One-to-One Grinds
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2019 Ordinary Level - Paper One
Section A
Question 1
Eimear earns a gross wage of €\(40{,}000\) per annum with Company \(A\).
(a) Eimear pays income tax at a rate of \(20\%\) on income up to the standard rate cut-off point of €\(35{,}300\). She pays tax at a rate of \(40\%\) on the remainder.
She has annual tax credits of €\(1{,}650\).
Find how much income tax she pays per annum.
\(7{,}290\mbox{ euro}\)
\begin{align}35{,}300\times0.2+(40{,}000-35{,}300)\times0.4-1{,}650=7{,}290\mbox{ euro}\end{align}
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(b) Eimear pays her health insurance which costs her €\(1500\) net. Find her annual income after paying income tax and health insurance (i.e. her net annual income).
\(31{,}210\mbox{ euro}\)
\begin{align}40{,}000-7{,}290-1{,}500=31{,}210\mbox{ euro}\end{align}
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(c) Eimear is planning to change jobs. She is offered a job by Company \(B\) with a gross wage of €\(38{,}000\) and a bonus of €\(1500\) (tax free to Eimear) to be paid by the company, which she would use to pay her health insurance. Her tax rates and credits would remain the same.
Find by how much Eimear’s net annual income (after paying income tax and health insurance) will increase if she accepts the job with Company \(B\).
\(300\mbox{ euro}\)
\begin{align}[38{,}000-35{,}300\times0.2-(38{,}000-35{,}300)\times0.4+1{,}650]-31{,}210=300\mbox{ euro}\end{align}
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Question 2
(a) The complex number \(z_1=2+i\), where \(i^2=-1\), is shown on the Argand Diagram below.
(i) \(z_2=2z_1\).
Find the value of \(z_2\), and plot and label it on the Argand Diagram.
(ii) \(\bar{z_1}\) is the complex conjugate of \(z_1\).
Write down the value of \(\bar{z_1}\), and plot and label it on the Argand Diagram.
(iii) Investigate if \(|z_2|=|z_1+\bar{z_1}|\).
(i) \(z_2=4+2i\)
(ii) \(\bar{z_1}=2-i\)
(iii) They are not equal.
(i)
\begin{align}z_2&=2z_1\\&=2(2+i)\\&=4+2i\end{align}
(ii)
\begin{align}\bar{z_1}=2-i\end{align}
(iii)
\begin{align}|\bar{z_2}|&=\sqrt{4^2+2^2}\\&=\sqrt{20}\end{align}
and
\begin{align}|z_1+\bar{z_1}|&=|(2+i)+(2-i)|\\&=|4|\\&=4\end{align}
Therefore, they are not equal.
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(b) Show that \(z_1=2+i\) is a solution of the equation \(z^2-4z+4=0\).
The answer is already in the question!
\begin{align}(2+i)^2-4(2+i)+5&=4+4i-1-8-4i+5\\&=0\end{align}
as required.
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Question 3
The function \(f\) is defined as \(f(x)=-x^3+4x^2+x-2\), where \(x\in\mathbb{R}\).
(a)
(i) Complete the table below for the values of \(F\) in the domain \(-1\leq x \leq 4\) and hence draw the graph of the function \(f(x)\) in the domain \(-1\leq x\leq4\), \(x\in\mathbb{R}\).
| \(x\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|---|
|
\(f(x)\) |
|
|
|
|
\(10\) |
|
(ii) Use your graph to estimate the two roots of \(f(x)\) which are in the domain \(-1\leq x \leq 4\).
(i)
| \(x\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|---|
|
\(f(x)\) |
\(2\) |
\(-2\) |
\(2\) |
\(8\) |
\(10\) |
\(2\) |
(ii) \(-0.7\) and \(0.7\)
(i)
| \(x\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|---|
|
\(f(x)\) |
\(2\) |
\(-2\) |
\(2\) |
\(8\) |
\(10\) |
\(2\) |
(ii) \(-0.7\) and \(0.7\)
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(b) Find the value of \(x\) for which \(f^{\prime\prime}(x)=0\), where \(f^{\prime\prime}(x)\) is the second derivative of \(f(x)\).
