One-to-One Grinds
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2019 Higher Level - Paper One
Section A
Question 1
(a) In the expansion of \((2x+1)(x^2+px+4)\), where \(p\in\mathbb{N}\), the coefficient of \(x\) is twice the coefficient of \(x^2\). Find the value of \(p\).
\(p=2\)
\begin{align}(2x+1)(x^2+px+4)&=2x^3+2px^2+8x+x^2+px+4\\&=2x^3+(2p+1)x^2+(8+p)x+4\end{align}
\begin{align}\downarrow\end{align}
\begin{align}8+p=2(2p+1)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}8+p=4p+2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3p=6\end{align}
\begin{align}\downarrow\end{align}
\begin{align}p=2\end{align}
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(b) Solve the equation \(\dfrac{3}{2x+1}+\dfrac{2}{5}=\dfrac{2}{3x-1}\) where \(x\neq-\dfrac{1}{2},\dfrac{1}{3}\), and \(x\in\mathbb{R}\).
\(x=-3\) or \(x=\dfrac{3}{4}\)
\begin{align}\frac{3}{2x+1}+\frac{2}{5}=\frac{2}{3x-1}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{3(5)+2(2x+1)}{(2x+1)(5)}=\frac{2}{3x-1}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{4x+17}{10x+5}=\frac{2}{3x-1}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(4x+17)(3x-1)=2(10x+5)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}12x^2-4x+51x-17=20x+10\end{align}
\begin{align}\downarrow\end{align}
\begin{align}12x^2+27x-27=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3x^2+9x-9=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x+3)(4x-3)=0\end{align}
\begin{align}\downarrow\end{align}
\(x=-3\) or \(x=\dfrac{3}{4}\)
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Question 2
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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(b) Prove, using induction, that \(f(n)\geq g(n)\), where \(n\geq2\) and \(n\in\mathbb{N}\).
The answer is already in the question!
The inequality is true for \(n=2\) as \(3^2\geq4(2)+1\).
Therefore let, us assume that it is true for \(n=k\), i.e. that
\begin{align}3^k\geq4k+1\end{align}
Using this assumption, we now wish to prove that it is also true for \(n=k+1\), i.e. that
\begin{align}3^{k+1}\geq4k+5\end{align}
First, we will multiply our assumption by \(3\):
\begin{align}3(3^k)\geq3(4k+1)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3^{k+1}\geq12k+3\end{align}
Since \(12k+3>4k+5\) for \(k\geq2\), it follows that
\begin{align}3^{k+1}\geq4k+5\end{align}
as required.
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Question 3
(a) Factorise fully: \(3xy-9x+4y-12\)
\((3x+4)(y-3)\)
\begin{align}3xy-9x+4y-12&=3x(y-3)+4(y-3)\\&=(3x+4)(y-3)\end{align}
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(b) \(g(x)=3x\ln x-9x+4\ln x-12\).
Using your answer to part (a) or otherwise, solve \(g(x)=0\).
\(x=-\dfrac{4}{3}\) or \(x=e^3\)
\begin{align}(3x+4)(\ln x-3)=0\end{align}
\begin{align}\downarrow\end{align}
\(3x+4=0\) or \(\ln x-3=0\)
\begin{align}\downarrow\end{align}
\(x=-\dfrac{4}{3}\) or \(x=e^3\)
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(c) Evaluate \(g'(e)\) correct to \(2\) decimal places.
\(-1.53\)
\begin{align}g'(x)=3x\left(\frac{1}{x}\right)+3(\ln x)-9+4\left(\frac{1}{x}\right)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}g'(e)&=3e\left(\frac{1}{e}\right)+3(\ln e)-9+4\left(\frac{1}{e}\right)\\&\approx-1.53\end{align}
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Question 4
(a) Find \(\int(4x^3-6x+10)\,dx\).
\(x^4-3x^2+10x+C\)
\begin{align}\int(4x^3-6x+10)\,dx&=\frac{4x^4}{4}-\frac{6x^2}{2}+10x+C\\&=x^4-3x^2+10x+C\end{align}
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(b) Part of the graph of a cubic function \(f(x)\) is shown below (graph not to scale).
