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Functions
NOTE: Clicking an entry in the right column will take you directly to that question!
(BLUE = Paper 1, RED = Paper 2)
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Linear Functions |
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Quadratic Functions |
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Cubic Functions |
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Trigonometric Functions |
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Logarithmic Functions |
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Exponential Functions |
2022 Paper 1 Question 7(a), (f) & (g)
Hannah is doing a training session. During this session, her heart-rate, \(h(x)\), is measured in beats per minute (BPM), where \(x\) is the time in minutes from the start of the session, \(x\in\mathbb{R}\).
For the first 8 minutes of the session, Hannah does a number of exercises.
As she does these exercises, her heart-rate changes. In this time, \(h(x)\) is given by:
\begin{align}h(x)=2x^3-28.5x^2+105x+70\end{align}
(a) Work out Hannah’s heart-rate \(4\) minutes after the start of the session.
\(162\mbox{ BPM}\)
\begin{align}h(4)&=2(4^3)-28.5(4^2)+105(4)+70\\&=162\mbox{ BPM}\end{align}
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Bruno, Karen, and Martha start a training session at the same time as Hannah.
All of their heart-rates are measured in BPM.
(f)
(i) For the first \(8\) minutes of the session, Bruno’s heart-rate, \(b(x)\), is always \(15\) BPM more
than Hannah’s heart-rate.
Use this information to write \(b'(x)\) in terms of \(\mathbf{h'(x)}\), where \(0\leq x\leq 8\), \(x\in\mathbb{R}\).
(ii) For the first \(8\) minutes of the session, Karen’s heart-rate, \(k(x)\), is always \(10\%\) less than Hannah’s heart-rate.
Use this information to write \(k'(x)\) in terms of \(\mathbf{h'(x)}\), where \(0\leq x\leq 8\), \(x\in\mathbb{R}\).
(i) \(b'(x)=h'(x)\)
(ii) \(k'(x)=0.9h'(x)\)
(i)
\begin{align}b(x)=h(x)+15\end{align}
\begin{align}\downarrow\end{align}
\begin{align}b'(x)=h'(x)\end{align}
(ii)
\begin{align}k(x)=0.9h(x)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}k'(x)=0.9h'(x)\end{align}
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(g) Martha does each exercise for a longer time than Hannah.
For \(0\leq x \leq 10\), Martha’s heart-rate, \(m(x)\), is:
\begin{align}m(x)=h(0.8x)\end{align}
Use \(h(x)=2x^3-28.5x^2+105x+70\) to write \(m(x)\) in the form:
\begin{align}m(x)=ax^3+bx^2+cx+d\end{align}
where \(a,b,c,d\in\mathbb{R}\), for \(0\leq x\leq 10\).
\(m(x)=1.024x^3-18.24x^2+84x+70\)
\begin{align}m(x)&=h(0.8x)\\&=2(0.8x)^3-28.5(0.8x^2)+(105(0.8x)+70\\&=1.024x^3-18.24x^2+84x+70\end{align}
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2022 Paper 1 Question 8(a)-(d)
A Ferris wheel has a diameter of \(120\mbox{ m}\).
When it is turning, it completes exactly \(10\) full rotations in one hour.
The diagram above shows the Ferris wheel before it starts to turn.
At this stage, the point \(A\) is the lowest point on the circumference of the wheel, and it is at a
height of \(12\mbox{ m}\) above ground level.
The height, \(h\), of the point \(A\) after the wheel has been turning for \(t\) minutes is given by:
\begin{align}h(t)=72-60\cos\left(\frac{\pi}{3}t\right)\end{align}
where \(h\) is in metres, \(t\in\mathbb{R}\), and \(\dfrac{\pi}{3}\) is in radians.