\(\dfrac{4}{3}\)
\begin{align}f(x)=-x^3+4x^2+x-2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}f'(x)=-3x^2+8x+1-2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}f^{\prime\prime}(x)=-6x+8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}f^{\prime\prime}(x)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-6x+8=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x=\frac{4}{3}\end{align}
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Question 4
(a) Solve for \(x\):
\begin{align}\frac{3x+1}{5}+\frac{x-2}{2}=\frac{47}{10}\end{align}
\(x=5\)
\begin{align}\frac{2(3x+1)}{10}+\frac{5(x-2)}{10}=\frac{47}{10}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2(3x+1)+5(x-2)=47\end{align}
\begin{align}\downarrow\end{align}
\begin{align}6x+2+5x-10=47\end{align}
\begin{align}\downarrow\end{align}
\begin{align}11x=55\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x&=\frac{55}{11}\\&=5\end{align}
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(b) Solve the simultaneous equations:
\begin{align}x-5y&=-13\\x^2+y^2&=13\end{align}
\(x=-3\) and \(y=2\) or \(x=2\) and \(y=3\)
\begin{align}x-5y&=-13\\x^2+y^2&=13\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x=5y-13\end{align}
\begin{align}x^2+y^2=13\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(5y-13)^2+y^2=13\end{align}
\begin{align}\downarrow\end{align}
\begin{align}25y^2-130y+169+y^2=13\end{align}
\begin{align}\downarrow\end{align}
\begin{align}26y^2-130y+156=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y^2-5y+6=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(y-2)(y-3)=0\end{align}
\begin{align}\downarrow\end{align}
\(y=2\) or \(y=3\)
\begin{align}\downarrow\end{align}
\begin{align}x&=5(2)-13\\&=-3\end{align}
or
\begin{align}x&=5(3)-13\\&=2\end{align}
\[\,\]
Solutions
\(x=-3\) and \(y=2\)
or
\(x=2\) and \(y=3\)
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Question 5
Harry draws a scale diagram of the portion of his garden that is covered in lawn.
His diagram is shown below.
Each box on the grid is \(1\mbox{ cm}\times 0.5\mbox{ cm}\).
Each \(\mbox{cm}\) on Harry’s diagram represents \(3\mbox{ m}\).
In order to estimate the area of the lawn Harry divides the diagram into eight sections.
(a) Use the trapezoidal rule to estimate the area of the lawn using the scale: \(1\mbox{ cm}=3\mbox{ cm}\).
\(324\mbox{ m}^2\)
\begin{align}A&=\frac{3}{2}[0+0+2(15+18+15+12+15+18+15)]\\&=324\mbox{ m}^2\end{align}
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(b) Nuala can walk at a speed of \(1.6\) metres per second.
Write this speed in kilometres per hour.
\(5.76\mbox{ km/hr}\)
\begin{align}1.6\frac{\mbox{metres}}{\mbox{second}}&=1.6\frac{\frac{1}{1{,}000}\mbox{kilometres}}{\frac{1}{60\times60}\mbox{hour}}\\&=\frac{1.6(60\times60)}{1{,}000}\frac{\mbox{kilometres}}{\mbox{hour}}\\&=5.76\frac{\mbox{kilometres}}{\mbox{hour}}\end{align}
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Question 6
(a) Solve the following inequality for \(x\in\mathbb{R}\) and show your solution on the numberline below:
\begin{align}2(3-x)<8\end{align}
\(x>-1\)
\begin{align}2(3-x)<8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}6-2x<8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-2x<2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-x<1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x>-1\end{align}
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(b) Solve for \(x\):
\begin{align}2^{2x-1}=64\end{align}
\(x=\dfrac{7}{2}\)
\begin{align}2^{2x-1}=64\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2^{2x-1}=2^6\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2x-1=6\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2x=7\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x=\frac{7}{2}\end{align}
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Section B
Question 7
A camogie goalkeeper, on a level pitch, hitaball straight up into the air.
The path that the ball travelled can be modelled by the function:
\(f(t)=-4t^2+16t+1\), \(t\in\mathbb{R}\)
where \(t\) is the time, in seconds, from when the ball is hit and \(f(t)\) was the height of the ball, in metres, above the pitch. The ball landed on the ground without being hit again.
(a) At what height was the ball when it was hit by the goalkeeper?
