L.C. MATHS

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Question 1

In a competition Mary has a probability of $$\dfrac{1}{20}$$ of winning, a probability of $$\dfrac{1}{10}$$ of finishing in second place, and a probability of $$\dfrac{1}{4}$$ of finishing in third place. If she wins the competition she gets €$$9000$$. If she comes second she gets €$$7000$$ and if she comes third she gets €$$3000$$. In all other cases she gets nothing. Each participant in the competition must pay €$$2000$$ to enter.

(a) Find the expected value of Mary’s loss if she enters the competition.

$$100\mbox{ euro}$$

Solution

\begin{align}E(X)&=2000-\left[\frac{1}{20}(9000)+\frac{1}{10}(7000)+\frac{1}{4}(3000)\right]\\&=100\mbox{ euro}\end{align}

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(b) Each of the 3 prizes in the competition above is increased by the same amount (€$$x$$) but the entry fee is unchanged.
For example, if Mary wins the competition now, she would get €($$9000+x$$).
Mary now expects to break even.
Find the value of $$x$$.

$$250\mbox{ euro}$$

Solution

\begin{align}\frac{1}{20}(9000+x)+\frac{1}{10}(7000+x)+\frac{1}{4}(3000+x)=2000\mbox{ euro}\end{align}

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

\begin{align}1900+\frac{8x}{20}=2000\mbox{ euro}\end{align}

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

\begin{align}\frac{8x}{20}=100\end{align}

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

\begin{align}8x=2000\end{align}

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

\begin{align}x&=\frac{2000}{8}\\&=250\mbox{ euro}\end{align}

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Question 2

(a) The diagram shows the standard normal curve. The shaded area represents $$67\%$$ of the data.
Find the value of $$z_1$$.

$$z_1=0.44$$

Solution

\begin{align}P(z<z_1)=0.67\end{align}

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

\begin{align}z_1=0.44\end{align}

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(b) The percentage results in a Maths exam for a class had a mean mark of $$70$$ with a standard deviation of $$15$$. The percentage results in an English exam for the same class had a mean mark of $$72$$ with a standard deviation of $$10$$. The results in both exams were normally distributed.

(i) Mary got $$65$$ in Maths and $$68$$ in English. In which exam did Mary do better relative to the other students in the class? Justify your answer.

(ii) In English the top $$15\%$$ of students were awarded an A grade.
Find the least whole number mark that merited the award of an A grade in English.

(iii) Using the empirical rule, or otherwise, estimate the percentage of students in the class who scored between $$52$$ and $$82$$ in the English test.

(i) Since her $$z$$-score was larger for maths, Mary did better in maths.

(ii) $$83\%$$

(iii) $$81.5\%$$

Solution

(i)

Maths

\begin{align}z&=\frac{x-\mu}{\sigma}\\&=\frac{65-70}{15}=-\frac{1}{3}\end{align}

$\,$

English

\begin{align}z&=\frac{x-\mu}{\sigma}\\&=\frac{68-72}{10}=-\frac{2}{5}\end{align}

Since her $$z$$-score was larger for maths, Mary did better in maths.

(ii)

\begin{align}P(z>z_1)=0.15\end{align}

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

\begin{align}z=1.04\end{align}

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

\begin{align}\frac{x-72}{10}=1.04\end{align}

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

\begin{align}x&=72+(10)(1.04)\\&=82.4\%\end{align}

Therefore, the least whole number score is $$83\%$$.

(iii) As $$52$$ is $$\dfrac{95}{2}=47.5\%$$ below the mean, and as $$82$$ is $$\dfrac{68}{2}=34\%$$ above the mean, the percentage of students is

\begin{align}47.5+34=81.5\%\end{align}

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Question 3

(a) A security code consists of six digits chosen at random from the digits $$0$$ to $$9$$.
The code may begin with zero and digits may be repeated.
For example $$071737$$ is a valid code.

(i) Find how many of the possible codes will end with a zero.

(ii) Find how many of the possible codes will contain the digits $$2$$ $$0$$ $$1$$ $$8$$ together and in this order.

