The Student Room Group

AQA FP2 June 2015 Unofficial Mark Scheme

Below is my unofficial mark scheme for today's FP2 paper; I would appreciate it if anyone who sat the paper could contribute to the mark scheme, particularly if you think one or more of my answers are wrong. Marks are in emboldened, underlined brackets, like so [x]. Corrected answers that are still being debated are simply underlined, as so x.

I didn't manage to get down all my answers so it would be great if people could contribute with their answers and the marks for each question. I couldn't do the two mark circle question so that needs an answer!



Unparseable latex formula:

[br]\begin{enumerate}[br]\item[br]\begin{enumerate}[br]\item A = 1, B = -1 \hfill \textbf{\underline{[3]}} \\[br]\item \dfrac{1}{2} - \dfrac{1}{(n+2)!} \hfill \textbf{\underline{[2]}} \\[br]\end{enumerate}[br]\end{enumeate}[br]





Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{1}[br]\item [br]\begin{enumerate}[br]\item \text{Sketch} y = tanhx \text{asymptotes are} y = \pm 1 \hfill \textbf{\underline{[3]}} \\[br]\item \text{Show} sech^{2}x + tanh^{2}x = 1 \hfill \textbf{\underline{[3]}} \\[br]\item x = \dfrac{ln 3}{2}, \dfrac{-ln 5}{2} \hfill \textbf{\underline{[5]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]





Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{2}[br]\item[br]\begin{enumerate}[br]\item \text{Show something} \hfill \textbf{\underline{[4]}} \\[br]\item S_{x} = \pi (6ln 2 - 2) \hfill \textbf{\underline{[5]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]





Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{3}[br]\item[br]\begin{enumerate}[br]\item f(k+1) - 16f(k) = 33(3^{3k}) \hfill \textbf{\underline{[3]}} \\[br]\item \text{Inductive proof.} \hfill \textbf{\underline{[4]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]





Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{4}[br]\item[br]\begin{enumerate}[br]\item \text{Sketch} \left|z - 2 + 4i\right| = \left|z\right| \hfill \textbf{\underline{[3]}} \\[br]\item[br]\begin{enumerate}[br]\item \text{Show midpoint} = \dfrac{5}{2} - \dfrac{5i}{4} \hfill \textbf{\underline{[4]}} \\[br]\item \text{Equation of circle} is \left|z-\left(\dfrac{5}{2} - \dfrac{5i}{4}\right)\right| = \dfrac{5\sqrt{5}}{4} \hfill \textbf{\underline{[2]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]\end{enumerate}[br]





Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{5}[br]\item[br]\begin{enumerate}[br]\item \dfrac{dy}{dx} = 2\sqrt{5 + 4x - x^{2}} \hfill \textbf{\underline{[5]}} \\[br]\item I = \dfrac{9 \sqrt{3} }{8} + \dfrac{3\pi}{4} \hfill \textbf{\underline{[3]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]





Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{6}[br]\item [br]\begin{enumerate}[br]\item[br]\begin{enumerate}[br]\item \alpha = \dfrac{-1}{3}, \beta = \gamma = \dfrac{2}{3} \hfill \textbf{\underline{[5]}} \\[br]\item k = 1 \hfill \textbf{\underline{[1]}} \\[br]\end{enumerate}[br]\item[br]\begin{enumerate}[br]\item \alpha^{2} = -2i, \alpha^{3} = -2 - 2i \hfill \textbf{\underline{[2]}} \\[br]\item k = 25 - 27i \hfill \textbf{\underline{[2]}} \\[br]\end{enumerate}[br]\item 4x^{3} - 12x^{2} + 35 = 0 \hfill \textbf{\underline{[7]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]





Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{7}[br]\item [br]\begin{enumerate}[br]\item[br]\begin{enumerate}[br]\item \text{Show} \omega \text{ is a root of } z^{5} = 1 \hfill \textbf{\underline{[1]}} \\[br]\item \text{Show} 1 + \omega + \omega^{2} + \omega^{3} + \omega^{4} = 0 \hfill \textbf{\underline{[1]}} \\[br]\item \texts{Roots in form of} \omega^{a} = cos\dfrac{k\pi}{5} + isin\dfrac{k\pi}{5} \text{ for } a = 2,3,4; k = 4,-1,-3 \hfill \textbf{\underline{[1]}} \\[br]\end{enumerate}[br]\item[br]\begin{enumerate}[br]\item \text{Show that } \left(\omega + \dfrac{1}{\omega}\right)^{2} + \left(\omega + \dfrac{1}{\omega}\right) + 1 = 0 \hfill \textbf{\underline{[2]}} \\[br]\item \text{Show} cos \dfrac{2\pi}{5} = \dfrac{\sqrt{5} - 1}{4} \hfill \textbf{\underline{[4]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]\end{enumerate}[br]

(edited 8 years ago)

Scroll to see replies

Reply 1
8. show cos2pi/5 = (root5-1)/4

1+ w + w^2 +w^3 +w^4 =0
(edited 8 years ago)
Reply 2
6(a) the constant was 2


5bii) [z-(5/2-5/4i)]= 5root5/4

8 was all show that questions pretty much and I just posted 7 on the other thread
so something was show (xdot)^2 + (ydot)^2 = (1+1/(t^2))^2
What was question 9?
Reply 5
How much will I get for drawing tanhx properly but stating stupidly asymptotes x=+-1 instead of y?
I got 4x^3 -12x^2 + 15
Reply 7
It'd be nice if my LaTeX would work my god.
Reply 8
Original post by ScienceGeek!
What was question 9?


Ignore that, my syntax is ****ed; I'm trying to fix it! :tongue:
Original post by Doomlar
Ignore that, my syntax is ****ed; I'm trying to fix it! :tongue:


Oh thank god, mini heart attack over! haha
Reply 10
Original post by ScienceGeek!
Oh thank god, mini heart attack over! haha


Really sorry about that hahaha! :tongue:
Reply 11
Wasn't 6 the inverse sin differentiation?
Reply 12
Original post by ubisoft
Wasn't 6 the inverse sin differentiation?


Can you remember the individual sections? I'll update the MS when I get home!
Reply 13
Original post by Doomlar
Can you remember the individual sections? I'll update the MS when I get home!


I done that wrong, but k was 2. it was 2root(quadratic), then it was intergrating that with limits, again can't remember.

Don't tell me you're using latex on your phone?!
Reply 14
Original post by ubisoft
Wasn't 6 the inverse sin differentiation?


Ye: differentiate: y= (x-2)(5+4x-x^2) + sin^-1((x-2)/3)

then integrate root 5+4x -x^2 between 2 and 7/2 giving 9root3/8 +3pi/4
Wasn't It 6 x-2 square root of 5+4x-x^2 +9sin^-1(x-2)/3
(edited 8 years ago)
Reply 16
Original post by ubisoft
I done that wrong, but k was 2. it was 2root(quadratic), then it was intergrating that with limits, again can't remember.

Don't tell me you're using latex on your phone?!


Nah don't worry I did it in the library at college :wink: I tried to latex on my phone once and ended up just walking away from my phone lmao

Posted from TSR Mobile
I got 33/4root3 + 3pi/4 for 6
I got k=-27? And radius as something root5/4??
(edited 8 years ago)
Original post by andywells
I got k=-27? And radius as something root5/4??


I got -(1+i)

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