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Past Papers

## Functions

NOTE: Clicking an entry in the right column will take you directly to that question!

(BLUE = Paper 1, RED = Paper 2)

Since 2015, the following subtopic... Has explicitly appeared in...

Linear Functions

Cubic Functions

Trigonometric Functions

Logarithmic Functions

Exponential Functions

## 2022 Paper 1 Question 7(a), (f) & (g)

Hannah is doing a training session. During this session, her heart-rate, $$h(x)$$, is measured in beats per minute (BPM), where $$x$$ is the time in minutes from the start of the session, $$x\in\mathbb{R}$$.
For the first 8 minutes of the session, Hannah does a number of exercises.
As she does these exercises, her heart-rate changes. In this time, $$h(x)$$ is given by:

\begin{align}h(x)=2x^3-28.5x^2+105x+70\end{align}

(a) Work out Hannah’s heart-rate $$4$$ minutes after the start of the session.

$$162\mbox{ BPM}$$

Solution

\begin{align}h(4)&=2(4^3)-28.5(4^2)+105(4)+70\\&=162\mbox{ BPM}\end{align}

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Bruno, Karen, and Martha start a training session at the same time as Hannah.
All of their heart-rates are measured in BPM.

(f)

(i) For the first $$8$$ minutes of the session, Bruno’s heart-rate, $$b(x)$$, is always $$15$$ BPM more
than Hannah’s heart-rate.

Use this information to write $$b'(x)$$ in terms of $$\mathbf{h'(x)}$$, where $$0\leq x\leq 8$$, $$x\in\mathbb{R}$$.

(ii) For the first $$8$$ minutes of the session, Karen’s heart-rate, $$k(x)$$, is always $$10\%$$ less than Hannah’s heart-rate.

Use this information to write $$k'(x)$$ in terms of $$\mathbf{h'(x)}$$, where $$0\leq x\leq 8$$, $$x\in\mathbb{R}$$.

(i) $$b'(x)=h'(x)$$

(ii) $$k'(x)=0.9h'(x)$$

Solution

(i)

\begin{align}b(x)=h(x)+15\end{align}

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

\begin{align}b'(x)=h'(x)\end{align}

(ii)

\begin{align}k(x)=0.9h(x)\end{align}

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

\begin{align}k'(x)=0.9h'(x)\end{align}

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(g) Martha does each exercise for a longer time than Hannah.
For $$0\leq x \leq 10$$, Martha’s heart-rate, $$m(x)$$, is:

\begin{align}m(x)=h(0.8x)\end{align}

Use $$h(x)=2x^3-28.5x^2+105x+70$$ to write $$m(x)$$ in the form:

\begin{align}m(x)=ax^3+bx^2+cx+d\end{align}

where $$a,b,c,d\in\mathbb{R}$$, for $$0\leq x\leq 10$$.

$$m(x)=1.024x^3-18.24x^2+84x+70$$

Solution

\begin{align}m(x)&=h(0.8x)\\&=2(0.8x)^3-28.5(0.8x^2)+(105(0.8x)+70\\&=1.024x^3-18.24x^2+84x+70\end{align}

Video Walkthrough

## 2022 Paper 1 Question 8(a)-(d)

A Ferris wheel has a diameter of $$120\mbox{ m}$$.
When it is turning, it completes exactly $$10$$ full rotations in one hour.
The diagram above shows the Ferris wheel before it starts to turn.
At this stage, the point $$A$$ is the lowest point on the circumference of the wheel, and it is at a
height of $$12\mbox{ m}$$ above ground level.

The height, $$h$$, of the point $$A$$ after the wheel has been turning for $$t$$ minutes is given by:

\begin{align}h(t)=72-60\cos\left(\frac{\pi}{3}t\right)\end{align}

where $$h$$ is in metres, $$t\in\mathbb{R}$$, and $$\dfrac{\pi}{3}$$ is in radians.

