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Calculus
NOTE: Clicking an entry in the right column will take you directly to that question!
(BLUE = Paper 1, RED = Paper 2)
| Since 2015, the following subtopic... | Has explicitly appeared in... |
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Derivatives from First Principles |
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Product/Quotient/Chain Rules |
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Polynomial Derivatives |
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Trigonometric Derivatives |
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Exponential Derivatives |
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Logarithmic Derivatives |
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Polynomial Integrals |
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Trigonometric Integrals |
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Exponential Integrals |
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Integrals and Areas |
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Average Value of a Function |
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Calculus and Curves |
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Rate of Change |
2022 Paper 1 Question 2
(a) \(g(x)=2x^2+5x+6\), where \(x\in\mathbb{R}\).
Find \(\int g(x)\,dx\).
\(\dfrac{2x^3}{3}+\dfrac{5x^2}{2}+6x+C\)
\begin{align}\int g(x)\,dx&=\int(2x^2+5x+6)\,dx\\&=\int 2x^2\,dx+\int5x\,dx+\int6\,dx\\&=2\int x^2\,dx+5\int x\,dx+6\int dx\\&=\frac{2x^3}{3}+\frac{5x^2}{2}+6x+C\end{align}
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(b) The diagram shows the graph of a function \(f(x)=ax^2+bx+c\), where \(a,b,c\in\mathbb{Z}\).
Three regions on the diagram are marked K, L, and N.
Each of these regions is bounded by the \(x\)-axis, the graph of\(f(x)\), and two vertical lines.
(i) The area of region K is \(538\) square units. Use integration of \(f(x)\) to show that:
\begin{align}4a+3b+3c=807\end{align}
(ii) The areas of the three regions K, L, and N give the following three equations (including the equation from part (b)(i)):
\begin{align}4a+3b+3c&=807\\28a+9b+3c&=879\\76a+15b+3c&=663\end{align}
Solve these equations to find the values of \(a\), \(b\) and \(c\).
(i) The answer is already in the question!
(ii) \(a=-12\), \(b=60\) and \(c=225\)
(i)
\begin{align}\int_0^2 f(x)\,dx=538\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\int_0^2 (ax^2+bx+c)\,dx=538\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\int_0^2ax^2\,dx+\int_0^2bx\,dx+\int_0^2c\,dx=538\end{align}
\begin{align}\downarrow\end{align}
\begin{align}a\int_0^2x^2\,dx+b\int_0^2x\,dx+c\int_0^2\,dx=538\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\left.\frac{ax^3}{3}\right|_0^2+\left.\frac{bx^2}{2}\right|_0^2+\left.cx\right|_0^2=538\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\left[\frac{a(2^3)}{3}-\frac{a(0^3)}{3})\right]+\left[\frac{b(2^2)}{2}-\frac{b(0^2)}{2}\right]+[c(2)-c(0)]=538\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{8a}{3}+\frac{4b}{2}+2c=538\end{align}
\begin{align}\downarrow\end{align}
\begin{align}8a+6b+6c=1614\end{align}
\begin{align}\downarrow\end{align}
\begin{align}4a+3b+3c=807\end{align}
as required.
(ii)
\begin{align}4a+3b+3c&=807\\28a+9b+3c&=879\\76a+15b+3c&=663\end{align}
\begin{align}\downarrow\end{align}
\begin{align}24a+6b&=72\\48a+6b&=-216\end{align}
\begin{align}\downarrow\end{align}
\begin{align}24a=-288\end{align}
\begin{align}\downarrow\end{align}
\begin{align}a=-12\end{align}
and
\begin{align}24a+6b=72\end{align}
\begin{align}\downarrow\end{align}
\begin{align}b&=\frac{72-24a}{6}\\&=\frac{72-24(-12)}{6}\\&=60\end{align}
and
\begin{align}4a+3b+3c&=807\end{align}
\begin{align}\downarrow\end{align}
\begin{align}c&=\frac{807-4a-3b}{3}\\&=\frac{807-4(-12)-3(60)}{3}\\&=225\end{align}
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2022 Paper 1 Question 5
(a) \(g(x)=x^2-\dfrac{1}{x}\) where \(x\in\mathbb{R}\).
