Original questions in the style of Papers 1 and 2, each with a full worked mark scheme. Tick questions to build a printable test. Topic 5 at Higher Level →
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Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 14 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →
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1AASL5-q00015.1 Limits and the derivativePaper 1Easy5 marks
Let \(f(x)=x^{2}+3x\). The point \(\text{A}(1,\,4)\) lies on the graph of \(f\). The point \(\text{B}\) also lies on the graph of \(f\) and has \(x\)-coordinate \(1+h\), where \(h\neq0\).
(a)
Show that the gradient of the chord \([\text{AB}]\) is \(5+h\).
(2)
(b)
Write down the limit of the gradient of \([\text{AB}]\) as \(h\to0\).
(1)
(c)
Find \(f'(x)\) and hence verify that \(f'(1)\) is equal to your answer to part (b).
2AASL5-q00025.1 Limits and the derivativePaper 1Easy6 marks
Let \(f(x)=\dfrac{x^{3}-1}{x-1}\), for \(x\neq1\). The following diagram shows part of the graph of \(f\). The open circle shows that the point \((1,\,3)\) does not lie on the graph.
(a)
Write down the value of \(\displaystyle\lim_{x\to1}f(x)\).
(1)
(b)
Explain why \(f(1)\) does not exist.
(1)
(c)
By expanding \((x-1)(x^{2}+x+1)\), show that \(f(x)=x^{2}+x+1\) for \(x\neq1\).
(2)
(d)
Hence find the gradient of the graph of \(f\) at the point where \(x=2\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(3\)
A1
(b)
substituting \(x=1\) gives \(\dfrac{0}{0}\) (division by zero is undefined)
R1
Accept “\(x=1\) is not in the domain” or “the denominator is zero when \(x=1\)”.
(c)
\((x-1)(x^{2}+x+1)=x^{3}+x^{2}+x-x^{2}-x-1\)
M1
\(=x^{3}-1\), so \(\dfrac{x^{3}-1}{x-1}=\dfrac{(x-1)(x^{2}+x+1)}{x-1}\) and the factor \(x-1\neq0\) cancels
5AASL5-q00115.2 Increasing and decreasing functionsPaper 1Easy7 marks
The following diagram shows the graph of a cubic function \(y=f(x)\). The graph has a local maximum point where \(x=-2\) and a local minimum point where \(x=3\).
(a)
Write down the set of values of \(x\) for which \(f'(x)\lt 0\).
(1)
(b)
Write down the value of \(f'(3)\).
(1)
(c)
It is given that \(f'(x)=x^{2}+bx+c\). Find the value of \(b\) and the value of \(c\).
(2)
(d)
Sketch the graph of \(y=f'(x)\), showing clearly the values of any intercepts with the axes.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(-2\lt x\lt 3\)
A1
Award A0 for \(-2\le x\le 3\), since \(f'(-2)=f'(3)=0\).
(b)
\(0\)
A1
(c)
\(f'(x)=(x+2)(x-3)\)
(M1)
Or \(4-2b+c=0\) and \(9+3b+c=0\).
\(b=-1,\ c=-6\)
A1
(d)
concave-up parabola
A1
\(x\)-intercepts \(-2\) and \(3\)
A1
\(y\)-intercept \(-6\)
A1
FT from (c). The vertex \(\left(\tfrac12,\,-\tfrac{25}{4}\right)\) need not be labelled.
6AASL5-q00125.2 Increasing and decreasing functionsPaper 1Easy4 marks
Let \(f(x)=ax^{2}+bx+3\), where \(a,b\in\mathbb{R}\). The function \(f\) is decreasing for \(x\lt 2\) and increasing for \(x\gt 2\). The gradient of the graph of \(f\) at the point where \(x=0\) is \(-8\).
15AASL5-q00465.6 Differentiation rulesPaper 1Easy6 marks
Let \(f(x)=\ln(x^{2}+5)\), for \(x\in\mathbb{R}\).
(a)
Find \(f'(x)\).
(2)
(b)
Find the gradient of the graph of \(f\) at \(x=1\).
(1)
(c)
Let \(h(x)=\ln\left(\dfrac{(x^{2}+5)^{3}}{x}\right)\), for \(x\gt 0\). By first writing \(h(x)\) in terms of \(f(x)\), show that the graph of \(h\) has a stationary point at \(x=1\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
attempt to use the chain rule, e.g. \(\dfrac{1}{x^{2}+5}\times\ldots\)
17AASL5-q00565.7 The second derivative and graphsPaper 1Easy4 marks
The diagram shows part of the graph of \(y=f(x)\). The points P, Q, R and S lie on the graph. Q is a local maximum point and R is a point of inflexion.
Copy and complete the table below, writing “positive”, “negative” or “zero” in each cell.
P
Q
R
S
\(f'(x)\)
\(f''(x)\)
(a)
Complete the row for \(f'(x)\).
(2)
(b)
Complete the row for \(f''(x)\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
at least two of: P positive, Q zero, R negative, S negative
A1
all four correct: positive, zero, negative, negative
A1
(b)
at least two of: P negative, Q negative, R zero, S positive
A1
all four correct: negative, negative, zero, positive
20AASL5-q00605.7 The second derivative and graphsPaper 1Easy6 marks
The diagram shows the graphs labelled A, B and C. One of them is the graph of \(y=f(x)\), one is the graph of \(y=f'(x)\) and one is the graph of \(y=f''(x)\).
(a)
State which graph is the graph of \(f\), which is the graph of \(f'\) and which is the graph of \(f''\).
(2)
(b)
Write down the number of points of inflexion on the graph of \(f\). Give a reason for your answer.
(2)
(c)
Given that \(f(x)=\dfrac{x^{4}}{4}-2x^{2}\), find the exact \(x\)-coordinates of the points of inflexion on the graph of \(f\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
C is the graph of \(f\)
A1
A is the graph of \(f'\) and B is the graph of \(f''\)
A1
Reasoning: the zeros of A are at the turning points of C; the zeros of B are at the turning points of A.
(b)
\(2\)
A1
graph B (\(f''\)) crosses the \(x\)-axis twice, changing sign each time
R1
Accept: graph A (\(f'\)) has two turning points. Do not award A1R0.
(c)
\(f''(x)=3x^{2}-4=0\)
M1
\(x=\pm\dfrac{2}{\sqrt3}\)
A1
Accept \(\pm\dfrac{2\sqrt3}{3}\).
Answers: (a) \(f\): C, \(f'\): A, \(f''\): B · (b) 2 · (c) \(x=\pm\dfrac{2}{\sqrt3}\)
21AASL5-q00615.7 The second derivative and graphsPaper 1Easy5 marks
Let \(y=\mathrm{e}^{-2x}+\sin(2x)\), for \(x\in\mathbb{R}\).
24AASL5-q00735.8 Stationary points, inflexion and optimizationPaper 1Easy8 marks
An open-topped water tank is a cuboid with a square base of side \(x\) metres and height \(h\) metres. Its volume is \(32\ \text{m}^{3}\). Let \(S\) be the area, in m2, of the metal used to make the base and the four sides.
(a)
Write down an expression for \(h\) in terms of \(x\).
(1)
(b)
Show that \(S=x^{2}+\dfrac{128}{x}\).
(2)
(c)
Find \(\dfrac{\mathrm{d}S}{\mathrm{d}x}\).
(1)
(d)
Find the value of \(x\) for which \(S\) is a minimum, and the minimum value of \(S\).
(3)
(e)
Use the second derivative to justify that this value of \(x\) gives a minimum.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(h=\dfrac{32}{x^{2}}\)
A1
(b)
\(S=x^{2}+4xh\)
M1
Base plus four sides; no top.
\(x^{2}+4x\times\dfrac{32}{x^{2}}\)
A1
\(S=x^{2}+\dfrac{128}{x}\)
AG
(c)
\(2x-\dfrac{128}{x^{2}}\)
A1
(d)
\(2x-\dfrac{128}{x^{2}}=0\Rightarrow x^{3}=64\)
M1
\(x=4\)
A1
\(S=16+32=48\ \text{m}^{2}\)
A1
(e)
\(\dfrac{\mathrm{d}^{2}S}{\mathrm{d}x^{2}}=2+\dfrac{256}{x^{3}}\), which is \(6\gt 0\) at \(x=4\), so a minimum
A particle moves in a straight line so that its displacement, \(s\) metres, from a fixed point O after \(t\) seconds is given by \(s(t)=2\sin(3t)\), for \(0\le t\le\pi\).
(a)
Find expressions for the velocity, \(v\), and the acceleration, \(a\), of the particle in terms of \(t\).
(3)
(b)
Show that \(a=-9s\).
(1)
(c)
Find the speed of the particle when \(t=\dfrac{\pi}{9}\).
(2)
(d)
Find the first time \(t\gt 0\) at which the particle is at rest.
(2)
Show mark scheme
Working
Mark
Notes
(a)
attempt to differentiate using the chain rule (factor of 3 seen)
A cyclist moves along a straight path. The graph shows her velocity, \(v\ \text{m s}^{-1}\), after \(t\) seconds, for \(0\le t\le 8\). The graph consists of three straight line segments joining the points \((0,\,0)\), \((2,\,6)\), \((5,\,6)\) and \((8,\,-3)\).
