L.C. MATHS

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## Question 1

(a) $$\dfrac{(4-2i)}{(2+4i)}=0+ki$$, where $$k\in\mathbb{Z}$$, and $$i^2=-1$$. Find the value of $$k$$.

$$k=-1$$

Solution

\begin{align}\frac{(4-2i)}{(2+4i)}&=\frac{4-2i}{2+4i}\times\frac{2-4i}{2-4i}\\&=\frac{8-16i-4i+8i^2}{4-8i+8i-16i^2}\\&=\frac{-20i}{20}\\&=-i\end{align}

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

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

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(b) Find $$\sqrt{-5+12i}$$.
Give both of your answers in the form $$a+bi$$, where $$a,b\in\mathbb{R}$$.

$$2+3i$$ and $$-2-3i$$

Solution

\begin{align}\sqrt{-5+12i}=a+bi\end{align}

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

\begin{align}-5+12i=(a+bi)^2\end{align}

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

\begin{align}-5+12i=(a^2-b^2)+2abi\end{align}

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

\begin{align}-5=a^2-b^2\end{align}

\begin{align}12=2ab\end{align}

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

\begin{align}a^2-b^2=-5\end{align}

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

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

\begin{align}\left(\frac{6}{b}\right)^2-b^2=-5\end{align}

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

\begin{align}\frac{36}{b^2}-b^2=-5\end{align}

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

\begin{align}36-b^4=-5b^2\end{align}

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

\begin{align}b^4-5b^2-36=0\end{align}

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

\begin{align}(b^2+4)(b^2-9)=0\end{align}

and therefore $$b=\pm 3$$ (since $$b\in\mathbb{R}$$). Hence

\begin{align}a&=\frac{6}{b}\\&=\frac{6}{\pm3}\\&=\pm2\end{align}

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

\begin{align}\sqrt{-5+12i}=2+3i\end{align}

or

\begin{align}\sqrt{-5+12i}=-2-3i\end{align}

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(c) Use De Moivre’s theorem to find the three roots of $$z^3=-8$$.
Give each of your answers in the form $$a+bi$$, where $$a,b\in\mathbb{R}$$, and $$i^2=-1$$.

$$1+\sqrt{3}i$$, $$-2$$ and $$1-\sqrt{3}i$$

Solution

\begin{align}r=8\end{align}

and

\begin{align}\theta=\pi+2\pi n\end{align}

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

\begin{align}z^3=8[\cos(\theta+2\pi n)+i\sin(\theta+2\pi n)]\end{align}

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

\begin{align}z&=8^{1/3}[\cos(\theta+2\pi n)+i\sin(\theta+2\pi n)]^{1/3}\\&=2\left[\cos\left(\frac{\pi+2\pi n}{3}\right)+i\sin\left(\frac{2\pi n}{3}\right)\right]\end{align}

$\,$

$$\mathbf{n=0}$$

\begin{align}z&=2\left[\cos\left(\frac{\pi+2\pi (0)}{3}\right)+i\sin\left(\frac{\pi+2\pi (0)}{3}\right)\right]\\&=1+\sqrt{3}i\end{align}

$\,$

$$\mathbf{n=1}$$

\begin{align}z&=2\left[\cos\left(\frac{\pi+2\pi (1)}{3}\right)+i\sin\left(\frac{\pi+2\pi (1)}{3}\right)\right]\\&=-2\end{align}

$\,$

$$\mathbf{n=2}$$

\begin{align}z&=2\left[\cos\left(\frac{\pi+2\pi (2)}{3}\right)+i\sin\left(\frac{\pi+2\pi (2)}{3}\right)\right]\\&=1-\sqrt{3}i\end{align}

Video Walkthrough

## Question 2

(a) Given that $$x=-3$$ is a solution to $$|x+p|=5$$, find the two values of $$p$$, where $$p\in\mathbb{Z}$$.

$$p=-2$$ or $$p=8$$

Solution

\begin{align}|-3+p|=5\end{align}

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

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

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

\begin{align}9-6p+p^2=25\end{align}

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

\begin{align}p^2-6p-16=0\end{align}

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

\begin{align}(p+2)(p-8)=0\end{align}

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

$$p=-2$$ or $$p=8$$

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(b) $$(x+4)$$ is a factor of $$f(x)=x^3+qx^2-22x+56$$, where $$x\in\mathbb{R}$$ and $$q\in\mathbb{Z}$$.

Show that $$q=-5$$, and find the three roots of $$f(x)$$.

$$x=-4$$, $$x=2$$ and $$x=7$$.