(a) Complete the table below. The value of \(h(1)\) is given.
| \(t\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) | \(8\) |
|---|---|---|---|---|---|---|---|---|---|
|
\(h(t)\) |
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\(42\) |
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| \(t\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) | \(8\) |
|---|---|---|---|---|---|---|---|---|---|
|
\(h(t)\) |
\(12\) |
\(42\) |
\(102\) |
\(132\) |
\(102\) |
\(42\) |
\(12\) |
\(42\) |
\(102\) |
| \(t\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) | \(8\) |
|---|---|---|---|---|---|---|---|---|---|
|
\(h(t)\) |
\(12\) |
\(42\) |
\(102\) |
\(132\) |
\(102\) |
\(42\) |
\(12\) |
\(42\) |
\(102\) |
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(b) Draw the graph of \(y=h(t)\) for \(0\leq t\leq 8\), \(t\in\mathbb{R}\).
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(c) Find the period and range of \(h(t)\).
Period: \(6\) minutes
Range: \([12,132]\).
According to the graph, the period is \(6\) minutes and the range is \([12,132]\).
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(d) During a \(50\)-minute period, what is the greatest number of minutes for which the point \(A\) could be higher than \(42\mbox{ m}\)?
\(34\) minutes
For each period, the point \(A\) spends at most \(4\) minutes at a height greater than \(42\mbox{ m}\).
\(50\) minutes is equivalent to \(8\) periods and \(2\) minutes.
Therefore, in \(8\) periods, the point \(A\) spends at most \(8\times4+2=34\) minutes at a height greater than \(42\mbox{ m}\).
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2022 Paper 1 Question 9(a)-(b)
Alex gets injections of a medicinal drug. Each injection has \(15\mbox{ mg}\) of the drug.
Each day, the amount of the drug left in Alex’s body from an injection decreases by \(40\%\).
So, the amount of the drug (in mg) left in Alex’s body \(t\) days after a single injection is given by:
\begin{align}15(0.6)^t\end{align}
where \(t\in\mathbb{R}\).
(a) Find the amount of the drug left in Alex’s body \(2.5\) days after a single \(15\mbox{ mg}\) injection.
Give your answer in \(\mbox{mg}\), correct to \(2\) decimal places.
\(4.18\mbox{ mg}\)
\begin{align}15(0.6^{2.5})&=4.182…\\&\approx 4.18\mbox{ mg}\end{align}
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(b) How long after a single \(15\mbox{ mg}\) injection will there be exactly \(1\mbox{ mg}\) of the drug left in Alex’s body? Give your answer in days, correct to \(1\) decimal place.
\(5.3\mbox{ days}\)
\begin{align}15(0.6^t)=1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}0.6^t=\frac{1}{15}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t&=\log_{0.6}\frac{1}{15}\\&\approx 5.3\mbox{ days}\end{align}
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2022 Paper 1 Question 10(a) & (b)
A student is asked to memorise a long list of digits, and then write down the list some time later.
The proportion, \(P\), of the digits recalled correctly after \(t\) hours can be modelled by the function:
\begin{align}0.82-0.12\ln(t+1)\end{align}
for \(0\leq t\leq 12\), \(t\in\mathbb{R}\).
(a) Find the proportion of the digits recalled correctly after \(3\) hours, according to this model.
Give your answer correct to \(2\) decimal places.
\(0.65\)
\begin{align}P(3)&=0.82-0.12\ln(3+1)\\&\approx0.65\end{align}
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(b) After how many hours would exactly \(55\%\) of the digits be recalled correctly, according to this model? Give your answer correct to \(2\) decimal places.
\(8.49\mbox{ hours}\)
\begin{align}P(t)=0.55\end{align}
\begin{align}\downarrow\end{align}
\begin{align}0.82-0.12\ln(t+1)=0.55\end{align}
\begin{align}\downarrow\end{align}
\begin{align}0.12\ln(t+1)=0.27\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\ln(t+1)&=\frac{0.27}{0.12}\\&=2.25\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t+1=e^{2.25}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t&=e^{2.25}-1\\&\approx8.49\mbox{ hours}\end{align}
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2021 Paper 1 Question 8(a)
(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.
| \(x\) | \(0\) | \(10\) | \(20\) | \(30\) | \(40\) | \(50\) | \(60\) | \(70\) | \(75\) |
|---|---|---|---|---|---|---|---|---|---|
|
\(h(x)\) |
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\(30\) |
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\(21\) |
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\(5\) |
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\(21.875\) |
(i) The answer is already in the question!
