434 free, high-quality past-paper-style questions on Calculus for IB Maths AA HL, each with a full mark scheme. Covers: Limits and the derivative; Increasing and decreasing functions; Differentiating polynomials and integer powers; and more. For when you have finished the past papers, or want to practise one topic at a time.Topic 5 at Standard Level →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.3,557 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
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 \(x=b\).
Diagram not to scale
(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 \(x=1\).
Diagram not to scale
(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 \(x\gt 0\).
Diagram not to scale
(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 \(x=0\).
R1 \(y\) is greatest when \(2-\sin x\) is least, i.e. when \(\sin x=1\) ; A1 maximum value \(y=1\)
✓ 2
Common error: Forgetting the minus sign when integrating \(y^{-2}\) leads to \(\frac{1}{y}=\sin x+c\), which gives the wrong constant and the wrong function.
33IBMA2H-053-00-045.3 Differentiating polynomials and integer powersPaper 1Easy6 marks
No calculator
Let \(f(x)=x^3+3x^2+6x-4\), for \(x\in\mathbb{R}\).
(a)
Find \(f'(x)\).
(1)
(b)
By writing \(f'(x)\) in the form \(a(x+b)^2+c\), show that f is increasing for all values of x.
(3)
(c)
Hence explain why the equation \(f(x)=0\) has exactly one real solution.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(f'(x)=3x^2+6x+6\)
✓ 1
Part (b)
M1 attempt to complete the square ; A1 \(f'(x)=3(x+1)^2+3\) ; R1 \(3(x+1)^2\geq0\) so \(f'(x)\geq3>0\) for all x, hence f is increasing
✓ 3
Part (c)
R1 \(f(0)=-4<0\) and \(f=6>0\), and f is continuous, so there is at least one root ; R1 f is increasing for all x, so the graph crosses the x-axis only once
✓ 3
Exam tip: A 'hence' explanation needs two ingredients: existence of a root (a sign change) and uniqueness (the function is strictly increasing).
34IBMA2H-053-10-015.8 Stationary points, inflexion and optimizationPaper 1Easy8 marks
No calculator
The function f is defined by \(f(x)=x^3+ax^2+bx\), where \(a,b\in\mathbb{Z}\). The graph of f has a stationary point at \((1,-4)\).
(a)
Find the value of a and the value of b.
(4)
(b)
Use the second derivative to determine the nature of the stationary point at \((1,-4)\).
(2)
(c)
Find the exact x-coordinate of the other stationary point of the graph of f.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 differentiates: \(f'(x)=3x^2+2ax+b\)
✓ 1
M1 uses both conditions: substituting \(x=1\) into f gives \(1+a+b=-4\), and substituting \(x=1\) into f' gives \(3+2a+b=0\)
✓ 1
A1 \(a=2\)
✓ 1
A1 \(b=-7\)
✓ 1
Part (b)
M1 \(f''(x)=6x+4\), so the value of f'' at \(x=1\) is \(10\)
✓ 1
R1 this is positive, hence \((1,-4)\) is a local minimum
✓ 1
Part (c)
M1 solves \(3x^2+4x-7=0\), e.g. \((3x+7)(x-1)=0\)
✓ 1
A1 \(x=-\frac{7}{3}\)
✓ 1
Exam tip: A stationary point gives you two facts: the point lies on the curve, and the gradient there is zero. With two unknowns, you need both equations.
35IBMA2H-053-10-045.15 Further derivatives and integrals; partial fractionsPaper 1Easy6 marks
No calculator
A battery with internal resistance 2 Ω is connected to a variable resistor of resistance R Ω. The power delivered to the resistor is \(P(R)=\frac{144R}{(R+2)^2}\) watts, \(R>0\).
(a)
Show that \(P'(R)=\frac{144(2-R)}{(R+2)^3}\).
(3)
(b)
Find the value of R for which the power is a maximum, justifying that it is a maximum.
(2)
(c)
Find the maximum power.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use the quotient rule (or product rule)
✓ 1
A1 \(P'(R)=\frac{144(R+2)^2-288R(R+2)}{(R+2)^4}\)
✓ 1
A1 factorises \((R+2)\) to obtain \(\frac{144(R+2)(2-R)}{(R+2)^4}=\frac{144(2-R)}{(R+2)^3}\) AG
✓ 1
Part (b)
M1 \(P'(R)=0\Rightarrow R=2\)
✓ 1
R1 \(P'(R)>0\) for \(0<R<2\) and \(P'(R)<0\) for \(R>2\), so maximum
✓ 1
Part (c)
A1 \(P(2)=\frac{288}{16}=18\) W
✓ 1
Exam tip: In a 'show that', factor out the common \((R+2)\) explicitly before cancelling; skipping that line can cost the final mark.
36IBMA2H-053-10-125.14 Implicit differentiation, related rates and optimisationPaper 1Easy6 marks
No calculator
A rectangular frame for a greenhouse vent is made from 12 m of aluminium strip. The frame has width w m and height h m, and two extra vertical bars of height h divide it into three panels. All 12 m of strip is used.
Diagram not to scale
(a)
Show that the area enclosed by the frame is \(A=6h-2h^2\).
(2)
(b)
Find the value of h that maximises the area, justifying that it gives a maximum.
(3)
(c)
Find the maximum area.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(2w+4h=12\)
✓ 1
A1 \(w=6-2h\), so \(A=wh=6h-2h^2\) AG
✓ 1
Part (b)
A1 \(\frac{dA}{dh}=6-4h\)
✓ 1
M1 sets \(6-4h=0\Rightarrow h=1.5\) m
✓ 1
R1 \(\frac{d^2A}{dh^2}=-4<0\), so maximum
✓ 1
Part (c)
A1 \(A=6(1.5)-2(1.5)^2=4.5\) m2
✓ 1
Common error: Count every piece of strip: there are four vertical lengths of \(h\) (two sides and two bars), not two.
37IBMA2H-053-04-015.9 KinematicsPaper 1Easy5 marks
No calculator
A delivery drone takes off and moves vertically upwards. Its velocity, in m s−1, t seconds after take-off is modelled by v(t) = 6 / ((t + 1)(t + 3)), 0 ≤ t ≤ 3.
Exam tip: When an answer is given, show every log law you use. Writing 3 ln(2/3) + 3 ln 3 = 3 ln 2 explicitly is what earns the final mark.
38IBMA2H-053-04-025.9 KinematicsPaper 1Easy6 marks
No calculator
A camera dolly used in filming moves along a straight track. Its velocity, in m s−1, at time t seconds is given by v(t) = t2 − 6t + 8, for 0 ≤ t ≤ 4.
(a)
Find the values of t for which the dolly is instantaneously at rest.
(2)
(b)
Find the total distance travelled by the dolly from t = 0 to t = 4.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 solves (t − 2)(t − 4) = 0
✓ 1
A1 t = 2 and t = 4
✓ 1
Part (b)
M1 recognises the velocity changes sign at t = 2 and splits the integral (or integrates |v|)
✓ 1
A1 ∫02 v dt = 20/3
✓ 1
A1 ∫24 v dt = −4/3
✓ 1
A1 total distance = 20/3 + 4/3 = 8 m
✓ 1
Common error: Integrating v from 0 to 4 in one go gives 16/3, the displacement, not the distance. Always check where v = 0 before finding a distance.
39IBMA2H-053-04-045.9 KinematicsPaper 1Easy7 marks
No calculator
An autonomous delivery cart moves along a straight pavement. Its acceleration, in m s−2, at time t seconds is a(t) = 4 − 2t. When t = 0 its velocity is 5 m s−1.
(a)
Find an expression for the velocity v(t).
(2)
(b)
Find the maximum velocity of the cart.
(2)
(c)
The cart comes to rest at time t = T, where T > 0. Find the displacement of the cart from t = 0 to t = T.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 integrates: v = 4t − t2 + c
✓ 1
A1 uses v = 5 at t = 0: v(t) = 5 + 4t − t2
✓ 1
Part (b)
M1 sets a(t) = 0 (or v′(t) = 0), giving t = 2
✓ 1
A1 maximum velocity v(2) = 9 m s−1
✓ 1
Part (c)
M1 sets 5 + 4T − T2 = 0 and solves: (T − 5)(T + 1) = 0, so T = 5 (rejecting T = −1)
✓ 1
M1 integrates v with limits 0 and 5: ∫05 (5 + 4t − t2) dt = [5t + 2t2 − t3/3]05
✓ 1
A1 100/3 m
✓ 1
Common error: Forgetting the constant of integration in part (a) gives v = 4t − t². This has a maximum of only 4 m s⁻¹, and the error carries into every later part.
40IBMA2H-053-09-015.1 Limits and the derivativePaper 1Easy7 marks
No calculator
After an oral dose, the concentration of a drug in a patient's bloodstream is modelled by \(C(t)=20te^{-0.5t}\), where \(C\) is measured in mg L−1 and \(t\) is the time in hours after the dose, \(t\ge 0\).
(a)
Find \(C'(t)\).
(3)
(b)
Hence find the time at which the concentration is greatest, justifying that it is a maximum, and find the exact value of this maximum concentration.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use the product rule with \(u=20t\), \(v=e^{-0.5t}\)
✓ 1
A1 derivative of \(e^{-0.5t}\) is \(-0.5e^{-0.5t}\)
R1 justification: \(C'(t)>0\) for \(0\le t<2\) and \(C'(t)<0\) for \(t>2\), so \(C'\) changes sign from positive to negative and \(t=2\) gives a maximum. Accept \(C''(2)=-10e^{-1}<0\)
✓ 1
A1 \(C(2)=40e^{-1}=\frac{40}{e}\) mg L−1
✓ 1
Common error: Forgetting the factor −0.5 from the chain rule when differentiating \(e^{-0.5t}\) gives the wrong stationary time. Factorise out the exponential before solving \(C'(t)=0\), since an exponential is never zero. Then justify the maximum with a sign change of \(C'\) or with \(C''<0\).
41IBMA2H-053-09-035.15 Further derivatives and integrals; partial fractionsPaper 1Easy6 marks
No calculator
Let \(f(x)=\arctan(2x)+3^x\) and \(g(x)=\log_2(x^2+1)\), for \(x\in\mathbb{R}\).
(a)
Find \(f'(x)\) and hence write down the exact value of \(f'(0)\).
(4)
(b)
Find the exact value of \(g'(1)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of chain rule on \(\arctan(2x)\)
✓ 1
A1 derivative of \(\arctan(2x)\) is \(\dfrac{2}{1+4x^2}\)
✓ 1
A1 derivative of \(3^x\) is \(3^x\ln 3\), giving \(f'(x)=\dfrac{2}{1+4x^2}+3^x\ln 3\)
✓ 1
A1 substituting \(x=0\) gives \(2+\ln 3\)
✓ 1
Part (b)
M1 \(g'(x)=\dfrac{2x}{(x^2+1)\ln 2}\), or change of base to \(\dfrac{\ln(x^2+1)}{\ln 2}\) and differentiate
Common error: Writing the derivative of \(3^x\) as \(x\cdot 3^{x-1}\) applies the power rule to an exponential. The power rule is only for a variable base raised to a constant power; \(a^x\) differentiates to \(a^x\ln a\).
42IBMA2H-053-09-045.14 Implicit differentiation, related rates and optimisationPaper 1Easy7 marks
No calculator
A designer models the curved edge of a glass panel by the curve with equation \(x^2+xy+y^3=7\).
(a)
Show that the point \(P(2,1)\) lies on the curve.
(1)
(b)
Find the gradient of the curve at \(P\).
(4)
(c)
The tangent to the curve at \(P\) meets the \(x\)-axis at \(Q\). Find the \(x\)-coordinate of \(Q\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 substitutes \(x=2\), \(y=1\): \(2^2+2\times 1+1^3=4+2+1=7\), so \(P\) lies on the curve
✓ 1
Part (b)
M1 attempt at implicit differentiation
✓ 1
A1 \(x\frac{dy}{dx}+y\) from the product \(xy\)
✓ 1
A1 \(3y^2\frac{dy}{dx}\) from \(y^3\), giving \(2x+y+x\frac{dy}{dx}+3y^2\frac{dy}{dx}=0\)
✓ 1
A1 \(\frac{dy}{dx}=-\frac{2x+y}{x+3y^2}=-\frac{5}{5}=-1\) at \(P\)
✓ 1
Part (c)
M1 tangent \(y-1=-(x-2)\), i.e. \(y=3-x\)
✓ 1
A1 setting \(y=0\) gives \(x=3\)
✓ 1
Common error: Differentiating \(xy\) as just \(\frac{dy}{dx}\) or just \(y\) loses a term. The product rule is needed whenever \(x\) and \(y\) are multiplied together.
43IBMA2H-053-06-015.8 Stationary points, inflexion and optimizationPaper 1Easy5 marks
No calculator
The energy used per kilometre by a delivery drone flying at a constant speed v m s−1 is modelled by \(E(v) = v^2 + \frac{54}{v}\), for 1 ≤ v ≤ 10, where E is measured in energy units.
(a)
Find E′(v).
(2)
(b)
Find the speed at which the energy used per kilometre is a minimum.
(2)
(c)
Find this minimum energy per kilometre.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes \(\frac{54}{v}\) as 54v−1 and differentiates term by term
✓ 1
A1 E′(v) = 2v − 54v−2
✓ 1
Part (b)
M1 sets 2v − 54v−2 = 0, giving v3 = 27
✓ 1
A1 v = 3 m s−1
✓ 1
Part (c)
A1 E(3) = 9 + 18 = 27 energy units
✓ 1
Common error: The derivative of 54/v is −54/v², not 54 or 54/v². Rewrite it as 54v⁻¹ before differentiating so the power rule handles the sign.
44IBMA2H-053-06-025.13 Limits and l’Hôpital’s rulePaper 1Easy4 marks
No calculator
Use l'Hôpital's rule to find the exact value of \(\lim_{x\to 0} \frac{e^{3x}-1-3x}{x^2}\).
(a)
Find the limit, showing each application of l'Hôpital's rule.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 notes the form 0/0 and differentiates the numerator and denominator separately
✓ 1
A1 \(\frac{3e^{3x}-3}{2x}\)
✓ 1
M1 notes this is still 0/0 and applies the rule again
✓ 1
A1 \(\frac{9e^{3x}}{2} \to \frac{9}{2}\)
✓ 1
Exam tip: Before each application of l'Hôpital's rule, state that the limit is of indeterminate form 0/0 by substituting the limiting value. This check earns credit and stops you applying the rule when it is not valid.
45IBMA2H-053-06-035.4 Tangents and normalsPaper 1Easy6 marks
No calculator
The cross-section of a cable guide on a stage lighting rig is modelled by the curve \(y = kx^2 + \frac{6}{x}\), for 0.5 ≤ x ≤ 3, where k is a constant. The gradient of the curve at x = 1 is 10.
(a)
Find the value of k.
(3)
(b)
Find the equation of the normal to the curve at x = 1, giving your answer in the form y = mx + c.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 differentiates, using 6x−1
✓ 1
A1 \(\frac{dy}{dx} = 2kx - 6x^{-2}\)
✓ 1
A1 2k − 6 = 10, so k = 8
✓ 1
Part (b)
A1 y = 8 + 6 = 14 at x = 1
✓ 1
M1 normal gradient = \(-\frac{1}{10}\)
✓ 1
A1 \(y = -\frac{1}{10}x + \frac{141}{10}\)
✓ 1
Common error: The normal is perpendicular to the tangent, so its gradient is the negative reciprocal −1/10. Using −10 or 10 is a frequent slip.
46IBMA2H-053-06-045.13 Limits and l’Hôpital’s rulePaper 1Easy5 marks
No calculator
Use l'Hôpital's rule to find \(\lim_{x\to 0} \frac{1-\cos 2x}{x\sin x}\).
(a)
Find the limit.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 notes the form 0/0 and applies l'Hôpital's rule
Common error: The denominator x sin x is a product, so its derivative needs the product rule: sin x + x cos x. Writing just cos x gives a wrong answer.
47IBMA2H-053-07-015.12 Limits, first principles and higher derivativesPaper 1Easy5 marks
No calculator
The lift, in mm, of a cam follower is modelled by \(f(x)=kx^2+3x\), where \(x\) is the angle of rotation of the cam in radians and \(k\) is a constant. It is known that \(f'(2)=11\). Use differentiation from first principles to find \(f'(x)\) in terms of \(k\) and \(x\), and hence find the value of \(k\).
(a)
Find \(f'(x)\) from first principles and hence find \(k\).
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 correct substitution into \(\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\) with \(f(x+h)=k(x+h)^2+3(x+h)\)
✓ 1
A1 numerator simplified to \(2kxh+kh^2+3h\)
✓ 1
M1 dividing by \(h\) to get \(2kx+kh+3\) and letting \(h\to0\)
✓ 1
A1 \(f'(x)=2kx+3\)
✓ 1
A1 \(4k+3=11\) so \(k=2\)
✓ 1
Exam tip: Write 'lim h→0' on every line until you actually let h tend to zero; dropping the limit notation early can cost the method mark.
48IBMA2H-053-07-035.14 Implicit differentiation, related rates and optimisationPaper 1Easy6 marks
No calculator
A hiking trail climbs a hillside. The height of the trail above a lake is \(y\) metres at a horizontal distance \(x\) metres from a marker post, where \(\frac{dy}{dx}=3\sqrt{x}-2\) for \(1\le x\le 16\). The trail passes through the point \((4,12)\).
(a)
Find an expression for \(y\) in terms of \(x\).
(5)
(b)
Find the height of the trail above the lake at \(x=9\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to integrate, at least one power increased by 1
✓ 1
A1 \(2x^{3/2}\)
✓ 1
A1 \(-2x+c\)
✓ 1
M1 substituting \(x=4,\ y=12\): \(16-8+c=12\)
✓ 1
A1 \(y=2x^{3/2}-2x+4\)
✓ 1
Part (b)
A1 \(54-18+4=40\) m
✓ 1
Common error: Forgetting the constant of integration means the given point cannot be used, and every later value will be wrong.
49IBMA2H-053-07-045.12 Limits, first principles and higher derivativesPaper 1Easy6 marks
No calculator
The size of a bacterial culture in a laboratory is modelled by \(f(x)=xe^{2x}\), where \(x\ge0\) is the time in days.
(a)
Show that \(f'(x)=(1+2x)e^{2x}\).
(2)
(b)
Find \(f''(x)\), giving your answer in the form \(a(x+b)e^{2x}\).
(2)
(c)
Find the value of \(f^{(3)}(0)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of the product rule with \(u=x\), \(v=e^{2x}\) and \(\frac{d}{dx}e^{2x}=2e^{2x}\)
✓ 1
A1 \(e^{2x}+2xe^{2x}=(1+2x)e^{2x}\)
✓ 1
Part (b)
M1 product rule applied to \((1+2x)e^{2x}\): \(2e^{2x}+2(1+2x)e^{2x}\)
✓ 1
A1 \(f''(x)=4(x+1)e^{2x}\)
✓ 1
Part (c)
M1 product rule applied to \(4(x+1)e^{2x}\): \(4e^{2x}+8(x+1)e^{2x}=(8x+12)e^{2x}\)
✓ 1
A1 \(f^{(3)}(0)=12\)
✓ 1
Exam tip: Factorise out \(e^{2x}\) at each stage. This keeps the next differentiation short and makes the pattern in higher derivatives easy to see.
50IBMA2H-053-01-015.14 Implicit differentiation, related rates and optimisationPaper 1Easy6 marks
No calculator
A drone flies in a vertical plane. Its path is modelled by \(y = x^2 - 6x + 11\) for \(0 \le x \le 6\), where \(x\) is the horizontal distance and \(y\) is the height, both in metres.
Diagram not to scale
(a)
Find \(\frac{dy}{dx}\).
(1)
(b)
Find the equation of the tangent to the path at the point where \(x = 4\).
(3)
(c)
At the point P the drone is travelling parallel to the line \(y = -4x + 1\). Find the coordinates of P.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(\frac{dy}{dx} = 2x - 6\)
✓ 1
Part (b)
A1 \(y = 16 - 24 + 11 = 3\) when \(x = 4\)
✓ 1
M1 substitutes \(x = 4\) into \(\frac{dy}{dx}\) to get gradient \(2\)
Exam tip: When substituting \(3x\) for \(x\), put the whole of \(3x\) in brackets before raising it to a power: \((3x)^3 = 27x^3\), not \(3x^3\).
52IBMA2H-053-01-045.19 Maclaurin seriesPaper 1Easy6 marks
No calculator
The envelope of a modulated radio signal is modelled by \(f(x) = e^{\sin x}\).
(a)
Show that \(f''(x) = (\cos^2 x - \sin x)\,e^{\sin x}\).
(2)
(b)
Hence find the Maclaurin series for \(f(x)\) up to and including the term in \(x^2\).
(3)
(c)
Use your answer to the previous part to estimate \(f(0.2)\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 chain rule: \(f'(x) = \cos x\, e^{\sin x}\)
✓ 1
A1 product rule: \(f''(x) = -\sin x\, e^{\sin x} + \cos^2 x\, e^{\sin x}\), leading to the given answer
✓ 1
Part (b)
A1 \(f(0) = 1\), \(f'(0) = 1\) and \(f''(0) = 1\)
✓ 1
M1 use of \(f(0) + x f'(0) + \frac{x^2}{2!} f''(0)\)
✓ 1
A1 \(f(x) \approx 1 + x + \frac{x^2}{2}\)
✓ 1
Part (c)
A1 \(1 + 0.2 + 0.02 = 1.22\)
✓ 1
Common error: Remember to divide by \(2!\) in the \(x^2\) term: the coefficient is \(\frac{f''(0)}{2}\), not \(f''(0)\).
53IBMA2H-053-08-015.1 Limits and the derivativePaper 1Easy6 marks
No calculator
Rainwater flows into an initially empty storage tank. The rate at which the volume of water \(V\) m3 in the tank changes is modelled by \(\frac{dV}{dt}=\frac{6t}{t^2+4}\), where \(t\) is the time in hours after the rain starts, \(t \ge 0\).
(a)
Find \(\int \frac{6t}{t^2+4}\,dt\).
(3)
(b)
Hence find the exact volume of water in the tank after 2 hours.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 recognises that the numerator is a multiple of the derivative of \(t^2+4\), e.g. by substituting \(u=t^2+4\)
✓ 1
A1 \(3\ln(t^2+4)\) (accept \(3\ln|t^2+4|\))
✓ 1
A1 \(+C\) included
✓ 1
Part (b)
M1 uses \(V(0)=0\) to find \(C\), e.g. \(0=3\ln 4+C\); OR sets up \(V(2)=\int_0^2 \frac{6t}{t^2+4}\,dt\)
✓ 1
A1 \(C=-3\ln 4\); OR \(\left[3\ln(t^2+4)\right]_0^2=3\ln 8-3\ln 4\)
✓ 1
A1 \(V(2)=3\ln 2\) m3
✓ 1
Exam tip: When the numerator is a constant multiple of the derivative of the denominator, the integral is a multiple of the log of the denominator. Differentiate your answer to check the multiple, and combine logs to give a single exact value.
54IBMA2H-053-08-025.15 Further derivatives and integrals; partial fractionsPaper 1Easy5 marks
No calculator
A delivery drone takes off vertically from the ground, so that \(h=0\) when \(t=0\). Its height \(h\) metres above the ground at time \(t\) seconds satisfies \(\frac{dh}{dt}=\frac{8t}{\sqrt{t^2+9}}\), for \(0 \le t \le 6\).
(a)
Find the height of the drone when \(t=4\).
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempts the reverse chain rule (or the substitution \(u=t^2+9\)), obtaining the form \(k(t^2+9)^{1/2}\)
✓ 1
A1 \(8\sqrt{t^2+9}\) (+C)
✓ 1
M1 uses the initial condition \(h=0\) when \(t=0\), or sets up \(h(4)=\int_0^4 \frac{8t}{\sqrt{t^2+9}}\,dt\)
✓ 1
A1 \(C=-24\), or \([8\sqrt{t^2+9}]_0^4=8\times5-8\times3\)
✓ 1
A1 \(h(4)=16\) m
✓ 1
Common error: Taking the constant of integration as zero because the drone starts on the ground. At \(t=0\) the expression \(8\sqrt{t^2+9}\) equals 24, not 0, so \(C=-24\). Equivalently, the definite integral from 0 to 4 subtracts this 24.
55IBMA2H-053-08-035.5 Introduction to integrationPaper 1Easy4 marks
No calculator
A designer cuts a decorative panel from sheet metal. The panel is the region bounded by the curve \(y=x^3\) for \(x\ge0\), the \(y\)-axis, and the lines \(y=1\) and \(y=8\). Lengths are in centimetres.
Diagram not to scale
(a)
Find the exact area of the panel.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 rearranges to \(x=y^{1/3}\), OR identifies the \(x\)-limits 1 and 2 (from \(x^3=1\) and \(x^3=8\))
✓ 1
M1 sets up a correct area integral: \(\int_1^8 y^{1/3}\,dy\), OR \(2\times8-1\times1-\int_1^2 x^3\,dx\)
✓ 1
A1 correct integration: \(\left[\frac{3}{4}y^{4/3}\right]_1^8\), OR \(\left[\frac{x^4}{4}\right]_1^2=\frac{15}{4}\)
✓ 1
A1 area \(=\frac{45}{4}\) cm2, from \(\frac{3}{4}(16-1)\) or \(15-\frac{15}{4}\)
✓ 1
Exam tip: For an area against the \(y\)-axis, write \(x\) as a function of \(y\) and use the \(y\)-values 1 and 8 as limits, not the \(x\)-values 1 and 2.
56IBMA2H-053-02-015.14 Implicit differentiation, related rates and optimisationPaper 1Easy6 marks
No calculator
The edge of a cam in a small engine is modelled by the curve \(x^3 + xy + y^3 = 11\), where \(x\) and \(y\) are measured in centimetres. The point P(1, 2) lies on the curve.
(a)
Find an expression for \(\frac{dy}{dx}\) in terms of \(x\) and \(y\).
(3)
(b)
The tangent to the curve at P meets the x-axis at Q. Find the exact x-coordinate of Q.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt at implicit differentiation, using the product rule on \(xy\)
✓ 1
A1 \(3x^2 + y + x\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0\)
✓ 1
A1 \(\frac{dy}{dx} = -\frac{3x^2 + y}{x + 3y^2}\)
✓ 1
Part (b)
M1 substitute (1, 2) to get gradient \(-\frac{5}{13}\)
Common error: Differentiating \(xy\) as just \(\frac{dy}{dx}\) or just \(y\) loses the product rule. Every term containing both \(x\) and \(y\) needs the product rule.
57IBMA2H-053-02-025.9 KinematicsPaper 1Easy6 marks
No calculator
The height of a delivery drone above the ground is modelled by \(h(t) = t^3 - 6t^2 + 9t + 2\), where \(h\) is in metres and \(t\) is the time in seconds, \(0 \le t \le 4\).
(a)
Find \(h'(t)\) and \(h''(t)\).
(2)
(b)
Find the value of \(t\) at which the vertical velocity \(h'(t)\) takes its minimum value. Justify your answer.
(3)
(c)
Find the vertical velocity of the drone at this time.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(h'(t) = 3t^2 - 12t + 9\)
✓ 1
A1 \(h''(t) = 6t - 12\)
✓ 1
Part (b)
M1 set \(h''(t) = 0\)
✓ 1
A1 \(t = 2\)
✓ 1
R1 valid justification that \(h'\) is a minimum at \(t = 2\), e.g. \(h''\) changes from negative to positive at \(t = 2\), or \(h'''(t) = 6 > 0\), or \(h'\) is a quadratic with positive leading coefficient so \(t = 2\) is its vertex
✓ 1
Part (c)
A1 \(h'(2) = 12 - 24 + 9 = -3\) m s−1, i.e. descending at 3 m s−1
✓ 1
Exam tip: The minimum (most negative) velocity occurs where the acceleration \(h''\) is zero and changes sign. This is a point of inflection of \(h\), not a stationary point of \(h\).
58IBMA2H-053-02-045.7 The second derivative and graphsPaper 1Easy7 marks
No calculator
Consider the function \(f(x) = x^4 - 4x^3 + 10\).
(a)
Find the interval on which the graph of \(f\) is concave down.
(3)
(b)
Find the coordinates of the points of inflection of the graph of \(f\).
(3)
(c)
Show that one of these points of inflection is also a stationary point.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(f''(x) = 12x^2 - 24x\)
✓ 1
M1 solve \(12x(x - 2) < 0\)
✓ 1
A1 \(0 < x < 2\)
✓ 1
Part (b)
A1 (0, 10)
✓ 1
A1 (2, −6)
✓ 1
R1 \(f''\) changes sign at both \(x = 0\) and \(x = 2\)
✓ 1
Part (c)
R1 \(f'(x) = 4x^3 - 12x^2\), so \(f'(0) = 0\); hence (0, 10) is a stationary point of inflection
✓ 1
Exam tip: \(f''(x) = 0\) alone does not prove a point of inflection. State that \(f''\) changes sign either side of the point.
59IBMA2H-053-03-015.14 Implicit differentiation, related rates and optimisationPaper 1Easy5 marks
No calculator
A security camera is mounted at point C, 12 m from a straight running track, and swivels to follow a sprinter. The point P is the point on the track nearest to C, and the sprinter is x metres from P. The angle between CP and the camera's line of sight is \(\theta = \arctan\left(\frac{x}{12}\right)\) radians.
Diagram not to scale
(a)
Find \(\frac{d\theta}{dx}\), giving your answer as a single fraction in terms of x.
(2)
(b)
The sprinter runs away from P at a constant 8 m s−1. Find the exact rate at which the camera is rotating when x = 16.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use the chain rule with the derivative of arctan, e.g. \(\frac{1}{1+(x/12)^2}\times\frac{1}{12}\)
✓ 1
A1 \(\frac{d\theta}{dx}=\frac{12}{144+x^2}\)
✓ 1
Part (b)
M1 use of \(\frac{d\theta}{dt}=\frac{d\theta}{dx}\times\frac{dx}{dt}\) with \(\frac{dx}{dt}=8\)
✓ 1
A1 \(\frac{d\theta}{dx}=\frac{12}{400}=\frac{3}{100}\) at x = 16
✓ 1
A1 \(\frac{d\theta}{dt}=\frac{6}{25}\) rad s−1 (= 0.24 rad s−1)
✓ 1
Common error: Forgetting the inner derivative \(\frac{1}{12}\) gives \(\frac{144}{144+x^2}\), which is too large by a factor of 12. Always differentiate the argument of arctan.
60IBMA2H-053-03-025.17 Areas with the y-axis and volumes of revolutionPaper 1Easy4 marks
No calculator
Find the exact value of \(\displaystyle\int_0^{3/4}\frac{1}{\sqrt{9-4x^2}}\,dx\).
(a)
Find the exact value of the integral.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 rewrite in a standard arcsin form, e.g. \(\frac{1}{2}\int\frac{1}{\sqrt{\frac{9}{4}-x^2}}\,dx\)
Common error: Writing \(\arcsin\left(\frac{2x}{3}\right)\) without the factor \(\frac{1}{2}\): differentiate your answer to check that you recover the integrand exactly.
61IBMA2H-053-03-035.17 Areas with the y-axis and volumes of revolutionPaper 1Easy5 marks
No calculator
In a hospital laboratory, the number of bacteria in a culture t hours after it is prepared is modelled by \(N = 500\times 3^{t/2}\).
(a)
Show that \(\frac{dN}{dt}=kN\), stating the exact value of the constant k.
(3)
(b)
Find the exact rate of growth of the culture when t = 4.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(\frac{d}{dt}a^{f(t)}=a^{f(t)}\ln a\, f'(t)\)
A1 hence \(\frac{dN}{dt}=kN\) with \(k=\frac{\ln 3}{2}\)
✓ 1
Part (b)
M1 substitute t = 4: \(N=500\times 9=4500\)
✓ 1
A1 \(2250\ln 3\) bacteria per hour
✓ 1
Exam tip: The derivative of \(a^x\) is \(a^x\ln a\), not \(x a^{x-1}\). The power rule applies only when the variable is in the base.
62IBMA2H-053-05-015.12 Limits, first principles and higher derivativesPaper 1Easy6 marks
No calculator
A model rocket is launched vertically. Its height above the launch pad, s metres, t seconds after launch is modelled by \(s(t)=24t-4t^2\) for \(0\le t\le 6\).
(a)
Use differentiation from first principles to show that \(s'(t)=24-8t\).
(4)
(b)
Hence find the maximum height reached by the rocket.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(s'(t)=\lim_{h\to0}\frac{s(t+h)-s(t)}{h}\) A1 correct expansion \(s(t+h)=24t+24h-4t^2-8th-4h^2\) A1 numerator simplifies to \(24h-8th-4h^2\), and dividing by h gives \(24-8t-4h\) A1 letting \(h\to0\) gives \(s'(t)=24-8t\) (AG)
✓ 4
Part (b)
M1 sets \(s'(t)=0\), giving \(t=3\) A1 \(s(3)=72-36=36\) m
✓ 2
Exam tip: In a first-principles proof, write 'lim' on every line until you actually let h tend to 0. Only cancel h after it is a common factor of the whole numerator.
63IBMA2H-053-05-025.16 Integration by substitution and by partsPaper 1Easy5 marks
No calculator
After an injection, the concentration of a drug in a patient's blood is modelled by \(C(t)=8te^{-2t}\) mg L−1, where t is the time in hours. Pharmacologists measure total drug exposure by the area under the concentration–time curve.
The shaded area is the drug exposure over the first hour
(a)
Find \(\int te^{-2t}\,dt\).
(3)
(b)
Hence find the exact total exposure over the first hour, \(\int_0^1 C(t)\,dt\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempts integration by parts with \(u=t\) and \(\frac{dv}{dt}=e^{-2t}\) A1 \(-\frac12te^{-2t}+\int\frac12e^{-2t}\,dt\) A1 \(-\frac12te^{-2t}-\frac14e^{-2t}+c\)
✓ 3
Part (b)
M1 substitutes the limits into 8 times the result of the previous part: \(8\left[\left(-\frac12e^{-2}-\frac14e^{-2}\right)-\left(-\frac14\right)\right]\) A1 \(2-6e^{-2}\) mg h L−1
✓ 2
Common error: The lower limit does not give zero here: at t = 0 the term −¼e⁻²t equals −¼. Always evaluate every term at the lower limit.
64IBMA2H-053-05-035.16 Integration by substitution and by partsPaper 1Easy4 marks
No calculator
A tapered steel rod lies along the x-axis from \(x=0\) to \(x=4\), where x is measured in metres. Its linear density is \(\rho(x)=x\sqrt{x^2+9}\) kg m−1, so its mass is \(\int_0^4 \rho(x)\,dx\).
(a)
Use the substitution \(u=x^2+9\) to find the exact mass of the rod.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(u=x^2+9\) with \(du=2x\,dx\) A1 integral becomes \(\frac12\int_9^{25}u^{1/2}\,du\), with limits changed correctly A1 \(\left[\frac13u^{3/2}\right]_9^{25}\) A1 \(\frac{125-27}{3}=\frac{98}{3}\) kg
✓ 4
Exam tip: Change the limits to u-values as soon as you substitute. Then you never need to convert back to x.
65IBMA2H-053-05-045.13 Limits and l’Hôpital’s rulePaper 1Easy5 marks
No calculator
An electronics engineer needs the error in the linear approximation \(e^{3x}\approx1+3x\) for small x. This leads to the limit \(L=\lim_{x\to0}\frac{e^{3x}-1-3x}{x^2}\).
(a)
Explain why l'Hôpital's rule may be applied to L.
(1)
(b)
Hence, using l'Hôpital's rule, find the value of L.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 as \(x\to0\), the numerator \(\to1-1-0=0\) and the denominator \(\to0\) (both differentiable), so the limit has the indeterminate form \(\frac00\)
✓ 1
Part (b)
M1 differentiates numerator and denominator separately: \(\lim_{x\to0}\frac{3e^{3x}-3}{2x}\) R1 notes this is still of the form \(\frac00\) M1 applies l'Hôpital's rule again: \(\lim_{x\to0}\frac{9e^{3x}}{2}\) A1 \(L=\frac92\)
✓ 4
Common error: L'Hôpital's rule differentiates the numerator and denominator separately. It is not the quotient rule. Check that the form is 0/0 before each application.
66IBMA3H-5-18r9-015.1 Limits and the derivativePaper 1Easy5 marks
No calculator
The mass \(m\) kg of honey stored in a hive super changes at a rate \(\dfrac{dm}{dt}=6t^2-4t+1\) kg per day, where \(t\) is the time in days. After 1 day the mass is 5 kg.
(a)
Find an expression for \(m\) in terms of \(t\).
(4)
(b)
Find the mass of honey after 2 days.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 integrates the rate (at least one term correct)
✓ 1
A1 \(m=2t^3-2t^2+t+c\)
✓ 1
M1 uses \(m\left(1\right)=5\): \(1+c=5\)
✓ 1
A1 \(m=2t^3-2t^2+t+4\)
✓ 1
Part (b)
A1 \(m(2)=16-8+2+4=14\) kg
✓ 1
Common error: Do not forget the constant of integration; the given condition fixes it.
67IBMA3H-5-18r9-025.11 Definite integrals and areasPaper 1Easy6 marks
No calculator
The number of patients waiting in an emergency department is modelled by \(N(t)=t^3-9t^2+24t+10\) for \(0\le t\le 8\), where \(t\) is the time in hours after the start of an observation period.
(a)
Find \(N''(t)\).
(2)
(b)
Find the value of \(t\) at which the graph of \(N\) has a point of inflexion, justifying that it is a point of inflexion.
(2)
(c)
Explain what the point of inflexion means for the rate at which the number of waiting patients is changing.
(1)
(d)
Find the number of patients waiting at this time.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 differentiates twice
✓ 1
A1 \(N''(t)=6t-18\)
✓ 1
Part (b)
A1 \(N''(t)=0\) gives \(t=3\)
✓ 1
R1 \(N''\) changes sign (negative for \(t<3\), positive for \(t>3\))
✓ 1
Part (c)
R1 the rate of change \(N'(t)\) is at a minimum (the number is falling fastest) at \(t=3\)
✓ 1
Part (d)
A1 \(N(3)=27-81+72+10=28\)
✓ 1
Exam tip: A point of inflexion of N is a turning point of N', so it marks where the rate of change is greatest or least.
68IBMA3H-5-18r9-075.16 Integration by substitution and by partsPaper 1Easy5 marks
No calculator
A saline drip delivers fluid at a rate of \(r(t)=3\sqrt t-\dfrac{2}{t^2}\) millilitres per minute, where \(t\ge1\) is the time in minutes.
Exam tip: Check the \(\frac00\) or \(\frac\infty\infty\) form before applying l'Hôpital's rule, and differentiate the top and bottom separately (not as a quotient).
70IBMA3H-5-18r18-035.3 Differentiating polynomials and integer powersPaper 1Easy6 marks
No calculator
A curved rail that guides a theatre spotlight follows the graph of \(y=f(x)\), where \(f(x)=x^4-8x^3+18x^2\) and \(x\), \(f(x)\) are in suitable units.
(a)
Find \(f''(x)\).
(3)
(b)
Find the values of \(x\) for which the graph of \(f\) is concave down.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to differentiate
✓ 1
A1 \(f'(x)=4x^3-24x^2+36x\)
✓ 1
A1 \(f''(x)=12x^2-48x+36\)
✓ 1
Part (b)
M1 \(f''(x)<0\), i.e. \(12x^2-48x+36<0\)
✓ 1
A1 critical values \(x=1\) and \(x=3\) (from \(12(x-1)(x-3)\))
✓ 1
A1 \(1<x<3\)
✓ 1
Common error: Confusing 'concave down' with 'decreasing': concavity depends on the sign of \(f''(x)\), whereas increasing or decreasing depends on the sign of \(f'(x)\).
71IBMA3H-5-18r18-045.8 Stationary points, inflexion and optimizationPaper 1Easy5 marks
No calculator
A shipping company models the fuel cost per nautical mile, in hundreds of pounds, of a ship travelling at \(v\) knots as \(C(v)=0.02v+\dfrac{8}{v}\), for \(5\le v\le40\).
(a)
Find \(C'(v)\).
(2)
(b)
Find the speed at which the cost per nautical mile is a minimum, and justify that it is a minimum.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to differentiate (\(\frac8v\to-8v^{-2}\))
✓ 1
A1 \(C'(v)=0.02-\dfrac{8}{v^2}\)
✓ 1
Part (b)
M1 set \(0.02-\dfrac8{v^2}=0\) (ECF from (a))
✓ 1
A1 \(v^2=400\), so \(v=20\) knots
✓ 1
R1 \(C''(v)=\dfrac{16}{v^3}>0\), so a minimum
✓ 1
Exam tip: Give a reason for the minimum (sign of the second derivative, or sign change of the first) every time an optimisation question says 'justify'.
72IBMA3H-5-18r19-015.8 Stationary points, inflexion and optimizationPaper 1Easy4 marks
No calculator
The energy cost of a warehouse robot moving at speed \(v\) m s−1 is modelled by \(E(v) = v^3 + \frac{8}{v^2} - 12\sqrt{v}\) joules per metre, for \(v > 0\).
(a)
Find \(E'(v)\).
(3)
(b)
Find the rate of change of \(E\) with respect to \(v\) when \(v = 4\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to differentiate using the power rule, with \(\frac{8}{v^2} = 8v^{-2}\) and \(\sqrt{v} = v^{1/2}\)
✓ 1
A1 \(3v^2 - 16v^{-3}\)
✓ 1
A1 \(-6v^{-1/2}\), so \(E'(v) = 3v^2 - \frac{16}{v^3} - \frac{6}{\sqrt{v}}\)
Exam tip: Rewrite every term as a power of \(v\) before differentiating: \(\frac{8}{v^2} = 8v^{-2}\) and \(\sqrt{v} = v^{1/2}\).
73IBMA3H-5-18r19-025.4 Tangents and normalsPaper 1Easy5 marks
No calculator
A fashion designer cuts a curved neckline along the curve \(y = 2x^2 - 3x + 1\), where \(x\) and \(y\) are in centimetres.
(a)
Find the gradient of the curve at the point where \(x = -1\).
(2)
(b)
Find the equation of the normal at this point. Give your answer in the form \(ax + by + c = 0\) where \(a\), \(b\), \(c\) are integers.
(2)
(c)
Find the y-coordinate of the point where this normal crosses the y-axis.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{dy}{dx} = 4x - 3\)
✓ 1
A1 gradient \(= -7\)
✓ 1
Part (b)
M1 \(y = 6\) at \(x = -1\) and normal gradient \(\frac{1}{7}\), e.g. \(y - 6 = \frac{1}{7}(x + 1)\)
✓ 1
A1 \(x - 7y + 43 = 0\)
✓ 1
Part (c)
A1 \(y = \frac{43}{7}\) (ECF from (b))
✓ 1
Common error: The normal gradient is the negative reciprocal of the tangent gradient: from \(-7\) it is \(+\frac{1}{7}\), with the sign changed as well as the number inverted.
74IBMA3H-5-18r19-035.14 Implicit differentiation, related rates and optimisationPaper 1Easy5 marks
No calculator
The net rate of change of the stock of fish in a managed lake is \(\frac{dS}{dt} = 3\sqrt{t} - \frac{2}{t^2}\) tonnes per day, for \(t \ge 1\), where \(t\) is the time in days. At \(t = 1\) the stock is 40 tonnes.
(a)
Find the change in the stock between \(t = 1\) and \(t = 4\).
Exam tip: Integrating \(-2t^{-2}\) gives \(+2t^{-1}\): the sign changes because the new power is negative. Check by differentiating.
75IBMA3H-5-18r19-085.9 KinematicsPaper 1Easy4 marks
No calculator
The speed of a robotic gripper is \(v = 6t(1 + t^2)^2\) cm s−1, where \(t\) is the time in seconds. The gripper moves in a straight line in one direction. Use the substitution \(u = 1 + t^2\) to find the distance travelled in the first second.
(a)
Find the distance travelled by the gripper from \(t = 0\) to \(t = 1\).
A1 \(\int 3u^2\,du = u^3\) with limits changed to \(u = 1\) and \(u = 2\)
✓ 1
M1 evaluate \([u^3]_1^2\)
✓ 1
A1 \(8 - 1 = 7\) cm
✓ 1
Exam tip: When you substitute in a definite integral, change the limits to values of \(u\) and do not go back to \(t\) afterwards.
76IBMA3H-5-19r20-015.13 Limits and l’Hôpital’s rulePaper 1Easy4 marks
No calculator
The weekly profit, in thousands of euros, of a small shop that sells x hundred units of a seasonal product is modelled by \(P(x)=x^{3}e^{-x}\), for \(0\le x\le 8\).
(a)
Find \(P'(x)\).
(3)
(b)
Hence find the value of x for which \(P'(x)=0\), where \(x>0\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use the product rule
✓ 1
A1 \(3x^{2}e^{-x}\)
✓ 1
A1 \(-x^{3}e^{-x}\)
✓ 1
Part (b)
A1 \(P'(x)=x^{2}e^{-x}(3-x)=0\), and since \(x^{2}e^{-x}>0\) for \(x>0\), \(x=3\)
✓ 1
Common error: Differentiating \(e^{-x}\) as \(e^{-x}\) and losing the minus sign. Always apply the chain rule to the exponent.
77IBMA3H-5-19r20-035.6 Differentiation rulesPaper 1Easy6 marks
No calculator
During an outbreak, the cumulative number of infections after t days is modelled by \(I(t)=40+30t^{2}-t^{3}\), for \(0\le t\le 20\).
(a)
Find \(I'(t)\).
(2)
(b)
Find the time at which the number of infections is increasing most rapidly, justifying that it is a maximum rate, and state this greatest rate in cases per day.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to differentiate (at least one correct term)
✓ 1
A1 \(I'(t)=60t-3t^{2}\)
✓ 1
Part (b)
M1 \(I''(t)=60-6t=0\)
✓ 1
A1 \(t=10\) days
✓ 1
R1 justification, e.g. \(I'''(t)=-6<0\), or \(I''\) changes sign from positive to negative at \(t=10\)
✓ 1
A1 \(I'(10)=600-300=300\) cases per day
✓ 1
Exam tip: The rate of change of \(I\) is \(I'\), so the greatest rate occurs where \(I'\) is stationary, that is where \(I''=0\).
78IBMA3H-5-19r20-145.9 KinematicsPaper 1Easy5 marks
No calculator
An autonomous cleaning robot moves along a supermarket aisle. Its displacement from a charging point is \(s=2t^{3}-9t^{2}+12t\) metres, where t is the time in seconds, for \(0\le t\le3\).
(a)
Find an expression for the velocity of the robot and the times at which it is instantaneously at rest.
(3)
(b)
Find the acceleration of the robot at each of these times.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(v=6t^{2}-18t+12\)
✓ 1
M1 sets \(v=0\) and attempts to solve, e.g. \(6(t-1)(t-2)=0\)
✓ 1
A1 \(t=1\) and \(t=2\)
✓ 1
Part (b)
A1 \(a=12t-18\), so at \(t=1\), \(a=-6\ \mathrm{m\,s^{-2}}\)
✓ 1
A1 at \(t=2\), \(a=6\ \mathrm{m\,s^{-2}}\)
✓ 1
Common error: Acceleration is the derivative of velocity, not of displacement. Differentiate \(s\) once for \(v\) and twice for \(a\).
79IBMA3H-5-19r26-015.12 Limits, first principles and higher derivativesPaper 1Easy5 marks
No calculator
The curved edge of a theatre backdrop rail is modelled by the function \(f(x)=2x^2-7x\).
(a)
Use differentiation from first principles to find \(f'(x)\).
