L.C. MATHS

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## Question 1

(a) $$(-4+3i)$$ is one root of the equation $$az^2+bz+c=0$$, where $$a,b,c\in\mathbb{R}$$, and $$i^2=-1$$.
Write the other root.

$$-4-3i$$

Solution

$$-4-3i$$

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(b) Use De Moivre’s Theorem to express $$(1+i)^8$$ in its simplest form.

$$16$$

Solution

\begin{align}z=1+i\end{align}

$\,$

Modulus

\begin{align}r&=\sqrt{1^2+1^2}\\&=\sqrt{2}\end{align}

$\,$

Argument

\begin{align}\tan\theta&=\frac{1}{1}\\&=1\end{align}

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

\begin{align}\theta&=\tan^{-1}(1)\\&=\frac{\pi}{4}\end{align}

$\,$

Polar Form

\begin{align}(i+1)=\sqrt{2}\left[\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\right]\end{align}

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

\begin{align}(i+1)^8&=\sqrt{2}^8\left[\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\right]^8\\&=16\left[\cos\left(\frac{8\pi}{4}\right)+i\sin\left(8\frac{\pi}{4}\right)\right]\\&=16[\cos(2\pi)+i\sin(2\pi)]\\&=16(1=)i)\\&=16\end{align}

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(c) $$(1+i)$$ is a root of the equation $$z^2+(-2+i)z+3-i=0$$.
Find its other root in the form $$m+ni$$, where $$m,n\in\mathbb{R}$$, and $$i^2=-1$$.

$$1-2i$$

Solution

\begin{align}z&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=\frac{-(-2+i)\pm\sqrt{(-2+i)^2-4(1)(3-i)}}{2(1)}\\&=\frac{2-i\pm\sqrt{4-4i+i^2-12+4i}}{2}\\&=\frac{2-i\pm\sqrt{-9}}{2}\\&=\frac{2-i\pm3i}{2}\end{align}

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

$$z=1-2i$$ or $$z=1+i$$

Therefore, the other root is $$z=1-2i$$.

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## Question 2

(a) Find the range of values of $$x$$ for which $$|x-4|\geq2$$, where $$x\in\mathbb{R}$$.

$$x\leq2$$ and $$x\geq6$$

Solution

\begin{align}|x-4|\geq 2\end{align}

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

\begin{align}|x-4|^2\geq 4\end{align}

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

\begin{align}(x-4)^2\geq 4\end{align}

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

\begin{align}(x-4)^2\geq 4\end{align}

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

\begin{align}x^2-8x+16\geq4\end{align}

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

\begin{align}x^2-8x+12\geq0\end{align}

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

\begin{align}(x-2)(x-6)\geq0\end{align}

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

$$x\leq2$$ and $$x\geq6$$

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(b) Solve the simultaneous equations:

\begin{align}x^2+xy+2y^2&=4\\2x+3y&=-1\end{align}

$$x=\dfrac{17}{11}$$ and $$y=-\dfrac{15}{11}$$ or $$x=-2$$ and $$y=1$$

Solution

\begin{align}x^2+xy+2y^2&=4\\2x+3y&=-1\end{align}

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

\begin{align}x^2+xy+2y^2=4\end{align}

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

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

\begin{align}\left(\frac{-1-3y}{2}\right)^2+\left(\frac{-1-3y}{2}\right)y+2y^2=4\end{align}

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

\begin{align}\frac{1+6y+9y^2}{4}+\frac{-y-3y^2}{2}+2y^2=4\end{align}

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

\begin{align}1+6y+9y^2-2y-6y^2+8y^2=16\end{align}

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

\begin{align}11y^2+4y-15=0\end{align}

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

\begin{align}(11y+15)(y-1)=0\end{align}

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

$$y=-\dfrac{15}{11}$$ or $$y=1$$

and

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

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

\begin{align}x=\frac{-1-3\left(-\dfrac{15}{11}\right)}{2}\end{align}

or

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

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

\begin{align}x=\frac{-1+\frac{45}{11}}{2}\end{align}

or

\begin{align}x=\frac{-4}{2}\end{align}

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

\begin{align}x=\frac{\frac{34}{11}}{2}\end{align}

or

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

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

\begin{align}x=\frac{17}{11}\end{align}

or

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

Therefore, the solutions are $$x=\dfrac{17}{11}$$ and $$y=-\dfrac{15}{11}$$ or $$x=-2$$ and $$y=1$$.