(a)
Write down the acceleration of the cyclist for \(0\lt t\lt 2\).
(1)
(b)
Write down the value of \(t\) at which the cyclist changes direction.
(1)
(c)
Find the distance travelled by the cyclist in the first 7 seconds.
(2)
(d)
Find the total distance travelled by the cyclist in the 8 seconds.
(1)
(e)
Find the displacement of the cyclist from her starting point when \(t=8\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(3\ \text{m s}^{-2}\)
A1
(b)
\(t=7\)
A1
(c)
area under the graph from \(t=0\) to \(t=7\), e.g. \(\tfrac12(7+3)\times6\), or \(6+18+6\)
(M1)
\(30\) m
A1
(d)
\(30+\tfrac12(1)(3)=31.5\) m
A1
FT from (c).
(e)
\(30-1.5=28.5\) m
A1
FT from (c). Award A0 for 31.5.
Answers: (a) \(3\ \text{m s}^{-2}\) · (b) \(t=7\) · (c) 30 m · (d) 31.5 m · (e) 28.5 m
27AASL5-q00935.10 Indefinite integrals and reverse chain rulePaper 1Easy6 marks
30AASL5-q00985.10 Indefinite integrals and reverse chain rulePaper 1Easy5 marks
A function \(f\) has derivative \(f'(x)=4\sin(2x)\), for \(x\in\mathbb{R}\). The graph of \(f\) passes through the point \(\left(\dfrac{\pi}{2},\,5\right)\).
(a)
Find \(f(x)\).
(4)
(b)
Write down the maximum value of \(f\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
attempt to integrate \(f'(x)\)
M1
\(-2\cos(2x)+C\)
A1
\(5=-2\cos\pi+C\)
M1
Substituting the point into an expression containing C.
\(f(x)=3-2\cos(2x)\)
A1
(b)
\(5\)
A1
FT from their f.
Answers: (a) \(f(x)=3-2\cos(2x)\) · (b) 5
31AASL5-q01045.10 Indefinite integrals and reverse chain rulePaper 1Easy5 marks
33AASL5-q01095.11 Definite integrals and areasPaper 1Easy6 marks
The functions \(f\) and \(g\) are continuous for all \(x\in\mathbb{R}\). It is given that \(\displaystyle\int_{1}^{5}f(x)\,\mathrm{d}x=8\), \(\displaystyle\int_{3}^{5}f(x)\,\mathrm{d}x=10\) and \(\displaystyle\int_{1}^{5}g(x)\,\mathrm{d}x=-3\).
34AASL5-q00045.1 Limits and the derivativePaper 1Medium7 marks
Water drains from a tank. The volume of water in the tank, \(V\) litres, \(t\) minutes after draining starts is given by \(V(t)=1600-80t+t^{2}\), for \(0\le t\le 40\).
(a)
Find \(\dfrac{\mathrm{d}V}{\mathrm{d}t}\).
(1)
(b)
Find the value of \(\dfrac{\mathrm{d}V}{\mathrm{d}t}\) when \(t=10\), and interpret its sign in context.
(2)
(c)
Find the time at which the water is draining at a rate of 30 litres per minute.
(2)
(d)
Show that the tank is empty when \(t=40\), and write down the rate at which the volume is changing at that instant.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(-80+2t\)
A1
(b)
\(-60\)
A1
negative, so the volume is decreasing (by 60 litres per minute)
R1
(c)
\(-80+2t=-30\)
M1
Award M0 for \(-80+2t=30\) (this gives \(t=55\), outside the domain).
\(t=25\) (minutes)
A1
(d)
\(V(40)=1600-3200+1600=0\)
A1
\(\dfrac{\mathrm{d}V}{\mathrm{d}t}=-80+80=0\) litres per minute
35AASL5-q00105.2 Increasing and decreasing functionsPaper 1Medium6 marks
The following diagram shows the graph of \(y=f'(x)\), the derivative of a function \(f\), for \(-1.6\le x\le 3.2\). The graph of \(f'\) crosses the \(x\)-axis at \((-1,\,0)\), has a local maximum point at \((0,\,4)\) and touches the \(x\)-axis at \((2,\,0)\).
(a)
Write down the \(x\)-coordinate of the local minimum point on the graph of \(f\). Justify your answer.
(2)
(b)
State, with a reason, the nature of the point on the graph of \(f\) where \(x=2\).
(2)
(c)
Find the set of values of \(x\) for which the graph of \(f\) is concave-down.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(x=-1\)
A1
\(f'(x)\) changes sign from negative to positive at \(x=-1\)
R1
Award R0 for “\(f'(x)=0\)” alone.
(b)
\(f'(2)=0\) but \(f'(x)\) does not change sign (positive on both sides of \(x=2\))
R1
a point of inflexion with zero gradient (stationary point of inflexion)
A1
Award A1 only if R1 is awarded. Also accept a statement that \(f'\) has a local minimum at \(x=2\) so \(f''\) changes sign.
(c)
concave-down where \(f''(x)\lt 0\), i.e. where the graph of \(f'\) is decreasing
R1
\(0\lt x\lt 2\)
A1
Accept \(0\le x\le 2\).
Answers: (a) \(x=-1\) · (b) a point of inflexion with zero gradient · (c) \(0\lt x\lt 2\)
36AASL5-q00135.2 Increasing and decreasing functionsPaper 1Medium6 marks
Let \(f(x)=x^{3}+3x^{2}+5x-7\), for \(x\in\mathbb{R}\).
(a)
Find \(f'(x)\).
(1)
(b)
Show that \(f'(x)=3(x+1)^{2}+2\).
(2)
(c)
Hence explain why \(f\) is increasing for all values of \(x\).
(1)
(d)
By considering \(f(0)\) and \(f(1)\), show that the equation \(f(x)=0\) has exactly one real solution, and that this solution lies between \(0\) and \(1\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(3x^{2}+6x+5\)
A1
(b)
\(3(x^{2}+2x)+5\) or expanding \(3(x+1)^{2}+2=3x^{2}+6x+3+2\)
M1
\(3\big((x+1)^{2}-1\big)+5\) or \(3x^{2}+6x+5\)
A1
\(f'(x)=3(x+1)^{2}+2\)
AG
(c)
\(3(x+1)^{2}\ge0\), so \(f'(x)\ge2\gt 0\) for all \(x\)
R1
(d)
\(f(0)=-7\lt 0\) and \(f(1)=2\gt 0\), so the graph crosses the \(x\)-axis between \(0\) and \(1\)
A1
\(f\) is increasing for all \(x\), so the graph cannot cross the \(x\)-axis more than once
R1
Answers: (a) \(3x^{2}+6x+5\) · (c) \(f'(x)\ge2\gt 0\) · (d) one root, between 0 and 1
37AASL5-q00185.3 Differentiating polynomials and integer powersPaper 1Medium8 marks
Let \(f(x)=\dfrac{x^{3}-4}{2x^{2}}\), for \(x\neq0\).
(a)
Write \(f(x)\) in the form \(ax+bx^{-2}\), where \(a,b\in\mathbb{Q}\).
(1)
(b)
Find \(f'(x)\).
(2)
(c)
Find the coordinates of the point on the graph of \(f\) where the gradient is zero.
(3)
(d)
Find the set of values of \(x\) for which \(f\) is increasing.
45AASL5-q00365.5 Introduction to integrationPaper 1Medium7 marks
The gradient of a curve is given by \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=3x^{2}-8x+k\), where \(k\in\mathbb{R}\). The curve has a horizontal tangent at the point where \(x=2\), and passes through the point \((0,\,5)\).
(a)
Find the value of \(k\).
(2)
(b)
Find the equation of the curve.
(3)
(c)
Find the equation of the tangent to the curve at the point where it meets the \(y\)-axis.
\((\ln x)^{2}\gt 0\); for \(1\lt x\lt\mathrm{e}\), \(\ln x-1\lt 0\) so \(f'(x)\lt 0\); for \(x\gt\mathrm{e}\), \(f'(x)\gt 0\); \(f'\) changes from negative to positive
R1
Accept a correct sign diagram, or a valid second-derivative argument.
(d)
\(f\left(\mathrm{e}^{2}\right)=\dfrac{\mathrm{e}^{2}}{2}\) and \(f'\left(\mathrm{e}^{2}\right)=\dfrac{2-1}{4}=\dfrac14\)
51AASL5-q00595.7 The second derivative and graphsPaper 1Medium6 marks
The function \(f\) is defined for \(x\in\mathbb{R}\). The diagram shows the graph of its second derivative, \(y=f''(x)\). The graph of \(f''\) crosses the \(x\)-axis at \(x=-2\) and touches it at \(x=1\).
(a)
Write down the \(x\)-coordinate of the point of inflexion on the graph of \(f\).
(1)
(b)
Explain why the graph of \(f\) does not have a point of inflexion at \(x=1\).
(1)
(c)
Write down the set of values of \(x\) for which the graph of \(f\) is concave-down.