Find \(g'(x)\), the derivative of \(g(x)\).
\(g'(x)=2x+\dfrac{1}{x^2}\)
\begin{align}g(x)&=x^2-\frac{1}{x}\\&=x^2-x^{-1}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}g'(x)&=2x-(-1)x^{-2}\\&=2x+\frac{1}{x^2}\end{align}
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(b) \(f(x)=2x^3-21x^2+40x+63\), where \(x\in\mathbb{R}\).
(i) \(x+1\) is a factor of \(f(x)\). Find the three values of \(x\) for which \(f(x)=0\).
(ii) Find the range of values of \(x\) for which \(f'(x)\) is negative, correct to \(2\) decimal places.
(i) \(x=-1\), \(x=4.5\) or \(x=7\)
(ii) \(1.14<x<5.86\)
(i)
\[
\require{enclose}
\begin{array}{rll}
2x^2-23x+63\phantom{000000}\, \\[-3pt]
x+1 \enclose{longdiv}{\,2x^3-21x^2+40x+63} \\[-3pt]
\underline{2x^3+2x^2\phantom{00000000000}\,\,} \\[-3pt]
-23x^2+40x\phantom{000}\,\\[-3pt]\phantom{0000}\underline{-23x^2-23x\phantom{00}\,}\\[-3pt]\phantom{00}63x+63\\[-3pt]\phantom{00}\underline{63x+63}\\[-3pt]0
\end{array}
\]
\begin{align}\downarrow\end{align}
\begin{align}2x^3-21x^2+40x+63=(x+1)(2x^2-23x+63)\end{align}
\[\,\]
Factors
\begin{align}(x+1)(2x^2-23x+63)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x+1)(2x-9)(x-7)=0\end{align}
\begin{align}\downarrow\end{align}
\(x=-1\), \(x=4.5\) or \(x=7\)
(ii)
Derivative
\begin{align}f'(x)=2x^3-21x^2+40x+63\end{align}
\begin{align}\downarrow\end{align}
\begin{align}f^{\prime\prime}(x)=6x^2-42x+40\end{align}
\[\,\]
Inequality
\begin{align}6x^2-42x+40<0\end{align}
Roots:
\(a=6\), \(b=-42\) and \(c=40\)
\begin{align}\frac{-b\pm\sqrt{b^2-4ac}}{2a}&=\frac{-(-42)\pm\sqrt{(-42)^2-4(6)(40)}}{2(6)}\\&=1.137…\mbox{ or }5.862…\\&\approx1.14\mbox{ or } 5.86\end{align}
\begin{align}\downarrow\end{align}
\begin{align}1.14<x<5.86\end{align}
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2022 Paper 1 Question 6
(a) Differentiate \(f(x)=2x^2+4x\) with respect to \(x\), from first principles.
\(4x+4\)
\begin{align}f(x)=2x^2+4x\end{align}
and
\begin{align}f(x+h)=2(x+h)^2+4(x+h)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{df}{dx}&=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}\\&=\lim_{h\rightarrow0}\frac{[2(x+h)^2+4(x+h)]-[2x^2+4x]}{h}\\&=\lim_{h\rightarrow0}\frac{2(x+h)^2+4(x+h)-2x^2-4x}{h}\\&=\lim_{h\rightarrow0}\frac{2x^2+4xh+2h^2+4x+4h-2x^2-4x}{h}\\&=\lim_{h\rightarrow0}\frac{4xh+2h^2+4h}{h}\\&=\lim_{h\rightarrow0}4x+2h+4\\&=4x+4\end{align}
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(b) A rectangle is expanding in area. Its width is \(x\mbox{ cm}\), where \(x\in\mathbb{R}\) and \(x>0\).
Its length is always four times its width.
Find the rate of change of the area of the rectangle with respect to its width, \(x\), when the area of the rectangle is \(225\mbox{ cm}^2\).
