Course Content
Higher Level (by year)
0/17
Higher Level (by topic)
0/13
Ordinary Level (by year)
0/17
Ordinary Level (by topic)
0/12
Past Papers

## Question 1

(a) In the expansion of $$(2x+1)(x^2+px+4)$$, where $$p\in\mathbb{N}$$, the coefficient of $$x$$ is twice the coefficient of $$x^2$$. Find the value of $$p$$.

$$p=2$$

Solution

\begin{align}(2x+1)(x^2+px+4)&=2x^3+2px^2+8x+x^2+px+4\\&=2x^3+(2p+1)x^2+(8+p)x+4\end{align}

\begin{align}\downarrow\end{align}

\begin{align}8+p=2(2p+1)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}8+p=4p+2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3p=6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}p=2\end{align}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b) Solve the equation $$\dfrac{3}{2x+1}+\dfrac{2}{5}=\dfrac{2}{3x-1}$$ where $$x\neq-\dfrac{1}{2},\dfrac{1}{3}$$, and $$x\in\mathbb{R}$$.

$$x=-3$$ or $$x=\dfrac{3}{4}$$

Solution

\begin{align}\frac{3}{2x+1}+\frac{2}{5}=\frac{2}{3x-1}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{3(5)+2(2x+1)}{(2x+1)(5)}=\frac{2}{3x-1}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{4x+17}{10x+5}=\frac{2}{3x-1}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(4x+17)(3x-1)=2(10x+5)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}12x^2-4x+51x-17=20x+10\end{align}

\begin{align}\downarrow\end{align}

\begin{align}12x^2+27x-27=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3x^2+9x-9=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(x+3)(4x-3)=0\end{align}

\begin{align}\downarrow\end{align}

$$x=-3$$ or $$x=\dfrac{3}{4}$$

Video Walkthrough

## Question 2

The graph of the function $$f(x)=3^x$$, where $$x\in\mathbb{R}$$, cuts the $$y$$‐axis at $$(0,1)$$ as shown in the
diagram below.

(a)

(i) Draw the graph of the function $$g(x)=4x+1$$ on the diagram.

(ii) Use substitution to verify that $$f(x)<g(x)$$, for $$x=1.9$$.

(i)

Solution

(i)

(ii)

\begin{align}f(1.9)&=3^{1.9}\\&=8.06…\end{align}

and

\begin{align}g(1.9)&=4(1.9)+1\\&=8.6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}f(1.9)<g(1.9)\end{align}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b) Prove, using induction, that $$f(n)\geq g(n)$$, where $$n\geq2$$ and $$n\in\mathbb{N}$$.

Solution

The inequality is true for $$n=2$$ as $$3^2\geq4(2)+1$$.

Therefore let, us assume that it is true for $$n=k$$, i.e. that

\begin{align}3^k\geq4k+1\end{align}

Using this assumption, we now wish to prove that it is also true for $$n=k+1$$, i.e. that

\begin{align}3^{k+1}\geq4k+5\end{align}

First, we will multiply our assumption by $$3$$:

\begin{align}3(3^k)\geq3(4k+1)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3^{k+1}\geq12k+3\end{align}

Since $$12k+3>4k+5$$ for $$k\geq2$$, it follows that

\begin{align}3^{k+1}\geq4k+5\end{align}

as required.

Video Walkthrough

## Question 3

(a) Factorise fully: $$3xy-9x+4y-12$$

$$(3x+4)(y-3)$$

Solution

\begin{align}3xy-9x+4y-12&=3x(y-3)+4(y-3)\\&=(3x+4)(y-3)\end{align}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b) $$g(x)=3x\ln x-9x+4\ln x-12$$.
Using your answer to part (a) or otherwise, solve $$g(x)=0$$.

$$x=-\dfrac{4}{3}$$ or $$x=e^3$$

Solution

\begin{align}(3x+4)(\ln x-3)=0\end{align}

\begin{align}\downarrow\end{align}

$$3x+4=0$$ or $$\ln x-3=0$$

\begin{align}\downarrow\end{align}

$$x=-\dfrac{4}{3}$$ or $$x=e^3$$

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(c) Evaluate $$g'(e)$$ correct to $$2$$ decimal places.

