Original questions in the style of Papers 1, 2 and 3, each with a full worked mark scheme. Tick questions to build a printable test. Topic 5 at Standard 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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1AAHL5-q00015.1 Limits and the derivativePaper 1Easy8 marks
Water flows into an empty tank. The volume of water in the tank, \(V\) litres, \(t\) minutes after the water starts to flow is given by \[V(t)=40-\dfrac{40}{(t+1)^{2}},\quad t\ge0.\]
(a)
Find \(\dfrac{\mathrm{d}V}{\mathrm{d}t}\).
(2)
(b)
Find the rate at which the volume is increasing when \(t=1\), stating the units of your answer.
(2)
(c)
Find the value of \(t\) at which water flows in at a rate of \(0.08\) litres per minute.
(2)
(d)
Describe what happens to \(V\) and to \(\dfrac{\mathrm{d}V}{\mathrm{d}t}\) as \(t\) becomes very large, and interpret your answer in context.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(V=40-40(t+1)^{-2}\), so \(\dfrac{\mathrm{d}V}{\mathrm{d}t}=-40(-2)(t+1)^{-3}\)
(M1)
Award for writing as a negative power and attempting the power rule.
4AAHL5-q00115.5 Introduction to integrationPaper 1Easy7 marks
Let \(f(x)=\dfrac{4}{x^{2}}\), for \(x\gt 0\). For \(b\gt 1\), let \(A(b)\) be the area of the region enclosed by the graph of \(f\), the \(x\)-axis and the lines \(x=1\) and \(x=b\).
(a)
Show that \(A(b)=4-\dfrac{4}{b}\).
(3)
(b)
Find the value of \(b\) for which \(A(b)=3.9\).
(2)
(c)
Write down \(\displaystyle\lim_{b\to\infty}A(b)\), and interpret this value in terms of areas.
Must see the integral with correct limits; \(f(x)\gt 0\) so the integral equals the area.
\(\left[-4x^{-1}\right]_{1}^{b}\)
A1
\(-\dfrac{4}{b}-(-4)\)
A1
\(=4-\dfrac{4}{b}\)
AG
(b)
\(\dfrac{4}{b}=0.1\)
(M1)
\(b=40\)
A1
(c)
\(4\)
A1
however far the region extends to the right, its area never exceeds 4 (it approaches 4)
R1
Accept equivalent.
Answers: (b) \(b=40\) · (c) 4
5AAHL5-q00125.5 Introduction to integrationPaper 1Easy8 marks
The derivative of a function \(f\) is given by \(f'(x)=3x^{2}+bx\), where \(b\in\mathbb{R}\). The line \(y=5x-3\) is the tangent to the graph of \(f\) at the point where \(x=1\).
(a)
Find the value of \(b\).
(2)
(b)
Find \(f(x)\).
(3)
(c)
The graph of \(f\) meets the \(x\)-axis at \(x=-1\) and at \(x=0\). Find the area of the region enclosed by the graph of \(f\) and the \(x\)-axis.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(f'(1)=5\Rightarrow3+b=5\)
(M1)
\(b=2\)
A1
(b)
\(f(x)=x^{3}+x^{2}+c\)
M1
Award for integration of their \(f'\), constant may be missing.
\(f(1)=5(1)-3=2\Rightarrow1+1+c=2\)
M1
Award for using the point of contact on the tangent.
7AAHL5-q00225.8 Stationary points, inflexion and optimizationPaper 1Easy10 marks
Let \(f(x)=3x^{5}-5x^{3}\), for \(x\in\mathbb{R}\).
(a)
Find the \(x\)-coordinates of the stationary points of the graph of \(f\).
(3)
(b)
Use the second derivative to determine the nature of the stationary points at \(x=1\) and \(x=-1\).
(3)
(c)
Explain why the second derivative test cannot be used at \(x=0\), and determine the nature of the stationary point there.
(2)
(d)
Find the coordinates of the points of inflexion with non-zero gradient.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(f'(x)=15x^{4}-15x^{2}\)
A1
\(15x^{2}\left(x^{2}-1\right)=0\)
M1
\(x=0,\ x=1,\ x=-1\)
A1
(b)
\(f''(x)=60x^{3}-30x\)
A1
\(f''(1)=30\gt 0\): local minimum at \((1,\,-2)\)
A1
Coordinates not required.
\(f''(-1)=-30\lt 0\): local maximum at \((-1,\,2)\)
A1
(c)
\(f''(0)=0\), which does not determine the nature
R1
\(f'(x)=15x^{2}\left(x^{2}-1\right)\lt 0\) on both sides of \(x=0\) (no change of sign of \(f'\)), and \(f''(x)=30x\left(2x^{2}-1\right)\) changes sign at \(x=0\): stationary point of inflexion
R1
(d)
\(f''(x)=0\Rightarrow x^{2}=\dfrac12\), and \(f''\) changes sign there
M1
\(\left(\dfrac{1}{\sqrt2},\,-\dfrac{7\sqrt2}{8}\right)\) and \(\left(-\dfrac{1}{\sqrt2},\,\dfrac{7\sqrt2}{8}\right)\)
A1
Accept \(\mp\dfrac{7}{4\sqrt2}\).
Answers: (a) \(0,\ \pm1\) · (b) min at \(x=1\), max at \(x=-1\) · (c) stationary point of inflexion · (d) \(\left(\pm\tfrac{1}{\sqrt2},\ \mp\tfrac{7\sqrt2}{8}\right)\)
8AAHL5-q00305.11 Definite integrals and areasPaper 1Easy7 marks
Consider the graph of \(y=\dfrac{1}{x}\), for \(x\gt 0\).
(a)
Show that, for \(0\lt a\lt b\), the area under the graph between \(x=a\) and \(x=b\) is \(\ln\dfrac{b}{a}\).
(2)
(b)
Hence show that, for any \(k\gt 0\), the area between \(x=ka\) and \(x=kb\) is equal to the area between \(x=a\) and \(x=b\).
(1)
(c)
Find the value of \(c\), where \(c\gt 2\), such that the area between \(x=2\) and \(x=c\) is equal to the area between \(x=1\) and \(x=3\).
(2)
(d)
The line \(x=p\) divides the region under the graph between \(x=1\) and \(x=9\) into two regions of equal area. Find the value of \(p\).
Consider the series \(\displaystyle\sum_{k=0}^{\infty}\left(\dfrac{2x-1}{3}\right)^{k}\). Find the set of values of \(x\) for which the series converges.
(3)
(c)
For the values of \(x\) found in part (b), find the sum of the series, giving your answer in the form \(\dfrac{a}{b-cx}\), where \(a,b,c\in\mathbb{Z}^{+}\) have no common factor.
(2)
(d)
State, with a reason, whether the sequence \(v_{n}=(-1.2)^{n}\), \(n\in\mathbb{Z}^{+}\), converges or diverges.
(1)
Show mark scheme
Working
Mark
Notes
(a)
dividing numerator and denominator by \(n^{2}\): \(\dfrac{6-\frac{1}{n}}{3+\frac{4}{n^{2}}}\)
M1
\(2\)
A1
(b)
\(\left|\dfrac{2x-1}{3}\right|\lt1\)
M1
\(-3\lt2x-1\lt3\)
A1
\(-1\lt x\lt2\)
A1
Award A0 for non-strict inequalities.
(c)
\(S_{\infty}=\dfrac{1}{1-\frac{2x-1}{3}}\)
M1
\(\dfrac{3}{4-2x}\)
A1
(d)
diverges, because \(|-1.2|\gt1\) so \(|v_{n}|\) increases without bound (the terms oscillate in sign with growing size)
11AAHL5-q00385.12 Limits, first principles and higher derivativesPaper 1Easy5 marks
The graph of \(y=f(x)\), for \(-2\le x\le5\), is shown. For \(-2\le x\lt1\) the graph is part of the curve \(y=x^{2}\); for \(1\le x\le5\) it consists of two line segments meeting at \((3,\,4)\). The point \((1,\,1)\) is not on the graph.
(a)
Write down the value of \(x\) at which \(f\) is not continuous.
(1)
(b)
Write down the value of \(x\), with \(1\lt x\lt5\), at which \(f\) is continuous but not differentiable. Give a reason based on the graph.
(2)
(c)
Write down the value of \(f'(4)\).
(1)
(d)
Write down the value of \(f'(0)\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(x=1\)
A1
(b)
\(x=3\)
A1
the graph has a sharp corner there: the gradient is \(1\) to the left and \(-1\) to the right, so there is no single tangent
R1
Accept any equivalent informal reason referring to the different left and right gradients.
18AAHL5-q00645.15 Further derivatives and integrals; partial fractionsPaper 1Easy7 marks
(a)
Find the derivative of \(f(x)=3^{2x}\).
(2)
(b)
Find the derivative of \(g(x)=\log_{2}(x^{2}+1)\).
(2)
(c)
Find the equation of the tangent to the curve \(y=\log_{3}x\) at the point where \(x=9\). Give your answer in the form \(y=mx+c\), where \(m\) and \(c\) are exact.
(3)
Show mark scheme
Working
Mark
Notes
(a)
use of \(\dfrac{\mathrm{d}}{\mathrm{d}x}a^{u}=a^{u}\ln a\,\dfrac{\mathrm{d}u}{\mathrm{d}x}\)
(M1)
Or writing \(3^{2x}=9^{x}\).
\(f'(x)=2\ln3\times3^{2x}\)
A1
Accept \(9^{x}\ln9\).
(b)
use of \(\dfrac{\mathrm{d}}{\mathrm{d}x}\log_{a}u=\dfrac{1}{u\ln a}\dfrac{\mathrm{d}u}{\mathrm{d}x}\)
(M1)
\(g'(x)=\dfrac{2x}{(x^{2}+1)\ln2}\)
A1
(c)
\(y=\log_{3}9=2\)
A1
gradient \(=\dfrac{1}{9\ln3}\)
(M1)
\(y=\dfrac{x}{9\ln3}+2-\dfrac{1}{\ln3}\)
A1
Award A0 for \(y-2=\dfrac{1}{9\ln3}(x-9)\) left in this form, as the form \(y=mx+c\) was required.