\(60\mbox{ cm}\)
\begin{align}A&=lw\\&=lx\\&=\\&=(4x)(x)\\&=4x^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{dA}{dx}=8x\end{align}
When \(A=4x^2=225\), i.e. when \(x= 7.5\), this gives
\begin{align}\frac{dA}{dx}&=8x\\&=8(7.5)\\&=60\mbox{ cm}\end{align}
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(c) The graph of a cubic function \(p(x)\) is shown in the first diagram below, for \(0\leq x\leq 4\), \(x\in\mathbb{R}\). The maximum value of \(p'(x)\) in this domain is \(1\), and \(p'(0)=-3\), where \(p'(x)\) is the derivative of \(p(x)\).
Use this information to draw the graph of \(p'(x)\) on the second set of axes below, for \(0\leq x\leq4\), \(x\in\mathbb{R}\).
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2022 Paper 1 Question 7(b)-(e)
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}
(b) Find \(h'(x)\).
\(h'(x)=6x^2-57x+105\)
\begin{align}h'(x)&=6x^2-57x+105\end{align}
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(c) Find \(h'(2)\), and explain what this value means in the context of Hannah’s heart-rate.
\(h'(2)=15\) is the rate of increase of Hannah’s heart rate at \(2\) minutes.
\begin{align}h'(2)&=6(2^2)-57(2)+105\\&=15\end{align}
This is the rate of increase of Hannah’s heart rate at \(2\) minutes.
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The graph below shows \(y=h(x)\), where \(0\leq x\leq 8\), \(x\in\mathbb{R}\).
(d) Find the least value and the greatest value of \(h(x)\), for \(0\leq x\leq 8\), \(x\in\mathbb{R}\).
Use calculus in your solution. You may also use information from the graph above, which is to scale.
Least value: \(70\).
Greatest value: \(185.625\).
Least Value
\begin{align}h(0)&=2(0^3)-28.5(0^2)+105(0)+70\\&=70\end{align}
\[\,\]
Greatest Value
\begin{align}h'(x)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}6x^2-57x+105=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2x^2-19x+35=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(2x-5)(x-7)=0\end{align}
\(x=2.5\) or \(x=7\)
\begin{align}\downarrow\end{align}
\begin{align}h(2.5)&=2(2.5^3)-28.5(2.5^2)+105(2.5)+70\\&=185.625\end{align}
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(e) How long after the start of the session is Hannah’s heart-rate decreasing most quickly, within the first \(8\) minutes? Give your answer in minutes and seconds.
Remember that \(h(x)=2x^3-28.5x^2+105x+70\).
\(4\mbox{ min } 45\mbox{ secs}\)
The second derivative is zero.
\begin{align}\downarrow\end{align}
\begin{align}12x-57=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x&=4.75\mbox{ min}\\&=4\mbox{ min } 45\mbox{ secs}\end{align}
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2022 Paper 1 Question 8(f)
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.
(f) Use integration to find the average height of the point \(A\) over the first \(8\) minutes that the wheel is turning. Give your answer correct to \(1\) decimal place.
\(65.8\mbox{ m}\)
\begin{align}\mbox{Average height }&=\frac{1}{8-0}\int_0^8\left[72-60\cos\left(\frac{\pi}{3}t\right)\right]\,dt\\&=\frac{1}{8}\left[72\int_0^8dt-60\int_0^8\cos\left(\frac{\pi}{3}t\right)\,dt\right]\\&=\frac{1}{8}\left[\left.72t\right|_0^8-\left.\frac{180}{\pi}\sin\left(\frac{\pi}{3}t\right)\right|_0^8\right]\\&=\frac{1}{8}\left[(72(8)-72(0))-\frac{180}{\pi}\sin\left(\frac{\pi}{3}(8)\right)+\frac{180}{\pi}\sin\left(\frac{\pi}{3}(0)\right)\right]\\&=72-6.20…\\&=65.8\mbox{ m}\end{align}
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2022 Paper 1 Question 10(c)-(d)
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 \(0\leq t\leq 12\), \(t\in\mathbb{R}\).
