AQA C2 Unofficial Markscheme 24th May 2017

I hope this is a joke or god help u on ur results

For 6)b)I) I'm pretty sure the translation was with the vector (4/3,0) and for the last question it asked to fine k in the form (c+1)^2 therefore k=3(c+1)^(2) +1

I think the boundaries will be slightly lower if not exactly the same as C2 2015, around 73 for 100 i'd say .

If they asked you to find and angle in terms of pi would you write pi=...

An angle in terms of pi would be

theta = ...
not
pi = ...

9 was (c+1)^2 = (k-1)/3

This is my exact point, if they asked for theta in terms of pi, you have agreed that it would be correct to write tge answer as theta=...
Therefore as they have asked for k in terms of (c+1)^2, the answer is k=...
Hence k=3(c+1)^(2) +1

Why is the translation -4/3 and not +4/3?

They didn't ask for it in terms of (c+1)^2, they asked for it in terms of k...

"9. Given that , express in terms of . [7 marks]"

I'm so confused, why was the translation not (4,0)???

should the stretch translation be -1/16??? i checked it on my calculator in the graph function in exam and it was right

AQA MPC2 June 2017 Unofficial Markscheme

1. A sector has a perimeter of 22cm and an angle of $\theta$ radians. The radius is 8cm.
(a) Find the value of $\theta$ [2 marks]
$16 + 8\theta = 22$
$\theta = \frac{3}{4}$

(b) Find the area of the sector [2 marks]
$\frac{1}{2}\times 8^2 \times \frac{3}{4} = 24$

2. A triangle ABC has an angle BAC of . There are sides of 6cm and 16cm.

(a) Show that the angle ACB is to the nearest degree [3 marks]
$\frac{\sin (ACB)}{6} = \frac{\sin 120}{16} \Rightarrow ACB = \arcsin \left(\frac{6 \sin 120}{16}\right) = 18.95\dots$, which is 19 degrees to the nearest degree.

(b) Find the area of the triangle. [3 marks]
Other angle is 41 degrees. Hence the area is $\frac{1}{2}\times 16 \times 6 \times \sin 41 = 31.5$cm²

3.
(a) Express $\frac{\sqrt{27^x}}{3^{2x-1}}$ in the form $3^p$ where $p$ is an expression in $x$. [3 marks]

$\frac{\sqrt{27^x}}{3^{2x-1}} = \frac{3^{\frac{3x}{2}}}{3^{2x-1}} = 3^{1 - \frac{x}{2}}$

(b) Hence solve the equation $\frac{\sqrt{27^x}}{3^{2x-1}} = \sqrt[3]{81}$. [2 marks]

$3^{1 - \frac{x}{2}} = 3^{\frac{4}{3}} \Rightarrow 1 - \frac{x}{2} = \frac{4}{3} \Rightarrow x = -\frac{2}{3}$

4. A geometric series has the $n$th term $u_n = 162\left(\frac{2}{3}\right)^n$.
(a) Find $u_1$ and $u_2$. [2 marks]
$u_1 = 108, \;u_2 = 72$.

(b) Find the sum to infinity of the series [3 marks]
First term is 108, common ratio is 2/3.
Hence sum to infinity is $\frac{108}{1 - \frac{2}{3}} = 324$

(c) Find the smallest value of $k$ such that $\sum_{n=k}^{\infty} u_n < 2.5$. [3 marks]
$\sum_{n=k}^{\infty} u_n = 486\left(\frac{2}{3}\right)^k$

$486\left(\frac{2}{3}\right)^k < 2.5 \Rightarrow k > \log_{\frac{2}{3}}\frac{2.5}{486}$.

Hence $k>12.997$ so $k$ is at least 13.

5. A curve satisfies $\frac{\mathrm{d}y}{\mathrm{d}x} = x^{\frac{3}{2}} - 2x$, where $x>0$.
(a) Show that there is only one value of $x$ for which there is a stationary point. [2 marks]
$\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow x(x^\frac{1}{2} - 2) = 0 \Rightarrow x^{1/2} = 2 \Rightarrow x= 4$. Hence there is only one value of $x$ for which there is a stationary point.

(b) Find $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ and show that this is a minimum point. [3 marks]
$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{3}{2}\sqrt{x} - 2$.
When $x = 4$, this is equal to 1. Hence it is positive, so the point is a minimum.

(c) The line $y=2$ is a tangent to the curve. Find the equation of the curve. [4 marks]
Must be a tangent at the stationary point. Hence the point (4, 2) lies on the curve (the stationary point). Integrating gives $y = \frac{2}{5}x^{\frac{5}{2}} - x^2 + c$. As $y(4) = 2$, $c = \frac{26}{5}$ so $y = \frac{2}{5}x^{\frac{5}{2}} - x^2 + \frac{26}{5}$.

6. There is a curve $y = 2^{3x}$.
(a) (i) Use the trapezium rule with five ordinates (four strips) to approximate $\int^{1}_0 2^{3x}\;\mathrm{d}x$ to two decimal places. [4 marks]
Use the trapezium rule formula to find that it is 3.44 to two decimal places.

(a) (ii) State how the approximation may be improved. [1 mark]
Use more strips/ordinates.

(a) (iii) The point (1, $k$) lies on the curve. Use your answer to part (a)(i) to find the area bounded by the curve, the line $x=0$ and the line $y=k$, to two decimal places. [3 marks]
$k=8$ so, by a sketch, the area is a rectangle with part (a)(i) subtracted. The area is $8 - 3.44 = 4.56$ to two decimal places.

