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Past Papers

## Calculus

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...

Derivatives from First Principles

Product/Quotient/Chain Rules

Polynomial Derivatives

Trigonometric Derivatives

Exponential Derivatives

Logarithmic Derivatives

Polynomial Integrals

Trigonometric Integrals

Exponential Integrals

Integrals and Areas

Average Value of a Function

Calculus and Curves

Rate of Change

## 2022 Paper 1 Question 2

(a) $$g(x)=2x^2+5x+6$$, where $$x\in\mathbb{R}$$.

Find $$\int g(x)\,dx$$.

$$\dfrac{2x^3}{3}+\dfrac{5x^2}{2}+6x+C$$

Solution

\begin{align}\int g(x)\,dx&=\int(2x^2+5x+6)\,dx\\&=\int 2x^2\,dx+\int5x\,dx+\int6\,dx\\&=2\int x^2\,dx+5\int x\,dx+6\int dx\\&=\frac{2x^3}{3}+\frac{5x^2}{2}+6x+C\end{align}

Video Walkthrough
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(b) The diagram shows the graph of a function $$f(x)=ax^2+bx+c$$, where $$a,b,c\in\mathbb{Z}$$.
Three regions on the diagram are marked K, L, and N.
Each of these regions is bounded by the $$x$$-axis, the graph of$$f(x)$$, and two vertical lines.

(i) The area of region K is $$538$$ square units. Use integration of $$f(x)$$ to show that:

\begin{align}4a+3b+3c=807\end{align}

(ii) The areas of the three regions K, L, and N give the following three equations (including the equation from part (b)(i)):

\begin{align}4a+3b+3c&=807\\28a+9b+3c&=879\\76a+15b+3c&=663\end{align}

Solve these equations to find the values of $$a$$, $$b$$ and $$c$$.

(ii) $$a=-12$$, $$b=60$$ and $$c=225$$

Solution

(i)

\begin{align}\int_0^2 f(x)\,dx=538\end{align}

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

\begin{align}\int_0^2 (ax^2+bx+c)\,dx=538\end{align}

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

\begin{align}\int_0^2ax^2\,dx+\int_0^2bx\,dx+\int_0^2c\,dx=538\end{align}

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

\begin{align}a\int_0^2x^2\,dx+b\int_0^2x\,dx+c\int_0^2\,dx=538\end{align}

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

\begin{align}\left.\frac{ax^3}{3}\right|_0^2+\left.\frac{bx^2}{2}\right|_0^2+\left.cx\right|_0^2=538\end{align}

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

\begin{align}\left[\frac{a(2^3)}{3}-\frac{a(0^3)}{3})\right]+\left[\frac{b(2^2)}{2}-\frac{b(0^2)}{2}\right]+[c(2)-c(0)]=538\end{align}

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

\begin{align}\frac{8a}{3}+\frac{4b}{2}+2c=538\end{align}

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

\begin{align}8a+6b+6c=1614\end{align}

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

\begin{align}4a+3b+3c=807\end{align}

as required.

(ii)

\begin{align}4a+3b+3c&=807\\28a+9b+3c&=879\\76a+15b+3c&=663\end{align}

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

\begin{align}24a+6b&=72\\48a+6b&=-216\end{align}

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

\begin{align}24a=-288\end{align}

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

\begin{align}a=-12\end{align}

and

\begin{align}24a+6b=72\end{align}

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

\begin{align}b&=\frac{72-24a}{6}\\&=\frac{72-24(-12)}{6}\\&=60\end{align}

and

\begin{align}4a+3b+3c&=807\end{align}

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

\begin{align}c&=\frac{807-4a-3b}{3}\\&=\frac{807-4(-12)-3(60)}{3}\\&=225\end{align}

Video Walkthrough

## 2022 Paper 1 Question 5

(a) $$g(x)=x^2-\dfrac{1}{x}$$ where $$x\in\mathbb{R}$$.

Find $$g'(x)$$, the derivative of $$g(x)$$.

$$g'(x)=2x+\dfrac{1}{x^2}$$

Solution

\begin{align}g(x)&=x^2-\frac{1}{x}\\&=x^2-x^{-1}\end{align}

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

\begin{align}g'(x)&=2x-(-1)x^{-2}\\&=2x+\frac{1}{x^2}\end{align}

Video Walkthrough
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(b) $$f(x)=2x^3-21x^2+40x+63$$, where $$x\in\mathbb{R}$$.

