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# Subject: Maths

## C3 EXAM PRACTICE QUESTIONS

### To improve your C3 Exam performance, attempt our C3 Exam questions, then view our C3 Exam Solutions and work on your weaknesses.

Question

For the curve with equation

$\dpi{100}&space;y=x^{3}\sqrt{\left&space;(&space;2x-1&space;\right&space;)},\:&space;\:&space;\:&space;\:&space;\:&space;\frac{dy}{dx}=\frac{x^{2}\left&space;(ax+b&space;\right&space;)}{\sqrt{\left&space;(&space;2x-1&space;\right&space;)}}.$

Find the values of $\dpi{100}&space;a$ and $\dpi{100}&space;b.$

#### Solution

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Question

Differentiate with respect to $\dpi{100}&space;x$,

(a)   $\dpi{100}&space;\cos&space;\left&space;(&space;\ln&space;x&space;\right&space;).$

(b)  $\dpi{100}&space;\frac{e^{x^{3}}}{\cos&space;x}.$

#### Solution

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Question

(a) Prove the identity

$\dpi{100}&space;\sec&space;^{2}\theta&space;=\frac{\textup{cosec}\,&space;\theta&space;}{\textup{cosec}\,&space;\theta-\sin&space;\theta&space;},\:&space;\:&space;\:&space;\:&space;\:&space;\:&space;\theta&space;\neq&space;\frac{2n+1}{2}\pi,&space;\:&space;\:&space;\:&space;\:&space;\:&space;\:&space;\:&space;n\in&space;\mathbb{Z}.$

(b) Solve the equation

$\dpi{100}&space;\tan&space;\theta&space;+3\cot&space;\theta&space;=5\sec&space;\theta&space;,$  for  $\dpi{100}&space;0\leq&space;\theta&space;\leq&space;180^{\circ}.$

#### Solution

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Question

(a) Differentiate $\dpi{100}&space;\ln&space;x\cos&space;x$ with respect to $\dpi{100}&space;x.$

(b) Use your answer from part (a) to differentiate with respect to $\dpi{100}&space;x,$

$\dpi{100}&space;\frac{\ln&space;x\cos&space;x}{x^{2}}.$

#### Solution

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Question

(i) Using the chain rule and an appropriate substitution differentiate $\dpi{100}&space;\sqrt{x^{2}-1},$ with respect to $\dpi{100}&space;x.$

(ii) Hence show that

$\dpi{100}&space;\frac{d}{dx}\left&space;(&space;\frac{\sqrt{x^{2}-1}}{x^{2}+1}&space;\right&space;)=-\frac{x\left&space;(&space;x^{2}&space;+3\right&space;)}{\sqrt{x^{2}-1}\left&space;(&space;x^{2}+1&space;\right&space;)^{2}}.$

#### Solution

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Question

Differentiate with respect to $\dpi{100}&space;x$,

i.   $\dpi{100}&space;-\textup{cosec}\,&space;x\cot&space;x.$

ii.   $\dpi{100}&space;\sec&space;x\tan&space;x.$

simplify your answer where possible.

#### Solution

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Question

Show that

(a)  $\dpi{100}&space;\frac{d}{dx}\left&space;(&space;x^{2}\sec&space;^{2}x&space;\right&space;)=2x\left&space;(&space;1+\tan&space;^{2}x&space;\right&space;)\left&space;(&space;x\tan&space;x+1&space;\right&space;).$

(b)  $\dpi{100}&space;\frac{d}{dx}\left&space;(&space;\frac{\tan&space;x}{e^{x}}&space;\right&space;)=\frac{\tan&space;^{2}x-\tan&space;x+1}{e^{x}}.$

#### Solution

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Question

(a) Differentiate  $\dpi{100}&space;\frac{1+\sqrt{x}}{1-\sqrt{x}},$ with respect to $\dpi{100}&space;x,$ and simplify where possible.

(b) show that

$\dpi{100}&space;\frac{d}{dx}\left&space;(&space;e^{x}\left&space;(&space;x+1&space;\right&space;)^{2}&space;\right&space;)=e^{x}\left&space;(&space;x+1&space;\right&space;)\left&space;(&space;x+3&space;\right&space;).$

#### Solution

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Question

show that

$\dpi{100}&space;\frac{d}{dx}\left&space;(&space;\frac{e^{2x}}{x^{2}+e^{2x}}&space;\right&space;)=\frac{2x\left&space;(&space;x-1&space;\right&space;)e^{2x}}{\left&space;(&space;x^{2}+e^{2x}&space;\right&space;)^{2}}.$

#### Solution

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Question

A curve has equation

$\dpi{100}&space;y=x^{2}e^{\frac{1}{2}x}.$

(a) Using the product rule for differentiation find $\small&space;\dpi{100}&space;\frac{\mathrm{d}&space;y}{\mathrm{d}&space;x}$,

and hence the coordinates of the turning points of $\dpi{100}&space;C.$

(b) Determine the nature of these turning points.