(1)
(d)
It is given that \(f'(1)=0\). Determine the nature of the stationary point on the graph of \(f\) at \(x=1\). Justify your answer.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(x=-2\)
A1
(b)
\(f''(x)\ge0\) on both sides of \(x=1\): \(f''\) does not change sign, so there is no change of concavity
R1
(c)
\(x\lt -2\)
A1
(d)
\(f''(x)\gt 0\) for \(-2\lt x\lt1\) and \(x\gt1\), so \(f'\) is increasing on an interval containing \(x=1\)
R1
as \(f'(1)=0\), \(f'(x)\lt 0\) just to the left of \(x=1\) and \(f'(x)\gt 0\) just to the right
R1
local minimum
A1
Award A1 only if at least one R1 is awarded. The second-derivative test fails here, since \(f''(1)=0\).
53AASL5-q00675.8 Stationary points, inflexion and optimizationPaper 1Medium8 marks
Let \(f(x)=3x^{4}-4x^{3}\), for \(x\in\mathbb{R}\). The graph of \(f\) has stationary points at \(x=0\) and \(x=1\).
(a)
Show that \(f'(x)=12x^{2}(x-1)\).
(2)
(b)
Find \(f''(x)\) and hence show that \((1,\,-1)\) is a local minimum point.
(2)
(c)
Given that \(f''(0)=0\), determine the nature of the stationary point at \(x=0\). Justify your answer.
(2)
(d)
Find the \(x\)-coordinate of the point of inflexion on the graph of \(f\) that has a non-zero gradient.
(2)
Show mark scheme
Working
Mark
Notes
(a)
attempt to differentiate, e.g. one term correct
M1
\(f'(x)=12x^{3}-12x^{2}\)
A1
Must see this expanded form before the factorized answer.
\(f'(x)=12x^{2}(x-1)\)
AG
(b)
\(f''(x)=36x^{2}-24x\)
A1
\(f''(1)=12\gt 0\) (and \(f(1)=3-4=-1\)), so a local minimum
R1
(c)
\(f'(x)\lt 0\) on both sides of \(x=0\), e.g. \(f'(-1)=-24\), \(f'(0.5)=-1.5\) (or \(f''\) changes sign at \(x=0\))
R1
Since \(12x^{2}\ge0\) and \(x-1\lt 0\) near \(x=0\).
a point of inflexion with zero gradient
A1
Award A1 only if R1 is awarded.
(d)
\(12x(3x-2)=0\)
M1
Setting their \(f''(x)=0\).
\(x=\tfrac{2}{3}\)
A1
\(f''\) changes sign at \(x=\tfrac23\) as it is a single root.
Answers: (b) \(f''(x)=36x^{2}-24x\) · (c) point of inflexion with zero gradient · (d) \(x=\tfrac23\)
54AASL5-q00685.8 Stationary points, inflexion and optimizationPaper 1Medium7 marks
Let \(f(x)=x^{3}+bx^{2}+cx+d\), where \(b,\,c,\,d\in\mathbb{R}\). The graph of \(f\) has a point of inflexion at \((1,\,3)\), and the gradient of the graph at this point is \(-2\).
(a)
Find \(f'(x)\) and \(f''(x)\).
(2)
(b)
Show that \(b=-3\).
(1)
(c)
Find the value of \(c\) and the value of \(d\).
(3)
(d)
Write down the set of values of \(x\) for which the graph of \(f\) is concave-down.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(f'(x)=3x^{2}+2bx+c\)
A1
\(f''(x)=6x+2b\)
A1
(b)
\(f''(1)=0\Rightarrow6+2b=0\)
A1
\(b=-3\)
AG
(c)
\(f'(1)=-2\Rightarrow3-6+c=-2\)
M1
\(c=1\)
A1
\(f(1)=1-3+1+d=3\Rightarrow d=4\)
A1
FT from their \(c\).
(d)
\(x\lt 1\)
A1
\(f''(x)=6x-6\lt 0\).
Answers: (c) \(c=1,\ d=4\) · (d) \(x\lt1\)
55AASL5-q00715.8 Stationary points, inflexion and optimizationPaper 1Medium9 marks
Let \(f(x)=2x+4\cos x\), for \(0\le x\le2\pi\).
(a)
Find \(f'(x)\).
(1)
(b)
Find the \(x\)-coordinates of the stationary points on the graph of \(f\).
(3)
(c)
Use the second derivative to determine the nature of each stationary point.
(3)
(d)
Find the greatest value of \(f(x)\) for \(0\le x\le2\pi\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(f'(x)=2-4\sin x\)
A1
(b)
\(2-4\sin x=0\Rightarrow\sin x=\dfrac12\)
M1
\(x=\dfrac{\pi}{6}\)
A1
\(x=\dfrac{5\pi}{6}\)
A1
(c)
\(f''(x)=-4\cos x\)
A1
\(f''\left(\tfrac{\pi}{6}\right)=-2\sqrt3\lt 0\), so a local maximum at \(x=\tfrac{\pi}{6}\)
R1
\(f''\left(\tfrac{5\pi}{6}\right)=2\sqrt3\gt 0\), so a local minimum at \(x=\tfrac{5\pi}{6}\)
R1
(d)
comparing the local maximum \(f\left(\tfrac{\pi}{6}\right)=\tfrac{\pi}{3}+2\sqrt3\) with the endpoint value \(f(2\pi)=4\pi+4\)
M1
\(f\) is increasing for \(\tfrac{5\pi}{6}\lt x\le2\pi\).
\(4\pi+4\)
A1
Award M0A0 for \(\tfrac{\pi}{3}+2\sqrt3\).
Answers: (b) \(x=\dfrac{\pi}{6},\ \dfrac{5\pi}{6}\) · (c) local max at \(\tfrac{\pi}{6}\), local min at \(\tfrac{5\pi}{6}\) · (d) \(4\pi+4\)
56AASL5-q00745.8 Stationary points, inflexion and optimizationPaper 1Medium9 marks
Let \(f(x)=x\mathrm{e}^{-x}\), for \(x\in\mathbb{R}\).
(a)
Find \(f'(x)\).
(2)
(b)
Show that \(f''(x)=(x-2)\mathrm{e}^{-x}\).
(1)
(c)
Show that the graph of \(f\) has a point of inflexion at \(x=2\).
(2)
(d)
Find the equation of the tangent to the graph of \(f\) at the point of inflexion.
(3)
(e)
Write down the \(x\)-coordinate of the point where this tangent meets the \(x\)-axis.
A particle moves in a straight line. Its displacement, \(s\) metres, from a fixed point O after \(t\) seconds is given by \(s(t)=t^{3}-6t^{2}+9t\), for \(0\le t\le5\).
(a)
Find an expression for the velocity, \(v\), of the particle.
(1)
(b)
Find the values of \(t\) when the particle is at rest.
(2)
(c)
Find the total distance travelled by the particle in the 5 seconds.
(3)
(d)
Find the speed of the particle when its acceleration is zero.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(v=3t^{2}-12t+9\)
A1
(b)
\(3(t-1)(t-3)=0\)
M1
\(t=1,\ t=3\)
A1
(c)
attempt to find \(s\) at \(t=0,1,3,5\): \(0,\ 4,\ 0,\ 20\)
(M1)
Or \(\int_{0}^{5}|v|\,\mathrm{d}t\) split at \(t=1\) and \(t=3\).
A particle moves in a straight line. Its acceleration, in m s−2, after \(t\) seconds is given by \(a(t)=4-2t\), for \(0\le t\le 6\). When \(t=0\) its velocity is \(5\ \text{m s}^{-1}\).
(a)
Find an expression for the velocity, \(v(t)\).
(3)
(b)
Find the value of \(t\) at which the particle changes direction.
(2)
(c)
Find the total distance travelled by the particle between \(t=0\) and \(t=6\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
attempt to integrate \(a(t)\)
M1
\(4t-t^{2}+c\)
A1
\(v(t)=5+4t-t^{2}\)
A1
(b)
\(5+4t-t^{2}=0\Rightarrow(5-t)(1+t)=0\)
M1
\(t=5\)
A1
Must reject \(t=-1\). \(v\) changes sign at \(t=5\) since \(v(6)=-7\).
Or \(\int_{0}^{6}|v|\,\mathrm{d}t\) split at their \(t\).
\(\dfrac{100}{3}\) and \(\dfrac{10}{3}\)
A1
Using \(5t+2t^{2}-\tfrac{t^{3}}{3}\): values \(\tfrac{100}{3}\) at \(t=5\) and \(30\) at \(t=6\).
\(\dfrac{110}{3}\) m
A1
Accept \(36\tfrac23\) m.
Answers: (a) \(v=5+4t-t^{2}\) · (b) \(t=5\) · (c) \(\tfrac{110}{3}\) m
59AASL5-q00835.9 KinematicsPaper 1Medium8 marks
A particle starts from a fixed point O and moves in a straight line. Its velocity, \(v\ \text{m s}^{-1}\), after \(t\) seconds is given by \(v(t)=4-\mathrm{e}^{t}\), for \(0\le t\le 2\).
(a)
Find the exact value of \(t\) at which the particle is instantaneously at rest.
(2)
(b)
Show that the displacement of the particle from O at this time is \(8\ln 2-3\) metres.