Solution

\begin{align}f(-4)=0\end{align}

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

\begin{align}(-4)^3+q(-4)^2-22(-4)+56=0\end{align}

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

\begin{align}-64+16q+88+56=0\end{align}

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

\begin{align}16q+80=0\end{align}

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

\begin{align}q&=-\frac{80}{16}\\&=-5\end{align}

and

$\require{enclose} \begin{array}{rll} x^2-9x+14\phantom{000000}\, \\[-3pt] x+4 \enclose{longdiv}{\,x^3-5x^2-22x+56} \\[-3pt] \underline{x^3+4x^2\phantom{0000000000}\,\,} \\[-3pt] -9x^2-22x\phantom{0000}\,\\[-3pt]\phantom{00000}\underline{-9x^2-36x\phantom{0000}\,}\\[-3pt]\phantom{00}14x+56\\[-3pt]\phantom{00}\underline{14x+56}\\[-3pt]0 \end{array}$

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

\begin{align}x^3-5x^2-22x+56&=(x+4)(x^2-9x+14)\\&=(x+4)(x-2)(x-7)\end{align}

Therefore, the roots are $$x=-4$$, $$x=2$$ and $$x=7$$.

Video Walkthrough

## Question 3

The diagram shows a cuboid with dimensions $$x$$, $$y$$ and $$z \mbox{ cm}$$.
The areas, in $$\mbox{cm}^2$$, of three of its faces are also shown.

(a) Find the volume of the cuboid in the form $$a\sqrt{b}\mbox{ cm}^3$$, where $$a,b\in\mathbb{N}$$.

$$8\sqrt{6}\mbox{ cm}^3$$

Solution

\begin{align}xy=4\sqrt{3}&&xz=2\sqrt{2}&&yz=8\sqrt{6}\end{align}

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

\begin{align}(xy)(xz)(yz)=(2\sqrt{2})(8\sqrt{6})(4\sqrt{3})\end{align}

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

\begin{align}(xyz)^2=384\end{align}

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

\begin{align}V^2=384\end{align}

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

\begin{align}V&=\sqrt{384}\\&=\sqrt{64\times6}\\&=\sqrt{64}\times\sqrt{6}\\&=8\sqrt{6}\mbox{ cm}^3\end{align}

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(b)

(i) Given that $$f(x)=3x^2+8x-35$$, where $$x\in\mathbb{R}$$, find the two roots of $$f(x)=0$$.

(ii) Hence or otherwise, solve the equation $$3^{2m+1}=35-8(3^m)$$, where $$m\in\mathbb{R}$$.
Give your answer in the form $$m=\log_3p-q$$, where $$p,q\in\mathbb{N}$$.

$$x=\dfrac{7}{3}$$ or $$x=-5$$

$$\log_3 7-1$$

Solution

(i)

\begin{align}3x^2+8x-35=0\end{align}

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

\begin{align}(3x-7)(x+5)=0\end{align}

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

$$x=\dfrac{7}{3}$$ or $$x=-5$$

(ii)

$$3^m=\dfrac{7}{3}$$ or $$3^m=-5$$

Since it is not possible for $$3^m=-5$$, the first root is the only possibility.

\begin{align}3^m=\frac{7}{3}\end{align}

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

\begin{align}m&=\log_3\left(\frac{7}{3}\right)\\&=\log_3 7-\log_3 3\\&=\log_3 7-1\end{align}

Video Walkthrough

## Question 4

(a) Prove using induction that $$2^{3n-1}$$ is divisible by $$7$$ for all $$n\in\mathbb{N}$$.

Solution

When $$n=1$$:

\begin{align}2^{3n-1}+3&=2^{3(1)-1}+3\\&=2^2+3\\&=7\end{align}

which is divisible by $$7$$.

Therefore, let us assume that it is true for $$n=k$$, i.e. that $$2^{3k-1}+3$$ is divisible by $$7$$.

Give that it is true for $$n=k$$ , we wish to now prove that it is also true for $$n=k+1$$, i.e. that $$2^{3(k+1)-1}+3$$ is divisible by $$7$$.

\begin{align}2^{3(k+1)-1}+3&=2^{3k+3-1}+3\\&=2^{3k+2}+3\\&=2^{3+(3k-1)}+3\\&=2^32^{3k-1}+3\\&=(8)2^{3k-1}+3\\&=(7+1)2^{3k-1}+3\\&=7^{3k-1}+(2^{3k-1}+3)\end{align}

As both terms are divisible by $$7$$, $$2^{3(k+1)-1}+3$$ is divisible by $$7$$, as required.

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(b) $$p,p+7,p+14,p+21,…$$ is an arithmetic sequence, where $$p\in\mathbb{N}$$.

(i) Find the $$n^{\mbox{th}}$$ term, $$T_n$$, in terms of $$n$$ and $$p$$, where $$n\in\mathbb{N}$$.

(ii) Find the smallest value of $$p$$ for which $$2021$$ is a term in the sequence.