(i) $$x+1$$ is a factor of $$f(x)$$. Find the three values of $$x$$ for which $$f(x)=0$$.

(ii) Find the range of values of $$x$$ for which $$f'(x)$$ is negative, correct to $$2$$ decimal places.

(i) $$x=-1$$, $$x=4.5$$ or $$x=7$$

(ii) $$1.14<x<5.86$$

Solution

(i)

$\require{enclose} \begin{array}{rll} 2x^2-23x+63\phantom{000000}\, \\[-3pt] x+1 \enclose{longdiv}{\,2x^3-21x^2+40x+63} \\[-3pt] \underline{2x^3+2x^2\phantom{00000000000}\,\,} \\[-3pt] -23x^2+40x\phantom{000}\,\\[-3pt]\phantom{0000}\underline{-23x^2-23x\phantom{00}\,}\\[-3pt]\phantom{00}63x+63\\[-3pt]\phantom{00}\underline{63x+63}\\[-3pt]0 \end{array}$

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

\begin{align}2x^3-21x^2+40x+63=(x+1)(2x^2-23x+63)\end{align}

$\,$

Factors

\begin{align}(x+1)(2x^2-23x+63)=0\end{align}

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

\begin{align}(x+1)(2x-9)(x-7)=0\end{align}

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

$$x=-1$$, $$x=4.5$$ or $$x=7$$

(ii)

Derivative

\begin{align}f'(x)=2x^3-21x^2+40x+63\end{align}

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

\begin{align}f^{\prime\prime}(x)=6x^2-42x+40\end{align}

$\,$

Inequality

\begin{align}6x^2-42x+40<0\end{align}

Roots:

$$a=6$$, $$b=-42$$ and $$c=40$$

\begin{align}\frac{-b\pm\sqrt{b^2-4ac}}{2a}&=\frac{-(-42)\pm\sqrt{(-42)^2-4(6)(40)}}{2(6)}\\&=1.137…\mbox{ or }5.862…\\&\approx1.14\mbox{ or } 5.86\end{align}

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

\begin{align}1.14<x<5.86\end{align}

Video Walkthrough

## 2022 Paper 1 Question 6

(a) Differentiate $$f(x)=2x^2+4x$$ with respect to $$x$$, from first principles.

$$4x+4$$

Solution

\begin{align}f(x)=2x^2+4x\end{align}

and

\begin{align}f(x+h)=2(x+h)^2+4(x+h)\end{align}

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

\begin{align}\frac{df}{dx}&=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}\\&=\lim_{h\rightarrow0}\frac{[2(x+h)^2+4(x+h)]-[2x^2+4x]}{h}\\&=\lim_{h\rightarrow0}\frac{2(x+h)^2+4(x+h)-2x^2-4x}{h}\\&=\lim_{h\rightarrow0}\frac{2x^2+4xh+2h^2+4x+4h-2x^2-4x}{h}\\&=\lim_{h\rightarrow0}\frac{4xh+2h^2+4h}{h}\\&=\lim_{h\rightarrow0}4x+2h+4\\&=4x+4\end{align}

Video Walkthrough
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(b) A rectangle is expanding in area. Its width is $$x\mbox{ cm}$$, where $$x\in\mathbb{R}$$ and $$x>0$$.
Its length is always four times its width.

Find the rate of change of the area of the rectangle with respect to its width, $$x$$, when the area of the rectangle is $$225\mbox{ cm}^2$$.

$$60\mbox{ cm}$$

Solution

\begin{align}A&=lw\\&=lx\\&=\\&=(4x)(x)\\&=4x^2\end{align}

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

\begin{align}\frac{dA}{dx}=8x\end{align}

When $$A=4x^2=225$$, i.e. when $$x= 7.5$$, this gives

\begin{align}\frac{dA}{dx}&=8x\\&=8(7.5)\\&=60\mbox{ cm}\end{align}

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(c) The graph of a cubic function $$p(x)$$ is shown in the first diagram below, for $$0\leq x\leq 4$$, $$x\in\mathbb{R}$$. The maximum value of $$p'(x)$$ in this domain is $$1$$, and $$p'(0)=-3$$, where $$p'(x)$$ is the derivative of $$p(x)$$.