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to form \(\frac{f(x+h)-f(x)}{h}\) with \(f(x+h)\) substituted
✓ 1
A1 \(f(x+h)=2x^2+4xh+2h^2-7x-7h\)
✓ 1
A1 numerator simplified to \(4xh+2h^2-7h\)
✓ 1
M1 divide through by \(h\) to obtain \(4x+2h-7\)
✓ 1
A1 let \(h\to 0\), so \(f'(x)=4x-7\)
✓ 1
Common error: Writing \(\lim_{h\to0}\) before \(h\) has been cancelled gives \(\frac{0}{0}\). Simplify the numerator and divide by \(h\) first, then take the limit.
80IBMA3H-5-19r26-025.12 Limits, first principles and higher derivativesPaper 1Easy6 marks
No calculator
The profile of a ship's keel is modelled by \(y=3x^4-\frac{2}{x^3}+\frac{5}{x}\) for \(x>0\), where \(x\) and \(y\) are in metres.
(a)
Find \(\frac{dy}{dx}\).
(3)
(b)
Find \(\frac{d^2y}{dx^2}\).
(2)
(c)
Find the value of \(\frac{d^2y}{dx^2}\) when \(x=1\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 rewrite as \(3x^4-2x^{-3}+5x^{-1}\)
✓ 1
A1 \(12x^3+6x^{-4}\)
✓ 1
A1 \(-5x^{-2}\), so \(\frac{dy}{dx}=12x^3+6x^{-4}-5x^{-2}\)
✓ 1
Part (b)
M1 differentiate again, attempting the power rule on each term
✓ 1
A1 \(36x^2-24x^{-5}+10x^{-3}\)
✓ 1
Part (c)
A1 \(36-24+10=22\)
✓ 1
Exam tip: Rewrite terms such as \(\frac{2}{x^3}\) as \(2x^{-3}\) before differentiating; the power rule then applies to every term in the same way.
81IBMA3H-5-19r26-135.3 Differentiating polynomials and integer powersPaper 1Easy7 marks
No calculator
The cross-section of a theatre ramp is modelled by the curve \(y=x^3-3x^2+2\) for \(0\le x\le4\), where \(x\) and \(y\) are in metres.
(a)
Find \(\frac{dy}{dx}\).
(2)
(b)
Find the gradient of the curve at the point where \(x=3\).
(1)
(c)
Find the equation of the tangent to the curve at the point where \(x=3\).
(3)
(d)
Find the \(x\)-coordinate of the point where this tangent crosses the \(x\)-axis.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to differentiate using the power rule
✓ 1
A1 \(3x^2-6x\)
✓ 1
Part (b)
A1 \(27-18=9\)
✓ 1
Part (c)
A1 \(y=2\) when \(x=3\)
✓ 1
M1 \(y-2=9(x-3)\)
✓ 1
A1 \(y=9x-25\)
✓ 1
Part (d)
A1 \(9x-25=0\), so \(x=\frac{25}{9}\)
✓ 1
Common error: Find the \(y\)-coordinate on the curve before writing the tangent equation. The gradient comes from \(\frac{dy}{dx}\), but the point must come from the original function.
82IBMA3H-5-20r27-065.17 Areas with the y-axis and volumes of revolutionPaper 1Easy6 marks
No calculator
The cross-section of a parabolic dish is the curve \(y=\frac12x^2\) for \(0\le x\le2\), with lengths in metres. The rim of the dish is at \(y=2\). The region R is enclosed by the curve, the \(y\)-axis and the line \(y=2\).
Region R is shaded
(a)
Find the exact area of R.
(3)
(b)
The region R is rotated through \(360°\) about the \(y\)-axis to form the inside of the dish. Find the exact volume of the dish.
Common error: For rotation about the \(y\)-axis, write \(x^2\) in terms of \(y\) and integrate with respect to \(y\); using \(\int y^2\,dx\) gives the volume about the \(x\)-axis instead.
83IBMA3H-5-20r27-085.4 Tangents and normalsPaper 1Easy6 marks
No calculator
The curve \(y=f(x)\) has equation \(f(x)=4x^3-\dfrac6x+2x^2\) for \(x\ne0\).
(a)
Find \(f'(x)\).
(2)
(b)
Find the gradient of the curve at the point where \(x=1\).
(2)
(c)
Find the equation of the tangent to the curve at the point where \(x=1\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to differentiate, with at least one correct term
✓ 1
A1 \(f'(x)=12x^2+6x^{-2}+4x\)
✓ 1
Part (b)
M1 substitute \(x=1\) into \(f'(x)\)
✓ 1
A1 22
✓ 1
Part (c)
M1 the curve passes through \((1,0)\), so \(y-0=22(x-1)\)
✓ 1
A1 \(y=22x-22\)
✓ 1
Common error: Differentiating \(-\frac6x\) as \(-\frac{6}{1}\) or \(\frac{6}{x^2}\) with the wrong sign: rewrite as \(-6x^{-1}\) first, giving \(+6x^{-2}\).
84IBMA3H-5-20r27-095.3 Differentiating polynomials and integer powersPaper 1Easy7 marks
No calculator
The altitude \(h\) of a hiking trail, in units of 100 m, is modelled by \(h(x)=x^3-6x^2+9x+2\) for \(0\le x\le5\), where \(x\) is the distance in km along the trail.
(a)
Find \(h'(x)\).
(2)
(b)
Find the values of \(x\) for which the altitude is decreasing.
(3)
(c)
Find the total drop in altitude, in metres, between \(x=1\) and \(x=3\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to differentiate
✓ 1
A1 \(h'(x)=3x^2-12x+9\)
✓ 1
Part (b)
M1 \(h'(x)<0\), so \(3(x-1)(x-3)<0\)
✓ 1
A1 critical values \(x=1\) and \(x=3\)
✓ 1
A1 \(1<x<3\)
✓ 1
Part (c)
M1 altitude at \(x=1\) is 6 and at \(x=3\) is 2
✓ 1
A1 400 m
✓ 1
Exam tip: A function is decreasing where \(f'(x)<0\); solve the inequality using a sketch of the quadratic rather than just its roots.
85IBMA3H-5-20r31-015.4 Tangents and normalsPaper 1Easy7 marks
No calculator
Part of a suspension cable is modelled by the curve \(y=2x^2-3x+1\), where \(x\) and \(y\) are in metres. The point P on the cable has \(x=2\).
(a)
Find the equation of the tangent to the curve at P, in the form \(y=mx+c\).
(3)
(b)
Find the equation of the normal at P, in the form \(ax+by+c=0\) where \(a\), \(b\), \(c\) are integers.
(2)
(c)
Find the distance between the points where the tangent and the normal cross the \(x\)-axis.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 P is \((2,3)\)
✓ 1
M1 \(\frac{dy}{dx}=4x-3\), so gradient at P is 5
✓ 1
A1 \(y=5x-7\)
✓ 1
Part (b)
M1 normal gradient \(=-\frac{1}{5}\), so \(y-3=-\frac{1}{5}(x-2)\)
✓ 1
A1 \(x+5y-17=0\)
✓ 1
Part (c)
M1 tangent crosses at \(x=\frac{7}{5}\) and normal at \(x=17\)
✓ 1
A1 distance \(=\frac{78}{5}=15.6\) m
✓ 1
Common error: The normal gradient is the negative reciprocal of the tangent gradient: \(-\frac{1}{5}\), not \(-5\) or \(\frac{1}{5}\).
86IBMA3H-5-20r31-055.9 KinematicsPaper 1Easy4 marks
No calculator
The rate at which beans leave a hopper is \(R(t)=3\sqrt{t}-\frac{2}{t^2}\) kg per minute, for \(t\ge1\), where \(t\) is the time in minutes. Find the total mass of beans that leave the hopper between \(t=1\) and \(t=4\).
(a)
Find the total mass.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 mass \(=\int_1^4\left(3t^{1/2}-2t^{-2}\right)dt\)
✓ 1
A1 antiderivative \(2t^{3/2}+\frac{2}{t}\)
✓ 1
M1 evaluates at 4 and at 1: \(16.5-4\)
✓ 1
A1 12.5 kg
✓ 1
Common error: Writing \(\int -2t^{-2}\,dt\) as \(-\frac{2}{t}\) loses a sign; integrating \(t^{-2}\) gives \(-t^{-1}\), so the term becomes \(+\frac{2}{t}\).
87IBMA3H-5-20r31-135.19 Maclaurin seriesPaper 1Easy5 marks
No calculator
The Maclaurin series for \(e^x\) is \(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots\)
(a)
Find the Maclaurin series for \(e^{-2x}\) up to and including the term in \(x^3\).
(3)
(b)
Use your series with \(x=0.1\) to estimate \(e^{-0.2}\), giving your answer to 4 significant figures.
Common error: When replacing \(x\) by \(-2x\), the term in \(x^3\) becomes \(\frac{(-2x)^3}{3!}\), so the coefficient is \(-\frac{8}{6}=-\frac{4}{3}\), not \(-\frac{2}{6}\).
88IBMA3H-5-19r33-015.15 Further derivatives and integrals; partial fractionsPaper 1Easy5 marks
No calculator
A beekeeper models the mass of nectar stored in a hive by \(N(t)=t^{2}e^{-t/2}\) kg, where \(t\ge 0\) is the time in days since the start of a nectar flow.
(a)
Find \(N'(t)\).
(3)
(b)
Find the value of \(t>0\) at which the mass of nectar stored is stationary.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use the product rule on \(t^{2}e^{-t/2}\)
✓ 1
A1 correct derivative of the exponential factor, \(-\frac{1}{2}e^{-t/2}\)
M1 set \(N'(t)=0\) and factorise, noting \(e^{-t/2}>0\), e.g. \(te^{-t/2}\left(2-\frac{t}{2}\right)=0\)
✓ 1
A1 \(t=4\) days
✓ 1
Exam tip: An exponential factor is never zero, so it can be cancelled when solving for stationary points.
89IBMA3H-5-19r33-025.18 First order differential equationsPaper 1Easy5 marks
No calculator
The rate at which patients arrive at a hospital emergency department is modelled by \(R(t)=3t\sqrt{t^{2}+16}\) patients per hour, where \(t\) is the number of hours after 08:00, for \(0\le t\le 3\).
(a)
Using the substitution \(u=t^{2}+16\), show that the number of patients arriving between 08:00 and 11:00 is \(\frac{3}{2}\int_{16}^{25}u^{1/2}\,du\).
(3)
(b)
Hence find the number of patients who arrive in this period.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{du}{dt}=2t\), so \(t\,dt=\frac{1}{2}du\)
M1 integrate to obtain \(\frac{3}{2}\cdot\frac{2}{3}u^{3/2}\) (or \(u^{3/2}\)) and substitute the limits, e.g. \(25^{3/2}-16^{3/2}\)
✓ 1
A1 \(125-64=61\) patients
✓ 1
Common error: When changing variable in a definite integral, either change the limits to the new variable or return to the original variable before substituting; do not mix the two.
90IBMA3H-5-20r36-015.1 Limits and the derivativePaper 1Easy6 marks
No calculator
The volume V of ash, in thousands of cubic metres, released from a volcanic vent t hours after an eruption begins satisfies dV/dt = 12t(3t2 + 1)3, with V = 0 when t = 0.
(a)
Find an expression for V in terms of t.
(4)
(b)
Find the volume of ash released in the first hour.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 recognises the reverse chain rule, e.g. substitution u = 3t2 + 1
✓ 1
A1 antiderivative ½(3t2 + 1)4 + C
✓ 1
M1 uses V(0) = 0 to find the constant
✓ 1
A1 V = ½(3t2 + 1)4 − ½
✓ 1
Part (b)
M1 substitutes t = 1 (ECF from (a))
✓ 1
A1 127.5 thousand cubic metres
✓ 1
Common error: Forgetting the constant of integration: an initial condition such as V = 0 at t = 0 is what fixes it.
91IBMA3H-5-20r36-025.5 Introduction to integrationPaper 1Easy5 marks
No calculator
A theatre backdrop panel is bounded by the stage floor (the x-axis), the lines x = 0 and x = 4, and a curved top edge with equation y = k√x, where k > 0 and lengths are in metres. The area of the panel is 16 m2.
The panel is shaded
(a)
Find the value of k.
(3)
(b)
Find the area of the part of the panel between x = 1 and x = 4.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempts ∫ kx1/2 dx = (2k/3)x3/2
✓ 1
A1 evaluates between 0 and 4 to get 16k/3 = 16
✓ 1
A1 k = 3
✓ 1
Part (b)
M1 uses limits 1 and 4 with 2x3/2 (ECF from (a))
✓ 1
A1 14 m2
✓ 1
Exam tip: Write powers as x^(1/2) before integrating; then 'add one to the power and divide by the new power' applies directly.
92IBMA3H-5-20r37-015.5 Introduction to integrationPaper 1Easy7 marks
No calculator
The height, h metres, of a glider above a ridge is modelled by h(t) = 2t3 + 12t−1 + 50, for t > 0, where t is the time in minutes after release.
(a)
Find h′(t).
(2)
(b)
Find the equation of the tangent to the graph of h at t = 2. Give your answer in the form h = mt + c.
(3)
(c)
Find the value of h″(2).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 6t2
✓ 1
A1 −12t−2
✓ 1
Part (b)
A1 h(2) = 72 and h′(2) = 21
✓ 1
M1 h − 72 = 21(t − 2)
✓ 1
A1 h = 21t + 30
✓ 1
Part (c)
A1 h″(t) = 12t + 24t−3
✓ 1
A1 h″(2) = 27
✓ 1
Common error: Write 12/t as 12t⁻¹ before differentiating; the power falls to −2, not 0 or −1.
93IBMA3H-5-20r37-035.15 Further derivatives and integrals; partial fractionsPaper 1Easy6 marks
No calculator
The following limits are each the gradient of a curve at a point.
(a)
Evaluate limh→0 [(3 + h)4 − 81]/h.
(3)
(b)
Evaluate limh→0 [2(5 + h)2 − 50]/h.
(2)
(c)
State what the limit in part (a) represents geometrically.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 recognise as the derivative of f(x) = x4 at x = 3
✓ 1
M1 f′(x) = 4x3
✓ 1
A1 108
✓ 1
Part (b)
M1 recognise as the derivative of 2x2 at x = 5, or expand the numerator to 20h + 2h2
✓ 1
A1 20
✓ 1
Part (c)
R1 the gradient of the tangent to y = x4 at x = 3
✓ 1
Exam tip: A limit of the form [f(a + h) − f(a)]/h is f′(a); identify f and a first, and the evaluation is quick.
94IBMA3H-5-20r37-045.8 Stationary points, inflexion and optimizationPaper 1Easy7 marks
No calculator
A farmer has 240 m of fencing to enclose a rectangular field. One side of the field lies along a straight river and needs no fence. The two sides perpendicular to the river each have length x metres.
Fence on three sides (solid lines). Diagram not to scale
(a)
Show that the area of the field is A = 240x − 2x2 square metres.
(2)
(b)
Find the value of x that maximises the area.
(3)
(c)
Find the maximum area, and justify that it is a maximum.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 side along the river = 240 − 2x
✓ 1
A1 A = x(240 − 2x) = 240x − 2x2
✓ 1
Part (b)
A1 A′ = 240 − 4x
✓ 1
M1 set A′ = 0
✓ 1
A1 x = 60
✓ 1
Part (c)
A1 maximum area = 7200 m2
✓ 1
R1 A″ = −4 < 0
✓ 1
Exam tip: Use the constraint to write the area in one variable before differentiating, and always justify the maximum with the second derivative or a sign change.
95IBMA3H-5-19r38-015.12 Limits, first principles and higher derivativesPaper 1Easy7 marks
No calculator
The function \(f\) is defined by \(f(x)=4x^{2}-3x\).
(a)
Use differentiation from first principles to show that \(f'(x)=8x-3\).
(5)
(b)
The gradient of the graph of \(y=f(x)\) at the point \(P\) is 13. Find the \(x\)-coordinate of \(P\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 forms \(\frac{f(x+h)-f(x)}{h}\) with \(f(x+h)=4(x+h)^2-3(x+h)\)
✓ 1
A1 expands: \(4x^2+8xh+4h^2-3x-3h\)
✓ 1
M1 subtracts \(f(x)\) and divides by \(h\)
✓ 1
A1 \(8x+4h-3\)
✓ 1
A1 takes the limit as \(h\to0\) to obtain \(8x-3\)
✓ 1
Part (b)
M1 \(8x-3=13\)
✓ 1
A1 \(x=2\)
✓ 1
Common error: When expanding \(f(x+h)\), squaring only the \(x\) and forgetting the cross term \(2xh\) loses the \(h\) that must cancel before the limit is taken.
96IBMA3H-5-19r38-025.1 Limits and the derivativePaper 1Easy5 marks
No calculator
The concentration, \(c\) mg per litre, of a flavour compound in a slow-cooked sauce is modelled by \(c(t)=2t^{3}-\frac{9}{t^{2}}+6\sqrt{t}\) for \(t\ge1\), where \(t\) is the time in hours since cooking began.
(a)
Find \(\frac{dc}{dt}\).
(3)
(b)
Find the rate of change of the concentration when \(t=1\). Include units in your answer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 rewrites as \(2t^3-9t^{-2}+6t^{\frac12}\)
✓ 1
A1 two terms correct, e.g. \(6t^2\) and \(+18t^{-3}\)
✓ 1
A1 \(\frac{dc}{dt}=6t^2+18t^{-3}+3t^{-\frac12}\)
✓ 1
Part (b)
M1 substitutes \(t=1\): \(6+18+3\)
✓ 1
A1 27 mg per litre per hour
✓ 1
Exam tip: Write roots and reciprocals as powers of \(t\) before differentiating; the power rule then applies term by term.
97IBMA3H-5-19r38-035.7 The second derivative and graphsPaper 1Easy5 marks
No calculator
A mountain ridge has height \(h\) metres relative to a base camp (negative values are below the camp), modelled by \(h(x)=2x^{4}-8x^{3}+5\) for \(-1\le x\le3\), where \(x\) is the horizontal distance in hundreds of metres from the camp.
(a)
Find \(h'(x)\) and \(h''(x)\).
(2)
(b)
Find the values of \(x\) at which the graph of \(y=h(x)\) has a point of inflexion. Justify your answer.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(h'(x)=8x^3-24x^2\)
✓ 1
A1 \(h''(x)=24x^2-48x\)
✓ 1
Part (b)
M1 sets \(h''(x)=0\): \(24x(x-2)=0\)
✓ 1
A1 \(x=0\) and \(x=2\)
✓ 1
R1 \(h''\) changes sign at each: positive for \(x<0\), negative for \(0<x<2\), positive for \(x>2\)
✓ 1
Common error: Setting \(h''(x)=0\) only gives candidates; a point of inflexion also needs the sign of \(h''\) to change.
98IBMA3H-5-20r45-015.14 Implicit differentiation, related rates and optimisationPaper 1Easy5 marks
No calculator
The upper edge of a curved hemline on a dress pattern is modelled by a curve with gradient \(\frac{dy}{dx} = 6x^2 - \frac{4}{x^2}\) for \(x > 0\), where \(x\) and \(y\) are measured in decimetres. The curve passes through the point \((1, 5)\). Find the equation of the curve.
(a)
Find an expression for y in terms of x.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to integrate, with at least one correct term
M1 substitute \((1, 5)\) into \(y = 2x^3 + \frac{4}{x} + c\) to find \(c\)
✓ 1
A1 \(y = 2x^3 + \frac{4}{x} - 1\)
✓ 1
Common error: Write \(\frac{4}{x^2}\) as \(4x^{-2}\) before integrating; integrating it as a logarithm is a frequent slip.
99IBMA3H-5-20r45-035.6 Differentiation rulesPaper 1Easy5 marks
No calculator
The angular velocity of a robot joint is \(\omega(t) = 8\cos 2t\) radians per second, where \(t\) is the time in seconds. A positive value means the joint turns anticlockwise.
(a)
Find the angle through which the joint turns, anticlockwise, between \(t = 0\) and \(t = \frac{\pi}{4}\).
(3)
(b)
Find the net angle turned by the joint between \(t = \frac{\pi}{4}\) and \(t = \frac{\pi}{2}\) and explain the sign of your answer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\int_0^{\pi/4} 8\cos 2t\,dt\) with an attempt to integrate
✓ 1
A1 \(4\sin 2t\)
✓ 1
A1 \(4\sin\frac{\pi}{2} - 4\sin 0 = 4\) radians
✓ 1
Part (b)
A1 \(4\sin\pi - 4\sin\frac{\pi}{2} = -4\) radians (ECF from the antiderivative in (a))
✓ 1
R1 the negative sign shows that the joint turns clockwise, as \(\omega < 0\) for \(\frac{\pi}{4} < t < \frac{\pi}{2}\)
✓ 1
Common error: When integrating \(\cos kt\), remember to divide by \(k\); leaving it out doubles or triples the answer.
100IBMA3H-5-20r45-045.13 Limits and l’Hôpital’s rulePaper 1Easy4 marks
No calculator
Use l'Hôpital's rule to find \(\lim_{x \to 0} \frac{e^{3x} - 1}{\sin 2x}\).
(a)
Evaluate the limit.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 both numerator and denominator tend to 0 as \(x \to 0\), so the form is \(\frac{0}{0}\) and l'Hôpital's rule applies
✓ 1
M1 differentiate numerator and denominator separately: \(\frac{3e^{3x}}{2\cos 2x}\)
Exam tip: Check that the limit is in an indeterminate form before differentiating, and differentiate the numerator and denominator separately, not as a quotient.
101IBMA3H-5-19r47-015.14 Implicit differentiation, related rates and optimisationPaper 1Easy5 marks
No calculator
An archaeologist models the floor of a trench. At a horizontal distance \(x\) metres from one end, the gradient of the floor is given by \(\frac{\mathrm{d}d}{\mathrm{d}x}=3x^2-8x+5\), where \(d\) metres is the depth of the floor below ground level. At the end of the trench, \(x=0\), the depth is 2 m.
(a)
Find an expression for \(d\) in terms of \(x\).
(3)
(b)
Find the depth of the trench floor when \(x=3\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to integrate (at least one power increased by one)
✓ 1
A1 \(d=x^3-4x^2+5x+c\)
✓ 1
A1 use of \(d(0)=2\) to obtain \(c=2\), so \(d=x^3-4x^2+5x+2\)
✓ 1
Part (b)
M1 substitution of \(x=3\) into their \(d\) (ECF from (a))
✓ 1
A1 \(d=27-36+15+2=8\) m
✓ 1
Exam tip: After integrating an indefinite integral, use the given condition straight away to find the constant; an unevaluated constant loses the final accuracy mark.
102IBMA3H-5-19r47-035.3 Differentiating polynomials and integer powersPaper 1Easy7 marks
No calculator
The concentration \(C\) mg L−1 of a pollutant in a stream is modelled by \(C=t^3-6t^2+9t+20\), for \(0\le t\le 4\), where \(t\) is the time in hours after a spill is detected.
(a)
Find the values of \(t\) for which \(\frac{\mathrm{d}C}{\mathrm{d}t}=0\).
(3)
(b)
Use the second derivative to determine the nature of each stationary point.
(2)
(c)
Find the least value of \(\frac{\mathrm{d}C}{\mathrm{d}t}\) for \(0\le t\le 4\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to differentiate
✓ 1
A1 \(\frac{\mathrm{d}C}{\mathrm{d}t}=3t^2-12t+9\)
✓ 1
A1 \(t=1\) and \(t=3\)
✓ 1
Part (b)
M1 \(\frac{\mathrm{d}^2C}{\mathrm{d}t^2}=6t-12\) evaluated at both values of \(t\)
✓ 1
A1 at \(t=1\) the value is \(-6<0\) so a local maximum; at \(t=3\) the value is \(6>0\) so a local minimum
✓ 1
Part (c)
M1 setting \(\frac{\mathrm{d}^2C}{\mathrm{d}t^2}=0\) to obtain \(t=2\) (accept completing the square or the vertex of the quadratic)
Exam tip: The rate of change of a quantity is itself a function; its extreme values occur where the second derivative of the original function is zero.
103IBMA3H-5-19r48-015.6 Differentiation rulesPaper 1Easy6 marks
No calculator
The function \(f\) is defined by \(f(x)=x^2e^{3x}\).
(a)
Find \(f'(x)\).
(3)
(b)
Find the exact \(x\)-coordinate of the stationary point of the graph of \(f\) for which \(x<0\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt at the product rule, with terms \(x^2\times(\ldots)\) and \(e^{3x}\times 2x\)
✓ 1
A1 \(\frac{d}{dx}e^{3x}=3e^{3x}\) seen
✓ 1
A1 \(f'(x)=2xe^{3x}+3x^2e^{3x}\)
✓ 1
Part (b)
M1 set \(f'(x)=0\) and factorise: \(xe^{3x}(2+3x)=0\)
✓ 1
R1 \(e^{3x}\neq 0\) for all \(x\), so \(x=0\) or \(2+3x=0\)
✓ 1
A1 \(x=-\frac{2}{3}\) (\(\approx -0.667\))
✓ 1
Common error: The derivative of \(e^{3x}\) is \(3e^{3x}\), not \(e^{3x}\); forgetting the factor 3 is the usual slip with the chain rule.
104AAHL5-q00035.2 Increasing and decreasing functionsPaper 1Medium7 marks
Let \(f(x)=2x+\dfrac{k}{x^{2}}\), for \(x\gt 0\), where \(k\in\mathbb{R}\).
(a)
Find \(f'(x)\).
(1)
(b)
Explain why \(f\) is increasing for all \(x\gt 0\) when \(k\le0\).
(2)
(c)
For \(k\gt 0\), find, in terms of \(k\), the interval on which \(f\) is decreasing.
(2)
(d)
Find the largest value of \(k\) for which \(f\) is increasing for all \(x\gt 2\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(f'(x)=2-\dfrac{2k}{x^{3}}\)
A1
(b)
for \(x\gt 0\), \(x^{3}\gt 0\), so \(-\dfrac{2k}{x^{3}}\ge0\) when \(k\le0\)
M1
hence \(f'(x)\ge2\gt 0\) for all \(x\gt 0\), so \(f\) is increasing
105AAHL5-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.
106AAHL5-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 \(\left(a,\,a^{2}\right)\).
Diagram not to scale
(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 \(y=4x-4\)
107AAHL5-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).
110AAHL5-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
111AAHL5-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.
Diagram not to scale
(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 \(\text{P}(1,\,1)\)
112AAHL5-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 \[I(x)=\dfrac{8}{x^{2}}+\dfrac{1}{(10-x)^{2}},\quad 0\lt x\lt 10.\]
Diagram not to scale
(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 \(u=1+\mathrm{e}^{x}\)
117AAHL5-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. \(\displaystyle\lim_{x\to0}\dfrac{2x\mathrm{e}^{x^{2}}+\sin x}{p\sin px}=\lim_{x\to0}\dfrac{(2+4x^{2})\mathrm{e}^{x^{2}}+\cos x}{p^{2}\cos px}=\dfrac{3}{p^{2}}\).
128AAHL5-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
129AAHL5-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.
139AAHL5-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: \(\displaystyle\int(\ln x)^{2}\,\mathrm{d}x=x(\ln x)^{2}-\displaystyle\int2\ln x\,\mathrm{d}x\)
140AAHL5-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) \(\pi\left(\ln2-\frac12\right)\)
141AAHL5-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.
145AAHL5-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 \(x=\dfrac{\pi}{6}\).
149AAHL5-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 \(\ln(1-x)=-x-\dfrac{x^{2}}{2}-\dfrac{x^{3}}{3}-\cdots\)
M1 multiplies through by \(x^2\) and recognises \(\frac{d}{dx}(x^2y)=4x^3\) ; A1 \(x^2y=x^4+c\) ; M1 uses \(x=1\), \(y=3\) to find \(c=2\) ; A1 \(y=x^2+\frac{2}{x^2}\)
✓ 4
Part (c)
M1 differentiates to get \(\frac{dy}{dx}=2x-\frac{4}{x^3}\) and considers \(\frac{dy}{dx}<0\) (or \(=0\)) ; A1 obtains \(2x^4<4\), i.e. \(x^4<2\), giving the critical value \(x=\sqrt[4]{2}\) ; A1 \(0<x<\sqrt[4]{2}\) (accept \(0<x\le\sqrt[4]{2}\))
Common error: After multiplying by the integrating factor, the right-hand side must also be multiplied: integrating \(4x\) instead of \(4x^3\) is a frequent slip.
152IBMA2H-053-00-065.10 Indefinite integrals and reverse chain rulePaper 1Medium9 marks
No calculator
Consider the differential equation \(\frac{dy}{dx}=\frac{x^2+y^2}{xy}\), for \(x>0\), \(y>0\), given that \(y=2\) when \(x=1\).
(a)
Use the substitution \(y=vx\) to show that \(x\frac{dv}{dx}=\frac{1}{v}\).
(3)
(b)
Hence find y in terms of x.
(4)
(c)
Find the exact value of y when \(x=e\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{dy}{dx}=v+x\frac{dv}{dx}\) ; A1 right-hand side becomes \(\frac{x^2+v^2x^2}{vx^2}=\frac{1+v^2}{v}\) ; A1 \(v+x\frac{dv}{dx}=\frac{1}{v}+v\), so \(x\frac{dv}{dx}=\frac{1}{v}\)
Exam tip: Convert the initial condition into terms of v before finding the constant: at \(x=1\), \(y=2\) gives \(v=2\).
153IBMA2H-053-00-095.3 Differentiating polynomials and integer powersPaper 1Medium7 marks
No calculator
An engineer models the profile of a skateboard ramp by \(f(x)=x^3-3kx^2+12x\), where \(k\in\mathbb{R}\). For safety, the ramp must rise continuously, so f must be increasing for all x.
(a)
Find \(f'(x)\).
(1)
(b)
Find the set of values of k for which f is increasing for all x.
(4)
(c)
For \(k=3\), find the exact interval on which f is decreasing.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(f'(x)=3x^2-6kx+12\)
✓ 1
Part (b)
R1 requires \(f'(x)\geq0\) for all x, a quadratic with positive leading coefficient that never goes below zero ; M1 uses discriminant \(\leq0\) ; A1 \((-6k)^2-4(3)(12)\leq0\), i.e. \(36k^2-144\leq0\) ; A1 \(-2\leq k\leq2\)
✓ 4
Part (c)
M1 sets \(f'(x)=3x^2-18x+12<0\), i.e. \(x^2-6x+4<0\), and solves to obtain the roots \(x=3\pm\sqrt{5}\) ; A1 \(3-\sqrt{5}<x<3+\sqrt{5}\) (accept \(\leq\))
✓ 2
Common error: Setting the discriminant greater than zero gives the values of k for which the gradient changes sign, which is the opposite of what is required.
154IBMA2H-053-00-115.13 Limits and l’Hôpital’s rulePaper 1Medium12 marks
No calculator
A software company releases a security update. The proportion p of its smartphones that have installed the update t days after release is modelled by \(\frac{dp}{dt}=2p(1-p)\), where \(p=0.1\) when \(t=0\).
(a)
Express \(\frac{1}{p(1-p)}\) in partial fractions.
(2)
(b)
Hence solve the differential equation, showing that \(p=\frac{1}{1+9e^{-2t}}\).
(5)
(c)
Explain why p is increasing for all \(t\geq0\).
(2)
(d)
Find the exact time at which the rate of installation is greatest.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes \(\frac{A}{p}+\frac{B}{1-p}\) and attempts to find A and B, e.g. from \(1=A(1-p)+Bp\) ; A1 \(\frac{1}{p}+\frac{1}{1-p}\)
✓ 2
Part (b)
M1 separates variables: \(\int\left(\frac{1}{p}+\frac{1}{1-p}\right)dp=\int2\,dt\) ; A1 \(\ln|p|-\ln|1-p|=2t+c\) ; M1 substitutes \(t=0\), \(p=0.1\) to obtain \(c=\ln\frac{1}{9}\) ; A1 \(\frac{p}{1-p}=\frac{1}{9}e^{2t}\), since \(0<p<1\) ; A1 rearranges correctly, e.g. \(9p=e^{2t}(1-p)\) so \(p=\frac{e^{2t}}{9+e^{2t}}=\frac{1}{1+9e^{-2t}}\)
✓ 5
Part (c)
R1 \(9e^{-2t}>0\), so \(1+9e^{-2t}>1\) and therefore \(0<p<1\) for all t ; R1 hence \(\frac{dp}{dt}=2p(1-p)>0\), so p is increasing
✓ 2
Part (d)
M1 maximises the rate \(\frac{dp}{dt}=2p-2p^2\), either by setting \(\frac{d}{dp}(2p-2p^2)=2-4p=0\) or by setting \(\frac{d^2p}{dt^2}=2(1-2p)\frac{dp}{dt}=0\), or by differentiating \(p(t)\) twice directly ; A1 \(p=\frac{1}{2}\), with a reason that this gives a maximum, e.g. \(2p-2p^2\) is a quadratic in p with negative leading coefficient, or \(\frac{d^2p}{dt^2}\) changes from positive to negative ; A1 \(\frac{1}{1+9e^{-2t}}=\frac{1}{2}\) gives \(9e^{-2t}=1\), so \(t=\ln3\) days
✓ 3
Common error: The integral of \(\frac{1}{1-p}\) is \(-\ln|1-p|\). Dropping the minus sign gives \(\ln p+\ln(1-p)\), which leads to an incorrect solution.
155IBMA2H-053-10-035.1 Limits and the derivativePaper 1Medium7 marks
No calculator
After an injection, the concentration of an anaesthetic in a patient's blood is modelled by \(C(t)=20te^{-0.4t}\) mg L−1, where t is the time in hours, \(t\geq0\).
(a)
Show that \(C'(t)=20e^{-0.4t}(1-0.4t)\).
(2)
(b)
Find the exact maximum concentration.
(2)
(c)
Find the time at which the concentration is decreasing most rapidly, justifying your answer.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use the product rule
✓ 1
A1 \(C'(t)=20e^{-0.4t}-8te^{-0.4t}=20e^{-0.4t}(1-0.4t)\) AG
M1 \(C''(t)=20e^{-0.4t}(0.16t-0.8)\) and sets \(C''(t)=0\)
✓ 1
A1 \(t=5\) hours
✓ 1
R1 \(C''<0\) for \(t<5\) and \(C''>0\) for \(t>5\), so \(C'\) has its minimum (most negative) value at \(t=5\)
✓ 1
Exam tip: 'Decreasing most rapidly' asks for the minimum of \(C'(t)\), so differentiate again and look for where \(C''(t)=0\) with a sign change.
156IBMA2H-053-10-055.8 Stationary points, inflexion and optimizationPaper 1Medium7 marks
No calculator
Two street lamps stand 30 m apart on a straight path. The brighter lamp is 8 times as strong as the other. At a point on the path x metres from the brighter lamp, the illumination is modelled by \(I(x)=k\left(\frac{8}{x^2}+\frac{1}{(30-x)^2}\right)\), \(0<x<30\), where k is a positive constant.
Diagram not to scale
(a)
Find \(I'(x)\).
(2)
(b)
Show that the illumination has a stationary value when \(x=20\).
(3)
(c)
Justify that this stationary value is the minimum illumination on the path, and find it in terms of k.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(-\frac{16k}{x^3}\)
✓ 1
A1 \(+\frac{2k}{(30-x)^3}\) (correct sign from the chain rule)
Common error: Differentiating \((30-x)^{-2}\) gives \(+2(30-x)^{-3}\), not \(-2(30-x)^{-3}\): the inner derivative \(-1\) flips the sign.
157IBMA2H-053-10-065.14 Implicit differentiation, related rates and optimisationPaper 1Medium7 marks
No calculator
The outline of a decorative metal bracket is modelled by the curve \(x^3+y^3=6xy\), for \(x,y\geq0\).
(a)
Show that \(\frac{dy}{dx}=\frac{2y-x^2}{y^2-2x}\).
(3)
(b)
Find the exact coordinates of the point, other than the origin, where the tangent to the curve is horizontal.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt at implicit differentiation, including the product rule on \(6xy\)
✓ 1
A1 \(3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}\)
✓ 1
A1 \(\frac{dy}{dx}(3y^2-6x)=6y-3x^2\Rightarrow\frac{dy}{dx}=\frac{2y-x^2}{y^2-2x}\) AG
✓ 1
Part (b)
M1 sets numerator to zero: \(y=\frac{x^2}{2}\)
✓ 1
M1 substitutes into the curve: \(x^3+\frac{x^6}{8}=3x^3\), so \(x^6=16x^3\)
✓ 1
A1 \(x^3=16\Rightarrow x=2\sqrt[3]{2}\)
✓ 1
A1 \(y=\frac{(2\sqrt[3]{2})^2}{2}=2\sqrt[3]{4}\)
✓ 1
Exam tip: For a horizontal tangent set only the numerator of \(\frac{dy}{dx}\) to zero, then substitute back into the original curve equation, not into the derivative.
158IBMA2H-053-10-085.16 Integration by substitution and by partsPaper 1Medium11 marks
No calculator
A straight steel beam is carried horizontally from a loading bay 5.4 m wide into a corridor 1.6 m wide that meets the bay at right angles. When the beam touches both outer walls and the inner corner, its length is \(L(\theta)=\frac{5.4}{\sin\theta}+\frac{1.6}{\cos\theta}\) metres, where \(0<\theta<\frac{\pi}{2}\).
Diagram not to scale
(a)
Find \(L'(\theta)\).
(2)
(b)
Show that \(L'(\theta)=0\) when \(\tan\theta=\frac{3}{2}\).
(2)
(c)
Find the exact value of L at this stationary point.
(3)
(d)
Given that \(L''(\theta)=\frac{5.4(1+\cos^2\theta)}{\sin^3\theta}+\frac{1.6(1+\sin^2\theta)}{\cos^3\theta}\), show that this stationary value is a minimum.
(2)
(e)
Explain why the minimum value of L is the length of the longest beam that can be carried round the corner. Using \(\sqrt{13}\approx3.6056\), state this length to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(-\frac{5.4\cos\theta}{\sin^2\theta}\)
✓ 1
A1 \(+\frac{1.6\sin\theta}{\cos^2\theta}\)
✓ 1
Part (b)
M1 sets \(L'(\theta)=0\) and rearranges: \(1.6\sin^3\theta=5.4\cos^3\theta\)
✓ 1
A1 \(\tan^3\theta=\frac{27}{8}\Rightarrow\tan\theta=\frac{3}{2}\) AG
✓ 1
Part (c)
M1 uses a right-angled triangle with sides 2, 3, \(\sqrt{13}\): \(\sin\theta=\frac{3}{\sqrt{13}}\), \(\cos\theta=\frac{2}{\sqrt{13}}\)
A1 \(L=2.6\sqrt{13}\) \(\left(=\frac{13\sqrt{13}}{5}\right)\) m
✓ 1
Part (d)
M1 considers the sign of each term for \(0<\theta<\frac{\pi}{2}\): \(\sin\theta>0\), \(\cos\theta>0\)
✓ 1
R1 both terms positive so \(L''(\theta)>0\), hence minimum
✓ 1
Part (e)
R1 a beam passes only if at every angle θ during the turn its length does not exceed \(L(\theta)\), so it must be no longer than the smallest value of \(L(\theta)\)
✓ 1
A1 longest beam \(=2.6\sqrt{13}\approx9.37\) m
✓ 1
Exam tip: Once you know \(\tan\theta\), draw the right-angled triangle to get exact \(\sin\theta\) and \(\cos\theta\). This keeps the answer in surd form without a calculator.
159IBMA2H-053-04-065.18 First order differential equationsPaper 1Medium13 marks
No calculator
In a chemical reaction, the concentration x mol dm−3 of a product after t minutes satisfies dx/dt = (2 − x)(4 − x), 0 ≤ x < 2, with x = 0 when t = 0.
Find the limiting value of the concentration as t → ∞.
(2)
(e)
Find the exact time at which the concentration is 1 mol dm−3.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 1 = A(4 − x) + B(2 − x) with substitution or comparing coefficients
✓ 1
A1 1/(2(2 − x)) − 1/(2(4 − x))
✓ 1
Part (b)
M1 separates variables and uses the partial fractions
✓ 1
A1 −½ ln(2 − x) + ½ ln(4 − x) = t + c, with correct signs
✓ 1
M1 substitutes x = 0, t = 0 to find c = ½ ln 2
✓ 1
A1 multiplies by 2 and combines logs to reach the given result, AG
✓ 1
Part (c)
M1 exponentiates: (4 − x)/(2 − x) = 2e2t
✓ 1
M1 rearranges: 4 − x = 4e2t − 2xe2t, collecting x terms
✓ 1
A1 x(2e2t − 1) = 4(e2t − 1), giving the result, AG
✓ 1
Part (d)
M1 divides numerator and denominator by e2t: 4(1 − e−2t)/(2 − e−2t)
✓ 1
A1 2 mol dm−3
✓ 1
Part (e)
M1 substitutes x = 1 into part (b) or (c): 3 = 2e2t
✓ 1
A1 t = ½ ln(3/2) minutes
✓ 1
Common error: ∫ 1/(2 − x) dx is −ln|2 − x|, not +ln|2 − x|. Missing the negative from the chain rule reverses the fraction inside the logarithm.
160IBMA2H-053-04-085.19 Maclaurin seriesPaper 1Medium10 marks
No calculator
Let f(x) = 3 / ((1 − x)(1 + 2x)).
(a)
Express f(x) in partial fractions.
(2)
(b)
Hence find the Maclaurin series for f(x) up to and including the term in x3.
(4)
(c)
State the set of values of x for which the full series converges.
(2)
(d)
Find the coefficient of x6 in the series.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 3 = A(1 + 2x) + B(1 − x) and substitutes x = 1 and x = −½
✓ 1
A1 f(x) = 1/(1 − x) + 2/(1 + 2x)
✓ 1
Part (b)
M1 expands 1/(1 − x) = 1 + x + x2 + x3 + …
✓ 1
M1 expands 2(1 + 2x)−1 = 2 − 4x + 8x2 − 16x3 + …
✓ 1
A1 adds corresponding terms, at least two coefficients correct
✓ 1
A1 3 − 3x + 9x2 − 15x3
✓ 1
Part (c)
R1 requires both |x| < 1 and |2x| < 1
✓ 1
A1 −½ < x < ½
✓ 1
Part (d)
M1 general coefficient 1 + 2(−2)n
✓ 1
A1 1 + 128 = 129
✓ 1
Common error: The interval of convergence is set by the more restrictive expansion. Stating |x| < 1 ignores that 1/(1 + 2x) only converges for |x| < ½.
161IBMA2H-053-04-115.16 Integration by substitution and by partsPaper 1Medium11 marks
No calculator
During a defibrillator pulse, the current I amperes flowing through a circuit t milliseconds after it is switched on is modelled by I(t) = 2 / ((t + 1)(t + 2)), t ≥ 0. The charge, in millicoulombs, that passes between t = a and t = b is ∫ab I(t) dt.
(a)
Express I(t) in partial fractions.
(2)
(b)
Show that the total charge that passes, ∫0∞ I(t) dt, is 2 ln 2 millicoulombs.
(4)
(c)
Show that half of the total charge has passed when t = √2.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 2 = A(t + 2) + B(t + 1) with substitution or comparing coefficients
✓ 1
A1 2/(t + 1) − 2/(t + 2)
✓ 1
Part (b)
M1 integrates to 2 ln(t + 1) − 2 ln(t + 2)
✓ 1
A1 writes as 2 ln((t + 1)/(t + 2)) and evaluates the lower limit as 2 ln(1/2)
✓ 1
M1 considers the limit as the upper limit → ∞, with (t + 1)/(t + 2) → 1 so the log term → 0
Exam tip: With an improper integral, write the upper limit as a variable and take the limit explicitly. Simply 'substituting ∞' does not earn the method mark.
162IBMA2H-053-09-065.14 Implicit differentiation, related rates and optimisationPaper 1Medium7 marks
No calculator
A rocket is launched vertically from a pad on level ground. A radar station R on the ground is 4 km from the pad. At time \(t\) seconds the rocket is at height \(h\) km, its angle of elevation from R is \(\theta\) radians and its distance from R is \(r\) km. When \(h=4\), the rocket is rising at 0.6 km s−1.
Diagram not to scale
(a)
Find the rate at which \(\theta\) is increasing when \(h=4\).
(4)
(b)
Find the exact rate at which \(r\) is increasing when \(h=4\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\tan\theta=\frac{h}{4}\)
✓ 1
M1 differentiate with respect to \(t\): \(\sec^2\theta\frac{d\theta}{dt}=\frac{1}{4}\frac{dh}{dt}\)
✓ 1
A1 when \(h=4\), \(\theta=\frac{\pi}{4}\) so \(\sec^2\theta=2\)
✓ 1
A1 \(\frac{d\theta}{dt}=\frac{0.6}{8}=\frac{3}{40}\) rad s−1
✓ 1
Part (b)
M1 \(r^2=16+h^2\)
✓ 1
M1 \(2r\frac{dr}{dt}=2h\frac{dh}{dt}\)
✓ 1
A1 \(r=\sqrt{32}=4\sqrt2\), so \(\frac{dr}{dt}=\frac{4(0.6)}{4\sqrt2}=\frac{3\sqrt2}{10}\) km s−1
✓ 1
Exam tip: Differentiate the general relationship with respect to time before substituting the particular values; substituting \(h=4\) first turns a variable into a constant and its rate becomes zero.
163IBMA2H-053-09-075.6 Differentiation rulesPaper 1Medium8 marks
No calculator
The brightness of a pulsing LED on a smart speaker is modelled by \(B(t)=e^{\sin\left(\frac{\pi t}{2}\right)}\), \(0\le t\le 4\), where \(t\) is time in seconds and \(B\) is measured in arbitrary units.
(a)
Find \(B'(t)\).
(2)
(b)
Find the values of \(t\) for which \(B'(t)=0\).
(3)
(c)
Show that the maximum brightness on \(0\le t\le 4\) occurs when \(t=1\) and the minimum brightness occurs when \(t=3\). Hence find the exact ratio of the maximum brightness to the minimum brightness.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 chain rule used, including the derivative of the inner function \(\frac{\pi t}{2}\)
R1 \(B'(t)\) changes from positive to negative at \(t=1\) and from negative to positive at \(t=3\) (or an equivalent argument using \(-1\le\sin\le 1\) with \(e^x\) increasing)
✓ 1
A1 values \(e\) at \(t=1\), \(e^{-1}\) at \(t=3\) and \(1\) at both endpoints \(t=0\) and \(t=4\), so \(e\) is the greatest and \(e^{-1}\) the least
✓ 1
A1 ratio \(=\frac{e}{e^{-1}}=e^2\)
✓ 1
Common error: Leaving out the inner factor \(\frac{\pi}{2}\) is a frequent slip in nested chain rules. On a closed interval, also compare the endpoint values before stating a global maximum or minimum.
164IBMA2H-053-09-095.15 Further derivatives and integrals; partial fractionsPaper 1Medium7 marks
No calculator
Let \(f(x)=x^x\), for \(x>0\).
(a)
Use logarithmic differentiation to show that \(f'(x)=x^x(1+\ln x)\).
(3)
(b)
Find the exact \(x\)-coordinate of the stationary point of the graph of \(f\) and justify that it is a minimum.