(i) $$100{,}000$$

(ii) $$300$$

Solution

(i)

\begin{align}N&=10\times10\times10\times10\times10\times1\\&=100{,}000\end{align}

(ii)

\begin{align}N&=3\times1\times10\times10\\&=300\end{align}

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(b) Find $$a$$, $$b$$, $$c$$ and $$d$$, if $$\dfrac{(n+3)!(n+2)!}{(n+1)!(n+1)!}=an^3+bn^2+cn+d$$, where $$a$$, $$b$$, $$c$$, and $$d\in\mathbb{N}$$.

$$a=1$$, $$b=7$$, $$c=16$$ and $$d=12$$

Solution

\begin{align}\frac{(n+3)!(n+2)!}{(n+1)!(n+1)!}&=\frac{(n+3)(n+2)(n+1)!(n+2)(n+1)!}{(n+1)!(n+1)!}\\&=(n+3)(n+2)(n+2)\\&=(n+3)(n^2+4n+4)\\&=n^3+4n^2+4n+3n^2+12n+12\\&=n^3+7n^2+16n+12\end{align}

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

\begin{align}a=1 && b=7 && c=16 && d=12\end{align}

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Question 4

(a) Find all the values of $$x$$ for which $$\cos(2x)=-\dfrac{\sqrt{3}}{2}$$, where $$0^{\circ}\leq x\leq 360^{\circ}$$.

$$x=75^{\circ}$$, $$x=105^{\circ}$$, $$x=255^{\circ}$$ and $$x=285^{\circ}$$

Solution

Let $$\theta=2x$$.

\begin{align}\cos\theta=-\frac{\sqrt{3}}{2}\end{align}

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

Reference Angle:

\begin{align}\cos A=\frac{\sqrt{3}}{2}\end{align}

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

\begin{align}A=30^{\circ}\end{align}

$\,$

\begin{align}\theta&=(180^{\circ}-30^{\circ})\\&=150^{\circ}+360^{\circ}n\end{align}

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

\begin{align}x&=\frac{\theta}{2}\\&=\frac{150^{\circ}+360^{\circ}n}{2}\\&=75^{\circ}+180^{\circ}n\end{align}

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

$$x=75^{\circ}$$ and $$x=255^{\circ}$$

$\,$

\begin{align}\theta&=(180^{\circ}+30^{\circ})+360^{\circ}n\\&=210^{\circ}+360^{\circ}n\end{align}

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

\begin{align}x&=\frac{\theta}{2}\\&=\frac{210^{\circ}+360^{\circ}n}{2}\\&=105^{\circ}+180^{\circ}n\end{align}

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

$$x=105^{\circ}$$ and $$x=285^{\circ}$$

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(b) Let $$\cos A=\dfrac{y}{2}$$, where $$0^{\circ}<A<90^{\circ}$$. Write $$\sin2A$$ in terms of $$y$$.

$$\dfrac{y\sqrt{4-y^2}}{2}$$

Solution

\begin{align}\sin A=\frac{\sqrt{4-y^2}}{2}\end{align}

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

\begin{align}\sin(2A)&=2\sin A\cos A\\&=2\left(\frac{\sqrt{4-y^2}}{2}\right)\left(\frac{y}{2}\right)\\&=\frac{y\sqrt{4-y^2}}{2}\end{align}

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Question 5

The line ݉$$m:2x+3y+1=0$$ is parallel to the line ݊$$n:2x+3y-51=0$$.

(a) Verify that $$A(-2,1)$$ is on $$m$$.

Solution

\begin{align}2x+3y+1&=2(-2)+3(1)+1\\&=-4+3+1\\&=0\end{align}

as required.

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(b) Find the coordinates of $$B$$, the point on the line ݊$$n$$ closest to $$A$$, as shown below.

$$(6,13)$$

Solution

\begin{align}2x+3y-51=0\end{align}

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

\begin{align}y=-\frac{2}{3}x+17\end{align}

Therefore, the slope of both parallel lines is $$-\dfrac{2}{3}$$.

The slope of $$AB$$ is therefore $$\dfrac{3}{2}$$.

