# Cambridge Additional Mathematics 2011 Past Paper Oct Nov Paper 12

## Question 1

Show that $$\frac{1}{\tan \theta+\cot \theta}=\sin \theta \cos \theta$$. [3]

\begin{aligned} & \frac{1}{\tan \theta+\cot \theta} \\=& \frac{1}{\left(\frac{\sin \theta}{\cos \theta}\right)+\left(\frac{\cos \theta}{\sin \theta}\right)} \\=& \frac{1}{\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta}} \\=& \frac{\sin \theta \cos \theta}{\sin ^{2} \theta+\cos ^{2} \theta} \\=& \sin \theta \cos \theta \text { (shown) } \end{aligned}

## Question 2

Find the coordinates of the points where the line $$2 y=x-1$$ meets the curve $$x^{2}+y^{2}=29$$.  [5]

$$2 y=x-1$$
$$y=\frac{x-1}{2}$$

Substitute into $$x^{2}+y^{2}=29$$

$$x^{2}+\left(\frac{x-1}{2}\right)^{2}=29$$
$$\frac{4 x^{2}+(x-1)^{2}}{4}=29$$
$$4 x^{2}+x^{2}-2 x+1=116$$
$$5 x^{2}-2 x-115=0$$
$$(5 x+23)(x-5)=0$$
$$x=\frac{-23}{5}, x=5$$

Substitute into $$y=\frac{x-1}{2}$$

\begin{aligned} x &=-\frac{33}{5} \\ y &=\frac{\left(-\frac{23}{5}\right)-1}{2} \\ &=-\frac{14}{5} \\ x &=5 \\ y &=\frac{5-1}{2} \\ &=2 \end{aligned}

The coordinate of the point are
$$\left(-\frac{23}{5},-\frac{14}{5}\right)$$ and (5,2)

## Question 3

(i) Express $$\log _{x} 2$$ in terms of a logarithm to base 2. [1]

$$\log _{x} 2$$
$$=\frac{\log _{2} 2}{\log _{2} x}$$
$$=\frac{1}{\log _{2} x}$$

(ii) Using the result of part (i), and the substitution $$u=\log _{2} x$$, find the values of $$x$$ which satisfy the equation $$\log _{2} x=3-2 \log _{x} 2$$ [4]

$$\log _{2} x=3-2 \log _{2} 2$$
$$\log _{2} x=3-2\left(\frac{1}{\log _{2} x}\right)$$

Substitute $$u=\log _{2} x$$

$$u=3-\frac{2}{u}$$
$$u-3=-\frac{2}{u}$$
$$u^{2}-3 u=-2$$
$$u^{2}-3 x+2=0$$
$$(u-1)(u-2)=0$$
$$u=1, u=2$$

\begin{aligned} \log _{2} x_{1} &=1 \\ x_{1} &=2 \end{aligned}

$$\log _{2} x_{2}=2$$
$$x_{2}=4$$

## Question 4

A curve has equation $$y=\left(3 x^{2}+15\right)^{\frac{2}{3}}$$. Find the equation of the normal to the curve at the point where $$x=2$$. [6]

when $$x=2,$$
$$y=\left(3(2)^{2}+15\right)^{\frac{2}{3}}$$
$$y=9$$

$$\frac{d y}{d x}=\frac{2}{3}\left(3 x^{2}+15\right)^{-\frac{1}{3}}(6 x)$$
$$\frac{d y}{d x}=4 x\left(3 x^{2}+15\right)^{-\frac{1}{3}}$$

\begin{aligned} & \text { When } x=2, \\ \frac{d y}{d x} &=4(2)\left(3(2)^{2}+(15)\right)^{-\frac{1}{3}}.\\ &=\frac{8}{\sqrt[3]{(12+(15)}} \\ &=\frac{8}{\sqrt[3]{27}} \\ &=\frac{8}{3} \end{aligned}

$$m=-\frac{3}{8}$$

Equation of the normal
\begin{aligned} y-9 &=-\frac{3}{8}(x-2) \\ y-9 &=-\frac{3}{8} x+\frac{3}{4} \\ y &=\frac{39}{4}-\frac{3}{8} x \end{aligned}

## Question 5

Variables $$x$$ and $$y$$ are such that, when $$y^{2}$$ is plotted against $$2^{x}$$, a straight line graph is obtained. This line has a gradient of 5 and passes through the point (16,81) .