$$-1.53$$

Solution

\begin{align}g'(x)=3x\left(\frac{1}{x}\right)+3(\ln x)-9+4\left(\frac{1}{x}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}g'(e)&=3e\left(\frac{1}{e}\right)+3(\ln e)-9+4\left(\frac{1}{e}\right)\\&\approx-1.53\end{align}

Video Walkthrough

## Question 4

(a) Find $$\int(4x^3-6x+10)\,dx$$.

$$x^4-3x^2+10x+C$$

Solution

\begin{align}\int(4x^3-6x+10)\,dx&=\frac{4x^4}{4}-\frac{6x^2}{2}+10x+C\\&=x^4-3x^2+10x+C\end{align}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b) Part of the graph of a cubic function $$f(x)$$ is shown below (graph not to scale).
The graph cuts the $$x$$‐axis at the three points $$A(2,0)$$, $$B$$, and $$C$$.

(i) Given that $$f'(x)=6x^2-54x+109$$, show that $$f(x)=2x^3-27x^2+109x-126$$.

(ii) Find the co‐ordinates of the point $$B$$ and the point $$C$$.

(ii) $$B:(4.5,0)$$ and $$C:(7,0)$$

Solution

(i)

\begin{align}f(x)&=\int f'(x)\,dx\\&=\int(6x^2-54x+109)\,dx\\&=\frac{6x^3}{3}-\frac{54x^2}{2}+109x+C\\&=2x^3-27x^2+109x+C\end{align}

and

\begin{align}f(2)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2(2^3)-27(2^2)+109(2)+C=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}126+C=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}C=-126\end{align}

\begin{align}\downarrow\end{align}

\begin{align}f(x)=2x^3-27x^2+109x-126\end{align}

as required.

(ii)

$\require{enclose} \begin{array}{rll} 2x^2-23x+63\phantom{00000000}\, \\[-3pt] x-2 \enclose{longdiv}{\,2x^3-27x^2+109x-126} \\[-3pt] \underline{2x^3-4x^2\phantom{000000000000}\,\,} \\[-3pt] -23x^2+109x\phantom{000}\,\\[-3pt]\phantom{0000}\underline{-23x^2+46x\phantom{000}\,}\\[-3pt]\phantom{00}63x-126\\[-3pt]\phantom{00}\underline{63x-126}\\[-3pt]0 \end{array}$

\begin{align}\downarrow\end{align}

\begin{align}f(x)&=(x-2)(2x^2-23x+63)\\&=(x-2)(2x-9)(x-7)\end{align}

\begin{align}\downarrow\end{align}

$$B:(4.5,0)$$ and $$C:(7,0)$$

Video Walkthrough

## Question 5

(a) $$3+2i$$ is a root of $$z^2+pz+q=0$$, where $$p,q\in\mathbb{R}$$, and $$i^2=-1$$.
Find the value of $$p$$ and the value of $$q$$.

$$p=-6$$ and $$q=13$$

Solution

\begin{align}(3+2i)^2+p(3+2i)+q=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}9+12i-4+3p+2pi+q=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(5+3p+q)+(12+2p)i=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(5+3p+q)+(12+2p)i=0\end{align}

\begin{align}\downarrow\end{align}

We therefore have the following two equations for two unknowns:

\begin{align}5+3p+q=0\end{align}

\begin{align}12+2p=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}5+3p+q=0\end{align}

\begin{align}p=-6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}5+3(-6)+q=0\end{align}

\begin{align}p=-6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}q=13\end{align}

\begin{align}p=-6\end{align}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b)

(i) $$v=2-2\sqrt{3}i$$. Write $$v$$ in the form $$r(\cos\theta+i\sin\theta)$$, where $$r\in\mathbb{R}$$ and $$0\leq\theta\leq2\pi$$.

(ii) Use your answer to part (b)(i) to find the two possible values of $$w$$, where $$w^2=v$$.
Give your answers in the form $$a+ib$$, where $$a,b\in\mathbb{R}$$.

