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IB Maths Analysis & Approaches · Standard Level · Topic 5

Calculus

480 free, high-quality past-paper-style questions on Calculus for IB Maths AA SL, 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 Higher Level →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.3,638 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked

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+ Add to test✓ In your test1AASL5-q00015.1 Limits and the derivative Paper 1Easy5 marks

Let $f(x)=x^{2}+3x$. The point $\text{A}(1,\,4)$ lies on the graph of $f$. The point $\text{B}$ also lies on the graph of $f$ and has $x$-coordinate $1+h$, where $h\neq0$.

(a)

Show that the gradient of the chord $[\text{AB}]$ is $5+h$.

(2)

(b)

Write down the limit of the gradient of $[\text{AB}]$ as $h\to0$.

(1)

(c)

Find $f'(x)$ and hence verify that $f'(1)$ is equal to your answer to part (b).

(2)

Show mark scheme

Working	Mark	Notes
(a)
$f(1+h)=(1+h)^{2}+3(1+h)=4+5h+h^{2}$	M1	Award for a correct expansion of $f(1+h)$.
gradient $=\dfrac{4+5h+h^{2}-4}{(1+h)-1}=\dfrac{5h+h^{2}}{h}$	A1
$=5+h$	AG
(b)
$5$	A1
(c)
$f'(x)=2x+3$	A1
$f'(1)=2(1)+3=5$, which equals the limit in (b)	A1	Must see the substitution.

Answers: (b) $5$ · (c) $f'(x)=2x+3$, $f'(1)=5$

+ Add to test✓ In your test2AASL5-q00025.1 Limits and the derivative Paper 1Easy6 marks

Let $f(x)=\dfrac{x^{3}-1}{x-1}$, for $x\neq1$. The following diagram shows part of the graph of $f$. The open circle shows that the point $(1,\,3)$ does not lie on the graph.

(a)

Write down the value of $\displaystyle\lim_{x\to1}f(x)$.

(1)

(b)

Explain why $f(1)$ does not exist.

(1)

(c)

By expanding $(x-1)(x^{2}+x+1)$, show that $f(x)=x^{2}+x+1$ for $x\neq1$.

(2)

(d)

Hence find the gradient of the graph of $f$ at the point where $x=2$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$3$	A1
(b)
substituting $x=1$ gives $\dfrac{0}{0}$ (division by zero is undefined)	R1	Accept “$x=1$ is not in the domain” or “the denominator is zero when $x=1$”.
(c)
$(x-1)(x^{2}+x+1)=x^{3}+x^{2}+x-x^{2}-x-1$	M1
$=x^{3}-1$, so $\dfrac{x^{3}-1}{x-1}=\dfrac{(x-1)(x^{2}+x+1)}{x-1}$ and the factor $x-1\neq0$ cancels	A1
$f(x)=x^{2}+x+1$ for $x\neq1$	AG
(d)
$f'(x)=2x+1$	(M1)
$f'(2)=5$	A1

Answers: (a) $3$ · (b) $\frac{0}{0}$ is undefined · (d) $5$

+ Add to test✓ In your test3AASL5-q00035.1 Limits and the derivative Paper 1Easy5 marks

A spherical balloon is being inflated. When its radius is $r$ cm, its volume is $V\ \text{cm}^{3}$, where $V=\dfrac{4}{3}\pi r^{3}$.

(a)

Find $\dfrac{\mathrm{d}V}{\mathrm{d}r}$.

(1)

(b)

Find the value of $\dfrac{\mathrm{d}V}{\mathrm{d}r}$ when $r=3$. Give your answer in terms of $\pi$.

(1)

(c)

Interpret your answer to part (b) in context, including its units.

(1)

(d)

Find the radius of the balloon when $\dfrac{\mathrm{d}V}{\mathrm{d}r}=100\pi$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$4\pi r^{2}$	A1
(b)
$36\pi$	A1
(c)
when the radius is 3 cm, the volume is increasing at a rate of $36\pi\ \text{cm}^{3}$ per cm increase in the radius	R1	Must refer to rate of change of volume with respect to radius and give units cm³ per cm (or cm²).
(d)
$4\pi r^{2}=100\pi\Rightarrow r^{2}=25$	(M1)
$r=5$ cm	A1	Award A0 if $r=-5$ is also given and not rejected.