(3)
(c)
Find the exact total distance travelled by the particle in the 2 seconds.
Must see \(\mathrm{e}^{\ln 4}=4\) used and \(4\ln 4=8\ln 2\).
\(=8\ln 2-3\)
AG
(c)
displacement at \(t=2\): \(\displaystyle\int_{0}^{2}\left(4-\mathrm{e}^{t}\right)\mathrm{d}t=9-\mathrm{e}^{2}\); or \(\displaystyle\int_{\ln 4}^{2}\left(\mathrm{e}^{t}-4\right)\mathrm{d}t=\mathrm{e}^{2}-12+8\ln 2\)
(M1)
distance \(=2(8\ln 2-3)-(9-\mathrm{e}^{2})\), or \((8\ln 2-3)+(\mathrm{e}^{2}-12+8\ln 2)\)
(M1)
Adding the two stages as positive quantities.
\(\mathrm{e}^{2}+16\ln 2-15\) m
A1
\(=3.479\ldots\) Accept equivalent exact forms, e.g. \(\mathrm{e}^{2}+8\ln 4-15\).
Answers: (a) \(t=\ln 4\) · (c) \(\mathrm{e}^{2}+16\ln 2-15\) m
60AASL5-q00845.9 KinematicsPaper 1Medium9 marks
A particle moves in a straight line, starting from a fixed point O. Its velocity, \(v\ \text{m s}^{-1}\), after \(t\) seconds is given by \(v(t)=3t^{2}-kt\), for \(0\le t\le 4\), where \(k\) is a positive constant. When \(t=4\) the particle is back at O.
(a)
Show that \(k=8\).
(2)
(b)
Find the value of \(t\), for \(0\lt t\lt 4\), at which the particle changes direction.
(2)
(c)
Find the total distance travelled by the particle in the 4 seconds.
66AASL5-q01065.11 Definite integrals and areasPaper 1Medium8 marks
Let \(f(x)=x^{3}-2x^{2}-3x\). The diagram shows part of the graph of \(f\). The shaded region A lies above the \(x\)-axis and the shaded region B lies below the \(x\)-axis.
(a)
Find the \(x\)-coordinates of the points where the graph of \(f\) meets the \(x\)-axis.
(2)
(b)
Find the area of region A.
(2)
(c)
Find the area of region B.
(3)
(d)
Hence write down the value of \(\displaystyle\int_{-1}^{3}f(x)\,\mathrm{d}x\).
Correct expression for the area must be seen; FT from their (a).
\(1\) and \(\left|-\tfrac14\right|=\tfrac14\)
A1
\(\dfrac54\)
A1
(d)
between \(\tfrac{\pi}{2}\) and \(\tfrac{2\pi}{3}\) the curve is below the \(x\)-axis, so that part contributes a negative value to the integral but a positive value to the area
79AASL5-q00725.8 Stationary points, inflexion and optimizationPaper 1Hard11 marks
The point \(\text{P}\left(a,\,\mathrm{e}^{-a}\right)\), where \(a\ge0\), lies on the curve \(y=\mathrm{e}^{-x}\). The line L is the tangent to the curve at P. The triangle formed by L and the coordinate axes is shaded in the diagram.
Diagram not to scale
(a)
Show that the equation of L is \(y=-\mathrm{e}^{-a}x+(a+1)\mathrm{e}^{-a}\).
(2)
(b)
Find, in terms of \(a\), the coordinates of the points where L meets the \(x\)-axis and the \(y\)-axis.
(2)
(c)
Hence show that the area of the shaded triangle is \(T=\dfrac12(a+1)^{2}\mathrm{e}^{-a}\).
(1)
(d)
Find \(\dfrac{\mathrm{d}T}{\mathrm{d}a}\).
(2)
(e)
Find the exact maximum value of \(T\), justifying that it is a maximum.
\(\sin^{3}(x+\pi)=-\sin^{3}x\), so the part of the graph from \(\pi\) to \(2\pi\) is below the axis with the same area as the part from \(0\) to \(\pi\)
R1
Accept \(\int_{\pi}^{2\pi}\sin^{3}x\,\mathrm{d}x=-\tfrac43\) found by symmetry or by their (b). Do not award A1R0.
82AASL5-q01055.10 Indefinite integrals and reverse chain rulePaper 1Hard15 marks
Let \(f(x)=2x\,\mathrm{e}^{-x^{2}}\), for \(x\in\mathbb{R}\).
(a)
Show that \(f'(x)=2\left(1-2x^{2}\right)\mathrm{e}^{-x^{2}}\).
(2)
(b)
Find the exact coordinates of the local maximum point on the graph of \(f\).
(3)
(c)
Find \(\displaystyle\int f(x)\,\mathrm{d}x\).
(2)
(d)
Let \(A(k)\) be the area of the region enclosed by the graph of \(f\), the \(x\)-axis and the line \(x=k\), where \(k\gt 0\). Show that \(A(k)=1-\mathrm{e}^{-k^{2}}\).
(2)
(e)
Find the exact value of \(k\) for which \(A(k)=\dfrac12\).
(2)
(f)
Explain why \(A(k)\lt 1\) for all \(k\gt 0\).
(1)
(g)
Hence find the exact value of \(\displaystyle\int_{-1}^{1}|f(x)|\,\mathrm{d}x\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
attempt to use the product rule, with \(\dfrac{\mathrm{d}}{\mathrm{d}x}\mathrm{e}^{-x^{2}}=-2x\,\mathrm{e}^{-x^{2}}\)
84AASL5-q01115.11 Definite integrals and areasPaper 1Hard10 marks
Let \(f'(x)=\tfrac12x(x-3)(x-5)=\tfrac12\left(x^{3}-8x^{2}+15x\right)\), for \(0\le x\le5\). The diagram shows the graph of \(y=f'(x)\). Region A is enclosed by the graph of \(f'\) and the \(x\)-axis for \(0\le x\le3\), and region B is enclosed by the graph of \(f'\) and the \(x\)-axis for \(3\le x\le5\).
It is given that \(f(0)=2\) and that the area of region B is \(\dfrac{8}{3}\).
(a)
Find the \(x\)-coordinate of the local maximum point on the graph of \(f\). Justify your answer.
(2)
(b)
Show that the area of region A is \(\dfrac{63}{8}\).
(3)
(c)
Find the value of \(f(3)\) and the value of \(f(5)\).
(3)
(d)
Find the least value of \(f(x)\) for \(0\le x\le5\). Justify your answer.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(x=3\)
A1
\(f'\) changes sign from positive to negative at \(x=3\)
Or a symmetry argument: \(6x-x^{2}=9-(x-3)^{2}\) is symmetric about \(x=3\).
\(27-9=18=\tfrac12(36)\)
A1
so \(x=3\) divides R into two regions of equal area
AG
Answers: (a) \(x=0,\ x=k\) · (c) \(k=6\)
86AASL5-q01185.11 Definite integrals and areasPaper 1Hard16 marks
Let \(f(x)=\dfrac{4x}{x^{2}+1}\), for \(x\in\mathbb{R}\). The diagram shows part of the graph of \(f\) and the line L, which is the tangent to the graph of \(f\) at the origin.
(a)
Show that \(f'(x)=\dfrac{4(1-x^{2})}{(x^{2}+1)^{2}}\).
(3)
(b)
Find the coordinates of the local maximum point and of the local minimum point on the graph of \(f\).
(3)
(c)
The second derivative is \(f''(x)=\dfrac{8x(x^{2}-3)}{(x^{2}+1)^{3}}\). Find the \(x\)-coordinates of the points of inflexion on the graph of \(f\). Justify your answer.
(3)
(d)
Find the equation of L.
(2)
(e)
The shaded region is enclosed by L, the graph of \(f\) and the line \(x=1\). Find the exact area of the shaded region.
Answers: (b) max \((1,2)\), min \((-1,-2)\) · (c) \(x=0,\ \pm\sqrt3\) · (d) \(y=4x\) · (e) \(2-2\ln2\)
87AASL5-q01195.11 Definite integrals and areasPaper 1Hard16 marks
Let \(f(x)=x\sqrt{9-x^{2}}\), for \(0\le x\le3\). The diagram shows the graph of \(f\) and the line \(y=x\). The graph of \(f\) has a maximum point at M. The line meets the graph of \(f\) at the origin O and at the point P. The shaded region S is enclosed by the graph of \(f\) and the line.
(a)
Show that \(f'(x)=\dfrac{9-2x^{2}}{\sqrt{9-x^{2}}}\), for \(0\le x\lt 3\).
88AASL5-q00055.1 Limits and the derivativePaper 2Easy5 marks
Let \(g(h)=\dfrac{3^{h}-1}{h}\), where \(h\neq0\). The table shows some values of \(g(h)\), correct to four decimal places.
\(h\)
\(-0.1\)
\(-0.01\)
\(-0.001\)
\(0.001\)
\(0.01\)
\(0.1\)
\(g(h)\)
1.0404
\(p\)
1.0980
1.0992
\(q\)
1.1612
(a)
Find the value of \(p\) and the value of \(q\).
