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Functions
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(BLUE = Paper 1, RED = Paper 2)
2022 Paper 1 Question 5(c)
(c) The diagram below shows the graphs of the functions \(k(x)\) and \(m(x)\), for \(0\leq x\leq 5\), \(x\in\mathbb{R}\).
Use the graphs to estimate each of the following, for \(0\leq x\leq 5\):
(i) the two values of \(x\) for which \(m(x)=0\)
(ii) the range of values of \(x\) for which \(k(x)\) is less than \(m(x)\).
(i) \(x=1\) and \(x=4.5\)
(ii) \(2<x<3.5\)
(i) \(x=1\) and \(x=4.5\)
(ii) \(2<x<3.5\)
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2022 Paper 1 Question 7(a) - (d)
Joseph is doing a training session. During the session, his heart-rate, \(h(x)\), is measured in beats
per minute (BPM). For part of the session, \(h(x)\) can be modelled using the following function:
\begin{align}h(x)=-0.38x^3+2.6x^2-0.13x+158\end{align}
where \(x\) is the time, in minutes, from the start of the session, and \(0\leq x \leq 6\), \(x\in\mathbb{R}\).
(a)
(i) Complete the table below to show the values of \(h(x)\) for the given values of \(x\).
Give each value of \(h(x)\) correct to the nearest whole number.
| Time (minutes) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
|
Heart-rate (BPM) |
|
\(160\) |
|
\(171\) |
|
|
\(169\) |
(ii) Draw the graph of \(y=h(x)\) on the axes below, for \(0\leq x\leq 6\), \(x\in\mathbb{R}\).
Note that the point A \((6,169)\) is on the graph.
(i)
| Time (minutes) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
|
Heart-rate (BPM) |
\(158\) |
\(160\) |
\(165\) |
\(171\) |
\(175\) |
\(175\) |
\(169\) |
(ii)
(i)
| Time (minutes) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
|
Heart-rate (BPM) |
\(158\) |
\(160\) |
\(165\) |
\(171\) |
\(175\) |
\(175\) |
\(169\) |
(ii)
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(b) Explain what the co-ordinates of the point A \((6,169)\) represent, in the context of Joseph’s heart-rate.
\(6\) minutes from the start of the session, Joseph’s heart-rate is \(169\) BPM.
\(6\) minutes from the start of the session, Joseph’s heart-rate is \(169\) BPM.
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(c) Using the same axes and scales, continue your graph on the previous page to show the following information. From the point represented by A \((6,169)\), Joseph’s heart-rate:
- stays at the same level for the next \(2\) minutes, and then
- decreases at a steady rate of \(10\) BPM per minute for \(2\) minutes.
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(d) During his training session, the number of calories per minute that Joseph is burning after \(x\) minutes can be modelled by \(c(x)\) as follows, where \(h(x)\) is Joseph’s heart-rate at that time:
\begin{align}c(x)=0.1h(x)-7\end{align}
Using the information in the table or graph, work out \(c(6)\), the number of calories per minute that Joseph is burning \(6\) minutes after the start of the session.
\(9.9\mbox{ calories per minute}\)
\begin{align}c(6)&=0.1h(6)-7\\&=0.1(169)-7\\&=9.9\mbox{ calories per minute}\end{align}
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2022 Paper 1 Question 9(a) - (c)
Brian buys a new car.
The graph below represents a model that can be used to predict the value of this car, \(V\), for the next
number of years. This model assumes that the value of the car reduces (depreciates) by a fixed percentage each year.
(a)
(i) Use the graph to write down \(V(0)\), the initial value of Brian’s car, and \(V(1)\), the value of Brian’s car after \(1\) year.
(ii) Show that the value of the car will reduce by \(20\%\) in its first year, according to this model.
(i) \(V(0)=30{,}000\mbox{ euro}\) and \(V(1)=24{,}000\mbox{ euro}\)
(ii) The answer is already in the question!
(i)
\begin{align}V(0)=30{,}000\mbox{ euro} && V(1)=24{,}000\mbox{ euro}\end{align}
(ii)
\begin{align}\frac{30{,}000-24{,}000}{30{,}000}\times 100=20\%\end{align}
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(b)
(i) Based on this model, write a formula for \(V(t)\), the value of Brian’s car after \(t\) years, in terms of the age of the car (\(t\)).
Use the fact that the value decreases by \(20\%\) each year.
(ii) Hence, or otherwise, work out the value of Brian’s car after \(4\) years, according to this model. Show your working out.
(i) \(V(t)=30{,}000(0.8)^t\)
(ii) \(12{,}288\mbox{ euro}\)
(i)
\begin{align}V(t)=30{,}000(0.8)^t\end{align}
(ii)
\begin{align}V(4)&=30{,}000(0.8)^4\\&=12{,}288\mbox{ euro}\end{align}
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(c) The graph from part (a) is shown again below.