(ii)
| \(x\) | \(0\) | \(10\) | \(20\) | \(30\) | \(40\) | \(50\) | \(60\) | \(70\) | \(80\) |
|---|---|---|---|---|---|---|---|---|---|
|
\(h(x)\) |
\(5\) |
\(30\) |
\(37\) |
\(32\) |
\(21\) |
\(10\) |
\(5\) |
\(12\) |
\(21.875\) |
(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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2021 Paper 1 Question 9(a)-(b)
(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.
Give your answer correct to the nearest whole number.
(i) The answer is already in the question!
(ii) The temperature of the coffee after an infinite amount of time has passed.
(iii) \(42^{\circ}\mbox{ C}\)
(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.
\(2\mbox{ min }21\mbox{ sec}\)
\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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2021 Paper 1 Question 10(b)
(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) \(k=6.25\)
(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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2020 Paper 1 Question 3(a)
(a) \(f(x)=6x-5\) and \(g(x)=\dfrac{x+5}{6}\). Investigate if \(f(g(x))=g(f(x))\).
\(f(g(x))=g(f(x))\)
\begin{align}f(g(x))&=6\left(\frac{x+5}{6}\right)-5\\&=x+5-5\\&=x\end{align}
and
\begin{align}g(f(x))&=\frac{(6x-5)+5}{6}\\&=\frac{6x}{6}\\&=x\end{align}
Therefore, \(f(g(x))=g(f(x))\).
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2020 Paper 1 Question 9(a)
The number of bacteria in the early stages of a growing colony of bacteria can be approximated
using the function:
\begin{align}N(t)=450e^{0.065t}\end{align}
where \(t\) is the time, measured in hours, since the colony started to grow, and \(N(t)\) is the number of bacteria in the colony at time \(t\).
(a)
(i) Find the number of bacteria in the colony after \(4.5\) hours.
Give your answer correct to the nearest whole number.
(ii) Find the time, in hours, that it takes the colony to grow to \(790\) bacteria.
Give your answer correct to \(1\) decimal place.
(i) \(603\)
(ii) \(8.7\mbox{ h}\)
(i)
\begin{align}N(4.5)&=450e^{0.065(4.5)}\\&\approx603\end{align}
(ii)
\begin{align}450e^{0.065t}=790\end{align}
\begin{align}\downarrow\end{align}
\begin{align}e^{0.065t}=\frac{790}{450}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}0.065t=\ln\left(\frac{790}{450}\right)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t&=\frac{1}{0.065}\ln\left(\frac{790}{450}\right)\\&=8.7\mbox{ h}\end{align}
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2019 Paper 1 Question 2(a)
The graph of the function \(f(x)=3^x\), where \(x\in\mathbb{R}\), cuts the \(y\)‐axis at \((0,1)\) as shown in the
diagram below.
(a)
(i) Draw the graph of the function \(g(x)=4x+1\) on the diagram.
(ii) Use substitution to verify that \(f(x)<g(x)\), for \(x=1.9\).
(i)
(ii) The answer is already in the question!
(i)
(ii)
\begin{align}f(1.9)&=3^{1.9}\\&=8.06…\end{align}
and
\begin{align}g(1.9)&=4(1.9)+1\\&=8.6\end{align}
\begin{align}\downarrow\end{align}
\begin{align}f(1.9)<g(1.9)\end{align}
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2019 Paper 1 Question 8(a) & (b)
The weekly revenue produced by a company manufacturing air conditioning units is seasonal.
The revenue (in euro) can be approximated by the function:
\(r(t)=22{,}500\cos\left(\frac{\pi}{26}t\right)+37{,}500\), \(t\geq0\)
where \(t\) is the number of weeks measured from the beginning of July and \(\left(\dfrac{\pi}{26}t\right)\) is in radians.
(a) Find the approximate revenue produced \(20\) weeks after the beginning of July.