(2)
(b)
Use the table to write down an estimate, correct to three significant figures, for \(\displaystyle\lim_{h\to0}\dfrac{3^{h}-1}{h}\).
(1)
(c)
The value of \(\dfrac{3^{h}-1}{h}\) is the gradient of the chord joining the points \((0,\,1)\) and \((h,\,3^{h})\) on the curve \(y=3^{x}\). State what the limit in part (b) represents for the curve \(y=3^{x}\).
(1)
(d)
Hence write down the equation of the tangent to \(y=3^{x}\) at the point \((0,\,1)\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(p=1.0926\)
A1
\(1.092599\ldots\)
\(q=1.1047\)
A1
\(1.104669\ldots\) Award A1A0 for both values correct but not to four decimal places.
(b)
\(1.10\)
A1
Do not accept \(1.099\) or \(1.1\) (not to three significant figures). The exact value is \(\ln 3=1.0986\ldots\), which is not required.
(c)
the gradient of the tangent to the curve at \((0,\,1)\)
A1
Accept “the gradient of the curve at \(x=0\)” or “the instantaneous rate of change of \(3^{x}\) at \(x=0\)”.
(d)
\(y=1.10x+1\)
A1
FT from (b). Accept \(y=1.1x+1\) or \(y=(\ln3)x+1\).
Answers: (a) \(p=1.0926\), \(q=1.1047\) · (b) \(1.10\) · (c) the gradient of the tangent at \((0,1)\) · (d) \(y=1.10x+1\)
89AASL5-q00075.1 Limits and the derivativePaper 2Easy6 marks
A ball is thrown vertically upwards. Its height, \(s\) metres, above the point of release after \(t\) seconds is given by \(s(t)=20t-4.9t^{2}\). The average velocity of the ball between \(t=1\) and \(t=1+h\) is \(\dfrac{s(1+h)-s(1)}{h}\). The table shows some values of this average velocity.
\(h\)
\(0.1\)
\(0.01\)
\(0.001\)
Average velocity (m s−1)
9.71
\(p\)
\(q\)
(a)
Show that the average velocity between \(t=1\) and \(t=1.1\) is \(9.71\) m s−1.
(2)
(b)
Find the value of \(p\) and the value of \(q\).
(2)
(c)
Use the table to estimate the velocity of the ball at the instant \(t=1\).
(1)
(d)
Write down, using derivative notation, the quantity that you have estimated in part (c).
90AASL5-q00385.5 Introduction to integrationPaper 2Easy5 marks
Let \(f(x)=2+3x-x^{2}\). The graph of \(f\) meets the \(x\)-axis at the points where \(x=p\) and \(x=q\), where \(p\lt q\).
(a)
Find the value of \(p\) and the value of \(q\).
(2)
(b)
Write down an integral that represents the area of the region enclosed by the graph of \(f\) and the \(x\)-axis.
(1)
(c)
Find the area of this region.
(1)
(d)
Explain why \(\displaystyle\int_{0}^{5}f(x)\,\mathrm{d}x\) does not give the area of the region enclosed by the graph of \(f\), the \(x\)-axis and the lines \(x=0\) and \(x=5\).
\(f(x)\lt 0\) for \(3.56\lt x\le 5\), so part of the region is below the \(x\)-axis and that part contributes a negative value to the integral
R1
Answers: (a) \(p=-0.562\), \(q=3.56\) · (c) \(11.7\) · (d) \(f(x)\lt 0\) for part of the interval
91AASL5-q00865.9 KinematicsPaper 2Easy7 marks
A particle moves in a straight line. Its velocity, \(v\ \text{m s}^{-1}\), after \(t\) seconds is given by \(v(t)=\mathrm{e}^{0.5t}-3\), for \(0\le t\le 4\).
(a)
Write down the initial velocity of the particle.
(1)
(b)
Find the value of \(t\) at which the particle changes direction.
(2)
(c)
Find the value of \(t\), for \(t\gt 0\), at which the speed of the particle is again equal to its initial speed.
(2)
(d)
Find the total distance travelled by the particle in the 4 seconds.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(-2\ \text{m s}^{-1}\)
A1
(b)
\(\mathrm{e}^{0.5t}-3=0\), or graph of \(v\) crossing the \(t\)-axis
(M1)
\(t=2.20\)
A1
\(2\ln 3=2.19722\ldots\)
(c)
\(\mathrm{e}^{0.5t}-3=2\)
(M1)
Recognizing that the velocity must now be \(+2\); \(v=-2\) only at \(t=0\).
92AASL5-q00065.1 Limits and the derivativePaper 2Medium6 marks
The number of bees, \(N\), in a colony \(t\) weeks after it is established is modelled by \(N(t)=\dfrac{2000}{1+49\mathrm{e}^{-0.6t}}\), for \(t\ge0\).
(a)
Write down the initial number of bees in the colony.
(1)
(b)
Find the rate at which the number of bees is increasing when \(t=4\).
(2)
(c)
Find the value of \(t\) at which the number of bees is increasing most rapidly.
(2)
(d)
Write down the number of bees in the colony at this time.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(40\)
A1
(b)
\(N'(4)\), e.g. GDC derivative at \(t=4\)
(M1)
\(180\) bees per week
A1
\(179.906\ldots\) Award A0 for \(N(4)=367\).
(c)
maximum of the graph of \(\dfrac{\mathrm{d}N}{\mathrm{d}t}\) on GDC
93AASL5-q00085.1 Limits and the derivativePaper 2Medium7 marks
A furniture maker produces \(x\) tables per week. The weekly cost, in euros, of producing \(x\) tables is modelled by \(C(x)=0.001x^{3}-0.3x^{2}+45x+2000\), for \(0\le x\le 250\). The marginal cost is defined as \(C'(x)\).
(a)
Find \(C'(x)\).
(1)
(b)
Find \(C'(150)\) and interpret this value in context.
(2)
(c)
Find the actual increase in the weekly cost when production increases from 150 to 151 tables, and comment on your answer in relation to part (b).
(2)
(d)
Find the values of \(x\) for which the marginal cost is €20 per table.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(C'(x)=0.003x^{2}-0.6x+45\)
A1
(b)
\(C'(150)=22.5\)
A1
when 150 tables are made, the cost is increasing at about €22.50 per extra table
R1
Must refer to rate of change of cost per table.
(c)
\(C(151)-C(150)=5397.651-5375=\)€\(22.65\)
A1
\(22.651\) Accept \(22.7\).
this is close to \(C'(150)\), so the marginal cost is a good estimate of the cost of one extra table
R1
(d)
\(0.003x^{2}-0.6x+45=20\), e.g. graph of \(C'\) and \(y=20\)
(M1)
\(x=59.2\) or \(x=141\)
A1
\(59.1751\ldots,\ 140.824\ldots\) Accept \(59\) and \(141\). Award A0 for only one value.
94AASL5-q00165.2 Increasing and decreasing functionsPaper 2Medium8 marks
The height, \(h\) metres, of a drone above the ground \(t\) seconds after it is switched on is modelled by \(h(t)=0.0004t^{3}-0.036t^{2}+0.8t+5\), for \(0\le t\le 60\).
(a)
Find \(h'(t)\).
(1)
(b)
Find \(h'(10)\) and state what it tells you about the motion of the drone at that time.
(2)
(c)
Find the interval of time during which the drone is descending.
(3)
(d)
Find the height of the drone at the instant it starts to descend.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(h'(t)=0.0012t^{2}-0.072t+0.8\)
A1
(b)
\(0.2\)
A1
positive, so the drone is rising (at 0.2 m s−1)
R1
(c)
\(h'(t)\lt 0\), e.g. solving \(0.0012t^{2}-0.072t+0.8=0\)
96AASL5-q00375.5 Introduction to integrationPaper 2Medium5 marks
Let \(f(x)=1+\ln(x^{2}+1)\), for \(x\ge0\). The shaded region R is enclosed by the graph of \(f\), the \(x\)-axis, the \(y\)-axis and the line \(x=3\).
(a)
Write down an integral that represents the area of R.
(1)
(b)
Find the area of R.
(1)
(c)
The line \(x=k\) divides R into two regions of equal area. Find the value of \(k\).
\(1.91052\ldots\) Award A1 only for a value with \(0\lt k\lt 3\).
Answers: (b) \(6.41\) · (c) \(k=1.91\)
97AASL5-q00395.5 Introduction to integrationPaper 2Medium7 marks
Let \(f(x)=kx(4-x)^{2}\), where \(k\gt 0\). The following diagram shows part of the graph of \(f\). The shaded region R is enclosed by the graph of \(f\) and the \(x\)-axis, for \(0\le x\le 4\).
Diagram not to scale
(a)
Write down an integral, in terms of \(k\), that represents the area of R.
(1)
(b)
The area of R is 32 square units. Find the value of \(k\).
(3)
(c)
By expanding \(f(x)\), find the equation of the tangent to the graph of \(f\) at the origin.
98AASL5-q00405.5 Introduction to integrationPaper 2Medium14 marks
During a storm, the rate of rainfall, \(r\) millimetres per hour, \(t\) hours after the storm begins is modelled by \(r(t)=0.1t^{2}(9-t)\), for \(0\le t\le 9\).
