AQA C2 Unofficial Markscheme 24th May 2017

Equation of curve is 2/5x^5/2-x^2+ 26/5

If I am not wrong

One of the trig equations were

48, 312

(a) (iii) The point (1, ) lies on the curve. Use your answer to part (a)(i) to find the area bounded by the curve, the line and the line , to two decimal places. [3 marks]
so, by a sketch, the area is a rectangle with part (a)(i) subtracted. The area is to two decimal places.

k=8 not 2?

Do you think slightly different methods for 7 will still be fully credited? I did the binomial expansion while it was still in log form and found an equation equal to k-1, then factored out 3

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.
(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]

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

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

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

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

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

(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]

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

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

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

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

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

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

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

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

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

Yeah, the translation should be +4/3

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