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## Question 1

(a) Find the two values of $$m\in\mathbb{Z}$$ for which the following equation in $$x$$ has exactly one solution:

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

$$m=-6$$ or $$m=6$$

Solution

\begin{align}b^2-4ac=0\end{align}

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

\begin{align}(-m)^2-4(3)(3)=0\end{align}

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

\begin{align}m^2=36\end{align}

$$m=-6$$ or $$m=6$$

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(b) Explain why the following equation in $$x$$ has no real solutions:

\begin{align}(2x+3)^2+7=0\end{align}

The discriminant is less than zero.

Solution

\begin{align}(2x+3)^2+7=0\end{align}

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

\begin{align}4x^2+12x+9+7=0\end{align}

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

\begin{align}4x^2+12x+16=0\end{align}

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

\begin{align}b^2-4ac&=12^2-4(4)(16)\\&=-112\end{align}

As the discriminant is less than zero, the roots are complex.

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(c)

(i) Show that $$x=-1$$ is not a solution of $$3x^2+2x+5=0$$.

(ii) Find the remainder when $$3x^2+2x+5$$ is divided by $$x+1$$.

That is, find the value of $$c$$ when $$3x^2+2x+5$$ is written in the form

\begin{align}3x^2+2x+5=(x+1)(ax+b)+c\end{align}

where $$a,b,\in\mathbb{Z}$$.

(ii) $$6$$

Solution

(i)

\begin{align}3x^2+2x+5\end{align}

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

\begin{align}3(-1)^2+2(-1)+5\end{align}

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

\begin{align}3-2+5\end{align}

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

\begin{align}6\end{align}

As this is not equal to zero, it is not a solution.

(ii)

\begin{align}3x^2+2x+5=(x+1)(ax+b)+c\end{align}

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

\begin{align}3x^2+2x+5=ax^2+bx+ax+b+c\end{align}

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

\begin{align}3x^2+2x+5=ax^2+(a+b)x+(b+c)\end{align}

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

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

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

\begin{align}5=b+c\end{align}

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

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

\begin{align}2=3+b\end{align}

\begin{align}5=b+c\end{align}

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

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

\begin{align}-1=b\end{align}

\begin{align}5=b+c\end{align}

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

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

\begin{align}-1=b\end{align}

\begin{align}5=-1+c\end{align}

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

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

\begin{align}-1=b\end{align}

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

Therefore, the remainder is $$6$$.

Video Walkthrough

## 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}

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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

## Question 3

(a) $$z=6+2i$$, where $$i^2=-1$$.

(i) Show that $$z-iz=8-4i$$.

(ii) Show that $$|z|^2+|iz|^2=|z-iz|^2$$.

(iii) The circle $$c$$ passes through the points $$z$$, $$iz$$ and $$0$$, as shown in the diagram below (not to scale). $$z$$ and $$iz$$ are endpoints of a diameter of the circle.

Find the area of the circle $$c$$ in terms of $$\pi$$.

(iii) $$20\pi\mbox{ square units}$$

Solution

(i)

\begin{align}z-iz&=(6+2i)-i(6+2i)\\&=6+2i-6i-2i^2\\&=6-4i+2\\&=8-4i\end{align}

as required.

(ii)

\begin{align}iz&=i(6+2i)\\&=6i+2i^2\\&=-2+6i\end{align}

\begin{align}|z|^2+|iz|^2=|z-iz|^2\end{align}

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

\begin{align}(6^2+2^2)+(2^2+6^2)=8^2+4^2\end{align}

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

\begin{align}36+4+4+36=64+16\end{align}

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

\begin{align}80=80\end{align}

Therefore, the relation is true.

(iii)

\begin{align}r&=\frac{|z-iz|}{2}\\&=\frac{\sqrt{8^2+4^2}}{2}\\&=\frac{\sqrt{80}}{2}\end{align}

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

\begin{align}A&=\pi r^2\\&=\pi\left(\frac{\sqrt{80}}{2}\right)^2\\&=\pi\left(\frac{80}{4}\right)\\&=20\pi\mbox{ square units}\end{align}

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(b) $$(\sqrt{3}-i)^9$$ can be written in the form $$a+ib$$, where $$a,b\in\mathbb{Z}$$ and $$i^2=-1$$.