Answers: (a) $4\pi r^{2}$ · (b) $36\pi$ · (c) volume increases at $36\pi$ cm³ per cm of radius · (d) $r=5$ cm

+ Add to test✓ In your test4AASL5-q00095.2 Increasing and decreasing functions Paper 1Easy5 marks

Let $f(x)=2x^{3}-9x^{2}+12x-5$, for $x\in\mathbb{R}$.

(a)

Find $f'(x)$.

(2)

(b)

Find the values of $x$ for which $f'(x)=0$.

(2)

(c)

Hence write down the interval on which $f$ is decreasing.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$6x^{2}-18x$	A1	Award for any two correct terms.
$f'(x)=6x^{2}-18x+12$	A1
(b)
$6x^{2}-18x+12=0\Rightarrow 6(x-1)(x-2)=0$	M1	Or correct use of the quadratic formula.
$x=1,\ x=2$	A1
(c)
$1\lt x\lt 2$	A1	FT from (b). Accept $1\le x\le 2$ or $]1,\,2[$. Award A0 for $f'(x)\lt 0$ alone.

Answers: (a) $f'(x)=6x^{2}-18x+12$ · (b) $x=1,\ 2$ · (c) $1\lt x\lt 2$

+ Add to test✓ In your test5AASL5-q00115.2 Increasing and decreasing functions Paper 1Easy7 marks

The following diagram shows the graph of a cubic function $y=f(x)$. The graph has a local maximum point where $x=-2$ and a local minimum point where $x=3$.

(a)

Write down the set of values of $x$ for which $f'(x)\lt 0$.

(1)

(b)

Write down the value of $f'(3)$.

(1)

(c)

It is given that $f'(x)=x^{2}+bx+c$. Find the value of $b$ and the value of $c$.

(2)

(d)

Sketch the graph of $y=f'(x)$, showing clearly the values of any intercepts with the axes.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$-2\lt x\lt 3$	A1	Award A0 for $-2\le x\le 3$, since $f'(-2)=f'(3)=0$.
(b)
$0$	A1
(c)
$f'(x)=(x+2)(x-3)$	(M1)	Or $4-2b+c=0$ and $9+3b+c=0$.
$b=-1,\ c=-6$	A1
(d)
concave-up parabola	A1
$x$-intercepts $-2$ and $3$	A1
$y$-intercept $-6$	A1	FT from (c). The vertex $\left(\tfrac12,\,-\tfrac{25}{4}\right)$ need not be labelled.

Answers: (a) $-2\lt x\lt 3$ · (b) $0$ · (c) $b=-1,\ c=-6$ · (d) parabola through $(-2,0)$, $(3,0)$, $(0,-6)$

Model sketch

+ Add to test✓ In your test6AASL5-q00125.2 Increasing and decreasing functions Paper 1Easy4 marks

Let $f(x)=ax^{2}+bx+3$, where $a,b\in\mathbb{R}$. The function $f$ is decreasing for $x\lt 2$ and increasing for $x\gt 2$. The gradient of the graph of $f$ at the point where $x=0$ is $-8$.

(a)

Find $f'(x)$.

(1)

(b)

Write down the value of $f'(2)$.

(1)

(c)

Find the value of $a$ and the value of $b$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$f'(x)=2ax+b$	A1
(b)
$0$	A1
(c)
$f'(0)=b=-8$	A1
$4a-8=0\Rightarrow a=2$	A1	FT from their $b$.

Answers: (a) $2ax+b$ · (b) $0$ · (c) $a=2,\ b=-8$

+ Add to test✓ In your test7AASL5-q00195.3 Differentiating polynomials and integer powers Paper 1Easy6 marks

Find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ for each of the following.

(a)

$y=4x^{5}-3x^{2}+x-9$

(2)

(b)

$y=\dfrac{3}{x^{4}}$, for $x\neq0$

(2)

(c)

$y=(x-3)(2x+5)$

(2)

Show mark scheme

Working	Mark	Notes
(a)
$20x^{4}-6x$	A1	Award for any two correct terms.
$\dfrac{\mathrm{d}y}{\mathrm{d}x}=20x^{4}-6x+1$	A1
(b)
$y=3x^{-4}$	(M1)
$-12x^{-5}$	A1	Accept $-\dfrac{12}{x^{5}}$.
(c)
$y=2x^{2}-x-15$	(M1)
$4x-1$	A1

Answers: (a) $20x^{4}-6x+1$ · (b) $-12x^{-5}$ · (c) $4x-1$

+ Add to test✓ In your test8AASL5-q00255.4 Tangents and normals Paper 1Easy6 marks

A curve has equation $y=x^{3}-2x+1$. The point $\text{P}(2,\,5)$ lies on the curve.