(c) Given that the equation of the tangent to $\dpi{100}&space;C$ at the point where $\dpi{100}&space;x=1$

can be expressed in the form $\dpi{100}&space;ax+by+c=0,$ find $\dpi{100}&space;a,b$ and $\dpi{100}&space;c.$

#### Solution

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Question

The curve shown in the figure has equation

$\dpi{100}&space;y=x^{3}\left&space;(&space;x-1&space;\right&space;)^{2}.$

(a) Clearly from the figure the curve has three turning points. Find them all.

(b) Using your answer from part (a) find $\small&space;\dpi{100}&space;\frac{\mathrm{d}^{2}&space;y}{\mathrm{d}&space;x^{2}}$.

(c) Hence determine the nature of the turning points of $\dpi{100}&space;C.$

#### Solution

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Question

A curve has equation

$\dpi{100}&space;f\left&space;(&space;x&space;\right&space;)=2^{x}\sin&space;x-1,\:&space;\:&space;\:&space;\:&space;\:&space;0\leq&space;x\leq&space;2\pi&space;.$

(a) show that in the interval  $\dpi{100}&space;\left&space;[&space;0,\frac{\pi&space;}{3}&space;\right&space;]$  there is a root $\dpi{100}&space;\alpha$ to the equation $\dpi{100}&space;f\left&space;(&space;x&space;\right&space;)=0.$

(b) show that $\dpi{100}&space;f\left&space;(&space;x&space;\right&space;)=0$ can be written in the form

$\dpi{100}&space;x=\arcsin&space;\left&space;(&space;\frac{1}{2^{x}}&space;\right&space;).$

(c) Hence using an appropriate iterative formula for $\dpi{100}&space;x_{n+1}$ and an appropriate value for

$\dpi{100}&space;x_{0},$ find the values of $\dpi{100}&space;x_{1},x_{2},x_{3},$ and $\dpi{100}&space;x_{4}$ to $\dpi{100}&space;3$ decimal places.

(d) Find, to $\dpi{100}&space;3$ decimal places, the coordinates of the turning points of $\dpi{100}&space;f\left&space;(&space;x&space;\right&space;).$

#### Solution

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Question

(a) Express  $\dpi{100}&space;\frac{\tan&space;x+\cot&space;x}{\textup{cosec}\,&space;x}$   in terms of $\dpi{100}&space;\cos&space;x.$

(b) Prove the identity

$\dpi{100}&space;\cos&space;^{4}x-\sin&space;^{4}x=\cos&space;2x.$

#### Solution

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Question

(a) Solve, for $\dpi{100}&space;0\leq&space;\theta&space;\leq&space;180^{\circ},$ the equation

$\dpi{100}&space;2\cos&space;\theta&space;-4\sin&space;^{2}\theta&space;+2=0.$

(b) Show that

$\dpi{100}&space;\left&space;(&space;\sec&space;^{2}\theta&space;+\tan&space;^{2}\theta&space;\right&space;)\left&space;(&space;\textup{cosec}^{2}\,&space;\theta&space;+\cot&space;^{2}\theta&space;\right&space;)=1+2\sec&space;^{2}\theta&space;\textup{cosec}^{2}\,&space;\theta&space;.$

#### Solution

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Question

(a) Show that

$\dpi{100}&space;\cot&space;2\theta&space;+\tan&space;\theta&space;=\textup{cosec}\,&space;2\theta&space;.$

(b) Solve, for $\dpi{100}&space;0\leq&space;\theta&space;\leq&space;360^{\circ},$ the equation

$\dpi{100}&space;\sin&space;\theta&space;=2\sin&space;\left&space;(&space;60^{\circ}-\theta&space;\right&space;),$

giving your answer/s to one decimal place.