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

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

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

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

\begin{align}2y-2=3x+6\end{align}

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

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

We will now solve the following two equations:

\begin{align}2x+3y-51=0\end{align}

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

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

\begin{align}6x+9y-153=0\end{align}

\begin{align}6x-4y+16=0\end{align}

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

\begin{align}13y-169=0\end{align}

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

\begin{align}y=13\end{align}

and

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

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

\begin{align}x&=\frac{2y-8}{3}\\&=\frac{2(13)-8}{3}\\&=6\end{align}

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

\begin{align}B:(6,13)\end{align}

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(c) Two touching circles, $$s$$ and $$t$$, are shown in the diagram. ݉$$m$$ is a tangent to $$s$$ at $$A$$ and ݊$$n$$ is a tangent to $$t$$ at $$B$$. The ratio of the radius of $$s$$ to the radius of $$t$$ is $$1:3$$.
Find the equation of s.

$$(x+1)^2+(y-2.5)^2=3.25$$

Solution

The point at which both circles intersect is:

\begin{align}\left(\frac{bx_1+ax_2}{b+a},\frac{by_1+ay_2}{b+a}\right)&=\left(\frac{3(-2)+1(6)}{3+1},\frac{3(1)+1(13)}{3+1}\right)\\&=(0,4)\end{align}

Circle $$s$$ therefore has a centre of

\begin{align}\left(\frac{0-2}{2},\frac{4+1}{2} \right)=(-1,2.5)\end{align}

\begin{align}r&=\sqrt{(-1-(-2))^2+(2.5-1)^2}\\&=\sqrt{3.25}\end{align}

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

\begin{align}(x+1)^2+(y-2.5)^2=3.25\end{align}

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Question 6

(a) Let $$\Delta ABC$$ be a triangle. Prove that if a line ݈$$l$$ is parallel to $$BC$$ and cuts $$[AB]$$ in the ratio $$s:t$$, where $$s,t\in\mathbb{N}$$, then it also cuts $$[AC]$$ in the same ratio.

Solution

Diagram

Given

Triangle $$\Delta ABC$$ and line $$XY$$, parallel to $$BC$$, that cuts $$AB$$ in the ratio $$s:t$$ where $$s,t\in\mathbb{N}$$.

$\,$

To Prove

\begin{align}[AY]:[YC]=s:t\end{align}

$\,$

Construction

Divide $$[AB]$$ into $$s+t$$ equal segments, with $$s$$ segments on $$[AX]$$ and $$t$$ segments on $$[XB]$$.

At each point of division, draw a line parallel to $$XY$$.

$\,$

Proof

According to another theorem, all of the parallel lines cut off segments of equal length along $$[AC]$$.

Therefore, $$[AC]$$ has been divided into $$s+t$$ equal segments of length $$k$$, with $$s$$ segments on $$[AY]$$ and $$t$$ segments on $$[YC]$$.

\begin{align}[AY]:[YC]&=ks:kt\\&=s:t\end{align}

as required.

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(b) In the triangle $$ABC$$ shown below:
$$|\angle CAB|=90^{\circ}$$, $$|AX=4\mbox{ cm}$$, $$|AY|=3\mbox{ cm}$$, $$XY\parallel BC$$, $$XZ\parallel AC$$, and $$|AX|:|XB|=1:2$$.

Find $$|BZ|$$.

$$10\mbox{ cm}$$

Solution

\begin{align}|XY|&=\sqrt{3^2+4^2}\\&=5\mbox{ cm}\end{align}

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

\begin{align}|ZC|=5\mbox{ cm}\end{align}

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

\begin{align}|AX|&=2|ZC|\\&=2(5)\\&=10\mbox{ cm}\end{align}

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Question 7

A section of a garden railing is shown below. This section consists of nine cylindrical bars, labelled $$A$$ to $$I$$, with a solid sphere attached to the centre of the top of each bar.
The volume of each sphere from $$B$$ to $$E$$ is $$1.75$$ times the volume of the previous sphere.

(a) The radius of sphere $$A$$ is $$3\mbox{ cm}$$. Find the sum of the volumes of the five spheres $$A$$, $$B$$, $$C$$, $$D$$ and $$E$$.
Give your answer correct to the nearest $$\mbox{cm}^3$$.

$$2324\mbox{ cm}^3$$

Solution

\begin{align}V_A&=\frac{4}{3}\pi r^3\\&=\frac{4}{3}\pi(3^3)\\&=36\pi\end{align}

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

\begin{align}S_5&=\frac{36\pi(1-1.75^5)}{1-1.75}\\&\approx2324\mbox{ cm}^3\end{align}

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(b)

(i) The surface area of sphere $$E$$ can be taken to be $$503\mbox{ cm}^2$$.
The height of the railing at $$E$$ (i.e. the sum of the heights of bar $$E$$ and sphere $$E$$) is $$1.2$$ metres.
Find the height of bar $$E$$, in $$\mbox{cm}$$, correct to $$1$$ decimal place.