The graph cuts the \(x\)‐axis at the three points \(A(2,0)\), \(B\), and \(C\).
(i) Given that \(f'(x)=6x^2-54x+109\), show that \(f(x)=2x^3-27x^2+109x-126\).
(ii) Find the co‐ordinates of the point \(B\) and the point \(C\).
(i) The answer is already in the question!
(ii) \(B:(4.5,0)\) and \(C:(7,0)\)
(i)
\begin{align}f(x)&=\int f'(x)\,dx\\&=\int(6x^2-54x+109)\,dx\\&=\frac{6x^3}{3}-\frac{54x^2}{2}+109x+C\\&=2x^3-27x^2+109x+C\end{align}
and
\begin{align}f(2)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2(2^3)-27(2^2)+109(2)+C=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}126+C=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}C=-126\end{align}
\begin{align}\downarrow\end{align}
\begin{align}f(x)=2x^3-27x^2+109x-126\end{align}
as required.
(ii)
\[
\require{enclose}
\begin{array}{rll}
2x^2-23x+63\phantom{00000000}\, \\[-3pt]
x-2 \enclose{longdiv}{\,2x^3-27x^2+109x-126} \\[-3pt]
\underline{2x^3-4x^2\phantom{000000000000}\,\,} \\[-3pt]
-23x^2+109x\phantom{000}\,\\[-3pt]\phantom{0000}\underline{-23x^2+46x\phantom{000}\,}\\[-3pt]\phantom{00}63x-126\\[-3pt]\phantom{00}\underline{63x-126}\\[-3pt]0
\end{array}
\]
\begin{align}\downarrow\end{align}
\begin{align}f(x)&=(x-2)(2x^2-23x+63)\\&=(x-2)(2x-9)(x-7)\end{align}
\begin{align}\downarrow\end{align}
\(B:(4.5,0)\) and \(C:(7,0)\)
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Question 5
(a) \(3+2i\) is a root of \(z^2+pz+q=0\), where \(p,q\in\mathbb{R}\), and \(i^2=-1\).
Find the value of \(p\) and the value of \(q\).
\(p=-6\) and \(q=13\)
\begin{align}(3+2i)^2+p(3+2i)+q=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}9+12i-4+3p+2pi+q=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(5+3p+q)+(12+2p)i=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(5+3p+q)+(12+2p)i=0\end{align}
\begin{align}\downarrow\end{align}
We therefore have the following two equations for two unknowns:
\begin{align}5+3p+q=0\end{align}
\begin{align}12+2p=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}5+3p+q=0\end{align}
\begin{align}p=-6\end{align}
\begin{align}\downarrow\end{align}
\begin{align}5+3(-6)+q=0\end{align}
\begin{align}p=-6\end{align}
\begin{align}\downarrow\end{align}
\begin{align}q=13\end{align}
\begin{align}p=-6\end{align}
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(b)
(i) \(v=2-2\sqrt{3}i\). Write \(v\) in the form \(r(\cos\theta+i\sin\theta)\), where \(r\in\mathbb{R}\) and \(0\leq\theta\leq2\pi\).
(ii) Use your answer to part (b)(i) to find the two possible values of \(w\), where \(w^2=v\).
Give your answers in the form \(a+ib\), where \(a,b\in\mathbb{R}\).
(i) \(v=4\left[\cos\dfrac{5\pi}{3}+i\sin\dfrac{5\pi}{3}\right]\)
(ii) \(w=-\sqrt{3}+i\) and \(w=\sqrt{3}-i\)
(i)
Modulus
\begin{align}r&=\sqrt{2^2+(-2\sqrt{3})^2}\\&=\sqrt{4+12}\\&=\sqrt{16}\\&=4\end{align}
\[\,\]
Argument
Reference Angle:
\begin{align}A&=\tan^{-1}\left(\frac{2\sqrt{3}}{2}\right)\\&=\frac{\pi}{3}^{\circ}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\theta&=2\pi-\frac{\pi}{3}\\&=\frac{\pi}{3}\end{align}
\[\,\]
Polar Form
\begin{align}v=4\left[\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3}\right]\end{align}
(ii)
\begin{align}w&=\pm\left[4\left(\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3}\right)\right]^{1/2}\\&=\pm2\left[\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right]\\&=\pm(-\sqrt{3}+i)\end{align}
\begin{align}\downarrow\end{align}
\(w=-\sqrt{3}+i\) and \(w=\sqrt{3}-i\)
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Question 6
(a)
(i) Given that \(x-\sqrt{32}=\sqrt{128}-5x\), find the value of \(x\), where \(x\in\mathbb{R}\).