#### Solution

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Question

(a) By writing $\dpi{100}&space;3\theta&space;=\left&space;(&space;2\theta&space;+\theta&space;\right&space;),$ show that

$\dpi{100}&space;\cos&space;3\theta&space;=4\cos&space;^{3}\theta&space;-3\cos&space;\theta&space;.$

(b) Show that

$\dpi{100}&space;\sin&space;^{2}\theta&space;=\frac{1}{2}\left&space;(&space;1-\cos&space;2\theta&space;\right&space;).$

#### Solution

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Question

$\dpi{100}&space;f\left&space;(&space;x&space;\right&space;)=\frac{1}{4}\cos&space;x+\frac{1}{3}\sin&space;x.$

Given that $\dpi{100}&space;f\left&space;(&space;x&space;\right&space;)$ can be expressed in the form $\dpi{100}&space;R\cos&space;\left&space;(&space;x-\alpha&space;\right&space;),$

where $\dpi{100}&space;R>&space;0$ and $\dpi{100}&space;0<&space;\alpha&space;<&space;\frac{\pi&space;}{2},$

(a) Find the values of $\dpi{100}&space;R$ and $\dpi{100}&space;\alpha$ to $\dpi{100}&space;3$ decimal places.

(b) Hence solve

$\dpi{100}&space;\frac{1}{4}\cos&space;x+\frac{1}{3}\sin&space;x=\frac{1}{3},$  for $\dpi{100}&space;0\leq&space;x\leq&space;2\pi&space;.$

(c) Write down the maximum value of $\dpi{100}&space;f\left&space;(&space;x&space;\right&space;)$ and the smallest possible value of $\dpi{100}&space;x$ for which this occurs.

#### Solution

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Question

$\dpi{100}&space;f\left&space;(&space;\theta&space;\right&space;)=8\cos&space;\theta&space;-15\sin&space;\theta&space;.$

(a) By expressing $\dpi{100}&space;f\left&space;(&space;x&space;\right&space;)$ in the form $\dpi{100}&space;R\cos&space;\left&space;(&space;x+\alpha&space;\right&space;),$ where $\dpi{100}&space;R>&space;0$ and $\dpi{100}&space;0<&space;\alpha&space;<&space;\frac{\pi&space;}{2},$

solve $\dpi{100}&space;f\left&space;(&space;\theta&space;\right&space;)=7,$ for $\dpi{100}&space;0<&space;x<&space;2\pi&space;.$

(b)  i.  Write down the maximum value of $\dpi{100}&space;f\left&space;(&space;\theta&space;\right&space;).$

ii. Find the smallest negative value of $\dpi{100}&space;\theta$ for which this maximum occurs.

#### Solution

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Question

The figure shows a sketch of the curve $\dpi{100}&space;C$ with equation $\dpi{100}&space;y=x\sqrt{2-x^{2}}.$

(a) Find  $\dpi{100}&space;\frac{dy}{dx}$.

(b) Hence find the coordinates of the turning points of $\dpi{100}&space;C.$

(c) Show that

$\dpi{100}&space;\frac{d^{2}y}{dx^{2}}=\frac{x\left&space;(&space;2-x&space;\right&space;)\left&space;(&space;2+x&space;\right&space;)}{\left&space;(&space;2-x^{2}&space;\right&space;)^{\frac{3}{2}}},$

and hence determine the nature of the turning points.

#### Solution

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Question

Differentiate the following functions with respect to $\dpi{100}&space;x$ and simplify where possible.

(a)  $\dpi{100}&space;\sec&space;\left&space;(&space;\frac{1}{x}&space;\right&space;)+\textup{cosec}^{2}\,&space;x.$

(b)  $\dpi{100}&space;sin\left&space;(&space;\frac{x}{5}&space;\right&space;).$

#### Solution

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Question

$\dpi{100}&space;f\left&space;(&space;x&space;\right&space;)=x^{3}+2x-1.$

(a) Show that there lies a root of $\dpi{100}&space;f\left&space;(&space;x&space;\right&space;)=0$  in the interval  $\dpi{100}&space;\left&space;[&space;0.45,0.46&space;\right&space;].$

(b) Show that $\dpi{100}&space;f\left&space;(&space;x&space;\right&space;)=0$  is equivalent to  $\dpi{100}&space;x=\frac{1}{2+x^{2}}.$

(c) Use the iterative formula  $\dpi{100}&space;x_{n+1}=\frac{1}{2+\left&space;(&space;x_{n}&space;\right&space;)^{2}},\:&space;\:&space;\:&space;\:&space;\:&space;x_{0}=0.455,$

to calculate the values of  $\dpi{100}&space;x_{1},x_{2},x_{3}$ and $\dpi{100}&space;x_{4}$ to five decimal places.