A different (linear) model assumes that the value of the car reduces (depreciates) by a fixed amount each year. The value of the car will also reduce by \(20\%\) in its first year, according to this model.
(i) Draw a line on the diagram above, passing through the first two points on the graph
with whole-number values of \(t\) (\(t=0\) and \(t=1\)). Continue your line until it reaches the horizontal axis.
(ii) Hence, or otherwise, estimate \(T\), the age of Brian’s car when its value would be €\(0\), according to this new model.
(i)
(ii) \(T=5\mbox{ years}\)
(i)
(ii) \(T=5\mbox{ years}\)
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2022 Paper 1 Question 10(a)-(b)
Keith plays hurling.
(a) During a match, Keith hits the ball with his hurl.
The height of the ball could be modelled by the following quadratic function:
\begin{align}h=-2t^2+5t+1.2\end{align}
where \(h\) is the height of the ball, in metres, \(t\) seconds after being hit, and \(t\in\mathbb{R}\).
(i) How high, in metres, was the ball when it was hit (when \(t=0\))?
(ii) The ball was caught after \(2.4\) seconds.
How high, in metres, was the ball when it was caught?
(iii) When the ball passed over the halfway line, it was at a height of \(3.2\) metres and its height was decreasing.
How many seconds after it was hit did the ball pass over the halfway line?
Remember that \(h=-2t^2+5t+1.2\).
(iv) Find \(\dfrac{dh}{dt}\) and hence find how long it took the ball to reach its greatest height.
Give your answer in seconds.
(i) \(1.2\mbox{ m}\)
(ii) \(1.68\mbox{ m}\)
(iii) \(2\mbox{ s}\)
(iv) \(\dfrac{5}{4}\mbox{ s}\)
(i)
\begin{align}h(0)&=-2(0^2)+5(0)+1.2\\&=1.2\mbox{ m}\end{align}
(ii)
\begin{align}h(2.4)&=-2(2.4^2)+5(2.4)+1.2\\&=1.68\mbox{ m}\end{align}
(iii)
\begin{align}h(t)&=3.2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-2t^2+5t+1.2=3.2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2t^2-5t+2=0\end{align}
\begin{align}\downarrow\end{align}
\(t=\dfrac{1}{2}\) or \(t=2\)
As the height was decreasing, it must be the second time, i.e. \(2\) seconds.
(iv)
\begin{align}\frac{dh}{dt}=-4t+5\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-4t+5=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t=\frac{5}{4}\mbox{ s}\end{align}
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(b) Later in the game, Keith hit the ball again. This time, the height of the ball \(t\) seconds after it was hit could be modelled by a different quadratic function, \(y=k(t)\), where \(k\) is in metres.
This time, the ball was \(1\) metre high when Keith hit it.
Its greatest height was \(5\) metres, which it reached after \(2\) seconds.
It hit the ground without being caught.
Using the information above, write down the co-ordinates of three points that must be on the graph of \(y=k(t)\), and draw the graph of \(y=k(t)\) on the axes below, from when the ball is hit until it hits the ground.
\begin{align}(0,1)&&(2,5)&&(4,1)\end{align}
\begin{align}(0,1)&&(2,5)&&(4,1)\end{align}
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2021 Paper 1 Question 5(a) & (b)
The function, \(f\) is defined as \(f(x)=3x^2-6x+7\), where \(x\in\mathbb{R}\).
(a) Find \(f(0.67)\), correct to \(2\) decimal places.
\(4.33\)
\begin{align}f(0.67)&=3(0.67)^2-6(0.67)+7\\&\approx4.33\end{align}
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(b) Find the value of \(x\) when \(f(x)=4\).
\(x=1\)
\begin{align}f(x)=4\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3x^2-6x+7=4\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3x^2-6x+3=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x^2-2x+1=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x-1)^2=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x=1\end{align}
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2021 Paper 1 Question 8(d) - (f)
A square sheet of cardboard, of side \(10\) units, is used to make an open box.
Squares of side \(x\) units, where \(x\in\mathbb{R}\), are removed from each corner of the cardboard and it is then folded along the dotted lines, as shown in the diagram below, in order to create the box.
It can be shown that the volume of the box can be written as
\begin{align}V(x)=4x^3-40x^2+100x\end{align}
(d) Complete the table below to show the values of \(V(x)=4x^3-40x^2+100x\), where \(x\in\mathbb{R}\),
for the given values of \(x\) in the domain \(0\leq c\leq5\).