(ii) The radius of each bar is $$1$$ cm. The volume of bar $$A$$ is $$71.3\pi\mbox{ cm}^3$$.
The heights of the bars $$A$$, $$B$$, $$C$$, $$D$$ and $$E$$ form an arithmetic sequence.
Find, in $$\mbox{cm}$$, the height of each bar.

(i) $$107.3\mbox{ cm}$$

(ii) $$71.3\mbox{ cm}, 80.3\mbox{ cm}, 89.3\mbox{ cm}, 98.3\mbox{ cm}, 107.3\mbox{ cm}$$

Solution

(i)

\begin{align}4\pi r^2=503\end{align}

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

\begin{align}r&=\sqrt{\frac{503}{4\pi}}\\&=6.33…\end{align}

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

\begin{align}h&=120-r\\&=120-2(6.33…)\\&\approx107.3\mbox{ cm}\end{align}

(ii)

\begin{align}V_A=\pi r^2 h_A\end{align}

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

\begin{align}h_A&=\frac{V_A}{\pi r^2}\\&=\frac{71.3\pi}{\pi(1^2)}\\&=71.3\mbox{ cm}\end{align}

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

\begin{align}d&=\frac{107.3-71.3}{4}=9\end{align}

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

\begin{align}71.3\mbox{ cm}&&80.3\mbox{ cm}&&89.3\mbox{ cm}&&98.3\mbox{ cm}&&107.3\mbox{ cm}\end{align}

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(c) There is a wall on each side of the section of railing, as shown in the diagram below which is
not to drawn to scale. The distance from wall to wall is $$1.5\mbox{ m}$$. The distance from the wall to
bar $$A$$ is $$20\mbox{ cm}$$ and similarly from the other wall to bar $$I$$ is $$20\mbox{ cm}$$.
The radius of each bar is $$1\mbox{ cm}$$. The gap between each bar is identical.
Find the size of this gap.

$$11.5\mbox{ cm}$$

Solution

\begin{align}\mbox{Sum of gaps}&=150-20-20-9(2\times1)\\&=92\mbox{ cm}\end{align}

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

\begin{align}\mbox{Gap}&=\frac{92}{8}\\&=11.5\mbox{ cm}\end{align}

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(d) The sphere on bar $$A$$ and the sphere on bar $$B$$ are to be joined by a straight rod as shown in the diagram below which is not to drawn to scale.
Find the length of the shortest rod that will join sphere $$A$$ to sphere $$B$$.
Give your answer in $$\mbox{cm}$$, correct to $$1$$ decimal place.

$$10.0\mbox{ cm}$$

Solution

\begin{align}V_B&=1.75V_A\\&=1.75(36\pi)\\&=63\pi\end{align}

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

\begin{align}\frac{4}{3}\pi r_B^3=63\pi\end{align}

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

\begin{align}r_B&=\sqrt[3]{\frac{3(63)}{4}}\\&=3.62…\mbox{ cm}\end{align}

The base $$b$$ of the triangle shown has a length of

\begin{align}b&=11.5+(2\times1)\\&=13.5\mbox{ cm}\end{align}

The height $$h$$ of the triangle shown is instead

\begin{align}h&=(9-r_A)+r_B\\&=(9-3)+3.62\\&=9.62\mbox{ cm}\end{align}

The hypotenuse $$H$$ of the triangle is therefore

\begin{align}H&=\sqrt{b^2+h^2}\\&=\sqrt{13.5^2+9.62^2}\\&16.576…\mbox{ cm}\end{align}

Therefore, the rod length $$L$$ is

\begin{align}L&=H-r_A-r_B\\&=16.576…-3-3.62\\&\approx10.0\mbox{ cm}\end{align}

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Question 8

Acme Confectionery makes cakes and chocolate bars.

(a)

(i) Acme Confectionery has launched a new bar called Chocolate Crunch. The weights of these new bars are normally distributed with a mean of $$4.64\mbox{ g}$$ and a standard deviation
of $$0.12\mbox{ g}$$. A sample of $$10$$ bars is selected at random and the mean weight of the sample is found.
Find the probability that the mean weight of the sample is between $$4.6\mbox{ g}$$ and $$4.7\mbox{ g}$$.