Use this information to draw the graph of $$p'(x)$$ on the second set of axes below, for $$0\leq x\leq4$$, $$x\in\mathbb{R}$$.

Solution
Video Walkthrough

## 2022 Paper 1 Question 7(b)-(e)

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}

(b) Find $$h'(x)$$.

$$h'(x)=6x^2-57x+105$$

Solution

\begin{align}h'(x)&=6x^2-57x+105\end{align}

Video Walkthrough
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(c) Find $$h'(2)$$, and explain what this value means in the context of Hannah’s heart-rate.

$$h'(2)=15$$ is the rate of increase of Hannah’s heart rate at $$2$$ minutes.

Solution

\begin{align}h'(2)&=6(2^2)-57(2)+105\\&=15\end{align}

This is the rate of increase of Hannah’s heart rate at $$2$$ minutes.

Video Walkthrough
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The graph below shows $$y=h(x)$$, where $$0\leq x\leq 8$$, $$x\in\mathbb{R}$$.

(d) Find the least value and the greatest value of $$h(x)$$, for $$0\leq x\leq 8$$, $$x\in\mathbb{R}$$.
Use calculus in your solution. You may also use information from the graph above, which is to scale.

Least value: $$70$$.

Greatest value: $$185.625$$.

Solution

Least Value

\begin{align}h(0)&=2(0^3)-28.5(0^2)+105(0)+70\\&=70\end{align}

$\,$

Greatest Value

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

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

\begin{align}6x^2-57x+105=0\end{align}

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

\begin{align}2x^2-19x+35=0\end{align}

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

\begin{align}(2x-5)(x-7)=0\end{align}

$$x=2.5$$ or $$x=7$$

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

\begin{align}h(2.5)&=2(2.5^3)-28.5(2.5^2)+105(2.5)+70\\&=185.625\end{align}

Video Walkthrough
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(e) How long after the start of the session is Hannah’s heart-rate decreasing most quickly, within the first $$8$$ minutes? Give your answer in minutes and seconds.

Remember that $$h(x)=2x^3-28.5x^2+105x+70$$.

$$4\mbox{ min } 45\mbox{ secs}$$

Solution

The second derivative is zero.

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

\begin{align}12x-57=0\end{align}

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

\begin{align}x&=4.75\mbox{ min}\\&=4\mbox{ min } 45\mbox{ secs}\end{align}

Video Walkthrough

## 2022 Paper 1 Question 8(f)

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.

(f) Use integration to find the average height of the point $$A$$ over the first $$8$$ minutes that the wheel is turning. Give your answer correct to $$1$$ decimal place.

$$65.8\mbox{ m}$$

Solution

\begin{align}\mbox{Average height }&=\frac{1}{8-0}\int_0^8\left[72-60\cos\left(\frac{\pi}{3}t\right)\right]\,dt\\&=\frac{1}{8}\left[72\int_0^8dt-60\int_0^8\cos\left(\frac{\pi}{3}t\right)\,dt\right]\\&=\frac{1}{8}\left[\left.72t\right|_0^8-\left.\frac{180}{\pi}\sin\left(\frac{\pi}{3}t\right)\right|_0^8\right]\\&=\frac{1}{8}\left[(72(8)-72(0))-\frac{180}{\pi}\sin\left(\frac{\pi}{3}(8)\right)+\frac{180}{\pi}\sin\left(\frac{\pi}{3}(0)\right)\right]\\&=72-6.20…\\&=65.8\mbox{ m}\end{align}

Video Walkthrough

## 2022 Paper 1 Question 10(c)-(d)

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

(c)

(i) Find the value of $$P'(1)$$.

(ii) $$P'(t)$$ is always negative for $$0\leq t\leq 12$$, $$t\in\mathbb{R}$$. What does this tell you about the proportion of digits recalled correctly after $$t$$ hours, according to this model?

(i) $$-0.06$$

(ii) As time advances, the amount of digits that the student can recall decreases.