(i) Express $$y^{2}$$ in terms of $$2^{x}$$. [6]

$$Y=m X+C$$
$$At (16,81)$$
$$81=5(16)+c$$
$$c=1$$

$$Y=y^{2}$$
$$X=2^{x}$$
$$m=5$$
$$c=1$$
Therefore
$$y^{2}=5\left(2^{2}\right)+1$$

(ii) Find the value of x when y = 6. [3]

$$y^{2}=5\left(2^{x}\right)+1$$
$$(6)^{2}=5\left(2^{x}\right)+1$$
$$5\left(2^{x}\right)=36-1$$
$$2^{x}=\frac{35}{5}=7$$
$$\lg 2^{x}=\lg 7$$
$$x=\frac{\lg 7}{\lg 2}=2.81$$

## Question 6

(i) Given that $$(3+x)^{5}+(3-x)^{5}=A+B x^{2}+C x^{4},$$ find the value of $$A,$$ of $$B$$ and of $$C$$. [4]

(ii) Hence, using the substitution $$y=x^{2},$$ solve, for $$x,$$ the equation $$(3+x)^{5}+(3-x)^{5}=1086$$. [4]

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## Question 7

(i) Show that $$\frac{(4-\sqrt{x})^{2}}{\sqrt{x}}$$ can be written in the form $$p x^{-\frac{1}{2}}+q+r x^{\frac{1}{2}},$$ where $$p, q$$ and $$r$$ are integers to be found. [3]

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(ii) A curve is such that $$\frac{ d y}{ d x}=\frac{(4-\sqrt{x})^{2}}{\sqrt{x}}$$ for $$x>0$$. Given that the curve passes through the point (9,30) , find the equation of the curve. [5]

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## Question 8

The line $$C D$$ is the perpendicular bisector of the line joining the point $$A(-1,-5)$$ and the point $$B(5,3)$$

(i) Find the equation of the line $$C D$$. [4]

\begin{aligned} \operatorname{Mid}_{A B} &=\left(\frac{5-1}{2}, \frac{3-5}{2}\right) \\ &=(2,-1) \end{aligned}

$$\begin{array}{l} m_{A B}=\frac{3+5}{5+1} \\ m_{A B}=\frac{4}{3} \end{array}$$

\begin{aligned} m_{A B} \times & m_{C D}=-1 \\ & m_{C D}=-\frac{3}{4} \end{aligned}

\begin{aligned} y+1 &=-\frac{3}{4}(x-2) \\ y &=-\frac{3}{4} x+\frac{1}{2} \end{aligned}

(ii) Given that $$M$$ is the midpoint of $$A B$$, that $$2 C M=M D$$, and that the $$x$$ -coordinate of $$C$$ is -2 , find the coordinates of $$D .$$ [3]

$$M(2,-1), C\left(-2, y_{c}\right), D\left(x_{d}, y_{d}\right)$$

Equation of line $$C D$$
$$y=-\frac{3}{4} x+\frac{1}{2}$$

$$A+C\left(-2, y_{c}\right)$$
$$y_{c}=-\frac{3}{4}(-2)+\frac{1}{2}=2$$

$$\left(\frac{(1) x_{d}+2(-2)}{1+2}, \frac{(1) y_{d}+2(2)}{1+2}\right)=(2,-1)$$

$$\frac{x_{d}-4}{3}=2$$
$$x_{d}=10$$

$$\frac{y_{d} +2(2)}{3}=-1$$
$$y_{d}=-7$$

(iii) Find the area of the triangle $$C A D$$. [2]

$$A=\frac{1}{2}\left|\begin{array}{cccc}-2 & -1 & 10 & -2 \\ 2 & -5 & -7 & 2\end{array}\right|$$
$$A=\frac{1}{2}|10+7+20+2+50-14|$$
$$A=\frac{1}{2}|75|$$
$$A=37.5$$ unit $$^{2}$$

## Question 9

(i) Given that $$y=x \sin 4 x,$$ find $$\frac{ d y}{ d x}$$. [3]

\begin{aligned} y &=x \sin 4 x \\ \frac{d y}{d x} &=(1) \sin 4 x+\cos 4 x(4) x \\ &=\sin 4 x+4 x \cos 4 x \end{aligned}