(3)
(c)
Find the exact minimum value of \(f\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 let \(y=x^x\), so \(\ln y=x\ln x\)
✓ 1
M1 differentiate implicitly, using the product rule on the right: \(\frac{1}{y}\frac{dy}{dx}=\ln x+1\)
✓ 1
A1 \(\frac{dy}{dx}=y(1+\ln x)=x^x(1+\ln x)\)
✓ 1
Part (b)
M1 \(1+\ln x=0\), since \(x^x>0\)
✓ 1
A1 \(x=\frac{1}{e}\)
✓ 1
R1 EITHER sign change: for \(0<x<\frac1e\), \(\ln x<-1\) so \(f'(x)<0\), and for \(x>\frac1e\), \(f'(x)>0\), hence a minimum; OR second derivative: \(f''(x)=x^x(1+\ln x)^2+x^{x-1}\), so \(f''\left(\frac1e\right)=e\left(\frac1e\right)^{1/e}>0\), hence a minimum
✓ 1
Part (c)
A1 \(\left(\frac1e\right)^{1/e}=e^{-1/e}\)
✓ 1
Common error: \(x^x\) is neither a power function nor an exponential with a constant base, so neither \(x\cdot x^{x-1}\) nor \(x^x\ln x\) is correct. Taking logarithms first turns it into a product that can be differentiated.
165IBMA2H-053-09-105.14 Implicit differentiation, related rates and optimisationPaper 1Medium8 marks
No calculator
A 12 V battery with internal resistance 2 Ω is connected to a variable resistor of resistance \(R\) ohms. The power, in watts, delivered to the resistor is \(P(R)=\dfrac{144R}{(R+2)^2}\), \(R>0\).
(a)
Show that \(P'(R)=\dfrac{144(2-R)}{(R+2)^3}\).
(3)
(b)
Find the maximum power delivered to the resistor, justifying that it is a maximum.
(3)
(c)
The resistance is increased at a constant rate of 0.5 Ω s−1. Find the rate of change of the power when \(R=6\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt at quotient rule (or product rule with \((R+2)^{-2}\))
✓ 1
A1 \(\dfrac{144(R+2)^2-144R\cdot 2(R+2)}{(R+2)^4}\) or equivalent
✓ 1
A1 factor \((R+2)\) cancelled: numerator \(144[(R+2)-2R]=144(2-R)\), giving the result
✓ 1
Part (b)
M1 \(P'(R)=0\Rightarrow R=2\)
✓ 1
A1 \(P(2)=\frac{288}{16}=18\) W
✓ 1
R1 \(P'(R)>0\) for \(0<R<2\) and \(P'(R)<0\) for \(R>2\), so maximum. Also accept a correct second-derivative argument, e.g. \(P''(2)<0\)
✓ 1
Part (c)
M1 \(\frac{dP}{dt}=P'(R)\frac{dR}{dt}\) with \(P'(6)=\frac{144(-4)}{512}=-\frac98\)
✓ 1
A1 \(\frac{dP}{dt}=-\frac{9}{16}=-0.5625\) W s−1 (power decreasing)
✓ 1
Exam tip: When the denominator is a power of a bracket, cancel a common factor of that bracket from the numerator before expanding; the algebra becomes much shorter.
166IBMA2H-053-06-065.3 Differentiating polynomials and integer powersPaper 1Medium9 marks
No calculator
Let \(f(x) = \frac{(2x-1)(x+3)}{x^2}\), for x > 0.
(a)
Express f(x) in the form a + bx−1 + cx−2, where a, b, c ∈ ℤ.
(2)
(b)
Hence find f′(x).
(2)
(c)
Find the exact coordinates of the stationary point of the graph of f.
(3)
(d)
Use the second derivative to determine the nature of this stationary point.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 expands the numerator to 2x2 + 5x − 3 and divides each term by x2
R1 at \(x = \frac{6}{5}\) the numerator is 12 − 18 = −6 and the denominator is positive, so f″ < 0 and the point is a maximum
✓ 1
Common error: You cannot differentiate a quotient by differentiating the top and bottom separately. Splitting it into a sum of integer powers first makes the power rule valid.
167IBMA2H-053-06-075.13 Limits and l’Hôpital’s rulePaper 1Medium6 marks
No calculator
It is given that \(\lim_{x\to 0} \frac{\ln(1+ax) - 2x}{x^2} = L\), where a and L are real constants and L is finite.
(a)
Find the value of a and the value of L.
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 at x = 0 the numerator and denominator are both 0
✓ 1
M1 applies l'Hôpital's rule to get \(\frac{\frac{a}{1+ax} - 2}{2x}\)
✓ 1
R1 the denominator → 0, so for a finite limit the numerator must → 0, giving a − 2 = 0
✓ 1
A1 a = 2
✓ 1
M1 applies the rule again: \(\frac{-\frac{a^2}{(1+ax)^2}}{2}\)
✓ 1
A1 L = \(-\frac{4}{2}\) = −2
✓ 1
Exam tip: If a limit is known to be finite but the denominator tends to 0, the numerator must also tend to 0. This reasoning is what produces the equation for the unknown constant.
168IBMA2H-053-06-105.8 Stationary points, inflexion and optimizationPaper 1Medium7 marks
No calculator
A closed packaging box has a square base of side x dm and height h dm. Its volume is 12 dm3. The material for the base and the top costs $3 per dm2, and the material for the four sides costs $2 per dm2.
Diagram not to scale
(a)
Show that the total cost of material, $C, is given by $C = 6x^2 + \frac{96}{x}$.
(2)
(b)
Find the value of x for which C is a minimum.
(3)
(c)
Justify that this value gives a minimum, and find the minimum cost.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses x2h = 12, so \(h = \frac{12}{x^2}\)
✓ 1
A1 C = 3(2x2) + 2(4xh) = 6x2 + \(\frac{96}{x}\)
✓ 1
Part (b)
M1 differentiates, using 96x−1
✓ 1
A1 \(\frac{dC}{dx}\) = 12x − 96x−2
✓ 1
A1 setting this to 0 gives x3 = 8, so x = 2
✓ 1
Part (c)
R1 \(\frac{d^2C}{dx^2}\) = 12 + 192x−3 > 0 at x = 2, so it is a minimum
✓ 1
A1 C = 24 + 48 = $72
✓ 1
Common error: A closed box has two square faces, the base and the top. Counting only one of them changes the cost function and moves the optimum.
169IBMA2H-053-07-055.12 Limits, first principles and higher derivativesPaper 1Medium7 marks
No calculator
Let \(f(x)=\sqrt{2x+5}\) for \(x>-\frac{5}{2}\).
(a)
Use differentiation from first principles to show that \(f'(x)=\frac{1}{\sqrt{2x+5}}\).
(5)
(b)
Hence find the equation of the tangent to the graph of \(f\) at \(x=2\).
M1 multiplying numerator and denominator by \(\sqrt{2x+2h+5}+\sqrt{2x+5}\)
✓ 1
A1 numerator simplifies to \(2h\)
✓ 1
M1 cancelling \(h\) and letting \(h\to0\) to obtain \(\frac{2}{2\sqrt{2x+5}}\)
✓ 1
A1 \(f'(x)=\frac{1}{\sqrt{2x+5}}\)
✓ 1
Part (b)
M1 point \((2,3)\) and gradient \(\frac{1}{3}\) used in a line equation
✓ 1
A1 \(y=\frac{1}{3}x+\frac{7}{3}\)
✓ 1
Exam tip: For roots, the conjugate is the key step: once the numerator becomes a multiple of h, the h cancels and the limit can be taken directly.
170IBMA2H-053-07-065.13 Limits and l’Hôpital’s rulePaper 1Medium9 marks
No calculator
The profile of a skate-park ramp is modelled by \(h(x)=\begin{cases}px^2, & 0\le x\le2\\ 6-\frac{q}{x}, & x>2\end{cases}\) where \(h\) is the height in metres at horizontal distance \(x\) metres and \(p,q\) are positive constants. The profile is continuous and differentiable at \(x=2\).
(a)
Use the continuity of \(h\) at \(x=2\) to write down an equation in \(p\) and \(q\).
(2)
(b)
Use the differentiability of \(h\) at \(x=2\) to write down a second equation in \(p\) and \(q\).
(2)
(c)
Find the values of \(p\) and \(q\).
(2)
(d)
By considering second derivatives, determine whether \(h''(x)\) is continuous at \(x=2\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 equating left and right limits of \(h\) at \(x=2\)
✓ 1
A1 \(4p=6-\frac{q}{2}\)
✓ 1
Part (b)
M1 differentiating both pieces: \(2px\) and \(\frac{q}{x^2}\)
✓ 1
A1 \(4p=\frac{q}{4}\)
✓ 1
Part (c)
M1 solving simultaneously, e.g. \(\frac{q}{4}=6-\frac{q}{2}\)
✓ 1
A1 \(q=8,\ p=\frac{1}{2}\)
✓ 1
Part (d)
M1 finding the second derivative of both pieces: \(2p\) and \(-\frac{2q}{x^3}\)
✓ 1
A1 the one-sided limits at \(x=2\) are \(1\) from the left and \(-\frac{16}{8}=-2\) from the right
✓ 1
R1 the one-sided limits differ, so \(h''\) is not continuous at \(x=2\) (the ramp's curvature changes abruptly)
✓ 1
Common error: Continuity alone does not give differentiability: both the function values and the one-sided derivatives must match at the join.
171IBMA2H-053-07-085.12 Limits, first principles and higher derivativesPaper 1Medium8 marks
No calculator
Each of the following limits can be identified as a derivative. Evaluate each limit, showing your reasoning.
A function \(g\) is differentiable with \(g(1)=4\) and \(g'(1)=-2\). Find \(\lim_{h\to0}\frac{g(1+3h)-4}{h}\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 recognising this as \(f'(2)\) for \(f(x)=x^5\), since \(2^5=32\)
✓ 1
A1 \(5\times2^4=80\)
✓ 1
Part (b)
M1 recognising this as the derivative of \(\sin x\) at \(x=\frac{\pi}{3}\), since \(\sin\frac{\pi}{3}=\frac{\sqrt3}{2}\)
✓ 1
A1 \(\cos\frac{\pi}{3}=\frac{1}{2}\)
✓ 1
Part (c)
M1 recognising this as the derivative of \(e^{3x}\) at \(x=0\)
✓ 1
A1 \(3e^0=3\)
✓ 1
Part (d)
M1 rewriting as \(3\times\frac{g(1+3h)-4}{3h}\) and recognising the limit of the fraction as \(g'\) at \(x=1\), using the given value 4 for \(g\) at \(x=1\)
✓ 1
A1 \(3\times(-2)=-6\)
✓ 1
Common error: In part (d) the step is 3h, not h, so the difference quotient must have 3h in its denominator. Omitting the factor of 3 gives −2 instead of −6.
172IBMA2H-053-07-095.1 Limits and the derivativePaper 1Medium9 marks
No calculator
A smartphone is plugged in to charge. Its battery charge, \(C\) %, is modelled so that \(\frac{dC}{dt}=27-9\sqrt{t}\), where \(t\) is the time in hours after the phone is plugged in. After 4 hours the charge is 79 %.
(a)
Find an expression for \(C(t)\).
(5)
(b)
Write down the charge when the phone was plugged in.
(1)
(c)
The charge stops increasing when \(\frac{dC}{dt}=0\). Find when this occurs and the charge at this time.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to integrate
✓ 1
A1 \(27t\)
✓ 1
A1 \(-6t^{3/2}+c\)
✓ 1
M1 substituting \(C(4)=79\): \(108-48+c=79\)
✓ 1
A1 \(C(t)=27t-6t^{3/2}+19\)
✓ 1
Part (b)
A1 19 %
✓ 1
Part (c)
M1 \(27-9\sqrt t=0\)
✓ 1
A1 \(t=9\) hours
✓ 1
A1 \(C(9)=243-162+19=100\) %
✓ 1
Exam tip: Here the unknown found from the boundary condition is the initial value; the constant c is exactly C(0), so no further working is needed for part (b).
173IBMA2H-053-07-105.9 KinematicsPaper 1Medium10 marks
No calculator
Engineers design the motion of a lift travelling 24 m between floors. Its displacement, \(s\) metres, is modelled by \(s(t)=12\left(1-\cos\frac{\pi t}{8}\right)\) for \(0\le t\le8\), with \(t\) in seconds. Passenger comfort depends on the jerk, defined as \(j=\frac{d^3s}{dt^3}\).
(a)
Find the velocity \(v(t)\).
(2)
(b)
Find the acceleration \(a(t)\) and the jerk \(j(t)\).
(2)
(c)
Find the maximum magnitude of the jerk and the time at which it occurs.
(3)
(d)
The journey time is changed to \(T\) seconds, with \(s(t)=12\left(1-\cos\frac{\pi t}{T}\right)\), \(0\le t\le T\). The maximum magnitude of jerk must not exceed 0.5 m s−3. Find the least value of \(T\), giving your answer in exact form.
M1 divides numerator and denominator by \(x^4\) and lets \(x \to 0\)
✓ 1
A1 limit \(= 2\)
✓ 1
Exam tip: Only the lowest-power terms matter in the limit; once both numerator and denominator start with \(x^4\), divide through by \(x^4\) before letting \(x \to 0\).
175IBMA2H-053-01-065.7 The second derivative and graphsPaper 1Medium7 marks
No calculator
The cross-section of a hill is modelled by \(y = 6\sqrt{x}\) for \(0 \le x \le 16\), where \(x\) and \(y\) are in metres. A spotlight at ground level at L\((-9, 0)\) shines a straight beam that just touches the hill at the point P\((a, 6\sqrt{a})\), so that the beam lies along the tangent to the curve at P.
A1 P\((9, 18)\) with gradient \(\frac{3}{\sqrt{9}} = 1\)
✓ 1
A1 \(y = x + 9\)
✓ 1
Part (d)
A1 \(\tan\theta = 1\), so \(\theta = 45°\)
✓ 1
Exam tip: When a tangent must pass through a point not on the curve, write the tangent at a general point \((a, f(a))\) and impose the condition; do not assume the given point lies on the curve.
176IBMA2H-053-01-085.14 Implicit differentiation, related rates and optimisationPaper 1Medium10 marks
No calculator
The boundary of an elliptical cam in a machine is modelled by the curve \(x^2 + xy + y^2 = 7\). The point A\((1, 2)\) lies on the curve.
(a)
Find an expression for \(\frac{dy}{dx}\) in terms of \(x\) and \(y\).
(3)
(b)
Find the equation of the normal to the curve at A, giving your answer in the form \(ax + by + c = 0\) where \(a, b, c \in \mathbb{Z}\).
(3)
(c)
The normal meets the curve again at B. Find the exact coordinates of B.
Exam tip: You already know one intersection of the normal with the curve (the point A), so use it as a factor or via the product of roots instead of solving the quadratic from scratch.
177IBMA2H-053-08-055.17 Areas with the y-axis and volumes of revolutionPaper 1Medium7 marks
No calculator
The inside of a parabolic solar-cooker bowl is formed by rotating the curve \(y=\frac{x^2}{8}\), for \(0\le y\le 2\), through \(2\pi\) radians about the \(y\)-axis. All lengths are in decimetres.
Diagram not to scale
(a)
Find the exact volume of the bowl.
(4)
(b)
The bowl is filled with water to a depth of \(h\) dm and contains half its full volume. Find the exact value of \(h\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(V=\pi\int x^2\,dy\)
✓ 1
M1 substitutes \(x^2=8y\): \(\pi\int_0^2 8y\,dy\)
✓ 1
A1 \(\left[4\pi y^2\right]_0^2\)
✓ 1
A1 \(16\pi\) dm3
✓ 1
Part (b)
M1 volume to depth \(h\) is \(4\pi h^2\)
✓ 1
M1 sets \(4\pi h^2=8\pi\)
✓ 1
A1 \(h=\sqrt2\)
✓ 1
Exam tip: Half the volume is not at half the depth. The bowl is wider near the top, so the half-volume level lies above \(y=1\).
178IBMA2H-053-08-065.16 Integration by substitution and by partsPaper 1Medium7 marks
No calculator
An engineer models the profile of a rotating cam by a function \(f\). Its derivative is \(f'(x)=\sin x\cos^4 x\), and \(f\left(\frac{\pi}{2}\right)=\frac{1}{5}\).
(a)
Find \(\int \sin x\cos^4x\,dx\).
(3)
(b)
Hence find the exact value of \(f\left(\frac{\pi}{3}\right)\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 recognises \(\sin x\) as a multiple of the derivative of \(\cos x\), e.g. substitutes \(u=\cos x\)
✓ 1
A1 \(-\frac{1}{5}\cos^5x\)
✓ 1
A1 \(+C\) included
✓ 1
Part (b)
M1 uses \(f\left(\frac{\pi}{2}\right)=\frac15\) with \(\cos\frac{\pi}{2}=0\)
✓ 1
A1 \(C=\frac15\)
✓ 1
M1 substitutes \(x=\frac{\pi}{3}\) using \(\cos\frac{\pi}{3}=\frac12\)
Common error: Losing the negative sign. The derivative of \(\cos x\) is \(-\sin x\), so the antiderivative is \(-\frac15\cos^5x\). Differentiate your answer to check.
179IBMA2H-053-08-075.17 Areas with the y-axis and volumes of revolutionPaper 1Medium6 marks
No calculator
The cross-section of a wind-turbine fairing is the region \(R\) bounded by the curve \(y=x^3+x\) for \(x\ge0\), the \(y\)-axis and the line \(y=10\). Lengths are in centimetres.
Diagram not to scale
(a)
Show that the curve passes through the point \((2,10)\).
(1)
(b)
Find \(\int_0^2 (x^3+x)\,dx\).
(2)
(c)
Hence find the area of \(R\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(2^3+2=8+2=10\), so \((2,10)\) lies on the curve
✓ 1
Part (b)
M1 integrates correctly to \(\frac{x^4}{4}+\frac{x^2}{2}\)
✓ 1
A1 \(4+2=6\)
✓ 1
Part (c)
R1 \(\frac{dy}{dx}=3x^2+1>0\), so the curve increases from \((0,0)\) to \((2,10)\) and lies below \(y=10\) for \(0\le x\le2\); hence \(R\) is the rectangle \(0\le x\le2\), \(0\le y\le10\) minus the area under the curve
✓ 1
M1 \(2\times10-6\), or equivalently \(\int_0^2 (10-x^3-x)\,dx\)
✓ 1
A1 \(14\) cm2
✓ 1
Exam tip: When \(x\) cannot easily be written in terms of \(y\), find an area against the \(y\)-axis by subtracting the area under the curve from a rectangle.
180IBMA2H-053-08-095.9 KinematicsPaper 1Medium13 marks
No calculator
A transdermal patch delivers a drug into the bloodstream. The rate at which the drug is absorbed is modelled by \(\frac{dA}{dt}=40te^{-0.25t^2}\), where \(A\) mg is the total amount absorbed \(t\) hours after the patch is applied, \(t\ge0\), and \(A=0\) when \(t=0\).
(a)
Find an expression for \(A\) in terms of \(t\).
(4)
(b)
State, with a reason, the total amount of drug absorbed in the long term.
(2)
(c)
Find the exact time at which 60 mg has been absorbed.
(3)
(d)
Find the exact time at which the rate of absorption is greatest, justifying that it is a maximum.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 integrates by reverse chain rule (or substitution \(u=-0.25t^2\)), obtaining the form \(ke^{-0.25t^2}\)
✓ 1
A1 \(A=-80e^{-0.25t^2}+C\)
✓ 1
M1 substitutes \(t=0\), \(A=0\) to find \(C\)
✓ 1
A1 \(A=80\left(1-e^{-0.25t^2}\right)\)
✓ 1
Part (b)
R1 as \(t\to\infty\), \(e^{-0.25t^2}\to0\)
✓ 1
A1 80 mg
✓ 1
Part (c)
M1 \(80\left(1-e^{-0.25t^2}\right)=60\), so \(e^{-0.25t^2}=\frac14\)
✓ 1
M1 takes logarithms: \(0.25t^2=\ln4\)
✓ 1
A1 \(t=2\sqrt{\ln4}\) hours (accept equivalent exact forms, e.g. \(2\sqrt{2\ln2}\) or \(\sqrt{4\ln4}\))
A1 sets the derivative equal to zero: \(t=\sqrt2\) hours, rejecting \(t=-\sqrt2\) since \(t\ge0\)
✓ 1
R1 the derivative is positive for \(0<t<\sqrt2\) and negative for \(t>\sqrt2\) (or an equivalent second-derivative argument), so the rate is a maximum
✓ 1
Common error: Dividing by the wrong factor. Differentiating \(e^{-0.25t^2}\) gives a factor of \(-0.5t\), so the constant in front is \(\frac{40}{-0.5}=-80\), not \(-10\).
181IBMA2H-053-08-115.16 Integration by substitution and by partsPaper 1Medium7 marks
No calculator
Answer the following.
(a)
Find \(\int\frac{x}{\sqrt{1-x^4}}\,dx\).
(3)
(b)
Find \(\int\frac{e^x}{1+e^{2x}}\,dx\) and hence evaluate \(\int_0^{\ln\sqrt3}\frac{e^x}{1+e^{2x}}\,dx\) exactly.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes \(u=x^2\), with \(du=2x\,dx\)
✓ 1
A1 \(\frac12\int\frac{1}{\sqrt{1-u^2}}\,du\)
✓ 1
A1 \(\frac12\arcsin(x^2)+C\)
✓ 1
Part (b)
M1 recognises the arctan form using \(u=e^x\)
✓ 1
A1 \(\arctan(e^x)+C\)
✓ 1
M1 substitutes the limits using \(e^{\ln\sqrt3}=\sqrt3\) and \(e^0=1\)
✓ 1
A1 \(\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}\)
✓ 1
Exam tip: Write \(x^4\) as \((x^2)^2\) and \(e^{2x}\) as \((e^x)^2\) to reveal the standard forms \(\frac{1}{\sqrt{1-u^2}}\) and \(\frac{1}{1+u^2}\).
182IBMA2H-053-02-065.14 Implicit differentiation, related rates and optimisationPaper 1Medium10 marks
No calculator
The boundary of an oval cycling track is modelled by the curve \(x^2 - xy + y^2 = 12\), where \(x\) and \(y\) are measured in hundreds of metres.
(a)
Show that \(\frac{dy}{dx} = \frac{2x - y}{x - 2y}\).
(3)
(b)
Find the coordinates of the points on the curve where the tangent is parallel to the x-axis.
(4)
(c)
Find the coordinates of the points on the curve where the tangent is parallel to the y-axis.
(2)
(d)
Hence state the greatest width of the track in the x-direction, in metres.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt at implicit differentiation
✓ 1
A1 \(2x - y - x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0\)
✓ 1
A1 rearranges to \(\frac{dy}{dx} = \frac{2x - y}{x - 2y}\)
A1 extreme x-values are ±4, so width = 8 hundred metres = 800 m
✓ 1
Common error: Vertical tangents come from the denominator of \(\frac{dy}{dx}\) being zero, not the numerator. Check that the numerator and denominator are not both zero at the same point.
183IBMA2H-053-02-085.8 Stationary points, inflexion and optimizationPaper 1Medium10 marks
No calculator
A function \(f\) has derivative \(f'(x) = (x - 1)^2(x + 3)\).
(a)
Find the x-coordinates of the stationary points of the graph of \(f\) and determine their nature, justifying your answer.
(4)
(b)
Show that \(f''(x) = (x - 1)(3x + 5)\).
(2)
(c)
Find the x-coordinates of the points of inflection of the graph of \(f\), justifying your answer.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(x = -3\) and \(x = 1\)
✓ 1
R1 \(f'\) changes from negative to positive at \(x = -3\)
✓ 1
A1 local minimum at \(x = -3\)
✓ 1
R1 \(f' > 0\) on both sides of \(x = 1\), so stationary point of inflection at \(x = 1\)
R1 both are simple roots so \(f''\) changes sign at each
✓ 1
Common error: A repeated factor in \(f'(x)\), such as \((x - 1)^2\), means \(f'\) does not change sign there, so the stationary point is not a maximum or minimum.
184IBMA2H-053-02-105.14 Implicit differentiation, related rates and optimisationPaper 1Medium7 marks
No calculator
The cross-section of a curved telescope mirror is modelled by part of the ellipse \(x^2 + 4y^2 = 8\).
(a)
Find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\).
(2)
(b)
Show that \(\frac{d^2y}{dx^2} = -\frac{1}{2y^3}\).
(4)
(c)
Hence state, with a reason, whether the part of the curve with \(y > 0\) is concave up or concave down.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(2x + 8y\frac{dy}{dx} = 0\)
✓ 1
A1 \(\frac{dy}{dx} = -\frac{x}{4y}\)
✓ 1
Part (b)
M1 quotient rule on \(-\frac{x}{4y}\), treating \(y\) as a function of \(x\)
M1 substitute \(\frac{dy}{dx} = -\frac{x}{4y}\) to get \(-\frac{x^2 + 4y^2}{16y^3}\)
✓ 1
A1 use \(x^2 + 4y^2 = 8\) to obtain \(-\frac{1}{2y^3}\)
✓ 1
Part (c)
R1 for \(y > 0\), \(\frac{d^2y}{dx^2} < 0\), so concave down
✓ 1
Common error: When differentiating \(\frac{dy}{dx}\) a second time, the \(y\) in the denominator still depends on \(x\), so its derivative brings in another \(\frac{dy}{dx}\).
185IBMA2H-053-03-055.15 Further derivatives and integrals; partial fractionsPaper 1Medium6 marks
No calculator
In a magnetic actuator, the force on a moving component when it is x metres from a reference point is \(F(x)=\frac{5x+1}{(x-1)(x+2)}\) newtons, for \(x>1\).
(a)
Express \(F(x)\) in partial fractions.
(3)
(b)
The work done as the component moves from x = 2 to x = 4 is \(\int_2^4 F(x)\,dx\) joules. Find the work done, giving your answer in the form \(\ln k\), where k is a rational number.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 write \(\frac{A}{x-1}+\frac{B}{x+2}\) and form \(5x+1=A(x+2)+B(x-1)\)
✓ 1
A1 \(A=2\)
✓ 1
A1 \(B=3\)
✓ 1
Part (b)
A1 \(2\ln(x-1)+3\ln(x+2)\)
✓ 1
M1 substitute limits and combine logs: \(2\ln 3+3\ln 6-3\ln 4\)
✓ 1
A1 \(\ln\frac{243}{8}\) J
✓ 1
Exam tip: Substituting the roots of each linear factor (x = 1 and x = −2) into the identity finds A and B in one line each, avoiding simultaneous equations.
186IBMA2H-053-03-065.16 Integration by substitution and by partsPaper 1Medium8 marks
No calculator
An autonomous survey boat leaves a harbour at time t = 0 hours. Its speed is modelled by \(v(t)=\frac{20}{t^2-4t+8}\) km h−1, \(t\geq 0\).
(a)
Write \(t^2-4t+8\) in the form \((t-p)^2+q\).
(2)
(b)
Find the exact distance travelled by the boat in the first 2 hours.
(4)
(c)
Assuming the model holds for all \(t\geq 0\), find the exact total distance the boat would travel.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(p=2\)
✓ 1
A1 \(q=4\)
✓ 1
Part (b)
M1 recognise arctan form \(\int\frac{1}{(t-2)^2+2^2}\,dt\)
R1 as \(t\to\infty\), \(\arctan\left(\frac{t-2}{2}\right)\to\frac{\pi}{2}\)
✓ 1
A1 \(10\left(\frac{\pi}{2}+\frac{\pi}{4}\right)=\frac{15\pi}{2}\) km
✓ 1
Common error: Taking \(\arctan(-1)\) as \(\frac{3\pi}{4}\) instead of \(-\frac{\pi}{4}\): the range of arctan is \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\).
187IBMA2H-053-03-075.16 Integration by substitution and by partsPaper 1Medium6 marks
No calculator
The cross-section of a skateboard ramp is the region bounded by the curve \(y=\arctan x\), the x-axis and the line x = 1, where x and y are in metres. Let \(f(x)=x\arctan x-\frac{1}{2}\ln(1+x^2)\).
The shaded region is the cross-section
(a)
Show that \(f'(x)=\arctan x\).
(3)
(b)
Hence find the exact area of the cross-section.
(2)
(c)
The ramp has uniform width 1.5 m. Write down the exact volume of the ramp.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 product rule on \(x\arctan x\)
✓ 1
A1 \(\arctan x+\frac{x}{1+x^2}\)
✓ 1
A1 \(-\frac{1}{2}\times\frac{2x}{1+x^2}=-\frac{x}{1+x^2}\), so the terms cancel to give \(\arctan x\) AG
✓ 1
Part (b)
M1 area \(=\left[x\arctan x-\frac{1}{2}\ln(1+x^2)\right]_0^1\), with the lower limit giving 0
✓ 1
A1 \(\frac{\pi}{4}-\frac{1}{2}\ln 2\) m2
✓ 1
Part (c)
A1 \(\frac{3\pi}{8}-\frac{3}{4}\ln 2\) m3
✓ 1
Exam tip: "Hence" signals that the first part has given you an antiderivative. You do not need to integrate \(\arctan x\) directly: just evaluate f at the limits, remembering to check the value at the lower limit.
188IBMA2H-053-03-095.16 Integration by substitution and by partsPaper 1Medium6 marks
No calculator
A test sled moves along a straight rail. Its velocity t seconds after release is \(v(t)=2\sec\left(\frac{\pi t}{6}\right)\tan\left(\frac{\pi t}{6}\right)\) m s−1, for \(0\leq t\leq 2\).
(a)
Find the exact acceleration of the sled when t = 1.
(3)
(b)
Find the exact distance travelled by the sled in the first 2 seconds.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 product rule with \(\frac{d}{d\theta}\sec\theta=\sec\theta\tan\theta\) and \(\frac{d}{d\theta}\tan\theta=\sec^2\theta\)
Exam tip: Since \(v\geq0\) on the interval, distance equals displacement; state this briefly to justify integrating v directly.
189IBMA2H-053-05-055.16 Integration by substitution and by partsPaper 1Medium7 marks
No calculator
After a car hits a bump, the vertical displacement of its suspension is modelled by \(y=e^{-t}\sin t\), where t is in seconds. An engineer needs the quantity \(Q=\int_0^{\pi}e^{-t}\sin t\,dt\).
The shaded area is Q
(a)
Use integration by parts twice to show that \(\int e^{-t}\sin t\,dt=-\frac12e^{-t}(\sin t+\cos t)+c\).
(5)
(b)
Find the exact value of Q.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 first application of parts, e.g. with \(u=\sin t\) and \(\frac{dv}{dt}=e^{-t}\) A1 \(I=-e^{-t}\sin t+\int e^{-t}\cos t\,dt\) M1 second application of parts: \(\int e^{-t}\cos t\,dt=-e^{-t}\cos t-\int e^{-t}\sin t\,dt\) A1 \(I=-e^{-t}\sin t-e^{-t}\cos t-I\) M1 rearranges \(2I=-e^{-t}(\sin t+\cos t)\) to obtain the given result (AG)
Exam tip: In cyclic integrals, keep the same choice of u type (the trig factor) in both applications. If you swap, you return to where you started. Call the integral I so that you can solve for it.
190IBMA2H-053-05-065.12 Limits, first principles and higher derivativesPaper 1Medium10 marks
No calculator
A manufacturer models the range, d km, of an electric scooter by \(d(x)=\sqrt{4x+1}\), where x is the energy stored in the battery in units of 100 Wh.
(a)
Use differentiation from first principles to show that \(d'(x)=\frac{2}{\sqrt{4x+1}}\).
(5)
(b)
Find the equation of the tangent to the graph of d at the point where \(x=2\).
(3)
(c)
Use your tangent to estimate the range when \(x=2.3\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes \(d'(x)=\lim_{h\to0}\frac{\sqrt{4x+4h+1}-\sqrt{4x+1}}{h}\) M1 multiplies numerator and denominator by the conjugate \(\sqrt{4x+4h+1}+\sqrt{4x+1}\) A1 numerator simplifies to \(4h\) A1 \(\lim_{h\to0}\frac{4}{\sqrt{4x+4h+1}+\sqrt{4x+1}}\) A1 letting \(h\to0\) gives \(\frac{4}{2\sqrt{4x+1}}=\frac{2}{\sqrt{4x+1}}\) (AG)
✓ 5
Part (b)
A1 \(d(2)=3\) and gradient \(d'(2)=\frac23\) M1 uses \(y-3=\frac23(x-2)\) A1 \(y=\frac23x+\frac53\)
✓ 3
Part (c)
M1 substitutes \(x=2.3\) into the tangent equation A1 3.2 km
✓ 2
Common error: With a square root, expanding the numerator directly leads nowhere. Multiply by the conjugate so that the h in the denominator can cancel.
191IBMA2H-053-05-075.16 Integration by substitution and by partsPaper 1Medium7 marks
No calculator
A phone downloads a large file. The download rate is modelled by \(r(t)=e^{\sqrt t}\) megabytes per second for \(1\le t\le4\), where t is in seconds.
(a)
Use the substitution \(u=\sqrt t\) to show that \(\int_1^4e^{\sqrt t}\,dt=\int_1^2 2ue^u\,du\).
(3)
(b)
Hence find the exact amount of data downloaded between \(t=1\) and \(t=4\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(t=u^2\), so \(dt=2u\,du\) A1 limits: \(t=1\) gives \(u=1\) and \(t=4\) gives \(u=2\) A1 obtains \(\int_1^2 2ue^u\,du\) (AG)
Exam tip: A substitution can turn an awkward integrand into one that needs integration by parts. Look for this two-stage structure when the first part says 'show that'.
192IBMA2H-053-05-095.16 Integration by substitution and by partsPaper 1Medium6 marks
No calculator
In a model of the magnetic field near a straight current-carrying wire segment, the field strength at a point 2 cm from the wire is proportional to \(I=\int_0^2\frac{1}{(x^2+4)^{3/2}}\,dx\).
(a)
Use the substitution \(x=2\tan\theta\) to show that \(I=\frac14\int_0^{\pi/4}\cos\theta\,d\theta\).
(4)
(b)
Hence find the exact value of I.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(dx=2\sec^2\theta\,d\theta\) A1 \(x^2+4=4\sec^2\theta\), so \((x^2+4)^{3/2}=8\sec^3\theta\) A1 limits: \(x=0\) gives \(\theta=0\) and \(x=2\) gives \(\theta=\frac{\pi}{4}\) A1 \(\frac{2\sec^2\theta}{8\sec^3\theta}=\frac14\cos\theta\), giving the result (AG)
Exam tip: Use the identity \(1+\tan^2\theta=\sec^2\theta\) straight away. Then the power 3/2 acts on a perfect square and simplifies cleanly.
193IBMA2H-053-05-105.13 Limits and l’Hôpital’s rulePaper 1Medium8 marks
No calculator
A roller-coaster track is designed in two sections. Its profile is given by \(f(x)=ax^2+bx\) for \(0\le x\le1\), \(f(x)=4\ln x+3\) for \(1<x\le3\), where \(a,b\in\mathbb{R}\) and distances are in tens of metres. The track must be smooth at the join, meaning f is continuous and differentiable at \(x=1\).
(a)
Use the condition for continuity at \(x=1\) to write an equation in a and b.
(2)
(b)
Use the condition for differentiability at \(x=1\) to write a second equation in a and b.
(2)
(c)
Find the values of a and b.
(2)
(d)
Determine whether \(f''\) is continuous at \(x=1\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 equates one-sided limits: \(\lim_{x\to1^-}f(x)=a+b\) and \(\lim_{x\to1^+}f(x)=4\ln1+3=3\) A1 \(a+b=3\)
✓ 2
Part (b)
M1 left derivative is \(2a+b\); right derivative is \(\frac4x\), which equals 4 at \(x=1\) A1 \(2a+b=4\)
✓ 2
Part (c)
M1 attempts to solve the two equations simultaneously A1 \(a=1\) and \(b=2\)
✓ 2
Part (d)
M1 \(f''(x)=2\) for \(x<1\) and \(f''(x)=-\frac{4}{x^2}\to-4\) as \(x\to1^+\) R1 the one-sided limits differ, so \(f''\) is not continuous at \(x=1\) and the curvature of the track changes abruptly
✓ 2
Common error: Continuity alone does not make a join smooth. You also need the left and right derivatives to agree at the join.
194IBMA3H-5-18r9-035.9 KinematicsPaper 1Medium6 marks
No calculator
A Formula 1 car brakes for a corner. Its acceleration is \(a(t)=6t-30\) m s−2 for \(0\le t\le 4\), where \(t\) is the time in seconds after braking begins. The speed at \(t=0\) is \(90\) m s−1.
(a)
Find an expression for the speed \(v(t)\) of the car.
(3)
(b)
The speed is positive for \(0\le t\le 3\). Find the distance travelled in the first 3 seconds.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 integrates \(a(t)\)
✓ 1
A1 \(v=3t^2-30t+c\)
✓ 1
A1 \(c=90\), so \(v(t)=3t^2-30t+90\)
✓ 1
Part (b)
M1 \(\int_0^3(3t^2-30t+90)\,dt\)
✓ 1
A1 \(\left[t^3-15t^2+90t\right]_0^3\)
✓ 1
A1 \(27-135+270=162\) m
✓ 1
Exam tip: Distance is the integral of speed only while the speed stays positive; check this before integrating.
195IBMA3H-5-18r18-065.13 Limits and l’Hôpital’s rulePaper 1Medium6 marks
No calculator
This question is about limits of functions involving \(\sin x\) as \(x\to0\).
(a)
Use l'Hôpital's rule to find \(\displaystyle\lim_{x\to0}\dfrac{\sin3x-3x}{x^3}\).
M1 write as \(\dfrac{\sin3x-3x}{x^3}\times\dfrac{x}{\sin x}\) and use \(\displaystyle\lim_{x\to0}\dfrac{x}{\sin x}=1\)
✓ 1
A1 \(-\dfrac92\)
✓ 1
Common error: Applying l'Hôpital's rule when the limit is not of the form \(\frac00\) or \(\frac\infty\infty\), or stopping after one application while the new quotient is still \(\frac00\).
196IBMA3H-5-18r18-075.14 Implicit differentiation, related rates and optimisationPaper 1Medium7 marks
No calculator
The curve \(C\) has equation \(x^2y+y^3=10\).
(a)
Show that \(\dfrac{dy}{dx}=-\dfrac{2xy}{x^2+3y^2}\).
(3)
(b)
Find the equation of the normal to \(C\) at the point \((1,2)\), in the form \(ax+by+c=0\) where \(a,b,c\in\mathbb{Z}\).
(2)
(c)
Find the coordinates of the point on \(C\) at which the tangent is horizontal.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 implicit differentiation attempted
✓ 1
A1 \(2xy+x^2\dfrac{dy}{dx}+3y^2\dfrac{dy}{dx}=0\) (product rule on \(x^2y\))
✓ 1
A1 rearrange to \(\dfrac{dy}{dx}=-\dfrac{2xy}{x^2+3y^2}\) (AG)
✓ 1
Part (b)
M1 tangent gradient \(=-\dfrac{4}{13}\), so normal gradient \(\dfrac{13}{4}\), and line through \((1,2)\)
✓ 1
A1 \(13x-4y-5=0\) (accept integer multiples)
✓ 1
Part (c)
M1 \(\dfrac{dy}{dx}=0\Rightarrow xy=0\); \(y=0\) is impossible (it gives \(0=10\)) so \(x=0\)
✓ 1
A1 \((0,\sqrt[3]{10})\) (\(y\approx2.15\))
✓ 1
Common error: Forgetting the product rule on terms such as \(x^2y\), or forgetting the factor \(\frac{dy}{dx}\) when differentiating a term in \(y\).
197IBMA3H-5-18r18-085.7 The second derivative and graphsPaper 1Medium7 marks
No calculator
The brightness of a stage light at distance \(x\) metres along a lighting rail is modelled by \(f(x)=xe^{-x}\), for \(x\ge0\).
(a)
Find \(f'(x)\).
(2)
(b)
Show that \(f\) has a maximum at \(x=1\).
(3)
(c)
Find the coordinates of the point of inflexion of the graph of \(f\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 product rule applied to \(x\cdot e^{-x}\)
✓ 1
A1 \(f'(x)=(1-x)e^{-x}\)
✓ 1
Part (b)
M1 \(f'(x)=0\Rightarrow x=1\) (ECF from (a))
✓ 1
A1 \(f''(x)=(x-2)e^{-x}\)
✓ 1
R1 at \(x=1\), \(f''=-e^{-1}<0\), so a maximum
✓ 1
Part (c)
M1 \(f''(x)=0\Rightarrow x=2\), with a sign change of \(f''\) (\(f''<0\) for \(x<2\) and \(f''>0\) for \(x>2\))
✓ 1
A1 \((2,2e^{-2})\) (\(y\approx0.271\))
✓ 1
Exam tip: A point of inflexion needs \(f''=0\) and a change of sign of \(f''\); solving \(f''=0\) alone does not prove it.
198IBMA3H-5-18r19-045.14 Implicit differentiation, related rates and optimisationPaper 1Medium6 marks
No calculator
A fish farm uses an open-topped storage crate with a square base of side \(x\) metres and height \(h\) metres. The volume of the crate is \(4\) m3.
Open-topped crate (the top face is not part of the crate). Diagram not to scale
(a)
Show that the total outer surface area of the crate (base and four sides) is \(S = x^2 + \frac{16}{x}\) m2.
(2)
(b)
Find the value of \(x\) that minimises \(S\), and hence find the minimum surface area.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(x^2h = 4\), so \(h = \frac{4}{x^2}\)
✓ 1
A1 \(S = x^2 + 4xh = x^2 + \frac{16}{x}\) (AG)
✓ 1
Part (b)
M1 \(S' = 2x - \frac{16}{x^2} = 0\)
✓ 1
A1 \(x^3 = 8\), so \(x = 2\)
✓ 1
A1 \(S = 4 + 8 = 12\) m2
✓ 1
R1 \(S''(2) = 2 + \frac{32}{8} = 6 > 0\), so a minimum
✓ 1
Exam tip: In optimisation, first eliminate all but one variable using the constraint, then differentiate. Justify the minimum with the second derivative or a sign change.
199IBMA3H-5-18r19-055.14 Implicit differentiation, related rates and optimisationPaper 1Medium7 marks
No calculator
The sustainable yield \(Y\) tonnes of a fishery depends on the fishing effort \(x\) (in thousands of boat-hours) and is modelled by \(Y = 0.02x^2(60 - x)\), for \(0 \le x \le 60\).
(a)
Find the value of \(x\) for which \(Y\) is a maximum.
(3)
(b)
Find the maximum yield and justify that it is a maximum.
(2)
(c)
Find the value of \(x\) at which the rate of change of \(Y\) with respect to \(x\) is greatest.
R1 \(\frac{d^2Y}{dx^2} = 2.4 - 0.12x\), which is \(-2.4 < 0\) at \(x = 40\)
✓ 1
Part (c)
M1 set \(\frac{d^2Y}{dx^2} = 0\)
✓ 1
A1 \(x = 20\)
✓ 1
Common error: The greatest rate of change of \(Y\) occurs where \(\frac{dY}{dx}\) is largest, so differentiate \(\frac{dY}{dx}\) and set that to zero, not \(\frac{dY}{dx}\) itself.
200IBMA3H-5-18r19-095.14 Implicit differentiation, related rates and optimisationPaper 1Medium6 marks
No calculator
A designer's outline is the curve \(x^2 + xy + y^2 = 7\), where \(x\) and \(y\) are in centimetres. The point P\((1, 2)\) lies on the curve.
(a)
Find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\), and hence its value at P.
(3)
(b)
Find the equation of the normal to the curve at P. Give your answer in the form \(ax + by + c = 0\) where \(a\), \(b\), \(c\) are integers.
(2)
(c)
Find the x-coordinate of the point where the normal meets the x-axis.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 implicit differentiation, including the product rule on \(xy\)
✓ 1
A1 \(2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0\), so \(\frac{dy}{dx} = -\frac{2x + y}{x + 2y}\)
✓ 1
A1 \(-\frac{4}{5}\) at P
✓ 1
Part (b)
M1 normal gradient \(\frac{5}{4}\) through \((1, 2)\)
✓ 1
A1 \(5x - 4y + 3 = 0\)
✓ 1
Part (c)
A1 \(x = -\frac{3}{5}\) (ECF from (b))
✓ 1
Common error: Differentiating \(xy\) requires the product rule: \(\frac{d}{dx}(xy) = y + x\frac{dy}{dx}\). Missing the first term is the most common slip.
201IBMA3H-5-18r19-115.16 Integration by substitution and by partsPaper 1Medium5 marks
No calculator
The velocity of a robotic arm is \(v = 4xe^{2x}\), where \(x\) is the position of the arm in metres along its rail, for \(0 \le x \le 1\). Use integration by parts to find \(\int_0^1 4xe^{2x}\,dx\), giving your answer in exact form.
(a)
Find the exact value of \(\int_0^1 4xe^{2x}\,dx\).
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 integration by parts with \(u = 4x\), \(\frac{dv}{dx} = e^{2x}\)
Common error: Take \(u\) as the factor that gets simpler on differentiating (here \(x\)) and \(\frac{dv}{dx}\) as the factor you can integrate. Remember the \(\frac{1}{2}\) from integrating \(e^{2x}\).
202IBMA3H-5-18r19-125.17 Areas with the y-axis and volumes of revolutionPaper 1Medium5 marks
No calculator
A fashion designer models a vase by rotating the region bounded by \(y = x^2 + 1\), the x-axis, the y-axis and the line \(x = 2\) through \(360°\) about the x-axis. Lengths are in decimetres.
M1 identifies the form \(\frac00\) and differentiates numerator and denominator (l'Hôpital's rule)
✓ 1
A1 \(\lim_{x\to0}\dfrac{3e^{3x}}{2\cos 2x}\)
✓ 1
A1 \(\dfrac32\)
✓ 1
Part (b)
M1 identifies \(\frac00\) and applies l'Hôpital's rule
✓ 1
A1 \(\lim_{x\to0}\dfrac{4\sin 4x}{2x}\)
✓ 1
M1 recognises the form is still \(\frac00\) and applies the rule again
✓ 1
A1 \(\dfrac{16\cos 4x}{2}\to8\)
✓ 1
Common error: l'Hôpital's rule differentiates the numerator and denominator separately. It is not the quotient rule. Check the form is \(\frac00\) or \(\frac{\infty}{\infty}\) before each use.
205IBMA3H-5-19r26-035.13 Limits and l’Hôpital’s rulePaper 1Medium6 marks
No calculator
Let \(L=\lim_{x\to0}\frac{e^{3x}-1-3x}{x^2}\).
(a)
Use l'Hôpital's rule to find the value of \(L\).
(4)
(b)
Use the Maclaurin series for \(e^{3x}\) up to and including the \(x^2\) term to verify your answer to part (a).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 identify the form \(\frac{0}{0}\) and differentiate numerator and denominator
✓ 1
A1 \(\frac{3e^{3x}-3}{2x}\)
✓ 1
M1 recognise \(\frac{0}{0}\) again and apply the rule a second time
✓ 1
A1 \(\frac{9e^{3x}}{2}\to\frac{9}{2}\), so \(L=\frac{9}{2}\)
✓ 1
Part (b)
M1 \(e^{3x}\approx1+3x+\frac{9x^2}{2}\), so numerator \(\approx\frac{9x^2}{2}\)
✓ 1
A1 dividing by \(x^2\) and letting \(x\to0\) gives \(\frac{9}{2}\)
✓ 1
Exam tip: Check the form at every stage: l'Hôpital's rule may only be applied while the limit is \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), and here it must be used twice.
206IBMA3H-5-19r26-045.19 Maclaurin seriesPaper 1Medium6 marks
No calculator
A theatre lighting designer models the brightness ramp of a lamp by \(f(x)=e^x\sin x\).
(a)
Find \(f(0)\), \(f'(0)\), \(f''(0)\) and \(f'''(0)\).