24AAHL5-q00855.17 Areas with the y-axis and volumes of revolutionPaper 1Easy6 marks
The following diagram shows part of the curve \(x=(y-1)(5-y)\). The shaded region R is enclosed by the curve and the \(y\)-axis. The point V is the point on the curve furthest from the \(y\)-axis.
Diagram not to scale
(a)
Write down the \(y\)-coordinates of the points where the curve meets the \(y\)-axis.
(1)
(b)
Find the coordinates of V.
(2)
(c)
Find the area of R.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(y=1\) and \(y=5\)
A1
(b)
\(\dfrac{\mathrm{d}x}{\mathrm{d}y}=6-2y=0\) or symmetry: \(y=\dfrac{1+5}{2}\)
25AAHL5-q00875.17 Areas with the y-axis and volumes of revolutionPaper 1Easy6 marks
The following diagram shows part of the graph of \(y=\sec x\), for \(0\le x\le\dfrac{\pi}{3}\). Region A is enclosed by the graph, the \(x\)-axis, the \(y\)-axis and the line \(x=\dfrac{\pi}{3}\). Region B is enclosed by the graph, the \(y\)-axis and the line \(y=2\).
Diagram not to scale
(a)
Region A is rotated through \(2\pi\) about the \(x\)-axis. Find the exact volume of the solid formed.
(3)
(b)
Region B is rotated through \(2\pi\) about the \(x\)-axis. Find the exact volume of the solid formed.
28AAHL5-q00965.18 First order differential equationsPaper 1Easy6 marks
Consider the differential equation \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=(2x+1)\cos^{2}y\), where \(-\dfrac{\pi}{2}\lt y\lt\dfrac{\pi}{2}\) and \(y=\dfrac{\pi}{4}\) when \(x=0\).
33AAHL5-q00065.3 Differentiating polynomials and integer powersPaper 1Medium9 marks
A curve has equation \(y=x^{3}-\dfrac{12}{x}\), for \(x\neq0\).
(a)
Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\).
(2)
(b)
Find the \(x\)-coordinates of the points on the curve where the gradient is 13.
(3)
(c)
By writing \(\dfrac{\mathrm{d}y}{\mathrm{d}x}-12\) in the form \(a\left(x-\dfrac{b}{x}\right)^{2}\), show that the gradient of the curve is never less than 12.
(2)
(d)
Find the coordinates of the points on the curve where the gradient is 12.
34AAHL5-q00085.4 Tangents and normalsPaper 1Medium7 marks
The curves \(C_{1}\) and \(C_{2}\) have equations \(y=x^{2}\) and \(y=-x^{2}+4x-4\) respectively. A line \(L\) is a tangent to \(C_{1}\) at the point \(\left(a,\,a^{2}\right)\).
(a)
Show that the equation of \(L\) is \(y=2ax-a^{2}\).
(2)
(b)
Show that \(L\) is also a tangent to \(C_{2}\) if and only if \(a(a-2)=0\).
(3)
(c)
Hence write down the equations of the two lines that are tangents to both curves.
tangent \(\Leftrightarrow\) exactly one point of contact \(\Leftrightarrow\) discriminant \(=0\): \((2a-4)^{2}-4\left(4-a^{2}\right)=0\)
R1
\(8a^{2}-16a=0\), i.e. \(8a(a-2)=0\)
A1
\(a(a-2)=0\)
AG
(c)
\(y=0\)
A1
from \(a=0\)
\(y=4x-4\)
A1
from \(a=2\)
Answers: (c) \(y=0\) and \(y=4x-4\)
35AAHL5-q00145.6 Differentiation rulesPaper 1Medium9 marks
Let \(f(x)=\dfrac{x}{\sqrt{x^{2}+4}}\), for \(x\in\mathbb{R}\).
(a)
Show that \(f'(x)=\dfrac{4}{\left(x^{2}+4\right)^{\frac32}}\).
(3)
(b)
Hence explain why \(f\) is an increasing function.
(1)
(c)
Find the exact values of \(x\) for which \(f'(x)=\dfrac{1}{16}\).
(3)
(d)
By considering \(f(x)\) for large positive and large negative values of \(x\), write down the equations of the horizontal asymptotes of the graph of \(f\).
in degrees, \(\dfrac{\mathrm{d}}{\mathrm{d}x}\sin\left(x^{\circ}\right)=\dfrac{\pi}{180}\cos\left(x^{\circ}\right)\neq\cos\left(x^{\circ}\right)\); the factor \(\dfrac{\pi}{180}\) disappears only when the angle is in radians
R1
Must refer to the extra factor (or the conversion factor).
38AAHL5-q00185.7 The second derivative and graphsPaper 1Medium10 marks
Let \(f(x)=\sqrt{x}\), for \(x\gt 0\).
(a)
Find \(f''(x)\), and hence state whether the graph of \(f\) is concave-up or concave-down.
(2)
(b)
Find the equation of the tangent to the graph of \(f\) at the point where \(x=9\).
(2)
(c)
Use the tangent to find an estimate for \(\sqrt{10}\), giving your answer as a fraction.
(1)
(d)
With reference to your answer to part (a), explain why your estimate is greater than \(\sqrt{10}\).
(1)
(e)
Show algebraically that \(\sqrt{x}\le\dfrac{x}{6}+\dfrac32\) for all \(x\gt 0\).
(2)
(f)
The chord joining the points \((9,\,3)\) and \((16,\,4)\) on the graph of \(f\) is used to estimate \(\sqrt{10}\). Find this estimate, and explain why it is less than \(\sqrt{10}\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(f''(x)=-\dfrac14x^{-\frac32}\)
A1
\(f''(x)\lt 0\) for all \(x\gt 0\), so the graph is concave-down
A1
(b)
\(f'(9)=\dfrac{1}{2\sqrt9}=\dfrac16\)
(M1)
\(y-3=\dfrac16(x-9)\), i.e. \(y=\dfrac{x}{6}+\dfrac32\)
A1
(c)
\(\dfrac{10}{6}+\dfrac32=\dfrac{19}{6}\)
A1
FT from their tangent.
(d)
the graph is concave-down, so it lies below its tangent (the tangent lies above the curve), hence the estimate is too large
39AAHL5-q00205.8 Stationary points, inflexion and optimizationPaper 1Medium9 marks
The point \(\text{P}\left(x,\,x^{2}\right)\) lies on the curve \(y=x^{2}\). The point A has coordinates \((3,\,0)\). Let \(D\) be the distance AP.
(a)
Show that \(D^{2}=x^{4}+x^{2}-6x+9\).
(1)
(b)
Show that \(\dfrac{\mathrm{d}\left(D^{2}\right)}{\mathrm{d}x}=0\) has exactly one real solution, \(x=1\).
(3)
(c)
Explain why the value of \(x\) that minimizes \(D^{2}\) also minimizes \(D\), and hence find the shortest distance from A to the curve, justifying that it is a minimum.
(3)
(d)
For this position of P, show that AP is perpendicular to the tangent to the curve at P.
discriminant of \(2x^{2}+2x+3\) is \(4-24=-20\lt 0\), so it has no real roots
R1
\(x=1\) is the only real solution
AG
(c)
\(D\ge0\) and \(D=\sqrt{D^{2}}\) is an increasing function of \(D^{2}\)
R1
\(\dfrac{\mathrm{d}\left(D^{2}\right)}{\mathrm{d}x}\lt 0\) for \(x\lt 1\) and \(\gt 0\) for \(x\gt 1\) (since \(2x^{2}+2x+3\gt 0\)), so minimum
R1
Or second derivative \(12x^{2}+2=14\gt 0\) at \(x=1\).
\(D_{\min}=\sqrt{1+1-6+9}=\sqrt5\)
A1
(d)
P is \((1,\,1)\); gradient of AP \(=\dfrac{1-0}{1-3}=-\dfrac12\); gradient of tangent \(=2\)
A1
\(-\dfrac12\times2=-1\), so AP is perpendicular to the tangent
R1
Answers: (c) shortest distance \(\sqrt5\), at \(\text{P}(1,\,1)\)
40AAHL5-q00215.8 Stationary points, inflexion and optimizationPaper 1Medium8 marks
Two lamps are placed at points A and B, 10 m apart. The illumination at a point on the line segment AB that is \(x\) m from A is modelled by \[I(x)=\dfrac{8}{x^{2}}+\dfrac{1}{(10-x)^{2}},\quad 0\lt x\lt 10.\]
(a)
Find \(I'(x)\).
(2)
(b)
Find the value of \(x\) at which \(I'(x)=0\).
(3)
(c)
Show that this value of \(x\) gives the minimum illumination.
(2)
(d)
Find the minimum illumination.
(1)
Show mark scheme
Working
Mark
Notes
(a)
writing as \(8x^{-2}+(10-x)^{-2}\) and differentiating, with the chain rule for the second term
A particle moves in a straight line, starting from a fixed point O. Its velocity, \(v\) m s−1, after \(t\) seconds is given by \[v(t)=2+\dfrac{6}{(t+1)^{2}},\quad t\ge0.\]
(a)
Write down the initial velocity of the particle and the value that the velocity approaches as \(t\) becomes large.
(2)
(b)
Find the acceleration of the particle when \(t=1\).
(2)
(c)
Show that the displacement of the particle from O after \(t\) seconds is \(s(t)=2t+6-\dfrac{6}{t+1}\).