Solution

(i)

\begin{align}P'(t)=-\frac{0.12}{t+1}\end{align}

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

\begin{align}P'(1)&=-\frac{0.12}{1+1}\\&=-0.06\end{align}

(ii) As time advances, the amount of digits that the student can recall decreases.

Video Walkthrough
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(d) Use calculus to show that the graph of $$y=P(t)$$ has no points of inflection, for $$0\leq t\leq 12$$, $$t\in\mathbb{R}$$.

As the second derivative can never be zero, there are no points of inflection.

Solution

\begin{align}\frac{d^2P}{dt^2}=\frac{0.12}{(t+2)^2}\end{align}

As this can never be zero, there are no points of inflection.

Video Walkthrough

## 2021 Paper 1 Question 5

(a) The derivative of $$f(x)=2x^3+6x^2-12x+3$$ can be expressed in the form $$f'(x)=a(x+b)^2$$, where $$a,b,c\in\mathbb{Z}$$ and $$x\in\mathbb{R}$$.

(i) Find the value of $$a$$, the value of $$b$$, and the value of $$c$$.

(ii) If $$g(x)=36x+5$$, find the range of values of $$x$$ for which $$f'(x)>g'(x)$$.

(i) $$a=6$$, $$b=1$$ and $$c=-18$$

(ii) $$x<-4$$ or $$x>2$$

Solution

(i)

\begin{align}f'(x)&=6x^2+12x-12\\&=6(x^2+2x-2)\\&=6(x^2+2x+1-1-2)\\&=6[(x+1)^2-3]\\&=6(x+1)^2-18\end{align}

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

\begin{align}a=6&&b=1&&c=-18\end{align}

(ii)

\begin{align}f'(x)>g'(x)\end{align}

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

\begin{align}6x^2+12x-12>36\end{align}

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

\begin{align}6x^2+12x-48>0\end{align}

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

\begin{align}x^2+2x-8>0\end{align}

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

\begin{align}(x+4)(x-2)>0\end{align}

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

$$x<-4$$ or $$x>2$$

Video Walkthrough
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(b) The diagram below shows $$t$$, the tangent line to $$h(x)=2\sin(2x)$$ , where $$0\leq x\leq \pi$$, at the point where $$x=\dfrac{\pi}{6}$$.
$$A(0,k)$$, where $$k\in\mathbb{R}$$, is the point where $$t$$ cuts the $$y$$-axis.
Find the value of $$k$$ correct to two decimal places.

$$k=0.68$$

Solution

Slope of Tangent

\begin{align}h'(x)=4\cos(2x)\end{align}

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

\begin{align}h’\left(\frac{\pi}{6}\right)&=4\cos\left[2\left(\frac{\pi}{6}\right)\right]\\&=2\end{align}

$\,$

Point on Tangent

\begin{align}h\left(\frac{\pi}{6}\right)&=2\sin\left[2\left(\frac{\pi}{6}\right)\right]\\&=\sqrt{3}\end{align}

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

\begin{align}(x,y)=\left(\frac{\pi}{6},\sqrt{3}\right)\end{align}

$\,$

Equation of Tangent

\begin{align}y=mx+c\end{align}

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

\begin{align}\sqrt{3}=2\left(\frac{\pi}{6}\right)+k\end{align}

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

\begin{align}k&=\sqrt{3}-\frac{\pi}{3}\\&\approx0.68\end{align}

Video Walkthrough

## 2021 Paper 1 Question 6

The diagram below shows the graph of $$h'(x)$$, the derivative of a cubic function $$h(x)$$.

(a) Show that $$h'(x)=-2x^2+4x+6$$.