Video Walkthrough

## Question 3

(a)

(i) $$f(x)=\dfrac{2}{e^x}$$ and $$g(x)=e^x-1$$, where $$x\in\mathbb{R}$$.

Complete the table below. Write your values correct to two decimal places where necessary.

$$x$$ $$0$$ $$0.5$$ $$1$$ $$\ln(4)$$

$$f(x)=\dfrac{2}{e^x}$$

$$g(x)=e^x-1$$

(ii) In the grid shown, use the table to draw the graphs of $$f(x)$$ and $$g(x)$$ in the domain $$0\leq x\leq \ln(4)$$. Label each graph clearly.

(iii) Use your graphs to estimate the value of $$x$$ for which $$f(x)=g(x)$$.

(i)

$$x$$ $$0$$ $$0.5$$ $$1$$ $$\ln(4)$$

$$f(x)=\dfrac{2}{e^x}$$

$$2$$

$$1.21$$

$$0.74$$

$$0.5$$

$$g(x)=e^x-1$$

$$0$$

$$0.65$$

$$1.72$$

$$3$$

(ii)

(iii) $$x\approx0.7$$

Solution

(i)

$$x$$ $$0$$ $$0.5$$ $$1$$ $$\ln(4)$$

$$f(x)=\dfrac{2}{e^x}$$

$$2$$

$$1.21$$

$$0.74$$

$$0.5$$

$$g(x)=e^x-1$$

$$0$$

$$0.65$$

$$1.72$$

$$3$$

(ii)

(iii) $$x\approx0.7$$

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(b) ݂Solve $$f(x)=g(x)$$ using algebra.

$$x=\ln2$$

Solution

\begin{align}\frac{2}{e^x}=e^x-1\end{align}

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

\begin{align}e^x(e^x-1)=2\end{align}

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

\begin{align}(e^x)^2-e^x-2=0\end{align}

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

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

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

\begin{align}(y+1)(y-2)=0\end{align}

$$y=-1$$ or $$y=2$$

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

$$e^x=-1$$ or $$e^x=2$$

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

$$x=\ln(-1)$$ or $$x=\ln2$$

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

$$x=\ln2$$

Video Walkthrough

## Question 4

(a) Prove by induction that $$8^n-1$$ is divisible by $$7$$ for all $$n\in\mathbb{N}$$.

The answer is already in the question!

Solution

The statement is true for $$n=1$$ as $$8^1-1=7$$ is divisible by $$7$$.

Therefore, let us assume that it is true for $$n=k$$, i.e. that $$8^k-1$$ is divisible by $$7$$.

Using this assumption, we wish to now prove that $$8^{k+1}-1$$ is also divisible by $$7$$.

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

As both terms are divisible by $$7$$, $$8^{k+1}-1$$ is divisible by $$7$$, as required.

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(b) Given $$\log_a2=p$$ and $$\log_a3=q$$, where ܽ$$a>0$$, write each of the following in terms of $$p$$ and $$q$$:

(i) $$\log_a\dfrac{8}{3}$$

(ii) $$\log_a\dfrac{9a^2}{16}$$

(i) $$3p-q$$

(ii) $$2(q+1)-4p$$

Solution

(i)

\begin{align}\log_a\frac{8}{3}&=\log_a8-\log_a3\\&=\log_a2^3-\log_a3\\&=3\log_a2-\log_a3\\&=3p-q\end{align}

(ii)

\begin{align}\log_a\frac{9a^2}{16}&=\log_a9a^2-\log_a16\\&=\log_a(3a)^2-\log_a2^4\\&=2\log_a3a-4\log_a2\\&=2(\log_a3+\log_aa)-4\log_a2\\&=2(q+1)-4p\end{align}

Video Walkthrough

## Question 5

(a)

(i) The lengths of the sides of a right-angled triangle are given by the expressions $$x-1$$, $$4x$$, and $$5x-9$$, as shown in the diagram.
Find the value of $$x$$.