Give your answer in the form \(a\sqrt{2}\), where \(a\in\mathbb{N}\).
(ii) \(A=\{\sqrt{32k^2},\sqrt{50k^2},\sqrt{128k^2},\sqrt{98k^2}\}\), where \(k\in\mathbb{N}\).
Show that the mean of set \(A\) is equal to the median of set \(A\).
(i) \(2\sqrt{2}\)
(ii) The answer is already in the question!
(i)
\begin{align}x-\sqrt{32}=\sqrt{128}-5x\end{align}
\begin{align}\downarrow\end{align}
\begin{align}6x=\sqrt{128}+\sqrt{32}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}6x=8\sqrt{2}+4\sqrt{2}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}6x=12\sqrt{2}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x=2\sqrt{2}\end{align}
(ii)
Mean
\begin{align}\frac{\sqrt{32k^2}+\sqrt{50k^2}+\sqrt{128k^2}+\sqrt{98k^2}}{4}&=\frac{4\sqrt{2}k+8\sqrt{2}k+7\sqrt{2}k+5\sqrt{2}k}{4}\\&=\frac{24\sqrt{2}k}{4}\\&=6\sqrt{2}k\end{align}
\[\,\]
Median
\begin{align}\frac{1}{2}(\sqrt{50k^2}+\sqrt{50k^2})&=\frac{1}{2}(5\sqrt{2}k+7\sqrt{2}k)\\&=\frac{1}{2}(12\sqrt{2}k)\\&=6\sqrt{2}k\end{align}
Therefore, the mean and the median are the same.
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(b) Prove, using contradiction, that \(\sqrt{2}\) is not a rational number.
The answer is already in the question!
Assume \(\sqrt{2}\) is a rational number, i.e. that
\begin{align}\sqrt{2}=\frac{a}{b}\end{align}
where \(a,b\in\mathbb{Z}\) and the fraction is written in its simplest form. Hence
\begin{align}a^2=2b^2\end{align}
and therefore \(a\) is even, i.e. \(a=2m\) for some \(m\in\mathbb{Z}\).
\begin{align}(2m)^2=2b^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}b^2=2m^2\end{align}
and therefore \(b\) is even, i.e. \(b=2n\) for some \(n\in\mathbb{Z}\).
\begin{align}\sqrt{2}&=\frac{a}{b}\\&=\frac{2m}{2n}\end{align}
However, we stated that the fraction was already in its simplest form – a contraction.
Hence, \(\sqrt{2}\) is irrational.
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Section B
Question 7
The closed line segment \([0,1]\) is shown below. The first three steps in the construction of the
Cantor Set are also shown:
- Step \(1\) removes the open middle third of the line segment \([0,1]\) leaving two closed line segments (i.e. the end points of the segments remain in the Cantor Set)
- Step \(2\) removes the middle third of the two remaining segments leaving four closed line segments
- Step \(3\) removes the middle third of the four remaining segments leaving eight closed line segments.
The process continues indefinitely. The set of points in the line segment \([0,1]\) that are not removed during the process is the Cantor Set.
(a)
(i) Complete the table below to show the length of the line segment(s) removed at each step for the first \(5\) steps. Give your answers as fractions.
| Step | Step \(1\) | Step \(2\) | Step \(3\) | Step \(4\) | Step \(5\) |
|---|---|---|---|---|---|
|
Length Removed |
\(\dfrac{1}{3}\) |
\(\dfrac{2}{9}\) |
|
|
|
(ii) Find the total length of all of the line segments removed from the initial line segment of length \(1\) unit, after a finite number (\(n\)) of steps in the process.
Give your answer in terms of \(n\).
(iii) Find the total length removed, from the initial line segment, after an infinite number of steps of the process.