(i) $$v=4\left[\cos\dfrac{5\pi}{3}+i\sin\dfrac{5\pi}{3}\right]$$

(ii) $$w=-\sqrt{3}+i$$ and $$w=\sqrt{3}-i$$

Solution

(i)

Modulus

\begin{align}r&=\sqrt{2^2+(-2\sqrt{3})^2}\\&=\sqrt{4+12}\\&=\sqrt{16}\\&=4\end{align}

$\,$

Argument

Reference Angle:

\begin{align}A&=\tan^{-1}\left(\frac{2\sqrt{3}}{2}\right)\\&=\frac{\pi}{3}^{\circ}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\theta&=2\pi-\frac{\pi}{3}\\&=\frac{\pi}{3}\end{align}

$\,$

Polar Form

\begin{align}v=4\left[\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3}\right]\end{align}

(ii)

\begin{align}w&=\pm\left[4\left(\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3}\right)\right]^{1/2}\\&=\pm2\left[\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right]\\&=\pm(-\sqrt{3}+i)\end{align}

\begin{align}\downarrow\end{align}

$$w=-\sqrt{3}+i$$ and $$w=\sqrt{3}-i$$

Video Walkthrough

## Question 6

(a)

(i) Given that $$x-\sqrt{32}=\sqrt{128}-5x$$, find the value of $$x$$, where $$x\in\mathbb{R}$$.
Give your answer in the form $$a\sqrt{2}$$, where $$a\in\mathbb{N}$$.

(ii) $$A=\{\sqrt{32k^2},\sqrt{50k^2},\sqrt{128k^2},\sqrt{98k^2}\}$$, where $$k\in\mathbb{N}$$.

Show that the mean of set $$A$$ is equal to the median of set $$A$$.

(i) $$2\sqrt{2}$$

Solution

(i)

\begin{align}x-\sqrt{32}=\sqrt{128}-5x\end{align}

\begin{align}\downarrow\end{align}

\begin{align}6x=\sqrt{128}+\sqrt{32}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}6x=8\sqrt{2}+4\sqrt{2}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}6x=12\sqrt{2}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=2\sqrt{2}\end{align}

(ii)

Mean

\begin{align}\frac{\sqrt{32k^2}+\sqrt{50k^2}+\sqrt{128k^2}+\sqrt{98k^2}}{4}&=\frac{4\sqrt{2}k+8\sqrt{2}k+7\sqrt{2}k+5\sqrt{2}k}{4}\\&=\frac{24\sqrt{2}k}{4}\\&=6\sqrt{2}k\end{align}

$\,$

Median

\begin{align}\frac{1}{2}(\sqrt{50k^2}+\sqrt{50k^2})&=\frac{1}{2}(5\sqrt{2}k+7\sqrt{2}k)\\&=\frac{1}{2}(12\sqrt{2}k)\\&=6\sqrt{2}k\end{align}

Therefore, the mean and the median are the same.

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b) Prove, using contradiction, that $$\sqrt{2}$$ is not a rational number.

Solution

Assume $$\sqrt{2}$$ is a rational number, i.e. that

\begin{align}\sqrt{2}=\frac{a}{b}\end{align}

where $$a,b\in\mathbb{Z}$$ and the fraction is written in its simplest form. Hence

\begin{align}a^2=2b^2\end{align}

and therefore $$a$$ is even, i.e. $$a=2m$$ for some $$m\in\mathbb{Z}$$.

\begin{align}(2m)^2=2b^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}b^2=2m^2\end{align}

and therefore $$b$$ is even, i.e. $$b=2n$$ for some $$n\in\mathbb{Z}$$.

\begin{align}\sqrt{2}&=\frac{a}{b}\\&=\frac{2m}{2n}\end{align}

However, we stated that the fraction was already in its simplest form – a contraction.

Hence, $$\sqrt{2}$$ is irrational.