(ii) Verify, with this value of $$x$$, that the lengths of the sides of the triangle above form a pythagorean triple.

(i) $$x=10$$

(ii) The answer is already in the question!

Solution

(i)

\begin{align}(x-1)^2+(4x)^2=(5x-9)^2\end{align}

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

\begin{align}x^2-2x+1+16x^2=25x^2-90x+81\end{align}

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

\begin{align}8x^2-88x+80=0\end{align}

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

\begin{align}x^2-11x+10=0\end{align}

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

\begin{align}(x-1)(x-10)=0\end{align}

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

$$x=1$$ or $$x=10$$

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

$$x=10$$

(as $$5(1)-9<0$$).

(ii)

\begin{align}(x-1)^2+(4x)^2=(5x-9)^2\end{align}

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

\begin{align}(10-1)^2+(4(10))^2=(5(10)-9)^2\end{align}

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

\begin{align}9^2+40^2=41^2\end{align}

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

\begin{align}1681=1681\end{align}

as required.

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(b)

(i) Show that $$f(x)=3x-2$$, where $$x\in\mathbb{R}$$, is an injective function.

(ii) Given that $$f(x)=3x-2$$, where $$x\in\mathbb{R}$$, find a formula for $$f^{-1}$$, the inverse function of $$f$$.

(i) The answer is already in the question!

(ii) $$f^{-1}(x)=\dfrac{x+2}{3}$$

Solution

(i)

\begin{align}f(a)=f(b)\end{align}

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

\begin{align}3a-2=3b-2\end{align}

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

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

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

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

as required.

(ii)

\begin{align}f(x)=3x-2\end{align}

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

\begin{align}x=3f^{-1}(x)-2\end{align}

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

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

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## Question 6

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

$$\dfrac{df}{dx}=8x+16$$

Solution

\begin{align}f(x)=(2x+4)^2=4x^2+16x+16\end{align}

and

\begin{align}f(x+h)&=4(x+h)^2+16(x+h)+16\\&=4x^2+8hx+4h^2+16x+16h+16\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{(4x^2+8hx+4h^2+16x+16h+16)-(4x^2+16x+16)}{h}\\&=\lim_{h\rightarrow0}\frac{8hx+4h^2+16h}{h}\\&=\lim_{h\rightarrow0}8x+4h+16\\&=8x+16\end{align}

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(b)

(i) If $$y=x\sin\left(\dfrac{1}{x}\right)$$, find $$\dfrac{dy}{dx}$$.

(ii) Find the slope of the tangent to the curve $$y=x\sin\left(\dfrac{1}{x}\right)$$.
Give your answer correct to two decimal places.

(i) $$\dfrac{dy}{dx}=-\dfrac{1}{x}\cos\left(\dfrac{1}{x}\right)+\sin\left(\dfrac{1}{x}\right)$$

(ii) $$0.15$$

Solution

(i)

$$u=x$$ and $$v=\sin\left(\dfrac{1}{x}\right)$$

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

\begin{align}\frac{dy}{dx}&=u\frac{dv}{dx}+v\frac{du}{dx}\\&=(x)\left(-\frac{1}{x^2}\right)\cos\left(\dfrac{1}{x}\right)+\sin\left(\dfrac{1}{x}\right)(1)\\&=-\frac{1}{x}\cos\left(\dfrac{1}{x}\right)+\sin\left(\dfrac{1}{x}\right)\end{align}

(ii)

\begin{align}\frac{dy}{dx}\left(\frac{4}{\pi}\right)&=-\frac{1}{\frac{4}{\pi}}\cos\left(\dfrac{1}{\frac{4}{\pi}}\right)+\sin\left(\dfrac{1}{\frac{4}{\pi}}\right)\\&=-\frac{\pi}{4}\cos\left(\dfrac{\pi}{4}\right)+\sin\left(\dfrac{\pi}{4}\right)\\&\approx0.15\end{align}

Video Walkthrough

## Question 7

(a)

(i) Air is pumped into a spherical exercise ball at the rate of $$250\mbox{ cm^3}$$
per second.
Find the rate at which the radius is increasing when the radius of the ball is $$20\mbox{cm}$$.
Give your answer in terms of $$\pi$$.