(4)
(b)
Hence write down the Maclaurin series for \(f(x)\) up to and including the term in \(x^3\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 product rule: \(f'(x)=e^x\sin x+e^x\cos x\)
✓ 1
A1 \(f''(x)=2e^x\cos x\)
✓ 1
A1 \(f'''(x)=2e^x\cos x-2e^x\sin x\)
✓ 1
A1 \(f(0)=0,\ f'(0)=1,\ f''(0)=2,\ f'''(0)=2\)
✓ 1
Part (b)
M1 use of \(f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3\)
✓ 1
A1 \(f(x)\approx x+x^2+\frac{x^3}{3}\) (ECF from (a))
✓ 1
Common error: Remember to divide by the factorials: the \(x^3\) coefficient is \(\frac{f'''(0)}{3!}\), not \(f'''(0)\).
207IBMA3H-5-19r26-065.15 Further derivatives and integrals; partial fractionsPaper 1Medium7 marks
No calculator
The rate at which cargo is loaded onto a ship is modelled by \(R(x)=\frac{5x+1}{(x-1)(x+2)}\) tonnes per hour, where \(x\ge2\) is the time in hours since loading began.
(a)
Express \(R(x)\) in the form \(\frac{A}{x-1}+\frac{B}{x+2}\), where \(A\) and \(B\) are constants to be found.
(3)
(b)
Hence find the exact mass of cargo loaded between \(x=2\) and \(x=4\). Give your answer in the form \(\ln k\), where \(k\in\mathbb{Q}\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(5x+1=A(x+2)+B(x-1)\)
✓ 1
A1 \(A=2\)
✓ 1
A1 \(B=3\)
✓ 1
Part (b)
M1 integrate to obtain logarithmic terms
✓ 1
A1 \(2\ln|x-1|+3\ln|x+2|\)
✓ 1
M1 substitute limits 4 and 2 into an expression of this form
Common error: When integrating \(\frac{A}{x-1}\) the result is \(A\ln|x-1|\); do not forget the constant multiplier or try to integrate the whole fraction as one logarithm.
208IBMA3H-5-19r26-085.15 Further derivatives and integrals; partial fractionsPaper 1Medium6 marks
No calculator
The flare of a ship's hull is modelled by the curve \(y=x^2\arctan(3x)\), for \(x\ge0\).
(a)
Find \(\frac{dy}{dx}\).
(4)
(b)
Find the exact gradient of the curve at \(x=\frac13\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 product rule with \(u=x^2\) and \(v=\arctan(3x)\)
M1 substitute \(x=\frac13\) using \(\arctan1=\frac{\pi}{4}\) and \(1+9x^2=2\)
✓ 1
A1 \(\frac{\pi}{6}+\frac16=\frac{\pi+1}{6}\)
✓ 1
Common error: The derivative of \(\arctan(3x)\) is \(\frac{3}{1+9x^2}\); the factor 3 from the chain rule is often omitted, and the 3 inside the bracket must also be squared in the denominator.
209IBMA3H-5-19r26-095.19 Maclaurin seriesPaper 1Medium7 marks
No calculator
Let \(f(x)=\ln(\sec x+\tan x)\) for \(0\le x<\frac{\pi}{2}\).
(a)
Show that \(f'(x)=\sec x\).
(5)
(b)
Hence find the exact value of \(\int_0^{\pi/4}\sec x\,dx\).
M1 factorise the numerator as \(\sec x(\tan x+\sec x)\)
✓ 1
A1 cancel the common factor to reach \(\sec x\)
✓ 1
Part (b)
M1 use of \([\ln(\sec x+\tan x)]_0^{\pi/4}\)
✓ 1
A1 \(\ln(\sqrt2+1)-\ln1=\ln(1+\sqrt2)\)
✓ 1
Exam tip: In a 'show that' question, every line of working must be written down. State each standard derivative you use and show the factorisation that leads to the cancellation.
210IBMA3H-5-19r26-105.14 Implicit differentiation, related rates and optimisationPaper 1Medium10 marks
No calculator
A shipping company designs an open-topped storage crate with a square base of side \(x\) metres and height \(h\) metres. The total external surface area (the base and the four sides) is 12 m2. The volume of the crate is \(V\) m3.
Open-topped crate (the top face is not part of the crate). Diagram not to scale
(a)
Show that \(V=3x-\frac{x^3}{4}\).
(3)
(b)
Find \(\frac{dV}{dx}\) and hence find the value of \(x\) for which \(V\) is stationary, \(x>0\).
(3)
(c)
Show that this stationary value is a maximum.
(2)
(d)
Find the height of the crate when \(V\) is a maximum.
(1)
(e)
Find the maximum volume of the crate.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 surface area equation \(x^2+4xh=12\)
✓ 1
M1 \(h=\frac{12-x^2}{4x}\) substituted into \(V=x^2h\)
✓ 1
A1 \(V=\frac{x(12-x^2)}{4}=3x-\frac{x^3}{4}\)
✓ 1
Part (b)
A1 \(\frac{dV}{dx}=3-\frac{3x^2}{4}\)
✓ 1
M1 set \(\frac{dV}{dx}=0\)
✓ 1
A1 \(x=2\) (reject \(x=-2\))
✓ 1
Part (c)
A1 \(\frac{d^2V}{dx^2}=-\frac{3x}{2}\)
✓ 1
R1 at \(x=2\) the second derivative is \(-3<0\), so a maximum
✓ 1
Part (d)
A1 \(h=\frac{12-4}{8}=1\) m
✓ 1
Part (e)
A1 \(V=6-2=4\) m3
✓ 1
Exam tip: In optimisation questions, eliminate the second variable using the constraint first, then differentiate. Always check the nature of the stationary point and use the value of \(x\) in the original context.
211IBMA3H-5-20r27-035.18 First order differential equationsPaper 1Medium7 marks
No calculator
A quantity \(y\) in a model of a cable signal satisfies the differential equation \(\dfrac{dy}{dx}+\dfrac{2}{x}y=6x\) for \(x>0\), with \(y=2\) when \(x=1\).
(a)
Find the integrating factor for this equation, simplifying your answer.
(2)
(b)
Solve the differential equation, giving \(y\) in terms of \(x\).
Exam tip: Multiply the whole equation by the integrating factor; the left side then becomes the derivative of (integrating factor × \(y\)).
212IBMA3H-5-20r27-105.2 Increasing and decreasing functionsPaper 1Medium5 marks
No calculator
A scientist models the total sugar dissolved after \(x\) minutes by \(f(x)=x^3+kx^2+12x+5\), where \(k\) is a constant. The model is required to give a function that is increasing for all real \(x\).
(a)
Find \(f'(x)\).
(1)
(b)
Find the set of possible values of \(k\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(f'(x)=3x^2+2kx+12\)
✓ 1
Part (b)
M1 \(f'(x)\ge0\) for all \(x\), so the discriminant is at most 0
✓ 1
M1 \((2k)^2-4(3)(12)\le0\)
✓ 1
A1 \(k^2\le36\)
✓ 1
A1 \(-6\le k\le6\)
✓ 1
Exam tip: 'Increasing for all \(x\)' means \(f'(x)\ge0\) for every \(x\), which for an upward quadratic means its discriminant is not positive.
213IBMA3H-5-20r27-125.3 Differentiating polynomials and integer powersPaper 1Medium6 marks
No calculator
The height \(f(x)\) of a ridge profile is modelled by \(f(x)=x^3+px^2+qx\), where \(p\) and \(q\) are constants. The function \(f\) is decreasing only for \(-1<x<3\).
(a)
Find the values of \(p\) and \(q\).
(4)
(b)
Find the value of \(f(-1)-f(3)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(f'(x)=3x^2+2px+q\)
✓ 1
M1 \(f'(-1)=0\) and \(f'(3)=0\) (or \(f'(x)=3(x+1)(x-3)\))
✓ 1
A1 \(p=-3\)
✓ 1
A1 \(q=-9\)
✓ 1
Part (b)
M1 \(f(-1)=5\) and \(f(3)=-27\), using \(f(x)=x^3-3x^2-9x\)
✓ 1
A1 32
✓ 1
Exam tip: If \(f\) changes from increasing to decreasing, or back, then \(f'(x)=0\) at those points; use both values to set up equations.
214IBMA3H-5-20r31-035.3 Differentiating polynomials and integer powersPaper 1Medium6 marks
No calculator
A brace on a bridge is modelled by the line \(y=3x+c\), which is tangent to the curve \(y=-\frac{12}{x}\), where \(x\ne0\). Find the two possible values of \(c\).
Exam tip: A tangent must share both the gradient and a point with the curve; use the gradient to find where, then use the point to find the missing constant.
215IBMA3H-5-19r33-055.9 KinematicsPaper 1Medium7 marks
No calculator
A drone filming a Formula 1 race moves along a straight line above the track. Its velocity at time \(t\) seconds is \(v(t)=3t^{2}-12t+9\) m s−1, for \(t\ge0\). At \(t=0\) the displacement of the drone from a fixed point \(O\) is \(s=2\) m.
(a)
Find the times at which the drone is instantaneously at rest.
(2)
(b)
Find an expression for the displacement \(s\) of the drone from \(O\) at time \(t\).
(3)
(c)
Find the total distance travelled by the drone during the first 3 seconds.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 set \(v(t)=0\), e.g. \(3(t-1)(t-3)=0\)
✓ 1
A1 \(t=1\) s and \(t=3\) s
✓ 1
Part (b)
M1 attempt to integrate \(v\)
✓ 1
A1 \(t^{3}-6t^{2}+9t\) (+C)
✓ 1
A1 use of \(s(0)=2\) to obtain \(s=t^{3}-6t^{2}+9t+2\)
✓ 1
Part (c)
M1 evaluate \(s=2\) at \(t=0\), \(s=6\) at \(t=1\) and \(s=2\) at \(t=3\), and recognise that the drone reverses direction at \(t=1\)
✓ 1
A1 distance \(=(6-2)+(6-2)=8\) m
✓ 1
Common error: Displacement and distance differ when the object changes direction: split the motion where the velocity is zero.
216IBMA3H-5-19r33-075.16 Integration by substitution and by partsPaper 1Medium6 marks
No calculator
Along a hedgerow, bees collect pollen at the rate \(p(x)=x^{2}\ln x\) grams per tens of metres, where \(x\) is the distance in tens of metres from the hive, for \(1\le x\le e\).
(a)
Use integration by parts to find \(\int x^{2}\ln x\,dx\).
(3)
(b)
Hence find the exact mass of pollen collected between \(x=1\) and \(x=e\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 integration by parts with \(u=\ln x\) and \(\frac{dv}{dx}=x^{2}\)
M1 substitute limits \(e\) and \(1\) into the antiderivative
✓ 1
A1 \(\left(\frac{e^{3}}{3}-\frac{e^{3}}{9}\right)-\left(0-\frac{1}{9}\right)\) using \(\ln e=1\), \(\ln 1=0\)
✓ 1
A1 \(\frac{2e^{3}+1}{9}\) g
✓ 1
Exam tip: For integrals of a logarithm multiplied by a power, let \(u=\ln x\) so that the integral left over is a simple power.
217IBMA3H-5-19r33-085.14 Implicit differentiation, related rates and optimisationPaper 1Medium7 marks
No calculator
During a race, the temperature rise of a Formula 1 tyre is modelled by \(\theta(t)=\dfrac{120t}{t^{2}+9}\) °C, where \(t\ge 0\) is the time in minutes after leaving the pit lane.
(a)
Use the quotient rule to show that \(\theta'(t)=\dfrac{120(9-t^{2})}{(t^{2}+9)^{2}}\).
(3)
(b)
Given that the temperature rise has a maximum value, find this maximum value.
(3)
(c)
Find the rate at which the temperature rise is changing when \(t=1\).
A1 simplifies to \(\dfrac{120(9-t^{2})}{(t^{2}+9)^{2}}\) (AG)
✓ 1
Part (b)
M1 set \(\theta'(t)=0\), so \(9-t^{2}=0\)
✓ 1
A1 \(t=3\) (rejecting \(t=-3\))
✓ 1
A1 \(\theta(3)=20\) °C
✓ 1
Part (c)
A1 \(\theta'\) at \(t=1\) is \(\dfrac{120\times 8}{100}=9.6\) °C per minute
✓ 1
Common error: Take care with the sign of the middle term in the quotient rule: it is \(vu'-uv'\), not \(uv'-vu'\).
218IBMA3H-5-19r33-095.16 Integration by substitution and by partsPaper 1Medium6 marks
No calculator
The flow rate of blood through a heart valve during one beat is modelled by \(Q(t)=240\sin t\cos^{2}t\) mL s−1, for \(0\le t\le \dfrac{\pi}{3}\), where \(t\) is in seconds.
(a)
Use the substitution \(u=\cos t\) to find the exact volume of blood that passes through the valve during this time.
(4)
(b)
Find the mean flow rate over this time interval.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(u=\cos t\), so \(du=-\sin t\,dt\)
✓ 1
A1 limits \(u=1\) and \(u=\frac{1}{2}\), giving \(240\int_{1/2}^{1}u^{2}\,du\)
✓ 1
M1 integrate to obtain \(80u^{3}\) and substitute the limits
✓ 1
A1 \(80\left(1-\frac{1}{8}\right)=70\) mL
✓ 1
Part (b)
M1 divide the volume by the length of the interval, \(\dfrac{70}{\pi/3}\)
✓ 1
A1 \(\dfrac{210}{\pi}\) mL s−1 (accept 66.8)
✓ 1
Exam tip: After a substitution, check that the integrand contains only the new variable; the factor \(\sin t\,dt\) is absorbed into \(-du\).
219IBMA3H-5-20r36-055.16 Integration by substitution and by partsPaper 1Medium6 marks
No calculator
Answer both parts using integration by inspection or substitution.
(a)
Find ∫ 4x ex2 dx.
(2)
(b)
Use the substitution u = 1 + cos x to show that ∫0π/2 sin x/(1 + cos x)2 dx = ½.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 recognises the reverse chain rule (or substitutes u = x2)
✓ 1
A1 2ex2 + C
✓ 1
Part (b)
M1 du = −sin x dx, giving the integral of −u−2
✓ 1
A1 antiderivative 1/u = 1/(1 + cos x)
✓ 1
M1 substitutes limits: u = 1 (at π/2) and u = 2 (at 0)
✓ 1
A1 1/1 − 1/2 = ½
✓ 1
Common error: When changing variable in a definite integral, either change the limits to u-values or return to x before substituting; mixing the two gives wrong answers.
220IBMA3H-5-20r36-065.9 KinematicsPaper 1Medium8 marks
No calculator
A cargo container moves along a straight rail on the deck of a ship. Its velocity is v(t) = 3t2 − 18t + 24 m s−1, for 0 ≤ t ≤ 5. Its displacement from its starting point is s metres, with s = 0 when t = 0.
(a)
Find the times at which the container is instantaneously at rest.
(2)
(b)
Find the displacement of the container from its starting point when t = 5.
(3)
(c)
Find the total distance travelled by the container in the first 5 seconds.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets v = 0 and factorises 3(t − 2)(t − 4)
✓ 1
A1 t = 2 and t = 4
✓ 1
Part (b)
M1 integrates v
✓ 1
A1 s = t3 − 9t2 + 24t (using s(0) = 0)
✓ 1
A1 s(5) = 20 m
✓ 1
Part (c)
M1 finds the positions at the turning points (ECF from (b))
✓ 1
A1 s(2) = 20 and s(4) = 16
✓ 1
A1 total distance = 20 + 4 + 4 = 28 m
✓ 1
Common error: Displacement and distance differ when velocity changes sign; find the turning points first and add the sizes of the separate movements.
221IBMA3H-5-20r36-095.19 Maclaurin seriesPaper 1Medium6 marks
No calculator
Use standard series results to answer both parts.
(a)
Use standard Maclaurin series to find the series for ex cos x up to and including the term in x3.
(4)
(b)
Hence find limx→0 (ex cos x − 1 − x)/x3.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes ex = 1 + x + x2/2 + x3/6 and cos x = 1 − x2/2
✓ 1
M1 multiplies the two series and collects terms
✓ 1
A1 the coefficient of x2 is ½ − ½ = 0
✓ 1
A1 1 + x − x3/3
✓ 1
Part (b)
M1 substitutes the series, giving (−x3/3 + higher powers)/x3
✓ 1
A1 −1/3
✓ 1
Exam tip: When asked for a limit 'hence', substitute the series, divide through by the lowest power of x remaining, and let x tend to zero.
222IBMA3H-5-20r37-025.6 Differentiation rulesPaper 1Medium8 marks
No calculator
The depth, d metres, of water in an irrigation channel is modelled by f(x) = √x, where x is the flow rate in m3 per minute and x > 0.
(a)
Use differentiation from first principles to show that f′(x) = 1/(2√x).
(5)
(b)
Find the y-intercept of the tangent to the graph of f at the point where x = 16.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 f′(x) = limh→0 (√(x + h) − √x)/h
✓ 1
M1 multiply numerator and denominator by √(x + h) + √x
✓ 1
A1 numerator becomes (x + h) − x = h
✓ 1
A1 cancel h to obtain 1/(√(x + h) + √x)
✓ 1
A1 let h → 0 to obtain 1/(2√x)
✓ 1
Part (b)
A1 f(16) = 4 and f′(16) = 1/8
✓ 1
M1 y − 4 = (1/8)(x − 16)
✓ 1
A1 y-intercept = 2
✓ 1
Exam tip: For a difference of square roots, multiply by the conjugate; the brackets then collapse to a difference of squares with no roots left.
223IBMA3H-5-20r37-075.11 Definite integrals and areasPaper 1Medium8 marks
No calculator
The vertical path of a drone is modelled by y = x3 + px2 + qx, where x is the horizontal distance in kilometres and p and q are constants. The graph has a point of inflection at x = 1 and a stationary point at x = 3.
(a)
Find the value of p.
(3)
(b)
Find the value of q.
(2)
(c)
Find the y-coordinate of the stationary point and show that it is a minimum.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 d2y/dx2 = 6x + 2p
✓ 1
M1 set equal to 0 at x = 1
✓ 1
A1 p = −3
✓ 1
Part (b)
M1 dy/dx = 3x2 + 2px + q = 0 at x = 3: 27 − 18 + q = 0 (ECF from (a))
✓ 1
A1 q = −9
✓ 1
Part (c)
A1 y = 27 − 27 − 27 = −27
✓ 1
M1 second derivative at x = 3: 18 − 6
✓ 1
R1 12 > 0, so a minimum
✓ 1
Exam tip: "Point of inflection at x = 1" gives an equation in the second derivative; "stationary at x = 3" gives one in the first derivative.
224IBMA3H-5-20r37-085.3 Differentiating polynomials and integer powersPaper 1Medium12 marks
No calculator
The boundary of a crop plot is modelled by the curve x2 + xy + y2 = 7, where x and y are in hundreds of metres.
(a)
Find an expression for dy/dx in terms of x and y.
(4)
(b)
Show that the point (1, 2) lies on the curve, and find the equation of the tangent at this point in the form ax + by = c.
(4)
(c)
Find the coordinates of the points on the curve where the tangent is horizontal.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 differentiate each term implicitly
✓ 1
A1 d/dx(xy) = y + x dy/dx
✓ 1
A1 2x + y + (x + 2y) dy/dx = 0
✓ 1
A1 dy/dx = −(2x + y)/(x + 2y)
✓ 1
Part (b)
A1 1 + 2 + 4 = 7
✓ 1
A1 gradient = −4/5 (ECF from (a))
✓ 1
M1 y − 2 = (−4/5)(x − 1)
✓ 1
A1 4x + 5y = 14
✓ 1
Part (c)
M1 numerator 2x + y = 0 so y = −2x
✓ 1
M1 substitute: x2 − 2x2 + 4x2 = 7 so 3x2 = 7
✓ 1
A1 x = ±√(7/3) (≈ ±1.5275)
✓ 1
A1 (√(7/3), −2√(7/3)) and (−√(7/3), 2√(7/3))
✓ 1
Common error: When differentiating xy, use the product rule; omitting the y term loses the accuracy marks for the whole expression.
225IBMA3H-5-19r38-055.8 Stationary points, inflexion and optimizationPaper 1Medium7 marks
No calculator
The altitude, \(A\) hundred metres, of a mountain trail at horizontal distance \(d\) km from the trailhead is modelled by \(A(d)=d^{3}-9d^{2}+24d+3\) for \(0\le d\le6\).
(a)
Find the values of \(d\) at which the trail has zero gradient.
(3)
(b)
Use the second derivative to determine the nature of each of these points.
(2)
(c)
Find the altitude of the trail at the local minimum, in metres.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets \(A'(d)=0\)
✓ 1
A1 \(A'(d)=3d^2-18d+24=3(d-2)(d-4)\)
✓ 1
A1 \(d=2\) and \(d=4\)
✓ 1
Part (b)
M1 \(A''(d)=6d-18\), evaluated at \(d=2\) and \(d=4\) giving \(-6\) and \(6\)
✓ 1
A1 local maximum at \(d=2\) since \(A''<0\); local minimum at \(d=4\) since \(A''>0\)
✓ 1
Part (c)
M1 substitutes \(d=4\): \(64-144+96+3\)
✓ 1
A1 \(A=19\) hundred metres \(=1900\) m
✓ 1
Exam tip: Altitude is given in hundreds of metres, so convert at the end: multiply the function value by 100.
226IBMA3H-5-19r38-075.13 Limits and l’Hôpital’s rulePaper 1Medium6 marks
No calculator
Evaluate each limit using l'Hôpital's rule, showing each step.
(a)
\(\lim_{x\to0}\frac{\sin 5x}{3x}\)
(2)
(b)
\(\lim_{x\to0}\frac{e^{3x}-1-3x}{x^{2}}\)
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 form is \(\frac00\), differentiates numerator and denominator: \(\frac{5\cos5x}{3}\)
✓ 1
A1 limit \(=\frac53\)
✓ 1
Part (b)
R1 form is \(\frac00\), so l'Hôpital's rule applies
✓ 1
A1 first application: \(\frac{3e^{3x}-3}{2x}\)
✓ 1
M1 recognises \(\frac00\) again and differentiates a second time
✓ 1
A1 \(\frac{9e^{3x}}{2}\to\frac92\)
✓ 1
Exam tip: Check the \(\frac00\) form before every application of l'Hôpital's rule; if it is no longer indeterminate, substitute instead.
227IBMA3H-5-19r38-085.18 First order differential equationsPaper 1Medium7 marks
No calculator
A pie is removed from an oven at temperature \(180\,{}^\circ\)C and left in a kitchen at constant temperature \(20\,{}^\circ\)C. Its temperature \(T\,{}^\circ\)C, \(t\) minutes after removal, satisfies \(\frac{dT}{dt}=-k(T-20)\), where \(k>0\) is a constant.
(a)
Solve the differential equation to show that \(T=20+160e^{-kt}\).
(3)
(b)
After 10 minutes the temperature of the pie is \(100\,{}^\circ\)C. Show that \(k=\frac{\ln2}{10}\).
(2)
(c)
Find the temperature of the pie 30 minutes after removal.
A1 uses \(T=180\) at \(t=0\) (so \(T-20>0\)) to obtain \(T=20+160e^{-kt}\)
✓ 1
Part (b)
M1 \(80=160e^{-10k}\)
✓ 1
A1 \(e^{-10k}=\frac12\), so \(k=\frac{\ln2}{10}\)
✓ 1
Part (c)
M1 \(e^{-30k}=\left(e^{-10k}\right)^3=\frac18\)
✓ 1
A1 \(T=20+160\times\frac18=40\,{}^\circ\)C
✓ 1
Exam tip: Work with \(e^{-10k}\) as a single quantity: tripling the time cubes it, which avoids needing a decimal value of \(k\).
228IBMA3H-5-19r38-105.3 Differentiating polynomials and integer powersPaper 1Medium7 marks
No calculator
The curve \(C\) has equation \(y=x^{3}-3x^{2}+2x+4\). A radio link is sent in a straight line along the tangent to \(C\) at the point where \(x=2\).
(a)
Find the equation of the tangent to \(C\) at the point where \(x=2\).
(4)
(b)
The tangent meets \(C\) again at the point \(P\). Find the coordinates of \(P\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(y'=3x^2-6x+2\)
✓ 1
A1 when \(x=2\), \(y=4\)
✓ 1
M1 gradient \(=3(4)-12+2=2\), used in \(y-4=2(x-2)\)
✓ 1
A1 \(y=2x\)
✓ 1
Part (b)
M1 equates: \(x^3-3x^2+2x+4=2x\), so \(x^3-3x^2+4=0\)
✓ 1
M1 uses the repeated root \(x=2\) to factorise: \((x-2)^2(x+1)=0\)
✓ 1
A1 \(P=(-1,-2)\)
✓ 1
Exam tip: A tangent touches the curve with a repeated root, so the equation for the intersection has a squared factor; use it to find the remaining root quickly.
229IBMA3H-5-19r38-115.19 Maclaurin seriesPaper 1Medium6 marks
No calculator
The function \(f\) is given by \(f(x)=xe^{2x}\).
(a)
Find \(f''(x)\) and \(f'''(x)\).
(3)
(b)
Hence find the Maclaurin series for \(f(x)\) up to and including the term in \(x^3\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses the product rule to find \(f'(x)=(1+2x)e^{2x}\)
Exam tip: Subtract the lower function from the upper function, and sketch or test a point to be sure which is on top.
231IBMA3H-5-20r45-075.1 Limits and the derivativePaper 1Medium5 marks
No calculator
A fish farm's demand for feed is modelled by \(F(t) = 20e^{0.5t}\) tonnes per month, where \(t\) is the time in months since the farm opened. The total feed needed during the first \(k\) months is 120 tonnes.
(a)
Show that the total feed needed in the first \(k\) months is \(40e^{0.5k} - 40\) tonnes.
(2)
(b)
Find the exact value of \(k\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\int_0^k 20e^{0.5t}\,dt\) with an attempt to integrate
✓ 1
A1 \(\left[40e^{0.5t}\right]_0^k\) and so \(40e^{0.5k} - 40\) (AG)
M1 recognising the form \(\frac00\) and differentiating numerator and denominator
✓ 1
A1 \(\frac{2\sin 2x}{2x}\)
✓ 1
M1 still \(\frac00\), differentiating again (or using \(\frac{\sin 2x}{x}\to2\))
✓ 1
A1 \(\frac{4\cos 2x}{2}\to2\)
✓ 1
Part (b)
M1 rewriting as \(\frac{x^2}{\mathrm{e}^{x}}\), form \(\frac{\infty}{\infty}\), and differentiating
✓ 1
A1 \(\frac{2x}{\mathrm{e}^{x}}\), which is still \(\frac{\infty}{\infty}\)
✓ 1
A1 \(\frac{2}{\mathrm{e}^{x}}\to0\)
✓ 1
Common error: Check that the limit is of the form 0/0 or ∞/∞ before every application of l'Hôpital's rule, and differentiate the numerator and denominator separately, not as a quotient.
235IBMA3H-5-19r47-065.3 Differentiating polynomials and integer powersPaper 1Medium7 marks
No calculator
Let \(f(x)=x\mathrm{e}^{-2x}\), for \(x\in\mathbb{R}\).
(a)
Find \(f'(x)\) and \(f''(x)\), giving each in the form \((ax+b)\mathrm{e}^{-2x}\).
(3)
(b)
Given that \(f'''(x)=(12-8x)\mathrm{e}^{-2x}\), show that the graph of \(f\) has exactly one point of inflection, and find its coordinates.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of the product rule
✓ 1
A1 \(f'(x)=(1-2x)\mathrm{e}^{-2x}\)
✓ 1
A1 \(f''(x)=(4x-4)\mathrm{e}^{-2x}\)
✓ 1
Part (b)
M1 setting \(f''(x)=0\)
✓ 1
A1 \(x=1\) (the only solution because \(\mathrm{e}^{-2x}>0\))
✓ 1
R1 the third derivative at \(x=1\) is \(4\mathrm{e}^{-2}\neq0\), so \(f''\) changes sign and the point is an inflection
✓ 1
A1 coordinates \((1,\mathrm{e}^{-2})\)
✓ 1
Exam tip: A point of inflection needs f'' = 0 together with a sign change of f''; showing that the third derivative is non-zero is a quick way to prove the sign change.
236IBMA3H-5-19r47-095.8 Stationary points, inflexion and optimizationPaper 1Medium11 marks
No calculator
Consider the function \(f(x)=x^4-8x^3+18x^2\), for \(x\in\mathbb{R}\).
(a)
Find \(f'(x)\) and \(f''(x)\).
(3)
(b)
Find the coordinates of the points where the graph has a horizontal tangent.
(3)
(c)
Determine the nature of each of the points found in part (b).
(3)
(d)
Find the values of \(x\) for which the graph of \(f\) is concave down.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to differentiate
✓ 1
A1 \(f'(x)=4x^3-24x^2+36x\)
✓ 1
A1 \(f''(x)=12x^2-48x+36\)
✓ 1
Part (b)
M1 setting \(f'(x)=0\) and factorising, \(4x(x-3)^2=0\)
✓ 1
A1 \(x=0\) and \(x=3\)
✓ 1
A1 \((0,0)\) and \((3,27)\)
✓ 1
Part (c)
A1 \(f''(0)=36>0\), so \((0,0)\) is a local minimum
✓ 1
A1 \(f''(3)=0\), so the second derivative test is inconclusive
✓ 1
R1 \(f'(x)=4x(x-3)^2\) is positive on both sides of \(x=3\) (for example \(f'(2)=8\), \(f'(4)=16\)), so \((3,27)\) is a horizontal point of inflection
✓ 1
Part (d)
M1 solving \(f''(x)<0\), that is \(12(x-1)(x-3)<0\)
✓ 1
A1 \(1<x<3\)
✓ 1
Exam tip: If f''(x) = 0 at a stationary point, the second-derivative test says nothing; check the sign of f'(x) on each side instead.
237IBMA3H-5-19r47-125.13 Limits and l’Hôpital’s rulePaper 1Medium7 marks
No calculator
The strength of a radio signal at a distance \(x\) metres from a transmitter in a data hall is modelled by \(f(x)=\frac{1}{x^2}\), for \(x>0\).
(a)
Use differentiation from first principles to show that \(f'(x)=-\frac{2}{x^3}\).
(5)
(b)
Find the equation of the tangent to the graph of \(y=f(x)\) at the point where \(x=2\).
M1 gradient \(f'(2)=-\frac14\) and point \(\left(2,\frac14\right)\) used in a correct equation
✓ 1
A1 \(y=-\frac14x+\frac34\) (accept \(x+4y=3\))
✓ 1
Common error: Subtract the two fractions before dividing by h; cancelling h at the start is not possible until the numerator has been simplified.
238IBMA3H-5-19r48-095.3 Differentiating polynomials and integer powersPaper 1Medium7 marks
No calculator
The curve \(y=f(x)\) has equation \(f(x)=x^3+px^2+qx\), where \(p\) and \(q\) are constants. The gradient of the curve is 0 at \(x=1\) and 4 at \(x=2\).
(a)
Find the value of \(p\) and the value of \(q\).
(4)
(b)
Hence find the set of values of \(x\) for which \(f\) is increasing.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(f'(x)=3x^2+2px+q\)
✓ 1
A1 \(3+2p+q=0\)
✓ 1
A1 \(12+4p+q=4\)
✓ 1
A1 \(p=-\frac{5}{2}\), \(q=2\)
✓ 1
Part (b)
M1 \(f'(x)=3x^2-5x+2>0\) (ECF from (a))
✓ 1
A1 \(f'(x)=(3x-2)(x-1)\), with critical values \(\frac{2}{3}\) and \(1\)
✓ 1
A1 \(x<\frac{2}{3}\) or \(x>1\)
✓ 1
Common error: 'Increasing' gives a strict inequality in two separate intervals here; writing \(\frac{2}{3}<x<1\) is the opposite region.
239AAHL5-q00045.2 Increasing and decreasing functionsPaper 1Hard9 marks
Let \(g(x)=x-\sin x\) and \(h(x)=\cos x-1+\dfrac{x^{2}}{2}\), for \(x\ge0\).
(a)
By considering \(g'(x)\), show that \(\sin x\lt x\) for all \(x\gt 0\).
(3)
(b)
Hence show that \(\cos x\gt 1-\dfrac{x^{2}}{2}\) for all \(x\gt 0\).
(2)
(c)
Hence show that \(\sin x\gt x-\dfrac{x^{3}}{6}\) for all \(x\gt 0\).
(3)
(d)
Deduce that \(0.479\lt\sin0.5\lt 0.5\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(g'(x)=1-\cos x\ge0\)
A1
\(g'(x)=0\) only at isolated points (\(x=2\pi,4\pi,\ldots\)), so \(g\) is increasing for \(x\ge0\)
R1
Accept a statement that \(g'(x)\gt 0\) except at isolated points.
\(g(0)=0\), so \(g(x)\gt 0\) for \(x\gt 0\), i.e. \(x-\sin x\gt 0\)
R1
\(\sin x\lt x\)
AG
(b)
\(h'(x)=-\sin x+x=g(x)\)
M1
\(h'(x)\gt 0\) for \(x\gt 0\) (by part (a)) and \(h(0)=0\), so \(h(x)\gt 0\) for \(x\gt 0\)
R1
\(\cos x\gt 1-\dfrac{x^{2}}{2}\)
AG
(c)
let \(k(x)=\sin x-x+\dfrac{x^{3}}{6}\); \(k'(x)=\cos x-1+\dfrac{x^{2}}{2}\)
M1
\(k'(x)=h(x)\gt 0\) for \(x\gt 0\) (by part (b))
A1
\(k(0)=0\) and \(k\) is increasing, so \(k(x)\gt 0\) for \(x\gt 0\)
R1
\(\sin x\gt x-\dfrac{x^{3}}{6}\)
AG
(d)
\(0.5-\dfrac{0.5^{3}}{6}=0.5-0.0208\overline{3}=0.4791\overline{6}\gt 0.479\), and \(\sin0.5\lt 0.5\) by part (a)
240AAHL5-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.
Diagram not to scale
(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 \(k-\dfrac12\gt 0\)
241AAHL5-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
242AAHL5-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
243AAHL5-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 \(\dfrac{1+\ln(\ln2)}{\ln2}\gt0\).
251AAHL5-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 \(x=\mathrm{e}^{\pi}\)
254AAHL5-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.
255AAHL5-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.
256AAHL5-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 \(x\to\mathrm{e}^{-}\)
258AAHL5-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.
259AAHL5-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 \(x^{3}\).
261IBMA2H-053-00-125.16 Integration by substitution and by partsPaper 1Hard13 marks
No calculator
Consider the differential equation \(\frac{dy}{dx}+y\tan x=\cos^2x\), for \(-\frac{\pi}{2}<x<\frac{\pi}{2}\), given that \(y=1\) when \(x=0\).
(a)
Show that an integrating factor is \(\sec x\).
(2)
(b)
Hence show that \(y=\cos x(1+\sin x)\).
(4)
(c)
Show that \(\frac{dy}{dx}=(1-2\sin x)(1+\sin x)\).
(3)
(d)
Hence find the intervals on which y is increasing and decreasing, and find the exact maximum value of y.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(e^{\int\tan x\,dx}=e^{-\ln(\cos x)}\) ; A1 \(=\frac{1}{\cos x}=\sec x\), valid since \(\cos x>0\) on the domain
✓ 2
Part (b)
M1 \(\frac{d}{dx}(y\sec x)=\cos^2x\sec x\) ; A1 \(=\cos x\) ; A1 \(y\sec x=\sin x+c\) ; A1 uses \(x=0\), \(y=1\) to give \(c=1\), so \(y=\cos x(1+\sin x)\)
✓ 4
Part (c)
M1 attempt to differentiate using the product rule ; A1 \(\frac{dy}{dx}=-\sin x(1+\sin x)+\cos^2x\) ; A1 uses \(\cos^2x=1-\sin^2x\) to obtain \(1-\sin x-2\sin^2x=(1-2\sin x)(1+\sin x)\)
✓ 3
Part (d)
R1 \(1+\sin x>0\) on \(-\frac{\pi}{2}<x<\frac{\pi}{2}\), so the sign depends on \(1-2\sin x\) ; A1 increasing for \(-\frac{\pi}{2}<x<\frac{\pi}{6}\) ; A1 decreasing for \(\frac{\pi}{6}<x<\frac{\pi}{2}\) ; A1 maximum \(y=\frac{\sqrt{3}}{2}\times\frac{3}{2}=\frac{3\sqrt{3}}{4}\)
✓ 4
Exam tip: Check the open domain before analysing signs: on \(-\frac{\pi}{2}<x<\frac{\pi}{2}\) the factor \(1+\sin x\) never vanishes, so only one stationary point needs testing.
262IBMA2H-053-10-095.16 Integration by substitution and by partsPaper 1Hard10 marks
No calculator
A drone hovers at a height of h metres directly above a point O on level ground. Its camera films a straight strip of ground AB lying on a line through O, where OA = 4 m and OB = 9 m, with A between O and B. The angle subtended at the drone by AB is \(\theta(h)=\arctan\left(\frac{9}{h}\right)-\arctan\left(\frac{4}{h}\right)\), \(h>0\).
Diagram not to scale
(a)
Show that \(\theta'(h)=\frac{180-5h^2}{(h^2+16)(h^2+81)}\).
(4)
(b)
Find the height at which θ is a maximum, justifying your answer.
(3)
(c)
Find the exact maximum value of θ, giving your answer in the form \(\arctan\left(\frac{p}{q}\right)\) where \(p,q\in\mathbb{Z}^+\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses the chain rule with \(\frac{d}{du}\arctan u=\frac{1}{1+u^2}\)
Common error: The derivative of \(\arctan\left(\frac{9}{h}\right)\) needs the inner derivative \(-\frac{9}{h^2}\); leaving it out gives \(\frac{1}{1+81/h^2}\) and the wrong stationary point.
263IBMA2H-053-10-105.15 Further derivatives and integrals; partial fractionsPaper 1Hard10 marks
No calculator
The function f is defined by \(f(x)=x^{\frac{1}{x}}\), for \(x>0\).
(a)
Use logarithmic differentiation to show that \(f'(x)=\frac{x^{\frac{1}{x}}(1-\ln x)}{x^2}\).
(4)
(b)
Find the exact coordinates of the maximum point of the graph of f, justifying that it is a maximum.
(3)
(c)
Hence, without using a calculator, determine which is larger, \(\pi^e\) or \(e^\pi\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 takes logarithms: \(\ln y=\frac{\ln x}{x}\)
✓ 1
M1 differentiates implicitly, using the quotient rule on the right-hand side
A1 multiplies by \(y=x^{\frac{1}{x}}\) to obtain the result AG
✓ 1
Part (b)
M1 \(f'(x)=0\Rightarrow\ln x=1\) since \(x^{\frac{1}{x}}>0\) and \(x^2>0\)
✓ 1
A1 \(\left(e,\,e^{\frac{1}{e}}\right)\)
✓ 1
R1 \(1-\ln x>0\) for \(0<x<e\) and \(<0\) for \(x>e\), so \(f'\) changes from positive to negative: maximum
✓ 1
Part (c)
R1 f has its global maximum at \(x=e\) (f decreasing for \(x>e\)) and \(\pi\neq e\), so \(f(\pi)<f(e)\)
✓ 1
M1 \(\pi^{\frac{1}{\pi}}<e^{\frac{1}{e}}\), raising both (positive) sides to the power \(e\pi\)
✓ 1
A1 \(\pi^e<e^\pi\), so \(e^\pi\) is larger
✓ 1
Exam tip: Raising both sides of an inequality between positive numbers to a positive power keeps the direction; state that the power \(e\pi\) is positive to secure the reasoning.
264IBMA2H-053-04-125.16 Integration by substitution and by partsPaper 1Hard15 marks
No calculator
After a tremor, the mass inside a seismometer oscillates along a vertical line. Its velocity, in cm s−1, t seconds after the tremor is modelled by v(t) = e−t sin t, 0 ≤ t ≤ 2π. The mass is at its equilibrium position when t = 0.
(a)
Find the maximum velocity of the mass for 0 ≤ t ≤ π, giving your answer in exact form.
(4)
(b)
Use integration by parts to show that ∫ e−t sin t dt = −½e−t(sin t + cos t) + C.
(5)
(c)
Find the displacement of the mass from equilibrium when t = π.
(2)
(d)
Show that the total distance travelled by the mass for 0 ≤ t ≤ 2π is ½(1 + e−π)2 cm.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 differentiates v using the product rule
✓ 1
A1 v′(t) = e−t(cos t − sin t)
✓ 1
M1 sets v′(t) = 0, giving tan t = 1, so t = π/4
✓ 1
A1 maximum v = (√2/2)e−π/4 cm s−1; this is the maximum because v(0) = v(π) = 0
✓ 1
Part (b)
M1 first application of parts, with u = sin t or u = e−t
✓ 1
A1 −e−t sin t + ∫ e−t cos t dt (or −e−t cos t − ∫ e−t cos t dt if u = e−t)
✓ 1
M1 second application of parts on ∫ e−t cos t dt, choosing u consistently with the first application
✓ 1
M1 recognises that the original integral I reappears: I = −e−t sin t − e−t cos t − I
✓ 1
A1 2I = −e−t(sin t + cos t), giving the result, AG
✓ 1
Part (c)
M1 evaluates [−½e−t(sin t + cos t)]0π
✓ 1
A1 ½(1 + e−π) cm
✓ 1
Part (d)
R1 v < 0 for π < t < 2π since sin t < 0, so the integral must be split at t = π
Common error: In cyclic integration by parts, swapping the choice of u between the two applications, or slipping a sign, makes the integral cancel to give 0 = something. Keep the same choice of u both times, and track every negative sign that comes from e⁻t.
265IBMA2H-053-09-115.14 Implicit differentiation, related rates and optimisationPaper 1Hard14 marks
No calculator
The centre line of a cycling velodrome is modelled by the closed curve \(x^2-xy+y^2=12\), where \(x\) and \(y\) are measured in units of 10 m from the centre O of the track. The positive \(y\)-direction points north.
(a)
Show that \(\dfrac{dy}{dx}=\dfrac{2x-y}{x-2y}\).
(3)
(b)
Find the coordinates of the points on the curve where the tangent is horizontal.
(3)
(c)
Find the value of \(\dfrac{d^2y}{dx^2}\) at the point \((2,4)\) and hence explain why \((2,4)\) is the most northerly point of the track.
(4)
(d)
A cyclist passes through \((2,4)\) with \(\dfrac{dx}{dt}=-2\) units per second. Find \(\dfrac{dy}{dt}\) at this instant and the exact rate of change of the cyclist's distance \(D\) from O.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt at implicit differentiation, with the product rule on \(xy\)
A1 \(\frac{d^2y}{dx^2}=-\frac13<0\), so \((2,4)\) is a local maximum of \(y\). The closed curve has only two horizontal tangents, and the other, \((-2,-4)\), is lower, so \((2,4)\) is the most northerly point
M1 \(D^2=x^2+y^2\Rightarrow D\frac{dD}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}\), with \(D=\sqrt{20}=2\sqrt5\)
✓ 1
A1 \(\frac{dD}{dt}=\frac{2(-2)+4(0)}{2\sqrt5}=-\frac{2\sqrt5}{5}\) units per second
✓ 1
Common error: When differentiating implicitly a second time, \(\frac{dy}{dx}\) is itself a function of \(x\). Terms such as \(y\frac{dy}{dx}\) therefore need the product rule again. Substitute known values only after this differentiation is complete.
266IBMA2H-053-09-125.12 Limits, first principles and higher derivativesPaper 1Hard14 marks
No calculator
Let \(f(x)=xe^{2x}\), for \(x\in\mathbb{R}\), and let \(f^{(n)}(x)\) denote the \(n\)th derivative of \(f\).
(a)
Find \(f'(x)\) and \(f''(x)\).
(3)
(b)
Prove by mathematical induction that \(f^{(n)}(x)=2^{n-1}(2x+n)e^{2x}\) for all \(n\in\mathbb{Z}^+\).
(6)
(c)
Hence show that the graph of \(y=f^{(n)}(x)\) has exactly one stationary point, and find its \(x\)-coordinate in terms of \(n\).
(3)
(d)
For \(n=3\), find the exact \(y\)-coordinate of this stationary point.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt at product rule
✓ 1
A1 \(f'(x)=(1+2x)e^{2x}\)
✓ 1
A1 \(f''(x)=2e^{2x}+2(1+2x)e^{2x}=(4x+4)e^{2x}\)
✓ 1
Part (b)
A1 \(n=1\): \(2^0(2x+1)e^{2x}=f'(x)\), so true for \(n=1\)
✓ 1
M1 assume true for \(n=k\): \(f^{(k)}(x)=2^{k-1}(2x+k)e^{2x}\)
✓ 1
M1 attempt to differentiate \(f^{(k)}(x)\) using the product rule
R1 true for \(n=1\), and true for \(n=k\) implies true for \(n=k+1\), so true for all \(n\in\mathbb{Z}^+\)
✓ 1
Part (c)
M1 stationary points where \(f^{(n+1)}(x)=2^n(2x+n+1)e^{2x}=0\)
✓ 1
R1 \(2^n e^{2x}>0\) for all \(x\), so the only solution comes from the linear factor; exactly one stationary point
✓ 1
A1 \(x=-\frac{n+1}{2}\)
✓ 1
Part (d)
M1 substitute \(x=-2\) into \(f'''(x)=4(2x+3)e^{2x}\)
✓ 1
A1 \(y=4(-1)e^{-4}=-4e^{-4}\)
✓ 1
Exam tip: In the inductive step, aim to write your answer in exactly the form of the statement with \(n\) replaced by \(k+1\); showing \(2^{(k+1)-1}(2x+(k+1))e^{2x}\) explicitly makes the link unmistakable. The concluding sentence carries its own mark.
267IBMA2H-053-06-115.13 Limits and l’Hôpital’s rulePaper 1Hard14 marks
No calculator
The function g is defined by \(g(x) = \frac{e^x - 1 - x}{x^2}\) for x ≠ 0, and g(0) = k.
(a)
Find the value of k for which g is continuous at x = 0.
(3)
(b)
Using the definition of the derivative, show that \(g'(0) = \frac{1}{6}\).
(5)
(c)
For x ≠ 0, find g′(x). Hence show that \(\lim_{x\to 0} g'(x) = g'(0)\).
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 notes the form 0/0 and applies l'Hôpital's rule to get \(\frac{e^x - 1}{2x}\)
M1 applies l'Hôpital's rule repeatedly, checking 0/0 each time
✓ 1
A1 a correct intermediate stage, e.g. \(\frac{2e^h - 2 - 2h}{6h^2}\) or \(\frac{2e^h - 2}{12h}\)
✓ 1
A1 \(\frac{2e^h}{12} \to \frac{1}{6}\)
✓ 1
Part (c)
M1 uses the product rule on (ex − 1 − x)x−2, or the quotient rule
✓ 1
A1 \(g'(x) = \frac{xe^x - 2e^x + x + 2}{x^3}\)
✓ 1
R1 at x = 0 the numerator is 0 − 2 + 0 + 2 = 0, so the form is 0/0
✓ 1
M1 applies l'Hôpital's rule
✓ 1
A1 \(\frac{xe^x - e^x + 1}{3x^2}\), which is still 0/0, then \(\frac{xe^x}{6x}\)
✓ 1
A1 this equals \(\frac{e^x}{6} \to \frac{1}{6}\) = g′(0), so g′ is continuous at 0
✓ 1
Exam tip: When a show-that question asks for the definition of the derivative, set up the limit of the difference quotient explicitly. Differentiating the formula for x ≠ 0 does not answer it.
268IBMA2H-053-06-125.13 Limits and l’Hôpital’s rulePaper 1Hard13 marks
No calculator
This question explores derivatives and limits involving integer powers of x.