(ii) Hence find $$\int x \cos 4 x d x$$ and evaluate $$\int_{0}^{\frac{\pi}{8}} x \cos 4 x d x$$. [6]

$$\frac{d}{d x} x \sin 4 x=4 x \cos 4 x+\sin 4 x$$
$$\int(4 x \cos 4 x+\sin 4 x) d x=x \sin 4 x$$
$$\int 4 x \cos 4 x+\int \sin 4 x d x=x \sin 4 x$$
$$4 \int x \cos 4 x=x \sin 4 x-\int \sin 4 x d x$$
$$\int x \cos 4 x=\frac{1}{4} x \sin 4 x-\frac{1}{4}\left[-\frac{\cos 4 x}{4}\right]+c$$
$$=\frac{1}{4} x \sin 4 x+\frac{1}{16} \cos 4 x+c$$

\begin{aligned} & \int_{0}^{\frac{\pi}{8}} x \cos 4 x d x \\=&\left[\frac{1}{4} x \sin 4 x+\frac{1}{16} \cos 4 x\right]_{0}^{\frac{\pi}{8}} \\=&\left[\frac{1}{4}\left(\frac{\pi}{8}\right) \sin 4\left(\frac{\pi}{8}\right)+\frac{1}{16} \cos 4\left(\frac{\pi}{8}\right)\right]-\left[0+\frac{1}{16} \cos 0\right] \\=&\left[\frac{\pi}{32}(1)+0\right]-\frac{1}{16} \\=& \frac{\pi}{32}-\frac{1}{16} \end{aligned}

## Question 10

(i) Solve $$2 \sec ^{2} x=5 \tan x+5,$$ for $$0^{\circ}<x<360^{\circ}$$. [5]

$$2 \sec ^{2} x=5 \tan x+5$$
$$2\left(\tan ^{2} x+1\right)=5 \tan x+5$$
$$2 \tan ^{2} x+2=5 \tan x+5$$
$$2 \tan ^{2} x-5 \tan x-3=0$$

$$\tan x=y$$
$$2 y^{2}-5 y-3=0$$
$$(2 y+1)(y-3)=0$$
$$y=-\frac{1}{2}, y=3$$

\begin{aligned} \tan & x=-\frac{1}{2} \\ x &=180^{\circ}-26.57^{\circ} \\ &=153.43^{\circ} \end{aligned}
or
\begin{aligned} x &=360^{\circ}-26 \cdot 57^{\circ} \\ &=333.43^{\circ} \end{aligned}

\begin{aligned} \tan x &=3 \\ x &=71.57^{\circ} \end{aligned}
or
\begin{aligned} x &=180^{\circ}+71.57^{\circ} \\ &=251.57^{\circ} \end{aligned}

(ii) Solve $$\sqrt{2} \sin \left(\frac{y}{2}+\frac{\pi}{3}\right)=1,$$ for $$0<y<4 \pi$$ radians. [5]

$$0<y<4 \pi$$
$$0<\frac{y}{2}<2 \pi$$
$$\frac{\pi}{3}<\frac{y}{2}+\frac{\pi}{3}<2 \frac{1}{3} \pi$$

$$\sqrt{2} \sin \left(\frac{y}{2}+\frac{\pi}{3}\right)=1$$
$$\sin \left(\frac{y}{2}+\frac{\pi}{3}\right)=\frac{1}{\sqrt{2}}$$

$$\frac{y}{2}+\frac{\pi}{3}=\pi-\frac{\pi}{4}$$
$$\frac{y}{2}=\frac{3 \pi}{4}-\frac{\pi}{3}$$
$$\frac{y}{2}=\frac{5 \pi}{12}$$
$$y=\frac{5 \pi}{6}$$
or
$$\frac{y}{2}+\frac{\pi}{3}=2 \pi+\frac{\pi}{4}$$
$$\frac{y}{2}=\frac{9 \pi}{4}-\frac{\pi}{3}$$
$$y=\frac{23 \pi}{6}$$

## Question 11

Answer only one of the following two alternatives.
EITHER
A curve has equation $$y= e ^{-x}(A \cos 2 x+B \sin 2 x)$$. At the point (0,4) on the curve, the gradient of the tangent is 6

(i) Find the value of $$A$$. [1]

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(ii) Show that $$B=5$$. [5]

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(iii) Find the value of $$x$$, where $$0<x<\frac{\pi}{2}$$ radians, for which $$y$$ has a stationary value. [5]

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OR

A curve has equation $$y=\frac{\ln \left(x^{2}-1\right)}{x^{2}-1},$$ for $$x>1$$.

(i) Show that $$\frac{ d y}{ d x}=\frac{k x\left(1-\ln \left(x^{2}-1\right)\right)}{\left(x^{2}-1\right)^{2}},$$ where $$k$$ is a constant to be found. [4]

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(ii) Hence find the approximate change in $$y$$ when $$x$$ increases from $$\sqrt{5}$$ to $$\sqrt{5}+p$$, where $$p$$ is small. [2]

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(iii) Find, in terms of e, the coordinates of the stationary point on the curve. [5]

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