(ii) Find the rate at which the surface area of the ball is increasing when the radius of the ball is $$20\mbox{ cm}$$.

(i) $$\dfrac{5}{32\pi}\mbox{ cm/s}$$

(ii) $$25\mbox{ cm}^2\mbox{/s}$$

Solution

(i)

\begin{align}\frac{dr}{dt}&=\frac{dr}{dV}\frac{dV}{dt}\\&=\left(\frac{1}{4\pi r^2}\right)(250)\end{align}

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

\begin{align}\frac{dr}{dt}(20)&=\left(\frac{1}{4\pi (20^2)}\right)(250)\\&=\frac{5}{32\pi}\mbox{ cm/s}\end{align}

(ii)

\begin{align}\frac{dA}{dt}&=\frac{dA}{dr}\frac{dr}{dt}\\&=(8\pi r)\left(\frac{1}{4\pi r^2}\right)(250)\end{align}

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

\begin{align}\frac{dA}{dt}(20)&=\frac{dA}{dr}\frac{dr}{dt}\\&=(8\pi (20))\left(\frac{1}{4\pi (20^2)}\right)(250)\\&=25\mbox{ cm}^2\mbox{/s}\end{align}

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(b) The inflated ball is kicked into the air from a point $$O$$ on the ground. Taking $$O$$ as the origin, $$(x,f(x))$$ approximately describes the path followed by the ball in the air, where

\begin{align}f(x)=-x^2+10x\end{align}

and both $$x$$ and $$f(x)$$ are measured in metres.

(i) Find the values of $$x$$ when the ball is on the ground.

(ii) Find the average height of the ball above the ground, during the interval from when it is kicked until it hits the ground again.

(i) $$x=0\mbox{ m}$$ or $$x=10\mbox{ m}$$

(ii) $$\dfrac{50}{3}\mbox{ m}$$

Solution

(i)

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

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

\begin{align}-x^2+10x=0\end{align}

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

\begin{align}x(-x+10)=0\end{align}

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

$$x=0\mbox{ m}$$ or $$x=10\mbox{ m}$$

(ii)

\begin{align}\mbox{Average}&=\frac{1}{b-a}\int_a^bf(x)\,dx\\&=\frac{1}{10-0}\int_0^{10}(-x^2+10x)\,dx\\&=\frac{1}{10}\left(-\frac{x^3}{3}+5x^2\right)_0^{10}\\&=\frac{1}{10}\left(-\frac{1000}{3}+500-0-0\right)\\&=-\frac{100}{3}+50\\&=\frac{50}{3}\mbox{ m}\end{align}

Video Walkthrough

## Question 8

(a) The diagram shows Sarah’s first throw at the basket in a basketball game. The ball left her hands at $$A$$ and entered the basket at $$B$$. Using the co-ordinate plane with $$A(-0.5,2.565)$$ and $$B(4.5,3.05)$$, the equation of
the path of the centre of the ball is

\begin{align}f(x)=-0.274x^2+1.193x+3.23\end{align}

where both $$x$$ and $$f(x)$$ are measured in metres.

(i) Find the maximum height reached by the centre of the ball, correct to three decimal places.

(ii) Find the acute angle to the horizontal at which the ball entered the basket.
Give your answer correct to the nearest degree.

(iii) Sarah took a second throw. This throw followed the path of the parabola $$g(x)$$ as shown.
The ball left Sarah’s hands at the point $$C(0,2)$$.
The graph $$y=g(x)$$ is the image of the graph $$y=f(x)$$ under the translation which maps $$A$$ onto $$C$$.
Using your result from part a(i), show that the centre of this ball reached its maximum height at the point $$(2.677,3.964)$$, correct to three decimal places.