Video Walkthrough

## Question 7

The closed line segment $$[0,1]$$ is shown below. The first three steps in the construction of the
Cantor Set are also shown:

• Step $$1$$ removes the open middle third of the line segment $$[0,1]$$ leaving two closed line segments (i.e. the end points of the segments remain in the Cantor Set)
• Step $$2$$ removes the middle third of the two remaining segments leaving four closed line segments
• Step $$3$$ removes the middle third of the four remaining segments leaving eight closed line segments.

The process continues indefinitely. The set of points in the line segment $$[0,1]$$ that are not removed during the process is the Cantor Set.

(a)

(i) Complete the table below to show the length of the line segment(s) removed at each step for the first $$5$$ steps. Give your answers as fractions.

Step Step $$1$$ Step $$2$$ Step $$3$$ Step $$4$$ Step $$5$$

Length Removed

$$\dfrac{1}{3}$$

$$\dfrac{2}{9}$$

(ii) Find the total length of all of the line segments removed from the initial line segment of length $$1$$ unit, after a finite number ($$n$$) of steps in the process.
Give your answer in terms of $$n$$.

(iii) Find the total length removed, from the initial line segment, after an infinite number of steps of the process.

(i)

Step Step $$1$$ Step $$2$$ Step $$3$$ Step $$4$$ Step $$5$$

Length Removed

$$\dfrac{1}{3}$$

$$\dfrac{2}{9}$$

$$\dfrac{4}{27}$$

$$\dfrac{8}{81}$$

$$\dfrac{16}{243}$$

(ii) $$1-\left(\dfrac{2}{3}\right)^n$$

(iii) $$1$$

Solution

(i)

Step Step $$1$$ Step $$2$$ Step $$3$$ Step $$4$$ Step $$5$$

Length Removed

$$\dfrac{1}{3}$$

$$\dfrac{2}{9}$$

$$\dfrac{4}{27}$$

$$\dfrac{8}{81}$$

$$\dfrac{16}{243}$$

(ii)

\begin{align}S_n&=\frac{a(1-r^n)}{1-r} \\&=\frac{\frac{1}{3}\left[1-\left(\frac{2}{3}\right)^n\right]}{1-\frac{2}{3}}\\&=1-\left(\frac{2}{3}\right)^n\end{align}

(iii)

\begin{align}S_{\infty}&=\frac{a}{1-r}\\&=\frac{\frac{1}{3}}{1-\frac{2}{3}}\\&=1\end{align}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b)

(i) Complete the table below to identify the end‐points labelled in the diagram.

Label $$A$$ $$B$$ $$C$$ $$D$$ $$E$$ $$F$$

End-point

(ii) Give a reason why $$\dfrac{1}{3}-\dfrac{1}{9}+\dfrac{1}{27}-\dfrac{1}{81}$$ is a point in the Cantor Set.

(iii) The limit of the series $$\dfrac{1}{3}-\dfrac{1}{9}+\dfrac{1}{27}-…$$ is a point in the Cantor Set. Find this point.

(i)

Label $$A$$ $$B$$ $$C$$ $$D$$ $$E$$ $$F$$

End-point

$$\dfrac{2}{3}$$

$$\dfrac{2}{9}$$

$$\dfrac{7}{9}$$

$$\dfrac{8}{9}$$

$$\dfrac{7}{27}$$

$$\dfrac{25}{27}$$

(ii) It is the end point of a segment.

(iii) $$\dfrac{1}{4}$$

Solution

(i)

Label $$A$$ $$B$$ $$C$$ $$D$$ $$E$$ $$F$$

End-point

$$\dfrac{2}{3}$$

$$\dfrac{2}{9}$$

$$\dfrac{7}{9}$$

$$\dfrac{8}{9}$$

$$\dfrac{7}{27}$$

$$\dfrac{25}{27}$$

(ii) It is the end point of a segment.