(c)
(i) Find the value of \(P'(1)\).
(ii) \(P'(t)\) is always negative for \(0\leq t\leq 12\), \(t\in\mathbb{R}\). What does this tell you about the proportion of digits recalled correctly after \(t\) hours, according to this model?
(i) \(-0.06\)
(ii) As time advances, the amount of digits that the student can recall decreases.
(i)
\begin{align}P'(t)=-\frac{0.12}{t+1}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}P'(1)&=-\frac{0.12}{1+1}\\&=-0.06\end{align}
(ii) As time advances, the amount of digits that the student can recall decreases.
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(d) Use calculus to show that the graph of \(y=P(t)\) has no points of inflection, for \(0\leq t\leq 12\), \(t\in\mathbb{R}\).
As the second derivative can never be zero, there are no points of inflection.
\begin{align}\frac{d^2P}{dt^2}=\frac{0.12}{(t+2)^2}\end{align}
As this can never be zero, there are no points of inflection.
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2021 Paper 1 Question 5
(a) The derivative of \(f(x)=2x^3+6x^2-12x+3\) can be expressed in the form \(f'(x)=a(x+b)^2\), where \(a,b,c\in\mathbb{Z}\) and \(x\in\mathbb{R}\).
(i) Find the value of \(a\), the value of \(b\), and the value of \(c\).
(ii) If \(g(x)=36x+5\), find the range of values of \(x\) for which \(f'(x)>g'(x)\).
(i) \(a=6\), \(b=1\) and \(c=-18\)
(ii) \(x<-4\) or \(x>2\)
(i)
\begin{align}f'(x)&=6x^2+12x-12\\&=6(x^2+2x-2)\\&=6(x^2+2x+1-1-2)\\&=6[(x+1)^2-3]\\&=6(x+1)^2-18\end{align}
\begin{align}\downarrow\end{align}
\begin{align}a=6&&b=1&&c=-18\end{align}
(ii)
\begin{align}f'(x)>g'(x)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}6x^2+12x-12>36\end{align}
\begin{align}\downarrow\end{align}
\begin{align}6x^2+12x-48>0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x^2+2x-8>0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x+4)(x-2)>0\end{align}
\begin{align}\downarrow\end{align}
\(x<-4\) or \(x>2\)
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(b) The diagram below shows \(t\), the tangent line to \(h(x)=2\sin(2x)\) , where \(0\leq x\leq \pi\), at the point where \(x=\dfrac{\pi}{6}\).
\(A(0,k)\), where \(k\in\mathbb{R}\), is the point where \(t\) cuts the \(y\)-axis.
Find the value of \(k\) correct to two decimal places.
\(k=0.68\)
Slope of Tangent
\begin{align}h'(x)=4\cos(2x)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}h’\left(\frac{\pi}{6}\right)&=4\cos\left[2\left(\frac{\pi}{6}\right)\right]\\&=2\end{align}
\[\,\]
Point on Tangent
\begin{align}h\left(\frac{\pi}{6}\right)&=2\sin\left[2\left(\frac{\pi}{6}\right)\right]\\&=\sqrt{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x,y)=\left(\frac{\pi}{6},\sqrt{3}\right)\end{align}
\[\,\]
Equation of Tangent
\begin{align}y=mx+c\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\sqrt{3}=2\left(\frac{\pi}{6}\right)+k\end{align}
\begin{align}\downarrow\end{align}
\begin{align}k&=\sqrt{3}-\frac{\pi}{3}\\&\approx0.68\end{align}
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2021 Paper 1 Question 6
The diagram below shows the graph of \(h'(x)\), the derivative of a cubic function \(h(x)\).
(a) Show that \(h'(x)=-2x^2+4x+6\).