Solution

\begin{align}y=ax^2+bx+c\end{align}

$\,$

\begin{align}(x,y)=(-1,0)\end{align}

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

\begin{align}0=a(-1)^2+b(-1)+c\end{align}

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

\begin{align}a-b+c=0\end{align}

$\,$

\begin{align}(x,y)=(0,6)\end{align}

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

\begin{align}6=a(0^2)+b(0)+c\end{align}

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

\begin{align}c=6\end{align}

$\,$

\begin{align}(x,y)=(3,0)\end{align}

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

\begin{align}0=a(3^2)+b(3)+c\end{align}

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

\begin{align}9a+3b+c=0\end{align}

$\,$

We therefore have the following $$3$$ equations $$3$$ unknowns:

\begin{align}a-b+c=0\end{align}

\begin{align}c=6\end{align}

\begin{align}9a+3b+c=0\end{align}

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

\begin{align}a-b=-6\end{align}

\begin{align}9a+3b=-6\end{align}

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

\begin{align}3a-3b=-18\end{align}

\begin{align}9a+3b=-6\end{align}

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

\begin{align}12a=-24\end{align}

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

\begin{align}a=-2\end{align}

and

\begin{align}a-b=-6\end{align}

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

\begin{align}b&=a+6\\&=-2+6\\&=4\end{align}

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

\begin{align}a=-2&&b=4&&c=6\end{align}

\begin{align}h'(x)=-2x^2+4x+6\end{align}

as required.

Video Walkthrough
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(b) Use $$h'(x)$$ to find the maximum positive value of the slope of a tangent to $$h(x)$$.

$$8$$

Solution

Setting the second derivative to zero gives:

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

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

\begin{align}x=1\end{align}

As the third derivative is $$-4$$, this is the $$x$$ coordinate of a maximum of $$h'(x)$$ and therefore the maximum positive slope of a tangent to $$h(x)$$ is

\begin{align}h'(1)&=-2(1)^2+4(1)+6\\&=8\end{align}

Video Walkthrough
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(c) The graph of $$h(x)$$ passes through the point $$(0,-2)$$.
Find the equation of $$h(x)$$.

$$h(x)=-\dfrac{2x^3}{3}+2x^2+6x-2$$

Solution

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

and

\begin{align}(x,y)=(0,-2)\end{align}

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

\begin{align}-2-\frac{2(0^3)}{3}+2(0^2)+6(0)+C\end{align}

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

\begin{align}C=-2\end{align}

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

\begin{align}h(x)=-\frac{2x^3}{3}+2x^2+6x-2\end{align}

Video Walkthrough

## 2021 Paper 1 Question 8(b)-(c)

(b) The function $$h(x)=0.001x^3-0.12x^2+3.6x+5$$ can be used to model the height above level ground (in metres) of a section of the path followed by a rollercoaster track, where $$x$$ is the horizontal distance from a fixed point.

(i) Find $$h'(x)$$, the derivative of $$h(x)$$.

(ii) Show that this section of the track reaches its maximum height above level ground when $$x=20$$.

(iii) Find, using calculus, the height above ground, in metres, at the instant the track passes through an inflection point. The function $$h(x)=0.001x^3-0.12x^2+3.6x+5$$.

(i) $$h'(x)=0.003x^2-0.24x+3.6$$

(iii) $$21\mbox{ m}$$

Solution

(i)

\begin{align}h'(x)=0.003x^2-0.24x+3.6\end{align}

(ii)

\begin{align}h'(20)&=0.003(20^2)-0.24(20)+3.6\\&=0\end{align}

According to the graph, the turning point at $$x=20$$ is a maximum rather than a minimum.

(iii) Setting the second derivative to zero:

\begin{align}0.006x-0.24=0\end{align}

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

\begin{align}x&=\frac{0.24}{0.006}\\&=40\end{align}

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

\begin{align}h(40)&=0.001(40^3)-0.12(40^2)+3.6(40)+5\\&=21\mbox{ m}\end{align}

Video Walkthrough
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(c) Use the function $$h(x)=0.001x^3-0.12x^2+3.6x+5$$, $$x\in\mathbb{R}$$, to find the average height of this section of the track above level ground, from $$x=0$$ to $$x=75$$.
Give your answer in metres correct to $$2$$ decimal places.

$$20.47\mbox{ m}$$

Solution

\begin{align}\mbox{Average}&=\frac{1}{b-a}\int_a^bh(x)\,dx\\&=\frac{1}{75-0}\int_0^{75}(0.001x^3-0.12x^2+3.6x+5)\,dx\\&=\frac{1}{75}\left[\frac{0.001x^4}{4}-0.12\frac{x^3}{3}+\frac{3.6x^2}{2}+5x\right]_0^{75}\\&=\frac{1}{75}\left[\frac{0.001(75^4)}{4}-0.12\frac{(75^3)}{3}+\frac{3.6(75^2)}{2}+5(75)\right]\\&\approx20.47\mbox{ m}\end{align}