(3)
(d)
Find the exact value of \(t\) at which the particle is 10 m from O.
splitting at the turning times, or \(\displaystyle\int_{0}^{2\pi}|4\cos t-2|\,\mathrm{d}t\)
M1
\(\displaystyle\int_{0}^{\pi/3}v\,\mathrm{d}t=2\sqrt3-\dfrac{2\pi}{3}\) and \(\displaystyle\int_{5\pi/3}^{2\pi}v\,\mathrm{d}t=2\sqrt3-\dfrac{2\pi}{3}\)
recognizing \(\dfrac{\mathrm{e}^{x}}{1+\mathrm{e}^{x}}\) as \(\dfrac{h'(x)}{h(x)}\) with \(h(x)=1+\mathrm{e}^{x}\), or substitution \(u=1+\mathrm{e}^{x}\)
45AAHL5-q00295.11 Definite integrals and areasPaper 1Medium8 marks
The diagram shows part of the curve \(y=x^{3}\) and the line \(L\), which is the tangent to the curve at the point \(\text{P}(1,\,1)\). The line \(L\) meets the curve again at the point Q. The shaded region R is enclosed by the curve and \(L\).
Diagram not to scale
(a)
Find the equation of \(L\).
(2)
(b)
Find the coordinates of Q.
(3)
(c)
Find the area of R.
(3)
Show mark scheme
Working
Mark
Notes
(a)
gradient \(=3(1)^{2}=3\)
(M1)
\(y=3x-2\)
A1
(b)
\(x^{3}=3x-2\Rightarrow x^{3}-3x+2=0\)
M1
\((x-1)^{2}(x+2)=0\)
M1
Recognizing \((x-1)^{2}\) as a factor because \(L\) touches the curve at \(x=1\).
Substituting \(px\) into the series for \(\cos x\).
dividing by \(x^{2}\): limit \(=\dfrac{\frac{3}{2}}{\frac{p^{2}}{2}}=\dfrac{3}{p^{2}}\)
A1
Accept a correct method using l’Hôpital’s rule, e.g. \(\displaystyle\lim_{x\to0}\dfrac{2x\mathrm{e}^{x^{2}}+\sin x}{p\sin px}=\lim_{x\to0}\dfrac{(2+4x^{2})\mathrm{e}^{x^{2}}+\cos x}{p^{2}\cos px}=\dfrac{3}{p^{2}}\).
56AAHL5-q00565.14 Implicit differentiation, related rates and optimisationPaper 1Medium7 marks
A trough is 4 m long. Its cross-section is an isosceles triangle with its vertex at the bottom; the triangle is 1.2 m wide at the top and 0.8 m deep, as shown. Water flows into the trough at a constant rate of 0.05 m3 min−1. When the depth of water is \(h\) m, the width of the water surface is \(w\) m and the volume of water is \(V\) m3.
Diagram not to scale
(a)
Show that \(V=3h^{2}\).
(2)
(b)
Find the rate at which the depth of water is increasing when \(h=0.4\).
(3)
(c)
Find the rate at which \(w\) is increasing at this instant.
(2)
Show mark scheme
Working
Mark
Notes
(a)
similar triangles: \(\dfrac{w}{h}=\dfrac{1.2}{0.8}\), so \(w=1.5h\)
M1
\(V=4\times\dfrac{1}{2}(1.5h)h\)
A1
\(V=3h^{2}\)
AG
(b)
\(\dfrac{dV}{dt}=6h\dfrac{dh}{dt}\)
M1
\(0.05=6(0.4)\dfrac{dh}{dt}\)
(A1)
\(\dfrac{dh}{dt}=\dfrac{1}{48}\) m min−1
A1
Accept \(0.0208\).
(c)
\(\dfrac{dw}{dt}=1.5\dfrac{dh}{dt}\)
M1
\(\dfrac{1}{32}\) m min−1
A1
Accept \(0.03125\). FT from (b).
Answers: (b) \(\dfrac{1}{48}\) m min−1 · (c) \(\dfrac{1}{32}\) m min−1
57AAHL5-q00665.15 Further derivatives and integrals; partial fractionsPaper 1Medium6 marks
Let \(f(x)=\arctan x+\arctan\dfrac{1}{x}\), \(x\neq0\).
(a)
Show that \(f'(x)=0\).
(3)
(b)
Hence find the value of \(f(x)\) for \(x\gt0\).
(2)
(c)
Write down the value of \(f(x)\) for \(x\lt0\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
chain rule on \(\arctan\dfrac1x\): \(\dfrac{1}{1+\frac{1}{x^{2}}}\times\left(-\dfrac{1}{x^{2}}\right)\)
M1
\(=-\dfrac{1}{x^{2}+1}\)
A1
\(f'(x)=\dfrac{1}{1+x^{2}}-\dfrac{1}{1+x^{2}}\)
A1
\(=0\)
AG
(b)
\(f\) is constant for \(x\gt0\), so evaluate at a convenient point, e.g. \(f(1)=\dfrac{\pi}{4}+\dfrac{\pi}{4}\)
M1
Must use (a) to justify that \(f\) is constant.
\(f(x)=\dfrac{\pi}{2}\)
A1
(c)
\(-\dfrac{\pi}{2}\)
A1
From \(f(-1)\), or because \(f\) is an odd function. \(f\) is constant on each of the two intervals separately, as the domain is not connected.
67AAHL5-q00865.17 Areas with the y-axis and volumes of revolutionPaper 1Medium10 marks
The following diagram shows part of the graph of \(y=\ln x\). The shaded region R is enclosed by the graph, the \(y\)-axis, the \(x\)-axis and the line \(y=2\).
Diagram not to scale
(a)
By integrating with respect to \(y\), find the exact area of R.
(3)
(b)
Use integration by parts to find \(\displaystyle\int_{1}^{\mathrm{e}^{2}}\ln x\,\mathrm{d}x\), and hence verify your answer to part (a).
(3)
(c)
The region R is rotated through \(2\pi\) about the \(x\)-axis. Find the exact volume of the solid formed.
R and the region under the curve from \(x=1\) to \(x=\mathrm{e}^{2}\) together form the rectangle \(0\le x\le\mathrm{e}^{2}\), \(0\le y\le2\); \(2\mathrm{e}^{2}-(\mathrm{e}^{2}+1)=\mathrm{e}^{2}-1\)
Cylinder of radius 2 and length \(\mathrm{e}^{2}\), minus the solid formed by the region under the curve; or \(\pi\displaystyle\int_{0}^{1}4\,\mathrm{d}x+\pi\displaystyle\int_{1}^{\mathrm{e}^{2}}\left(4-(\ln x)^{2}\right)\mathrm{d}x\).
by parts: \(\displaystyle\int(\ln x)^{2}\,\mathrm{d}x=x(\ln x)^{2}-\displaystyle\int2\ln x\,\mathrm{d}x\)
68AAHL5-q00885.17 Areas with the y-axis and volumes of revolutionPaper 1Medium6 marks
The following diagram shows part of the graph of \(y=\dfrac{1}{1+x^{2}}\), for \(x\ge0\). The shaded region R is enclosed by the graph, the \(y\)-axis and the line \(y=\dfrac12\).
(a)
Show that, on this graph, \(x^{2}=\dfrac{1-y}{y}\).
(2)
(b)
The region R is rotated through \(2\pi\) about the \(y\)-axis. Find the exact volume of the solid formed.
(4)
Show mark scheme
Working
Mark
Notes
(a)
\(1+x^{2}=\dfrac1y\)
M1
\(x^{2}=\dfrac1y-1\)
A1
\(=\dfrac{1-y}{y}\)
AG
(b)
\(V=\pi\displaystyle\int x^{2}\,\mathrm{d}y\)
M1
limits \(y=\dfrac12\) and \(y=1\): \(\pi\displaystyle\int_{1/2}^{1}\left(\dfrac1y-1\right)\mathrm{d}y\)
A1
\(\pi\left[\ln y-y\right]_{1/2}^{1}\)
A1
\(\pi\left(\ln2-\dfrac12\right)\)
A1
Answers: (b) \(\pi\left(\ln2-\frac12\right)\)
69AAHL5-q00975.18 First order differential equationsPaper 1Medium9 marks
Consider the differential equation \(x\dfrac{\mathrm{d}y}{\mathrm{d}x}=y^{2}-1\), where \(x\gt0\) and \(y=3\) when \(x=1\).
(a)
Express \(\dfrac{1}{y^{2}-1}\) in partial fractions.
(2)
(b)
Hence show that \(y=\dfrac{2+x^{2}}{2-x^{2}}\).
(6)
(c)
State the largest interval of values of \(x\), containing \(x=1\), for which this solution is valid.
73AAHL5-q01035.18 First order differential equationsPaper 1Medium7 marks
Consider the differential equation \(\dfrac{\mathrm{d}y}{\mathrm{d}x}+y\tan x=\cos^{2}x\), where \(-\dfrac{\pi}{2}\lt x\lt\dfrac{\pi}{2}\) and \(y=2\) when \(x=0\).
(a)
Show that \(\sec x\) is an integrating factor for this differential equation.
(2)
(b)
Hence find \(y\) in terms of \(x\).
(4)
(c)
Find the exact value of \(y\) when \(x=\dfrac{\pi}{6}\).
77AAHL5-q01165.19 Maclaurin seriesPaper 1Medium8 marks
Let \(f(x)=\ln\left(1+x-2x^{2}\right)\), for \(-\dfrac{1}{2}\lt x\lt1\).
(a)
Factorise \(1+x-2x^{2}\).
(1)
(b)
Hence find the Maclaurin series for \(f(x)\) up to and including the term in \(x^{3}\).
(3)
(c)
Show that the coefficient of \(x^{n}\) in the Maclaurin series for \(f(x)\) is \(\dfrac{(-1)^{n+1}2^{n}-1}{n}\), for \(n\in\mathbb{Z}^{+}\).
(2)
(d)
Given that the Maclaurin series for \(\ln(1+x)\) converges to \(\ln(1+x)\) for \(-1\lt x\le1\), find the set of values of \(x\) for which the Maclaurin series for \(f(x)\) converges to \(f(x)\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\((1+2x)(1-x)\)
A1
(b)
\(f(x)=\ln(1+2x)+\ln(1-x)\)
M1
Use of the laws of logarithms.