The answer is already in the question!
\begin{align}y=ax^2+bx+c\end{align}
\[\,\]
\begin{align}(x,y)=(-1,0)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}0=a(-1)^2+b(-1)+c\end{align}
\begin{align}\downarrow\end{align}
\begin{align}a-b+c=0\end{align}
\[\,\]
\begin{align}(x,y)=(0,6)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}6=a(0^2)+b(0)+c\end{align}
\begin{align}\downarrow\end{align}
\begin{align}c=6\end{align}
\[\,\]
\begin{align}(x,y)=(3,0)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}0=a(3^2)+b(3)+c\end{align}
\begin{align}\downarrow\end{align}
\begin{align}9a+3b+c=0\end{align}
\[\,\]
We therefore have the following \(3\) equations \(3\) unknowns:
\begin{align}a-b+c=0\end{align}
\begin{align}c=6\end{align}
\begin{align}9a+3b+c=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}a-b=-6\end{align}
\begin{align}9a+3b=-6\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3a-3b=-18\end{align}
\begin{align}9a+3b=-6\end{align}
\begin{align}\downarrow\end{align}
\begin{align}12a=-24\end{align}
\begin{align}\downarrow\end{align}
\begin{align}a=-2\end{align}
and
\begin{align}a-b=-6\end{align}
\begin{align}\downarrow\end{align}
\begin{align}b&=a+6\\&=-2+6\\&=4\end{align}
\begin{align}\downarrow\end{align}
\begin{align}a=-2&&b=4&&c=6\end{align}
\begin{align}h'(x)=-2x^2+4x+6\end{align}
as required.
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(b) Use \(h'(x)\) to find the maximum positive value of the slope of a tangent to \(h(x)\).
\(8\)
Setting the second derivative to zero gives:
\begin{align}-4x+4=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x=1\end{align}
As the third derivative is \(-4\), this is the \(x\) coordinate of a maximum of \(h'(x)\) and therefore the maximum positive slope of a tangent to \(h(x)\) is
\begin{align}h'(1)&=-2(1)^2+4(1)+6\\&=8\end{align}
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(c) The graph of \(h(x)\) passes through the point \((0,-2)\).
Find the equation of \(h(x)\).
\(h(x)=-\dfrac{2x^3}{3}+2x^2+6x-2\)
\begin{align}h(x)&=\int h'(x)\,dx\\&=\int(-2x^2+4x+6)\,dx\\&=-\frac{2x^3}{3}+2x^2+6x+C\end{align}
and
\begin{align}(x,y)=(0,-2)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-2-\frac{2(0^3)}{3}+2(0^2)+6(0)+C\end{align}
\begin{align}\downarrow\end{align}
\begin{align}C=-2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}h(x)=-\frac{2x^3}{3}+2x^2+6x-2\end{align}
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2021 Paper 1 Question 8(b)-(c)
(b) The function \(h(x)=0.001x^3-0.12x^2+3.6x+5\) can be used to model the height above level ground (in metres) of a section of the path followed by a rollercoaster track, where \(x\) is the horizontal distance from a fixed point.
(i) Find \(h'(x)\), the derivative of \(h(x)\).
(ii) Show that this section of the track reaches its maximum height above level ground when \(x=20\).
(iii) Find, using calculus, the height above ground, in metres, at the instant the track passes through an inflection point. The function \(h(x)=0.001x^3-0.12x^2+3.6x+5\).
(i) \(h'(x)=0.003x^2-0.24x+3.6\)
(ii) The answer is already in the question!
(iii) \(21\mbox{ m}\)
(i)
\begin{align}h'(x)=0.003x^2-0.24x+3.6\end{align}
(ii)
\begin{align}h'(20)&=0.003(20^2)-0.24(20)+3.6\\&=0\end{align}
According to the graph, the turning point at \(x=20\) is a maximum rather than a minimum.
(iii) Setting the second derivative to zero:
\begin{align}0.006x-0.24=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x&=\frac{0.24}{0.006}\\&=40\end{align}
\begin{align}\downarrow\end{align}
\begin{align}h(40)&=0.001(40^3)-0.12(40^2)+3.6(40)+5\\&=21\mbox{ m}\end{align}
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(c) Use the function \(h(x)=0.001x^3-0.12x^2+3.6x+5\), \(x\in\mathbb{R}\), to find the average height of this section of the track above level ground, from \(x=0\) to \(x=75\).