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## 2021 Paper 1 Question 9(c)-(d)

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

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

(c) Find, to the nearest $$^{\circ}\mbox{C}$$, the temperature the coffee has reached when $$T'(t)=-4.05$$ is the rate at which the coffee is cooling, in $$^{\circ}\mbox{C}$$ per minute.

$$70^{\circ}\mbox{ C}$$

Solution

\begin{align}T'(t)=-6.075e^{-0.081t}\end{align}

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

\begin{align}-4.05=-6.075e^{-0.081t}\end{align}

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

\begin{align}e^{-0.081t}=\frac{4.05}{6.075}\end{align}

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

\begin{align}-0.081t=\ln\left(\frac{4.05}{6.075}\right)\end{align}

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

\begin{align}t&=-\frac{1}{0.081}\ln\left(\frac{4.05}{6.075}\right)\\&=5.0057…\end{align}

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

\begin{align}T(5.0057…)&=75e^{-0.081(5.0057)}+20\\&\approx70^{\circ}\mbox{ C}\end{align}

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(d) A sugar cube is put into the coffee.
The sugar keeps its cube shape as it dissolves.
As the sugar cube dissolves, its volume decreases at the constant rate of $$\dfrac{1}{20}\mbox{ cm}^3/\mbox{sec}$$.
Let $$x(t)$$ be the sidelength of the cube at time $$t$$.
Find the rate of change of $$x(t)$$ when the volume of the cube reaches $$\dfrac{1}{64}\mbox{ cm}^3$$.

$$-\dfrac{4}{15}\mbox{ cm/s}$$

Solution

\begin{align}V=x^3\end{align}

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

\begin{align}\frac{dV}{dx}=3x^2\end{align}

$\,$

\begin{align}\frac{dV}{dt}=-\frac{1}{20}\end{align}

and

\begin{align}\frac{dV}{dt}=\frac{dV}{dx}\times\frac{dx}{dt}\end{align}

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

\begin{align}\frac{dV}{dx}\times\frac{dx}{dt}=-\frac{1}{20}\end{align}

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

\begin{align}(3x^2)\frac{dx}{dt}=-\frac{1}{20}\end{align}

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

\begin{align}\frac{dx}{dt}=-\frac{1}{60x^2}\end{align}

When $$V=\dfrac{1}{64}$$, $$x=\sqrt{\dfrac{1}{64}}=\dfrac{1}{4}$$ and therefore

\begin{align}\frac{dx}{dt}&=-\frac{1}{60\left(\frac{1}{4}\right)^2}\\&=-\frac{4}{15}\mbox{ cm/s}\end{align}

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## 2021 Paper 1 Question 10(a)

(a) Water is flowing into and being removed from a tank.
The volume of water in the tank is measured on a daily basis, starting at day $$0$$.
The volume of water, in litres, in the tank can be modelled by the formula

$$V(t)=60+41t-3t^2$$, while $$V(t)\geq0$$,

where $$t\in\mathbb{R}$$ is the time in days, starting at day $$0$$.
Use the function $$V(t)$$ to answer the following $$4$$ questions.

(i) Find the value of $$t$$ when the tank empties.

(ii) Find the rate at which the volume of water in the tank is changing when $$t=5$$.

(iii) Find the value of $$t$$ when the volume of water in the tank is a maximum.

(iv) Find the maximum volume of the water in the tank, correct to the nearest litre.

(i) $$t=15\mbox{ days}$$

(ii) $$11\mbox{ litres/day}$$

(iii) $$t=\dfrac{41}{6}\mbox{ days}$$

(iv) $$200\mbox{ litres}$$

Solution

(i)

\begin{align}V(t)=0\end{align}

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

\begin{align}3t^2-41t-60=0\end{align}

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

\begin{align}(t-15)(3t+4)=0\end{align}

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

\begin{align}t=15\mbox{ days}\end{align}

(since $$t\geq0$$).