(a) Complete the table below. The value of $$h(1)$$ is given.

$$t$$ $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$

$$h(t)$$

$$42$$

$$t$$ $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$

$$h(t)$$

$$12$$

$$42$$

$$102$$

$$132$$

$$102$$

$$42$$

$$12$$

$$42$$

$$102$$

Solution
$$t$$ $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$

$$h(t)$$

$$12$$

$$42$$

$$102$$

$$132$$

$$102$$

$$42$$

$$12$$

$$42$$

$$102$$

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(b) Draw the graph of $$y=h(t)$$ for $$0\leq t\leq 8$$, $$t\in\mathbb{R}$$.

Solution
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(c) Find the period and range of $$h(t)$$.

Period: $$6$$ minutes

Range: $$[12,132]$$.

Solution

According to the graph, the period is $$6$$ minutes and the range is $$[12,132]$$.

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(d) During a $$50$$-minute period, what is the greatest number of minutes for which the point $$A$$ could be higher than $$42\mbox{ m}$$?

$$34$$ minutes

Solution

For each period, the point $$A$$ spends at most $$4$$ minutes at a height greater than $$42\mbox{ m}$$.

$$50$$ minutes is equivalent to $$8$$ periods and $$2$$ minutes.

Therefore, in $$8$$ periods, the point $$A$$ spends at most $$8\times4+2=34$$ minutes at a height greater than $$42\mbox{ m}$$.

Video Walkthrough

## 2022 Paper 1 Question 9(a)-(b)

Alex gets injections of a medicinal drug. Each injection has $$15\mbox{ mg}$$ of the drug.
Each day, the amount of the drug left in Alex’s body from an injection decreases by $$40\%$$.
So, the amount of the drug (in mg) left in Alex’s body $$t$$ days after a single injection is given by:

\begin{align}15(0.6)^t\end{align}

where $$t\in\mathbb{R}$$.

(a) Find the amount of the drug left in Alex’s body $$2.5$$ days after a single $$15\mbox{ mg}$$ injection.

Give your answer in $$\mbox{mg}$$, correct to $$2$$ decimal places.

$$4.18\mbox{ mg}$$

Solution

\begin{align}15(0.6^{2.5})&=4.182…\\&\approx 4.18\mbox{ mg}\end{align}

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(b) How long after a single $$15\mbox{ mg}$$ injection will there be exactly $$1\mbox{ mg}$$ of the drug left in Alex’s body? Give your answer in days, correct to $$1$$ decimal place.

$$5.3\mbox{ days}$$

Solution

\begin{align}15(0.6^t)=1\end{align}

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

\begin{align}0.6^t=\frac{1}{15}\end{align}

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

\begin{align}t&=\log_{0.6}\frac{1}{15}\\&\approx 5.3\mbox{ days}\end{align}

Video Walkthrough

## 2022 Paper 1 Question 10(a) & (b)

A student is asked to memorise a long list of digits, and then write down the list some time later.
The proportion, $$P$$, of the digits recalled correctly after $$t$$ hours can be modelled by the function:

\begin{align}0.82-0.12\ln(t+1)\end{align}

for $$0\leq t\leq 12$$, $$t\in\mathbb{R}$$.

(a) Find the proportion of the digits recalled correctly after $$3$$ hours, according to this model.
Give your answer correct to $$2$$ decimal places.

$$0.65$$

Solution

\begin{align}P(3)&=0.82-0.12\ln(3+1)\\&\approx0.65\end{align}

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(b) After how many hours would exactly $$55\%$$ of the digits be recalled correctly, according to this model? Give your answer correct to $$2$$ decimal places.

$$8.49\mbox{ hours}$$

Solution

\begin{align}P(t)=0.55\end{align}

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

\begin{align}0.82-0.12\ln(t+1)=0.55\end{align}

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

\begin{align}0.12\ln(t+1)=0.27\end{align}

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

\begin{align}\ln(t+1)&=\frac{0.27}{0.12}\\&=2.25\end{align}

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

\begin{align}t+1=e^{2.25}\end{align}

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

\begin{align}t&=e^{2.25}-1\\&\approx8.49\mbox{ hours}\end{align}

Video Walkthrough

## 2021 Paper 1 Question 8(a)

(a) The table in Part (a)(ii) below shows some of the values of the function:

$$h(x)=0.001x^3-0.12x^2+px+5$$, $$x\in\mathbb{R}$$,

in the domain $$0\leq x\leq 75$$.