Give your answer correct to the nearest euro.
\(20{,}659\mbox{ euro}\)
\begin{align}r(20)&=22{,}500\cos\left(\frac{\pi}{26}(20)\right)+37{,}500\\&\approx20{,}659\mbox{ euro}\end{align}
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(b) Find the two values of the time \(t\), within the first \(52\) weeks, when the revenue is approximately €\(26{,}250\).
\(t=\dfrac{52}{3}\) weeks and \(t=\dfrac{104}{3}\) weeks
\begin{align}22{,}500\cos\left(\frac{\pi}{26}t\right)+37{,}500=26{,}250\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\cos\left(\frac{\pi}{26}t\right)=-\frac{1}{2}\end{align}
\begin{align}\downarrow\end{align}
\(\dfrac{\pi}{26}t=\dfrac{2\pi}{3}\) and \(\dfrac{\pi}{26}t=\dfrac{4\pi}{3}\)
\begin{align}\downarrow\end{align}
\(t=\dfrac{52}{3}\) weeks and \(t=\dfrac{104}{3}\) weeks
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(c) Find \(r'(t)\), the derivative of \(r(t)=22{,}500\cos\left(\frac{\pi}{26}t\right)+37{,}500\).
\(r'(t)=-\dfrac{11{,}250\pi}{13}\sin\left(\dfrac{\pi}{26}t\right)\)
\begin{align}r'(t)&=\left(\frac{\pi}{26}\right)(22{,}500)\left[-\sin\left(\frac{\pi}{26}t\right)\right]\\&=-\frac{11{,}250\pi}{13}\sin\left(\frac{\pi}{26}t\right)\end{align}
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(d) Use calculus to show that the revenue is increasing \(30\) weeks after the beginning of July.
The answer is already in the question!
\begin{align}r'(30)&=-\frac{11{,}250\pi}{13}\sin\left(\frac{\pi}{26}(3)\right)\\&=1263.44…\end{align}
As \(r'(30)>0\), it is increasing.
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(e) Find a value for the time \(t\), within the first \(52\) weeks, when the revenue is at a minimum.
Use \(r^{\prime\prime}(t)\), to verify your answer.
The revenue is a minimum in the \(26\)th week.
\begin{align}-\frac{11{,}250\pi}{13}\sin\left(\frac{\pi}{26}t\right)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\sin\left(\frac{\pi}{26}t\right)=0\end{align}
\begin{align}\downarrow\end{align}
\(\left(\dfrac{\pi}{26}t\right)=0\) and \(\left(\dfrac{\pi}{26}t\right)=\pi\)
\begin{align}\downarrow\end{align}
\(t=0\) and \(t=26\)
\[\,\]
Verify
\begin{align}r^{\prime\prime}(t)=-\frac{11{,}250\pi}{13}\left(\frac{\pi}{26}\right)\cos\left(\frac{\pi}{26}t\right)\end{align}
\begin{align}\downarrow\end{align}
\(r^{\prime\prime}(0)>0\) and \(r^{\prime\prime}(26)<0\)
Therefore, the revenue is a minimum in the \(26\)th week.
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2019 Paper 1 Question 9(b) & (c)
(b)
(i) Complete the table shown.
| \(x\) | \(0\) | \(\dfrac{12}{2+\pi}\) |
|---|---|---|
|
\(y=\dfrac{12-(2+\pi)x}{2}\) |
|
|
(ii) On the diagram below, draw the graph of the linear function, \(y=\dfrac{12-(2+\pi)x}{2}\) for \(0\leq x\leq\dfrac{12}{2+\pi}\) where \(x\in\mathbb{R}\).
(iii) Find the slope of the graph of \(y\), correct to \(2\) decimal places.
Interpret this slope in the context of the question.
(i)
| \(x\) | \(0\) | \(\dfrac{12}{2+\pi}\) |
|---|---|---|
|
\(y=\dfrac{12-(2+\pi)x}{2}\) |
\(6\) |
\(0\) |
(ii)
(iii) The slope is \(-2.57\) and therefore, for a \(1\) metre increase in the radius, the height of the rectangle decreases by \(2.57\) metres.