\(\ln(1+2x)=2x-2x^{2}+\dfrac{8x^{3}}{3}-\cdots\) and \(\ln(1-x)=-x-\dfrac{x^{2}}{2}-\dfrac{x^{3}}{3}-\cdots\)
80AAHL5-q00095.4 Tangents and normalsPaper 1Hard9 marks
Consider the curve \(y=x^{2}\). Let \(\text{P}\left(a,\,a^{2}\right)\), where \(a\neq0\), be a point on the curve.
(a)
Show that the equation of the normal to the curve at P is \(y=-\dfrac{x}{2a}+a^{2}+\dfrac{1}{2}\).
(3)
(b)
The normal at P passes through the point \(\text{Q}(0,\,k)\). Find \(a^{2}\) in terms of \(k\).
(1)
(c)
Find the set of values of \(k\) for which exactly three distinct normals to the curve pass through Q.
(3)
(d)
For \(k=\dfrac{5}{2}\), find the area of the triangle whose vertices are Q and the two points on the curve whose normals pass through Q but are not the \(y\)-axis.
(2)
Show mark scheme
Working
Mark
Notes
(a)
gradient of tangent \(=2a\), so gradient of normal \(=-\dfrac{1}{2a}\)
A1
\(y-a^{2}=-\dfrac{1}{2a}(x-a)\)
M1
\(y=-\dfrac{x}{2a}+\dfrac{1}{2}+a^{2}\)
A1
\(y=-\dfrac{x}{2a}+a^{2}+\dfrac{1}{2}\)
AG
(b)
\(k=a^{2}+\dfrac12\Rightarrow a^{2}=k-\dfrac12\)
A1
(c)
the normal at the origin is the \(y\)-axis, which passes through Q for every \(k\)
R1
two further distinct normals require two distinct non-zero values \(a=\pm\sqrt{k-\frac12}\), so \(k-\dfrac12\gt 0\)
81AAHL5-q00155.6 Differentiation rulesPaper 1Hard10 marks
Let \(f(x)=x^{3}+3x-2\), for \(x\in\mathbb{R}\).
(a)
Show that \(f\) is increasing for all \(x\), and hence explain why \(f^{-1}\) exists.
(2)
(b)
By differentiating \(f\left(f^{-1}(x)\right)=x\) with respect to \(x\), find the gradient of the graph of \(y=f^{-1}(x)\) at the point where \(x=2\).
(3)
(c)
Find the equation of the tangent to the graph of \(y=f^{-1}(x)\) at the point where \(x=2\).
(2)
(d)
The tangent to the graph of \(y=f(x)\) at \((1,\,2)\) and the tangent found in part (c) meet at the point R. Find the coordinates of R, and explain why R lies on the line \(y=x\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(f'(x)=3x^{2}+3\ge3\gt 0\)
A1
\(f\) is increasing so it is one-to-one, hence \(f^{-1}\) exists
the graphs of \(f\) and \(f^{-1}\) are reflections of each other in \(y=x\), so the two tangents are reflections of each other in \(y=x\) and must meet on that line
82AAHL5-q00195.7 The second derivative and graphsPaper 1Hard13 marks
Let \(f(x)=ax^{3}+bx^{2}+cx+d\), where \(a,b,c,d\in\mathbb{R}\) and \(a\neq0\).
(a)
Show that the graph of \(f\) has exactly one point of inflexion, and that it occurs at \(x=p\), where \(p=-\dfrac{b}{3a}\).
(3)
(b)
Show that \(f(p+h)+f(p-h)=2f(p)\) for all \(h\in\mathbb{R}\).
(3)
(c)
Interpret the result of part (b) geometrically.
(1)
(d)
Suppose that the graph of \(f\) has two stationary points, at \(x=x_{1}\) and \(x=x_{2}\). Use the result of part (b) to show that the point of inflexion is the midpoint of the line segment joining the two stationary points.
(3)
(e)
Let \(f(x)=2x^{3}-3x^{2}-12x+5\). Find the coordinates of the two stationary points and of the point of inflexion of the graph of \(f\), and hence verify the result of part (d).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(f''(x)=6ax+2b\)
A1
\(f''(x)=0\Rightarrow x=-\dfrac{2b}{6a}=-\dfrac{b}{3a}\), the only solution since \(f''\) is linear
A1
\(f''\) is linear with non-zero gradient \(6a\), so it changes sign at \(x=p\); hence exactly one point of inflexion
R1
(b)
expanding \(a(p\pm h)^{3}+b(p\pm h)^{2}+c(p\pm h)+d\) and adding
M1
terms in \(h\) and \(h^{3}\) cancel: sum \(=2\left(ap^{3}+bp^{2}+cp+d\right)+h^{2}(6ap+2b)\)
A1
\(6ap+2b=2(3ap+b)=0\) since \(p=-\dfrac{b}{3a}\)
R1
\(f(p+h)+f(p-h)=2f(p)\)
AG
(c)
the graph has rotational symmetry of order 2 (a half-turn) about its point of inflexion \(\left(p,\,f(p)\right)\)
A1
Accept “\(\left(p,f(p)\right)\) is the midpoint of every chord joining points with \(x=p\pm h\)”.
(d)
\(x_{1},x_{2}\) are the roots of \(f'(x)=3ax^{2}+2bx+c=0\), so \(x_{1}+x_{2}=-\dfrac{2b}{3a}=2p\)
M1
so \(x_{1}=p-h\) and \(x_{2}=p+h\) for some \(h\); the midpoint has \(x\)-coordinate \(p\)
A1
by part (b), \(\dfrac{f(x_{1})+f(x_{2})}{2}=f(p)\), so the midpoint is \(\left(p,\,f(p)\right)\), the point of inflexion
R1
(e)
\(f'(x)=6x^{2}-6x-12=6(x-2)(x+1)\): stationary points \((-1,\,12)\) and \((2,\,-15)\)
A1
\(p=\dfrac{3}{6}=\dfrac12\), \(f\left(\dfrac12\right)=-\dfrac32\): point of inflexion \(\left(\dfrac12,\,-\dfrac32\right)\)
A1
midpoint \(\left(\dfrac{-1+2}{2},\,\dfrac{12-15}{2}\right)=\left(\dfrac12,\,-\dfrac32\right)\), as required
A1
Answers: (e) stationary points \((-1,12)\), \((2,-15)\); inflexion \(\left(\tfrac12,-\tfrac32\right)\), which is their midpoint
83AAHL5-q00325.11 Definite integrals and areasPaper 1Hard15 marks
Let \(f(x)=\mathrm{e}^{2x}-4\mathrm{e}^{x}+3\), for \(x\in\mathbb{R}\).
(a)
Find the exact solutions of \(f(x)=0\).
(3)
(b)
Find the coordinates of the minimum point of the graph of \(f\), justifying that it is a minimum.
(4)
(c)
Find \(f''(x)\), and hence show that the graph of \(f\) has a point of inflexion at the origin.
(3)
(d)
Write down the equation of the horizontal asymptote of the graph of \(f\).
(1)
(e)
Find the exact area of the region enclosed by the graph of \(f\) and the \(x\)-axis.
(4)
Show mark scheme
Working
Mark
Notes
(a)
\(\left(\mathrm{e}^{x}\right)^{2}-4\mathrm{e}^{x}+3=0\), a quadratic in \(\mathrm{e}^{x}\)
\(h'\) changes from positive to negative there; justification is not required.
(e)
form \(\dfrac{0}{0}\); differentiating with respect to \(k\): \(\displaystyle\lim_{k\to2}\dfrac{1/k}{1}\)
M1
\(\dfrac{1}{2}\)
A1
this equals the value of \(t\) at the maximum of \(h\) found in (d): as \(k\to2\) the maximum point of \(g_{k}\) moves to the maximum point of the limiting function \(h\)
R1
Accept any comment that links the limit to the answer \(t=\dfrac{1}{2}\) in part (d).
for \(x\gt0\): \(2\gt\mathrm{e}^{1/\mathrm{e}}\) (since \(\mathrm{e}\ln2\approx1.88\gt1\)), so \(2^{x}\gt\left(\mathrm{e}^{1/\mathrm{e}}\right)^{x}\ge x\), because the curve \(y=\left(\mathrm{e}^{1/\mathrm{e}}\right)^{x}\) is concave up and lies on or above its tangent \(y=x\)
R1
Accept a convincing equivalent argument, e.g. \(2^{x}-x\) has minimum value \(\dfrac{1+\ln(\ln2)}{\ln2}\gt0\).
91AAHL5-q00815.16 Integration by substitution and by partsPaper 1Hard11 marks
Let \(I=\displaystyle\int\sin(\ln x)\,\mathrm{d}x\), for \(x\gt0\).
(a)
By using integration by parts twice, show that \(I=\dfrac{x}{2}\left(\sin(\ln x)-\cos(\ln x)\right)+C\).
(5)
(b)
Hence find the exact value of \(\displaystyle\int_{1}^{\mathrm{e}^{\pi}}\sin(\ln x)\,\mathrm{d}x\).
(2)
(c)
Find the exact total area of the regions enclosed by the curve \(y=\sin(\ln x)\) and the \(x\)-axis for \(1\le x\le\mathrm{e}^{2\pi}\). Give your answer in the form \(\dfrac{(\mathrm{e}^{k\pi}+b)^{2}}{c}\), where \(k,b,c\in\mathbb{Z}^{+}\).
(4)
Show mark scheme
Working
Mark
Notes
(a)
first application of parts with \(u=\sin(\ln x)\), \(\dfrac{\mathrm{d}v}{\mathrm{d}x}=1\)
M1
Students who first substitute \(x=\mathrm{e}^{t}\) to obtain \(\int\mathrm{e}^{t}\sin t\,\mathrm{d}t\) and then use parts twice may earn full marks.
recognising that the curve is below the \(x\)-axis for \(\mathrm{e}^{\pi}\lt x\lt\mathrm{e}^{2\pi}\) (where \(\pi\lt\ln x\lt2\pi\)), so splitting the integral at \(x=\mathrm{e}^{\pi}\)
94AAHL5-q00895.17 Areas with the y-axis and volumes of revolutionPaper 1Hard10 marks
The shaded region R is enclosed by the graphs of \(y=x^{3}\) and \(y=4x\), for \(x\ge0\).