Give your answer in metres correct to \(2\) decimal places.
\(20.47\mbox{ m}\)
\begin{align}\mbox{Average}&=\frac{1}{b-a}\int_a^bh(x)\,dx\\&=\frac{1}{75-0}\int_0^{75}(0.001x^3-0.12x^2+3.6x+5)\,dx\\&=\frac{1}{75}\left[\frac{0.001x^4}{4}-0.12\frac{x^3}{3}+\frac{3.6x^2}{2}+5x\right]_0^{75}\\&=\frac{1}{75}\left[\frac{0.001(75^4)}{4}-0.12\frac{(75^3)}{3}+\frac{3.6(75^2)}{2}+5(75)\right]\\&\approx20.47\mbox{ m}\end{align}
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2021 Paper 1 Question 9(c)-(d)
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)=75e^{-0.081t}+20\end{align}
where \(t\) is time measured in minutes from when the coffee was brewed.
(c) Find, to the nearest \(^{\circ}\mbox{C}\), the temperature the coffee has reached when \(T'(t)=-4.05\) is the rate at which the coffee is cooling, in \(^{\circ}\mbox{C}\) per minute.
\(70^{\circ}\mbox{ C}\)
\begin{align}T'(t)=-6.075e^{-0.081t}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-4.05=-6.075e^{-0.081t}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}e^{-0.081t}=\frac{4.05}{6.075}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-0.081t=\ln\left(\frac{4.05}{6.075}\right)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t&=-\frac{1}{0.081}\ln\left(\frac{4.05}{6.075}\right)\\&=5.0057…\end{align}
\begin{align}\downarrow\end{align}
\begin{align}T(5.0057…)&=75e^{-0.081(5.0057)}+20\\&\approx70^{\circ}\mbox{ C}\end{align}
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(d) A sugar cube is put into the coffee.
The sugar keeps its cube shape as it dissolves.
As the sugar cube dissolves, its volume decreases at the constant rate of \(\dfrac{1}{20}\mbox{ cm}^3/\mbox{sec}\).
Let \(x(t)\) be the sidelength of the cube at time \(t\).
Find the rate of change of \(x(t)\) when the volume of the cube reaches \(\dfrac{1}{64}\mbox{ cm}^3\).
\(-\dfrac{4}{15}\mbox{ cm/s}\)
\begin{align}V=x^3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{dV}{dx}=3x^2\end{align}
\[\,\]
\begin{align}\frac{dV}{dt}=-\frac{1}{20}\end{align}
and
\begin{align}\frac{dV}{dt}=\frac{dV}{dx}\times\frac{dx}{dt}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{dV}{dx}\times\frac{dx}{dt}=-\frac{1}{20}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(3x^2)\frac{dx}{dt}=-\frac{1}{20}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{dx}{dt}=-\frac{1}{60x^2}\end{align}
When \(V=\dfrac{1}{64}\), \(x=\sqrt[3]{\dfrac{1}{64}}=\dfrac{1}{4}\) and therefore
\begin{align}\frac{dx}{dt}&=-\frac{1}{60\left(\frac{1}{4}\right)^2}\\&=-\frac{4}{15}\mbox{ cm/s}\end{align}
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2021 Paper 1 Question 10(a)
(a) Water is flowing into and being removed from a tank.
The volume of water in the tank is measured on a daily basis, starting at day \(0\).
The volume of water, in litres, in the tank can be modelled by the formula
\(V(t)=60+41t-3t^2\), while \(V(t)\geq0\),
where \(t\in\mathbb{R}\) is the time in days, starting at day \(0\).
Use the function \(V(t)\) to answer the following \(4\) questions.
(i) Find the value of \(t\) when the tank empties.
(ii) Find the rate at which the volume of water in the tank is changing when \(t=5\).
(iii) Find the value of \(t\) when the volume of water in the tank is a maximum.
(iv) Find the maximum volume of the water in the tank, correct to the nearest litre.