(a)
Find the rate of rainfall when \(t=5\).
(1)
(b)
Find \(r'(t)\).
(2)
(c)
Find the interval of time during which the rate of rainfall is increasing.
(3)
(d)
Find \(r'(7)\) and interpret your answer in context.
(2)
(e)
Find the total rainfall during the storm.
(2)
(f)
Find the rainfall during the last three hours of the storm.
(2)
(g)
Find the time at which half of the total rainfall has fallen.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(10\) mm per hour
A1
(b)
\(r(t)=0.9t^{2}-0.1t^{3}\) or \(1.8t\)
A1
Award for expanding correctly or for one correct term of \(r'(t)\).
\(r'(t)=1.8t-0.3t^{2}\)
A1
(c)
\(1.8t-0.3t^{2}=0\) or \(r'(t)\gt 0\)
M1
\(0.3t(6-t)=0\Rightarrow t=0,\ t=6\)
A1
\(0\lt t\lt 6\)
A1
Accept \(0\le t\le 6\).
(d)
\(r'(7)=-2.1\)
A1
the rate of rainfall is decreasing at 2.1 mm per hour, per hour
R1
Must refer to the rate of rainfall decreasing; units not required.
Answers: (a) 10 mm h−1 · (b) \(1.8t-0.3t^{2}\) · (c) \(0\lt t\lt 6\) · (d) \(-2.1\) · (e) 54.7 mm · (f) 22.3 mm · (g) \(5.53\) h
99AASL5-q00535.6 Differentiation rulesPaper 2Medium7 marks
The height, \(h\) metres, of a seat on a Ferris wheel above the ground \(t\) minutes after it passes the lowest point is modelled by \(h(t)=36-32\cos\left(\dfrac{\pi t}{15}\right)\), for \(0\le t\le30\).
(a)
Find \(h'(t)\).
(2)
(b)
Find the rate at which the height of the seat is changing when \(t=4\).
(1)
(c)
Find the maximum rate at which the height of the seat increases, and the value of \(t\) at which it occurs.
(2)
(d)
Find the values of \(t\) for which the height of the seat is increasing at a rate of 5 metres per minute.
(2)
Show mark scheme
Working
Mark
Notes
(a)
attempt to use the chain rule, e.g. \(\sin\left(\tfrac{\pi t}{15}\right)\times\tfrac{\pi}{15}\)
\(\dfrac{32\pi}{15}\sin\left(\dfrac{\pi t}{15}\right)=5\), e.g. intersection on GDC
(M1)
\(t=4.02\) and \(t=11.0\)
A1
\(4.02070\ldots\), \(10.9793\ldots\) Both values required.
Answers: (a) \(\dfrac{32\pi}{15}\sin\left(\dfrac{\pi t}{15}\right)\) · (b) 4.98 m min−1 · (c) 6.70 m min−1 at \(t=7.5\) · (d) \(t=4.02,\ 11.0\)
100AASL5-q00545.6 Differentiation rulesPaper 2Medium9 marks
Let \(f(x)=4\ln x\) and \(g(x)=\mathrm{e}^{\frac{x}{2}}\), for \(x\gt 0\). The graphs of \(f\) and \(g\) intersect at \(x=a\) and \(x=b\), where \(a\lt b\).
(a)
Find \(f'(x)\) and \(g'(x)\).
(2)
(b)
Find the value of \(x\) at which the tangents to the two graphs are parallel.
(2)
(c)
Find the value of \(a\) and the value of \(b\).
(2)
(d)
For \(a\le x\le b\), let \(d(x)=f(x)-g(x)\). Explain why \(d\) has its maximum value at your answer to part (b), and find this maximum value.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(f'(x)=\dfrac{4}{x}\)
A1
\(g'(x)=\dfrac12\mathrm{e}^{\frac{x}{2}}\)
A1
(b)
\(\dfrac{4}{x}=\dfrac12\mathrm{e}^{\frac{x}{2}}\), e.g. intersection on GDC
(M1)
\(x=2.40\)
A1
\(2.40433\ldots\)
(c)
\(a=1.92\)
A1
\(1.92306\ldots\)
\(b=2.89\)
A1
\(2.89429\ldots\)
(d)
\(d'(x)=f'(x)-g'(x)=0\) where the tangents are parallel
R1
Accept: this is the only stationary point of \(d\), and \(d(a)=d(b)=0\) with \(d\gt 0\) between.
102AASL5-q00755.8 Stationary points, inflexion and optimizationPaper 2Medium8 marks
The diagram shows part of the graph of \(y=6\cos\left(\dfrac{x}{2}\right)\), for \(-\pi\le x\le\pi\). The rectangle ABCD has vertices \(\text{A}(-x,\,0)\) and \(\text{B}(x,\,0)\) on the \(x\)-axis, where \(0\lt x\lt\pi\), and vertices C and D on the curve.
(a)
Show that the area of the rectangle is \(A=12x\cos\left(\dfrac{x}{2}\right)\).
(2)
(b)
Find the maximum area of the rectangle.
(3)
(c)
Find the maximum perimeter of the rectangle.
(3)
Show mark scheme
Working
Mark
Notes
(a)
width \(\text{AB}=2x\)
A1
height \(\text{BC}=6\cos\left(\tfrac{x}{2}\right)\), so \(A=2x\times6\cos\left(\tfrac{x}{2}\right)\)
A1
\(A=12x\cos\left(\tfrac{x}{2}\right)\)
AG
(b)
use of GDC to find the maximum of \(A\), e.g. graph, or solving \(\dfrac{\mathrm{d}A}{\mathrm{d}x}=0\)
use of GDC to find the maximum of \(P\) (at \(x=1.45946\ldots\))
(M1)
Or \(\dfrac{\mathrm{d}P}{\mathrm{d}x}=4-6\sin\left(\tfrac{x}{2}\right)=0\).
maximum perimeter \(=14.8\)
A1
\(14.7821\ldots\)
Answers: (b) \(13.5\) · (c) \(14.8\)
103AASL5-q00775.8 Stationary points, inflexion and optimizationPaper 2Medium8 marks
A small factory makes \(x\) bicycles per month, where \(0\le x\le100\). The monthly profit, in dollars, is modelled by \(P(x)=-0.01x^{3}+0.6x^{2}+30x-400\).
(a)
Find \(P'(x)\).
(1)
(b)
Find the value of \(x\) for which \(P'(x)=0\).
(2)
(c)
Use the second derivative to show that this value of \(x\) gives a maximum.
(2)
(d)
The factory can only make a whole number of bicycles. Find the number of bicycles that maximizes the monthly profit, and the maximum monthly profit.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(P'(x)=-0.03x^{2}+1.2x+30\)
A1
(b)
\(-0.03x^{2}+1.2x+30=0\), e.g. GDC
(M1)
Equivalent to \(x^{2}-40x-1000=0\).
\(x=57.4\)
A1
\(57.4165\ldots\ \left(=20+\sqrt{1400}\right)\); reject the negative root.
104AASL5-q00785.8 Stationary points, inflexion and optimizationPaper 2Medium9 marks
A lorry travels 600 km at a constant speed of \(v\) km h−1. The driver is paid $24 per hour. The cost of fuel is \(0.00006v^{2}\) dollars per kilometre. Let \(C\) be the total cost, in dollars, of the journey.
(a)
Show that \(C(v)=\dfrac{14\,400}{v}+0.036v^{2}\).
(2)
(b)
Find the speed for which the total cost is a minimum.
(3)
(c)
Justify that this speed gives a minimum.
(1)
(d)
Find the minimum total cost.
(1)
(e)
Regulations require the lorry to travel at a speed between 70 km h−1 and 90 km h−1. Find the minimum total cost under these regulations. Justify your answer.
(2)
Show mark scheme
Working
Mark
Notes
(a)
time \(=\dfrac{600}{v}\) hours, so driver’s pay \(=24\times\dfrac{600}{v}=\dfrac{14\,400}{v}\)
A1
fuel \(=600\times0.00006v^{2}=0.036v^{2}\)
A1
\(C(v)=\dfrac{14\,400}{v}+0.036v^{2}\)
AG
(b)
\(C'(v)=-\dfrac{14\,400}{v^{2}}+0.072v=0\)
M1
Or minimum of \(C\) on GDC.
\(v^{3}=200\,000\)
A1
\(v=58.5\) km h−1
A1
\(58.4803\ldots\)
(c)
\(C''(v)=\dfrac{28\,800}{v^{3}}+0.072\gt 0\) for \(v\gt 0\)
R1
Or \(C'\) changes from negative to positive.
(d)
$369.35
A1
\(369.354\ldots\) Accept $369.
(e)
\(C'(v)\gt 0\) for \(v\gt 58.5\), so \(C\) is increasing on \(70\le v\le90\) and the minimum is at \(v=70\)
R1
$382.11
A1
\(382.114\ldots\) Accept $382.
Answers: (b) 58.5 km h−1 · (d) $369.35 · (e) $382.11 at 70 km h−1
105AASL5-q00895.9 KinematicsPaper 2Medium8 marks
A particle starts from a fixed point O and moves in a straight line. Its velocity, \(v\ \text{m s}^{-1}\), after \(t\) seconds is given by \(v(t)=1-3\sin(1.2t)\), for \(0\le t\le5\).