\(1\mbox{ m}\)
\begin{align}f(0)&=-4(0^2)+16(0)+1\\&=1\mbox{ m}\end{align}
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(b)
(i) Complete the table below to show the height of the ball at various intervals during the
first \(4\) seconds of its flight.
| Time \(t\) | \(0\) | \(0.5\) | \(1\) | \(1.5\) | \(2\) | \(2.5\) | \(3\) | \(3.5\) | \(4\) |
|---|---|---|---|---|---|---|---|---|---|
|
Height (m) |
|
|
|
|
|
\(16\) |
|
|
|
(ii) On the grid below draw a graph to show the height of the ball while it was in the air.
(i)
| Time \(t\) | \(0\) | \(0.5\) | \(1\) | \(1.5\) | \(2\) | \(2.5\) | \(3\) | \(3.5\) | \(4\) |
|---|---|---|---|---|---|---|---|---|---|
|
Height (m) |
\(1\) |
\(8\) |
\(13\) |
\(16\) |
\(17\) |
\(16\) |
\(13\) |
\(8\) |
\(1\) |
(ii)
(i)
| Time \(t\) | \(0\) | \(0.5\) | \(1\) | \(1.5\) | \(2\) | \(2.5\) | \(3\) | \(3.5\) | \(4\) |
|---|---|---|---|---|---|---|---|---|---|
|
Height (m) |
\(1\) |
\(8\) |
\(13\) |
\(16\) |
\(17\) |
\(16\) |
\(13\) |
\(8\) |
\(1\) |
(ii)
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(c) Use your graph to estimate:
(Show your work on the graph above)
(i) the length of time the ball was in the air from the time it was hit until it landed on the ground
(ii) the length of time the ball was \(10\mbox{ m}\), or more, above the ground.
(i) \(4.1\mbox{ s}\)
(ii) \(2.8\mbox{ s}\)
(i) \(4.1\mbox{ s}\)
(ii)
\begin{align}3.4-0.6=2.8\mbox{ s}\end{align}
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(d)
(i) Find \(f'(t)\), the derivative of \(f(t)=-4t^2+16t+1\).
(ii) Use your answer from part (d)(i) to find the speed of the ball when it had been in the air for \(4\) seconds. Give your answer in metres per second.
(iii) Use your answer from part (d)(i) to find the value of \(t\) for which the ball was descending and travelling at a speed of \(8\) metres per second.
(i) \(f'(t)=-8t+16\)
(ii) \(16\mbox{ m/s}\)
(iii) \(3\mbox{ s}\)
(i)
\begin{align}f'(t)&=-8t+16\end{align}
(ii)
\begin{align}f'(4)&=-8(4)+16\\&=-16\mbox{ m/s}\end{align}
Therefore, the speed of the ball is \(16\mbox{ m/s}\).
(iii)
\begin{align}f'(t)=-8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-8t+16=-8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t&=\frac{-8-16}{-8}\\&=3\mbox{ s}\end{align}
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Question 8
(a) The power (\(P\)) of an engine is measured in horsepower using the formula:
\begin{align}P=\frac{R\times T}{5252}\end{align}
where \(R\) is the engine speed measured in revolutions per minute (\(\mbox{RPM}\)) and \(T\) is the torque measured in appropriate units.
(i) Find the power of an engine that generates \(480\) units of torque at \(2500\mbox{ RPM}\).
Give your answer correct to the nearest whole number.
(ii) Rearrange the formula to write \(R\) in terms of \(P\) and \(T\).
(i) \(228\mbox{ horsepower}\)
(ii) \(R=\dfrac{5{,}252P}{T}\)
(i)
\begin{align}P&=\frac{2,{}500\times480}{5{,}252}\\&\approx228\mbox{ horsepower}\end{align}
(ii)
\begin{align}R=\frac{5{,}252P}{T}\end{align}
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(b) A company was set up in January \(2016\) to repair engines. In the first month of its existence the company made a loss of €\(4000\). This loss reduced by €\(250\) a month for each month that the company traded.
(i) Complete the table below to show the company’s loss/profit for each of the first six months of trading.
| Month | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) |
|---|---|---|---|---|---|---|
|
\(\mathbf{h(t)}\) |
\(-4000\) |
\(-3750\) |
|
\(-3250\) |
|
|
(ii) Show that the profit the company makes in month \(n\) is given by the formula \(T_n=250n-4250\).
(iii) What profit does the company make in January 2018 (i.e. month 25)?