| \(x\mbox{ m}\) | \(0\) | \(0.5\) | \(1\) | \(1.5\) | \(2\) | \(2.5\) | \(3\) | \(3.5\) | \(4\) | \(4.5\) | \(5\) |
|---|---|---|---|---|---|---|---|---|---|---|---|
|
\(V(x)\mbox{ m}^3\) |
|
\(40.5\) |
|
|
|
|
|
|
|
\(4.5\) |
|
| \(x\mbox{ m}\) | \(0\) | \(0.5\) | \(1\) | \(1.5\) | \(2\) | \(2.5\) | \(3\) | \(3.5\) | \(4\) | \(4.5\) | \(5\) |
|---|---|---|---|---|---|---|---|---|---|---|---|
|
\(V(x)\mbox{ m}^3\) |
\(0\) |
\(40.5\) |
\(64\) |
\(73.5\) |
\(72\) |
\(62.5\) |
\(48\) |
\(31.5\) |
\(16\) |
\(4.5\) |
\(0\) |
| \(x\mbox{ m}\) | \(0\) | \(0.5\) | \(1\) | \(1.5\) | \(2\) | \(2.5\) | \(3\) | \(3.5\) | \(4\) | \(4.5\) | \(5\) |
|---|---|---|---|---|---|---|---|---|---|---|---|
|
\(V(x)\mbox{ m}^3\) |
\(0\) |
\(40.5\) |
\(64\) |
\(73.5\) |
\(72\) |
\(62.5\) |
\(48\) |
\(31.5\) |
\(16\) |
\(4.5\) |
\(0\) |
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(e) Draw the graph of the function \(V(x)\) on the grid below.
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(f) Use your graph to estimate each of the following values.
In each case show your work on the graph above.
(i) The maximum volume of the box.
(ii) The values of \(x\) which will create a box which has a volume of \(30\) units cubed.
(iii) The volume of the box when \(x\) is \(2.8\) units.
(i) \(74\)
(ii) \(x=0.35\) and \(x=3.55\)
(iii) \(54\)
(i) \(74\)
(ii) \(x=0.35\) and \(x=3.55\)
(iii) \(54\)
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2020 Paper 1 Question 5(b)
(b)
(i) Complete the table below to show the value of the function \(f(x)=-x^2-x+6\) for each of the given values of \(x\).
| \(x\) | \(-4\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) |
|---|---|---|---|---|---|---|---|
|
\(f(x)\) |
|
|
\(4\) |
|
|
|
|
(ii) Hence draw the graph of \(f\) in the domain \(-4\leq x\leq2\), where \(x\in\mathbb{R}\), on the diagram below.
(iii) On the same diagram as the function \(f\), draw the graph of the function \(g(x)=f(x)-2\), in the domain \(-2\leq x\leq 4\), where \(x\in\mathbb{R}\).
Label the graphs of \(f(x)\) and \(g(x)\) clearly.
(i)
| \(x\) | \(-4\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) |
|---|---|---|---|---|---|---|---|
|
\(f(x)\) |
\(-6\) |
\(0\) |
\(4\) |
\(6\) |
\(6\) |
\(4\) |
\(0\) |
(ii)
(iii)
(i)
| \(x\) | \(-4\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) |
|---|---|---|---|---|---|---|---|
|
\(f(x)\) |
\(-6\) |
\(0\) |
\(4\) |
\(6\) |
\(6\) |
\(4\) |
\(0\) |
(ii)
(iii)
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2020 Paper 1 Question 8(a) & (b)
A swimmer is on a starting block at the beginning of a race. When she dives off the block until she resurfaces, the level of the swimmer relative to the level of the water is given by the function:
\begin{align}h(x)=\frac{1}{60}x^2-\frac{1}{4}x+\frac{3}{5}\end{align}
In the function, \(x\) is the horizontal distance in metres of the swimmer from the block, \(0\leq x\leq 12\), where \(x\in\mathbb{R}\) and \(h(x)\) is measured in metres.
(a) Find the height of the block above the water.
\(\dfrac{3}{5}\mbox{ m}\)
\begin{align}h(0)&=\frac{1}{60}(0^2)-\frac{1}{4}(0)+\frac{3}{5}\\&=\frac{3}{5}\mbox{ m}\end{align}
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(b)
(i) Show that the swimmer is on the surface of the water (i.e. \(h(x)=0\)) when she is \(12\) metres from the starting block.
(ii) Find the horizontal distance, in metres, from the starting block to the point where the swimmer enters the water.
(i) The answer is already in the question!
(ii) \(x=3\mbox{ m}\)
(i)
\begin{align}h(12)&=\frac{1}{60}(12^2)-\frac{1}{4}(12)+\frac{3}{5}\\&=\frac{3}{5}\mbox{ m}\\&=0\end{align}
as required.
(ii)
\begin{align}h(x)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{1}{60}x^2-\frac{1}{4}x+\frac{3}{5}=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x^2-15x+36=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x-3)(x-12)=0\end{align}
\begin{align}\downarrow\end{align}
\(x=3\) or \(x=12\)
\begin{align}\downarrow\end{align}
\begin{align}x=3\mbox{ m}\end{align}
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2019 Paper 1 Question 3(a)
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 \(-1\leq x\leq4\), \(x\in\mathbb{R}\).
| \(x\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|---|
|
\(f(x)\) |
|
|
|
|
\(10\) |
|