(iii)

\begin{align}S_{\infty}&=\frac{a}{1-r}\\&=\frac{\frac{1}{3}}{1-\left(-\frac{1}{3}\right)}\\&=\frac{1}{4}\end{align}

Video Walkthrough

## Question 8

The weekly revenue produced by a company manufacturing air conditioning units is seasonal.
The revenue (in euro) can be approximated by the function:

$$r(t)=22{,}500\cos\left(\frac{\pi}{26}t\right)+37{,}500$$, $$t\geq0$$

where $$t$$ is the number of weeks measured from the beginning of July and $$\left(\dfrac{\pi}{26}t\right)$$ is in radians.

(a) Find the approximate revenue produced $$20$$ weeks after the beginning of July.

$$20{,}659\mbox{ euro}$$

Solution

\begin{align}r(20)&=22{,}500\cos\left(\frac{\pi}{26}(20)\right)+37{,}500\\&\approx20{,}659\mbox{ euro}\end{align}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b) Find the two values of the time $$t$$, within the first $$52$$ weeks, when the revenue is approximately €$$26{,}250$$.

$$t=\dfrac{52}{3}$$ weeks and $$t=\dfrac{104}{3}$$ weeks

Solution

\begin{align}22{,}500\cos\left(\frac{\pi}{26}t\right)+37{,}500=26{,}250\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\cos\left(\frac{\pi}{26}t\right)=-\frac{1}{2}\end{align}

\begin{align}\downarrow\end{align}

$$\dfrac{\pi}{26}t=\dfrac{2\pi}{3}$$ and $$\dfrac{\pi}{26}t=\dfrac{4\pi}{3}$$

\begin{align}\downarrow\end{align}

$$t=\dfrac{52}{3}$$ weeks and $$t=\dfrac{104}{3}$$ weeks

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(c) Find $$r'(t)$$, the derivative of $$r(t)=22{,}500\cos\left(\frac{\pi}{26}t\right)+37{,}500$$.

$$r'(t)=-\dfrac{11{,}250\pi}{13}\sin\left(\dfrac{\pi}{26}t\right)$$

Solution

\begin{align}r'(t)&=\left(\frac{\pi}{26}\right)(22{,}500)\left[-\sin\left(\frac{\pi}{26}t\right)\right]\\&=-\frac{11{,}250\pi}{13}\sin\left(\frac{\pi}{26}t\right)\end{align}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(d) Use calculus to show that the revenue is increasing $$30$$ weeks after the beginning of July.

Solution

\begin{align}r'(30)&=-\frac{11{,}250\pi}{13}\sin\left(\frac{\pi}{26}(3)\right)\\&=1263.44…\end{align}

As $$r'(30)>0$$, it is increasing.

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(e) Find a value for the time $$t$$, within the first $$52$$ weeks, when the revenue is at a minimum.
Use $$r^{\prime\prime}(t)$$, to verify your answer.

The revenue is a minimum in the $$26$$th week.

Solution

\begin{align}-\frac{11{,}250\pi}{13}\sin\left(\frac{\pi}{26}t\right)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\sin\left(\frac{\pi}{26}t\right)=0\end{align}

\begin{align}\downarrow\end{align}

$$\left(\dfrac{\pi}{26}t\right)=0$$ and $$\left(\dfrac{\pi}{26}t\right)=\pi$$

\begin{align}\downarrow\end{align}

$$t=0$$ and $$t=26$$

$\,$

Verify

\begin{align}r^{\prime\prime}(t)=-\frac{11{,}250\pi}{13}\left(\frac{\pi}{26}\right)\cos\left(\frac{\pi}{26}t\right)\end{align}

\begin{align}\downarrow\end{align}

$$r^{\prime\prime}(0)>0$$ and $$r^{\prime\prime}(26)<0$$

Therefore, the revenue is a minimum in the $$26$$th week.

Video Walkthrough

## Question 9 Norman windows consist of a rectangle topped by a semi‐circle as shown above.
Let the height of the rectangle be $$y$$ metres and the radius of the semi‐circle be $$x$$ metres as shown. The perimeter of the window is $$P$$.

(a)

(i) Write $$P$$ in terms of $$x$$, $$y$$ and $$\pi$$.

(ii) In a particular Norman window the perimeter $$P=12$$ metres.