(i) $$T_n=p+7n-7$$

(ii) $$p=5$$

Solution

(i)

\begin{align}T_n=a+(n-1)d\end{align}

\begin{align}a=p && d=7\end{align}

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

\begin{align}T_n&=p+(n-1)(7)\\&=p+7n-7\end{align}

(ii)

\begin{align}p+7n-7=2021\end{align}

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

\begin{align}p=2028-7n\end{align}

Since $$p\in\mathbb{N}$$, the smallest value of $$p$$ occurs for the largest value of $$n\in\mathbb{N}$$ that satisfies $$2028-7n>0$$, i.e. $$n=289$$, and therefore

\begin{align}p&=2028-7(289)\\&=5\end{align}

Video Walkthrough

## 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 $$x>2$$

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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

## 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.

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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}

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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

## Question 7

The tip of the pendulum of a grandfather clock swings initially through an arc length of $$45\mbox{ cm}$$.
On each successive swing the length of the arc is $$90\%$$ of the previous length.

(a)

(i) Complete the table below by filling in the missing lengths.

Swing $$1$$ $$2$$ $$3$$ $$4$$ $$5$$

Length of Arc (cm)

$$45$$

$$\dfrac{729}{20}$$

(ii) $$T_n=45(0.9)^{n-1}$$ is the arc length of swing $$n$$.
Find the arc length of swing $$25$$, correct to $$1$$ decimal place.

(iii) Find the total distance travelled by the tip of the pendulum when it has completed swing $$40$$. Give your answer, in $$\mbox{cm}$$, correct to the nearest whole number.

(iv) Swing $$p$$ is the first swing which has an arc length of less than $$2\mbox{ cm}$$. Find the value of $$p$$.

(i)

Swing $$1$$ $$2$$ $$3$$ $$4$$ $$5$$

Length of Arc (cm)

$$45$$

$$\dfrac{81}{2}$$

$$\dfrac{729}{20}$$

$$\dfrac{6{,}561}{200}$$

$$\dfrac{59{,}049}{2{,}000}$$

(ii) $$3.6\mbox{ cm}$$

(iii) $$443\mbox{ cm}$$

(iv) $$p=31$$

Solution

(i)

Swing $$1$$ $$2$$ $$3$$ $$4$$ $$5$$

Length of Arc (cm)

$$45$$

$$\dfrac{81}{2}$$

$$\dfrac{729}{20}$$

$$\dfrac{6{,}561}{200}$$

$$\dfrac{59{,}049}{2{,}000}$$

(ii)

\begin{align}T_n=45(0.9)^{n-1}\end{align}

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

\begin{align}T_{25}&=45(0.9)^{25-1}\\&\approx3.6\mbox{ cm}\end{align}

(iii)

\begin{align}S_n=\frac{a(1-r^n)}{1-r}\end{align}

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

\begin{align}S_{40}&=\frac{45(1-0.9^{40})}{1-0.9}\\&\approx443\mbox{ cm}\end{align}

(iv)

\begin{align}45(0.9)^{n-1}=2\end{align}

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

\begin{align}(0.9)^{n-1}=\frac{2}{45}\end{align}

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

\begin{align}n-1=\log_{0.9}\left(\frac{2}{45}\right)\end{align}

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

\begin{align}n&=\log_{0.9}\left(\frac{2}{45}\right)+1\\&=30.551…\end{align}

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

\begin{align}p=31\end{align}

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(b)

(i) If the length of the pendulum is $$1\mbox{ m}$$, show that the angle, $$\theta$$, of swing $$1$$ of the pendulum is $$26^{\circ}$$, correct to the nearest degree.

(ii) Hence, find the total accumulated angle that the pendulum swings through (i.e. the sum of all the angles it swings through until it stops swinging).

(iii) Hence, or otherwise, find the total distance travelled by the tip of the pendulum when it has moved through half of the total accumulated angle.
Give your answer, in $$\mbox{cm}$$, correct to the nearest integer.

(i) $$26^{\circ}$$

(ii) $$260^{\circ}$$

(iii) $$227\mbox{ cm}$$

Solution

(i)

\begin{align}l=2\pi r\left(\frac{\theta}{360^{\circ}}\right)\end{align}

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

\begin{align}\theta&=\frac{(360^{\circ})(l)}{2\pi r}\\&=\frac{(360^{\circ})(45)}{2\pi(100)}\\&\approx26^{\circ}\end{align}

(ii)

\begin{align}S_{\infty}&=\frac{a}{1-r}\\&=\frac{26}{1-0.9}\\&=260^{\circ}\end{align}

(iii)

\begin{align}l&=2\pi r\left(\frac{\theta}{360^{\circ}}\right)\\&=2\pi(100\frac{130^{\circ}}{360^{\circ}}\\&\approx227\mbox{ cm}\end{align}

Video Walkthrough

## Question 8

(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)

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

$$h(x)$$

$$5$$

$$30$$

$$37$$

$$32$$

$$21$$

$$10$$

$$5$$

$$12$$

$$21.875$$

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(b) The function $$h(x)$$ 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}

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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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## Question 9

(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}

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(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 $$V=\dfrac{1}{64}$$, $$x=\sqrt[3]{\dfrac{1}{64}}=\dfrac{1}{4}$$ 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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## Question 10

(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}

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(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}

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