(i) \(t=15\mbox{ days}\)
(ii) \(11\mbox{ litres/day}\)
(iii) \(t=\dfrac{41}{6}\mbox{ days}\)
(iv) \(200\mbox{ litres}\)
(i)
\begin{align}V(t)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3t^2-41t-60=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(t-15)(3t+4)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t=15\mbox{ days}\end{align}
(since \(t\geq0\)).
(ii)
\begin{align}V'(t)=41-6t\end{align}
\begin{align}\downarrow\end{align}
\begin{align}V'(5)&=41-6(5)\\&=11\mbox{ litres/day}\end{align}
(iii)
\begin{align}V'(t)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}41-6t=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t=\frac{41}{6}\mbox{ days}\end{align}
(Since the second derivative (\(-6\)) is negative, this is a maximum.)
(iv)
\begin{align}V\left(\frac{41}{6}\right)&=60+41\left(\frac{41}{6}\right)-3\left(\frac{41}{6}\right)^2\\&\approx200\mbox{ litres}\end{align}
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2020 Paper 1 Question 4
The diagram below shows two functions \(f(x)\) and \(g(x)\).
The function \(f(x)\) is given by the formula \(f(x)=x^3+kx^2+15x+8\), where \(k\in\mathbb{Z}\), and \(x\in\mathbb{R}\).
(a) Given that \(f'(3)=-12\), show that \(k=-9\), where \(f'(3)\) is the derivative of \(f(x)\) at \(x=3\).
The answer is already in the question!
\begin{align}f'(x)=3x^2+2kx+15\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-12=3(3^2)+2k(3)+15\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-12=27+6k+15\end{align}
\begin{align}\downarrow\end{align}
\begin{align}k&=\frac{-12-27-15}{6}\\&=-9\end{align}
as required.
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(b) The function \(g(x)\) is the line that passes through the two turning points of \(f(x)=x^3-9x^2+15x+8\).
Find the equation of \(g(x)\).
\(y=-8x+23\)
\begin{align}f'(x)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3x^2-18x+15=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x^2-6x+5=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x-1)(x-5)=0\end{align}
\begin{align}\downarrow\end{align}
\(x=1\) or \(x=5\)
\[\,\]
Points
\begin{align}f(1)&=1^3-9(1^2)+15(1)+8\\&=15\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x_1,y_1)=(1,15)\end{align}
and
\begin{align}f(5)&=5^3-9(5^2)+15(5)+8\\&=-17\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x_2,y_2)=(5,-17)\end{align}
\[\,\]
Slope
\begin{align}m&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{-17-15}{5-1}\\&=-8\end{align}
\[\,\]
Equation
\begin{align}y-y_1=m(x-x_1)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y-15=-8(x-1)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y-15=-8x+8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y=-8x+23\end{align}
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(c) Show that the graph of \(g(x)\) contains the point of inflection of \(f(x)\).
The answer is already in the question!
Setting the second derivative of \(f(x)\) to zero gives:
\begin{align}6x-18=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x=3\end{align}
and
\begin{align}f(3)&=3^3-9(3^2)+15(3)+8\\&=-1\end{align}
Hence, the point of inflection is \((3,-1)\).
\begin{align}g(x)=-8x+23\end{align}
\begin{align}\downarrow\end{align}
\begin{align}g(3)&=-8(3)+23\\&=-1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(3,-1)\in g(x)\end{align}
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2020 Paper 1 Question 6
(a) Differentiate \((3x-5)(2x+4)\) with respect to \(x\) from first principles.
\(12x+2\)
\begin{align}f(x)&=(3x-5)(2x+4)\\&=6x^2+12x-10x-20\\&=6x^2+2x-20\end{align}
and
\begin{align}f(x+h)&=6(x+h)^2+2(x+h)-20\\&=6x^2+12hx+6h^2+2x+2h-20\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{df}{dx}&=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}\\&=\lim_{h\rightarrow0}\frac{(6x^2+12hx+6h^2+2x+2h-20)-(6x^2+2x-20)}{h}\\&=\lim_{h\rightarrow0}\frac{12hx+6h^2+2h}{h}\\&=\lim_{h\rightarrow0}12x+6h+2\\&=12x+2\end{align}
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(b)
(i) \(h(x)=\dfrac{1}{2}\ln(2x+3)+C\), where \(C\) is a constant.