(ii)

\begin{align}V'(t)=41-6t\end{align}

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

\begin{align}V'(5)&=41-6(5)\\&=11\mbox{ litres/day}\end{align}

(iii)

\begin{align}V'(t)=0\end{align}

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

\begin{align}41-6t=0\end{align}

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

\begin{align}t=\frac{41}{6}\mbox{ days}\end{align}

(Since the second derivative ($$-6$$) is negative, this is a maximum.)

(iv)

\begin{align}V\left(\frac{41}{6}\right)&=60+41\left(\frac{41}{6}\right)-3\left(\frac{41}{6}\right)^2\\&\approx200\mbox{ litres}\end{align}

Video Walkthrough

## 2020 Paper 1 Question 4

The diagram below shows two functions $$f(x)$$ and $$g(x)$$.

The function $$f(x)$$ is given by the formula $$f(x)=x^3+kx^2+15x+8$$, where $$k\in\mathbb{Z}$$, and $$x\in\mathbb{R}$$.

(a) Given that $$f'(3)=-12$$, show that $$k=-9$$, where $$f'(3)$$ is the derivative of $$f(x)$$ at $$x=3$$.

Solution

\begin{align}f'(x)=3x^2+2kx+15\end{align}

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

\begin{align}-12=3(3^2)+2k(3)+15\end{align}

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

\begin{align}-12=27+6k+15\end{align}

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

\begin{align}k&=\frac{-12-27-15}{6}\\&=-9\end{align}

as required.

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(b) The function $$g(x)$$ is the line that passes through the two turning points of $$f(x)=x^3-9x^2+15x+8$$.
Find the equation of $$g(x)$$.

$$y=-8x+23$$

Solution

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

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

\begin{align}3x^2-18x+15=0\end{align}

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

\begin{align}x^2-6x+5=0\end{align}

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

\begin{align}(x-1)(x-5)=0\end{align}

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

$$x=1$$ or $$x=5$$

$\,$

Points

\begin{align}f(1)&=1^3-9(1^2)+15(1)+8\\&=15\end{align}

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

\begin{align}(x_1,y_1)=(1,15)\end{align}

and

\begin{align}f(5)&=5^3-9(5^2)+15(5)+8\\&=-17\end{align}

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

\begin{align}(x_2,y_2)=(5,-17)\end{align}

$\,$

Slope

\begin{align}m&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{-17-15}{5-1}\\&=-8\end{align}

$\,$

Equation

\begin{align}y-y_1=m(x-x_1)\end{align}

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

\begin{align}y-15=-8(x-1)\end{align}

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

\begin{align}y-15=-8x+8\end{align}

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

\begin{align}y=-8x+23\end{align}

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(c) Show that the graph of $$g(x)$$ contains the point of inflection of $$f(x)$$.

Solution

Setting the second derivative of $$f(x)$$ to zero gives:

\begin{align}6x-18=0\end{align}

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

\begin{align}x=3\end{align}

and

\begin{align}f(3)&=3^3-9(3^2)+15(3)+8\\&=-1\end{align}

Hence, the point of inflection is $$(3,-1)$$.

\begin{align}g(x)=-8x+23\end{align}

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

\begin{align}g(3)&=-8(3)+23\\&=-1\end{align}

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

\begin{align}(3,-1)\in g(x)\end{align}

Video Walkthrough

## 2020 Paper 1 Question 6

(a) Differentiate $$(3x-5)(2x+4)$$ with respect to $$x$$ from first principles.

$$12x+2$$

Solution

\begin{align}f(x)&=(3x-5)(2x+4)\\&=6x^2+12x-10x-20\\&=6x^2+2x-20\end{align}

and

\begin{align}f(x+h)&=6(x+h)^2+2(x+h)-20\\&=6x^2+12hx+6h^2+2x+2h-20\end{align}

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

\begin{align}\frac{df}{dx}&=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}\\&=\lim_{h\rightarrow0}\frac{(6x^2+12hx+6h^2+2x+2h-20)-(6x^2+2x-20)}{h}\\&=\lim_{h\rightarrow0}\frac{12hx+6h^2+2h}{h}\\&=\lim_{h\rightarrow0}12x+6h+2\\&=12x+2\end{align}

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(b)

(i) $$h(x)=\dfrac{1}{2}\ln(2x+3)+C$$, where $$C$$ is a constant.