Diagram not to scale
(a)
Find the coordinates of the points of intersection of the two graphs for \(x\ge0\).
(2)
(b)
The region R is rotated through \(2\pi\) about the \(y\)-axis. Show that the volume of the solid formed is \(\pi\displaystyle\int_{0}^{8}\left(y^{2/3}-\dfrac{y^{2}}{16}\right)\mathrm{d}y\).
(3)
(c)
Find the exact value of this volume.
(2)
(d)
The region R is instead rotated through \(2\pi\) about the \(x\)-axis. Find the exact volume of the solid formed.
95AAHL5-q00915.17 Areas with the y-axis and volumes of revolutionPaper 1Hard15 marks
Let \(f(x)=\arcsin\dfrac{x}{2}\), for \(0\le x\le2\). The following diagram shows the graph of \(y=f(x)\). Region R is enclosed by the graph, the \(x\)-axis and the line \(x=2\). Region S is enclosed by the graph, the \(y\)-axis and the line \(y=\dfrac{\pi}{2}\).
(a)
Show that \(f'(x)=\dfrac{1}{\sqrt{4-x^{2}}}\).
(2)
(b)
Use integration by parts to find \(\displaystyle\int\arcsin\dfrac{x}{2}\,\mathrm{d}x\).
(4)
(c)
Hence find the exact area of R.
(2)
(d)(i)
Express \(x\) in terms of \(y\) for this graph.
(1)
(d)(ii)
By integrating with respect to \(y\), find the area of S.
(2)
(d)(iii)
Explain why your answers to (c) and (d)(ii) are consistent.
(1)
(e)
Region S is rotated through \(2\pi\) about the \(y\)-axis. Find the exact volume of the solid formed.
96AAHL5-q00925.17 Areas with the y-axis and volumes of revolutionPaper 1Hard16 marks
Let \(f(x)=\dfrac{3}{\sqrt{x^{2}+3x}}\), for \(x\gt0\). The following diagram shows part of the graph of \(y=f(x)\). The shaded region R is enclosed by the graph, the \(x\)-axis and the lines \(x=1\) and \(x=3\).
Diagram not to scale
(a)
Write down the equations of the asymptotes of the graph of \(y=f(x)\).
(2)
(b)
Find \(f'(x)\) and hence explain why \(f\) is decreasing for \(x\gt0\).
(3)
(c)
Express \(\dfrac{9}{x(x+3)}\) in partial fractions.
(2)
(d)
The region R is rotated through \(2\pi\) about the \(x\)-axis. Show that the volume of the solid formed is \(3\pi\ln2\).
(4)
(e)
For \(b\gt1\), let \(V(b)\) be the volume of the solid formed when the region enclosed by the graph, the \(x\)-axis and the lines \(x=1\) and \(x=b\) is rotated through \(2\pi\) about the \(x\)-axis. Find the value of \(b\) for which \(V(b)=3\pi\ln3\).
(3)
(f)
Show that \(V(b)\lt3\pi\ln4\) for all \(b\gt1\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(x=0\)
A1
\(y=0\)
A1
(b)
\(f(x)=3(x^{2}+3x)^{-1/2}\), chain rule
M1
\(f'(x)=-\dfrac{3(2x+3)}{2(x^{2}+3x)^{3/2}}\)
A1
for \(x\gt0\), \(2x+3\gt0\) and \((x^{2}+3x)^{3/2}\gt0\), so \(f'(x)\lt0\)
Solution must be continuous on an interval containing \(x=1\); \(1-\ln x=0\) at \(x=\mathrm{e}\).
\(y\to+\infty\) as \(x\to\mathrm{e}^{-}\) (vertical asymptote \(x=\mathrm{e}\))
A1
Answers: (c) \(0\lt x\lt\mathrm{e}\); \(y\to\infty\) as \(x\to\mathrm{e}^{-}\)
98AAHL5-q01085.18 First order differential equationsPaper 1Hard13 marks
The current, \(I\) amperes, in an electrical circuit \(t\) seconds after it is switched on satisfies the differential equation
\(\dfrac{\mathrm{d}I}{\mathrm{d}t}+5I=26\sin t\), where \(I=0\) when \(t=0\).
(a)
Write down an integrating factor for this differential equation.
(1)
(b)
Use integration by parts twice to show that \(\displaystyle\int\mathrm{e}^{5t}\sin t\,\mathrm{d}t=\frac{\mathrm{e}^{5t}}{26}\left(5\sin t-\cos t\right)+C\).
(5)
(c)
Hence show that \(I=5\sin t-\cos t+\mathrm{e}^{-5t}\).
(3)
(d)
Describe the behaviour of \(I\) for large values of \(t\).
(1)
(e)
Hence find the maximum value of the current for large values of \(t\), giving your answer in exact form.
99AAHL5-q01175.19 Maclaurin seriesPaper 1Hard10 marks
Let \(y=\mathrm{e}^{\arctan x}\).
(a)
Show that \(\left(1+x^{2}\right)\dfrac{\mathrm{d}y}{\mathrm{d}x}=y\).
(1)
(b)
By differentiating the result in part (a), show that \(\left(1+x^{2}\right)\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}+(2x-1)\dfrac{\mathrm{d}y}{\mathrm{d}x}=0\).
(2)
(c)
Show that \(\left(1+x^{2}\right)\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}}+(4x-1)\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}+2\dfrac{\mathrm{d}y}{\mathrm{d}x}=0\).
(2)
(d)
Hence find the Maclaurin series for \(y\) up to and including the term in \(x^{3}\).
101AAHL5-q00025.1 Limits and the derivativePaper 2Easy8 marks
Two fully charged batteries, A and B, start to discharge at the same time. The charge remaining in each battery, as a percentage of its full charge, \(t\) hours later is modelled by \[Q_{\text{A}}(t)=100\,\mathrm{e}^{-0.08t}\quad\text{and}\quad Q_{\text{B}}(t)=100-3.5t,\qquad 0\le t\le28.\]
(a)
Find \(Q_{\text{A}}'(t)\).
(1)
(b)
Find the value of \(Q_{\text{A}}'(10)\), and interpret its sign in context.
(2)
(c)
Find the value of \(t\) at which the two batteries are discharging at the same rate.
(2)
(d)
Find the value of \(t\), where \(t\gt 0\), at which the two batteries have the same charge.
(2)
(e)
State which battery is discharging faster at the time found in part (d). Justify your answer.
102AAHL5-q01065.18 First order differential equationsPaper 2Easy7 marks
A parachutist falls from rest. Her velocity, \(v\) m s−1, \(t\) seconds after she starts to fall, is modelled by the differential equation \(\dfrac{\mathrm{d}v}{\mathrm{d}t}=9.81-0.0032v^{2}\), where \(v=0\) when \(t=0\). The parachute has not yet opened.
(a)
Use Euler’s method with a step length of \(0.5\) seconds to find an approximate value of \(v\) when \(t=2\).
(3)
(b)
Continuing the method, find the first value of \(t\) for which the approximate value of \(v\) exceeds \(30\).
(2)
(c)
Find the velocity that the parachutist approaches as \(t\) increases.
Answers: (a) \(19.1\) m s−1 · (b) \(t=3.5\) s · (c) \(55.4\) m s−1
103AAHL5-q01075.18 First order differential equationsPaper 2Easy7 marks
A cup of tea is left to cool in a room where the temperature is a constant \(21^{\circ}\mathrm{C}\). The temperature of the tea, \(T^{\circ}\mathrm{C}\), \(t\) minutes after it is made, satisfies \(\dfrac{\mathrm{d}T}{\mathrm{d}t}=-k(T-21)\), where \(k\gt0\). Initially \(T=86\).
(a)
Solve the differential equation to show that \(T=21+65\mathrm{e}^{-kt}\).
(3)
(b)
After 4 minutes the temperature of the tea is \(68^{\circ}\mathrm{C}\). Find the value of \(k\).
(2)
(c)
Find the time taken for the tea to cool to \(40^{\circ}\mathrm{C}\).
Or gradient from origin: \(\dfrac{a\sin a+2}{a}=\sin a+a\cos a\).
\(a\sin a+2=a\sin a+a^{2}\cos a\)
A1
\(a^{2}\cos a=2\)
AG
(c)
graph or solver for \(a^{2}\cos a-2=0\) on \([0,2\pi]\)
(M1)
There is no solution in \(\left[0,\tfrac{\pi}{2}\right]\) since \(a^{2}\cos a\lt 0.6\) there, and none in \(\left]\tfrac{\pi}{2},\tfrac{3\pi}{2}\right[\) since \(\cos a\lt 0\) there; the only solution lies in \(\left]\tfrac{3\pi}{2},2\pi\right[\).
106AAHL5-q00405.12 Limits, first principles and higher derivativesPaper 2Medium8 marks
Let \(S_{n}=\displaystyle\sum_{k=1}^{n}\dfrac{1}{k^{2}}\) and \(H_{n}=\displaystyle\sum_{k=1}^{n}\dfrac{1}{k}\), for \(n\in\mathbb{Z}^{+}\).
(a)
Find \(S_{10}\) and \(S_{100}\), giving your answers correct to four significant figures.
(2)
(b)
It is known that \(\displaystyle\lim_{n\to\infty}S_{n}=\dfrac{\pi^{2}}{6}\). Find the least value of \(n\) for which \(\dfrac{\pi^{2}}{6}-S_{n}\lt0.005\).
(2)
(c)
Find \(H_{100}\) and \(H_{1000}\), giving your answers correct to four significant figures.
(1)
(d)
Show that \(H_{2n}-H_{n}\ge\dfrac{1}{2}\) for all \(n\in\mathbb{Z}^{+}\).