(i)
| Step | Step \(1\) | Step \(2\) | Step \(3\) | Step \(4\) | Step \(5\) |
|---|---|---|---|---|---|
|
Length Removed |
\(\dfrac{1}{3}\) |
\(\dfrac{2}{9}\) |
\(\dfrac{4}{27}\) |
\(\dfrac{8}{81}\) |
\(\dfrac{16}{243}\) |
(ii) \(1-\left(\dfrac{2}{3}\right)^n\)
(iii) \(1\)
(i)
| Step | Step \(1\) | Step \(2\) | Step \(3\) | Step \(4\) | Step \(5\) |
|---|---|---|---|---|---|
|
Length Removed |
\(\dfrac{1}{3}\) |
\(\dfrac{2}{9}\) |
\(\dfrac{4}{27}\) |
\(\dfrac{8}{81}\) |
\(\dfrac{16}{243}\) |
(ii)
\begin{align}S_n&=\frac{a(1-r^n)}{1-r} \\&=\frac{\frac{1}{3}\left[1-\left(\frac{2}{3}\right)^n\right]}{1-\frac{2}{3}}\\&=1-\left(\frac{2}{3}\right)^n\end{align}
(iii)
\begin{align}S_{\infty}&=\frac{a}{1-r}\\&=\frac{\frac{1}{3}}{1-\frac{2}{3}}\\&=1\end{align}
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(b)
(i) Complete the table below to identify the end‐points labelled in the diagram.
Give your answers as fractions.
| Label | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) |
|---|---|---|---|---|---|---|
|
End-point |
|
|
|
|
|
|
(ii) Give a reason why \(\dfrac{1}{3}-\dfrac{1}{9}+\dfrac{1}{27}-\dfrac{1}{81}\) is a point in the Cantor Set.
(iii) The limit of the series \(\dfrac{1}{3}-\dfrac{1}{9}+\dfrac{1}{27}-…\) is a point in the Cantor Set. Find this point.
(i)
| Label | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) |
|---|---|---|---|---|---|---|
|
End-point |
\(\dfrac{2}{3}\) |
\(\dfrac{2}{9}\) |
\(\dfrac{7}{9}\) |
\(\dfrac{8}{9}\) |
\(\dfrac{7}{27}\) |
\(\dfrac{25}{27}\) |
(ii) It is the end point of a segment.
(iii) \(\dfrac{1}{4}\)
(i)
| Label | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) |
|---|---|---|---|---|---|---|
|
End-point |
\(\dfrac{2}{3}\) |
\(\dfrac{2}{9}\) |
\(\dfrac{7}{9}\) |
\(\dfrac{8}{9}\) |
\(\dfrac{7}{27}\) |
\(\dfrac{25}{27}\) |
(ii) It is the end point of a segment.
(iii)
\begin{align}S_{\infty}&=\frac{a}{1-r}\\&=\frac{\frac{1}{3}}{1-\left(-\frac{1}{3}\right)}\\&=\frac{1}{4}\end{align}
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Question 8
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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Question 9
Norman windows consist of a rectangle topped by a semi‐circle as shown above.
Let the height of the rectangle be \(y\) metres and the radius of the semi‐circle be \(x\) metres as shown. The perimeter of the window is \(P\).
(a)
(i) Write \(P\) in terms of \(x\), \(y\) and \(\pi\).
(ii) In a particular Norman window the perimeter \(P=12\) metres.
Show that \(y=\dfrac{12-(2+\pi)x}{2}\) for \(0\leq x\leq \dfrac{12}{2+\pi}\), where \(x\in\mathbb{R}\).
(i) \(P=2x+2y+\pi x\)
(ii) The answer is already in the question!
(i)
\begin{align}P&=2x+2y+\frac{1}{2}(2\pi x)\\&=2x+2y+\pi x\end{align}
(ii)
\begin{align}2x+2y+\pi x=12\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2y=12-2x-\pi x\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y&=\frac{12-2x-\pi x}{2}\\&=\frac{12-(2+\pi)x}{2}\end{align}
as required.
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(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.
Our Video Walkthroughs are in the final stages of editing and will go live during Summer 2023.
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!
(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.
