? Did well in I and II?
    I went well, II not well and III terrible so was really disappointed. Expecting S12 or S22 What about you?


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    for the question just before the hyperbolic integration, what did people get for the x value of the stationary point?

    i got x = ln(root34-5/3)

    that's definitely wrong, right?
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    (Original post by tanyapotter)
    for the question just before the hyperbolic integration, what did people get for the x value of the stationary point?

    i got x = ln(root34-5/3)

    that's definitely wrong, right?
    Ln2/3 I think
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    (Original post by jjsnyder)
    These are my answers to todays FP2 exam. Not all of them are necassarily correct, please feel free to correct me on any you think are wrong. Not sure about the question numbers.
    FP2 June 2016
    1.
    a) A=4
    b)  \frac{50}{609}
    2.
    a) i) B=1-2i
    ii) ab = 5
    b) Gamma is 1/3
    c) p=-5 q=-7
    3. a) Show that s is the required integral
    b) ln7-3/4
    4. a) (sqrt(3)/2)/(1+3x)sqrt(x)
    b) (sqrt3)pi/18
    5. Induction Q.
    Trivial in the case n=1 since clearly 1+p >= 1+p
    Assume true for n=k
    (1+p)^k+1 = (1+p)^k(1+p) = (1+kp)(1+p)
    = 1 + kp + p + kp^2 >= 1 + (k+1)p as kp^2>=0
    Thus true for n=k+1 so true for all positive integers n by induction.

    6.a) Show that Q
    b) 6sqrt3 + 3arcosh2 + 5
    7. a) 8
    b)i) Sketch with a circle with centre -4root3 + 4i radius 4
    ii) 2pi/3
    c) Roots are 2e^i(xpi/18) where x=-7,5,17
    8. a) Show that Q
    b) Roots were itan(xpi) where x= 1,3,5,7
    c) i) 1
    ii) 6

    can you remember the marks for each question. if so can you update the markcheme
 
 
 
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