(b) The curve $y=2^{3x}$ may be mapped onto the curve of $y = 2^{3x-4}$ by a translation and a stretch. Describe:
(i) the translation [2 marks]

Translation in the vector ($\frac{4}{3}$, 0)

(ii) stretch [2 marks]

Stretch parallel to the $y$-axis scale factor $\frac{1}{16}$.

(c) Use logarithms to solve the equation $2^{3x-4} = 7$, using your answer to three significant figures. [2 marks]

$x = \frac{1}{3}(4 + \log_2 7) = 2.27$ to three significant figures.

7. A curve has the equation $y = 7x + 6 - \frac{1}{x^2}$.
(a) The region bounded by the curve and the lines $x=1$ and $x=2$ is above the $x$-axis. Show that the area of this region is 16. [5 marks]

Area (as curve is above the $x$-axis) is:
$\int^{2}_{1} 7x + 6 - \frac{1}{x^2}\;\mathrm{d}x = \left[\frac{7x^2}{2} + 6x + \frac{1}{x}\right]^2_1$. Substitute in the limits to show that the area is 16.

(b) The normal to the curve is parallel to the line $2y + 8x = 3$. Find the equation of the normal to the curve. [6 marks]
Differentiating gives $\frac{\mathrm{d}y}{\mathrm{d}x} = 7 - \frac{2}{x^3}$. Now, the normal has a gradient of $-4$. Hence the gradient of the tangent to the curve is $\frac{1}{4}$. So

$7 - \frac{2}{x^3} = \frac{1}{4} \Rightarrow \frac{8}{x^3} = -27 \Rightarrow x = -\frac{2}{3}$. Hence $y = -\frac{11}{12}$.

So the equation of the normal is $y + \frac{11}{12} = -4(x + \frac{2}{3})$.

8.
(a) Solve the equation $\cos \theta = \frac{2}{3}$ in the range $0\leqslant \theta \leqslant 360$. Give your answers to the nearest degree. [2 marks]

The values of $\theta$ are 48 and 312.

(b) (i) Given that $4\sin \theta \tan \theta = 4 - \cos \theta$, show that $3\cos^2\theta + 4\cos \theta - 4 = 0$. [3 marks]

$4\sin\theta \frac{\sin\theta}{\cos\theta}= 4 - \cos \theta$

$4\frac{\sin^2\theta}{\cos\theta} = 4 - \cos \theta$

$4\sin^2\theta = 4\cos\theta - \cos^2 \theta$

$4 - 4\cos^2\theta = 4\cos\theta - \cos^2 \theta$

$3\cos^2\theta + 4\cos\theta - 4 = 0$

(ii) Show that there is only one possible value of $\cos \theta$. [2 marks]
Factorising gives $(3 \cos \theta - 2)(\cos \theta + 2) = 0$. As $|\cos\theta| \leqslant 1$, the only possible value of $\cos \theta$ is $\frac{2}{3}$.

(c) Hence solve the equation $4\sin 4x \tan 4x = 4 - \cos 4x$ in the range $0\leqslant x \leqslant 180$. [4 marks]

Solving $\cos 4x = \frac{2}{3}$ in the range $0 \leqslant 4x \leqslant 720$.

Use part (a) and then add on 360 to both solutions to give
$4x = \left\lbrace 48,\; 312,\;408,\;672\right\rbrace$

So
$x = \left\lbrace 12,\; 78,\;102,\;168\right\rbrace$

9. Given that $3\log_2\left(c+2\right) - \log_2\left(\frac{k}{2} + c^3\right) = 1$, express $(c+1)^2$ in terms of $k$. [7 marks]

$\log_2((c+2)^3) - \log_2\left( \frac{2k+c^3}{2} \right) = \log_2 2$

$\log_2(\frac{2(c+2)^3}{2k+c^3}) = \log_2 2$

$\frac{2(c+2)^3}{2k+c^3} = 2$

$2(c+2)^3 = 4k + 2c^3$

$2c^3 + 12c^2 + 24c + 16 = 4k + 2c^3$

$12c^2 + 24c + 16 = 4k$

$12(c+1)^2 = 4k - 4$

$(c+1)^2 = \frac{k - 1}{3}$

Because then it would be 2^3(x-4), which gives you 2^3x-12. But the question wanted 2^3x-4, hence a translation of [4/3, 0] gives you 2^3(x-4/3) -> 2^3x-4

Thank you so so much for this Integer123! I counted up and I think I got 53/75, so not too bad although I should have easily got full marks on the last question, in addition to the translation and stretch questions. Based off the grade boundaries from last year and a couple of years before that, I'm guessing the grade boundaries might be the following:

A = 59/75 (80/100)
B = 52/75 (70/100)
C = 45/75 (60/100)
D = 38/75 (50/100)
E = 32/75 (40/100)

If that's the case, then I think I'll just get a B, probably scoring around 71 UMS. What do you guys think?

Question 2.b) surely using the equation for area, 1/2absinc, if you use sides 16 and 6 you have to use the angle of 19 from previous answer, which gives the area of 15.6 cm^2 otherwise you are using sides a and b with angle b (not c) ????

Nooo! For question 3, I put -1 instead of +1 (forgot to minus the -1 so it's +1). How many marks would I lose and I think I got the next bit right using my value. Is there error carried forward marks?
Also knew it was either 12 or 13 but had to time to think about it so I just put 12 down. What is my luck

For the last question, I did everything up to here:



How many marks do you reckon i can pick up at this question?