(2)
(e)
Hence explain why the series \(\displaystyle\sum_{k=1}^{\infty}\dfrac{1}{k}\) diverges, even though \(\dfrac{1}{k}\to0\) as \(k\to\infty\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(S_{10}=1.550\)
A1
\(1.54976\ldots\) Accept \(1.55\).
\(S_{100}=1.635\)
A1
\(1.63498\ldots\)
(b)
\(\dfrac{\pi^{2}}{6}-S_{n}\lt0.005\), e.g. table: \(\dfrac{\pi^{2}}{6}-S_{199}=0.0050125\ldots\), \(\dfrac{\pi^{2}}{6}-S_{200}=0.0049875\ldots\)
(M1)
\(n=200\)
A1
(c)
\(H_{100}=5.187\) and \(H_{1000}=7.485\)
A1
Both required.
(d)
\(H_{2n}-H_{n}=\dfrac{1}{n+1}+\dfrac{1}{n+2}+\dots+\dfrac{1}{2n}\), which has \(n\) terms
M1
each term is at least \(\dfrac{1}{2n}\), so the sum is at least \(n\times\dfrac{1}{2n}\)
R1
\(H_{2n}-H_{n}\ge\dfrac{1}{2}\)
AG
(e)
\(H_{2^{m}}\ge1+\dfrac{m}{2}\) (each doubling of \(n\) adds at least \(\dfrac{1}{2}\)), so the partial sums increase without bound and the series diverges
R1
Accept any clear argument that repeated doubling adds at least \(\dfrac{1}{2}\) each time, so \(H_{n}\) has no finite limit.
107AAHL5-q00585.14 Implicit differentiation, related rates and optimisationPaper 2Medium8 marks
A camera at point C is 50 m from a straight road; the point on the road nearest to C is N. A car travels along the road away from N at a constant speed of 30 m s−1. When the car is \(x\) m from N, the angle between [CN] and the line from C to the car is \(\theta\) radians. The camera rotates so that it always points at the car.
(a)
Find \(\dfrac{d\theta}{dt}\) when the car is 50 m from N.
(3)
(b)
Find the rate at which the distance between the camera and the car is increasing at this instant.
(2)
(c)
The camera can rotate at a rate of at most 0.5 rad s−1. Find the values of \(x\) for which the camera cannot keep pointing at the car.
Answers: (a) 0.3 rad s−1 · (b) 21.2 m s−1 · (c) \(0\le x\lt22.4\)
108AAHL5-q00595.14 Implicit differentiation, related rates and optimisationPaper 2Medium9 marks
At 12:00 ship A is at a port O and ship B is 40 km due east of O. Ship A sails due north at 12 km h−1 and ship B sails due west at 16 km h−1. At time \(t\) hours after 12:00 the distance between the ships is \(s\) km, for \(0\le t\le2.5\).
(a)
Show that \(s^{2}=400t^{2}-1280t+1600\).
(2)
(b)
Find \(\dfrac{ds}{dt}\) when \(t=0.5\).
(3)
(c)
Interpret the sign of your answer to part (b).
(1)
(d)
Find the minimum distance between the ships and the time at which it occurs.
(3)
Show mark scheme
Working
Mark
Notes
(a)
A at \((0,\,12t)\), B at \((40-16t,\,0)\); \(s^{2}=(40-16t)^{2}+(12t)^{2}\)
M1
\(1600-1280t+256t^{2}+144t^{2}\)
A1
\(s^{2}=400t^{2}-1280t+1600\)
AG
(b)
\(2s\dfrac{ds}{dt}=800t-1280\)
M1
at \(t=0.5\): \(s=\sqrt{1060}=32.557\ldots\)
(A1)
\(\dfrac{ds}{dt}=-13.5\) km h−1
A1
\(-13.5144\ldots\)
(c)
the distance between the ships is decreasing
R1
(d)
\(\dfrac{ds}{dt}=0\Rightarrow800t-1280=0\)
M1
Or minimum of \(s^{2}\) from GDC.
\(t=1.6\) (13:36)
A1
\(s=24\) km
A1
Answers: (b) \(-13.5\) km h−1 · (d) 24 km at \(t=1.6\)
109AAHL5-q00605.14 Implicit differentiation, related rates and optimisationPaper 2Medium8 marks
A wire of length 20 cm is to be used to make a square and a circle. A length \(x\) cm is bent into the square and the remainder into the circle. The whole wire may be used for one shape only, so \(0\le x\le20\).
(a)
Show that the total area enclosed is \(A(x)=\dfrac{x^{2}}{16}+\dfrac{(20-x)^{2}}{4\pi}\).
(2)
(b)
Find the value of \(x\) for which the total area is a minimum.
(3)
(c)
Find the maximum total area, justifying your answer.
(3)
Show mark scheme
Working
Mark
Notes
(a)
square side \(\dfrac{x}{4}\); circle radius \(r=\dfrac{20-x}{2\pi}\)
the only stationary point is a minimum, so the maximum is at an end point; evaluating \(A(0)\) and \(A(20)\)
M1
\(A(0)=\dfrac{100}{\pi}=31.8\) and \(A(20)=25\)
A1
maximum area \(31.8\) cm2, when all the wire forms the circle (\(x=0\))
A1
\(31.8309\ldots\)
Answers: (b) \(x=11.2\) · (c) \(31.8\) cm2 when \(x=0\)
110AAHL5-q00845.16 Integration by substitution and by partsPaper 2Medium6 marks
Let \(f(x)=x\ln x\), for \(x\gt0\).
(a)
Use integration by parts to show that \(\displaystyle\int x\ln x\,\mathrm{d}x=\dfrac{x^{2}}{2}\ln x-\dfrac{x^{2}}{4}+C\).
(2)
(b)
The region in the first quadrant enclosed by the graph of \(f\), the \(x\)-axis and the line \(x=k\), where \(k\gt1\), has area 10. Find the value of \(k\).
Or \(\displaystyle\int_{1}^{k}x\ln x\,\mathrm{d}x=10\).
solving with GDC
(M1)
\(k=4.44\)
A1
\(4.43778\ldots\) Award A0 for a value from the lower limit 0.
Answers: (b) \(k=4.44\)
111AAHL5-q00905.17 Areas with the y-axis and volumes of revolutionPaper 2Medium7 marks
The graphs of \(y=3\sin x\) and \(y=x\), for \(x\ge0\), meet at the origin and at the point P, where \(x=a\). The shaded region R is enclosed by the two graphs.
(a)
Find the value of \(a\).
(2)
(b)
Find the area of R.
(2)
(c)
The region R is rotated through \(2\pi\) about the \(x\)-axis. Find the volume of the solid formed.
112AAHL5-q00935.17 Areas with the y-axis and volumes of revolutionPaper 2Medium14 marks
The inside of a vase is modelled by rotating the curve \(x=g(y)\), where \(g(y)=3+1.5\sin\left(\dfrac{\pi y}{12}\right)\), \(0\le y\le20\), through \(2\pi\) about the \(y\)-axis. Distances are measured in centimetres; \(y\) is the height above the base of the vase and \(x\) is the inner radius at that height. Region A is enclosed by the curve, the \(y\)-axis, the \(x\)-axis and the line \(y=20\).
(a)
Find the radius of the opening at the top of the vase.
(2)
(b)
Write down the maximum inner radius of the vase and the height at which it occurs.
(2)
(c)
Find the area of region A.
(2)
(d)
Find the capacity of the vase.
(3)
(e)
Water is poured into the empty vase until it holds \(500\ \text{cm}^{3}\). Find the depth of the water.
(3)
(f)
Find the additional volume of water needed to fill the vase when the depth of water is 10 cm.
\(2x-y=2x-\left(2x-2+3\mathrm{e}^{-x}\right)=2-3\mathrm{e}^{-x}\), so \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=2x-y\)
A1
\(x=0\): \(y=0-2+3=1\)
A1
\(y=2x-2+3\mathrm{e}^{-x}\) satisfies both
AG
(b)
\(y_{n+1}=y_{n}+0.1\,(2x_{n}-y_{n})\), with \(x_{0}=0,\ y_{0}=1\)
(M1)
\(y_{1}=0.9,\ y_{2}=0.83,\ y_{3}=0.787,\ \ldots\)
(A1)
At least two correct values, or ten steps seen on a GDC table.
\(y(1)\approx1.05\)
A1
\(1.046035\ldots\)
(c)
exact value \(3\mathrm{e}^{-1}=1.10363\ldots\); \(\dfrac{1.10363\ldots-1.04603\ldots}{1.10363\ldots}\times100\)
(M1)
\(5.22\%\)
A1
\(5.2193\ldots\%\). Accept \(5.19\%\) to \(5.23\%\). FT from (b).
(d)
\(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=3\mathrm{e}^{-x}\gt0\) (or \(2-\dfrac{\mathrm{d}y}{\mathrm{d}x}=2-2x+y\gt0\)), so the solution curve is concave up
A1
each Euler step follows the tangent, which lies below a concave-up curve, so every step (and hence the final value) underestimates \(y\)
R1
Answers: (b) \(1.05\) \((1.046035\ldots)\) · (c) \(5.22\%\) · (d) curve concave up, so tangent steps lie below it
114AAHL5-q01195.19 Maclaurin seriesPaper 2Medium8 marks
The function \(\dfrac{\sin x}{x}\) has no elementary antiderivative.
(a)
Using the Maclaurin series for \(\sin x\), find the first four non-zero terms of the series for \(\dfrac{\sin x}{x}\), \(x\neq0\).
(2)
(b)
Hence find the first four non-zero terms of the Maclaurin series for \(\displaystyle\int_{0}^{x}\frac{\sin t}{t}\,\mathrm{d}t\).
(2)
(c)
Use your answer to part (b) to find an approximation for \(\displaystyle\int_{0}^{1}\frac{\sin x}{x}\,\mathrm{d}x\), giving your answer to six decimal places.