Use de Moivre’s Theorem to find the value of $$a$$ and the value of $$b$$.

$$a=0$$ and $$b=512$$

Solution

Modulus

\begin{align}r&=\sqrt{(\sqrt{3})^2+1^2}\\&=\sqrt{3+1}\\&=\sqrt{4}\\&=2\end{align}

$\,$

Reference Angle

\begin{align}\tan A=\frac{1}{\sqrt{3}}\end{align}

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

\begin{align}A&=\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)\\&=30^{\circ}\end{align}

$\,$

Argument

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

\begin{align}\theta&=360^{\circ}-30^{\circ}=330^{\circ}\end{align}

$\,$

Polar Form

\begin{align}\sqrt{3}-i=2(\cos330^{\circ}+i\sin330^{\circ})\end{align}

$\,$

De Moivre’s Theorem

\begin{align}(\sqrt{3}-i)^9&=2^9[\cos9(330^{\circ})+i\sin9(330^{\circ})]\\&=512[\cos2970^{\circ}+i\sin2970^{\circ}]\\&=512(0+i)\\&=0+512i\end{align}

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

$$a=0$$ and $$b=512$$

Video Walkthrough

## Question 4

(a) A sequence $$u_1,u_2,u_3…$$ is defined as follows, for $$n\in\mathbb{N}$$:

\begin{align}u_1=2, && u_2=64, && u_{n+1}=\sqrt{\frac{u_n}{u_{n-1}}} \end{align}

Write $$u_3$$ in the form $$2^p$$, where $$p\in\mathbb{R}$$.

$$2^{5/2}$$

Solution

\begin{align}u_3&=\sqrt{\frac{u_2}{u_1}}\\&=\sqrt{\frac{64}{2}}\\&=\sqrt{32}\\&=\sqrt{2^5}\\&=2^{5/2}\end{align}

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(b) The first three terms in an arithmetic sequence are as follows, where $$k\in\mathbb{R}$$:

\begin{align}5e^{-k},&&13,&&5e^k\end{align}

(i) By letting $$y=e^k$$ in this arithmetic sequence, show that:

\begin{align}5y^2-26y+5=0\end{align}

(ii) Use the equation in $$y$$ in part (b)(i) to find the two possible values of $$k$$.
Give each value in the form $$\ln p$$ or $$-\ln p$$, where $$p\in\mathbb{N}$$.

(ii) $$k=-\ln5$$ or $$k=\ln5$$

Solution

(i)

\begin{align}13-5e^{-k}=5e^k-13\end{align}

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

\begin{align}13-\frac{5}{y}=5y-13\end{align}

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

\begin{align}13y-5=5y^2-13y\end{align}

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

\begin{align}5y^2-26y+5=0\end{align}

as required.

(ii)

\begin{align}5y^2-26y+5=0\end{align}

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

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

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

$$y=\dfrac{1}{5}$$ or $$y=5$$

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

$$e^k=\dfrac{1}{5}$$ or $$e^k=5$$

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

$$k=\ln\left(\dfrac{1}{5}\right)$$ or $$k=\ln5$$

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

$$k=-\ln5$$ or $$k=\ln5$$

Video Walkthrough

## 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}

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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

## 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}

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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

## Question 7

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

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

Solution

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

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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.

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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}

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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}

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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

## Question 8

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 $$42\mbox{ m}$$.

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(e) By solving the following equation, find the second time (value of $$t$$) that the point $$A$$ is at a height of $$110\mbox{ m}$$, after it starts turning:

\begin{align}72-60\cos\left(\frac{\pi}{3}t\right)=110\end{align}

Give your answer in minutes, correct to $$2$$ decimal places.

$$3.85\mbox{ minutes}$$

Solution

\begin{align}72-60\cos\left(\frac{\pi}{3}t\right)=110\end{align}

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

\begin{align}60\cos\left(\frac{\pi}{3}t\right)=-38\end{align}

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

\begin{align}\cos\left(\frac{\pi}{3}t\right)=-\frac{38}{60}\end{align}

$\,$

Reference Angle

\begin{align}\cos\left(A\right)=\frac{38}{60}\end{align}

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

Cosine is negative in 2nd quadrant (1st time) and 3rd quadrant (2nd time).

$\,$

\begin{align}\frac{\pi}{3}t=\pi+0.8849…\end{align}

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

\begin{align}t&=\frac{3(\pi+0.8849…)}{\pi}\\&\approx3.85\mbox{ min}\end{align}

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(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.
Remember that $$h(t)=72-60\cos\left(\frac{\pi}{3}t\right)$$.

$$65.8\mbox{ m}$$

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}

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## Question 9

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}

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Alex is given a $$15\mbox{ mg}$$ injection of the drug at the same time every day for a long period of time.