(a)
Using differentiation from first principles, prove that the derivative of x−2 is −2x−3.
(4)
(b)
Find \(\lim_{x\to 1}\frac{x^{-3} - 1}{x^2 - 1}\).
(3)
(c)
Let n be a positive integer with n ≥ 2. It is given that \(\lim_{x\to 1}\frac{x^n - nx + n - 1}{(x-1)^2} = 15\). Find the value of n.
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\lim_{h\to 0}\frac{(x+h)^{-2} - x^{-2}}{h}\)
✓ 1
M1 combines into a single fraction \(\frac{x^2 - (x+h)^2}{hx^2(x+h)^2}\)
✓ 1
A1 simplifies to \(\frac{-2x - h}{x^2(x+h)^2}\)
✓ 1
A1 as h → 0 this gives \(\frac{-2x}{x^4}\) = −2x−3
✓ 1
Part (b)
R1 at x = 1 the form is 0/0
✓ 1
M1 differentiates to get \(\frac{-3x^{-4}}{2x}\)
✓ 1
A1 \(-\frac{3}{2}\)
✓ 1
Part (c)
R1 at x = 1 the numerator is 1 − n + n − 1 = 0 and the denominator is 0
✓ 1
M1 applies l'Hôpital's rule to get \(\frac{nx^{n-1} - n}{2(x-1)}\)
✓ 1
R1 notes this is still 0/0 at x = 1
✓ 1
M1 applies the rule again to get \(\frac{n(n-1)x^{n-2}}{2}\)
✓ 1
A1 \(\frac{n(n-1)}{2} = 15\)
✓ 1
A1 n = 6, rejecting n = −5
✓ 1
Common error: l'Hôpital's rule differentiates the numerator and denominator separately. It is not the quotient rule. Applying the quotient rule to the whole fraction makes the working far longer and usually wrong.
269IBMA2H-053-07-115.12 Limits, first principles and higher derivativesPaper 1Hard13 marks
No calculator
Let \(f(x)=\ln(1+2x)\) for \(x>-\frac{1}{2}\).
(a)
Find \(f'(x)\), \(f''(x)\) and \(f^{(3)}(x)\).
(3)
(b)
Conjecture an expression for \(f^{(n)}(x)\), \(n\in\mathbb{Z}^+\).
(2)
(c)
Prove your conjecture using mathematical induction.
A1 \(=\frac{(-1)^{k}k!\,2^{k+1}}{(1+2x)^{k+1}}\), the required form for \(n=k+1\)
✓ 1
R1 true for \(n=1\), and true for \(n=k\) implies true for \(n=k+1\), hence true for all \(n\in\mathbb{Z}^+\)
✓ 1
Part (d)
M1 substituting \(n=5\), \(x=0\)
✓ 1
A1 \(4!\times2^5=768\)
✓ 1
Exam tip: The reasoning mark for induction needs all three ideas stated: the base case holds, the k to k+1 step holds, and therefore the result holds for all positive integers.
270IBMA2H-053-01-115.19 Maclaurin seriesPaper 1Hard13 marks
No calculator
In machine learning, the softplus activation function is \(f(x) = \ln(1 + e^x)\). Its derivative is the logistic function \(\sigma(x)\).
(a)
Show that \(f'(x) = \frac{1}{1 + e^{-x}}\).
(1)
(b)
Let \(\sigma(x) = \frac{1}{1 + e^{-x}}\). Show that \(\sigma'(x) = \sigma(x)\big(1 - \sigma(x)\big)\).
(2)
(c)
Hence show that \(f'''(x) = \sigma(1 - \sigma)(1 - 2\sigma)\) and deduce that \(f'''(0) = 0\).
(3)
(d)
Find \(f^{(4)}(0)\) and hence find the Maclaurin series for \(f(x)\) up to and including the term in \(x^4\).
(4)
(e)
Show that \(f(x) - f(-x) = x\) for all \(x\). Hence explain why the coefficients of \(x^3, x^5, x^7, \dots\) in the Maclaurin series of \(f(x)\) are all zero.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(f'(x) = \frac{e^x}{1 + e^x}\) and divides numerator and denominator by \(e^x\) to obtain the given result
A1 \(\frac{1 + e^x}{1 + e^{-x}} = e^x\), so the difference is \(\ln e^x = x\)
✓ 1
R1 even-power terms cancel in \(f(x) - f(-x)\), leaving twice the odd-power terms; since this equals \(x\) exactly, the odd part is \(\frac{x}{2}\) and every odd coefficient beyond \(x\) is zero
✓ 1
Exam tip: Expressing higher derivatives in terms of \(\sigma\) keeps the algebra short: you only need \(\sigma(0) = \frac{1}{2}\) to evaluate every derivative at \(x = 0\).
271IBMA2H-053-08-125.17 Areas with the y-axis and volumes of revolutionPaper 1Hard14 marks
No calculator
A heat-shield component for a small space probe is designed from the region \(R\). \(R\) lies in the first quadrant and is bounded by the curve \(y=\ln x\), the \(y\)-axis, the \(x\)-axis and the line \(y=2\). All lengths are in decimetres.
Diagram not to scale
(a)
Find the exact area of \(R\).
(3)
(b)
\(R\) is rotated through \(2\pi\) radians about the \(y\)-axis. Find the exact volume of the solid formed.
(3)
(c)
The line \(y=k\) divides \(R\) into two regions of equal area. Find \(k\) in exact form.
(3)
(d)
\(R\) is instead rotated through \(2\pi\) radians about the \(x\)-axis. Find the exact volume of the solid formed.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 rearranges to \(x=e^y\)
✓ 1
M1 integrates with respect to \(y\): \(\int_0^2e^y\,dy\)
✓ 1
A1 \(e^2-1\) dm2
✓ 1
Part (b)
M1 uses \(\pi\int x^2\,dy\)
✓ 1
A1 \(\pi\int_0^2e^{2y}\,dy\)
✓ 1
A1 \(\frac{\pi}{2}\left(e^4-1\right)\) dm3
✓ 1
Part (c)
M1 \(\int_0^ke^y\,dy=\frac{e^2-1}{2}\)
✓ 1
A1 \(e^k=\frac{e^2+1}{2}\)
✓ 1
A1 \(k=\ln\left(\frac{e^2+1}{2}\right)\)
✓ 1
Part (d)
M1 recognises the volume as a cylinder of radius 2 and length \(e^2\) minus the solid under \(y=\ln x\) from \(x=1\) to \(x=e^2\)
Common error: In (d), using \(\pi\int_1^{e^2}(\ln x)^2dx\) alone. That gives the solid below the curve, not \(R\). \(R\) lies to the left of and above the curve, so it must be subtracted from the cylinder.
272IBMA2H-053-02-125.14 Implicit differentiation, related rates and optimisationPaper 1Hard16 marks
No calculator
The orbit of a probe around a moon is modelled by the curve \(x^2 + 2xy + 2y^2 = 8\), where \(x\) and \(y\) are measured in thousands of kilometres.
(a)
Show that \(\frac{dy}{dx} = -\frac{x + y}{x + 2y}\).
(3)
(b)
Find the exact coordinates of the points on the orbit where \(\frac{dy}{dx} = 0\).
(3)
(c)
Show that at these points \(\frac{d^2y}{dx^2} = -\frac{1}{x + 2y}\).
(4)
(d)
Hence find the maximum value of \(y\) on the orbit, justifying your answer.
(2)
(e)
At one instant the probe is at the point with \(x = 2\) and \(y > 0\), and \(\frac{dx}{dt} = 3\) units per hour. Find the exact value of \(\frac{dy}{dt}\) at this instant.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 implicit differentiation with product rule on \(2xy\)
R1 at \((-2\sqrt{2}, 2\sqrt{2})\), \(x + 2y = 2\sqrt{2} > 0\) so \(\frac{d^2y}{dx^2} < 0\), a maximum
✓ 1
A1 maximum \(y = 2\sqrt{2}\)
✓ 1
Part (e)
M1 substitute \(x = 2\): \(y^2 + 2y - 2 = 0\)
✓ 1
A1 \(y = \sqrt{3} - 1\)
✓ 1
M1 \(\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}\) with \(\frac{dy}{dx} = -\frac{1 + \sqrt{3}}{2\sqrt{3}}\)
✓ 1
A1 \(\frac{dy}{dt} = -\frac{3 + \sqrt{3}}{2}\) units per hour
✓ 1
Exam tip: Simplify the second derivative using the conditions at the point (\(y' = 0\), \(x + y = 0\)) before any further algebra; it cuts the work considerably.
273IBMA2H-053-03-115.15 Further derivatives and integrals; partial fractionsPaper 1Hard15 marks
No calculator
Let \(f(x)=\arctan x+\arctan\left(\frac{1}{x}\right)\) for \(x>0\).
(a)
Show that \(f'(x)=0\).
(3)
(b)
Hence find the exact value of \(f(x)\) for all \(x>0\).
(2)
(c)
Write \(\frac{2x+7}{x^2+2x+5}\) in the form \(\frac{2x+2}{x^2+2x+5}+\frac{c}{x^2+2x+5}\), stating c.
(2)
(d)
Hence find the exact value of \(\displaystyle\int_{-1}^{1}\frac{2x+7}{x^2+2x+5}\,dx\).
(5)
(e)
Using part (b), show that \(\displaystyle\int_{-1}^{3}\frac{5}{x^2+2x+5}\,dx+\int_{-1}^{0}\frac{5}{x^2+2x+5}\,dx=\frac{5\pi}{4}\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 chain rule on \(\arctan\left(\frac{1}{x}\right)\): \(\frac{1}{1+x^{-2}}\times\left(-\frac{1}{x^2}\right)\)
R1 by part (b) with \(x=2\), \(\arctan2+\arctan\frac{1}{2}=\frac{\pi}{2}\), giving \(\frac{5\pi}{4}\) AG
✓ 1
Exam tip: When a quadratic denominator does not factorise, split the numerator into a multiple of the denominator's derivative (giving ln) plus a constant (giving arctan).
274IBMA2H-053-05-115.13 Limits and l’Hôpital’s rulePaper 1Hard15 marks
No calculator
A jeweller designs a glass bead by rotating the curve \(y=x\ln x\), \(0<x\le1\), through \(2\pi\) about the x-axis. All lengths are in centimetres. Let \(f(x)=x\ln x\) for \(x>0\).
Dashed curve is the reflection of y = x ln x. The solid is formed by rotation about the x-axis
(a)
Use l'Hôpital's rule to show that \(\lim_{x\to0^+}x\ln x=0\).
(3)
(b)
Explain why defining \(f(0)=0\) makes f continuous at \(x=0\) from the right.
(1)
(c)
Find \(\int_0^1x\ln x\,dx\).
(5)
(d)
Find the exact volume of the bead.
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 rewrites as \(\frac{\ln x}{1/x}\), which has the form \(\frac{-\infty}{\infty}\) M1 applies l'Hôpital's rule: \(\lim_{x\to0^+}\frac{1/x}{-1/x^2}\) A1 \(=\lim_{x\to0^+}(-x)=0\) (AG)
✓ 3
Part (b)
R1 \(\lim_{x\to0^+}f(x)=0=f(0)\), so the limit equals the function value
✓ 1
Part (c)
M1 integrates by parts with \(u=\ln x\) and \(\frac{dv}{dx}=x\) A1 \(\frac{x^2}{2}\ln x-\int\frac{x}{2}\,dx\) A1 \(\frac{x^2}{2}\ln x-\frac{x^2}{4}\) R1 lower limit: \(\frac{a^2}{2}\ln a=\frac{a}{2}\cdot a\ln a\to0\) as \(a\to0^+\), using part (a) A1 \(-\frac14\)
✓ 5
Part (d)
M1 \(V=\pi\int_0^1x^2(\ln x)^2\,dx\) M1 integrates by parts with \(u=(\ln x)^2\) and \(\frac{dv}{dx}=x^2\) A1 \(\frac{x^3}{3}(\ln x)^2-\frac23\int x^2\ln x\,dx\) A1 \(\int x^2\ln x\,dx=\frac{x^3}{3}\ln x-\frac{x^3}{9}\) R1 boundary terms at 0 vanish: \(x^3\ln x=x^2(x\ln x)\to0\) and \(x^3(\ln x)^2=x(x\ln x)^2\to0\) A1 \(V=\pi\cdot\frac{2}{27}=\frac{2\pi}{27}\) cm3
✓ 6
Exam tip: When an integrand is undefined at a limit, treat that limit as \(a\to0^+\). Justify each boundary term with a limit argument; do not just substitute 0 into ln x.
275IBMA2H-053-05-125.13 Limits and l’Hôpital’s rulePaper 1Hard15 marks
No calculator
For integers \(n\ge0\), let \(I_n=\int_0^1x^ne^x\,dx\).
(a)
Show that \(I_n=e-nI_{n-1}\) for \(n\ge1\).
(4)
(b)
Find \(I_0\) and hence find the exact value of \(I_3\).
(4)
(c)
Show that \(\frac{1}{n+1}\le I_n\le\frac{e}{n+1}\).
(4)
(d)
Hence find \(\lim_{n\to\infty}I_n\) and \(\lim_{n\to\infty}nI_{n-1}\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 integrates by parts with \(u=x^n\) and \(\frac{dv}{dx}=e^x\) A1 \(I_n=\left[x^ne^x\right]_0^1-\int_0^1nx^{n-1}e^x\,dx\) A1 \(\left[x^ne^x\right]_0^1=e-0=e\) for \(n\ge1\) A1 \(I_n=e-nI_{n-1}\) (AG)
✓ 4
Part (b)
A1 \(I_0=e-1\) M1 applies the reduction formula repeatedly A1 \(I_1=1\) and \(I_2=e-2\) A1 \(I_3=e-3(e-2)=6-2e\)
✓ 4
Part (c)
R1 for \(0\le x\le1\), \(1\le e^x\le e\) M1 multiplies by \(x^n\ge0\) and integrates over [0, 1] A1 \(\int_0^1x^n\,dx=\frac{1}{n+1}\) A1 hence \(\frac{1}{n+1}\le I_n\le\frac{e}{n+1}\)
✓ 4
Part (d)
R1 both bounds tend to 0, so by the squeeze principle \(I_n\to0\) M1 rearranges the reduction formula: \(nI_{n-1}=e-I_n\) A1 \(\lim_{n\to\infty}nI_{n-1}=e\)
✓ 3
Exam tip: Once you have a reduction formula, check it with a small case you can work directly, such as I₁ = 1. This catches sign errors early.
276IBMA3H-5-18r18-125.13 Limits and l’Hôpital’s rulePaper 1Hard15 marks
No calculator
The function \(f\) is defined by \(f(x)=\dfrac{\ln x}{x}\), for \(x>0\).
(a)
Use l'Hôpital's rule to find \(\displaystyle\lim_{x\to\infty}f(x)\), and write down the equation of the horizontal asymptote of the graph of \(f\).
(4)
(b)
Find the coordinates of the stationary point of the graph of \(f\).
(4)
(c)
Show that \(f''(x)=\dfrac{2\ln x-3}{x^3}\), and hence show that the stationary point is a maximum.
(3)
(d)
Find the interval on which the graph of \(f\) is concave down, and explain why the graph has a point of inflexion at the end of this interval.
(3)
(e)
Find the coordinates of the point of inflexion.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 state the form \(\frac\infty\infty\) and differentiate numerator and denominator
Exam tip: Because \(\ln x\) is only defined for \(x>0\), the sign of \(f''(x)\) depends only on the numerator \(2\ln x-3\); use this to describe the concavity quickly.
277IBMA3H-5-19r20-135.16 Integration by substitution and by partsPaper 1Hard10 marks
No calculator
A sensor in a solar inverter records a decaying current of \(I(t)=e^{-t}\sin t\) milliamperes, where t is the time in seconds, for \(t\ge0\).
(a)
Use integration by parts twice to show that \(\displaystyle\int e^{-x}\sin x\,dx=-\tfrac12e^{-x}(\sin x+\cos x)+C\).
(5)
(b)
Find the exact charge, in millicoulombs, that flows between \(t=0\) and \(t=\pi\), given by \(\int_0^{\pi}I(t)\,dt\).
(2)
(c)
Find the exact value of t at which the current first has a local maximum.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 first integration by parts with \(u=\sin x\), \(\frac{dv}{dx}=e^{-x}\)
R1 recognises the original integral recurs, so \(2\int e^{-x}\sin x\,dx=-e^{-x}(\sin x+\cos x)\) and divides by 2 (AG)
✓ 1
Part (b)
M1 substitutes the limits \(\pi\) and 0 into the antiderivative
✓ 1
A1 \(\tfrac12\left(1+e^{-\pi}\right)\) mC
✓ 1
Part (c)
M1 attempt to differentiate using the product rule
✓ 1
A1 \(I'(t)=e^{-t}(\cos t-\sin t)\)
✓ 1
A1 \(\tan t=1\), so \(t=\dfrac{\pi}{4}\) s
✓ 1
Exam tip: When the original integral reappears after two rounds of integration by parts, treat it as an unknown and solve the resulting equation for it.
278IBMA3H-5-19r26-075.16 Integration by substitution and by partsPaper 1Hard10 marks
No calculator
A theatre lighting designer models the intensity of light along a stage by \(f(x)=\frac{3x^2+4x+5}{(x+1)(x^2+1)}\) for \(0\le x\le1\).
(a)
Find the constants \(A\), \(B\) and \(C\) such that \(f(x)=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}\).
(4)
(b)
Hence find the exact value of \(\int_0^1 f(x)\,dx\).
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(3x^2+4x+5=A(x^2+1)+(Bx+C)(x+1)\)
✓ 1
A1 \(A=2\) (for example by substituting \(x=-1\))
✓ 1
A1 \(B=1\) (from the \(x^2\) coefficients: \(A+B=3\))
✓ 1
A1 \(C=3\) (from the constant terms: \(A+C=5\))
✓ 1
Part (b)
M1 split the integral using the result of (a)
✓ 1
A1 \(\int\frac{2}{x+1}dx=2\ln(x+1)\)
✓ 1
A1 \(\int\frac{x}{x^2+1}dx=\frac12\ln(x^2+1)\)
✓ 1
A1 \(\int\frac{3}{x^2+1}dx=3\arctan x\)
✓ 1
M1 substitute limits 1 and 0
✓ 1
A1 \(\frac52\ln2+\frac{3\pi}{4}\) (ECF from (a))
✓ 1
Exam tip: An irreducible quadratic factor needs a linear numerator \(Bx+C\). Split it afterwards into an \(\frac{x}{x^2+1}\) part (a logarithm) and a constant part (an arctangent).
279IBMA3H-5-20r27-055.10 Indefinite integrals and reverse chain rulePaper 1Hard9 marks
No calculator
For \(x>0\) and \(y>0\), a function \(y\) satisfies the differential equation \(\dfrac{dy}{dx}=\dfrac{x^2+y^2}{xy}\), with \(y=2\) when \(x=1\).
(a)
Use the substitution \(y=vx\) to show that \(x\dfrac{dv}{dx}=\dfrac1v\).
(3)
(b)
Solve the differential equation, giving \(y\) in terms of \(x\).
(4)
(c)
Find the exact value of \(y\) when \(x=\mathrm{e}\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{dy}{dx}=v+x\frac{dv}{dx}\)
✓ 1
M1 right-hand side becomes \(\frac{x^2+v^2x^2}{x\cdot vx}=\frac{1+v^2}{v}\)
M1 substitute \(x=\mathrm{e}\), with \(\ln\mathrm{e}=1\)
✓ 1
A1 \(y=\mathrm{e}\sqrt6\)
✓ 1
Exam tip: After the substitution \(y=vx\) the equation is separable in \(v\) and \(x\); remember to convert back to \(y\) at the end using \(v=\frac yx\).
280IBMA3H-5-20r27-135.5 Introduction to integrationPaper 1Hard13 marks
No calculator
Let \(f(x)=x^2+\dfrac{16}{x}\) for \(x>0\).
(a)
Find \(f'(x)\).
(2)
(b)
Show that \(f\) has a minimum at the point \((2,12)\).
(4)
(c)
Find the set of values of \(k\) for which \(f(x)=k\) has exactly two solutions.
(2)
(d)
The tangent to the curve at the point where \(x=4\) meets the coordinate axes at two points. Find the area of the triangle formed by the tangent and the two axes.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to differentiate
✓ 1
A1 \(f'(x)=2x-16x^{-2}\)
✓ 1
Part (b)
M1 \(f'(x)=0\), so \(2x^3=16\)
✓ 1
A1 \(x=2\) and \(f(2)=12\)
✓ 1
M1 sign of \(f'\) either side, e.g. \(f'=-14<0\) at \(x=1\) and \(f'>0\) at \(x=3\)
✓ 1
R1 \(f\) decreases then increases, so the point is a minimum
✓ 1
Part (c)
R1 \(f(x)\to\infty\) as \(x\to0\) and as \(x\to\infty\), and the minimum value is 12
✓ 1
A1 \(k>12\)
✓ 1
Part (d)
M1 \(f(4)=20\) and \(f'(4)=7\)
✓ 1
A1 tangent \(y=7x-8\)
✓ 1
A1 intercepts \(\left(\frac87,0\right)\) and \((0,-8)\)
✓ 1
M1 area \(=\frac12\times\frac87\times8\)
✓ 1
A1 \(\frac{32}{7}\)
✓ 1
Common error: Writing \(f'(x)=2x-\frac{16}{x}\) is a frequent slip: \(\frac{16}{x}=16x^{-1}\) differentiates to \(-16x^{-2}\).
281IBMA3H-5-20r31-115.10 Indefinite integrals and reverse chain rulePaper 1Hard8 marks
No calculator
Consider the differential equation \(\frac{dy}{dx}=\frac{x^2+y^2}{xy}\) for \(x>0\), \(y>0\), with \(y=2\) when \(x=1\).
(a)
Use the substitution \(y=vx\) to show that \(x\frac{dv}{dx}=\frac{1}{v}\).
Common error: After solving for \(v\), remember to multiply by \(x\) to return to \(y\); stopping at \(v\) is a frequent slip.
282IBMA3H-5-20r36-135.19 Maclaurin seriesPaper 1Hard9 marks
No calculator
Let f(x) = (1 + ax)n, where a ≠ 0 and n are constants. The Maclaurin series of f(x) begins 1 − 6x + 24x2 + …
(a)
Show that na = −6 and n(n − 1)a2 = 48.
(3)
(b)
Find the values of a and n.
(4)
(c)
Find the coefficient of x3 in the Maclaurin series of f(x).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 f′(x) = na(1 + ax)n−1, so f′(0) = na
✓ 1
A1 coefficient of x is f′(0) = −6, so na = −6
✓ 1
A1 f″(0) = n(n − 1)a2 and f″(0)/2 = 24, so n(n − 1)a2 = 48
✓ 1
Part (b)
M1 uses (na)2 = 36 in n2a2 − na2 = 48
✓ 1
A1 na2 = −12
✓ 1
M1 divides by na = −6
✓ 1
A1 a = 2 and n = −3
✓ 1
Part (c)
M1 f‴(0) = n(n − 1)(n − 2)a3 = −480, then divides by 3! (ECF from (b))
✓ 1
A1 −80
✓ 1
Exam tip: Match coefficients with f(0), f′(0) and f″(0)/2 to turn a given series into equations, then eliminate one unknown by dividing or substituting.
283IBMA3H-5-20r37-125.7 The second derivative and graphsPaper 1Hard9 marks
No calculator
A curve has equation yex + xy2 = 1. The point P(0, 1) lies on the curve.
(a)
Show that dy/dx = −2 at P.
(3)
(b)
Find the value of d2y/dx2 at P.
(5)
(c)
State, with a reason, whether the curve is concave up or concave down at P.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 differentiate implicitly using the product rule on both terms
✓ 1
A1 y′ex + yex + y2 + 2xyy′ = 0
✓ 1
A1 at (0, 1): y′ + 1 + 1 = 0, so y′ = −2
✓ 1
Part (b)
M1 differentiate the equation from part (a) implicitly
Exam tip: Differentiate the first-derivative equation again rather than rearranging for dy/dx first; substituting the known values at the point keeps the algebra short.
284IBMA3H-5-19r38-145.13 Limits and l’Hôpital’s rulePaper 1Hard14 marks
No calculator
The rate \(R\) at which a flavour compound is released from a dissolving tablet, \(t\) minutes after it is placed in water, is modelled by \(R(t)=te^{-kt}\) for \(t\ge0\), where \(k>0\) is a constant.
(a)
Find \(R'(t)\) and show that \(R\) has exactly one stationary point, at \(t=\frac1k\).
(3)
(b)
Find \(R''(t)\) and hence show that the stationary point is a maximum.
(3)
(c)
The maximum value of \(R\) is \(\frac{1}{2e}\). Show that \(k=2\).
(2)
(d)
Find the value of \(t\) at which the graph of \(R\) has a point of inflexion, justifying that it is a point of inflexion.
(3)
(e)
Using integration by parts, find the exact value of \(\int_0^\infty te^{-2t}\,dt\), which represents the total amount released.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses the product rule
✓ 1
A1 \(R'(t)=(1-kt)e^{-kt}\)
✓ 1
A1 \(e^{-kt}>0\), so \(R'(t)=0\) only when \(t=\frac1k\)
✓ 1
Part (b)
M1 differentiates again using the product rule
✓ 1
A1 \(R''(t)=(k^2t-2k)e^{-kt}\)
✓ 1
R1 at \(t=\frac1k\), \(R''=-ke^{-1}<0\), so a maximum
✓ 1
Part (c)
M1 \(R\left(\frac1k\right)=\frac1k e^{-1}\)
✓ 1
A1 \(\frac{1}{ke}=\frac{1}{2e}\), so \(k=2\)
✓ 1
Part (d)
M1 sets \(R''(t)=0\): \(4t-4=0\)
✓ 1
A1 \(t=1\)
✓ 1
R1 \(R''<0\) for \(t<1\) and \(R''>0\) for \(t>1\), so the sign changes
✓ 1
Part (e)
M1 integration by parts with \(u=t\), \(dv=e^{-2t}dt\)
A1 as the upper limit \(N\to\infty\), \(Ne^{-2N}\to0\) and \(e^{-2N}\to0\), so the value is \(\frac14\)
✓ 1
Common error: An improper integral must be treated as a limit: write the upper limit as \(N\), evaluate, then let \(N\to\infty\) rather than substituting infinity directly.
285IBMA3H-5-20r45-135.18 First order differential equationsPaper 1Hard10 marks
No calculator
Consider the differential equation \(\frac{dy}{dx} = \frac{x^2 + y^2}{xy}\) for \(x > 0\) and \(y > 0\), with \(y = 2\) when \(x = 1\).
(a)
Use the substitution \(y = vx\) to show that \(x\frac{dv}{dx} = \frac{1}{v}\).
(3)
(b)
Solve \(x\frac{dv}{dx} = \frac{1}{v}\) to find a general solution for \(y^2\) in terms of \(x\).
(3)
(c)
Use the initial condition to find the particular solution.
(2)
(d)
Find the exact value of \(y\) when \(x = e\), and its value to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{dy}{dx} = v + x\frac{dv}{dx}\)
✓ 1
M1 substitute to get \(v + x\frac{dv}{dx} = \frac{1 + v^2}{v}\)
M1 separate variables: \(\int v\,dv = \int\frac{1}{x}\,dx\)
✓ 1
A1 \(\frac{v^2}{2} = \ln x + c\)
✓ 1
A1 \(y^2 = 2x^2(\ln x + c)\) using \(v = \frac{y}{x}\)
✓ 1
Part (c)
M1 substitute \(x = 1\), \(y = 2\) to get \(4 = 2(0 + c)\)
✓ 1
A1 \(c = 2\) so \(y = x\sqrt{2\ln x + 4}\)
✓ 1
Part (d)
A1 \(y = e\sqrt{2 + 4} = e\sqrt{6}\)
✓ 1
A1 6.66
✓ 1
Exam tip: Replace v by y/x before substituting the initial condition, or substitute x = 1 and v = 2 together; do not mix the two forms.
286IBMA3H-5-19r47-135.13 Limits and l’Hôpital’s rulePaper 1Hard11 marks
No calculator
Use l'Hôpital's rule (repeated if necessary) in each part.
(a)
Given that \(\displaystyle\lim_{x\to0}\frac{\mathrm{e}^{2x}-1-ax}{x^2}\) exists and is finite, find the value of the constant \(a\) and the value of the limit.
M1 differentiating again (or using \(\frac{\sin x}{x}\to1\))
✓ 1
A1 limit \(=-\frac13\)
✓ 1
Exam tip: Simplify the quotient after each application of l'Hôpital's rule; cancelling a common factor can save a further differentiation.
287AAHL5-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.
288AAHL5-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
289AAHL5-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 \(40^{\circ}\mathrm{C}\).
291IBMA2H-053-00-035.18 First order differential equationsPaper 2Easy6 marks
Calculator
A smart water tank is filled by a pump while water leaks through a small outlet. The depth of water, h metres, after t hours satisfies \(\frac{dh}{dt}=0.5-0.2\sqrt{h}\), with \(h=1\) when \(t=0\).
(a)
Use Euler's method with a step length of 0.5 hours to estimate the depth of water when \(t=1.5\). Give your answer to three significant figures.
(4)
(b)
State, with a reason, whether your answer to part (a) is likely to be an overestimate or an underestimate of the true depth.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(h_{n+1}=h_n+0.5(0.5-0.2\sqrt{h_n})\) ; A1 at \(t=0.5\), \(h\approx1.15\) ; A1 at \(t=1\), \(h\approx1.2928\) ; A1 at \(t=1.5\), \(h\approx1.43\) m
✓ 4
Part (b)
R1 \(\frac{dh}{dt}>0\), so h increases with t, and \(\frac{dh}{dt}\) decreases as h increases. The gradient at the start of each step is therefore larger than the gradient over the rest of the step; equivalently, the curve is concave down ; A1 overestimate
✓ 2
Exam tip: Carry full calculator values from one Euler step to the next; rounding each intermediate value can change the final three significant figures.
292IBMA2H-053-10-115.8 Stationary points, inflexion and optimizationPaper 2Easy7 marks
Calculator
A section of a roller-coaster track is modelled by \(y=0.001x^3-0.09x^2+2.4x+30\), \(0\leq x\leq50\), where x is the horizontal distance and y the height of the track, both in metres.
(a)
Find the coordinates of the crest (local maximum) of the track, justifying that it is a maximum.
(3)
(b)
Between the crest and the following dip, the track descends. Find the x-coordinate at which the track is steepest on this descent, and the gradient there.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{dy}{dx}=0.003x^2-0.18x+2.4=0\), giving \(x=20\) or \(x=40\)
✓ 1
A1 crest at \((20,50)\)
✓ 1
R1 \(\frac{d^2y}{dx^2}=0.006x-0.18\); at \(x=20\) it equals \(-0.06<0\), so maximum
✓ 1
Part (b)
M1 sets \(\frac{d^2y}{dx^2}=0.006x-0.18=0\)
✓ 1
A1 \(x=30\)
✓ 1
R1 \(\frac{d^2y}{dx^2}\) changes from negative to positive at \(x=30\), so the gradient is a minimum (most negative) there
✓ 1
A1 gradient \(=0.003(900)-0.18(30)+2.4=-0.3\)
✓ 1
Common error: 'Steepest' refers to the gradient, not the height. Finding where \(y\) is smallest gives the dip, not the steepest point.
293IBMA2H-053-04-035.1 Limits and the derivativePaper 2Easy6 marks
Calculator
In the first 5 seconds of a 100 m race, a sprinter's velocity, in m s−1, is modelled by v(t) = 10(1 − e−0.8t), where t is the time in seconds after the start.
(a)
Find the sprinter's acceleration when t = 1.
(2)
(b)
Find the time at which the sprinter first reaches a speed of 9 m s−1.
(2)
(c)
Find the distance the sprinter runs in the first 5 seconds.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 differentiates: a(t) = 8e−0.8t
✓ 1
A1 3.59 m s−2
✓ 1
Part (b)
M1 sets 10(1 − e−0.8t) = 9, so e−0.8t = 0.1
✓ 1
A1 t = 2.88 s
✓ 1
Part (c)
M1 ∫05 10(1 − e−0.8t) dt
✓ 1
A1 37.7 m
✓ 1
Exam tip: Velocity is positive throughout, so the distance is simply the definite integral of v. Your GDC can evaluate it directly; write down the integral you entered.
294IBMA2H-053-07-025.1 Limits and the derivativePaper 2Easy5 marks
Calculator
After a tablet is swallowed, the rate at which a drug is absorbed into the bloodstream is modelled by \(r(t)=12te^{-0.4t}\) mg per hour, where \(t\) is the time in hours after the tablet is taken.
(a)
Write down an integral that represents the total amount of drug absorbed in the first 6 hours.
(1)
(b)
Find the amount of drug absorbed in the first 6 hours.
(2)
(c)
Find the amount of drug absorbed between \(t=6\) and \(t=10\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(\int_0^6 12te^{-0.4t}\,dt\)
✓ 1
Part (b)
M1 attempt to evaluate the definite integral using technology
✓ 1
A1 51.9 mg
✓ 1
Part (c)
M1 \(\int_6^{10} 12te^{-0.4t}\,dt\)
✓ 1
A1 16.3 mg
✓ 1
Common error: Evaluating r(6) or r(10) gives a rate, not an amount. The total absorbed requires the area under the rate curve.
295IBMA2H-053-01-035.14 Implicit differentiation, related rates and optimisationPaper 2Easy8 marks
Calculator
The path of a shot put is modelled by \(y = 2 + 0.8x - 0.05x^2\), where \(x\) is the horizontal distance from the athlete and \(y\) is the height above the ground, both in metres.
Diagram not to scale
(a)
Find the gradient of the path when \(x = 10\).
(2)
(b)
A camera on the ground lies on the normal to the path at the point where \(x = 10\). Find the equation of this normal and hence the horizontal position of the camera.
(4)
(c)
Find the angle, in degrees, between the direction of motion of the shot and the horizontal when \(x = 10\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{dy}{dx} = 0.8 - 0.1x\)
✓ 1
A1 gradient \(= -0.2\)
✓ 1
Part (b)
A1 \(y = 2 + 8 - 5 = 5\) when \(x = 10\)
✓ 1
M1 normal gradient \(= -\frac{1}{-0.2} = 5\)
✓ 1
A1 \(y - 5 = 5(x - 10)\), i.e. \(y = 5x - 45\)
✓ 1
A1 \(y = 0\) gives \(x = 9\) m
✓ 1
Part (c)
M1 \(\tan\theta = 0.2\)
✓ 1
A1 \(\theta = 11.3°\) (below the horizontal)
✓ 1
Exam tip: The normal gradient is the negative reciprocal of the tangent gradient; check that the product of the two gradients is −1.
296IBMA2H-053-08-045.17 Areas with the y-axis and volumes of revolutionPaper 2Easy4 marks
Calculator
The outer shape of a plastic sports-drink bottle is modelled by rotating the curve \(y=4-0.1x^2\), for \(0\le x\le5\), through \(2\pi\) radians about the \(x\)-axis. All lengths are in centimetres.
Diagram not to scale
(a)
Find the volume of the solid formed.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(V=\pi\int y^2\,dx\)
✓ 1
A1 correct integrand \((4-0.1x^2)^2\)
✓ 1
A1 correct limits 0 and 5
✓ 1
A1 \(V=\frac{635\pi}{12}\approx166\) cm3 (3 s.f.); accept the exact value or a GDC value of 166
✓ 1
Common error: Squaring term by term to get \(16-0.01x^4\). Expand the bracket in full, or let the calculator evaluate \(\pi\int_0^5(4-0.1x^2)^2\,dx\) directly.
297IBMA2H-053-02-035.14 Implicit differentiation, related rates and optimisationPaper 2Easy5 marks
Calculator
A spherical medicinal pellet dissolves in the stomach so that its volume decreases at a constant rate of 2.5 mm3 s−1. The pellet remains spherical with radius \(r\) mm.
(a)
Find the rate of change of the radius when \(r = 4\).
(3)
(b)
Find the rate of change of the surface area of the pellet at this instant.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\)
✓ 1
A1 \(-2.5 = 4\pi(16)\frac{dr}{dt}\)
✓ 1
A1 \(\frac{dr}{dt} = -0.0124\) mm s−1
✓ 1
Part (b)
M1 \(\frac{dA}{dt} = 8\pi r\frac{dr}{dt}\)
✓ 1
A1 \(-1.25\) mm2 s−1
✓ 1
Common error: A decreasing volume means \(\frac{dV}{dt}\) is negative. Using +2.5 gives the wrong sign for every rate that follows.
298IBMA2H-053-03-045.14 Implicit differentiation, related rates and optimisationPaper 2Easy5 marks
Calculator
A vertical wind turbine tower of height 30 m stands on level ground. When the angle of elevation of the Sun is θ radians, the length of the tower's shadow is \(s = 30\cot\theta\) metres. During one morning θ increases at a constant rate of 0.2 radians per hour.
Diagram not to scale
(a)
Find \(\frac{ds}{d\theta}\).
(2)
(b)
Find the rate of change of the shadow length when θ = 0.5.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(\frac{d}{d\theta}\cot\theta=-\csc^2\theta\)
✓ 1
A1 \(\frac{ds}{d\theta}=-30\csc^2\theta\)
✓ 1
Part (b)
M1 use of \(\frac{ds}{dt}=\frac{ds}{d\theta}\times\frac{d\theta}{dt}\)
A1 −26.1 m h−1 (shadow shortening at 26.1 m per hour)
✓ 1
Common error: Leaving the calculator in degree mode gives a wildly different value. Rates of change of trigonometric functions require θ in radians.
299IBMA3H-5-18r9-085.17 Areas with the y-axis and volumes of revolutionPaper 2Easy4 marks
Calculator
The funnel of a beekeeper's smoker is formed by rotating the region under the curve \(y=\sqrt x\), for \(0\le x\le4\), through \(360^\circ\) about the \(x\)-axis. Lengths are in centimetres.
Cross-section: the region under y = √x is rotated about the x-axis
(a)
Show that the volume of the funnel is \(8\pi\ \text{cm}^3\).
(3)
(b)
Find the volume as a decimal, to 3 significant figures.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(V=\pi\int_0^4y^2\,dx=\pi\int_0^4x\,dx\)
✓ 1
A1 \(\pi\left[\frac{x^2}{2}\right]_0^4\)
✓ 1
A1 \(8\pi\) (AG)
✓ 1
Part (b)
A1 \(25.1\ \text{cm}^3\)
✓ 1
Exam tip: For rotation about the x-axis, square the function before integrating; do not square the integral.
300IBMA3H-5-18r18-025.18 First order differential equationsPaper 2Easy6 marks
Calculator
After an eruption, the concentration \(A\) mg m−3 of ash at a monitoring station, \(t\) hours after the peak, satisfies the differential equation \(\dfrac{dA}{dt}=-0.4A\). At \(t=0\) the concentration is \(50\) mg m−3.
(a)
Solve the differential equation to find \(A\) in terms of \(t\).
(4)
(b)
Find the time at which the concentration has fallen to \(10\) mg m−3. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 separate variables: \(\int\dfrac1A\,dA=\int-0.4\,dt\)
✓ 1
A1 \(\ln A=-0.4t+c\)
✓ 1
M1 use \(A=50\) when \(t=0\) to find the constant (\(c=\ln50\))
Common error: Writing \(\int\frac1A\,dA=\frac{A^2}{2}\) or leaving out the constant of integration. Always add \(+c\) at the integration step, before using the initial condition.
301IBMA3H-5-19r26-145.9 KinematicsPaper 2Easy6 marks
Calculator
A stage lift moves vertically so that its displacement from the floor is \(s=t^3-6t^2+9t\) metres, where \(t\) is the time in seconds and \(0\le t\le5\).
(a)
Find an expression for the velocity \(v=\frac{ds}{dt}\).
(2)
(b)
Find the times at which the lift is instantaneously at rest.
(2)
(c)
Find the acceleration of the lift at each of these times.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to differentiate
✓ 1
A1 \(v=3t^2-12t+9\)
✓ 1
Part (b)
M1 set \(v=0\) and solve
✓ 1
A1 \(t=1\) and \(t=3\)
✓ 1
Part (c)
A1 \(a=6t-12\)
✓ 1
A1 \(a=-6\) m s-2 at \(t=1\) and \(a=6\) m s-2 at \(t=3\)
✓ 1
Exam tip: Acceleration is the derivative of velocity. Check each answer with its sign and unit: a negative value shows acceleration in the opposite direction to increasing \(s\).
302IBMA3H-5-20r27-015.18 First order differential equationsPaper 2Easy6 marks
Calculator
The mass \(m\) grams of a yeast culture, \(t\) hours after it is set to ferment, satisfies the differential equation \(\dfrac{dm}{dt}=0.2m\). Initially the mass is 5 g.
(a)
Solve the differential equation to find \(m\) in terms of \(t\).
(4)
(b)
Find the time taken for the mass to reach 20 g. Give your answer in hours to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 separate variables: \(\int\frac1m\,dm=\int0.2\,dt\)
✓ 1
A1 \(\ln m=0.2t+c\)
✓ 1
M1 use of \(m=5\) when \(t=0\), so \(c=\ln5\)
✓ 1
A1 \(m=5e^{0.2t}\)
✓ 1
Part (b)
M1 \(20=5e^{0.2t}\), so \(e^{0.2t}=4\)
✓ 1
A1 \(t=5\ln4=6.93\) hours
✓ 1
Exam tip: After integrating, use the initial condition straight away to find the constant, before you rearrange for \(m\).
303IBMA3H-5-20r31-085.14 Implicit differentiation, related rates and optimisationPaper 2Easy5 marks
Calculator
The distance \(y\) km of a probe from a beacon satisfies \(\frac{dy}{dx}=\frac{3x^2}{y}\), where \(x\) is the time in hours, \(y>0\) and \(y=2\) when \(x=0\).
(a)
Solve the differential equation, giving y in terms of x.
(4)
(b)
Find the distance when x = 2, correct to 3 significant figures.
Exam tip: State which root you take: the condition y > 0 selects the positive square root.
304IBMA3H-5-19r33-035.9 KinematicsPaper 2Easy7 marks
Calculator
A Formula 1 car is tested on a straight track. Its velocity \(t\) seconds after starting from rest is \(v(t)=60\left(1-e^{-0.25t}\right)\) m s−1, for \(t\ge 0\).
(a)
Find an expression for the acceleration of the car at time \(t\).
(2)
(b)
Find the time at which the acceleration is 5 m s−2. Give your answer to 3 significant figures.
(2)
(c)
Find the distance travelled by the car in the first 4 seconds. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 differentiate \(v\) with respect to \(t\) (chain rule on \(e^{-0.25t}\))
✓ 1
A1 \(a(t)=15e^{-0.25t}\) m s−2
✓ 1
Part (b)
M1 \(15e^{-0.25t}=5\) leading to \(e^{-0.25t}=\frac{1}{3}\) and use of logarithms
Exam tip: Distance from a velocity function is the area under the velocity-time graph, so integrate over the stated time interval.
305IBMA3H-5-20r36-035.9 KinematicsPaper 2Easy5 marks
Calculator
A harbour tug moves in a straight line with velocity v(t) = 1.2t − 0.06t2 m s−1, for 0 ≤ t ≤ 20, where t is in seconds.
(a)
Find the acceleration of the tug when t = 3.
(2)
(b)
Find the distance travelled by the tug in the first 10 seconds.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 differentiates: a = 1.2 − 0.12t
✓ 1
A1 0.84 m s−2
✓ 1
Part (b)
M1 sets up ∫010(1.2t − 0.06t2) dt (the velocity is positive, so distance equals displacement)
✓ 1
M1 antiderivative 0.6t2 − 0.02t3
✓ 1
A1 40 m
✓ 1
Exam tip: Acceleration is the derivative of velocity, and displacement is the integral of velocity; write which relationship you are using before calculating.
306IBMA3H-5-20r36-045.19 Maclaurin seriesPaper 2Easy6 marks
Calculator
Let f(x) = ln(1 + 2x).
(a)
Find the Maclaurin series of f(x) up to and including the term in x3.
(4)
(b)
Use your series with x = 0.1 to estimate ln 1.2, giving your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempts to find f′, f″ and f‴
✓ 1
A1 f(0) = 0, f′(0) = 2, f″(0) = −4, f‴(0) = 16
✓ 1
M1 uses f(0) + f′(0)x + f″(0)x2/2! + f‴(0)x3/3!
✓ 1
A1 2x − 2x2 + (8/3)x3
✓ 1
Part (b)
M1 substitutes x = 0.1: 0.2 − 0.02 + (8/3)(0.001) (ECF from (a))
✓ 1
A1 0.183
✓ 1
Exam tip: Evaluate each derivative at x = 0 before dividing by the factorial; this keeps the arithmetic small and reduces slips.
307IBMA3H-5-19r38-045.14 Implicit differentiation, related rates and optimisationPaper 2Easy5 marks
Calculator
The power, \(P\) milliwatts, of a signal in an optical cable at distance \(x\) km from the transmitter satisfies the differential equation \(\frac{dP}{dx}=-0.3P\). The power at the transmitter is 50 mW.
(a)
Solve the differential equation to show that \(P=50e^{-0.3x}\).
(3)
(b)
Find the distance from the transmitter at which the power has fallen to 10 mW. Give your answer to 3 significant figures.
A1 uses \(P=50\) when \(x=0\) to obtain \(P=50e^{-0.3x}\)
✓ 1
Part (b)
M1 \(e^{-0.3x}=0.2\), then takes natural logarithms
✓ 1
A1 \(x=\frac{\ln5}{0.3}=5.36\) km
✓ 1
Exam tip: After integrating, include the constant of integration immediately and use the initial condition to find it before rearranging.
308IBMA3H-5-20r45-025.16 Integration by substitution and by partsPaper 2Easy5 marks
Calculator
The harvest rate of a fishery is modelled by \(R(t) = 40 + 12\sin\left(\frac{\pi t}{6}\right)\) tonnes per month, where \(t\) is the time in months after the start of the season and \(0 \le t \le 6\) (angles in radians).
(a)
Write down an integral for the total harvest during the first 6 months.
(1)
(b)
Find the total harvest during the first 6 months, giving your answer to 3 significant figures.
(2)
(c)
Hence find the mean harvest rate over the first 6 months, giving your answer to 3 significant figures. Use the unrounded answer to part (b).
M1 attempt to evaluate the integral, e.g. use of a calculator or antiderivative \(40t - \frac{72}{\pi}\cos\frac{\pi t}{6}\)
✓ 1
A1 286 tonnes
✓ 1
Part (c)
M1 divide the total by 6, e.g. \(\frac{285.84\ldots}{6}\)
✓ 1
A1 47.6 tonnes per month
✓ 1
Exam tip: Check that your calculator is in radian mode before evaluating an integral containing a trigonometric function of a pure number.
309IBMA3H-5-19r47-025.9 KinematicsPaper 2Easy7 marks
Calculator
A cyclist accelerates along a straight, level road. Her acceleration at time \(t\) seconds is \(a=1.2-0.1t\) m s−2, for \(0\le t\le 20\). At \(t=0\) her speed is 3 m s−1.
(a)
Find an expression for her speed \(v\) m s−1 at time \(t\).
(3)
(b)
Find her greatest speed during the 20 seconds.