#### Solution

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Question

The figure shows the curve with equation   $\dpi{100}&space;y=x-2\sin&space;x,\:&space;\:&space;\:&space;\:&space;\:&space;0\leq&space;x\leq&space;2\pi&space;,$

which intersects the $\dpi{100}&space;x$-axis at point $\dpi{100}&space;A$ and has two turning points.

(a) Find the coordinates of the turning points of the curve.

Let point $\dpi{100}&space;A$ have $\dpi{100}&space;x$-coordinate $\dpi{100}&space;\alpha&space;.$ To find an approximation to $\dpi{100}&space;\alpha,$ the iterative formula

$\dpi{100}&space;x_{n+1}=2\sin&space;x_{n}$  is used.

(b) Taking $\dpi{100}&space;x_{0}=\frac{7\pi&space;}{12},$ find to five decimal places the values of $\dpi{100}&space;x_{1},x_{2},x_{3},x_{4}$ and $\dpi{100}&space;x_{5}.$

(c) Show that, correct to five decimal places, $\dpi{100}&space;\alpha&space;=1.89549$

#### Solution

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Question

The function $\dpi{100}&space;f$ is given by

$\dpi{100}&space;f\left&space;(&space;x&space;\right&space;)=1-\frac{3}{\left&space;(&space;x-1&space;\right&space;)}-\frac{x-10}{\left&space;(&space;x+2&space;\right&space;)\left&space;(&space;x-1&space;\right&space;)},\:&space;\:&space;\:&space;\:&space;\:&space;x\in&space;\mathbb{R},\:&space;\;&space;x\neq&space;-2,\:&space;\:&space;x\neq&space;1.$

(a) Show that

$\dpi{100}&space;f\left&space;(&space;x&space;\right&space;)=\frac{x-2}{x+2}.$

The function $\dpi{100}&space;g$ is given by

$\dpi{100}&space;g\left&space;(&space;x&space;\right&space;)=e^{2x}\ln&space;x,\:&space;\:&space;\:&space;\:&space;\:&space;\:x\in&space;\mathbb{R},\:&space;\:&space;\:&space;x\neq&space;0.$

(b) Show that $\dpi{100}&space;g\,^{'}\left&space;(&space;x&space;\right&space;)=e^{2x}\left&space;(&space;\frac{1}{x}+2\ln&space;x&space;\right&space;).$

#### Solution

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Express  $\dpi{100}&space;\frac{1-x}{2x^{2}-2}+\frac{1}{2x-2}$  as a single fraction in its simplest from.

#### Solution

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Question

(a) Simplify  $\dpi{100}&space;\frac{3x^{2}-7x-6}{x^{2}+x-12}.$

(b) Given that  $\dpi{100}&space;e^{2x}-3e^{x}-4=0,$

find $\dpi{100}&space;x$ in terms of $\dpi{100}&space;\ln&space;.$

#### Solution

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Question

(a) Differentiate  $\dpi{100}&space;\cos&space;^{2}\left&space;(&space;2x-4&space;\right&space;)$  with respect to $\dpi{100}&space;x.$

(b) Find  $\dpi{100}&space;\frac{dy}{dx}$  for  $\dpi{100}&space;y=\left&space;(&space;2x-1&space;\right&space;)\ln&space;x.$

#### Solution

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Question

Differentiate with respect to $\dpi{100}&space;x$

(a) $\dpi{100}&space;\left&space;(&space;x-2&space;\right&space;)^{2}\sqrt{x}.$

(b) $\dpi{100}&space;x^{2}\sin&space;5x,$

simplifying where possible.

#### Solution

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Question

Differentiate with respect to $\dpi{100}&space;x$

(a) $\dpi{100}&space;\frac{x^{3}}{\cos&space;x}.$

(b) $\dpi{100}&space;x^{3}\left&space;(&space;2x+1&space;\right&space;)^{3},$

simplifying where possible.

#### Solution

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Question

(a) Find  $\dpi{100}&space;\frac{dy}{dx}$  at the point $\dpi{100}&space;x=\frac{1}{2}$  on the curve with equation $\dpi{100}&space;y=\sqrt{1-x^{2}}.$

(b) Differentiate $\dpi{100}&space;\left&space;(&space;x-1&space;\right&space;)^{2}e^{x^{2}-2x}$  with respect to $\dpi{100}&space;x.$

#### Solution

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