(iv) Hence, or otherwise, find the equation of the parabola $$g(x)$$.

(i) $$4.529\mbox{ m}$$

(ii) $$52^{\circ}$$

(iii) The answer is already in the question!

(iv) $$g(x)=-0.273x^2+1.462x+2$$

Solution

(i)

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

\begin{align}-0.548x+1.193=0\end{align}

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

\begin{align}x&=\frac{-1.193}{-0.548}\\&=2.177…\end{align}

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

\begin{align}f(2.177…)&=-0.274(2.177…)^2+1.193(2.177…)+3.23\\&\approx4.529\mbox{ m}\end{align}

(ii)

\begin{align}\tan\theta&=f'(4.5)\\&=-0.548(4.5)+1.193\\&=-1.273\end{align}

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

\begin{align}\theta\approx52^{\circ}\end{align}

(iii)

\begin{align}(-0.5,2.565)\rightarrow(0,2)\end{align}

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

\begin{align}(x,y)\rightarrow(x+0.5,y-0.565)\end{align}

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

\begin{align}(2.177,4.529)\rightarrow(2.177+0.5,4.529-0.565)\end{align}

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

\begin{align}(2.177,4.529)\rightarrow(2.677,3.964)\end{align}

as required.

(iv)

\begin{align}g(x)=ax^2+bx+c\end{align}

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

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

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

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

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

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

\begin{align}g(x)=ax^2+bx+2\end{align}

and

\begin{align}g(4.5)=3.05\end{align}

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

\begin{align}a(4.5^2)+b(4.5)+2=3.05\end{align}

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

\begin{align}20.25a+4.5b=1.05\end{align}

and

\begin{align}g'(2.677)=0\end{align}

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

\begin{align}2a(2.677)+b=0\end{align}

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

\begin{align}5.354a+b=0\end{align}

We therefore have the following two equations for two unknowns:

\begin{align}20.25a+4.5b=1.05\end{align}

\begin{align}5.354a+b=0\end{align}

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

\begin{align}20.25a+4.5b=1.05\end{align}

\begin{align}24.093a+4.5b=0\end{align}

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

\begin{align}3.843a=-1.05\end{align}

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

\begin{align}a&=-\frac{1.05}{3.843}\\&\approx-0.273\end{align}

and

\begin{align}5.354a+b=0\end{align}

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

\begin{align}b&=-5.354a\\&=-5.354(-0.273)\\&=1.462\end{align}

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

\begin{align}g(x)=-0.273x^2+1.462x+2\end{align}

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(b) The heptathlon is an Olympic competition. It consists of seven events including the $$200\mbox{ m}$$ race and the javelin. The scoring system uses formulas to calculate a score for each event. The table below shows the formulas for two of the events and the values of constants used in these formulas, where $$x$$ is the time taken (in seconds) or distance achieved (in metres) by the competitor and $$y$$ is the number of points scored in the event.

Event $$x$$ Formula $$a$$ $$b$$ $$c$$

$$200$$ m race

Time (s)

$$y=a(b-x)^c$$

4.99087

$$42.5$$

$$1.81$$

Javelin

Distance (m)

$$y=a(x-b)^c$$

$$15.9803$$

$$3.8$$

$$1.04$$

(i) In the heptathlon, Jessica ran $$200\mbox{ m}$$ in $$23.8\mbox{ s}$$ and threw the javelin $$58.2\mbox{ m}$$.
Use the formulas in the table to find the number of points she scored in each of these events, correct to the nearest point.

(ii) The world record distance for the javelin, in the heptathlon, would merit a score of $$1295$$ points. Find the world record distance for the javelin, in the heptathlon, correct to two decimal places.