(2)
(c)
Find the distance she travels in the first 12 seconds.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to integrate \(a\) with respect to \(t\)
✓ 1
A1 \(v=1.2t-0.05t^2+c\)
✓ 1
A1 \(c=3\), so \(v=3+1.2t-0.05t^2\)
✓ 1
Part (b)
M1 setting \(a=0\) to obtain \(t=12\) (acceleration is positive before and negative after, so the speed is greatest at \(t=12\))
✓ 1
A1 \(v(12)=3+14.4-7.2=10.2\) m s−1
✓ 1
Part (c)
M1 \(\int_0^{12}(3+1.2t-0.05t^2)\,\mathrm{d}t\)
✓ 1
A1 \(93.6\) m
✓ 1
Common error: Distance is found by integrating the speed, not by multiplying the greatest speed by the time.
310IBMA3H-5-19r48-025.4 Tangents and normalsPaper 2Easy6 marks
Calculator
A water-slide support beam is modelled by the curve \(y=2x^2-5x+3\), where \(x\) and \(y\) are in metres. The point \(A\) on the curve has \(x\)-coordinate 3.
(a)
Find the gradient of the curve at \(A\).
(2)
(b)
Find the equation of the tangent to the curve at \(A\), giving your answer in the form \(y=mx+c\).
(2)
(c)
The normal to the curve at \(A\) meets the \(y\)-axis at \(B\). Find the \(y\)-coordinate of \(B\), correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{dy}{dx}=4x-5\)
✓ 1
A1 gradient \(=7\)
✓ 1
Part (b)
M1 \(y=2(3)^2-5(3)+3=6\) and use of \(y-6=7(x-3)\)
✓ 1
A1 \(y=7x-15\)
✓ 1
Part (c)
M1 normal gradient \(-\frac{1}{7}\) and \(y-6=-\frac{1}{7}(x-3)\) (ECF from (a))
✓ 1
A1 \(y=\frac{45}{7}=6.43\) (3 s.f.)
✓ 1
Exam tip: The normal gradient is the negative reciprocal of the tangent gradient, so the two gradients multiply to \(-1\).
311IBMA3H-5-19r48-045.14 Implicit differentiation, related rates and optimisationPaper 2Easy4 marks
Calculator
A circular ripple spreads out across the surface of a swimming pool. The radius \(r\) cm of the ripple increases at a constant rate of \(4\ \text{cm s}^{-1}\). The area of the ripple is \(A\ \text{cm}^2\).
(a)
Show that \(\frac{dA}{dt}=8\pi r\).
(2)
(b)
Find the rate at which the area is increasing at the instant when \(r=10\) cm. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(A=\pi r^2\) so \(\frac{dA}{dr}=2\pi r\), and use of \(\frac{dA}{dt}=\frac{dA}{dr}\times\frac{dr}{dt}\)
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 \(\left]\tfrac{3\pi}{2},2\pi\right[\).
313AAHL5-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.
314AAHL5-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.
Diagram not to scale
(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) \(0\le x\lt22.4\)
315AAHL5-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 \(0\le t\le2.5\).
Diagram not to scale
(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 \(t=1.6\)
316AAHL5-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 \(0\le x\le20\).
Diagram not to scale
(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 \(x=0\)
317AAHL5-q00845.16 Integration by substitution and by partsPaper 2Medium6 marks
Let \(f(x)=x\ln x\), for \(x\gt0\).
Diagram not to scale
(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) \(k=4.44\)
318AAHL5-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.
319AAHL5-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
321AAHL5-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 \(1-\dfrac{1}{18}+\dfrac{1}{600}-\dfrac{1}{35\,280}\).
322AAHL5-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
323IBMA2H-053-00-075.9 KinematicsPaper 2Medium9 marks
Calculator
A patient receives a drug by continuous intravenous infusion. The concentration of the drug in the blood, C mg L−1, t hours after the infusion starts satisfies \(\frac{dC}{dt}=3-0.25C\), with \(C=0\) when \(t=0\).
(a)
Solve the differential equation to show that \(C=12\left(1-e^{-0.25t}\right)\).
(5)
(b)
Show that the concentration is always increasing, and state the long-term concentration predicted by the model.
(2)
(c)
The drug becomes effective once the concentration reaches 9.5 mg L−1. Find the time taken for this to happen.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 separates variables: \(\int\frac{dC}{3-0.25C}=\int dt\) ; A1 \(-4\ln|3-0.25C|\) ; A1 \(=t+k\) ; M1 uses \(t=0\), \(C=0\) to obtain \(k=-4\ln3\) ; A1 rearranges \(3-0.25C=3e^{-0.25t}\) to give \(C=12(1-e^{-0.25t})\)
✓ 5
Part (b)
R1 \(\frac{dC}{dt}=3e^{-0.25t}>0\) for all t ; A1 \(C\to12\) mg L−1
Common error: The integral of \(\frac{1}{3-0.25C}\) is \(-4\ln|3-0.25C|\), not \(\ln|3-0.25C|\): divide by the coefficient of C, including its sign.
324IBMA2H-053-00-085.9 KinematicsPaper 2Medium8 marks
Calculator
During a time trial, the speed of a cyclist, v m s−1, is modelled by \(v(t)=8+3\ln(t+1)-0.5t\), for \(0\leq t\leq20\), where t is the time in seconds.
(a)
Find \(v'(t)\).
(2)
(b)
Find the interval of time during which the speed of the cyclist is increasing.
(2)
(c)
Find the maximum speed of the cyclist.
(2)
(d)
Find the distance travelled by the cyclist in the first 10 seconds.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 differentiates \(\ln(t+1)\) correctly to give \(\frac{1}{t+1}\) ; A1 \(v'(t)=\frac{3}{t+1}-0.5\)
✓ 2
Part (b)
M1 solves \(v'(t)=0\) (or \(v'(t)>0\)) to give \(t=5\) ; A1 \(0\leq t<5\) (accept \(0\leq t\leq5\))
✓ 2
Part (c)
M1 substitutes \(t=5\): \(v=5.5+3\ln6\) ; A1 10.9 m s−1
✓ 2
Part (d)
M1 \(\int_0^{10}v(t)\,dt\) ; A1 104 m
✓ 2
Exam tip: Give the interval of increase within the stated domain, starting at \(t=0\). Since \(v>0\) throughout, the distance travelled is simply \(\int v\,dt\); use your GDC to evaluate it.
325IBMA2H-053-10-075.14 Implicit differentiation, related rates and optimisationPaper 2Medium11 marks
Calculator
A heavy truck makes a 200 km journey at a constant speed of v km h−1. At this speed it uses fuel at a rate of \(\left(\frac{v^2}{200}+5\right)\) litres per hour. Fuel costs 2 dollars per litre and the driver is paid 30 dollars per hour.
(a)
Show that the total cost of the journey, in dollars, is \(C(v)=2v+\frac{8000}{v}\).
(3)
(b)
Find the speed that minimises the cost, justifying that it gives a minimum.
(4)
(c)
Find the minimum cost.
(1)
(d)
On a motorway the speed must satisfy \(70\leq v\leq90\). Find the minimum cost of the journey under this restriction, explaining your reasoning.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 journey time \(=\frac{200}{v}\) hours
✓ 1
M1 cost per hour \(=2\left(\frac{v^2}{200}+5\right)+30=0.01v^2+40\)
✓ 1
A1 \(C=\frac{200}{v}(0.01v^2+40)=2v+\frac{8000}{v}\) AG
✓ 1
Part (b)
M1 \(C'(v)=2-\frac{8000}{v^2}\)
✓ 1
M1 sets \(C'(v)=0\): \(v^2=4000\)
✓ 1
A1 \(v=63.2\) km h−1
✓ 1
R1 \(C''(v)=\frac{16000}{v^3}>0\) for \(v>0\), so minimum (accept a sign change of \(C'\) from negative to positive)
✓ 1
Part (c)
A1 \(C=253\) dollars (252.98…)
✓ 1
Part (d)
R1 \(C'(v)>0\) for \(v>63.2\), so \(C\) is increasing on \(70\leq v\leq90\) (accept comparison of \(C(70)\) and \(C(90)\) with this reasoning)
✓ 1
M1 minimum occurs at the endpoint \(v=70\): \(C(70)=140+\frac{8000}{70}\)
✓ 1
A1 254 dollars (254.29)
✓ 1
Common error: When the stationary point lies outside the allowed interval, the minimum is at an endpoint. Check the sign of \(C'\) on the interval rather than quoting the unrestricted optimum.
326IBMA2H-053-04-055.18 First order differential equationsPaper 2Medium10 marks
Calculator
A speedboat travelling at 30 m s−1 cuts its engine at time t = 0. While it coasts, its velocity v m s−1 satisfies the differential equation dv/dt = −0.002v2.
(a)
Show that v = 30 / (1 + 0.06t).
(4)
(b)
Find the time taken for the speed to fall to 10 m s−1.
(2)
(c)
Find the distance travelled by the boat during this time.
(3)
(d)
State one feature of this model that is unrealistic as t becomes large.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 separates variables: ∫ v−2 dv = ∫ −0.002 dt
✓ 1
A1 −1/v = −0.002t + c
✓ 1
M1 uses v = 30 at t = 0 to obtain c = −1/30
✓ 1
A1 1/v = 0.002t + 1/30, leading to v = 30/(1 + 0.06t), AG
✓ 1
Part (b)
M1 30/(1 + 0.06t) = 10 so 1 + 0.06t = 3
✓ 1
A1 t = 33.3 s
✓ 1
Part (c)
M1 ∫0100/3 30/(1 + 0.06t) dt
✓ 1
A1 500 ln(1 + 0.06t) evaluated, or GDC value seen
✓ 1
A1 500 ln 3 = 549 m
✓ 1
Part (d)
R1 v never reaches zero and the distance 500 ln(1 + 0.06t) increases without bound, so the boat would never stop
✓ 1
Exam tip: For a 'show that' differential equation, the constant must be found explicitly from the initial condition. Jumping from the integral to the given formula loses the method marks.
327IBMA2H-053-04-075.9 KinematicsPaper 2Medium10 marks
Calculator
A racing greyhound leaves the starting trap at t = 0. Its velocity, in m s−1, is modelled by v(t) = 20te−0.5t, for 0 ≤ t ≤ 10, where t is in seconds.
(a)
Find the maximum velocity of the greyhound and the time at which it occurs.
(3)
(b)
Find the acceleration of the greyhound when t = 6.
(2)
(c)
Find the time taken for the greyhound to run its first 50 m.
(3)
(d)
Find the distance run by the greyhound in the 10 seconds.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets v′(t) = 20e−0.5t(1 − 0.5t) = 0, or uses GDC maximum
✓ 1
A1 t = 2 s
✓ 1
A1 v = 40e−1 = 14.7 m s−1
✓ 1
Part (b)
M1 substitutes t = 6 into v′(t)
✓ 1
A1 −1.99 m s−2
✓ 1
Part (c)
M1 recognises distance as an integral of v from 0 to T
✓ 1
A1 ∫0T 20te−0.5t dt = 50
✓ 1
A1 T = 4.24 s
✓ 1
Part (d)
M1 ∫010 20te−0.5t dt
✓ 1
A1 76.8 m
✓ 1
Exam tip: In part (c) the unknown is a limit of integration. Set up the integral equation clearly, then solve it with the GDC's solver or by graphing.
328IBMA2H-053-04-095.9 KinematicsPaper 2Medium11 marks
Calculator
A self-driving car A starts from rest at a junction at t = 0 and moves along a straight road. Its velocity, in m s−1, is vA(t) = 3t − 0.1t2, for 0 ≤ t ≤ 20, where t is in seconds. At t = 0 a cyclist B is 20 m ahead of the junction, travelling in the same direction at a constant 8 m s−1.
(a)
Find an expression for the displacement sA(t) of car A from the junction.
(2)
(b)
Find the time at which car A catches up with the cyclist.
(3)
(c)
Find the speed of car A at this moment.
(2)
(d)
Find the greatest distance by which the cyclist is ahead of car A before being caught.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 integrates vA
✓ 1
A1 sA = 1.5t2 − t3/30, with constant 0
✓ 1
Part (b)
M1 equates displacements: sA = 20 + 8t
✓ 1
A1 1.5t2 − t3/30 = 20 + 8t
✓ 1
A1 t = 8.51 s
✓ 1
Part (c)
M1 substitutes their time into vA
✓ 1
A1 18.3 m s−1
✓ 1
Part (d)
R1 the gap 20 + 8t − sA is greatest when its derivative is zero, i.e. when vA = 8
✓ 1
A1 3t − 0.1t2 = 8 gives t = 2.96 s
✓ 1
M1 substitutes into 20 + 8t − 1.5t2 + t3/30
✓ 1
A1 31.4 m
✓ 1
Exam tip: The gap between two objects is greatest or least when their velocities are equal. Justifying this with the derivative of the gap earns the reasoning mark.
329IBMA2H-053-09-055.9 KinematicsPaper 2Medium7 marks
Calculator
During the first part of a 100 m race, the displacement of a sprinter from the start line is modelled by \(s(t)=11t+13.75\left(e^{-0.8t}-1\right)\), where \(s\) is in metres and \(t\) is the time in seconds after the starting signal.
(a)
Show that the velocity of the sprinter is \(v(t)=11-11e^{-0.8t}\).
(2)
(b)
Find an expression for the acceleration \(a(t)\).
(1)
(c)
Find the time at which the sprinter reaches a velocity of 10 m s−1.
(2)
(d)
Find the acceleration of the sprinter at this time.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 chain rule: derivative of \(e^{-0.8t}\) is \(-0.8e^{-0.8t}\)
M1 substitute \(e^{-0.8t}=\frac{1}{11}\) (or \(t=2.997...\)) into \(a(t)\)
✓ 1
A1 \(a=0.8\) m s−2
✓ 1
Exam tip: Using the exact relation \(e^{-0.8t}=\frac{1}{11}\) in the final part avoids rounding errors from an intermediate value of \(t\).
330IBMA2H-053-09-085.9 KinematicsPaper 2Medium12 marks
Calculator
After a car passes over a speed bump, the vertical displacement of its body from equilibrium is modelled by \(y=e^{-t}\sin(2t)\), \(t\ge 0\), where \(y\) is in centimetres and \(t\) is the time in seconds.
(a)
Show that \(\dfrac{dy}{dt}=e^{-t}(2\cos 2t-\sin 2t)\).
(2)
(b)
Show that \(\dfrac{d^2y}{dt^2}=-e^{-t}(3\sin 2t+4\cos 2t)\).
(3)
(c)
Hence show that \(\dfrac{d^2y}{dt^2}+2\dfrac{dy}{dt}+5y=0\).
(3)
(d)
Find the first time \(t>0\) at which the displacement is a maximum.
A1 sine terms \(-3-2+5=0\) and cosine terms \(-4+4=0\), so the sum is \(0\)
✓ 1
Part (d)
M1 \(2\cos2t=\sin2t\Rightarrow\tan2t=2\)
✓ 1
A1 \(t=\frac{1}{2}\arctan 2=0.554\) s
✓ 1
Part (e)
M1 substitute \(t=0.5535...\) into \(y\)
✓ 1
A1 \(y=0.514\) cm
✓ 1
Exam tip: For a 'show that', every intermediate line counts. Write the unsimplified product-rule result before collecting terms so that the given answer is clearly derived rather than asserted.
331IBMA2H-053-06-085.9 KinematicsPaper 2Medium6 marks
Calculator
During the first 20 seconds after launch, the height of a small research rocket above the launch pad is modelled by h(t) = 0.4t3 + 2t2, where h is in metres and t is in seconds.
(a)
Find an expression for the velocity of the rocket at time t.
(2)
(b)
Find the time at which the velocity of the rocket is 200 m s−1.
(2)
(c)
Find the acceleration of the rocket at this time.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 differentiates h(t)
✓ 1
A1 v = 1.2t2 + 4t
✓ 1
Part (b)
M1 solves 1.2t2 + 4t − 200 = 0
✓ 1
A1 t = 11.4 s, rejecting the negative root
✓ 1
Part (c)
M1 a = 2.4t + 4
✓ 1
A1 a = 31.2 m s−2
✓ 1
Common error: Substitute the unrounded value of t from your calculator when finding the acceleration. Using 11.4 gives 31.4, which is not correct to 3 significant figures.
332IBMA2H-053-06-095.13 Limits and l’Hôpital’s rulePaper 2Medium7 marks
Calculator
A biologist models the growth of a cell culture over one day. When growth is applied in discrete steps each of length x days, the population is multiplied over the day by \(f(x) = (1+3x)^{1/x}\), for 0 < x ≤ 1.
(a)
By considering ln f(x), use l'Hôpital's rule to show that \(\lim_{x\to 0^+} \ln f(x) = 3\).
(4)
(b)
Hence write down the exact value of \(\lim_{x\to 0^+} f(x)\).
(1)
(c)
Find the percentage by which f(0.01) is less than this limit.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\ln f(x) = \frac{\ln(1+3x)}{x}\)
✓ 1
R1 notes the form 0/0 as x → 0
✓ 1
M1 differentiates numerator and denominator to get \(\frac{3/(1+3x)}{1}\)
✓ 1
A1 → 3 as x → 0, so the limit is 3
✓ 1
Part (b)
A1 e3
✓ 1
Part (c)
M1 f(0.01) = 1.03100 = 19.2186… (unrounded value used) and e3 = 20.0855…
Exam tip: For limits of the form 1∞, take logarithms to turn the power into a quotient. Then apply l'Hôpital's rule and exponentiate at the end. In percentage calculations keep full calculator values: rounding 1.03¹⁰⁰ to 19.2 first gives 4.41% instead of 4.32%.
333IBMA2H-053-07-075.14 Implicit differentiation, related rates and optimisationPaper 2Medium6 marks
Calculator
The depth of a river, \(d\) metres, at a distance \(x\) metres from the left bank is modelled by \(d(x)=2.4\sin\left(\frac{\pi x}{15}\right)\) for \(0\le x\le15\).
Diagram not to scale
(a)
Find the area of the cross-section of the river.
(2)
(b)
The water flows through this cross-section at a uniform speed of 0.6 m s−1. Find the volume of water passing per second.
(2)
(c)
A sensor is placed at \(x=a\) so that the cross-sectional area between the left bank and the sensor is 5 m2. Find \(a\).
Exam tip: When the unknown is a limit of integration, set up the equation clearly and use the GDC's solver or graph intersection; state the equation you solved. Carry unrounded values forward between parts.
334IBMA2H-053-01-075.19 Maclaurin seriesPaper 2Medium8 marks
Calculator
For a space probe of mass \(m\) moving at speed \(v\), its kinetic energy is \(E = (\gamma - 1)mc^2\), where \(\gamma = \left(1 - \frac{v^2}{c^2}\right)^{-1/2}\) and \(c\) is the speed of light. Let \(f(x) = (1 - x)^{-1/2}\).
(a)
Find \(f'(x)\) and \(f''(x)\), and hence find the Maclaurin series for \(f(x)\) up to and including the term in \(x^2\).
(3)
(b)
Hence show that \(E \approx \frac{1}{2}mv^2 + \frac{3mv^4}{8c^2}\).
(2)
(c)
The probe travels at \(v = 0.2c\). Find the percentage error made when the kinetic energy is approximated by \(\frac{1}{2}mv^2\) instead of the exact value of \(E\).
Common error: Percentage error is measured relative to the exact value, so divide by the exact energy, not by the approximation.
335IBMA2H-053-01-095.19 Maclaurin seriesPaper 2Medium8 marks
In a model of the concentration \(y\) of a nutrient in a bioreactor, \(y\) satisfies the differential equation \(\frac{dy}{dx} = x + y^2\), with \(y = 1\) when \(x = 0\).
(a)
Write down the value of \(\frac{dy}{dx}\) when \(x = 0\).
(1)
(b)
Show that \(\frac{d^2y}{dx^2} = 1 + 2y\frac{dy}{dx}\) and find its value when \(x = 0\).
(2)
(c)
Find the value of \(\frac{d^3y}{dx^3}\) when \(x = 0\).
(2)
(d)
Hence find the Maclaurin series for \(y\) up to and including the term in \(x^3\).
(2)
(e)
Use your series to estimate \(y\) when \(x = 0.1\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(\frac{dy}{dx} = 0 + 1^2 = 1\)
✓ 1
Part (b)
M1 differentiates both terms: \(\frac{d}{dx}(x) = 1\) and \(\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}\), giving the result
✓ 1
A1 value at \(x = 0\): \(1 + 2\times 1\times 1 = 3\)
A1 \(y \approx 1 + x + \frac{3}{2}x^2 + \frac{4}{3}x^3\)
✓ 1
Part (e)
A1 \(y \approx 1.116\); accept \(1.12\) to 3 s.f.
✓ 1
Common error: When differentiating \(2y\frac{dy}{dx}\), both factors depend on \(x\), so the product rule is needed. Differentiating only one factor loses the \(2\left(\frac{dy}{dx}\right)^2\) term.
336IBMA2H-053-01-105.1 Limits and the derivativePaper 2Medium10 marks
Calculator
After an oral dose, the concentration of a drug in a patient's blood is modelled by \(C(t) = 12te^{-0.5t}\), where \(C\) is in mg L−1 and \(t\) is the time in hours, \(t \ge 0\).
(a)
Find \(C'(t)\).
(2)
(b)
Find the maximum concentration.
(2)
(c)
It is given that \(C''(t) = 3e^{-0.5t}(t - 4)\). Explain why the concentration is decreasing most rapidly at \(t = 4\), and find the equation of the tangent to the graph of \(C\) at \(t = 4\).
(4)
(d)
A pharmacist uses this tangent as a linear model for the elimination of the drug. Find the time at which this linear model predicts the concentration reaches zero.
M1 \(C'(t) = 0\) gives \(t = 2\) (or a GDC maximum at \(t = 2\))
✓ 1
A1 \(C(2) = 24e^{-1} = 8.83\) mg L−1
✓ 1
Part (c)
R1 \(C''(t)\) changes sign from negative to positive only at \(t = 4\), so \(C'\) has its least value there, and \(C'(4) < 0\), so this is the greatest rate of decrease
Exam tip: The greatest rate of decrease occurs where the derivative is at its minimum (and negative). Look at where the second derivative changes sign, not where the first derivative is zero.
337IBMA2H-053-08-105.17 Areas with the y-axis and volumes of revolutionPaper 2Medium7 marks
Calculator
A bushing for a bicycle suspension is modelled as follows. The region \(R\) lies between the curve \(y=xe^{-x/2}\) and the line \(y=\frac{x}{4}\), between the origin and their other point of intersection \(P\). \(R\) is rotated through \(2\pi\) radians about the \(x\)-axis. Lengths are in centimetres.
The region R between the curve and the line, from the origin to P. R is rotated about the x-axis. Diagram not to scale
(a)
Find the exact \(x\)-coordinate of \(P\).
(2)
(b)
Show that the curve lies above the line for \(0<x<2\ln4\).
(1)
(c)
Find the volume of the bushing.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 equates and divides by \(x\ne0\): \(e^{-x/2}=\frac14\)
✓ 1
A1 \(x=2\ln4\) \((=4\ln2)\)
✓ 1
Part (b)
R1 \(e^{-x/2}\) is decreasing and equals \(\frac14\) at \(x=2\ln4\), so \(e^{-x/2}>\frac14\) for \(0<x<2\ln4\); multiplying by \(x>0\) gives \(xe^{-x/2}>\frac{x}{4}\)
✓ 1
Part (c)
M1 uses a washer method: \(\pi\int\left(y_1^2-y_2^2\right)dx\)
✓ 1
A1 integrand \(x^2e^{-x}-\frac{x^2}{16}\)
✓ 1
A1 limits 0 and \(2\ln4\)
✓ 1
A1 \(1.90\) cm3
✓ 1
Common error: Using \(\pi\int(y_1-y_2)^2\,dx\) is wrong. Square each radius separately: outer radius squared minus inner radius squared.
338IBMA2H-053-02-055.14 Implicit differentiation, related rates and optimisationPaper 2Medium7 marks
Calculator
A rocket is launched vertically from a pad. A tracking camera is on level ground 2000 m from the pad. At time \(t\) seconds the rocket is \(h\) m above the pad, and the camera is pointed at an angle of elevation \(\theta\) radians. When \(h = 1500\), the rocket is rising at 300 m s−1.
Diagram not to scale
(a)
Find the rate at which \(\theta\) is increasing at this instant.
(4)
(b)
Find the rate at which the distance between the camera and the rocket is increasing at this instant.
M1 \(2s\frac{ds}{dt} = 2h\frac{dh}{dt}\) with \(s = 2500\)
✓ 1
A1 \(\frac{ds}{dt} = 180\) m s−1
✓ 1
Exam tip: Write the relationship between the variables before substituting any values; substituting \(h = 1500\) too early makes \(h\) a constant and its rate disappears.
339IBMA2H-053-02-075.8 Stationary points, inflexion and optimizationPaper 2Medium10 marks
Calculator
A shipping crate is a closed cuboid with a square base of side \(x\) m and height \(h\) m. Its volume is 2 m3. The base material costs $15 per m2, the top costs $10 per m2 and the four sides cost $8 per m2. The total material cost is $C.
Diagram not to scale
(a)
Show that \(C = 25x^2 + \frac{64}{x}\).
(3)
(b)
Find the value of \(x\) that minimises the cost.
(3)
(c)
Use the second derivative to justify that this value gives a minimum.
(2)
(d)
Find the minimum cost.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(x^2h = 2\)
✓ 1
M1 \(C = 15x^2 + 10x^2 + 8(4xh) = 25x^2 + 32xh\)
✓ 1
A1 substitute \(h = \frac{2}{x^2}\) to obtain \(25x^2 + \frac{64}{x}\)
✓ 1
Part (b)
M1 \(\frac{dC}{dx} = 50x - \frac{64}{x^2}\)
✓ 1
M1 set \(\frac{dC}{dx} = 0\), so \(x^3 = 1.28\)
✓ 1
A1 \(x = 1.09\) m
✓ 1
Part (c)
M1 \(\frac{d^2C}{dx^2} = 50 + \frac{128}{x^3}\)
✓ 1
R1 this is positive for \(x > 0\) (e.g. it equals 150 at \(x = 1.09\)), so the stationary point is a minimum
✓ 1
Part (d)
M1 substitute \(x = 1.0858...\) into \(C\)
✓ 1
A1 $88.4
✓ 1
Common error: The top and base have different costs and there are four sides, each of area \(xh\). Using one rate for the top and base, or counting only two sides, gives the wrong cost function. You then cannot reach the printed result in the first part.
340IBMA2H-053-02-095.14 Implicit differentiation, related rates and optimisationPaper 2Medium10 marks
Calculator
A hemispherical mould of radius 15 cm is being filled with melted chocolate at a rate of 20 cm3 s−1. When the depth of chocolate is \(h\) cm, its volume is \(V = \frac{\pi h^2(45 - h)}{3}\) cm3.
Cross-section of the mould. Diagram not to scale
(a)
Show that \(\frac{dV}{dh} = \pi(30h - h^2)\).
(2)
(b)
Find the rate at which the depth is increasing when \(h = 5\).
(3)
(c)
The surface of the chocolate is a circle of radius \(r\) cm. Show that \(r^2 = 30h - h^2\).
(2)
(d)
Find the rate at which the area of the chocolate surface is increasing when \(h = 5\).
M1 distance from centre of hemisphere to surface is \(15 - h\), use Pythagoras with radius 15
✓ 1
A1 \(r^2 = 225 - (15 - h)^2 = 30h - h^2\)
✓ 1
Part (d)
M1 \(A = \pi(30h - h^2)\)
✓ 1
M1 \(\frac{dA}{dt} = \pi(30 - 2h)\frac{dh}{dt}\)
✓ 1
A1 3.2 cm2 s−1
✓ 1
Exam tip: Link rates with the chain rule through the quantity whose rate you know; here every rate goes through \(\frac{dh}{dt}\).
341IBMA2H-053-03-085.14 Implicit differentiation, related rates and optimisationPaper 2Medium12 marks
Calculator
An audiologist models the sound from a small speaker of acoustic power P watts. At a distance r metres the intensity is \(I=\frac{P}{4\pi r^2}\) W m−2, and the sound level in decibels is \(L=10\log_{10}\left(\frac{I}{10^{-12}}\right)\).
(a)
Show that \(\frac{dL}{dr}=-\frac{20}{r\ln 10}\).
(3)
(b)
A listener walks directly away from the speaker at 1.4 m s−1. Find the rate of change of L when r = 8.
(3)
(c)
Given that P = 0.05, find the distance at which the sound level is 85 dB, the recommended limit for prolonged exposure.
(3)
(d)
Express I in terms of L, and hence find \(\frac{dI}{dL}\) when L = 85.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use log laws: \(L=10\log_{10}\left(\frac{P}{4\pi\times10^{-12}}\right)-20\log_{10}r\)
✓ 1
M1 use \(\frac{d}{dr}\log_{10}r=\frac{1}{r\ln 10}\)
Common error: Differentiating \(\log_{10}r\) as \(\frac{1}{r}\): change of base gives \(\frac{1}{r\ln 10}\). The factor \(\ln 10\) is easily dropped.
342IBMA2H-053-03-105.13 Limits and l’Hôpital’s rulePaper 2Medium13 marks
Calculator
In an industrial reactor, the concentration x mol dm−3 of a product t minutes after the start satisfies \(\frac{dx}{dt}=0.05(4-x)(2-x)\), with x = 0 when t = 0 and \(0\leq x<2\).
(a)
Express \(\frac{1}{(4-x)(2-x)}\) in partial fractions.
(3)
(b)
Show that \(t=10\ln\left(\frac{4-x}{4-2x}\right)\).
(4)
(c)
Express x in terms of t, and hence state the limiting concentration of the product.
(4)
(d)
Find the time taken for the concentration to reach 75% of its limiting value.
A1 \(B=-\frac{1}{2}\), so the result is \(\frac{1}{2(2-x)}-\frac{1}{2(4-x)}\)
✓ 1
Accept any equivalent correct form, e.g. \(\frac{1}{2(x-4)}-\frac{1}{2(x-2)}\).
—
Award the A1 marks for the corresponding constants in any valid form.
—
The final A1 may be awarded for the correct final expression even if the constants are not stated separately.
—
Do not award A1 for constants that are correct only for a different form from the one used.
—
Award M1 for any valid method of finding the constants (cover-up rule, equating coefficients or substitution) applied to a correct form, even if not written as an identity explicitly.
—
Do not award M1 if the form is incorrect (e.g. a missing factor).
—
Part (b)
M1 separate variables: \(\int\frac{dx}{(4-x)(2-x)}=\int0.05\,dt\)
✓ 1
A1 \(-\frac{1}{2}\ln(2-x)+\frac{1}{2}\ln(4-x)=0.05t+c\), i.e. \(\frac{1}{2}\ln\left(\frac{4-x}{2-x}\right)=0.05t+c\) (moduli not needed since \(0\leq x<2\))
✓ 1
M1 use x = 0, t = 0 to find \(c=\frac{1}{2}\ln 2\)
✓ 1
A1 combine logarithms and rearrange to \(t=10\ln\left(\frac{4-x}{4-2x}\right)\) AG
✓ 1
Part (c)
M1 exponentiate: \((4-2x)e^{t/10}=4-x\)
✓ 1
M1 collect x terms: \(x(2e^{t/10}-1)=4e^{t/10}-4\)
✓ 1
A1 \(x=\frac{4(e^{t/10}-1)}{2e^{t/10}-1}\)
✓ 1
R1 dividing numerator and denominator by \(e^{t/10}\) gives \(x=\frac{4(1-e^{-t/10})}{2-e^{-t/10}}\); as \(t\to\infty\), \(e^{-t/10}\to0\), so \(x\to2\) mol dm−3
✓ 1
Part (d)
M1 substitute x = 1.5: \(t=10\ln\frac{2.5}{1}\)
✓ 1
A1 9.16 minutes
✓ 1
Common error: Integrating \(\frac{1}{2-x}\) as \(\ln(2-x)\) instead of \(-\ln(2-x)\): the coefficient of x is −1, which flips the sign.
343IBMA2H-053-05-085.9 KinematicsPaper 2Medium8 marks
Calculator
A remote-controlled car moves in a straight line from rest. Its velocity is \(v(t)=12te^{-0.4t}\) m s−1 for \(0\le t\le5\), where t is in seconds.
(a)
Find the maximum velocity of the car.
(3)
(b)
Use integration by parts to show that the distance travelled in the first 5 seconds is \(75-225e^{-2}\) metres. Hence give this distance to 3 significant figures.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses the product rule: \(v'(t)=12e^{-0.4t}-4.8te^{-0.4t}\) A1 \(v'(t)=0\) gives \(t=2.5\) A1 \(v=30e^{-1}\approx11.0\) m s−1
✓ 3
Part (b)
M1 since \(v\ge0\), distance \(=\int_0^5v\,dt\); integrates by parts with \(u=t\) A1 \(\int te^{-0.4t}\,dt=-2.5te^{-0.4t}+\int2.5e^{-0.4t}\,dt\) A1 \(-2.5te^{-0.4t}-6.25e^{-0.4t}\) M1 substitutes limits 0 and 5 and multiplies by 12: \(12\left[(-12.5-6.25)e^{-2}+6.25\right]\) A1 \(75-225e^{-2}\approx44.5\) m
✓ 5
Common error: With \(e^{-0.4t}\), integrating divides by −0.4, so you multiply by −2.5. Students often multiply by −0.4 instead, which gives answers far too small.
344IBMA3H-5-18r9-045.15 Further derivatives and integrals; partial fractionsPaper 2Medium10 marks
Calculator
The number of bees in a hive is modelled by \(P(t)=\dfrac{60000}{1+9e^{-0.4t}}\), where \(t\ge 0\) is the time in weeks since the start of observations.
(a)
Find the number of bees at \(t=0\).
(1)
(b)
Use the chain rule to show that \(P'(t)=\dfrac{216000e^{-0.4t}}{\left(1+9e^{-0.4t}\right)^2}\).
(3)
(c)
Find the initial rate of growth of the bee population.
(1)
(d)
Find the time at which there are 30000 bees. Give your answer to 3 significant figures.
(3)
(e)
Find the rate of growth of the population at the time found in part (d).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(P(0)=\frac{60000}{10}=6000\)
✓ 1
Part (b)
M1 writes \(P=60000(1+9e^{-0.4t})^{-1}\) and applies the chain rule
A1 \(P'(0)=\frac{216000}{100}=2160\) bees per week
✓ 1
Part (d)
M1 \(1+9e^{-0.4t}=2\)
✓ 1
M1 \(e^{-0.4t}=\frac19\) and takes logarithms
✓ 1
A1 \(t=\frac{\ln9}{0.4}=5.49\) weeks
✓ 1
Part (e)
M1 substitutes \(e^{-0.4t}=\frac19\) into \(P'\)
✓ 1
A1 \(\frac{216000\times\frac19}{2^2}=6000\) bees per week
✓ 1
Exam tip: For a logistic model, substituting the exact value of the exponential is quicker and more accurate than using a rounded time.
345IBMA3H-5-18r9-095.16 Integration by substitution and by partsPaper 2Medium8 marks
Calculator
The nose of a Formula 1 car is modelled by rotating the curve \(y=0.2e^{-x/2}\), for \(0\le x\le1.5\), through \(360^\circ\) about the \(x\)-axis. Lengths are in metres.
Cross-section: the region under the curve is rotated about the x-axis
(a)
Write down an integral for the volume \(V\) of the nose.
(2)
(b)
Find \(V\) in cubic metres, to 3 significant figures.
(3)
(c)
The plane \(x=a\) cuts the nose into two pieces, the piece with \(0\le x\le a\) having half of the total volume. Find \(a\), to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(V=\pi\int y^2\,dx\)
✓ 1
A1 \(V=\pi\int_0^{1.5}0.04e^{-x}\,dx\)
✓ 1
Part (b)
M1 antiderivative \(-0.04\pi e^{-x}\)
✓ 1
M1 substitutes the limits: \(0.04\pi\left(1-e^{-1.5}\right)\)
✓ 1
A1 \(V=0.0976\ \text{m}^3\)
✓ 1
Part (c)
M1 \(0.04\pi(1-e^{-a})=\frac12(0.0976\ldots)\), so \(1-e^{-a}=\frac12\left(1-e^{-1.5}\right)\)
✓ 1
M1 \(e^{-a}=0.6116\ldots\)
✓ 1
A1 \(a=0.492\) m
✓ 1
Exam tip: Keep the unrounded exact expression for V when solving the later equation; rounding the volume first can shift the third significant figure.
346IBMA3H-5-18r9-105.17 Areas with the y-axis and volumes of revolutionPaper 2Medium9 marks
Calculator
A hospital medicine cup is modelled using the region R bounded by the curve \(y=x^2+1\) for \(x\ge0\), the \(y\)-axis and the lines \(y=1\) and \(y=5\). Lengths are in centimetres.
Region R is shaded
(a)
Express \(x^2\) in terms of \(y\).
(1)
(b)
Find the exact area of R.
(4)
(c)
The cup is formed by rotating R through \(360^\circ\) about the \(y\)-axis. Find its exact volume.
(3)
(d)
State the capacity of the cup in millilitres, to 3 significant figures. (\(1\ \text{cm}^3=1\) mL)
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(x^2=y-1\) (and \(x=\sqrt{y-1}\))
✓ 1
Part (b)
M1 \(\int_1^5\sqrt{y-1}\,dy\)
✓ 1
A1 \(\frac23(y-1)^{3/2}\)
✓ 1
M1 substitutes the limits \(y=5\) and \(y=1\)
✓ 1
A1 \(\frac{16}{3}\ \text{cm}^2\)
✓ 1
Part (c)
M1 \(V=\pi\int_1^5x^2\,dy=\pi\int_1^5(y-1)\,dy\)
✓ 1
A1 \(\pi\left[\frac{(y-1)^2}{2}\right]_1^5\)
✓ 1
A1 \(V=8\pi\ \text{cm}^3\)
✓ 1
Part (d)
A1 \(25.1\) mL
✓ 1
Common error: For rotation about the y-axis, rewrite the curve as x in terms of y and integrate with respect to y between y-limits.
347IBMA3H-5-18r9-115.5 Introduction to integrationPaper 2Medium8 marks
Calculator
A cross-section of an air duct on a Formula 1 car is the region S bounded by the curve \(y=\ln x\), the \(y\)-axis and the lines \(y=1\) and \(y=3\). Lengths are in centimetres.
Region S is shaded
(a)
Write \(x\) in terms of \(y\) for the curve.
(1)
(b)
Find the exact area of S.
(3)
(c)
Write the area from part (b) as a decimal, to 3 significant figures.
(1)
(d)
The line \(y=k\), where \(1<k<3\), divides S into two regions of equal area. Find \(k\), to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(x=e^y\)
✓ 1
Part (b)
M1 \(\int_1^3e^y\,dy\)
✓ 1
A1 \(\left[e^y\right]_1^3\)
✓ 1
A1 \(e^3-e\)
✓ 1
Part (c)
A1 \(17.4\ \text{cm}^2\)
✓ 1
Part (d)
M1 \(\int_1^ke^y\,dy=\frac12\left(e^3-e\right)\)
✓ 1
A1 \(e^k=\frac{e^3+e}{2}\)
✓ 1
A1 \(k=2.43\)
✓ 1
Exam tip: Areas bounded by the y-axis are found by integrating x with respect to y; the limits must be y-values.
348IBMA3H-5-18r9-125.17 Areas with the y-axis and volumes of revolutionPaper 2Medium7 marks
Calculator
A beekeeper's feeder reservoir is formed by rotating the region under the curve \(y=\dfrac{2}{\sqrt x}\), for \(1\le x\le4\), through \(360^\circ\) about the \(x\)-axis. Lengths are in decimetres, so \(1\ \text{dm}^3=1\) litre.
Cross-section: the region under the curve is rotated about the x-axis
(a)
Find the exact volume of the reservoir.
(3)
(b)
Write the volume in litres, to 3 significant figures.
(1)
(c)
A second reservoir is formed from the same curve for \(1\le x\le a\). Its volume is \(10\) litres. Find \(a\), to 3 significant figures.
Common error: The integrand is y squared: (2/√x)² = 4/x, not 2/x.
349IBMA3H-5-18r18-055.14 Implicit differentiation, related rates and optimisationPaper 2Medium7 marks
Calculator
A lava dome is modelled as a hemisphere of radius \(r\) metres. Its volume is \(V=\frac23\pi r^3\) m3 and its curved surface area is \(S=2\pi r^2\) m2. Lava is added so that \(V\) increases at a constant rate of \(150\) m3 per day.
Hemispherical lava dome. Diagram not to scale
(a)
Find the rate at which \(r\) is increasing when \(r=20\). Give your answer to 3 significant figures.
(4)
(b)
Find the rate at which the curved surface area is increasing when \(r=20\). Use the unrounded value of \(\dfrac{dr}{dt}\) from (a).
Exam tip: In related-rates questions, differentiate the formula with respect to the radius first, and substitute the particular value of \(r\) only after applying the chain rule.
350IBMA3H-5-18r18-095.5 Introduction to integrationPaper 2Medium7 marks
Calculator
A theatre poster has a printed area of \(480\) cm2. There are margins of \(4\) cm at the top and bottom and \(2.5\) cm at each side. The printed region has width \(x\) cm. The total area of paper used is \(A\) cm2.
Diagram not to scale
(a)
Show that \(A=520+8x+\dfrac{2400}{x}\).
(2)
(b)
Find the value of \(x\) that minimises \(A\). Give your answer to 3 significant figures.
(3)
(c)
Show that this is a minimum and find the minimum area of paper, to 3 significant figures. Use the unrounded value of \(x\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 printed height \(=\dfrac{480}{x}\), so the paper has dimensions \((x+5)\) by \(\left(\dfrac{480}{x}+8\right)\)
✓ 1
A1 expand: \(480+8x+\dfrac{2400}{x}+40\) giving the required form (AG)
✓ 1
Part (b)
A1 \(\dfrac{dA}{dx}=8-\dfrac{2400}{x^2}\)
✓ 1
M1 set \(\dfrac{dA}{dx}=0\)
✓ 1
A1 \(x=17.3\) (\(x^2=300\))
✓ 1
Part (c)
R1 \(\dfrac{d^2A}{dx^2}=\dfrac{4800}{x^3}>0\) for \(x>0\), so a minimum
✓ 1
A1 \(797\) cm2 (\(797.13\ldots\))
✓ 1
Common error: Adding the margin once instead of twice: a \(4\) cm margin at the top and bottom adds \(8\) cm to the height.
351IBMA3H-5-18r18-135.14 Implicit differentiation, related rates and optimisationPaper 2Medium7 marks
Calculator
A theatre prop is a closed cuboid with a square base of side \(x\) cm and height \(h\) cm. Its volume is \(4000\) cm3. Gold trim is fixed along all \(12\) edges of the prop, using a total length of \(L\) cm.
Diagram not to scale
(a)
Show that \(L=8x+\dfrac{16000}{x^2}\).
(2)
(b)
Find the value of \(x\) that minimises \(L\). Give your answer to 3 significant figures.
(3)
(c)
Show that this is a minimum and find the minimum length of trim, to 3 significant figures. Use the unrounded value of \(x\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(h=\dfrac{4000}{x^2}\) and \(L=8x+4h\)
✓ 1
A1 \(L=8x+\dfrac{16000}{x^2}\) (AG)
✓ 1
Part (b)
A1 \(\dfrac{dL}{dx}=8-\dfrac{32000}{x^3}\)
✓ 1
M1 set \(\dfrac{dL}{dx}=0\)
✓ 1
A1 \(x^3=4000\), so \(x=15.9\)
✓ 1
Part (c)
R1 \(\dfrac{d^2L}{dx^2}=\dfrac{96000}{x^4}>0\), so a minimum
✓ 1
A1 \(L=190\) cm (\(190.49\ldots\))
✓ 1
Common error: Counting only four or eight edges of the cuboid. A cuboid has 12 edges: 8 of length \(x\) (two squares) and 4 of length \(h\).
352IBMA3H-5-18r18-145.7 The second derivative and graphsPaper 2Medium7 marks
Calculator
The function \(f\) is defined by \(f(x)=x^2+\dfrac{16}{x}\), for \(x\ne0\).
(a)
Find the coordinates of the stationary point of the graph of \(f\).
(3)
(b)
Use the second derivative to classify the stationary point, and find the \(x\)-coordinate of the point of inflexion, correct to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(f'(x)=2x-\dfrac{16}{x^2}\)
✓ 1
M1 set \(f'(x)=0\), so \(x^3=8\)
✓ 1
A1 \((2,12)\)
✓ 1
Part (b)
A1 \(f''(x)=2+\dfrac{32}{x^3}\)
✓ 1
A1 \(f''(2)=6>0\), so a local minimum
✓ 1
M1 \(f''(x)=0\), so \(x^3=-16\)
✓ 1
A1 \(x=-2.52\), with a sign change of \(f''\) (e.g. \(f''(-3)>0\) and \(f''(-2)<0\)), so a point of inflexion
✓ 1
Common error: Taking the cube root of a negative number incorrectly (for example discarding it): \(x^3=-16\) has the real solution \(x=-\sqrt[3]{16}\).
353IBMA3H-5-18r19-065.3 Differentiating polynomials and integer powersPaper 2Medium6 marks
Calculator
A fashion pattern includes the curve \(y = x^3 - 3x + 2\), where \(x\) and \(y\) are in centimetres. The point A\((2, 4)\) lies on the curve.
(a)
Find the equation of the tangent to the curve at A.
(3)
(b)
The tangent at A meets the curve again at the point B. Find the coordinates of B.
M1 solve, using the repeated root \(x = 2\): \((x-2)^2(x+4) = 0\) (or GDC)
✓ 1
A1 B\((-4, -50)\)
✓ 1
Exam tip: A tangent touches the curve at its point of contact, so the equated equation has a repeated root there. Use this to factorise the cubic quickly.
354IBMA3H-5-18r19-105.14 Implicit differentiation, related rates and optimisationPaper 2Medium7 marks
Calculator
A conical hatchery tank is mounted with its vertex pointing downwards. When the water depth is \(h\) metres, the surface radius is \(r = \frac{h}{2}\) metres. Water drains from the tank at a constant rate of \(0.04\) m3 per minute. The volume of a cone of radius \(r\) and height \(h\) is \(V = \frac{1}{3}\pi r^2h\).
Conical tank, vertex downwards. Diagram not to scale
(a)
Show that \(V = \frac{\pi h^3}{12}\).
(2)
(b)
Find the rate at which the depth of water is changing when \(h = 1.2\) m. Give your answer to 3 significant figures.
(3)
(c)
Find the rate at which the surface radius is changing at this instant. Use the unrounded answer to (b) and give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitute \(r = \frac{h}{2}\) into \(V = \frac{1}{3}\pi r^2h\)
Exam tip: In related rates, write the quantity you know the rate of (\(\frac{dV}{dt}\)), the quantity you want (\(\frac{dh}{dt}\)), and link them with the chain rule. A draining tank gives a negative rate of change.
355IBMA3H-5-19r20-055.5 Introduction to integrationPaper 2Medium7 marks
Calculator
The hub of a wind turbine is modelled by rotating the region bounded by the curve \(y=\sqrt{x}\,e^{x/2}\), the x-axis and the line \(x=1\) through \(360^\circ\) about the x-axis. Lengths are in metres.
Cross-section: the region under the curve is rotated about the x-axis
(a)
Find the exact volume of the hub.
(5)
(b)
The hub is cast from metal that costs €40 per cubic metre. Find the cost of the metal, to the nearest euro.
M1 integration by parts with \(u=x\), \(\frac{dv}{dx}=e^{x}\)
✓ 1
A1 \(\pi\left[xe^{x}-e^{x}\right]_0^1\)
✓ 1
A1 \(V=\pi\,\mathrm{m^3}\)
✓ 1
Part (b)
M1 \(40\pi\)
✓ 1
A1 €126
✓ 1
Exam tip: For volumes of revolution about the x-axis, square the whole function before integrating. Here \((\sqrt{x}\,e^{x/2})^{2}=xe^{x}\).