(a)
Find the first time at which the particle is at rest.
(2)
(b)
Find the number of times the particle changes direction during the 5 seconds. Justify your answer.
(2)
(c)
Find the total distance travelled by the particle.
(2)
(d)
Find the distance of the particle from O when \(t=5\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(1-3\sin(1.2t)=0\), or first zero of \(v\) on the GDC
(M1)
\(t=0.283\) s
A1
\(0.283197\ldots\)
(b)
\(v\) changes sign at \(t=0.283\) and at \(t=2.33\) only; the next zero, \(t=5.52\), is outside the interval
R1
\(2.33479\ldots\) and \(5.51918\ldots\) Accept a sketch of \(v\) showing exactly two sign changes.
Answers: (a) \(t=0.283\) s · (b) 2 · (c) 10.2 m · (d) 4.90 m
106AASL5-q00905.9 KinematicsPaper 2Medium8 marks
A sprinter starts from rest at the start line of a 100 m race. Her velocity, \(v\ \text{m s}^{-1}\), \(t\) seconds after the start is modelled by \(v(t)=11\left(1-\mathrm{e}^{-0.8t}\right)\), for \(t\ge0\).
(a)
Find her initial acceleration.
(2)
(b)
Write down the velocity that the model predicts she approaches.
(1)
(c)
Find the distance she runs in the first 5 seconds.
(2)
(d)
Find the time she takes to complete the 100 m race.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(a(t)=8.8\,\mathrm{e}^{-0.8t}\), or GDC derivative at \(t=0\)
108AASL5-q01155.11 Definite integrals and areasPaper 2Medium7 marks
Let \(f(x)=3\ln(x+1)\) and \(g(x)=x^{2}-2x\), for \(x\ge0\). The graphs of \(f\) and \(g\) meet at the origin and at a second point. The shaded region R is enclosed by the two graphs.
(a)
Find the \(x\)-coordinate of the second point of intersection.
(2)
(b)
Write down an integral that represents the area of R.
(1)
(c)
Find the area of R.
(1)
(d)
The line \(x=a\) divides R into two regions of equal area. Find the value of \(a\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(3\ln(x+1)=x^{2}-2x\), or intersection on the GDC
Find the total area of the region enclosed by the graph of \(f\), the \(x\)-axis and the line \(x=4\).
(2)
(d)
Find the value of \(b\), where \(0\lt b\lt 4\), for which \(\displaystyle\int_{0}^{b}f(x)\,\mathrm{d}x=0\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(x=1.57\)
A1
\(\tfrac{\pi}{2}=1.57079\ldots\)
(b)
use of GDC integration with correct limits
(M1)
\(-4.68\)
A1
\(-4.68085\ldots\)
(c)
\(\displaystyle\int_{0}^{4}|x\cos x|\,\mathrm{d}x\), or \(0.570796\ldots+|-5.25164\ldots|\)
(M1)
\(5.82\)
A1
\(5.82244\ldots\)
(d)
\(\displaystyle\int_{0}^{b}x\cos x\,\mathrm{d}x=0\), e.g. graph of \(y=\displaystyle\int_{0}^{x}f(u)\,\mathrm{d}u\) and its zero, or equating the area below the axis from \(1.57\) to \(b\) with \(0.571\)
111AASL5-q00175.2 Increasing and decreasing functionsPaper 2Hard16 marks
The cross-section of a garden mound is modelled by \(d(x)=\dfrac{-x^{3}+8x^{2}+48x}{80}\), for \(0\le x\le 12\), where \(d\) is the height of the mound in metres at a horizontal distance \(x\) metres from one end.
(a)
Find \(d'(x)\).
(2)
(b)
Find the gradient of the mound at \(x=4\), and state whether the mound is rising or falling there.
(2)
(c)
Find the set of values of \(x\) for which the height of the mound is increasing.
(3)
(d)
Write down the maximum height of the mound.
(1)
(e)(i)
A straight wooden beam rests along the tangent to the mound at the point where \(x=10\).
Find the equation of this tangent.
(3)
(e)(ii)
Find the value of \(x\) at which the beam reaches the ground.
(1)
(f)
Find the area of the cross-section of the mound.
(2)
(g)
Find the horizontal width of the part of the mound that is more than 2 m high.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{-3x^{2}+16x}{80}\)
A1
Award for any two correct terms.
\(d'(x)=\dfrac{-3x^{2}+16x+48}{80}\)
A1
Accept \(-0.0375x^{2}+0.2x+0.6\).
(b)
\(d'(4)=0.8\)
A1
positive, so rising
R1
(c)
\(d'(x)\gt 0\), e.g. solving \(-3x^{2}+16x+48=0\)
M1
\(x=7.47\)
A1
\(7.47406\ldots\) (the other root, \(-2.14\), is outside the domain)
\(0\le x\lt 7.47\)
A1
Accept \(0\lt x\lt 7.47\).
(d)
\(4.85\) m
A1
\(4.85168\ldots\)
(e)(i)
\(d(10)=3.5\)
(A1)
\(d'(10)=-1.15\)
(A1)
\(y=-1.15x+15\)
A1
Accept \(y-3.5=-1.15(x-10)\).
(e)(ii)
\(x=13.0\)
A1
\(13.0434\ldots\) FT from (e)(i).
(f)
\(\displaystyle\int_{0}^{12}d(x)\,\mathrm{d}x\)
(M1)
\(36\ \text{m}^{2}\)
A1
(g)
\(d(x)=2\Rightarrow x=2.58\) or \(x=11.0\)
(M1)
\(2.58118\ldots,\ 11.0357\ldots\)
\(8.45\) m
A1
\(8.45455\ldots\)
Answers: (a) \(\frac{-3x^{2}+16x+48}{80}\) · (b) \(0.8\), rising · (c) \(0\le x\lt 7.47\) · (d) \(4.85\) m · (e)(i) \(y=-1.15x+15\) (ii) \(x=13.0\) · (f) \(36\) m2 · (g) \(8.45\) m
112AASL5-q00315.4 Tangents and normalsPaper 2Hard8 marks
Let \(f(x)=\dfrac{12}{1+x^{2}}\), for \(-5\le x\le 5\).
(a)
Use your graphic display calculator to find the gradient of the graph of \(f\) at the point where \(x=1\).
(1)
(b)
Find the equation of the normal to the graph of \(f\) at the point where \(x=1\).
(3)
(c)
The normal meets the graph of \(f\) again at the point A. Find the \(x\)-coordinate of A.
(2)
(d)
There is one other point on the graph of \(f\) at which the tangent is parallel to the tangent at \(x=1\). Find the \(x\)-coordinate of this point.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(-6\)
A1
(b)
\(f(1)=6\)
(A1)
gradient of normal \(=\dfrac{1}{6}\)
(M1)
\(y-6=\dfrac{1}{6}(x-1)\)
A1
Accept \(y=\dfrac{1}{6}x+\dfrac{35}{6}\) or \(y=0.167x+5.83\).
(c)
\(\dfrac{12}{1+x^{2}}=\dfrac{1}{6}x+\dfrac{35}{6}\), e.g. intersection of graphs
(M1)
\(x=-1.06\)
A1
\(-1.05888\ldots\)
(d)
graph of \(y=f'(x)\) and \(y=-6\), or solving \(f'(x)=-6\)
113AASL5-q00665.7 The second derivative and graphsPaper 2Hard13 marks
The concentration of a drug in a patient’s blood, \(C\) milligrams per litre (mg L−1), \(t\) hours after a tablet is taken is modelled by \(C(t)=12\left(\mathrm{e}^{-0.2t}-\mathrm{e}^{-0.8t}\right)\), for \(t\ge0\).
(a)
Write down the value of \(C(0)\).
(1)
(b)
Find \(C'(t)\).
(2)
(c)
Show that the concentration is a maximum when \(t=\dfrac{5}{3}\ln4\).
(3)
(d)
Find the maximum concentration.
(1)
(e)
Find the time at which the concentration is decreasing most rapidly, and the rate of decrease at this time.
(3)
(f)
The drug is effective while the concentration is at least 3 mg L−1. Find the length of time for which the drug is effective.
114AASL5-q00765.8 Stationary points, inflexion and optimizationPaper 2Hard9 marks
A closed cylindrical container of radius \(r\) cm and height \(h\) cm has a volume of \(500\ \text{cm}^{3}\). The material for the curved surface and the top costs 2 cents per cm2. The material for the base costs 4 cents per cm2. Let \(C\) be the total cost of the material, in cents.
(a)
Write down an expression for \(h\) in terms of \(r\).
(1)
(b)
Show that \(C=6\pi r^{2}+\dfrac{2000}{r}\).
(2)
(c)
Find \(\dfrac{\mathrm{d}C}{\mathrm{d}r}\).
(2)
(d)
Find the value of \(r\) for which the cost is a minimum. Justify that this value gives a minimum.