(i) Use $$h(10)=30$$ to show that $$p=3.6$$.

(ii) Complete the table below and hence draw the graph of $$h(x)$$ in the domain $$0\leq x \leq 75$$ on the grid below.

$$x$$ $$0$$ $$10$$ $$20$$ $$30$$ $$40$$ $$50$$ $$60$$ $$70$$ $$75$$

$$h(x)$$

$$30$$

$$21$$

$$5$$

$$21.875$$

(ii)

$$x$$ $$0$$ $$10$$ $$20$$ $$30$$ $$40$$ $$50$$ $$60$$ $$70$$ $$80$$

$$h(x)$$

$$5$$

$$30$$

$$37$$

$$32$$

$$21$$

$$10$$

$$5$$

$$12$$

$$21.875$$

Solution

(i)

\begin{align}30=0.001(10^3)-0.12(10^2)+p(10)+5\end{align}

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

\begin{align}30=1-12+10p+5\end{align}

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

\begin{align}p&=\frac{30-1+12-5}{10}\\&=3.6\end{align}

as required.

(ii)

$$x$$ $$0$$ $$10$$ $$20$$ $$30$$ $$40$$ $$50$$ $$60$$ $$70$$ $$80$$

$$h(x)$$

$$5$$

$$30$$

$$37$$

$$32$$

$$21$$

$$10$$

$$5$$

$$12$$

$$21.875$$

Video Walkthrough

## 2021 Paper 1 Question 9(a)-(b)

(a) A cup of coffee is freshly brewed to $$95^{\circ}\mbox{C}$$.
The temperature, $$T$$, in degrees centigrade, of the coffee as it cools is given by the formula

\begin{align}T(t)=Ae^{-0.081t}+20\end{align}

where $$A$$ is constant and $$t$$ is time measured in minutes from when the coffee was brewed.

(i) Show that $$A=75$$.

(ii) Explain what the value $$20$$ in the formula represents in the context of the coffee cooling.

(iii) Find the decrease in the temperature of the coffee $$10$$ minutes after brewing.

(ii) The temperature of the coffee after an infinite amount of time has passed.

(iii) $$42^{\circ}\mbox{ C}$$

Solution

(i)

\begin{align}T(0)=95^{\circ}\mbox{ C}\end{align}

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

\begin{align}95=Ae^{-0.081(0)}+20\end{align}

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

\begin{align}95=A+20\end{align}

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

\begin{align}A=75\end{align}

as required.

(ii) The temperature of the coffee after an infinite amount of time has passed.

(iii)

\begin{align}T(10)&=75e^{-0.081(10)}+20\\&53.364…\end{align}

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

\begin{align}\mbox{Decrease}&=95-53.364…\\&\approx42^{\circ}\mbox{ C}\end{align}

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(b) $$T(t)=Ae^{-0.081t}+20$$ gives the temperature of the coffee at time $$t$$.
If the ideal temperature to drink coffee is $$82^{\circ}\mbox{C}$$, find the time, to the nearest second, that it takes for the coffee to reach this temperature.

$$2\mbox{ min }21\mbox{ sec}$$

Solution

\begin{align}82=75e^{-0.081t}+20\end{align}

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

\begin{align}e^{-0.081t}=\frac{82-20}{75}\end{align}

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

\begin{align}-0.081t=\ln\left(\frac{62}{75}\right)\end{align}

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

\begin{align}t&=-\frac{1}{0.081}\ln\left(\frac{62}{75}\right)\\&=2.3500…\mbox{ min}\\&\approx2\mbox{ min }21\mbox{ sec}\end{align}