356IBMA3H-5-19r20-065.14 Implicit differentiation, related rates and optimisationPaper 2Medium7 marks
Calculator
The inside surface of a glass sweet bowl in a gift shop is formed by rotating the curve \(y=\dfrac{x^{2}}{20}\), for \(x\ge0\), about the y-axis. Lengths are in centimetres and the bowl is 8 cm deep.
Cross-section of the bowl: the curve is rotated about the y-axis
(a)
Calculate the volume of the bowl. Give your answer to three significant figures.
(4)
(b)
Sweets are poured in at a constant rate of 50 cm3 s−1. Treating the sweets as a liquid, find the rate at which the depth is increasing when the depth is 5.0 cm. Give your answer to three significant figures.
M1 \(V=10\pi y^{2}\) for depth y, so \(\frac{dV}{dy}=20\pi y\)
✓ 1
M1 \(\frac{dV}{dt}=\frac{dV}{dy}\times\frac{dy}{dt}\), so \(50=20\pi(5)\frac{dy}{dt}\)
✓ 1
A1 \(\frac{dy}{dt}=0.159\) cm s−1
✓ 1
Common error: When rotating about the y-axis, the integral is \(\pi\int x^{2}\,dy\). Writing the integrand in terms of x but integrating with respect to y loses the mark.
357IBMA3H-5-19r20-075.9 KinematicsPaper 2Medium7 marks
Calculator
A drone leaves a retail warehouse and flies along a straight path. Its velocity, v m s−1, at time t seconds is given by \(v(t)=t^{2}-6t+8\), for \(0\le t\le5\). Positive velocity is away from the warehouse.
(a)
Find the times at which the drone is instantaneously at rest.
(2)
(b)
Find the displacement of the drone from the warehouse at \(t=5\).
(2)
(c)
Find the total distance travelled by the drone in the first 5 seconds. Give your answer to three significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(t^{2}-6t+8=0\) solved, e.g. \((t-2)(t-4)=0\)
✓ 1
A1 \(t=2\) and \(t=4\)
✓ 1
Part (b)
M1 \(\int_0^5 (t^{2}-6t+8)\,dt\)
✓ 1
A1 \(\frac{20}{3}\approx6.67\) m
✓ 1
Part (c)
M1 splits the integral at \(t=2\) and \(t=4\), where the velocity changes sign
✓ 1
M1 adds the magnitudes: \(\frac{20}{3}+\left|-\frac{4}{3}\right|+\frac{4}{3}\)
✓ 1
A1 \(\frac{28}{3}\approx9.33\) m
✓ 1
Common error: Distance is not the same as displacement. Integrate over each interval where the velocity has one sign and add the absolute values.
358IBMA3H-5-19r20-085.16 Integration by substitution and by partsPaper 2Medium7 marks
Calculator
During an outbreak of a seasonal illness, the rate at which new cases are reported is modelled by \(r(t)=50\,t\,e^{-t/2}\) cases per day, where t is the time in days since the first report.
(a)
Show that \(\int_0^4 t\,e^{-t/2}\,dt=4-12e^{-2}\).
(5)
(b)
Hence find the total number of new cases reported during the first 4 days. Give your answer to the nearest whole number.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 integration by parts with \(u=t\), \(\frac{dv}{dt}=e^{-t/2}\)
✓ 1
A1 \(-2te^{-t/2}+\int 2e^{-t/2}\,dt\)
✓ 1
A1 \(-2te^{-t/2}-4e^{-t/2}\)
✓ 1
M1 substitutes limits 4 and 0: \(\left(-8e^{-2}-4e^{-2}\right)-(-4)\)
✓ 1
A1 \(4-12e^{-2}\) (AG)
✓ 1
Part (b)
M1 \(50\left(4-12e^{-2}\right)\)
✓ 1
A1 119 cases
✓ 1
Exam tip: In a "show that" question every line of working must be visible, including the substitution of both limits.
359IBMA3H-5-19r20-105.14 Implicit differentiation, related rates and optimisationPaper 2Medium7 marks
Calculator
A food company packages soup in closed cylindrical tins, each with volume 500 cm3. A tin has base radius r cm and height h cm. The company wants to minimise the total surface area S cm2 of metal used.
Closed cylindrical tin. Diagram not to scale
(a)
Show that \(S=2\pi r^{2}+\dfrac{1000}{r}\).
(2)
(b)
Find the value of r for which S is a minimum. Give your answer to three significant figures.
(4)
(c)
Find the minimum surface area. Use the unrounded value of r and give your answer to three significant figures.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(h=\dfrac{500}{\pi r^{2}}\)
✓ 1
A1 substitutes into \(S=2\pi r^{2}+2\pi rh\) to obtain \(2\pi r^{2}+\dfrac{1000}{r}\) (AG)
✓ 1
Part (b)
M1 attempt to differentiate
✓ 1
A1 \(\dfrac{dS}{dr}=4\pi r-\dfrac{1000}{r^{2}}\)
✓ 1
M1 sets \(\dfrac{dS}{dr}=0\) and attempts to solve
✓ 1
A1 \(r^{3}=\dfrac{250}{\pi}\), so \(r=4.30\) cm
✓ 1
Part (c)
A1 \(S_{\min}=349\ \mathrm{cm^2}\)
✓ 1
Exam tip: Store the unrounded value of r on your calculator and reuse it in later parts to avoid rounding errors.
360IBMA3H-5-19r26-115.14 Implicit differentiation, related rates and optimisationPaper 2Medium12 marks
Calculator
The height of a lava dome above its base is modelled by \(H(t)=0.002t^3-0.09t^2+1.2t+5\) for \(0\le t\le30\), where \(H\) is in metres and \(t\) is the time in days since the eruption began.
(a)
Find \(\frac{dH}{dt}\).
(2)
(b)
Find the times at which the height of the dome is stationary.
(3)
(c)
The rate of change of the height of the dome has a minimum value at time \(t=T\). Find \(\frac{d^2H}{dt^2}\) and hence find \(T\), justifying that this is a minimum.
(4)
(d)
Find the height of the dome at \(t=T\) and the value of \(\frac{dH}{dt}\) at \(t=T\), and interpret the sign of \(\frac{dH}{dt}\) in context.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to differentiate each term
✓ 1
A1 \(0.006t^2-0.18t+1.2\)
✓ 1
Part (b)
M1 set \(\frac{dH}{dt}=0\)
✓ 1
A1 \(t=10\)
✓ 1
A1 \(t=20\)
✓ 1
Part (c)
A1 \(\frac{d^2H}{dt^2}=0.012t-0.18\)
✓ 1
M1 set \(\frac{d^2H}{dt^2}=0\)
✓ 1
A1 \(T=15\)
✓ 1
R1 \(\frac{d^3H}{dt^3}=0.012>0\) (or the gradient of \(\frac{dH}{dt}\) changes from negative to positive), so a minimum
✓ 1
Part (d)
A1 \(H(15)=9.5\) m
✓ 1
A1 \(\frac{dH}{dt}=-0.15\) m/day
✓ 1
R1 the dome is shrinking, and its height is falling at its fastest rate of 0.15 m per day at this time
✓ 1
Common error: The minimum of the rate of change of \(H\) is found from \(\frac{d^2H}{dt^2}=0\), not from \(\frac{dH}{dt}=0\); the latter gives the times when the height itself is stationary.
361IBMA3H-5-20r27-025.18 First order differential equationsPaper 2Medium8 marks
Calculator
A bowl of custard cools in a kitchen of constant temperature 20 °C. Its temperature \(T\) °C, \(t\) minutes after it is removed from the heat, satisfies \(\dfrac{dT}{dt}=-k(T-20)\), where \(k>0\) is a constant. Initially \(T=90\).
(a)
Show that \(T=20+70e^{-kt}\).
(4)
(b)
After 5 minutes the temperature is 60 °C. Find the exact value of \(k\), and give it to 3 significant figures.
(2)
(c)
Find the time at which the custard reaches 30 °C. Give your answer in minutes to 3 significant figures, using the exact value of \(k\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 separate variables: \(\int\frac{1}{T-20}\,dT=\int-k\,dt\)
✓ 1
A1 \(\ln(T-20)=-kt+c\)
✓ 1
M1 use of \(T=90\) when \(t=0\), so \(c=\ln70\)
✓ 1
A1 \(T-20=70e^{-kt}\), so \(T=20+70e^{-kt}\) (AG)
✓ 1
Part (b)
M1 \(40=70e^{-5k}\)
✓ 1
A1 \(k=\frac15\ln\frac74=0.112\)
✓ 1
Part (c)
M1 \(10=70e^{-kt}\), so \(t=\frac{\ln7}{k}\)
✓ 1
A1 \(t=17.4\) minutes
✓ 1
Common error: When using the rounded \(k=0.112\) the answer in (c) drifts; keep \(k=\frac15\ln\frac74\) exact in your working.
362IBMA3H-5-20r27-045.18 First order differential equationsPaper 2Medium7 marks
Calculator
The concentration \(c\) g/L of salt in a brine tank, \(t\) minutes after a valve is opened, satisfies \(\dfrac{dc}{dt}=t^2-c\), with \(c=2\) when \(t=1\).
(a)
Use Euler's method with a step length of 0.2 to estimate the value of \(c\) when \(t=1.6\). Give your answer to 3 significant figures.
(4)
(b)
Using the unrounded estimate from (a), find the estimated value of \(\dfrac{dc}{dt}\) at \(t=1.6\), to 3 significant figures, and state whether the concentration is increasing or decreasing at that time.
(2)
(c)
State one change that would make the estimate in (a) more accurate.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(c_{n+1}=c_n+0.2(t_n^2-c_n)\)
✓ 1
A1 \(c(1.2)\approx1.8\)
✓ 1
A1 \(c(1.4)\approx1.728\)
✓ 1
A1 \(c(1.6)\approx1.7744\approx1.77\)
✓ 1
Part (b)
M1 \(1.6^2-1.7744\)
✓ 1
A1 0.786, so the concentration is increasing (ECF from (a))
✓ 1
Part (c)
R1 use a smaller step length
✓ 1
Common error: Each Euler step uses the \(t\) and \(c\) values from the start of that step; mixing in the new \(c\) gives a different (wrong) recurrence.
363IBMA3H-5-20r27-115.8 Stationary points, inflexion and optimizationPaper 2Medium8 marks
Calculator
The annual cost, £\(C\) thousand, of a microwave link whose relay masts are spaced \(x\) km apart is modelled by \(C(x)=2x^2+\dfrac{450}{x}\) for \(2\le x\le8\).
(a)
Find \(C'(x)\).
(2)
(b)
Find the value of \(x\) for which \(C'(x)=0\), to 3 significant figures.
(2)
(c)
Show that this value of \(x\) gives a minimum cost.
(2)
(d)
Find the minimum annual cost, in £ thousand to 3 significant figures, using the unrounded value of \(x\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to differentiate, with at least one correct term
✓ 1
A1 \(C'(x)=4x-450x^{-2}\)
✓ 1
Part (b)
M1 \(4x=\frac{450}{x^2}\), so \(x^3=112.5\)
✓ 1
A1 \(x=4.83\)
✓ 1
Part (c)
M1 evaluate or discuss the sign of \(C'\) either side, e.g. \(C'(4)=-12.1<0\) and \(C'(5)=2>0\)
✓ 1
R1 \(C\) is decreasing then increasing, so the stationary point is a minimum
✓ 1
Part (d)
M1 substitute \(x=112.5^{1/3}\) into \(C\)
✓ 1
A1 £140 thousand
✓ 1
Common error: Showing that \(C'(x)=0\) is not enough to prove a minimum; give evidence for the sign change of \(C'\) (or the sign of \(C''\)).
364IBMA3H-5-20r27-145.1 Limits and the derivativePaper 2Medium8 marks
Calculator
The number of subscribers to a broadband provider, in thousands, is modelled by \(N(t)=-0.5t^3+6t^2+2t+40\) for \(0\le t\le10\), where \(t\) is the time in years since the start of 2020.
(a)
Find \(N'(t)\).
(2)
(b)
Find the value of \(t\) at which the number of subscribers is growing fastest.
(2)
(c)
Find the greatest rate of growth, in thousands of subscribers per year.
(1)
(d)
Find the values of \(t\) for which \(N\) is increasing. Give any decimal answer to 3 significant figures.
Common error: The fastest growth occurs where \(N'\) is greatest, i.e. where \(N''=0\), not where \(N'=0\) (which marks the peak of \(N\)).
365IBMA3H-5-20r31-025.4 Tangents and normalsPaper 2Medium6 marks
Calculator
The path of a space probe in a simulation is modelled by \(y=\ln(2x+1)\) for \(x>-\frac{1}{2}\). The point P on the path has \(x=2\).
(a)
Show that the gradient of the path at P is \(\frac{2}{5}\).
(2)
(b)
Find the equation of the normal to the path at P.
(2)
(c)
The normal meets the \(y\)-axis at Q. Find the \(y\)-coordinate of Q, correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{dy}{dx}=\frac{2}{2x+1}\)
✓ 1
AG at \(x=2\) the gradient is \(\frac{2}{5}\)
✓ 1
Part (b)
M1 normal gradient \(=-\frac{5}{2}\)
✓ 1
A1 \(y-\ln 5=-\frac{5}{2}(x-2)\)
✓ 1
Part (c)
M1 substitute \(x=0\) into the normal equation
✓ 1
A1 y = 6.61
✓ 1
Exam tip: Keep ln 5 exact until the final step, then evaluate once to avoid rounding drift.
366IBMA3H-5-20r31-065.5 Introduction to integrationPaper 2Medium6 marks
Calculator
A logo for a coffee roaster is the region \(R\) bounded by the curve \(y=e^x\), the line \(y=6-x\) and the \(y\)-axis. Lengths are in centimetres.
Region R is shaded
(a)
Find the \(x\)-coordinate of the point where the curve and the line intersect, correct to 3 significant figures.
(2)
(b)
Find the area of R, correct to 3 significant figures. Use the unrounded intersection value.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 solves \(e^x=6-x\)
✓ 1
A1 x = 1.50
✓ 1
Part (b)
M1 area \(=\int_0^a\left((6-x)-e^x\right)dx\) with \(a\) the answer to (a)
✓ 1
A1 correct limits 0 and a
✓ 1
A1 antiderivative \(6x-\frac{x^2}{2}-e^x\)
✓ 1
A1 4.39 cm²
✓ 1
Exam tip: Sketch both graphs first so you know which one is above; the line is above the curve from x = 0 up to the intersection.
367IBMA3H-5-20r31-075.5 Introduction to integrationPaper 2Medium6 marks
Calculator
The cross-sectional area under a sensor's response curve \(y=xe^{x^2}\), for \(x\ge0\), between \(x=0\) and \(x=k\) is denoted by \(A(k)\).
The shaded area is A(k)
(a)
Use the substitution \(u=x^2\) to show that \(A(k)=\frac{e^{k^2}-1}{2}\).
(3)
(b)
Given that \(A(k)=4\), find the value of \(k\), correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(du=2x\,dx\), so \(x\,dx=\frac{1}{2}du\)
✓ 1
M1 \(\frac{1}{2}\int_0^{k^2}e^u\,du\) with limits changed
✓ 1
AG \(\frac{1}{2}\left(e^{k^2}-1\right)\)
✓ 1
Part (b)
M1 \(e^{k^2}-1=8\)
✓ 1
A1 \(k^2=\ln 9\)
✓ 1
A1 k = 1.48
✓ 1
Common error: When you substitute in a definite integral, change the limits to the new variable (here u from 0 to k²) or return to x before substituting limits.
368IBMA3H-5-20r31-095.18 First order differential equationsPaper 2Medium8 marks
Calculator
Roasted beans leaving a roaster at 210 °C cool in a room at 22 °C. The temperature \(T\) °C after \(t\) minutes satisfies \(\frac{dT}{dt}=-k(T-22)\), where \(k>0\) is a constant and \(T>22\).
(a)
Solve the differential equation to show that \(T=22+188e^{-kt}\).
(4)
(b)
After 5 minutes the temperature is 130 °C. Find the value of k, correct to 3 significant figures.
(2)
(c)
Find the time taken for the beans to reach 40 °C, correct to 3 significant figures. Use the unrounded value of k.
M1 \(18=188e^{-kt}\), so \(t=\frac{1}{k}\ln\frac{188}{18}\)
✓ 1
A1 t = 21.2 minutes
✓ 1
Common error: Newton's law of cooling involves the temperature difference \(T-22\), not \(T\); separating \(\frac{dT}{T}\) is a frequent error.
369IBMA3H-5-20r31-105.9 KinematicsPaper 2Medium9 marks
Calculator
The speed \(v\) m s−1 of a probe moving through a resistive gas at time \(t\) seconds, \(t\ge0\), satisfies \(\frac{dv}{dt}+0.2v=3e^{-0.2t}\) with \(v=4\) when \(t=0\).
(a)
Use an integrating factor to show that \(v=(3t+4)e^{-0.2t}\).
(5)
(b)
Find the greatest speed of the probe, correct to 3 significant figures, justifying that it is a maximum.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 integrating factor \(e^{0.2t}\)
✓ 1
M1 \(\frac{d}{dt}\left(ve^{0.2t}\right)=3\)
✓ 1
A1 \(ve^{0.2t}=3t+c\)
✓ 1
M1 uses v = 4 at t = 0 to find c = 4
✓ 1
A1 \(v=(3t+4)e^{-0.2t}\)
✓ 1
Part (b)
M1 sets \(\frac{dv}{dt}=0\), so \(0.2v=3e^{-0.2t}\), giving \(0.6t+0.8=3\)
✓ 1
A1 \(t=\frac{11}{3}\)
✓ 1
A1 greatest speed 7.20 m s⁻¹
✓ 1
R1 \(\frac{dv}{dt}=e^{-0.2t}(2.2-0.6t)\) changes from positive to negative at \(t=\frac{11}{3}\)
✓ 1
Exam tip: You can find the stationary point directly from the differential equation without differentiating the solution.
370IBMA3H-5-20r31-125.18 First order differential equationsPaper 2Medium6 marks
Calculator
The population \(y\) (in thousands) of microbes in a sample from a probe satisfies \(\frac{dy}{dt}=0.4y\left(1-\frac{y}{10}\right)\), where \(t\) is in hours, and \(y=2\) when \(t=0\).
(a)
Use Euler's method with a step length of 0.5 to estimate y when t = 1. Give your answer to 3 significant figures.
(4)
(b)
Explain why this estimate is less than the true value of y at t = 1.
R1 for \(y<5\), \(\frac{dy}{dt}=0.4y-0.04y^2\) increases as y increases, since its derivative with respect to y is \(0.4-0.08y>0\)
✓ 1
R1 so the gradient increases along the solution (the curve is concave up) and each tangent-line step lies below the curve
✓ 1
Exam tip: Euler's method gives an underestimate when the solution curve is concave up and an overestimate when it is concave down.
371IBMA3H-5-19r33-065.16 Integration by substitution and by partsPaper 2Medium6 marks
Calculator
A hospital pump delivers a drug at the rate \(r(t)=8te^{-t/2}\) mg per hour, where \(t\) is the time in hours since the start of the infusion, for \(0\le t\le 6\).
(a)
Use integration by parts to show that \(\int 8te^{-t/2}\,dt=-16(t+2)e^{-t/2}+C\).
(4)
(b)
Hence find the total mass of drug delivered during the 6 hours. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt at integration by parts with \(u=8t\) and \(\frac{dv}{dt}=e^{-t/2}\)
Exam tip: In integration by parts choose \(u\) to be the factor that simplifies when differentiated; here \(t\) becomes 1.
372IBMA3H-5-19r33-105.16 Integration by substitution and by partsPaper 2Medium6 marks
Calculator
A drug is absorbed from the stomach at the rate \(r(t)=e^{-t}\sin t\) mg per hour, where \(t\) is the time in hours, for \(0\le t\le\pi\). Let \(I=\int_0^{\pi}e^{-t}\sin t\,dt\).
(a)
By using integration by parts twice, show that \(I=\dfrac{1+e^{-\pi}}{2}\).
(5)
(b)
Find the total mass of drug absorbed. Give your answer to 3 significant figures.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 first integration by parts with \(u=\sin t\) and \(\frac{dv}{dt}=e^{-t}\)
✓ 1
A1 \(I=\left[-e^{-t}\sin t\right]_0^{\pi}+\int_0^{\pi}e^{-t}\cos t\,dt\), where the first term is \(0\)
✓ 1
M1 second integration by parts on \(\int_0^{\pi}e^{-t}\cos t\,dt\)
A1 \(2I=1+e^{-\pi}\), so \(I=\dfrac{1+e^{-\pi}}{2}\) (AG)
✓ 1
Part (b)
A1 \(0.522\) mg
✓ 1
Exam tip: When the original integral reappears after two rounds of integration by parts, treat it as an unknown and solve the resulting equation.
373IBMA3H-5-20r36-075.5 Introduction to integrationPaper 2Medium8 marks
Calculator
A theatre set designer cuts an arch from a flat board. In metres, the arch is the region between the curve y = 4 − x2 and the line y = 1.
The arch is shaded
(a)
Find the x-coordinates of the points where the curve meets the line.
(2)
(b)
Find the exact area of the region between the curve and the line.
(4)
(c)
The board costs €45 per square metre of arch. Find the cost, to the nearest euro, of the board needed for the arch itself.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets 4 − x2 = 1
✓ 1
A1 x = ±√3
✓ 1
Part (b)
M1 sets up ∫−√3√3 (3 − x2) dx
✓ 1
A1 antiderivative 3x − x3/3
✓ 1
M1 substitutes limits (or doubles the integral from 0 to √3)
✓ 1
A1 4√3 m2
✓ 1
Part (c)
M1 45 × 4√3 (ECF from (b))
✓ 1
A1 €312
✓ 1
Exam tip: Between two curves, integrate the upper function minus the lower function; subtract before integrating to keep the algebra simple.
374IBMA3H-5-20r36-085.9 KinematicsPaper 2Medium7 marks
Calculator
After its engines are cut, the speed of a cargo ship is modelled by v(t) = 12e−0.25t m s−1, where t is the time in seconds since the engines were cut.
(a)
Find the acceleration of the ship when t = 2. Give your answer to 3 significant figures.
(2)
(b)
Find the distance travelled by the ship in the first 8 seconds. Give your answer to 3 significant figures.
(3)
(c)
Show that the ship never travels further than 48 m after the engines are cut.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 differentiates: a = −3e−0.25t
✓ 1
A1 a = −1.82 m s−2
✓ 1
Part (b)
M1 sets up ∫08 12e−0.25t dt
✓ 1
A1 antiderivative −48e−0.25t
✓ 1
A1 48(1 − e−2) = 41.5 m
✓ 1
Part (c)
M1 distance after time T is 48(1 − e−0.25T)
✓ 1
R1 since e−0.25T > 0 for all T, the distance is always less than 48 m
✓ 1
Common error: A negative acceleration here means the ship is slowing down; do not drop the sign, and give the unit m s⁻².
375IBMA3H-5-20r36-105.19 Maclaurin seriesPaper 2Medium8 marks
Calculator
A model for the pressure ratio P in a magma chamber (current pressure divided by initial pressure) at scaled time t satisfies dP/dt = t + P2, with P = 1 when t = 0.
(a)
Find the Maclaurin series for P up to and including the term in t3.
(6)
(b)
Use your series to estimate P when t = 0.2. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 P′(0) = 0 + 12 = 1
✓ 1
M1 differentiates implicitly: P″ = 1 + 2PP′
✓ 1
A1 P″(0) = 3
✓ 1
M1 differentiates again: P‴ = 2(P′)2 + 2PP″
✓ 1
A1 P‴(0) = 8
✓ 1
A1 P = 1 + t + (3/2)t2 + (4/3)t3
✓ 1
Part (b)
M1 substitutes t = 0.2 (ECF from (a))
✓ 1
A1 1.27
✓ 1
Exam tip: To build a series from a differential equation, differentiate the equation repeatedly and evaluate at the starting point each time using the values you already found.
376IBMA3H-5-20r36-115.19 Maclaurin seriesPaper 2Medium5 marks
Calculator
The intensity of light from a theatre spotlight at distance x metres from the centre of its beam is modelled by I = e−x2.
(a)
Use the Maclaurin series for eu to find a series for I up to and including the term in x4.
(2)
(b)
Hence estimate ∫00.4 e−x2 dx. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes u = −x2 into 1 + u + u2/2
✓ 1
A1 1 − x2 + x4/2
✓ 1
Part (b)
M1 integrates the series term by term (ECF from (a))
✓ 1
A1 x − x3/3 + x5/10
✓ 1
A1 0.380
✓ 1
Exam tip: Replace u in a known series by the whole expression, here −x², using brackets so the sign and the square are both applied.
377IBMA3H-5-20r37-095.14 Implicit differentiation, related rates and optimisationPaper 2Medium8 marks
Calculator
An aircraft flies in a straight line at a constant height of 3000 m directly towards a radar station on the ground, at a constant horizontal speed of 240 m s−1. Let x be the horizontal distance of the aircraft from the point directly above the station, and let r be its straight-line distance from the station.
Diagram not to scale
(a)
Find the rate at which r is changing when x = 4000 m.
(4)
(b)
Let θ be the angle of elevation of the aircraft from the station, so that tan θ = 3000/x. Find the rate at which θ is increasing when x = 4000 m, in radians per second.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 r2 = x2 + 30002
✓ 1
M1 2r dr/dt = 2x dx/dt with dx/dt = −240
✓ 1
A1 r = 5000
✓ 1
A1 dr/dt = −192 m s−1, so r decreases at 192 m s−1
✓ 1
Part (b)
M1 sec2θ dθ/dt = −(3000/x2) dx/dt
✓ 1
A1 right-hand side = 0.045
✓ 1
A1 sec2θ = 1 + (3/4)2 = 25/16
✓ 1
A1 dθ/dt = 0.0288 rad s−1
✓ 1
Exam tip: Write a relationship between the variables first, then differentiate with respect to time, and substitute numbers only afterwards.
378IBMA3H-5-20r37-105.14 Implicit differentiation, related rates and optimisationPaper 2Medium8 marks
Calculator
Grain is poured onto a flat floor from a conveyor, forming a pile in the shape of a cone whose height h metres is always twice its base radius. Grain is added at a constant rate of 0.6 m3 per minute.
Conical pile of grain. Diagram not to scale
(a)
Show that the volume of the pile is V = (π/12)h3.
(2)
(b)
Find the rate at which the height of the pile is increasing when h = 3 m. Give your answer in m per minute to 3 significant figures.
(3)
(c)
Find the rate at which the area of the circular base is increasing when h = 3 m. Use the unrounded value of dh/dt from part (b).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 r = h/2 substituted into V = (1/3)πr2h
✓ 1
A1 V = (1/3)π(h2/4)h = (π/12)h3
✓ 1
Part (b)
M1 dV/dh = (π/4)h2 and dV/dt = dV/dh × dh/dt
✓ 1
M1 0.6 = (9π/4) dh/dt
✓ 1
A1 0.0849 m per minute
✓ 1
Part (c)
M1 A = πr2 = (π/4)h2
✓ 1
M1 dA/dt = (π/2)h × dh/dt
✓ 1
A1 0.4 m2 per minute (ECF from (b))
✓ 1
Common error: Substitute r = h/2 before differentiating so that the volume is a function of one variable; differentiating with both r and h present gives a wrong rate.
379IBMA3H-5-19r38-065.14 Implicit differentiation, related rates and optimisationPaper 2Medium7 marks
Calculator
A food manufacturer makes closed cylindrical tins of volume 500 cm3. A tin has base radius \(r\) cm and height \(h\) cm. The total outside surface area of a tin is \(A\) cm2.
Closed cylindrical tin. Diagram not to scale
(a)
Show that \(A=2\pi r^{2}+\frac{1000}{r}\).
(2)
(b)
Find the value of \(r\) that minimises \(A\). Give your answer to 3 significant figures.
(3)
(c)
Justify that this value of \(r\) gives a minimum, and find the minimum surface area to 3 significant figures. Use the unrounded value of \(r\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\pi r^2h=500\), so \(h=\frac{500}{\pi r^2}\)
✓ 1
A1 \(A=2\pi r^2+2\pi rh=2\pi r^2+\frac{1000}{r}\)
✓ 1
Part (b)
M1 \(\frac{dA}{dr}=4\pi r-\frac{1000}{r^2}\)
✓ 1
M1 sets \(\frac{dA}{dr}=0\), so \(r^3=\frac{250}{\pi}\)
✓ 1
A1 \(r=4.30\) cm
✓ 1
Part (c)
R1 \(\frac{d^2A}{dr^2}=4\pi+\frac{2000}{r^3}>0\) for \(r>0\), so minimum
✓ 1
A1 \(A=349\) cm2
✓ 1
Common error: Forgetting to rewrite \(\frac{1000}{r}\) as \(1000r^{-1}\) before differentiating gives the wrong derivative; also check the second derivative rather than assuming a stationary point is a minimum.
380IBMA3H-5-19r38-095.18 First order differential equationsPaper 2Medium6 marks
Calculator
The backlog of data waiting at a network router, \(y\) gigabytes at time \(x\) hours, satisfies \(\frac{dy}{dx}=2x-y\) with \(y=1\) when \(x=0\).
(a)
Use Euler's method with step length \(h=0.25\) to estimate \(y\) when \(x=1\). Show each step of your working, and give your final estimate to 3 significant figures.
(4)
(b)
The exact solution is \(y=2x-2+3e^{-x}\). Find the absolute error in the Euler estimate for \(y(1)\), to 3 significant figures. Use the unrounded Euler estimate \(0.94921875\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(y_{n+1}=y_n+h(2x_n-y_n)\)
✓ 1
A1 \(y(0.25)\approx0.75\)
✓ 1
A1 \(y(0.5)\approx0.6875\) and \(y(0.75)\approx0.765625\)
✓ 1
A1 \(y\approx0.949\) GB when \(x=1\)
✓ 1
Part (b)
M1 exact value at \(x=1\): \(3e^{-1}=1.10364\ldots\)
✓ 1
A1 error \(=1.10364-0.94922=0.154\) GB
✓ 1
Exam tip: Keep every Euler value unrounded until the end; rounding at each step builds up error that is not part of the method itself.
381IBMA3H-5-20r45-065.9 KinematicsPaper 2Medium9 marks
Calculator
A robot moves along a straight track. Its acceleration at time \(t\) seconds is \(a(t) = 6t - 12\) m s−2. At \(t = 0\) the robot is at the start position \(s = 0\) with velocity 9 m s−1.
(a)
Show that the velocity of the robot is \(v(t) = 3t^2 - 12t + 9\).
(3)
(b)
Find an expression for the displacement \(s(t)\) from the start position.
(2)
(c)
Find the total distance travelled by the robot in the first 4 seconds.
M1 solve \(v = 0\) to get \(t = 1\) and \(t = 3\) (the robot reverses direction at these times)
✓ 1
A1 positions 4 m at \(t = 1\), 0 m at \(t = 3\) and 4 m at \(t = 4\)
✓ 1
M1 add the magnitudes of the three movements, \(|4 - 0| + |0 - 4| + |4 - 0|\)
✓ 1
A1 12 m
✓ 1
Common error: Total distance is not the same as final displacement: split the motion where the velocity changes sign and add the absolute values.
382IBMA3H-5-20r45-115.9 KinematicsPaper 2Medium8 marks
Calculator
After fishing was restricted, the biomass \(S\) (thousands of tonnes) of a fish stock is modelled by \(\frac{dS}{dt} = 0.2(60 - S)\), where \(t\) is the time in years since the restriction began and \(S < 60\). Initially \(S = 20\).
(a)
Solve the differential equation to show that \(S = 60 - 40e^{-0.2t}\).
(5)
(b)
Find the time at which the biomass first reaches 50 thousand tonnes, giving your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 separate variables: \(\int\frac{1}{60 - S}\,dS = \int 0.2\,dt\)
✓ 1
A1 \(-\ln(60 - S) = 0.2t + c\)
✓ 1
M1 use of \(S = 20\) at \(t = 0\) to find the constant
Exam tip: Keep the form ln(60 − S) rather than ln|60 − S| only when you have stated that S < 60.
383IBMA3H-5-19r47-075.17 Areas with the y-axis and volumes of revolutionPaper 2Medium10 marks
Calculator
The cross-section of a burial mound is modelled by the curve \(h(x)=2.4-0.15x^2\) for \(-4\le x\le 4\), where \(x\) and \(h\) are measured in metres, with the \(x\)-axis along the level ground.
Cross-section of the mound; the shaded part is above the height 1.2 m. Diagram not to scale
(a)
Show that the mound is 8 m wide at ground level, and state its greatest height.
(2)
(b)
Find the area of the cross-section.
(3)
(c)
The mound has this same cross-section along a length of 15 m. Find its volume.
(2)
(d)
The part of the mound above a height of 1.2 m is removed. Find the cross-sectional area of the removed part. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(h(\pm4)=2.4-0.15\times16=0\), so the mound meets the ground at \(x=\pm4\) and its width is 8 m
✓ 1
A1 greatest height 2.4 m (at \(x=0\))
✓ 1
Part (b)
M1 \(\int_{-4}^{4}(2.4-0.15x^2)\,\mathrm{d}x\)
✓ 1
A1 antiderivative \(2.4x-0.05x^3\)
✓ 1
A1 area \(=12.8\) m2
✓ 1
Part (c)
M1 volume \(=12.8\times15\) (ECF from (b))
✓ 1
A1 \(192\) m3
✓ 1
Part (d)
M1 solving \(h(x)=1.2\) to obtain \(x=\pm2\sqrt2\) (\(\pm2.83\))
Common error: When finding the area above a horizontal line, integrate the difference between the curve and the line, not the curve alone.
384IBMA3H-5-19r47-085.9 KinematicsPaper 2Medium11 marks
Calculator
A robotic carriage moves along a straight rail in a data centre. Its displacement from the rail's start point at time \(t\) seconds is \(s=t^3-9t^2+24t\) metres, for \(0\le t\le5\).
(a)
Find the times at which the carriage is instantaneously at rest.
(3)
(b)
Find the total distance travelled by the carriage during the 5 seconds.
(3)
(c)
Find the velocity of the carriage at the instant when its acceleration is zero.
(2)
(d)
Find the intervals of time during which the speed of the carriage is increasing.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to differentiate \(s\)
✓ 1
A1 \(v=3t^2-18t+24\)
✓ 1
A1 \(t=2\) and \(t=4\)
✓ 1
Part (b)
M1 evaluating \(s\) at \(t=0,2,4,5\) (since the direction of motion changes at \(t=2\) and \(t=4\))
✓ 1
A1 \(s(2)=20\), \(s(4)=16\), \(s(5)=20\)
✓ 1
A1 total distance \(=20+4+4=28\) m
✓ 1
Part (c)
M1 \(a=6t-18=0\) so \(t=3\)
✓ 1
A1 \(v(3)=-3\) m s−1
✓ 1
Part (d)
M1 speed increases when \(v\) and \(a\) have the same sign
✓ 1
A1 \(2<t<3\) (both negative)
✓ 1
A1 \(4<t\le5\) (both positive)
✓ 1
Common error: Total distance is not the final displacement: split the motion at each time the velocity is zero and add the magnitudes of the displacements.
385IBMA3H-5-19r47-145.1 Limits and the derivativePaper 2Medium7 marks
Calculator
The electrical power drawn by a data centre is modelled by \(P(t)=40+6\sin\left(\frac{\pi t}{12}\right)\) kW, where \(t\) is the time in hours after midnight, for \(0\le t\le24\).
(a)
Find the energy used, in kWh, between \(t=0\) and \(t=12\). Give your answer to 3 significant figures.
(3)
(b)
Find the total energy used in the full 24 hours.
(2)
(c)
Electricity costs €0.18 per kWh. Find the cost of the energy used in the 24 hours.
M1 \(\int_0^{24}P(t)\,\mathrm{d}t\) (the sine term covers a whole period)
✓ 1
A1 \(960\) kWh
✓ 1
Part (c)
M1 \(960\times0.18\) (ECF from (b))
✓ 1
A1 €172.80
✓ 1
Exam tip: Over one full period the integral of a sine function is zero, so only the constant part contributes to the total.
386IBMA3H-5-19r48-055.8 Stationary points, inflexion and optimizationPaper 2Medium10 marks
Calculator
A rectangular compound for a wind farm control building is to have an area of 800 m2. The side facing the access road has length \(x\) m and the other side has length \(y\) m. Fencing along the road side costs €80 per metre, and fencing along each of the other three sides costs €30 per metre. The total cost of the fencing is €\(C\).
Diagram not to scale
(a)
Show that \(C=110x+\frac{48000}{x}\).
(2)
(b)
Find \(\frac{dC}{dx}\).
(2)
(c)
Find the value of \(x\) that minimises the cost, correct to 3 significant figures.
(2)
(d)
Use the unrounded value of \(x\) to find the minimum cost, to the nearest euro.
(2)
(e)
Justify that this value of \(x\) gives a minimum.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(C=80x+30x+60y\) (three other sides: one of length \(x\), two of length \(y\)) and \(xy=800\)
✓ 1
A1 substitute \(y=\frac{800}{x}\) to obtain \(C=110x+\frac{48000}{x}\) (AG)
✓ 1
Part (b)
M1 attempt to differentiate, with \(\frac{48000}{x}\) written as \(48000x^{-1}\)
✓ 1
A1 \(\frac{dC}{dx}=110-\frac{48000}{x^2}\)
✓ 1
Part (c)
M1 set \(\frac{dC}{dx}=0\), giving \(x^2=\frac{48000}{110}\)
✓ 1
A1 \(x=20.9\) (20.889...)
✓ 1
Part (d)
M1 substitute \(x=\sqrt{\frac{48000}{110}}\) into \(C\)
✓ 1
A1 €4596
✓ 1
Part (e)
M1 \(\frac{d^2C}{dx^2}=\frac{96000}{x^3}\) (or a sign change of \(\frac{dC}{dx}\) from negative to positive)
✓ 1
R1 positive for \(x>0\), hence a minimum
✓ 1
Exam tip: State the unrounded value used for a later part; carrying the full calculator value avoids rounding drift in the final cost.
387IBMA3H-5-19r48-065.14 Implicit differentiation, related rates and optimisationPaper 2Medium10 marks
Calculator
A curve has equation \(x^2+xy+y^2=7\).
(a)
Show that \(\frac{dy}{dx}=-\frac{2x+y}{x+2y}\).
(4)
(b)
Find the equation of the tangent to the curve at the point \((1,2)\), giving your answer in the form \(ax+by=c\) with integer values of \(a\), \(b\) and \(c\).
(3)
(c)
There are two points on the curve where the tangent is horizontal. Find the \(y\)-coordinate of the point where \(x>0\), correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt at implicit differentiation, with a \(\frac{dy}{dx}\) term appearing
A1 rearrange to \(\frac{dy}{dx}=-\frac{2x+y}{x+2y}\) (AG)
✓ 1
Part (b)
M1 substitute \(x=1\), \(y=2\) into the gradient expression
✓ 1
A1 gradient \(=-\frac{4}{5}\)
✓ 1
A1 \(4x+5y=14\)
✓ 1
Part (c)
M1 horizontal tangent means \(2x+y=0\), so \(y=-2x\)
✓ 1
M1 substitute into the curve: \(x^2-2x^2+4x^2=7\), so \(3x^2=7\)
✓ 1
A1 \(x=\sqrt{\frac{7}{3}}=1.53\), so \(y=-2x=-3.06\)
✓ 1
Common error: When differentiating \(xy\) implicitly, both factors depend on \(x\); the product rule gives \(y+x\frac{dy}{dx}\), not just \(y\).
388IBMA3H-5-19r48-075.14 Implicit differentiation, related rates and optimisationPaper 2Medium7 marks
Calculator
The wingtip of a wind turbine blade rotates in a vertical plane. Its height above the ground is \(h\) metres after \(t\) seconds, where \(h=45+30\sin\left(\frac{\pi t}{4}\right)\). Use radians throughout.
(a)
Find \(\frac{dh}{dt}\).
(2)
(b)
Find the rate of change of the height when \(t=1.5\). Give your answer to 3 significant figures.
(2)
(c)
Find the greatest speed at which the tip is moving downwards. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt at the chain rule, with the derivative of \(\sin\) giving \(\cos\left(\frac{\pi t}{4}\right)\)
M1 substitute \(t=1.5\) into the answer to (a): \(7.5\pi\cos\left(\frac{3\pi}{8}\right)\)
✓ 1
A1 \(9.02\ \text{m s}^{-1}\) (ECF from (a))
✓ 1
Part (c)
M1 greatest downward speed occurs when \(\cos\left(\frac{\pi t}{4}\right)=-1\)
✓ 1
A1 \(\frac{dh}{dt}=-7.5\pi\)
✓ 1
A1 \(23.6\ \text{m s}^{-1}\)
✓ 1
Exam tip: Check that the calculator is in radian mode before evaluating any trigonometric derivative.
389IBMA3H-5-19r48-085.14 Implicit differentiation, related rates and optimisationPaper 2Medium6 marks
Calculator
Gravel is poured onto level ground at a wind farm site at a constant rate of \(3\ \text{m}^3\) per minute. The gravel forms a cone of height \(h\) m and base radius \(r\) m, where \(h=2r\) at all times. The volume of a cone is \(V=\frac{1}{3}\pi r^2h\).
Conical heap of gravel. Diagram not to scale
(a)
Show that \(V=\frac{\pi h^3}{12}\).
(2)
(b)
Find the rate at which the height is increasing at the instant when \(h=4\) m. Give your answer to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitute \(r=\frac{h}{2}\) into \(V=\frac{1}{3}\pi r^2h\)
M1 area \(=\frac{1}{2}\times QR\times 5\) (height is the \(y\)-coordinate of \(P\))
✓ 1
A1 \(45.4\)
✓ 1
Exam tip: For a triangle with a vertex above the \(x\)-axis and the other two on it, the base lies on the axis and the height is the \(y\)-coordinate of the apex.
391IBMA3H-5-19r48-115.14 Implicit differentiation, related rates and optimisationPaper 2Medium7 marks
Calculator
A library ladder of length 4 m leans against a vertical wall. The foot of the ladder is \(x\) m from the wall and the top is \(y\) m above the ground. The foot slides away from the wall at a constant \(0.3\ \text{m s}^{-1}\).
Diagram not to scale
(a)
By differentiating \(x^2+y^2=16\) with respect to time, show that \(\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}\).
(2)
(b)
Find the value of \(y\) when \(x=2.4\).
(1)
(c)
Find the rate at which the top of the ladder is moving when \(x=2.4\), stating whether it is moving up or down.
(2)
(d)
Find the value of \(x\), correct to 3 significant figures, for which the top and the foot of the ladder move at the same speed.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 implicit differentiation with respect to \(t\): \(2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\)
✓ 1
A1 rearrange to \(\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}\) (AG)
✓ 1
Part (b)
A1 \(y=\sqrt{16-2.4^2}=3.2\)
✓ 1
Part (c)
M1 substitute \(x=2.4\), \(y=3.2\), \(\frac{dx}{dt}=0.3\) into the result of (a)
✓ 1
A1 \(\frac{dy}{dt}=-0.225\), so the top is moving down at \(0.225\ \text{m s}^{-1}\)
✓ 1
Part (d)
M1 equal speeds means \(\left|\frac{x}{y}\right|=1\), so \(x=y\) and \(2x^2=16\)
✓ 1
A1 \(x=2\sqrt{2}=2.83\)
✓ 1
Common error: A negative rate of change is not a mistake: the sign tells you the direction, so state 'moving down' with the speed.
392AAHL5-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
394AAHL5-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) \(a=2.46\), \(k=0.257\)
395AAHL5-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 \(\sin\theta=\dfrac{h}{s}\)
396AAHL5-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 (\(t=7.34\))
397IBMA2H-053-00-105.1 Limits and the derivativePaper 2Hard13 marks
Calculator
A tank initially contains 100 litres of pure water. Brine with a salt concentration of 0.2 kg per litre flows into the tank at 3 litres per minute. The mixture is kept uniform and flows out at 2 litres per minute. Let S kg be the mass of salt in the tank after t minutes.
(a)
Show that \(\frac{dS}{dt}=0.6-\frac{2S}{100+t}\).
(2)
(b)
Solve the differential equation to show that \(S=0.2(100+t)-\frac{2\times10^5}{(100+t)^2}\).
(5)
(c)
Find the mass of salt in the tank after 50 minutes.
(2)
(d)
Show that the mass of salt in the tank is increasing for all \(t\geq0\).
(2)
(e)
The tank has a capacity of 300 litres. Find the concentration of salt in the tank at the instant it begins to overflow.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 rate in \(=0.2\times3=0.6\) kg per minute and volume \(=100+t\) litres ; A1 rate out \(=2\times\frac{S}{100+t}\), giving the result
✓ 2
Part (b)
M1 integrating factor \(e^{\int\frac{2}{100+t}dt}\) ; A1 \((100+t)^2\) ; M1 \(\frac{d}{dt}\left[S(100+t)^2\right]=0.6(100+t)^2\) and integrates ; A1 \(S(100+t)^2=0.2(100+t)^3+c\) ; A1 uses \(S=0\) at \(t=0\) to give \(c=-2\times10^5\) and the result
✓ 5
Part (c)
M1 substitutes \(t=50\): \(30-\frac{2\times10^5}{150^2}\) ; A1 21.1 kg
✓ 2
Part (d)
M1 \(\frac{dS}{dt}=0.2+\frac{4\times10^5}{(100+t)^3}\) ; R1 both terms are positive for \(t\geq0\), so \(\frac{dS}{dt}>0\)
✓ 2
Part (e)
M1 overflow when \(100+t=300\), i.e. \(t=200\), and uses \(\frac{S}{300}\) ; A1 0.193 kg per litre
✓ 2
Exam tip: In mixing problems the volume changes when inflow and outflow rates differ; the outflow concentration must use the volume at time t, not the initial volume.
398IBMA2H-053-04-105.18 First order differential equationsPaper 2Hard15 marks
Calculator
A small probe is dropped by parachute. Once the parachute opens, at time t = 0, the probe's downward speed is 10 m s−1. For t ≥ 0 its speed v m s−1 satisfies dv/dt = 0.4(25 − v2), where t is in seconds.
(a)
Show that 10 / (25 − v2) = 1/(5 − v) + 1/(5 + v).
(2)
(b)
Hence show that v = 5(3e4t + 1) / (3e4t − 1).
(5)
(c)
Find the value that v approaches as t increases, justifying your answer.
(2)
(d)
Find the time taken for the speed to fall to 6 m s−1.
(2)
(e)
By writing the acceleration as v dv/ds, where s is the distance fallen since the parachute opened, find the distance fallen while the speed decreases from 10 m s−1 to 6 m s−1.
A1 since v > 5, (v + 5)/(v − 5) = 3e4t, rearranged to the given result, AG
✓ 1
Part (c)
R1 as t → ∞, (3e4t + 1)/(3e4t − 1) → 1 because the constant terms become negligible
✓ 1
A1 v → 5 m s−1
✓ 1
Part (d)
M1 (6 + 5)/(6 − 5) = 3e4t, so e4t = 11/3
✓ 1
A1 t = ¼ ln(11/3) = 0.325 s
✓ 1
Part (e)
M1 v dv/ds = 0.4(25 − v2) and separates: ∫ v/(25 − v2) dv = ∫ 0.4 ds
✓ 1
A1 −½ ln|25 − v2| = 0.4s + k
✓ 1
M1 applies v = 10 and v = 6 as limits: 0.4s = ½ ln(75/11)
✓ 1
A1 s = 2.40 m
✓ 1
Common error: Here v > 5, so 5 − v is negative. Dropping the modulus without checking the sign gives (v + 5)/(5 − v) = 3e^(4t), which does not satisfy v = 10 at t = 0.