(i)
| \(x\) | \(0\) | \(\dfrac{12}{2+\pi}\) |
|---|---|---|
|
\(y=\dfrac{12-(2+\pi)x}{2}\) |
\(6\) |
\(0\) |
(ii)
(iii)
\begin{align}y&=\frac{12-(2+\pi)x}{2}\\&=-\left(\frac{2+\pi}{2}\right)+6\\&=mx+c\end{align}
\begin{align}\downarrow\end{align}
\begin{align}m&=-\left(\frac{2+\pi}{2}\right)\\&\approx-2.57\mbox{ m}\end{align}
Therefore, for a \(1\) metre increase in the radius, the height of the rectangle decreases by \(2.57\) metres.
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(c)
(i) The Norman window shown below has a perimeter of \(12\) metres and \(y=\dfrac{12-(2+\pi)x}{2}\).
Show that the function \(a(x)=\dfrac{24x-(\pi+4)x^2}{2}\) represents the area of the
window, in terms of \(x\) and \(\pi\).
(ii) Find \(a'(x)\).
(iii) Find the relationship between \(x\) and \(y\) when the area of the window in part (c)(i) is at its
maximum.
(i) The answer is already in the question!
(ii) \(a'(x)=12-(\pi+4)x\)
(iii) The area is a maximum when the radius is equal to the height of the rectangle.
(i)
\begin{align}a(x)&=2xy+\frac{1}{2}(\pi x^2)\\&=2x\left(\frac{12-(2+\pi)x}{2}\right)+\frac{1}{2}(\pi x^2)\\&=\frac{24x-4x^2-2\pi x^2+\pi x^2}{2}\\&=\frac{24x-(\pi+4)x^2}{2}\end{align}
as required.
(ii)
\begin{align}a'(x)&=\frac{24-2(\pi+4)x}{2}\\&=12-(\pi+4)x\end{align}
(iii)
\begin{align}a'(x)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}12-(\pi+4)x=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x=\frac{12}{\pi+4}\end{align}
and
\begin{align}y&=\frac{12-(2+\pi)x}{2}\\&=\frac{12-(2+\pi)\left(\frac{12}{\pi+4}\right)}{2}\\&=\frac{12(\pi+4)-(2+\pi)(12)}{2(\pi+4)}\\&=\frac{6(\pi+4)-(2+\pi)(6)}{\pi+4}\\&=\frac{6\pi+24-12-6\pi}{\pi+4}\\&=\frac{12}{\pi+4}\\&=x\end{align}
Therefore, the area is a maximum when the radius is equal to the height of the rectangle.
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2018 Paper 1 Question 2(b)
(b) If ݂\(f(x)=x^3-17x^2+80x-64\), \(x\in\mathbb{R}\), show that ݂\(f(1)=0\), and find another value of \(x\) for which ݂\(f(x)=0\).
\(x=8\)
\begin{align}f(1)&=1^3-17(1^2)+80(1)-64\\&=1-17+80-64\\&=0\end{align}
as required.
\[
\require{enclose}
\begin{array}{rll}
x^2-16x+64\phantom{000000}\, \\[-3pt]
x-1 \enclose{longdiv}{\,x^3-17x^2+80x-64} \\[-3pt]
\underline{x^3-x^2\phantom{00000000000}\,\,} \\[-3pt]
-16x^2+80x\phantom{000}\,\\[-3pt]\phantom{0000}\underline{-16x^2+16x\phantom{00}\,}\\[-3pt]\phantom{00}64x-64\\[-3pt]\phantom{00}\underline{64x-64}\\[-3pt]0
\end{array}
\]
\begin{align}\downarrow\end{align}
\begin{align}f(x)&=x^3-17x^2+80x-64\\&=(x-1)(x^2-16x+64)\\&=(x-1)(x-8)^2\end{align}
Therefore, another value of \(x\) in which \(f(x)=0\) is \(x=8\).