(ii) A company surveyed $$400$$ people, chosen from the population of people who had bought at least one Chocolate Crunch bar.
Of those surveyed, $$324$$ of them said they liked the new bar.
Create the $$95\%$$ confidence interval for the population proportion who liked the new bar.
Give your answer correct to $$2$$ decimal places.

(i) $$0.796$$

(ii) $$0.77\leq p\leq0.85$$

Solution

(i)

\begin{align}z_1&=\frac{x_1-\mu}{\sigma/\sqrt{n}}\\&=\frac{4.6-4.64}{0.12/\sqrt{10}}\\&=-1.05…\end{align}

and

\begin{align}z_2&=\frac{x_2-\mu}{\sigma/\sqrt{n}}\\&=\frac{4.7-4.64}{0.12/\sqrt{10}}\\&=1.58…\end{align}

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

\begin{align}p(-1.05<z<1.58)&=0.9429-(1-0.8531)\\&=0.796\end{align}

(ii)

\begin{align}1.96\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}&=1.96\sqrt{\frac{(0.81)(1-0.81)}{400}}\\&=0.0384…\end{align}

$\,$

Confidence Interval

\begin{align}\hat{p}-0.0384…\leq p\leq\hat{p}+0.0384&=0.81-0.0384…\leq p\leq0.81+0.0384\\&\approx0.77\leq p\leq0.85\end{align}

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(b)

(i) State whether each of the following statements are Always True, Sometimes True or Never True.
In the statements, $$n$$ is the size of the sample and $$\hat{p}$$ is the sample proportion.

1. When forming confidence intervals (for fixed $$n$$ and $$\hat{p}$$), an increased confidence level implies a wider interval.
2. As the value of $$\hat{p}$$ increases (for fixed $$n$$), the estimated standard error of the population proportion increases.
3. As the value of $$\hat{p}(1-\hat{p})$$ increases (for fixed $$n$$), the estimated standard error of the population proportion increases.
4. As $$n$$, the number of people sampled, increases (for fixed $$\hat{p}$$, the estimated standard error of the population proportion increases.

(ii) Using calculus or otherwise, find the maximum value of $$\hat{p}(1-\hat{p})$$.

(iii) Hence, find the largest possible value of the radius of the $$95\%$$ confidence interval for a population proportion, given a random sample of size $$800$$.

(i)

1. Always true.
2. Sometimes true.
3. Always true
4. Never true

(ii) $$\dfrac{1}{4}$$

(ii) $$0.0346…$$

Solution

(i)

1. Always true.
2. Sometimes true.
3. Always true
4. Never true

(ii)

\begin{align}A(\hat{p})&=\hat{p}(1-\hat{p})\\&=\hat{p}-\hat{p}^2\end{align}

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

\begin{align}\frac{dA}{d\hat{p}}&=1-2\hat{p}\end{align}

$\,$

Maximum

\begin{align}1-2\hat{p}=0\end{align}

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

\begin{align}\hat{p}=\frac{1}{2}\end{align}

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

\begin{align}A\left(\frac{1}{2}\right)&=\frac{1}{2}-\left(\frac{1}{2}\right)^2\\&=\frac{1}{4}\end{align}

(ii)

\begin{align}r&=1.96\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\\&=1.96\sqrt{\frac{\frac{1}{4}\left(1-\frac{1}{4}\right)}{800}}\\&=0.0346…\end{align}

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(c) Acme Confectionery has an employee pension plan. For an employee who qualifies for the full pension, Acme Confectionery will pay a sum of €$$20{,}000$$ on the day of retirement. It will then pay a sum on the same date each subsequent year for the next $$25$$ years. Each year the employee is paid a sum that is $$1\%$$ more than the amount paid in the previous year.
What sum of money must the company have set aside on the day of retirement in order to fund this pension? Assume an annual interest rate (AER) of $$2.4\%$$.

$$440{,}132.40\mbox{ euro}$$

Solution

\begin{align}\mbox{Amount}=20{,}000S_{26}\end{align}

\begin{align}a=1&&r=\frac{1.01}{1.024}\end{align}

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

\begin{align}\mbox{Amount}&=20{,}000S_{26}\\&=20{,}000\left[\frac{a\left(1-\left(\frac{1.01}{1.024}\right)^{26}\right)}{1-\frac{1.01}{1.024}}\right]\\&\approx440{,}132.40\mbox{ euro}\end{align}

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Question 9

In engineering, a crank-and-slider mechanism can be used to change circular motion into motion back and forth in a straight line.