Video Walkthrough

## 2021 Paper 1 Question 10(b)

(b) Trees of the same age, size and type are growing at the same rate in a forest.
It is possible to measure the radius of the trunk of a tree at the end of each growing season.
The increase in the radius of the trunks of these trees each year, in $$\mbox{cm}$$, during a particular $$10$$ year period can be modelled by the formula:

\begin{align}I(t)=1.5+\sin\left(\frac{\pi t}{5}\right)\end{align}

where $$1\leq t\leq 10$$ and $$t\in\mathbb{N}$$ is measured in years. Therefore

$$r(1)=r(0)+I(1)$$,

$$r(2)=r(1)+I(2)$$, …,

$$r(t+1)=r(t)+I(t+1)$$

where $$r(t)$$ is the radius of the trunk of the tree after $$t$$ years and $$r(0)$$ is the radius of the trunk of a tree at the beginning of year $$1$$.

(i) In order for the model to be reasonable it must satisfy a number of conditions.
One condition is written below:

The radius of the trees is increasing year on year.

Show that $$r(t)$$ satisfies this condition.

(ii) Show that $$I(6)<I(5)$$ and explain what this means in the context of the growth of a tree.

(iii) Two identical trees are growing in the forest.
At the beginning of year $$1$$, a tree was cut down and a section of its trunk, in the shape of a cylinder of radius $$10\mbox{ cm}$$, standard length $$h\mbox{ cm}$$, and volume $$V_1\mbox{ cm}^3$$ was sent to a sawmill.
Given that $$r(0)=10$$, the formula $$r(t+1)=r(t)+I(t+1)$$, where $$t\in\mathbb{N}$$, can be used to find the radius of the second tree for subsequent values of $$t$$.
Find $$r(2)$$, the radius of the second tree at $$t=2$$.
Give your answer in the form $$a+\sin\left(\dfrac{b\pi}{5}\right)+\sin\left(\dfrac{c\pi}{5}\right)$$,
where $$a$$, $$b$$ and $$c\in\mathbb{N}$$.

(iv) At $$t=10$$ the second tree was also cut down.
A section of its trunk, in the shape of a cylinder, of radius $$r(10)\mbox{ cm}$$, standard length $$h\mbox{ cm}$$, and volume $$V_2\mbox{ cm}^3$$ was also sent to a sawmill.
If $$V_2=kV_1$$, where $$k\in\mathbb{R}$$, find the value of $$k$$.

(i) $$I(t)>0$$ for all $$t\geq0$$.

(ii) The tree grew less in the sixth year compared to the fifth year.

(iii) $$r(2)=13+\sin\left(\dfrac{\pi}{5}\right)+\sin\left(\dfrac{2\pi}{5}\right)$$

(iv) $$k=6.25$$

Solution

(i) The minimum value of a sine function is $$-1$$. Therefore, the smallest possible value $$I(t)$$ is $$1.5-1=0.5$$, which is positive. Therefore, the radius of the tree is always increasing.

(ii)

\begin{align}I(6)&=1.5+\sin\left(\frac{\pi(6)}{5}\right)\\&=0.9122…\end{align}

and

\begin{align}I(5)&=1.5+\sin\left(\frac{\pi(5)}{5}\right)\\&=1.5\end{align}

Therefore, $$I(6)<I(5)$$, as required.

This implies that the tree grew less in the sixth year compared to the fifth year.