(2)
(d)
Use your GDC to find the percentage error in the approximation \(1-\dfrac{1}{18}+\dfrac{1}{600}-\dfrac{1}{35\,280}\).
115AAHL5-q01205.19 Maclaurin seriesPaper 2Medium10 marks
The function \(y=f(x)\) satisfies the differential equation \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=x+y^{2}\), where \(y=1\) when \(x=0\).
(a)
Find \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\) and \(\dfrac{\mathrm{d}^{3}y}{\mathrm{d}x^{3}}\) in terms of \(y\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and \(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\).
(3)
(b)
Hence find the Maclaurin series for \(f(x)\) up to and including the term in \(x^{3}\).
(3)
(c)
Use the series to estimate \(f(0.1)\).
(1)
(d)
Use Euler’s method with a step length of \(0.05\) to estimate \(f(0.1)\).
(2)
(e)
Given that \(f(0.1)=1.11649\), correct to five decimal places, state which estimate is more accurate.
the Maclaurin series estimate (error about \(0.00016\), compared with about \(0.0089\) for Euler)
A1
Must be consistent with their values.
Answers: (b) \(1+x+\dfrac{3}{2}x^{2}+\dfrac{4}{3}x^{3}\) · (c) \(1.116\) · (d) \(1.107625\) · (e) the series estimate
116AAHL5-q00235.8 Stationary points, inflexion and optimizationPaper 2Hard13 marks
A straight boundary separates a field from a marsh. A power cable is to be laid from point A in the field, 40 m from the boundary, to point B in the marsh, 30 m from the boundary. The points A′ and B′ are the feet of the perpendiculars from A and B to the boundary, and A′B′ \(=100\) m.
The cable is laid in a straight line from A to a point P on the boundary, where A′P \(=x\) m and \(0\le x\le100\), and then in a straight line from P to B. Laying the cable costs €15 per metre in the field and €60 per metre in the marsh.
Diagram not to scale
(a)
Show that the total cost, \(C\) euros, is given by \[C(x)=15\sqrt{1600+x^{2}}+60\sqrt{900+(100-x)^{2}}.\]
(2)
(b)
Find the cost when P is at B′, giving your answer correct to the nearest cent.
(1)
(c)
Find \(C'(x)\).
(3)
(d)
Find the value of \(x\) that minimizes \(C\), and the minimum cost correct to the nearest cent.
(3)
(e)
Hence find the amount saved by using the cheapest route rather than the route in part (b).
(1)
(f)
The angles α and β are between the normal to the boundary at P and the lines PA and PB respectively, as shown in the diagram. Show that, for the cheapest route, \(\sin\alpha=4\sin\beta\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
AP \(=\sqrt{40^{2}+x^{2}}\) and PB \(=\sqrt{30^{2}+(100-x)^{2}}\) (Pythagoras)
M1
cost \(=\) length \(\times\) cost per metre for each section, and adding
At time \(t=0\) a car starts from rest at a point O and moves along a straight road. Its velocity, \(v\) m s−1, after \(t\) seconds is given by \[v(t)=20\left(1-\mathrm{e}^{-0.25t}\right),\quad t\ge0.\] At the same instant, a motorbike passes O travelling in the same direction at a constant velocity of 12 m s−1.
(a)
Find the initial acceleration of the car.
(2)
(b)
Write down the value that the velocity of the car approaches as \(t\) increases.
(1)
(c)
Find the time at which the car and the motorbike have the same velocity.
(2)
(d)
Show that the displacement of the car from O after \(t\) seconds is \(s(t)=20t+80\mathrm{e}^{-0.25t}-80\).
(3)
(e)
Find the greatest distance between the motorbike and the car before the car overtakes the motorbike. Justify that the distance is greatest at this time.
(3)
(f)
Find the time at which the car overtakes the motorbike.
(2)
(g)
Find the speed of the car at the moment it overtakes the motorbike.
gap \(d(t)=12t-s(t)\); \(d'(t)=12-v(t)=0\) at the time found in (c)
M1
\(d'(t)\gt 0\) before this time (car slower) and \(d'(t)\lt 0\) after it (car faster), so the gap is a maximum
R1
\(18.7\) m
A1
\(18.6787\ldots\)
(f)
\(20t+80\mathrm{e}^{-0.25t}-80=12t\), \(t\gt 0\)
(M1)
\(t=8.93\) s
A1
\(8.92644\ldots\)
(g)
\(17.9\) m s−1
A1
\(17.8528\ldots\) FT from (f).
Answers: (a) 5 m s−2 · (b) 20 m s−1 · (c) 3.67 s · (e) 18.7 m · (f) 8.93 s · (g) 17.9 m s−1
118AAHL5-q00315.11 Definite integrals and areasPaper 2Hard9 marks
The diagram shows the graph of \(y=\sin x\), for \(0\le x\le\pi\), and the line \(y=kx\), where \(k\gt 0\). The line meets the curve at the origin and at the point where \(x=a\), with \(0\lt a\lt\pi\). The shaded region R is enclosed by the curve and the line.
Diagram not to scale
(a)
Show that the area of the region enclosed by the graph of \(y=\sin x\), for \(0\le x\le\pi\), and the \(x\)-axis is 2.
(2)
(b)
Explain why the line \(y=kx\) meets the curve at a point with \(0\lt a\lt\pi\) only if \(k\lt 1\).
(1)
(c)
Show that the area of R is \(1-\cos a-\dfrac12a\sin a\).
(3)
(d)
The line divides the region in part (a) into two regions of equal area. Find the value of \(a\) and the value of \(k\).
the line \(y=x\) is the tangent to \(y=\sin x\) at the origin and \(\sin x\lt x\) for \(x\gt 0\); so for \(k\ge1\), \(kx\gt\sin x\) for all \(x\gt 0\) and there is no second intersection
R1
Accept an argument using the gradient \(1\) of \(\sin x\) at the origin and the concavity of the curve.
\(1-\cos a-\dfrac12a\sin a=1\), i.e. \(\cos a+\dfrac12a\sin a=0\), solved with GDC for \(0\lt a\lt\pi\)
M1
\(a=2.46\)
A1
\(2.45871\ldots\)
\(k=0.257\)
A1
\(0.256649\ldots\)
Answers: (d) \(a=2.46\), \(k=0.257\)
119AAHL5-q00625.14 Implicit differentiation, related rates and optimisationPaper 2Hard16 marks
A lamp L hangs directly above the centre O of a horizontal circular table of radius 0.8 m. The lamp is \(h\) m above the table. The point E is on the edge of the table, LE \(=s\) m, and \(\theta\) is the angle between [LE] and the table surface, as shown.
The brightness of the illumination at E is modelled by \(B=\dfrac{k\sin\theta}{s^{2}}\), where \(k\) is a positive constant.
Diagram not to scale
(a)
Show that \(B=\dfrac{kh}{\left(h^{2}+0.64\right)^{3/2}}\).
(2)
(b)(i)
Show that \(\dfrac{dB}{dh}=\dfrac{k\left(0.64-2h^{2}\right)}{\left(h^{2}+0.64\right)^{5/2}}\).
(3)
(b)(ii)
Find the value of \(h\) that maximises \(B\), justifying that it gives a maximum.
(3)
(c)(i)
The length of the cable holding the lamp can only be adjusted so that \(0.7\le h\le1.5\).
Explain why, with this restriction, \(B\) is greatest when \(h=0.7\).
(2)
(c)(ii)
Find this greatest value of \(B\) as a percentage of the value found in part (b)(ii).
(2)
(d)
The lamp is lowered at a constant rate of 0.1 m min−1. Find the rate of change of \(B\), in terms of \(k\), at the instant when \(h=1\).
(2)
(e)
The table is replaced by a circular table of radius \(r\) m, with the lamp still above its centre and \(0.7\le h\le1.5\). Find the set of values of \(r\) for which the greatest brightness at the edge of the table does not occur at an end point of the interval for \(h\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(s=\sqrt{h^{2}+0.64}\) and \(\sin\theta=\dfrac{h}{s}\)
120AAHL5-q01095.18 First order differential equationsPaper 2Hard16 marks
A school has 1200 students. At 08:00 one morning, 24 of them have heard a rumour. The number of students who have heard the rumour \(t\) hours after 08:00 is \(n\), and the spread of the rumour is modelled by the differential equation
\(\dfrac{\mathrm{d}n}{\mathrm{d}t}=kn(1200-n)\), where \(k\) is a positive constant and \(0\lt n\lt1200\).
(a)
Express \(\dfrac{1}{n(1200-n)}\) in partial fractions.
(2)
(b)
Hence show that \(n=\dfrac{1200}{1+49\mathrm{e}^{-1200kt}}\).
(5)
(c)
At 11:00, 300 students have heard the rumour. Find the value of \(k\).
(3)
(d)
By considering \(\dfrac{\mathrm{d}n}{\mathrm{d}t}\) as a function of \(n\), find the time of day at which the rumour is spreading most rapidly.
(3)
(e)
Find the time of day at which 95% of the students have heard the rumour.
(2)
(f)
Suggest one reason why this model may not be realistic for large values of \(t\).
\(7.34\) hours after 08:00. Accept 15:20 or 15:21.
(f)
eg students go home after school so the rumour stops spreading; some students may never hear it (or be absent); \(k\) may not stay constant
R1
Any sensible reason.
Answers: (a) \(\dfrac{1}{1200}\left(\dfrac{1}{n}+\dfrac{1}{1200-n}\right)\) · (c) \(k=7.76\times10^{-4}\) · (d) about 12:11 (\(t=4.18\)) · (e) about 15:21 (\(t=7.34\))
121AAHL5-q01105.18 First order differential equationsPaper 3Hard27 marks
This question investigates the error made when Euler’s method is used to approximate a solution of a differential equation.
Consider the differential equation \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=y\), where \(y=1\) when \(x=0\).
(a)(i)
Write down the exact value of \(y\) when \(x=1\).
(1)
(a)(ii)
Use Euler’s method with a step length of \(0.5\) to find an approximate value of \(y\) when \(x=1\).