399IBMA2H-053-07-125.12 Limits, first principles and higher derivativesPaper 2Hard12 marks
Calculator
An electric go-kart starts from rest. Its speed is modelled by \(v(t)=1.2t^2\) m s−1 for \(0\le t\le3\), with \(t\) in seconds. A data logger estimates the distance travelled by dividing \([0,3]\) into \(n\) equal intervals and, for each interval, multiplying the interval length by the speed at the end of that interval. Let \(U_n\) be the resulting estimate. You may use \(\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}\).
(a)
Find \(U_6\).
(2)
(b)
Show that \(U_n=\frac{5.4(n+1)(2n+1)}{n^2}\).
(3)
(c)
Find \(L=\lim_{n\to\infty}U_n\).
(2)
(d)
Find the least value of \(n\) for which \(U_n\) exceeds \(L\) by less than 1 %.
(3)
(e)
Use integration to confirm that \(L\) equals the exact distance travelled.
Common error: Using the speed at the start of each interval gives a lower estimate, with i running from 0 to n−1. Check which endpoint the question specifies before writing the sum.
400IBMA2H-053-01-125.4 Tangents and normalsPaper 2Hard14 marks
Calculator
A curved bypass road follows the curve \(y = x^2\), where \(x\) and \(y\) are in kilometres. An ambulance station is at H\((5, 0)\). An access road is to be built from H to the point P\((a, a^2)\) on the bypass so that its length is as short as possible.
Diagram not to scale
(a)
Write down the gradient of the normal to the curve at P.
(1)
(b)
Show that if the normal at P passes through H, then \(2a^3 + a - 5 = 0\).
(3)
(c)
Explain why this equation has exactly one real solution, and find it.
(3)
(d)
Let \(D(a)\) be the square of the distance HP. Show that \(D'(a) = 0\) leads to the same equation, justify that it gives a minimum, and find the shortest possible length of the access road.
(4)
(e)
The tangent to the curve at this point P meets the \(x\)-axis at Q. Find the area of triangle HPQ.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(-\frac{1}{2a}\)
✓ 1
Part (b)
M1 gradient of HP \(= \frac{a^2 - 0}{a - 5}\)
✓ 1
M1 equates to \(-\frac{1}{2a}\)
✓ 1
A1 \(2a^3 = 5 - a\), leading to \(2a^3 + a - 5 = 0\)
✓ 1
Part (c)
M1 considers \(g(a) = 2a^3 + a - 5\) with \(g'(a) = 6a^2 + 1\)
✓ 1
R1 \(g'(a) > 0\) for all \(a\), so \(g\) is strictly increasing and has exactly one root
✓ 1
A1 \(a = 1.23\) (1.23477...)
✓ 1
Part (d)
M1 \(D(a) = (a - 5)^2 + a^4\), so \(D'(a) = 2(a - 5) + 4a^3\)
✓ 1
A1 \(D'(a) = 0 \iff 2a^3 + a - 5 = 0\)
✓ 1
R1 \(D''(a) = 2 + 12a^2 > 0\), so this is a minimum
✓ 1
A1 shortest length \(= \sqrt{D(1.23477)} = 4.06\) km
A1 area \(= \frac{1}{2}\left(5 - \frac{a}{2}\right)a^2 = 3.34\) km2
✓ 1
Exam tip: The shortest distance from a point to a smooth curve is along a normal to the curve; this links the normal condition and the derivative of the distance to the same equation, which is a useful check.
401IBMA2H-053-08-085.17 Areas with the y-axis and volumes of revolutionPaper 2Hard12 marks
Calculator
The interior of a glass flower vase is modelled by rotating part of the curve \(y=2\ln(x+1)\), for \(0\le y\le 6\), through \(2\pi\) radians about the \(y\)-axis. Here \(y\) is the height above the lowest point of the interior and \(x\) is the radius, both in centimetres.
Cross-section of the vase (the curve rotated about the y-axis); full height 6 cm. Diagram not to scale
(a)
Express \(x\) in terms of \(y\).
(2)
(b)
Find the volume of water the vase holds when full, giving your answer to 3 significant figures.
(3)
(c)
The vase is filled to a depth of \(h\) cm and then contains half of its full volume. Find \(h\).
(4)
(d)
Water is poured into the empty vase at a constant rate of 20 cm3 s−1. Let \(h\) cm be the depth of water at time \(t\) seconds. Find the rate at which the water level is rising when the depth is 4 cm.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{y}{2}=\ln(x+1)\), so \(x+1=e^{y/2}\)
✓ 1
A1 \(x=e^{y/2}-1\)
✓ 1
Part (b)
M1 uses \(V=\pi\int x^2\,dy\)
✓ 1
A1 \(\pi\int_0^6\left(e^{y/2}-1\right)^2dy\)
✓ 1
A1 \(V=1040\) cm3 (1043.28…)
✓ 1
Part (c)
M1 sets an integral with upper limit \(h\) equal to half the full volume
Exam tip: The rate of change of volume with depth equals the area of the water surface, \(\pi x^2\). This gives \(\frac{dV}{dh}\) directly, without differentiating an integral.
402IBMA2H-053-02-115.14 Implicit differentiation, related rates and optimisationPaper 2Hard14 marks
Calculator
In rugby, the goalposts are 5.6 m apart on the goal line. A try is scored at point T on the goal line, outside the posts, 10 m from the nearer post and 15.6 m from the further post. The conversion kick is taken from a point P on the line through T perpendicular to the goal line, with TP = \(x\) m. Let \(\theta\) be the angle subtended at P by the two posts.
Diagram not to scale
(a)
Show that \(\theta = \arctan\left(\frac{15.6}{x}\right) - \arctan\left(\frac{10}{x}\right)\).
(2)
(b)
Show that \(\frac{d\theta}{dx} = \frac{10}{x^2 + 100} - \frac{15.6}{x^2 + 243.36}\).
(3)
(c)
Find the value of \(x\) that maximises \(\theta\), justifying that it is a maximum.
(4)
(d)
Find the maximum value of \(\theta\) in degrees.
(2)
(e)
The kicker, initially at \(x = 20\), walks directly away from T at 2 m s−1. Find the rate of change of \(\theta\) at this instant.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 angle between PT and line to further post is \(\arctan\frac{15.6}{x}\)
✓ 1
A1 subtract angle to nearer post \(\arctan\frac{10}{x}\)
✓ 1
Part (b)
M1 chain rule on \(\arctan\left(\frac{k}{x}\right)\)
M1 substitute, using \((0.8)^t=e^{t\ln0.8}=e^{-1}\)
✓ 1
A1 \(C=\frac{6}{\ln1.25}\times e^{-1}=\frac{6}{e\ln1.25}\) AG
✓ 1
Part (c)
M1 differentiate \(C'(t)\) using the product rule
✓ 1
A1 \(C''(t)=6(0.8)^t\ln0.8\,(2+t\ln0.8)\)
✓ 1
Part (d)
M1 set \(C''(t)=0\)
✓ 1
A1 \(t=\frac{2}{\ln1.25}=8.96\) hours
✓ 1
R1 \(\ln0.8<0\), so \(C''<0\) before and \(C''>0\) after, hence \(C'\) has a minimum there
✓ 1
A1 \(C'=6e^{-2}(1-2)=-0.812\) mg L−1 h−1
✓ 1
Common error: Finding where \(C''=0\) but not checking the sign change: a zero of the second derivative alone does not prove the rate of decrease is greatest.
404IBMA3H-5-18r9-065.3 Differentiating polynomials and integer powersPaper 2Hard12 marks
Calculator
The response of a heart-rate sensor is modelled by \(f(x)=xe^{-x^2/2}\) for \(x\ge0\), where \(x\) is the time in seconds.
(a)
Show that \(f'(x)=(1-x^2)e^{-x^2/2}\).
(3)
(b)
Show that \(f''(x)=x(x^2-3)e^{-x^2/2}\).
(3)
(c)
Find the exact coordinates of the local maximum of the graph of \(f\), and justify that it is a maximum.
(3)
(d)
The graph has one point of inflexion for \(x>0\). Find its \(x\)-coordinate, and its \(y\)-coordinate to 3 significant figures.
Common error: When the region lies between two curves, subtract the squares of the functions, not the square of the difference.
406IBMA3H-5-18r9-145.16 Integration by substitution and by partsPaper 2Hard9 marks
Calculator
The spout of a ceramic honey pot is modelled by rotating the region under the curve \(y=xe^{x}\), for \(0\le x\le1\), through \(360^\circ\) about the \(x\)-axis. Lengths are in decimetres, so \(1\ \text{dm}^3=1\) litre.
Cross-section: the region under the curve is rotated about the x-axis
(a)
Show that the volume of the spout is \(V=\pi\displaystyle\int_0^1x^2e^{2x}\,dx\).
(2)
(b)
Use integration by parts twice to find \(\displaystyle\int x^2e^{2x}\,dx\).
(4)
(c)
Find the exact value of \(V\).
(2)
(d)
Write \(V\) in litres, to 3 significant figures.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(V=\pi\int y^2\,dx\)
✓ 1
A1 \(y^2=x^2e^{2x}\) and limits \(0\) to \(1\) (AG)
Exam tip: After integrating by parts, differentiate your answer to check that you recover x squared times e to the 2x.
407IBMA3H-5-18r18-105.9 KinematicsPaper 2Hard12 marks
Calculator
Sediment in a volcanic crater lake is modelled by the mass \(M\) tonnes in the lake at time \(t\) hours, where \(\dfrac{dM}{dt}+0.2M=30e^{-0.1t}\), \(t\ge0\), and \(M=0\) when \(t=0\).
(a)
Use an integrating factor to show that \(M=300\left(e^{-0.1t}-e^{-0.2t}\right)\).
(5)
(b)
Find the time at which \(M\) is greatest, and the greatest mass. Give the time to 3 significant figures.
(3)
(c)
After reaching its greatest value, \(M\) decreases. Find the time at which \(M\) has fallen to \(50\) tonnes. Give your answer to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 integrating factor \(e^{0.2t}\)
✓ 1
M1 multiply through and recognise \(\dfrac{d}{dt}\left(Me^{0.2t}\right)=30e^{0.1t}\)
M1 let \(u=e^{-0.1t}\): \(300(u-u^2)=50\), i.e. \(6u^2-6u+1=0\)
✓ 1
A1 \(u=\dfrac{3\pm\sqrt3}{6}\) (\(0.7887\) or \(0.2113\))
✓ 1
M1 after the maximum \(u<\dfrac12\), so \(u=\dfrac{3-\sqrt3}{6}\); then \(t=-10\ln u\)
✓ 1
A1 \(t=15.5\) hours (\(15.54\ldots\))
✓ 1
Exam tip: After finding both roots of a quadratic in \(e^{-0.1t}\), use the context (here: after the maximum) to decide which one is the required time.
408IBMA3H-5-18r18-115.14 Implicit differentiation, related rates and optimisationPaper 2Hard10 marks
Calculator
A cargo ship sails along a straight path at a constant speed of \(18\) km h−1. The path is parallel to a straight coastline. A lighthouse \(L\) stands on the coast, with the nearest point of the ship's path to \(L\) at a distance of \(4\) km. Let \(x\) km be the distance the ship has sailed from that nearest point (so \(x\) increases at \(18\) km h−1), let \(\theta\) be the angle between the line from \(L\) to the nearest point and the line from \(L\) to the ship, and let \(s\) km be the distance from \(L\) to the ship.
Diagram not to scale
(a)
Show that \(\dfrac{d\theta}{dt}=4.5\cos^2\theta\).
(3)
(b)
Find the rate of change of \(\theta\) when \(x=3\).
(2)
(c)
Find the rate at which \(s\) is increasing when \(x=3\).
(3)
(d)
At a later time, \(\dfrac{d\theta}{dt}=1\) radian per hour. Find the value of \(\theta\) at this time, in radians, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(x=4\tan\theta\)
✓ 1
M1 differentiate with respect to \(t\) (chain rule): \(\dfrac{dx}{dt}=4\sec^2\theta\,\dfrac{d\theta}{dt}\)
✓ 1
A1 \(18=4\sec^2\theta\,\dfrac{d\theta}{dt}\), so \(\dfrac{d\theta}{dt}=4.5\cos^2\theta\) (AG)
✓ 1
Part (b)
M1 \(\tan\theta=\dfrac34\), so \(\cos\theta=\dfrac45\)
✓ 1
A1 \(\dfrac{d\theta}{dt}=4.5\times\dfrac{16}{25}=2.88\) radians per hour
A1 \(s=5\), so \(\dfrac{ds}{dt}=\dfrac{3\times18}{5}=10.8\) km h−1
✓ 1
Part (d)
M1 \(4.5\cos^2\theta=1\), so \(\cos\theta=\sqrt{\dfrac29}\) (\(\theta\) acute)
✓ 1
A1 \(\theta=1.08\) radians (\(1.0799\ldots\))
✓ 1
Common error: Substituting \(x=3\) before differentiating, which turns the variable into a constant and loses the rate of change. Differentiate the general relationship first.
409IBMA3H-5-18r19-075.8 Stationary points, inflexion and optimizationPaper 2Hard11 marks
Calculator
A fish hatchery needs a closed cylindrical tank of volume \(2\) m3, with base radius \(r\) metres and height \(h\) metres. The circular base and lid cost €30 per square metre and the curved side costs €18 per square metre.
Closed cylindrical tank. Diagram not to scale
(a)
Find an expression for \(h\) in terms of \(r\).
(2)
(b)
Show that the total cost of the tank is \(C = 60\pi r^2 + \frac{72}{r}\) euros.
(3)
(c)
Find the value of \(r\) that minimises \(C\). Give your answer to 3 significant figures.
(4)
(d)
Show that this value of \(r\) gives a minimum, and find the minimum cost. Use the unrounded value of \(r\) and give your answer in euros to 2 decimal places.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\pi r^2h = 2\)
✓ 1
A1 \(h = \frac{2}{\pi r^2}\)
✓ 1
Part (b)
A1 base and lid: \(30 \times 2\pi r^2 = 60\pi r^2\)
✓ 1
M1 curved area \(2\pi rh\) with \(h\) from (a), giving \(\frac{4}{r}\)
R1 \(\frac{d^2C}{dr^2} = 120\pi + \frac{144}{r^3} > 0\) for \(r > 0\), so a minimum
✓ 1
A1 \(C = \text{€}187.54\)
✓ 1
Common error: Remember that the lid is also costed: a closed tank has two circular ends. Forgetting one end changes the equation for the optimum radius.
410IBMA3H-5-18r19-145.3 Differentiating polynomials and integer powersPaper 2Hard10 marks
Calculator
A curve has equation \(x^3 + y^3 = 9xy\). The point A\((2, 4)\) lies on the curve.
(a)
Show that \(\frac{dy}{dx} = \frac{3y - x^2}{y^2 - 3x}\).
(3)
(b)
Find the equation of the tangent to the curve at A. Give your answer in the form \(ax + by + c = 0\) where \(a\), \(b\), \(c\) are integers.
(2)
(c)
The curve has a horizontal tangent at a point where \(x > 0\). Find the coordinates of this point to 3 significant figures, and verify that the denominator of \(\frac{dy}{dx}\) is not zero there.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 differentiate implicitly: \(3x^2 + 3y^2\frac{dy}{dx}\) on the left
✓ 1
A1 \(9y + 9x\frac{dy}{dx}\) on the right (product rule)
R1 \(y^2 - 3x = 22.68 - 11.34 \ne 0\) (using unrounded values), so the tangent is horizontal
✓ 1
Common error: \(\frac{dy}{dx} = 0\) only when the numerator is zero and the denominator is not. Check the denominator at the point you find.
411IBMA3H-5-19r20-115.14 Implicit differentiation, related rates and optimisationPaper 2Hard14 marks
Calculator
A solar heating system uses a hot-water store. Its inner shape is formed by rotating the region R about the y-axis through \(360^\circ\), where R is bounded by the curve \(y=\ln x\), the y-axis, and the lines \(y=0\) and \(y=1\). Lengths are in metres, and the y-axis is vertical with the base of the store at \(y=0\).
Region R is shaded
(a)
Find the area of R. Give your answer to three significant figures.
(3)
(b)
Show that the volume of the store is \(\dfrac{\pi(e^{2}-1)}{2}\) m3.
(4)
(c)
When the water has depth h metres, the volume of water is \(V(h)\). Find the depth at which the store is half full by volume. Give your answer to three significant figures.
(4)
(d)
Water is pumped into the store at a constant rate of 0.5 m3 per minute. Find the rate at which the depth is increasing when \(h=0.5\). Give your answer to three significant figures.
Common error: For the half-full depth, halve the volume, not the height. The relationship between \(V\) and h is not linear.
412IBMA3H-5-19r20-125.1 Limits and the derivativePaper 2Hard14 marks
Calculator
The number of people, N, infected in a town t days after an outbreak begins is modelled by \(N(t)=\dfrac{1000}{1+99e^{-0.5t}}\), for \(t\ge0\).
(a)
Find \(N(0)\).
(1)
(b)
Find an expression for \(\dfrac{dN}{dt}\).
(3)
(c)
Show that \(\dfrac{dN}{dt}=\dfrac{N(1000-N)}{2000}\).
(3)
(d)
Hence, or otherwise, find the greatest rate at which people become infected and the time at which this occurs. Give the time to three significant figures.
(4)
(e)
Find the time at which the number infected first reaches 900. Give your answer to three significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(N(0)=\dfrac{1000}{100}=10\)
✓ 1
Part (b)
M1 attempt at chain rule or quotient rule
✓ 1
A1 derivative of the denominator \(-49.5e^{-0.5t}\) used correctly
A1 \(\dfrac{N(1000-N)}{2000}=\dfrac{49500e^{-0.5t}}{\left(1+99e^{-0.5t}\right)^{2}}\), which equals \(\dfrac{dN}{dt}\) (AG)
✓ 1
Part (d)
M1 treats \(\frac{dN}{dt}=\frac{1000N-N^{2}}{2000}\) as a function of N and maximises, giving \(N=500\)
✓ 1
A1 greatest rate \(=\dfrac{500\times500}{2000}=125\) people per day
✓ 1
M1 solves \(1+99e^{-0.5t}=2\)
✓ 1
A1 \(t=2\ln 99=9.19\) days
✓ 1
Part (e)
M1 \(\dfrac{1000}{1+99e^{-0.5t}}=900\)
✓ 1
A1 \(e^{-0.5t}=\dfrac{1}{891}\)
✓ 1
A1 \(t=2\ln 891=13.6\) days
✓ 1
Exam tip: In "show that" questions it is often easier to start from the target form and simplify both sides to the same expression.
413IBMA3H-5-19r26-125.19 Maclaurin seriesPaper 2Hard10 marks
Calculator
Let \(g(x)=\frac{1}{(1-x)(1+2x)}\) for \(|x|<\frac12\).
(a)
Find constants \(A\) and \(B\) such that \(g(x)=\frac{A}{1-x}+\frac{B}{1+2x}\).
(3)
(b)
Hence find the Maclaurin series for \(g(x)\) up to and including the term in \(x^3\).
(4)
(c)
Use the first four terms of this series to estimate \(\int_0^{0.1}g(x)\,dx\). Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(1=A(1+2x)+B(1-x)\)
✓ 1
A1 \(A=\frac13\)
✓ 1
A1 \(B=\frac23\)
✓ 1
Part (b)
M1 use of the standard series for \((1+u)^{-1}\) on each fraction
✓ 1
A1 \(\frac{1}{1-x}=1+x+x^2+x^3+\ldots\)
✓ 1
A1 \(\frac{1}{1+2x}=1-2x+4x^2-8x^3+\ldots\)
✓ 1
A1 \(g(x)\approx1-x+3x^2-5x^3\) (ECF from (a))
✓ 1
Part (c)
M1 integrate the series term by term
✓ 1
A1 \(x-\frac{x^2}{2}+x^3-\frac{5x^4}{4}\)
✓ 1
A1 \(0.095875\ldots=0.0959\)
✓ 1
Exam tip: Use the unrounded series coefficients in the integration and round only the final answer to the accuracy requested.
414IBMA3H-5-20r27-075.17 Areas with the y-axis and volumes of revolutionPaper 2Hard11 marks
Calculator
A jelly mould is formed by rotating the curve \(x=2\mathrm{e}^{0.3y}\), for \(0\le y\le6\), through \(360°\) about the \(y\)-axis. Lengths are in cm, and \(y\) measures the height above the base of the mould.
The shaded region (right half of the cross-section) is rotated about the y-axis
(a)
Find the area of the region bounded by the curve, the \(y\)-axis and the lines \(y=0\) and \(y=6\). Give your answer in cm2 to 3 significant figures.
(3)
(b)
Find the volume of the mould, in cm3 to 3 significant figures.
(3)
(c)
The mould is filled to a depth of \(h\) cm. Show that the volume \(V\) of jelly in the mould satisfies \(\dfrac{dV}{dh}=4\pi\mathrm{e}^{0.6h}\).
(2)
(d)
Jelly is poured in at a constant rate of 30 cm3 s−1. Find the rate at which the depth is increasing when \(h=4\). Give your answer in cm s−1 to 3 significant figures.
Exam tip: Each new derivative is built from the previous one; evaluate at x = 0 using the values already found, rather than solving for y.
416IBMA3H-5-19r33-125.9 KinematicsPaper 2Hard15 marks
Calculator
A Formula 1 car travels in a straight line along the pit lane towards its pit box. For \(0\le t\le 15\), its velocity is modelled by \(v(t)=20te^{-t/3}\) m s−1, where \(t\) is in seconds. The car is at its garage door, \(s=0\), when \(t=0\).
(a)
Show that the acceleration of the car is \(a(t)=20e^{-t/3}\left(1-\dfrac{t}{3}\right)\).
(3)
(b)
Find the maximum speed of the car, giving your answer to 3 significant figures.
(2)
(c)
Use integration by parts to show that the distance of the car from its garage door is \(s(t)=180-60(t+3)e^{-t/3}\).
(4)
(d)
Find the distance travelled in the first 9 seconds. Give your answer to 3 significant figures.
(2)
(e)
The car decelerates for \(t>3\). Show that the magnitude of the deceleration is greatest when \(t=6\), and find its value to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use the product rule on \(20t\,e^{-t/3}\)
✓ 1
A1 \(a(t)=20e^{-t/3}-\frac{20}{3}te^{-t/3}\)
✓ 1
A1 factorises to \(20e^{-t/3}\left(1-\frac{t}{3}\right)\) (AG)
✓ 1
Part (b)
M1 \(a(t)=0\) giving \(t=3\)
✓ 1
A1 \(v(3)=60e^{-1}=22.1\) m s−1
✓ 1
Part (c)
M1 integration by parts on \(\int 20te^{-t/3}\,dt\) with \(u=20t\)
✓ 1
A1 \(-60te^{-t/3}+\int 60e^{-t/3}\,dt\)
✓ 1
A1 \(-60te^{-t/3}-180e^{-t/3}+C\)
✓ 1
A1 use of \(s(0)=0\) giving \(C=180\) and so \(s(t)=180-60(t+3)e^{-t/3}\) (AG)
✓ 1
Part (d)
M1 substitute \(t=9\): \(180-720e^{-3}\)
✓ 1
A1 \(144\) m
✓ 1
Part (e)
M1 attempt to differentiate \(a(t)\) using the product rule
✓ 1
A1 \(a'(t)=\frac{20}{9}e^{-t/3}(t-6)\), which is zero at \(t=6\)
✓ 1
R1 \(a'(t)<0\) for \(t<6\) and \(a'(t)>0\) for \(t>6\) (or \(a''(6)>0\)), so \(a\) has a minimum, i.e. greatest deceleration, at \(t=6\)
✓ 1
A1 \(a(6)=-20e^{-2}\), so the greatest deceleration is \(2.71\) m s−2
✓ 1
Exam tip: To show that a greatest value occurs, justify it with a sign change of the derivative or the sign of the second derivative; finding the stationary point alone is not enough.
417IBMA3H-5-19r33-135.15 Further derivatives and integrals; partial fractionsPaper 2Hard13 marks
Calculator
In a beekeeping study, the mass of honey produced per hive is modelled by \(Y(x)=\dfrac{4\ln x}{x^{2}}\) kg, where \(x\ge1\) is the number of hundreds of hives sharing the same meadow.
(a)
Show that \(Y'(x)=\dfrac{4(1-2\ln x)}{x^{3}}\).
(3)
(b)
Given that \(Y\) has a maximum value, find the exact value of \(x\) at which it occurs and the exact maximum value of \(Y\).
(3)
(c)
Show that \(Y''(x)=\dfrac{4(6\ln x-5)}{x^{4}}\).
(3)
(d)
Find the exact value of \(x\) at which the graph of \(Y\) has a point of inflection, and justify that it is a point of inflection.
(3)
(e)
Find the value of \(Y\) at the point of inflection, to 3 significant figures.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use the quotient rule (or product rule with \(x^{-2}\))
M1 integration by parts on \(\int ue^{-u}\,du\) with \(u\) differentiated
✓ 1
A1 \(-ue^{-u}+\int e^{-u}\,du=-(u+1)e^{-u}\)
✓ 1
A1 \(\left[-(u+1)e^{-u}\right]_0^1=1-2e^{-1}\)
✓ 1
A1 \(\frac{1}{2}-e^{-1}\)
✓ 1
Part (c)
M1 differentiate using the product rule
✓ 1
A1 \(I'(x)=x^{2}e^{-x^{2}}\left(3-2x^{2}\right)\)
✓ 1
A1 \(x=\sqrt{\frac{3}{2}}=1.22\) cm
✓ 1
Exam tip: For a substitution, write \(dx\) in terms of \(du\) and check that every \(x\) has been replaced, including in the limits.
419IBMA3H-5-20r36-125.19 Maclaurin seriesPaper 2Hard9 marks
Calculator
Let f(x) = ln(cos x), for values of x close to 0.
(a)
Find the Maclaurin series of f(x) up to and including the term in x4.
(5)
(b)
Hence find limx→0 (ln(cos x) + x2/2)/x4.
(2)
(c)
Use your series to estimate ln(cos 0.2). Give your answer to 4 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses cos x = 1 − x2/2 + x4/24
✓ 1
M1 applies ln(1 + u) = u − u2/2 + … with u = −x2/2 + x4/24
✓ 1
A1 u2 = x4/4 + higher powers
✓ 1
A1 coefficient of x2 is −1/2
✓ 1
A1 coefficient of x4 is 1/24 − 1/8 = −1/12
✓ 1
Part (b)
M1 substitutes the series, leaving −x4/12 + higher powers in the numerator
✓ 1
A1 −1/12
✓ 1
Part (c)
M1 substitutes x = 0.2: −0.02 − 0.0016/12 (ECF from (a))
✓ 1
A1 −0.02013
✓ 1
Common error: When squaring u = −x²/2 + x⁴/24, keep only terms up to x⁴; the cross term and (x⁴/24)² are of higher order and must not be included.
420IBMA3H-5-20r36-145.9 KinematicsPaper 2Hard11 marks
Calculator
A scenery piece on a theatre stage moves along a straight rail. Its velocity at time t seconds, for t ≥ 0, is v(t) = 6te−t2/2 m s−1. Its displacement from its starting point is s metres, with s = 0 when t = 0.
(a)
Find the time at which the acceleration of the scenery piece is zero.
(4)
(b)
Find an expression for s in terms of t.
(4)
(c)
Find the time at which the scenery piece is 5 m from its starting point. Give your answer to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses the product rule on v(t)
✓ 1
A1 a = 6e−t2/2 − 6t2e−t2/2 = 6(1 − t2)e−t2/2
✓ 1
M1 sets a = 0 (the exponential factor is never zero)
✓ 1
A1 t = 1
✓ 1
Part (b)
M1 recognises the reverse chain rule: ∫ 6te−t2/2 dt
✓ 1
A1 −6e−t2/2 + C
✓ 1
M1 uses s(0) = 0 to find C = 6
✓ 1
A1 s = 6 − 6e−t2/2
✓ 1
Part (c)
M1 sets 6 − 6e−t2/2 = 5 (ECF from (b))
✓ 1
M1 e−t2/2 = 1/6, so t2 = 2 ln 6
✓ 1
A1 t = 1.89 s
✓ 1
Exam tip: If the derivative of the exponent appears as a factor outside the exponential, think reverse chain rule: here the derivative of −t²/2 is −t, which matches the t in v(t).
421IBMA3H-5-20r37-135.8 Stationary points, inflexion and optimizationPaper 2Hard12 marks
Calculator
A grain silo is a vertical cylinder of radius r metres and wall height h metres, topped by a hemispherical roof of the same radius, and it stands on the ground with no floor. Its total volume is 512π/3 m3, so that πr2h + (2/3)πr3 = 512π/3. The curved wall costs €40 per m2 and the roof costs €80 per m2.
Cross-section of the silo. Diagram not to scale
(a)
Show that the total cost, in euros, is C = 40960π/(3r) + (320π/3)r2.
(4)
(b)
Find the value of r for which the cost is a minimum.
(4)
(c)
Justify that this value of r gives a minimum cost.
(2)
(d)
Find the minimum cost, correct to the nearest euro.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 C = 40(2πrh) + 80(2πr2)
✓ 1
M1 solve the volume equation for h, or for 2πrh = 2V/r − (4/3)πr2
✓ 1
M1 substitute into the cost expression
✓ 1
A1 C = 80V/r + (320π/3)r2 = 40960π/(3r) + (320π/3)r2
✓ 1
Part (b)
A1 dC/dr = −40960π/(3r2) + (640π/3)r
✓ 1
M1 set dC/dr = 0
✓ 1
A1 r3 = 64
✓ 1
A1 r = 4
✓ 1
Part (c)
A1 d2C/dr2 = 81920π/(3r3) + 640π/3
✓ 1
R1 positive for r = 4, so a minimum
✓ 1
Part (d)
M1 C(4) = 10240π/3 + 5120π/3 = 5120π
✓ 1
A1 €16085
✓ 1
Exam tip: Eliminate h using the volume condition before differentiating, and keep π exact until the final answer.
422IBMA3H-5-20r37-145.8 Stationary points, inflexion and optimizationPaper 2Hard13 marks
Calculator
Two aircraft, treated as points moving in the same horizontal plane, approach a crossing point O on perpendicular straight routes. At time t = 0 aircraft A is 30 000 m from O flying towards it at 240 m s−1, and aircraft B is 50 000 m from O flying towards it at 320 m s−1. Distances are measured positive before an aircraft reaches O. Let D metres be the distance between the aircraft at time t seconds.
Positions at t = 0. Diagram not to scale
(a)
Write down expressions for the distances of A and B from O at time t, and hence an expression for D2.
(2)
(b)
Find the time at which the aircraft are closest together.
(4)
(c)
Find the minimum distance between the aircraft and justify that it is a minimum.
(3)
(d)
Let a and b be the signed distances of A and B from O. When t = 125 s, A is at O. Use related rates to find dD/dt at this instant.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 A: 30000 − 240t and B: 50000 − 320t
✓ 1
A1 D2 = (30000 − 240t)2 + (50000 − 320t)2
✓ 1
Part (b)
M1 differentiate D2 with respect to t
✓ 1
A1 −480(30000 − 240t) − 640(50000 − 320t)
✓ 1
M1 set equal to zero
✓ 1
A1 t = 145 s
✓ 1
Part (c)
M1 substitute t = 145: distances −4800 and 3600
✓ 1
A1 D = 6000 m
✓ 1
R1 second derivative of D2 is 320 000 > 0, so a minimum
✓ 1
Part (d)
M1 D2 = a2 + b2, so D dD/dt = a da/dt + b db/dt
✓ 1
A1 a = 0, b = 10 000, db/dt = −320
✓ 1
A1 D = 10 000
✓ 1
A1 dD/dt = −320 m s−1
✓ 1
Exam tip: Minimising D² instead of D gives the same time and avoids a square root in the derivative.
423IBMA3H-5-19r38-125.8 Stationary points, inflexion and optimizationPaper 2Hard12 marks
Calculator
A mountaineer is at point \(A\) on a flat glacier, 4 km from a straight, level track. The nearest point on the track to \(A\) is \(F\). A mountain hut \(H\) is on the track, 10 km from \(F\). The mountaineer walks in a straight line over the glacier to a point \(P\) on the track, where \(FP=x\) km and \(0\le x\le10\), then walks along the track to \(H\). The speed on the glacier is \(3\) km h-1 and the speed on the track is \(5\) km h-1. The total time taken is \(T\) hours.
Diagram not to scale
(a)
Show that \(T=\frac{\sqrt{16+x^{2}}}{3}+\frac{10-x}{5}\).
(2)
(b)
Find \(\frac{dT}{dx}\).
(3)
(c)
Show that \(T\) has a stationary point when \(x=3\).
(3)
(d)
Show that this stationary point gives the minimum value of \(T\).
(2)
(e)
Find the minimum time, giving your answer in minutes.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(AP=\sqrt{16+x^2}\) by Pythagoras, with time \(=\frac{\text{distance}}{\text{speed}}\)
✓ 1
A1 glacier time \(\frac{\sqrt{16+x^2}}{3}\) plus track time \(\frac{10-x}{5}\)
Exam tip: After squaring to remove a square root, check each solution in the original equation, and confirm that it lies within the allowed domain.
424IBMA3H-5-19r38-135.1 Limits and the derivativePaper 2Hard12 marks
Calculator
A vat contains 100 litres of pure water. A syrup solution of concentration 0.4 kg per litre flows in at 3 litres per minute. The mixture is stirred continuously and leaves the vat at 1 litre per minute. Let \(S\) kg be the mass of dissolved sugar in the vat \(t\) minutes after the flow begins.
(a)
Show that \(\frac{dS}{dt}=1.2-\frac{S}{100+2t}\).
(2)
(b)
Use an integrating factor to solve the differential equation, expressing \(S\) in terms of \(t\).
(6)
(c)
Find the mass of sugar in the vat when \(t=150\).
(2)
(d)
State the value approached by the concentration \(\frac{S}{100+2t}\), in kg per litre, as \(t\to\infty\), and justify your answer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 rate in \(=3\times0.4=1.2\) kg per minute
✓ 1
A1 volume is \(100+2t\) litres, so rate out \(=1\times\frac{S}{100+2t}\)
✓ 1
Part (b)
M1 standard form \(\frac{dS}{dt}+\frac{1}{100+2t}S=1.2\)
A1 uses \(S=0\) at \(t=0\) to obtain \(C=-400\), so \(S=0.4(100+2t)-\frac{400}{\sqrt{100+2t}}\)
✓ 1
Part (c)
M1 \(100+2t=400\), so \(S=160-\frac{400}{20}\)
✓ 1
A1 \(S=140\) kg
✓ 1
Part (d)
A1 0.4 kg per litre
✓ 1
R1 concentration \(=0.4-\frac{400}{(100+2t)^{3/2}}\), and the second term tends to 0
✓ 1
Exam tip: In an integrating-factor question, the factor can be written without a constant: the constant of integration is included when you integrate the right-hand side.
425IBMA3H-5-20r45-125.18 First order differential equationsPaper 2Hard12 marks
Calculator
A robot arm's position error \(y\) (mm) at time \(x\) seconds satisfies \(\frac{dy}{dx} = x + y\), with \(y = 1\) when \(x = 0\).
(a)
Use Euler's method with a step length of 0.2 to estimate \(y\) when \(x = 0.6\).
(3)
(b)
Solve the differential equation, giving \(y\) in terms of \(x\).
(6)
(c)
Find the difference between the exact value of \(y\) at \(x = 0.6\) and the Euler estimate from part (a), giving your answer to 3 significant figures.
(2)
(d)
Suggest one change to the method that would reduce this difference.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(y_{n+1} = y_n + 0.2\,(x_n + y_n)\)
✓ 1
A1 \(y(0.2) \approx 1.2\) and \(y(0.4) \approx 1.48\)
✓ 1
A1 \(y(0.6) \approx 1.856\)
✓ 1
Part (b)
M1 rewrite as \(\frac{dy}{dx} - y = x\) and use the integrating factor \(e^{-x}\)
✓ 1
A1 \(\frac{d}{dx}\left(ye^{-x}\right) = xe^{-x}\)
✓ 1
M1 attempt to integrate \(xe^{-x}\) by parts
✓ 1
A1 \(ye^{-x} = -xe^{-x} - e^{-x} + C\)
✓ 1
M1 use of \(y = 1\) at \(x = 0\) to find \(C = 2\)
✓ 1
A1 \(y = 2e^x - x - 1\)
✓ 1
Part (c)
A1 exact value \(2e^{0.6} - 1.6 = 2.0442\ldots\)
✓ 1
A1 difference = 0.188 (ECF from (a))
✓ 1
Part (d)
R1 use a smaller step length
✓ 1
Common error: In Euler's method use the x-value at the start of each step in the gradient formula, not at the end.
426IBMA3H-5-20r45-145.1 Limits and the derivativePaper 2Hard11 marks
Calculator
A fishing fleet's catch rate is modelled by \(R(t) = Ae^{-0.3t}\) tonnes per day, where \(t\) is the time in days since the start of the season and \(A\) is a positive constant. The total catch in the first 10 days is 80 tonnes.
(a)
Show that \(A = \frac{24}{1 - e^{-3}}\).
(3)
(b)
Find the value of \(A\) to 4 significant figures.
(1)
(c)
Find the total catch between \(t = 3\) and \(t = 6\), giving your answer to 3 significant figures. Use the unrounded value of \(A\).
(2)
(d)
Find the time at which the total catch reaches 60 tonnes, giving your answer to 3 significant figures. Use the unrounded value of \(A\).
(3)
(e)
The fleet returns to port when the catch rate falls below 5 tonnes per day. Find the time at which this happens, giving your answer to 3 significant figures. Use the unrounded value of \(A\).
Common error: Keep the unrounded value of A for every later part; rounding it early can change the third significant figure of the final answers.
427IBMA3H-5-19r47-105.15 Further derivatives and integrals; partial fractionsPaper 2Hard10 marks
Calculator
Let \(f(x)=\frac{\ln x}{x^2}\), for \(x>0\).
(a)
Show that \(f'(x)=\frac{1-2\ln x}{x^3}\).
(3)
(b)
Find the coordinates of the stationary point of the graph of \(f\). Give your answers to 3 significant figures.
(3)
(c)
The second derivative of \(f\) is \(f''(x)=\frac{6\ln x-5}{x^4}\). Show that the stationary point is a local maximum.
(1)
(d)
Find the coordinates of the point of inflection. Give your answers to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of the quotient rule (or product rule with \(x^{-2}\))
✓ 1
A1 \(f'(x)=\frac{\frac1x\cdot x^2-2x\ln x}{x^4}\)
✓ 1
A1 simplification to \(\frac{x-2x\ln x}{x^4}=\frac{1-2\ln x}{x^3}\)
✓ 1
Part (b)
M1 setting \(f'(x)=0\) so \(\ln x=\frac12\)
✓ 1
A1 \(x=\mathrm{e}^{1/2}=1.65\)
✓ 1
A1 \(y=\frac{1}{2\mathrm{e}}=0.184\)
✓ 1
Part (c)
R1 \(f''(\mathrm{e}^{1/2})=\frac{3-5}{\mathrm{e}^{2}}<0\), so a local maximum
✓ 1
Part (d)
M1 setting \(f''(x)=0\) so \(\ln x=\frac56\)
✓ 1
A1 \(x=\mathrm{e}^{5/6}=2.30\)
✓ 1
A1 \(y=\frac{5/6}{\mathrm{e}^{5/3}}=0.157\)
✓ 1
Common error: Give coordinates as a pair: after finding x from the derivative, substitute it back into f (not into f') to find y.
428IBMA3H-5-19r47-115.9 KinematicsPaper 2Hard12 marks
Calculator
A survey drone flies along a straight line over an excavation site. Its velocity at time \(t\) seconds is \(v=4t\mathrm{e}^{-t/2}\) m s−1, for \(t\ge0\). At \(t=0\) the drone is at the start of the line, so its displacement is \(s=0\).
(a)
Find the time at which the velocity is greatest, and the greatest velocity.
(4)
(b)
Use integration by parts to show that the displacement is \(s=16-(8t+16)\mathrm{e}^{-t/2}\).
(4)
(c)
Find the displacement at \(t=4\). Give your answer to 3 significant figures.
(2)
(d)
Find the value that \(s\) approaches as \(t\to\infty\), and state what this means for the drone.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of the product rule to find \(\frac{\mathrm{d}v}{\mathrm{d}t}\)
R1 \(s(0)=0\) gives \(c=16\), hence the given result
✓ 1
Part (c)
M1 substitution of \(t=4\), \(s=16-48\mathrm{e}^{-2}\)
✓ 1
A1 \(9.50\) m
✓ 1
Part (d)
A1 \(s\to16\) m since \((8t+16)\mathrm{e}^{-t/2}\to0\)
✓ 1
A1 the drone approaches a point 16 m from its starting point and never goes beyond it
✓ 1
Exam tip: In a 'show that' question, quote the condition used to fix the constant of integration; that step carries the final reasoning mark.
429IBMA3H-5-19r48-125.8 Stationary points, inflexion and optimizationPaper 2Hard12 marks
Calculator
A swimmer is at a buoy \(B\) in a lake, 120 m from the nearest point \(P\) on a straight shoreline. A lifeguard hut \(H\) is on the shore, 500 m from \(P\). The swimmer swims in a straight line to a point \(Q\) on the shore, where \(PQ=x\) m and \(0\le x\le 500\), then runs along the shore from \(Q\) to \(H\). The swimmer swims at \(1.2\ \text{m s}^{-1}\) and runs at \(3\ \text{m s}^{-1}\). The total time taken is \(T\) seconds.
Diagram not to scale
(a)
Show that \(T=\frac{\sqrt{14400+x^2}}{1.2}+\frac{500-x}{3}\).
(2)
(b)
Find \(\frac{dT}{dx}\).
(3)
(c)
Find the value of \(x\) for which \(T\) is a minimum, correct to 3 significant figures.
(3)
(d)
Justify that this value of \(x\) gives a minimum value of \(T\).
(2)
(e)
Use the unrounded value of \(x\) to find the minimum time, to the nearest second.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 swimming distance \(BQ=\sqrt{120^2+x^2}\) and running distance \(500-x\)
✓ 1
A1 time \(=\frac{\text{distance}}{\text{speed}}\) for each stage, summed, giving the given expression (AG)
✓ 1
Part (b)
M1 attempt at the chain rule on \((14400+x^2)^{\frac{1}{2}}\)
✓ 1
A1 \(\frac{x}{1.2\sqrt{14400+x^2}}\)
✓ 1
A1 \(-\frac{1}{3}\) for the second term
✓ 1
Part (c)
M1 set \(\frac{dT}{dx}=0\): \(\frac{x}{\sqrt{14400+x^2}}=0.4\)
✓ 1
M1 square and solve: \(0.84x^2=2304\)
✓ 1
A1 \(x=52.4\) (52.37...)
✓ 1
Part (d)
M1 \(\frac{d^2T}{dx^2}=\frac{14400}{1.2(14400+x^2)^{3/2}}\) (or a sign change of \(\frac{dT}{dx}\) from negative to positive)
✓ 1
R1 positive for all \(x\), hence a minimum
✓ 1
Part (e)
M1 substitute \(x=52.372\ldots\) into \(T\)
✓ 1
A1 \(258\) s
✓ 1
Exam tip: State the interval of allowed values of \(x\) when setting up an optimisation problem, and check that your stationary point lies inside it.
430IBMA3H-5-19r48-135.14 Implicit differentiation, related rates and optimisationPaper 2Hard12 marks
Calculator
A curve has equation \(x^3+y^3=9xy\).
(a)
Show that \(\frac{dy}{dx}=\frac{3y-x^2}{y^2-3x}\).
(4)
(b)
The point \((2,4)\) lies on the curve. Find the equation of the tangent at this point, in the form \(y=mx+c\).
(3)
(c)
There is one point on the curve with \(x>0\) where the tangent is horizontal. Find the \(y\)-coordinate of this point, correct to 3 significant figures.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt at implicit differentiation, with a \(\frac{dy}{dx}\) term appearing
✓ 1
A1 \(3x^2+3y^2\frac{dy}{dx}\) on the left
✓ 1
A1 \(9y+9x\frac{dy}{dx}\) on the right (product rule)
✓ 1
A1 collect \(\frac{dy}{dx}\) terms and rearrange to the given result (AG)
✓ 1
Part (b)
M1 substitute \(x=2\), \(y=4\) into the expression for \(\frac{dy}{dx}\)
✓ 1
A1 gradient \(=\frac{12-4}{16-6}=0.8\)
✓ 1
A1 \(y=0.8x+2.4\)
✓ 1
Part (c)
M1 horizontal tangent means \(3y-x^2=0\), so \(y=\frac{x^2}{3}\)
✓ 1
M1 substitute into the curve: \(x^3+\frac{x^6}{27}=3x^3\)
✓ 1
A1 \(\frac{x^6}{27}=2x^3\)
✓ 1
R1 divide by \(x^3\) (valid as \(x>0\)) to obtain \(x^3=54\), so \(x=3.78\)
✓ 1
A1 \(y=\frac{x^2}{3}=4.76\)
✓ 1
Common error: After substituting into the curve equation, dividing by \(x^3\) loses the solution \(x=0\); state why that solution is excluded rather than dropping it silently.
431IBMA3H-5-19r48-145.14 Implicit differentiation, related rates and optimisationPaper 2Hard10 marks
Calculator
A drone flies horizontally at a constant height of 60 m and a constant speed of \(8\ \text{m s}^{-1}\) directly towards a point \(O\) on the ground. The drone is a horizontal distance \(x\) m from the point directly above \(O\), and the angle of elevation of the drone from \(O\) is \(\theta\) radians.
Diagram not to scale
(a)
Show that \(\frac{d\theta}{dt}=\frac{480}{3600+x^2}\).
(4)
(b)
Find the rate of change of \(\theta\) when \(x=80\).
(2)
(c)
Find the greatest value of \(\frac{d\theta}{dt}\), giving your answer to 3 significant figures.
(2)
(d)
Find the value of \(x\), correct to 3 significant figures, when \(\frac{d\theta}{dt}=0.1\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\tan\theta=\frac{60}{x}\)
✓ 1
M1 differentiate with respect to \(t\): \(\sec^2\theta\frac{d\theta}{dt}=-\frac{60}{x^2}\frac{dx}{dt}\), with \(\frac{dx}{dt}=-8\)
M1 greatest when the denominator is smallest, that is at \(x=0\)
✓ 1
A1 \(\frac{480}{3600}=0.133\ \text{rad s}^{-1}\)
✓ 1
Part (d)
M1 \(\frac{480}{3600+x^2}=0.1\), so \(x^2=1200\)
✓ 1
A1 \(x=34.6\) (taking \(x>0\))
✓ 1
Exam tip: For angle rates, relate the angle to the distances with a trig ratio first, differentiate implicitly, then substitute the values at the end.
432AAHL5-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 \(\dfrac{k^{2}\mathrm{e}^{k}}{2n}\)
433AAHL5-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 \(-\tfrac{4}{3}\)
434AAHL5-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 \(\mathrm{e}\)
We use essential cookies to make this site work. With your permission we'd also use analytics and advertising cookies to understand which resources help students most and how families find us. Nothing optional is set unless you accept. Privacy policy
EssentialNeeded for the site and your cookie choice to work.
Always on
AnalyticsAnonymous visit statistics that help improve the free resources.
MarketingMeasures whether ads reach the right families. Off unless you turn it on.