(iii)

\begin{align}r(t+1)=r(t)+I(t+1)\end{align}

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

\begin{align}r(2)&=r(1)+I(2)\\&=r(0)+I(1)+I(2)\\&=10+\left[1.5+\sin\left(\frac{\pi(1)}{5}\right)\right]+\left[1.5+\sin\left(\frac{\pi(2)}{5}\right)\right]\\&=13+\sin\left(\frac{\pi}{5}\right)+\sin\left(\frac{2\pi}{5}\right)\end{align}

(iv)

\begin{align}r(10)&=r(0)+[I(1)+I(2)+…+I(10)]\\&=10+10(1.5)+0\\&=25\mbox{ cm}\end{align}

\begin{align}V_2=kV_1\end{align}

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

\begin{align}\pi r_1^2h=k(\pi r_2^2h)\end{align}

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

\begin{align}r_1^2=kr_2^2\end{align}

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

\begin{align}k&=\frac{r_1^2}{r_2^2}\\&=\frac{25^2}{10^2}\\&=6.25\end{align}

Video Walkthrough

## 2020 Paper 1 Question 3(a)

(a) $$f(x)=6x-5$$ and $$g(x)=\dfrac{x+5}{6}$$. Investigate if $$f(g(x))=g(f(x))$$.

$$f(g(x))=g(f(x))$$

Solution

\begin{align}f(g(x))&=6\left(\frac{x+5}{6}\right)-5\\&=x+5-5\\&=x\end{align}

and

\begin{align}g(f(x))&=\frac{(6x-5)+5}{6}\\&=\frac{6x}{6}\\&=x\end{align}

Therefore, $$f(g(x))=g(f(x))$$.

Video Walkthrough

## 2020 Paper 1 Question 9(a)

The number of bacteria in the early stages of a growing colony of bacteria can be approximated
using the function:

\begin{align}N(t)=450e^{0.065t}\end{align}

where $$t$$ is the time, measured in hours, since the colony started to grow, and $$N(t)$$ is the number of bacteria in the colony at time $$t$$.

(a)

(i) Find the number of bacteria in the colony after $$4.5$$ hours.

(ii) Find the time, in hours, that it takes the colony to grow to $$790$$ bacteria.
Give your answer correct to $$1$$ decimal place.

(i) $$603$$

(ii) $$8.7\mbox{ h}$$

Solution

(i)

\begin{align}N(4.5)&=450e^{0.065(4.5)}\\&\approx603\end{align}

(ii)

\begin{align}450e^{0.065t}=790\end{align}

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

\begin{align}e^{0.065t}=\frac{790}{450}\end{align}

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

\begin{align}0.065t=\ln\left(\frac{790}{450}\right)\end{align}

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

\begin{align}t&=\frac{1}{0.065}\ln\left(\frac{790}{450}\right)\\&=8.7\mbox{ h}\end{align}

Video Walkthrough

## 2019 Paper 1 Question 2(a)

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

## 2019 Paper 1 Question 8(a) & (b)

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}

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(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

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(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}

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(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.

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(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

## 2019 Paper 1 Question 9(b) & (c)

(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.

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(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

## 2018 Paper 1 Question 2(b)

(b) If ݂$$f(x)=x^3-17x^2+80x-64$$, $$x\in\mathbb{R}$$, show that ݂$$f(1)=0$$, and find another value of $$x$$ for which ݂$$f(x)=0$$.

$$x=8$$

Solution

\begin{align}f(1)&=1^3-17(1^2)+80(1)-64\\&=1-17+80-64\\&=0\end{align}

as required.

$\require{enclose} \begin{array}{rll} x^2-16x+64\phantom{000000}\, \\[-3pt] x-1 \enclose{longdiv}{\,x^3-17x^2+80x-64} \\[-3pt] \underline{x^3-x^2\phantom{00000000000}\,\,} \\[-3pt] -16x^2+80x\phantom{000}\,\\[-3pt]\phantom{0000}\underline{-16x^2+16x\phantom{00}\,}\\[-3pt]\phantom{00}64x-64\\[-3pt]\phantom{00}\underline{64x-64}\\[-3pt]0 \end{array}$

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

\begin{align}f(x)&=x^3-17x^2+80x-64\\&=(x-1)(x^2-16x+64)\\&=(x-1)(x-8)^2\end{align}

Therefore, another value of $$x$$ in which $$f(x)=0$$ is $$x=8$$.

Video Walkthrough