(iv) Find the month in which the company breaks even (i.e. €\(0\) profit).
(v) Find \(S_n\), the general term for the total profit of the company after \(n\) months.
(vi) Hence, or otherwise, find the total profit of the company at the end of January \(2019\) (i.e. month \(37\)).
(i)
| Month | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) |
|---|---|---|---|---|---|---|
|
\(\mathbf{h(t)}\) |
\(-4000\) |
\(-3750\) |
\(-3500\) |
\(-3250\) |
\(-3000\) |
\(-2750\) |
(ii) The answer is already in the question!
(iii) \(2{,}000\mbox{ euro}\)
(iv) The \(17\)th month
(v) \(S_n=-4125n+125n^2\)
(vi) \(18{,}500\mbox{ euro}\)
(i)
| Month | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) |
|---|---|---|---|---|---|---|
|
\(\mathbf{h(t)}\) |
\(-4000\) |
\(-3750\) |
\(-3500\) |
\(-3250\) |
\(-3000\) |
\(-2750\) |
(ii)
\begin{align}T_n&=a+(n-1)d\\&=-4000+(n-1)(250)\\&=-4000+250n-250\\&=250n-4250\end{align}
as required.
(iii)
\begin{align}T_{25}&=250(25)-4250\\&=2{,}000\mbox{ euro}\end{align}
(iv)
\begin{align}T_n=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}250n-4250=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}n&=\frac{4250}{250}\\&=17\end{align}
(v)
\begin{align}S_n&=\frac{n}{2}[2a+(n-1)d]\\&=\frac{n}{2}[2(-4000)+(n-1)(250)]\\&=\frac{n}{2}(-8000+250n-250)\\&=-4125n+125n^2\end{align}
(vi)
\begin{align}S_{37}&=-4125(37)+125(37^2)\\&=18{,}500\mbox{ euro}\end{align}
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Question 9
Avril has a website. For a certain period of time, the total number of registered users of the website, \(U(m)\), can be estimated by using the formula:
\begin{align}U(m)=3000(1.8)^m\end{align}
where \(m\) is the number of months from when the website was launched.
(a) Use the formula to estimate the number of registered users\(8\) months after the website was launched.
\(330{,}599\)
\begin{align}U(8)&=3000(1.8)^8\\&\approx330{,}599\end{align}
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(b) There are \(31{,}493\) registered users at the end of a particular month.
Estimate how many users there will be one month later.
\(56{,}687\)
\begin{align}31{,}493\times1.8\approx56{,}687\end{align}
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(c) Users can register free of charge on the website.
Through advertising, Avril earns €\(0.0012\) each month, for every user who is registered.
(i) How much would she earn in a month when there were \(600{,}000\) registered users?
(ii) Avril earns €\(1285.37\) for a particular month.
How many registered users did the website have in that month.
(iii) A web design company charges Avril €\(80\) per month to maintain and host the site.
How much will Avril make or lose in month \(4\) and in month \(12\) from the website.
(i) \(720\mbox{ euro}\)
(ii) \(1{,}071{,}142\)
(iii) Month 4: Lose \(42.21\mbox{ euro}\). Month 12: Make \(4{,}084.59\mbox{ euro}\).
(i)
\begin{align}600{,}000\times0.0012=720\mbox{ euro}\end{align}
(ii)
\begin{align}\frac{1285.37}{0.0012}\approx1{,}071{,}142\end{align}
(iii)
Month 4
\begin{align}3000(1.8)^4\times0.0012-80=-42.21\mbox{ euro}\end{align}
i.e. she will lose \(42.21\mbox{ euro}\) in month \(4\).
\[\,\]
Month 12
\begin{align}3000(1.8)^{12}\times0.0012-80=4{,}084.59\mbox{ euro}\end{align}
i.e. she will make \(4{,}084.59\mbox{ euro}\) in month \(12\).
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(d) Another web design company in the UK offers to host and maintain her website for £\(55\) sterling per month. Avril uses the exchange rate on a particular day to find how much this would cost in euro. She finds that it would cost €\(62.70\).
Find the exchange rate for \(1\) euro on that day. Give your answer correct to 4 decimal places.
\(0.8772\mbox{ sterling}\)
\begin{align}\frac{55}{62.70}\approx0.8772\mbox{ sterling}\end{align}