(3)
(e)
Find the minimum cost, in dollars, correct to the nearest cent.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(h=\dfrac{500}{\pi r^{2}}\)
A1
(b)
\(C=2(2\pi rh)+2(\pi r^{2})+4(\pi r^{2})\)
M1
Award for a cost expression with the three surfaces and correct prices.
115AASL5-q00795.8 Stationary points, inflexion and optimizationPaper 2Hard13 marks
An open gutter is made by bending a sheet of metal 50 cm wide. The cross-section is a trapezium with a horizontal base of 20 cm and two sloping sides of 15 cm, each making an angle \(\theta\) with the horizontal, where \(0\lt\theta\le\dfrac{\pi}{2}\), as shown in the diagram.
Diagram not to scale
(a)
Show that the area of the cross-section, \(A\ \text{cm}^{2}\), is given by \(A=300\sin\theta+225\sin\theta\cos\theta\).
(3)
(b)
Find \(\dfrac{\mathrm{d}A}{\mathrm{d}\theta}\).
(3)
(c)
Show that \(\dfrac{\mathrm{d}A}{\mathrm{d}\theta}=0\) when \(6\cos^{2}\theta+4\cos\theta-3=0\).
(2)
(d)
Hence find the value of \(\theta\) for which the area of the cross-section is a maximum, and find this maximum area.
(3)
(e)
Find the percentage by which this maximum area exceeds the area of the cross-section when the sides are vertical.
(2)
Show mark scheme
Working
Mark
Notes
(a)
height \(=15\sin\theta\)
A1
width of the top \(=20+2(15\cos\theta)=20+30\cos\theta\)
Two particles, P and Q, leave a fixed point O at the same time, \(t=0\), and move along the same straight line. The velocity of P, \(v_{\text{P}}\ \text{m s}^{-1}\), after \(t\) seconds is given by \(v_{\text{P}}(t)=10\sin(0.8t)\,\mathrm{e}^{-0.2t}\), for \(0\le t\le 8\). Particle Q moves with a constant velocity of \(2\ \text{m s}^{-1}\).
(a)
Find the value of \(t\) at which P first changes direction.
(2)
(b)
Find the total distance travelled by P in the 8 seconds.
(2)
(c)
Determine whether P is speeding up or slowing down when \(t=5\). Justify your answer.
(3)
(d)
Find the values of \(t\), for \(0\lt t\le 8\), at which P and Q are at the same position.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\sin(0.8t)=0\), or first zero of \(v_{\text{P}}\) after \(t=0\) on the GDC
Equating the displacement of P with the displacement of Q; e.g. graphs of both displacements on the GDC.
\(t=0.546\)
A1
\(0.546089\ldots\)
\(t=6.07\)
A1
\(6.06781\ldots\)
Answers: (a) \(t=3.93\) · (b) 25.0 m · (c) speeding up · (d) \(t=0.546,\ 6.07\)
117AASL5-q00885.9 KinematicsPaper 2Hard9 marks
A particle starts from a fixed point O and moves in a straight line. Its velocity, \(v\ \text{m s}^{-1}\), after \(t\) seconds is given by \(v(t)=5-t\,\mathrm{e}^{0.4t}\), for \(t\ge0\). The particle first moves away from O, then turns and returns to O.
(a)
Find the value of \(t\) at which the particle turns.
(2)
(b)
Find the greatest distance of the particle from O before it returns.
(2)
(c)
Find the time at which the particle returns to O.
(3)
(d)
Find the speed of the particle when it returns to O.
Displacement equal to zero; accept use of a graph of the displacement function.
\(T\gt 2.13\) (rejects \(T=0\))
(A1)
\(T=3.61\) s
A1
\(3.61393\ldots\)
(d)
\(|v(3.61393\ldots)|\)
(M1)
\(10.3\ \text{m s}^{-1}\)
A1
\(v=-10.3386\ldots\) Award A0 for \(-10.3\).
Answers: (a) \(t=2.13\) · (b) 6.57 m · (c) \(t=3.61\) s · (d) \(10.3\ \text{m s}^{-1}\)
118AASL5-q00915.9 KinematicsPaper 2Hard15 marks
A particle P moves in a straight line, starting from a fixed point O. Its velocity, \(v\ \text{m s}^{-1}\), after \(t\) seconds is given by
\(v(t)=12t\,\mathrm{e}^{-0.5t}-3\), for \(0\le t\le10\). The graph of \(v\) is shown below.
(a)
Write down the initial speed of P.
(1)
(b)
Find the maximum velocity of P.
(2)
(c)
Find the values of \(t\) at which P changes direction.
(2)
(d)
Find the acceleration of P when \(t=1\).
(2)
(e)
Determine whether P is speeding up or slowing down when \(t=7\). Justify your answer.
(2)
(f)
Find the total distance travelled by P in the 10 seconds.
(3)
(g)
Find the greatest distance of P from O during the 10 seconds.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(3\ \text{m s}^{-1}\)
A1
Award A0 for \(-3\).
(b)
use of GDC to find the maximum of \(v\) (at \(t=2\))
(M1)
\(5.83\ \text{m s}^{-1}\)
A1
\(5.82910\ldots\ (=24\mathrm{e}^{-1}-3)\)
(c)
\(v(t)=0\), e.g. zeros on graph
(M1)
\(t=0.289\) and \(t=6.52\)
A1
\(0.288842\ldots\), \(6.52337\ldots\)
(d)
\(a=v'(1)\), e.g. GDC derivative, or \(a=12\mathrm{e}^{-0.5t}-6t\mathrm{e}^{-0.5t}\)
(M1)
\(3.64\ \text{m s}^{-2}\)
A1
\(3.63918\ldots\)
(e)
\(v(7)=-0.463\lt 0\) and \(a(7)=-0.906\lt 0\); velocity and acceleration have the same sign
A train leaves station A from rest at time \(t=0\) and travels along a straight track to station B, where it stops. Its velocity, \(v\ \text{m s}^{-1}\), \(t\) seconds after leaving A is modelled by \(v(t)=0.0005\,t^{2}(60-t)\), for \(0\le t\le60\).
(a)
Find the maximum velocity of the train.
(2)
(b)
Find the distance between station A and station B.
(2)
(c)
Find the maximum acceleration of the train.
(2)
(d)
Find the time taken for the train to travel half of the distance from A to B.
(3)
(e)
At \(t=0\), a cyclist passes station A riding at a constant velocity of \(8\ \text{m s}^{-1}\) in the same direction as the train. Find the time at which the train catches up with the cyclist, and the distance from A at which this happens.
(3)
(f)
Find the greatest distance between the cyclist and the train before the train catches up with the cyclist.
\(\dfrac{T^{3}}{100}-\dfrac{T^{4}}{8000}=270\), or graph of the integral
(A1)
\(T=36.9\) s
A1
\(36.8563\ldots\)
(e)
\(\displaystyle\int_{0}^{t}v(u)\,\mathrm{d}u=8t\)
M1
Equating the distances travelled.
\(t=40\) s
A1
Other solutions of \(t^{3}-80t^{2}+64\,000=0\) are outside \(0\lt t\le 60\).
\(320\) m
A1
(f)
the gap \(8t-s(t)\) is greatest when the velocities are equal, \(v(t)=8\)
R1
Or attempt to maximize the gap with technology.
\(t=20\)
A1
The other solution, \(t=54.6\), is after the train has caught up.
\(8(20)-60=100\) m
A1
\(s(20)=60\).
Answers: (a) \(16\ \text{m s}^{-1}\) · (b) 540 m · (c) \(0.6\ \text{m s}^{-2}\) · (d) 36.9 s · (e) \(t=40\) s, 320 m from A · (f) 100 m
120AASL5-q01205.11 Definite integrals and areasPaper 2Hard15 marks
At 00:00 a tank holds 2000 litres of water. For the next 24 hours, water flows into the tank at a rate of \(R(t)=60t\,\mathrm{e}^{-0.25t}\) litres per hour, where \(t\) is the time in hours after 00:00. At the same time, water is pumped out of the tank at a constant rate of 30 litres per hour.
(a)
Find the maximum rate at which water flows into the tank.
(2)
(b)
Find the total volume of water that flows into the tank during the 24 hours.
(2)
(c)
Find the values of \(t\) at which the rate of inflow is equal to the rate of outflow.
(2)
(d)
Explain why the volume of water in the tank is greatest when \(t\) is equal to the larger of your values in part (c).
(2)
(e)
Find the greatest volume of water in the tank during the 24 hours. Give your answer to the nearest litre.
(3)
(f)
Find the volume of water in the tank at the end of the 24 hours. Give your answer to the nearest litre.
(2)
(g)
Find the least volume of water in the tank during the 24 hours. Give your answer to the nearest litre.
(2)
Show mark scheme
Working
Mark
Notes
(a)
maximum of \(R\) on GDC, or \(R'(t)=0\Rightarrow t=4\)
for \(0.578\lt t\lt 13.0\), \(R(t)\gt 30\), so the volume is increasing
R1
for \(13.0\lt t\le 24\), \(R(t)\lt 30\), so the volume is decreasing
R1
Must refer to the comparison of rates on both sides of \(t=13.0\). The volume also decreases for \(0\lt t\lt 0.578\), but only by about 8 litres, so \(V(13.0)\gt V(0)\).
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