(c) Explain why the total amount of the drug, in mg, in Alex’s body immediately after the $$4$$th injection is given by:

\begin{align}15+15(0.6)+15(0.6)^2+15(0.6)^3\end{align}

Solution

On the fourth day:

• $$15(0.6^3)$$ is the amount left from the first injection.
• $$15(0.6^2)$$ is the amount left from the second injection.
• $$15(0.6)$$ is the amount left from the third injection.
• $$15$$ is the amount in the fourth injection that was just given.

The total amount on the fourth day is then the sum of the above.

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(d) Find the total amount of the drug in Alex’s body immediately after the $$10$$th injection.
Give your answer in $$\mbox{mg}$$, correct to $$2$$ decimal places.

$$37.27\mbox{ mg}$$

Solution

\begin{align}S_n&=\frac{a(1-r^n)}{1-r}\\&=\frac{15(1-0.6^{10})}{1-0.6}\\&\approx37.27\mbox{ mg}\end{align}

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(e) Use the formula for the sum to infinity of a geometric series to estimate the amount of the drug (in $$\mbox{mg}$$) in Alex’s body, after a long period of time during which he gets daily injections.

$$37.5\mbox{ mg}$$

Solution

\begin{align}S_\infty&=\frac{a}{1-r}\\&=\frac{15}{1-0.6}\\&=37.5\mbox{ mg}\end{align}

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(f) Jessica also gets daily injections of a medicinal drug at the same time every day.
She gets $$d\mbox{ mg}$$ of the drug in each injection, where $$d\in\mathbb{R}$$.
Each day, the amount of the drug left in Jessica’s body from an injection decreases by $$15\%$$.

(i) Use the sum of a geometric series to show that the total amount of the drug (in $$\mbox{mg}$$) in Jessica’s body immediately after the $$n$$th injection, where $$n\in\mathbb{N}$$, is:

\begin{align}\frac{20d(1-0.85^n)}{3}\end{align}

(ii) Immediately after the $$7$$th injection, there are $$50\mbox{ mg}$$ of the drug in Jessica’s body.

Find the amount of the drug in one of Jessica’s daily injections.
Give your answer correct to the nearest $$\mbox{mg}$$.

(ii) $$11\mbox{ mg}$$

Solution

(i)

\begin{align}S_n&=\frac{a(1-r^n)}{1-r}\\&=\frac{d(1-0.85^n)}{1-0.85}\\&\frac{20d(1-0.85^n)}{3}\end{align}

as required

(ii)

\begin{align}\frac{20d(1-0.85^7)}{3}=50\end{align}

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

\begin{align}d&=\frac{(3)(50)}{20(1-0.85^7)}\\&\approx11\mbox{ mg}\end{align}

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## Question 10

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}

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(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.

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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.

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If we learn a skill, and then don’t practise it, how well we can do it usually decreases over time.
For example, if you learn to play the guitar, and then don’t play the guitar for a number of months, you will probably not be as good at playing the guitar the first time you try it again.

This effect can be modelled by the following equation:

\begin{align}A=B(t+1)^c\end{align}

where $$A$$ is a measure of how well the skill can be done at a certain time ($$t=0$$);
$$B$$ is a measure of how well the skill can be done $$t$$ months later, without practising;
$$c$$ is a constant; and $$A,B,c,t\in\mathbb{R}$$.

(e)

(i) Write $$c$$ in terms of $$\log_{10} A$$, $$\log_{10} B$$ and $$\log_{10} (t+1)$$.

(ii) A student got $$80\%$$ on a guitar exam.
After two years of not playing the guitar, the student got $$47\%$$ on the same exam.

Use this to find the value of $$c$$ in the model above, correct to $$3$$ decimal places.

(i) $$c=\dfrac{\log_{10}A-\log_{10}B}{\log_{10}(t+1)}$$

(ii) $$0.165$$

Solution

(i)

\begin{align}A=B(t+1)^c\end{align}

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

\begin{align}\log_{10}A&=\log_{10}[B(t+1)^c]\\&=\log_{10}B+\log_{10}(t+1)^c\\&=\log_{10}B+c\log_{10}(t+1)\end{align}

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

\begin{align}c=\frac{\log_{10}A-\log_{10}B}{\log_{10}(t+1)}\end{align}

(ii)

\begin{align}c&=\frac{\log_{10}80-\log_{10}47}{\log_{10}(24+1)}\\&\approx0.165\end{align}

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