The Student Room Group

AQA C4 June 2014 Unofficial Mark Scheme

That paper was utterly horrid; worst C4 paper I've ever seen. I didn't have enough time to get all my answers down, so I'd appreciate it if people could contribute their answers, and we'll try and establish a full mark scheme. The links to my other mark schemes for this exam season are linked at the bottom of this post. Marks are in emboldened, underlined brackets, like so [x]. Corrected answers that are still being debated are simply underlined, as so x.

The exam paper is attached to the bottom of this post; thanks to maggiehodgson for it!

Unparseable latex formula:

[br]\begin{enumerate}[br]\item[br]\begin{enumerate}[br]\item m = \dfrac{-1}{2} \hfill \textbf{\underline{[3]}} \\[br]\item \text{Any correct cartesian of: } x = \dfrac{t^{2}}{2} + 1, y = \dfrac{4}{t} - 1 \hfill \textbf{\underline{[2]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]



Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{1}[br]\item [br]\begin{enumerate}[br]\item A = 2, B = 3 \hfill \textbf{\underline{[3]}} \\[br]\item x^2 + 3ln({\dfrac{2x^{2} - x + 2}{5}}) + 1 \hfill \textbf{\underline{[4]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]



Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{2}[br]\item[br]\begin{enumerate}[br]\item (1-4x)^{\frac{1}{4}} = 1 - x - \dfrac{3}{2} x^2 \hfill \textbf{\underline{[2]}} \\[br]\item (2+3x)^{-3} = \dfrac{1}{8} - \dfrac{9}{16} x + \dfrac{27}{16} x^2 \hfill \textbf{\underline{[3]}} \\[br]\item \dfrac{1}{8} - \dfrac{11}{16} x + \dfrac{33}{16} x^2 \hfill \textbf{\underline{[2]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]



Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{3}[br]\item[br]\begin{enumerate}[br]\item A = 5000 \hfill \textbf{\underline{[1]}} \\[br]\item[br]\begin{enumerate}[br]\item \text{Show } p^{10} = 5 \hfill \textbf{\underline{[1]}} \\[br]\item V = £75000 \text{ in } 2018 \hfill \textbf{\underline{[4]}} \\[br]\end{enumerate}[br]\item [br]\begin{enumerate}[br]\item \text{Show } T =\dfrac{ln( \dfrac{5}{2})}{ln (\dfrac{p}{q})} \hfill \textbf{\underline{[4]}} \\[br]\item V = W \text{ in } 2023 \hfill \textbf{\underline{[1]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]\end{enumerate}[br]



Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{4}[br]\item[br]\begin{enumerate}[br]\item[br]\begin{enumerate}[br]\item A = 53.1, R = 5 \hfill \textbf{\underline{[3]}} \\[br]\item \theta = 18.5, 198.5 \hfill \textbf{\underline{[3]}} \\[br]\end{enumerate}[br]\item [br]\begin{enumerate}[br]\item \text{Show } ( tan{2\theta}tan{\theta} = 2 ) \equiv ( 2tan{\theta}^2 = 1 ) \hfill \textbf{\underline{[2]}} \\[br]\item \theta = 35.3, 144.7 \hfill \textbf{\underline{[2]}} \\[br]\end{enumerate}[br]\item[br]\begin{enumerate}[br]\item \text{Show } (2x-1) \text{ is a factor of } 8x^{3} - 4x + 1 \hfill \textbf{\underline{[1]}} \\[br]\item \text{Show } 4cos{2\theta}cos{\theta} + 1 = 8x^{3} - 4x + 1 \text{ where } x = cos{\theta} \hfill \textbf{\underline{[1]}} \\[br]\item cos{72} = \dfrac{\sqrt{5} - 1}{4} \hfill \textbf{\underline{[3]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]



Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{5}[br]\item [br]\begin{enumerate}[br]\item \text{Show } \overrightarrow{PQ} \text{ is parallel to } \left(\ \begin{array}{c} 1 \\ -1 \\ 1 \end{array}\right)\ \hfill \textbf{\underline{[3]}} \\[br]\item[br]\begin{enumerate}[br]\item R(3, -2, 4) \hfill \textbf{\underline{[3]}} \\[br]\item S(3, -5, 1) \hfill \textbf{\underline{[4]}} \\[br]\end{enumeratE}[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 \dfrac{dy}{dx} = \dfrac{-3ye^{3x}}{e^{3x}-2sin{2y}} \hfill \textbf{\underline{[6]}} \\[br]\item m = -\pi \hfill \textbf{\underline{[1]}} \\[br]\end{enumerate}[br]\item \text{y-intercept} = \dfrac{\pi}{4} - \dfrac{ln{2}}{\pi} \hfill \textbf{\underline{[2]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]



Unparseable latex formula:

[br]\begin{enumerate}[br]\setcounter{enumi}{7}[br]\item [br]\begin{enumerate}[br]\item \dfrac{3}{1-3x} + \dfrac{1}{1+x} - \dfrac{4}{(1+x)^{2}} \hfill \textbf{\underline{[4]}} \\[br]\item \dfrac{-1}{2} e^{-2y} = ln(\dfrac{1+x}{1-3x}) + \dfrac{4}{1+x} - \dfrac{9}{2} \hfill \textbf{\underline{[7]}} \\[br]\end{enumerate}[br]\end{enumerate}[br]



C1: http://www.thestudentroom.co.uk/showthread.php?t=2685494
C2: http://www.thestudentroom.co.uk/showthread.php?t=2689168
C3: http://www.thestudentroom.co.uk/showthread.php?t=2709037
FP1: http://www.thestudentroom.co.uk/showthread.php?t=2709071
S1: http://www.thestudentroom.co.uk/showthread.php?t=2704980
(edited 9 years ago)

Scroll to see replies

Reply 1
Ooh - for the second part of Q1, I think I had something like (x-1)(y+1)^2 = 8? Does that ring any bells? :confused:
Reply 2
There was one question where you had to find the value of P. (root(5)-1)/p) P = 4
I think it was in question 5 b
(edited 9 years ago)
Reply 3
Or what Flauta said. :L
Reply 4
I wouldn't be able to recall my dy/dx but I made the gradient to be -pi for Q6
Vectors: (6,-6,6) leaving lambda to be 6,Point S (3,-5,1)
Original post by Vernish
Ooh - for the second part of Q1, I think I had something like (x-1)(y+1)^2 = 8? Does that ring any bells? :confused:

Yeah, that's what I got.
Reply 6
What was 2b?
Where the the /5 come from on part 2b??? I did not get that
I think I got theta = 53.1.
x=8/(y+1)^2 +1

and it was 2017!
Reply 10
Alpha was 53.1 for 5ai
and the angles were around 18.4 and 198.4
Reply 11
Original post by Fillly
I wouldn't be able to recall my dy/dx but I made the gradient to be -pi for Q6
Vectors: (6,-6,6) leaving lambda to be 6,Point S (3,-5,1)


Agree with these.
Original post by Phmat
Alpha was 53.1 for 5ai
and the angles were around 18.4 and 198.4


They will allow 18.5 and 198.5 i think because people used 53.1, I used 53.13 to avoid rounding errors but did put 53.1 previously.
Original post by Ollie_Brown
Where the the /5 come from on part 2b??? I did not get that

The constant was 1-3ln5, for me, so the 5 and the function are in the ln()
Original post by Sayonara
I think I got theta = 53.1.


I did too, one of few things that I'm pretty sure I got right
Also yeah, angle alpha was 53.1 getting 18.5 and 198.5
Reply 16
For 1 (b) I had

y=42x21 y = \frac{4}{\sqrt{2x-2}} -1

which would be
(y+1)2=162x2 (y+1)^2 = \frac{16}{2x-2}

The equations were:

x=t22+1 x = \frac{t^2}{2} + 1

and

y=4t1 y = \frac{4}{t} - 1
Reply 17
Original post by Math12345
x=8/(y+1)^2 +1

and it was 2017!


It was 2018. Because It started in April, 16.8 years ahead rolls over into 2018
Evil question
Original post by chilliBeanDream
The constant was 1-3ln5, for me, so the 5 and the function are in the ln()


ahhh, i see, scared me. Don't reckon they'll deduct any for just leaving the constant on the end will they?
Reply 19
What was question 2b???

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