(iii) The formula used to calculate the points for the $$800\mbox{ m}$$ race, in the heptathlon, is the same formula used for the $$200\mbox{ m}$$ race but with different constants.
Jessica ran the $$800\mbox{ m}$$ race in $$2$$ minutes and $$1.84$$ seconds which merited $$1087$$ points.
If ܽ$$a=0.11193$$ and $$b=254$$ for the $$800\mbox{ m}$$ race, find the value of $$c$$ for this event, correct to two decimal places.

(i) 200 m: $$1{,}000$$ points and javelin: $$1{,}020$$ points

(ii) $$72.23\mbox{ m}$$

(iii) $$1.88$$

Solution

(i)

200 m

\begin{align}y=a(b-x)^c\end{align}

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

\begin{align}y(23.8)&=4.99087(42.5-23.8)^{1.81}\\&\approx1{,}000\end{align}

$\,$

Javelin

\begin{align}y=a(x-b)^c\end{align}

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

\begin{align}y(58.2)&=15.9803(58.2-3.8)^{1.04}\\&\approx1{,}020\end{align}

(ii)

\begin{align}15.9803(x-3.8)^{1.04}=1295\end{align}

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

\begin{align}(x-3.8)^{1.04}=\frac{1295}{15.9803}\end{align}

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

\begin{align}\ln[(x-3.8)^{1.04}]=\ln\left(\frac{1295}{15.9803}\right)\end{align}

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

\begin{align}1.04\ln(x-3.8)=\ln\left(\frac{1295}{15.9803}\right)\end{align}

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

\begin{align}\ln(x-3.8)&=\frac{1}{1.04}\ln\left(\frac{1295}{15.9803}\right)\\&=4.225…\end{align}

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

\begin{align}x-3.8&=e^{4.225…}\\&=68.4343…\end{align}

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

\begin{align}x&=68.4343…+3\\&\approx72.23\mbox{ m}\end{align}

(iii)

\begin{align}y=a(b-x)^c\end{align}

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

\begin{align}(b-x)^c=\frac{y}{a}\end{align}

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

\begin{align}\ln[(b-x)^c]=\ln\left(\frac{y}{a}\right)\end{align}

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

\begin{align}c\ln(b-x)=\ln\left(\frac{y}{a}\right)\end{align}

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

\begin{align}c&=\frac{1}{\ln(b-x)}\ln\left(\frac{y}{a}\right)\\&=\frac{1}{\ln(254-121.84)}\ln\left(\frac{1087}{0.11193}\right)\\&\approx1.88\end{align}

Video Walkthrough

## Question 9

(a) At the first stage of a pattern, a point moves $$4$$ units from the origin in the positive direction along the $$x$$-axis. For the second stage, it turns left and moves $$2$$ units parallel to the $$y$$-axis. For the third stage, it turns left and moves $$1$$ unit parallel to the $$x$$-axis.
At each stage, after the first one, the point turns left and moves half the distance of the previous
stage, as shown

(i) How many stages has the point completed when the total distance it has travelled, along its path, is $$7.9375$$ units?

(ii) Find the maximum distance the point can move, along its path, if it continues in this pattern indefinitely.

(iii) Complete the second row of the table below showing the changes to the x co-ordinate, the first nine times the point moves to a new position. Hence, or otherwise, find the $$x$$ co-ordinate and the $$y$$ co-ordinate of the final position that the point is approaching, if it continues indefinitely in this pattern.

Stage 1st 2nd 3rd 4th 5th 6th 7th 8th 9th

Change in $$x$$

$$+4$$

$$0$$

$$-1$$

Change in $$y$$

(i) $$7$$

(ii) $$8\mbox{ units}$$

(iii)

Stage 1st 2nd 3rd 4th 5th 6th 7th 8th 9th

Change in $$x$$

$$+4$$

$$0$$

$$-1$$

$$0$$

$$\dfrac{1}{4}$$

$$0$$

$$-\dfrac{1}{16}$$

$$0$$

$$\dfrac{1}{64}$$

Change in $$y$$

$$0$$

$$2$$

$$0$$

$$-\dfrac{1}{2}$$

$$0$$

$$\dfrac{1}{8}$$

$$0$$

$$-\dfrac{1}{32}$$

$$0$$

$$\left(\dfrac{16}{5},\dfrac{8}{5}\right)$$

Solution

(i)

\begin{align}\frac{a(1-r^n)}{1-r}=7.9375\end{align}

\begin{align}a=4&&r=\frac{1}{2}\end{align}

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

\begin{align}\frac{4\left[1-\left(\frac{1}{2}\right)^n\right]}{1-\frac{1}{2}}=7.9375\end{align}

