AQA A2 Mathematics MPC4 Core 4 - 9th June 2015 [Discussion & unofficial markscheme]

Going of this mark scheme, I believe I only made 1 mistake (the angle of the vectors question). I obviously miscopied a number somewhere. So, I believe I will have got 72/73, which should still be 100 UMS.

Does anyone think average of 53 on both papers is enough for 140 ums?

How many marks would I get for 8A if i just left my answer in re-arranged form with X, but forgot to calculate the C (because didn't read question properly) ?

Do people think 73/75 on Core 3 and 64 or 65/75 on Core 4 will be enough for 180 on the two?


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*** OFFICIAL AQA CORE 4 MPC4 JUNE 2015 EXAM DISCUSSION THREAD ***1.A) Express $f(x)= \dfrac{19x-2}{(5-x)(1+6x)}$ in the form $f(x) = \dfrac{A}{(5-x)} + \dfrac{B}{(1+6x)}$

$\Rightarrow A=3, B=-1$ (3 marks)

B) Find $\displaystyle\int_0^4 f(x)\ dx$ in the form $p \ln (5)$.

$= \dfrac{8}{3}\ln (5)$ (6 marks)

2.
A) Express $2\cos x – 5 \sin x$ in the form $R \cos (x+\alpha), R>0, 0<\alpha<\frac{\pi}{2}$.

$\Rightarrow\sqrt{29} \cos (x+1.19)$ (3 Marks)

B) Find value of$x$ in interval $0<x<2 \pi$ where is at a maximum.

(i) $x=5.09$(2 Marks)

(ii) Solve equation $2 \cos x – 5 \sin x + 1 = 0$

$x=0.567,3.34$ (3 Marks)

3.
A) $g(x) =8x^3 - 12x^2 - 2x + d$. Given when $g(x)$is divided by $(2x+1)$ the remainder is $-2$, find $d$.

$\Rightarrow d=1$ (2 Marks)

B) Given $x= - \frac{1}{2}$ is a solution of $g(x)=0$, express $g(x)$ as 3 linear factors.

$\Rightarrow g(x) = (2x+1)(2x-1)(2x-3)$ (3 marks)

C) $h(x)$is given by $\dfrac{4x^2 – 1}{f(x)}$ where $x>2$. Simplify $h(x)$,showing it is a decreasing function.

$h(x) = \dfrac{1}{(2x-3)}$,

EITHER

$h'(x) =-2 (2x-3)^{-2}$. $\rightarrow h’(x)$ is always negative, therefore it is a decreasing function.

OR

Could plot graph of $h(x) = \dfrac{1}{2x-3}$ showing $h(x) \rightarrow 0$ when $x \rightarrow \infty$ (4 Marks)

4.
A) Find the binomial expansion of $(1+5x)^{\frac{1}{5}}$:

$= 1 + x -2x^2$ (2 Marks)

B)
(i) Find the binomial expansion of $(8 + 3x)^{\frac{-2}{3}}$:

$=\dfrac{1}{4} - \dfrac{1}{16} x + \dfrac{5}{256} x^2$ (3 Marks)

(ii) Find an estimate for $\sqrt[3]{\dfrac{1}{81}}$.

$= 0.2313$ (2 Marks)

5. $x = \cos(2t), y = \sin (t)$.

A) Find the gradient at $t = \dfrac{\pi}{6}$.

$\dfrac{dy}{dx}= \dfrac{-1}{2}$ (3 Marks)

B) Find the equation of the normal at $t=\dfrac{\pi}{6}$ in the form $y = mx + c$:

$y=2x - \frac{1}{2}$ (2 Marks)

C) Normal cuts curve at $Q(\cos (2q), \sin (q))$. Set up a quadratic in terms of $\sin(q)$ to find where the normal cuts the curve again. Then find the $x$co-ordinate of $Q$.

$\cos (2q)= \dfrac{-1}{8}$ (5 Marks)

6. $A =(3,2,10), B = (5,-2, 4)$
Line $l$passes through $A$ and has the equation $r = \begin{bmatrix} 3 \\ 2 \\ 10 \end{bmatrix} + \lambda \begin{bmatrix} 3 \\ 1 \\ -2 \end{bmatrix}$.

A) Find the acute angle between line $l$ and $\vec{AB}$.

$\theta= 60^{o}$ (4 Marks)

B) $C = (15,6, 2)$ (4 Marks)

C) $E = (11,0, 0)$ and $(23, 4, -8)$ (4 Marks)

7. $y^3 +2e^{-3x}y - x = k$.

A) Find $k$given point $P \left(\ln (2), \frac{1}{2}\right)$ lies on the curve.

$k=\dfrac{1}{4}- \ln(2)$ (1 Mark)

B) Find the gradient at the point $P$:

$\dfrac{dy}{dx}= \dfrac{11}{8}$ (6 Marks)

8. $\dfrac{dx}{dt}= \dfrac{\sqrt{4+5x}}{5 (1+t)^{2}}$.

A) Given $t=0$when $x=0$ find $x$ in terms of $t$.

$x = \dfrac{1}{5}[(\frac{-1}{2} (1+t)^{-1} + \frac{5}{2})^{2} - 4]$ (7 marks)

B) Another pond models a circle of radius $r$. The rate of change of radius $r$is inversely proportional to pond surface area. Write a differential equationto model this situation.

(i) $\dfrac{dr}{dt}= \dfrac{k}{\pi r^2}$ (3 Marks)

(ii) The rate of change of radius $r$ is $4.5$ when the radius is $1$ metre. Find the radius when the rate is $0.5$.

$r=3\ \text{metres}$ (2 Marks)
______________________________

Date: 9th June 2015
Time: 09:00am
Duration: 1h 30m
______________________________
CORE 3 EXAM DISCUSSION THREAD:
http://www.thestudentroom.co.uk/showthread.php?t=3041463

MECHANICS 2 EXAM DISCUSSION THREAD:
http://www.thestudentroom.co.uk/showthread.php?t=3047357

______________________________

RESOURCES:

definitely

Question 8bi - there should be no π in the answer. The question asked for a differential in terms of r, t and a constant of proportionality (e.g. k). While dr/dt = k/πr^2 is technically a correct answer, it is not the simplest form - 1/π is also a constant, and can therefore be included in the constant of proportionality. So the answer should be dr/dt = k/r^2.

I'm so worried. I think I lost 10+ marks on that paper due to stupid mistakes


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How many do you think you would get if you left the pi in?

I reckon they will only dock 1 mark at most for that. They'll probably just ignore it though, as k/π is still of constant value.

what overall grade would it be if I get 57/60 for c3, 48/60 for c4 (made loads of silly mistakes), and 61/75 for s1, considering that I got 240/300 last year?