Show that $$y=\dfrac{12-(2+\pi)x}{2}$$ for $$0\leq x\leq \dfrac{12}{2+\pi}$$, where $$x\in\mathbb{R}$$.

(i) $$P=2x+2y+\pi x$$

Solution

(i)

\begin{align}P&=2x+2y+\frac{1}{2}(2\pi x)\\&=2x+2y+\pi x\end{align}

(ii)

\begin{align}2x+2y+\pi x=12\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2y=12-2x-\pi x\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y&=\frac{12-2x-\pi x}{2}\\&=\frac{12-(2+\pi)x}{2}\end{align}

as required.

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b)

(i) Complete the table shown.

$$x$$ $$0$$ $$\dfrac{12}{2+\pi}$$

$$y=\dfrac{12-(2+\pi)x}{2}$$

(ii) On the diagram below, draw the graph of the linear function, $$y=\dfrac{12-(2+\pi)x}{2}$$ for $$0\leq x\leq\dfrac{12}{2+\pi}$$ where $$x\in\mathbb{R}$$.

(iii) Find the slope of the graph of $$y$$, correct to $$2$$ decimal places.
Interpret this slope in the context of the question.

(i)

$$x$$ $$0$$ $$\dfrac{12}{2+\pi}$$

$$y=\dfrac{12-(2+\pi)x}{2}$$

$$6$$

$$0$$

(ii)

(iii) The slope is $$-2.57$$ and therefore, for a $$1$$ metre increase in the radius, the height of the rectangle decreases by $$2.57$$ metres.

Solution

(i)

$$x$$ $$0$$ $$\dfrac{12}{2+\pi}$$

$$y=\dfrac{12-(2+\pi)x}{2}$$

$$6$$

$$0$$

(ii)

(iii)

\begin{align}y&=\frac{12-(2+\pi)x}{2}\\&=-\left(\frac{2+\pi}{2}\right)+6\\&=mx+c\end{align}

\begin{align}\downarrow\end{align}

\begin{align}m&=-\left(\frac{2+\pi}{2}\right)\\&\approx-2.57\mbox{ m}\end{align}

Therefore, for a $$1$$ metre increase in the radius, the height of the rectangle decreases by $$2.57$$ metres.

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(c)

(i) The Norman window shown below has a perimeter of $$12$$ metres and $$y=\dfrac{12-(2+\pi)x}{2}$$.

Show that the function $$a(x)=\dfrac{24x-(\pi+4)x^2}{2}$$ represents the area of the
window, in terms of $$x$$ and $$\pi$$.

(ii) Find $$a'(x)$$.

(iii) Find the relationship between $$x$$ and $$y$$ when the area of the window in part (c)(i) is at its
maximum.

(ii) $$a'(x)=12-(\pi+4)x$$

(iii) The area is a maximum when the radius is equal to the height of the rectangle.

Solution

(i)

\begin{align}a(x)&=2xy+\frac{1}{2}(\pi x^2)\\&=2x\left(\frac{12-(2+\pi)x}{2}\right)+\frac{1}{2}(\pi x^2)\\&=\frac{24x-4x^2-2\pi x^2+\pi x^2}{2}\\&=\frac{24x-(\pi+4)x^2}{2}\end{align}

as required.

(ii)

\begin{align}a'(x)&=\frac{24-2(\pi+4)x}{2}\\&=12-(\pi+4)x\end{align}

(iii)

\begin{align}a'(x)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}12-(\pi+4)x=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=\frac{12}{\pi+4}\end{align}

and

\begin{align}y&=\frac{12-(2+\pi)x}{2}\\&=\frac{12-(2+\pi)\left(\frac{12}{\pi+4}\right)}{2}\\&=\frac{12(\pi+4)-(2+\pi)(12)}{2(\pi+4)}\\&=\frac{6(\pi+4)-(2+\pi)(6)}{\pi+4}\\&=\frac{6\pi+24-12-6\pi}{\pi+4}\\&=\frac{12}{\pi+4}\\&=x\end{align}

Therefore, the area is a maximum when the radius is equal to the height of the rectangle.

Video Walkthrough