Find \(h'(x)\), the derivative of \(h(x)\).
(ii) The diagram below shows part of the graph of the function \(h'(x)\).
The shaded region in the diagram is between the graph and the \(x\)-axis,
from \(x=0\) to \(x=A\).
This shaded region has an area of \(\mathbf{\ln 3}\) square units. Find the value of \(A\).
(i) \(h'(x)=\dfrac{1}{2x+3}\)
(ii) \(A=12\)
(i)
\begin{align}h'(x)&=\frac{1}{2}\left(\frac{1}{2x+3}\right)(2)\\&=\frac{1}{2x+3}\end{align}
(ii)
\begin{align}\int_0^A\frac{1}{2x+3}\,dx=\ln3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\left.\frac{1}{2}\ln(2x+3)\right|_0^A=\ln3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{1}{2}[\ln(2A+3)-\ln3]=\ln3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{1}{2}\ln\left(\frac{2A+3}{3}\right)=\ln3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\ln\left(\frac{2A+3}{3}\right)^{1/2}=\ln3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\left(\frac{2A+3}{3}\right)^{1/2}=3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{2A+3}{3}=3^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2A+3=27\end{align}
\begin{align}\downarrow\end{align}
\begin{align}A&=\frac{27-3}{2}\\&=12\end{align}
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2020 Paper 1 Question 8
A rectangle is inscribed in a circle of radius \(5\) units and centre \(O(0,0)\) as shown below.
Let \(R(x,y)\), where \(x,y\in\mathbb{R}\), be the vertex of the rectangle in the first quadrant as shown.
Let \(\theta\) be the angle between \([OR]\) and the positive \(x\)-axis, where \(0\leq\theta\leq\dfrac{\pi}{2}\).
(a)
(i) The point \(R(x,y)\) can be written as \(a\cos \theta, b\sin\theta\), where \(a,b\in\mathbb{R}\).
Find the value of \(a\) and the value of \(b\).
(ii) Show that\(A(\theta)\), the area of the rectangle, measured in square units, can be written as \(A(\theta)=50\sin2\theta\).
(iii) Use calculus to show that the rectangle with maximum area is a square.
(iv) Find this maximum area.
(i) \(a=5\) and \(b=5\)
(ii) The answer is already in the question!
(iii) The answer is already in the question!
(iv) \(50\mbox{ square units}\)
(i)
\begin{align}x=5\cos\theta&&y=5\sin\theta\end{align}
\begin{align}\downarrow\end{align}
\(a=5\) and \(b=5\)
(ii)
\begin{align}A(\theta)&=(2\times5\sin\theta)(2\times5\cos\theta)\\&=100\sin\theta\cos\theta\\&=50(2\sin\theta\cos\theta)\\&=50\sin2\theta\end{align}
(iii)
Setting the derivative to zero, we obtain:
\begin{align}(50\cos2\theta)(2)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\cos2\theta=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2\theta&=\cos^{-1}(0)\\&=90^{\circ}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\theta=45^{\circ}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}l&=10\cos45^{\circ}\\&=\frac{10}{\sqrt{2}}\end{align}
and
\begin{align}w&=10\sin45^{\circ}\\&=\frac{10}{\sqrt{2}}\end{align}
As these are the same, the maximum area is a square.
(iv)
\begin{align}A&=l\times w\\&=\left(\frac{10}{\sqrt{2}}\right)\times\left(\frac{10}{\sqrt{2}}\right)\\&=50\mbox{ square units}\end{align}
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(b) A person who is \(2\mbox{ m}\) tall is walking towards a streetlight of height \(5\mbox{ m}\) at a speed of \(1.5\mbox{ m/s}\).
Find the rate, in \(\mbox{m/s}\), at which the length of the person’s shadow (\(x\)), cast by the streetlight, is changing.