(2)
(a)(iii)
Use Euler’s method with a step length of \(0.25\) to find an approximate value of \(y\) when \(x=1\).
(1)
(b)
Show that Euler’s method with a step length of \(\dfrac{1}{n}\), \(n\in\mathbb{Z}^{+}\), gives the approximation \(E_{n}=\left(1+\dfrac{1}{n}\right)^{n}\) for the value of \(y\) when \(x=1\).
(3)
(c)(i)
Let \(D_{n}=\mathrm{e}-E_{n}\) be the error in this approximation.
Find the values of \(nD_{n}\) for \(n=10,\ 100\) and \(1000\).
(2)
(c)(ii)
Hence suggest the value, to three significant figures, of \(\displaystyle\lim_{n\to\infty}nD_{n}\).
(1)
(d)
Use the Maclaurin series for \(\ln(1+x)\) to show that \(n\ln\left(1+\dfrac{1}{n}\right)=1-\dfrac{1}{2n}+\dfrac{1}{3n^{2}}-\cdots\), for \(n\gt1\).
(2)
(e)
Hence, using the Maclaurin series for \(\mathrm{e}^{x}\), show that \(E_{n}=\mathrm{e}\left(1-\dfrac{1}{2n}+\dfrac{11}{24n^{2}}-\cdots\right)\).
(4)
(f)
Hence find the exact value of \(\displaystyle\lim_{n\to\infty}nD_{n}\) and compare it with your answer to part (c)(ii).
(2)
(g)
Euler’s method with a step length of \(\dfrac{1}{n}\) is now applied to \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=ky\), \(k\gt0\), with \(y=1\) when \(x=0\), to approximate \(y\) when \(x=1\). Show that the error \(\mathrm{e}^{k}-\left(1+\dfrac{k}{n}\right)^{n}\) is approximately \(\dfrac{k^{2}\mathrm{e}^{k}}{2n}\) for large \(n\).
(4)
(h)(i)
An improved estimate of \(\mathrm{e}\) is \(F_{n}=2E_{2n}-E_{n}\).
Use part (e) to show that \(\mathrm{e}-F_{n}\approx\dfrac{11\mathrm{e}}{48n^{2}}\) for large \(n\).
(3)
(h)(ii)
Find \(F_{10}\) and compare its error with \(D_{10}\).
(2)
Show mark scheme
Working
Mark
Notes
(a)(i)
\(\mathrm{e}\)
A1
Accept \(2.72\).
(a)(ii)
\(y_{1}=1+0.5(1)=1.5\)
M1
\(y_{2}=1.5+0.5(1.5)=2.25\)
A1
(a)(iii)
\(1.25^{4}=2.44140625\approx2.44\)
A1
(b)
\(y_{r+1}=y_{r}+\dfrac{1}{n}y_{r}\)
M1
\(=\left(1+\dfrac{1}{n}\right)y_{r}\), so \(y_{r}\) is geometric with \(y_{0}=1\): \(y_{r}=\left(1+\dfrac{1}{n}\right)^{r}\)
A1
\(x=1\) is reached after \(n\) steps of length \(\dfrac{1}{n}\)
\(\mathrm{e}^{k}-\left(1+\dfrac{k}{n}\right)^{n}=\dfrac{k^{2}\mathrm{e}^{k}}{2n}+\) terms in \(\dfrac{1}{n^{2}},\ \dfrac{1}{n^{3}},\ldots\), so approximately \(\dfrac{k^{2}\mathrm{e}^{k}}{2n}\)
122AAHL5-q01115.18 First order differential equationsPaper 3Hard26 marks
This question investigates families of curves that cross each other at right angles.
Suppose every curve of a family satisfies a differential equation \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=g(x,y)\). A curve whose gradient at every point \((x,y)\) is \(-\dfrac{1}{g(x,y)}\) is called an orthogonal trajectory of the family: it meets every curve of the family at right angles.
The following diagram, which is referred to in part (c), shows the circles \(x^{2}+y^{2}=2ax\) for \(a=\pm1,\pm2\) and the circles \(x^{2}+y^{2}=2by\) for \(b=\pm1,\pm2\).
(a)(i)
Consider the family of parabolas \(y=cx^{2}\), where \(c\neq0\) and \(x\neq0\).
Show that every curve of the family satisfies \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{2y}{x}\).
(2)
(a)(ii)
Write down the differential equation satisfied by the orthogonal trajectories.
(1)
(a)(iii)
Solve your differential equation to show that the orthogonal trajectories are the ellipses \(x^{2}+2y^{2}=k\), where \(k\) is a positive constant.
(3)
(b)(i)
Now consider the family of curves \(y=cx^{m}\), where \(c\neq0\), \(x\neq0\) and \(m\in\mathbb{Z}^{+}\).
Show that the orthogonal trajectories of this family are the curves \(x^{2}+my^{2}=k\).
(3)
(b)(ii)
State the value of \(m\) for which the orthogonal trajectories are circles, and describe the original family of curves in this case.
(1)
(c)(i)
Now consider the family of circles \(x^{2}+y^{2}=2ax\), where \(a\neq0\). Some of these circles, and some of their orthogonal trajectories, are shown in the diagram above.
By eliminating \(a\), show that every circle of the family satisfies \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{y^{2}-x^{2}}{2xy}\).
(4)
(c)(ii)
Write down the differential equation satisfied by the orthogonal trajectories.
(1)
(c)(iii)
Use the substitution \(y=vx\), and the identity \(\dfrac{1-v^{2}}{v\left(1+v^{2}\right)}\equiv\dfrac{1}{v}-\dfrac{2v}{1+v^{2}}\), to show that the orthogonal trajectories are the circles \(x^{2}+y^{2}=2by\), where \(b\) is a constant.
(6)
(c)(iv)
Describe the orthogonal trajectories geometrically.
(1)
(d)
The circles \(x^{2}+y^{2}=4x\) and \(x^{2}+y^{2}=2y\) meet at the origin and at the point P. Find the coordinates of P and verify that the two circles meet at right angles at P.
gradients at P from (c)(i) and (c)(ii): \(\dfrac{\frac{64}{25}-\frac{16}{25}}{2\cdot\frac{4}{5}\cdot\frac{8}{5}}=\dfrac{3}{4}\) and \(\dfrac{2\cdot\frac{4}{5}\cdot\frac{8}{5}}{\frac{16}{25}-\frac{64}{25}}=-\dfrac{4}{3}\)
M1
Or by implicit differentiation of each circle.
\(\dfrac{3}{4}\times\left(-\dfrac{4}{3}\right)=-1\), so the tangents at P are perpendicular
R1
Answers: (a)(ii) \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=-\dfrac{x}{2y}\) · (b)(ii) \(m=1\), lines through the origin · (c)(ii) \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{2xy}{x^{2}-y^{2}}\) · (c)(iv) circles centred on the \(y\)-axis through O · (d) \(\mathrm{P}\left(\tfrac{4}{5},\tfrac{8}{5}\right)\), gradients \(\tfrac{3}{4}\) and \(-\tfrac{4}{3}\)
123AAHL5-q01235.19 Maclaurin seriesPaper 3Hard27 marks
This question uses a family of integrals to investigate how quickly the Maclaurin series for \(\mathrm{e}^{x}\) at \(x=1\) converges, and to prove that \(\mathrm{e}\) is irrational.
Let \(J_{n}=\dfrac{1}{n!}\displaystyle\int_{0}^{1}x^{n}\mathrm{e}^{-x}\,\mathrm{d}x\), for \(n\in\mathbb{N}\), and let \(S_{n}=\displaystyle\sum_{r=0}^{n}\frac{1}{r!}\).
(a)
Find the exact values of \(J_{0}\) and \(J_{1}\).
(3)
(b)
Show that \(J_{n}=J_{n-1}-\dfrac{1}{n!\,\mathrm{e}}\), for \(n\in\mathbb{Z}^{+}\).
(3)
(c)
Hence find \(J_{2}\) and \(J_{3}\), giving each answer in the form \(1-\dfrac{q}{\mathrm{e}}\), where \(q\in\mathbb{Q}\).
(2)
(d)
Prove by mathematical induction that \(J_{n}=1-\dfrac{S_{n}}{\mathrm{e}}\), for \(n\in\mathbb{N}\).
(5)
(e)
By considering the values of \(\mathrm{e}^{-x}\) for \(0\le x\le1\), show that \(\dfrac{1}{\mathrm{e}\,(n+1)!}\lt J_{n}\lt\dfrac{1}{(n+1)!}\).
(3)
(f)
Hence show that \(0\lt\mathrm{e}-S_{n}\lt\dfrac{\mathrm{e}}{(n+1)!}\), and explain how this confirms the value of the Maclaurin series for \(\mathrm{e}^{x}\) at \(x=1\).
(3)
(g)
Find the least value of \(n\) for which the inequality in part (f) guarantees that \(S_{n}\) differs from \(\mathrm{e}\) by less than \(10^{-8}\).
(3)
(h)
Suppose that \(\mathrm{e}=\dfrac{p}{q}\), where \(p,q\in\mathbb{Z}^{+}\). By considering \(q!\left(\mathrm{e}-S_{q}\right)\), prove that \(\mathrm{e}\) is irrational.
dividing by \(n!\): \(J_{n}=-\dfrac{1}{n!\,\mathrm{e}}+\dfrac{1}{(n-1)!}\displaystyle\int_{0}^{1}x^{n-1}\mathrm{e}^{-x}\,\mathrm{d}x\), and the second term is \(J_{n-1}\)
\(\dfrac{\mathrm{e}}{(n+1)!}\to0\) as \(n\to\infty\), so \(S_{n}\to\mathrm{e}\): the series \(1+1+\dfrac{1}{2!}+\dfrac{1}{3!}+\cdots\) (the Maclaurin series for \(\mathrm{e}^{x}\) with \(x=1\)) converges to \(\mathrm{e}\)
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