In the diagrams below, the crank $$[OD]$$ rotates about the fixed point $$O$$. The point $$C$$ slides back and forth in a horizontal line. $$[CD]$$ is the rod that connects $$C$$ to the crank. The diagrams below show three of the possible positions for $$C$$ and $$D$$. $$|OD|=10\mbox{ cm}$$ and $$|DC|=30\mbox{ cm}$$.

(a) The diagram below shows a particular position of the mechanism with $$|\angle DCO|=15^{\circ}$$.
Find $$|\angle COD|$$, correct to the nearest degree.

$$51^{\circ}$$

Solution

\begin{align}\frac{\sin|\angle COD|}{30}=\frac{\sin15^{\circ}}{10}\end{align}

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

\begin{align}\sin|\angle COD|&=30\left(\frac{\sin15^{\circ}}{10}\right)\\&=0.77645…\end{align}

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

\begin{align}|\angle COD|&=\sin^{-1}(0.77645)\\&\approx51^{\circ}\end{align}

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(b) As $$D$$ moves in a circle around $$O$$, the angle $$\alpha$$ in the diagram below increases.
The distance $$|CX|$$ can be considered to be a function of $$\alpha$$ and written as ݂$$f(\alpha)$$.

(i) Write down the period and range of ݂$$f$$.

(ii) Complete the table below for ݂$$f(\alpha)$$.
Give your answers correct to $$2$$ decimal places where appropriate.
(Note: Diagram 1 at the start of this question represents $$\alpha=0^{\circ}$$).

$$\alpha$$ $$0^{\circ}$$ $$90^{\circ}$$ $$180^{\circ}$$ $$270^{\circ}$$ $$360^{\circ}$$

$$f(\alpha)$$ (cm)

$$30$$

(iii) Use your values from the table to draw a rough sketch of $$f$$ in the domain $$0^{\circ}\leq\alpha\leq360^{\circ}$$.

(iv) Referring to Diagrams 1, 2, and 3 near the start of this question, for which of the three positions of the mechanism will a $$1$$ degree change in $$\alpha$$ cause the greatest change in the position of $$C$$? Explain your answer.

(i) The period is $$2\pi$$ and the range is $$[10,30]$$.

(ii)

$$\alpha$$ $$0^{\circ}$$ $$90^{\circ}$$ $$180^{\circ}$$ $$270^{\circ}$$ $$360^{\circ}$$

$$f(\alpha)$$ (cm)

$$30$$

$$18.28$$

$$10$$

$$18.28$$

$$30$$

(iii)

(iv) Diagram $$2$$

Solution

(i) The period is $$2\pi$$ and the range is $$[10,30]$$.

(ii)

$$\alpha$$ $$0^{\circ}$$ $$90^{\circ}$$ $$180^{\circ}$$ $$270^{\circ}$$ $$360^{\circ}$$

$$f(\alpha)$$ (cm)

$$30$$

$$18.28$$

$$10$$

$$18.28$$

$$30$$

(iii)

(iv) Diagram $$2$$ as the graph is steeper at that time than at the times of the other two diagrams.

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(c) The diagram below shows another crank-and-slider mechanism with different dimensions.
In the diagram, $$\|AB|=36\mbox{ cm}$$, $$|AX|=31\mbox{ cm}$$, and $$|\angle BAO|=10^{\circ}$$.
(Note: $$|\angle OBA|\neq 90^{\circ}$$)
Find $$r$$, the length of the crank. Give your answer in $$\mbox{cm}$$, correct to the nearest $$\mbox{cm}$$.

$$7\mbox{ cm}$$

Solution

\begin{align}r^2&=36^2+(31+r)^2-2(36)(31+r)(\cos10^{\circ})\\&=1296+961+62r+r^2-2232\cos10^{\circ}+72r\cos10^{\circ}\end{align}

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

\begin{align}r&=\frac{2232\cos10^{\circ}-1296-961}{62-72\cos10^{\circ}}\\&\approx7\mbox{ cm}\end{align}

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