Find $$h'(x)$$, the derivative of $$h(x)$$.

(ii) The diagram below shows part of the graph of the function $$h'(x)$$.
The shaded region in the diagram is between the graph and the $$x$$-axis,
from $$x=0$$ to $$x=A$$.
This shaded region has an area of $$\mathbf{\ln 3}$$ square units. Find the value of $$A$$.

(i) $$h'(x)=\dfrac{1}{2x+3}$$

(ii) $$A=12$$

Solution

(i)

\begin{align}h'(x)&=\frac{1}{2}\left(\frac{1}{2x+3}\right)(2)\\&=\frac{1}{2x+3}\end{align}

(ii)

\begin{align}\int_0^A\frac{1}{2x+3}\,dx=\ln3\end{align}

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

\begin{align}\left.\frac{1}{2}\ln(2x+3)\right|_0^A=\ln3\end{align}

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

\begin{align}\frac{1}{2}[\ln(2A+3)-\ln3]=\ln3\end{align}

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

\begin{align}\frac{1}{2}\ln\left(\frac{2A+3}{3}\right)=\ln3\end{align}

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

\begin{align}\ln\left(\frac{2A+3}{3}\right)^{1/2}=\ln3\end{align}

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

\begin{align}\left(\frac{2A+3}{3}\right)^{1/2}=3\end{align}

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

\begin{align}\frac{2A+3}{3}=3^2\end{align}

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

\begin{align}2A+3=27\end{align}

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

\begin{align}A&=\frac{27-3}{2}\\&=12\end{align}

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## 2020 Paper 1 Question 8

A rectangle is inscribed in a circle of radius $$5$$ units and centre $$O(0,0)$$ as shown below.
Let $$R(x,y)$$, where $$x,y\in\mathbb{R}$$, be the vertex of the rectangle in the first quadrant as shown.
Let $$\theta$$ be the angle between $$[OR]$$ and the positive $$x$$-axis, where $$0\leq\theta\leq\dfrac{\pi}{2}$$.

(a)

(i) The point $$R(x,y)$$ can be written as $$a\cos \theta, b\sin\theta$$, where $$a,b\in\mathbb{R}$$.
Find the value of $$a$$ and the value of $$b$$.

(ii) Show that$$A(\theta)$$, the area of the rectangle, measured in square units, can be written as $$A(\theta)=50\sin2\theta$$.

(iii) Use calculus to show that the rectangle with maximum area is a square.

(iv) Find this maximum area.

(i) $$a=5$$ and $$b=5$$

(iv) $$50\mbox{ square units}$$

Solution

(i)

\begin{align}x=5\cos\theta&&y=5\sin\theta\end{align}

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

$$a=5$$ and $$b=5$$

(ii)

\begin{align}A(\theta)&=(2\times5\sin\theta)(2\times5\cos\theta)\\&=100\sin\theta\cos\theta\\&=50(2\sin\theta\cos\theta)\\&=50\sin2\theta\end{align}

(iii)

Setting the derivative to zero, we obtain:

\begin{align}(50\cos2\theta)(2)=0\end{align}

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

\begin{align}\cos2\theta=0\end{align}

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

\begin{align}2\theta&=\cos^{-1}(0)\\&=90^{\circ}\end{align}

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

\begin{align}\theta=45^{\circ}\end{align}

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

\begin{align}l&=10\cos45^{\circ}\\&=\frac{10}{\sqrt{2}}\end{align}

and

\begin{align}w&=10\sin45^{\circ}\\&=\frac{10}{\sqrt{2}}\end{align}

As these are the same, the maximum area is a square.

(iv)

\begin{align}A&=l\times w\\&=\left(\frac{10}{\sqrt{2}}\right)\times\left(\frac{10}{\sqrt{2}}\right)\\&=50\mbox{ square units}\end{align}

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(b) A person who is $$2\mbox{ m}$$ tall is walking towards a streetlight of height $$5\mbox{ m}$$ at a speed of $$1.5\mbox{ m/s}$$.
Find the rate, in $$\mbox{m/s}$$, at which the length of the person’s shadow ($$x$$), cast by the streetlight, is changing.