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

\begin{align}\end{align}

\begin{align}8-8\left(\frac{1}{2}\right)^n=7.9375\end{align}

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

\begin{align}\left(\frac{1}{2}\right)^n&=-\frac{7.9375-8}{8}\\&=\frac{1}{128}\end{align}

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

\begin{align}n=7\end{align}

(ii)

\begin{align}S_{\infty}&=\frac{a}{1-r}\\&=\frac{4}{1-\frac{1}{2}}\\&=8\mbox{ units}\end{align}

(iii)

Stage 1st 2nd 3rd 4th 5th 6th 7th 8th 9th

Change in $$x$$

$$+4$$

$$0$$

$$-1$$

$$0$$

$$\dfrac{1}{4}$$

$$0$$

$$-\dfrac{1}{16}$$

$$0$$

$$\dfrac{1}{64}$$

Change in $$y$$

$$0$$

$$2$$

$$0$$

$$-\dfrac{1}{2}$$

$$0$$

$$\dfrac{1}{8}$$

$$0$$

$$-\dfrac{1}{32}$$

$$0$$

\begin{align}S_{x,\infty}&=\frac{a_x}{1-r_x}\\&=\frac{4}{1-\left(-\frac{1}{4}\right)}\\&=\frac{16}{5}\mbox{ units}\end{align}

and

\begin{align}S_{y,\infty}&=\frac{a_y}{1-r_y}\\&=\frac{2}{1-\left(-\frac{1}{4}\right)}\\&=\frac{8}{5}\mbox{ units}\end{align}

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

\begin{align}(x,y)=\left(\frac{16}{5},\frac{8}{5}\right)\end{align}

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

(b) A male bee comes from an unfertilised egg, i.e. he has a female parent but he does not have a male parent. A female bee comes from a fertilised egg, i.e. she has a female parent and a male parent.

(i) The following diagram shows the ancestors of a certain male bee. We identify his generation as $$G_1$$ and our diagram goes back to $$G_4$$. Continue the diagram to $$G_5$$.

(ii) The number of ancestors of this bee in each generation can be calculated by the formula

\begin{align}G_{n+2}=G_{n+1}+G_n\end{align}

where $$G_1=1$$ and $$G_2=1$$, as in the diagram.
Use this formula to calculate the number of ancestors in $$G_6$$ and in $$G_7$$.

(iii) The number of ancestors in each generation can also be calculated by using the formula

\begin{align}G_n=\frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2^n\sqrt{5}}\end{align}

Use this formula to verify the number of ancestors in $$G_3$$.

(i)

(ii) $$G_6=8$$ and $$G_7=13$$

(iii) $$G_3=2$$

Solution

(i)

(ii)

\begin{align}G_6&=G_5+G_4\\&=5+3\\&=8\end{align}

and

\begin{align}G_7&=G_6+G_5\\&=8+5\\&=13\end{align}

(iii)

\begin{align}G_3&=\frac{(1+\sqrt{5})^3-(1-\sqrt{5})^3}{2^3\sqrt{5}}\\&=\frac{(1+\sqrt{5})(1+2\sqrt{5}+5)-[(1-\sqrt{5})(1-2\sqrt{5}+5)]}{8\sqrt{5}}\\&=\frac{1+2\sqrt{5}+5+\sqrt{5}+10+5\sqrt{5}-1+2\sqrt{5}-5+\sqrt{5}-10+5\sqrt{5}}{8\sqrt{5}}\\&=\frac{16\sqrt{5}}{8\sqrt{5}}\\&=2\end{align}

as required.

Video Walkthrough