(a)

Find the gradient of the curve at P.

(2)

(b)

Find the equation of the tangent to the curve at P. Give your answer in the form $y=mx+c$.

(2)

(c)

Find the equation of the normal to the curve at P. Give your answer in the form $ax+by+d=0$, where $a,b,d\in\mathbb{Z}$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\dfrac{\mathrm{d}y}{\mathrm{d}x}=3x^{2}-2$	(M1)
$3(2)^{2}-2=10$	A1
(b)
$y-5=10(x-2)$	(M1)
$y=10x-15$	A1
(c)
gradient $-\dfrac{1}{10}$: $y-5=-\dfrac{1}{10}(x-2)$	(M1)
$x+10y-52=0$	A1	Accept any integer multiple.

Answers: (a) $10$ · (b) $y=10x-15$ · (c) $x+10y-52=0$

+ Add to test✓ In your test9AASL5-q00335.5 Introduction to integration Paper 1Easy6 marks

A function $f$ has derivative $f'(x)=4x^{3}+\dfrac{6}{x^{2}}$, for $x\neq0$. The graph of $f$ passes through the point $(1,\,3)$.

(a)

Find $f(x)$.

(4)

(b)

Find the equation of the tangent to the graph of $f$ at the point $(1,\,3)$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
attempt to integrate, e.g. one power increased by 1	M1
$x^{4}-6x^{-1}+C$	A1	Accept $x^{4}-\dfrac{6}{x}$. Condone the absence of $+C$ here.
substituting $(1,\,3)$: $1-6+C=3$	M1	Must include $C$.
$f(x)=x^{4}-\dfrac{6}{x}+8$	A1
(b)
gradient $f'(1)=4+6=10$	A1
$y=10x-7$	A1	Accept $y-3=10(x-1)$.

Answers: (a) $f(x)=x^{4}-\dfrac{6}{x}+8$ · (b) $y=10x-7$

+ Add to test✓ In your test10AASL5-q00345.5 Introduction to integration Paper 1Easy6 marks

Find each of the following integrals.

(a)

$\displaystyle\int(6x^{2}-4x+3)\,\mathrm{d}x$

(2)

(b)

$\displaystyle\int\left(x^{3}-\dfrac{2}{x^{2}}\right)\mathrm{d}x$, for $x\neq0$

(2)

(c)

$\displaystyle\int(2x+1)^{2}\,\mathrm{d}x$

(2)

Show mark scheme

Working	Mark	Notes
(a)
$2x^{3}-2x^{2}$	A1
$+3x+c$	A1	Penalize a missing $+c$ once only in this question.
(b)
$\dfrac{x^{4}}{4}$	A1
$+2x^{-1}+c$	A1	Accept $\dfrac{2}{x}$.
(c)
$\displaystyle\int(4x^{2}+4x+1)\,\mathrm{d}x$	(M1)
$\dfrac{4x^{3}}{3}+2x^{2}+x+c$	A1	Accept $\dfrac{(2x+1)^{3}}{6}+c$.

Answers: (a) $2x^{3}-2x^{2}+3x+c$ · (b) $\frac{x^{4}}{4}+\frac{2}{x}+c$ · (c) $\frac{4x^{3}}{3}+2x^{2}+x+c$

+ Add to test✓ In your test11AASL5-q00355.5 Introduction to integration Paper 1Easy6 marks

Let $F(x)=(x^{2}+1)^{2}$.

(a)

By first expanding $F(x)$, find $F'(x)$.

(2)

(b)

Hence find $\displaystyle\int(x^{3}+x)\,\mathrm{d}x$.

(2)

(c)

The function $G$ satisfies $G'(x)=x^{3}+x$ and $G(1)=3$. Find $G(x)$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$F(x)=x^{4}+2x^{2}+1$	(M1)
$F'(x)=4x^{3}+4x$	A1
(b)
$\dfrac{1}{4}(x^{2}+1)^{2}$	A1	Accept $\dfrac{x^{4}}{4}+\dfrac{x^{2}}{2}$.
$+c$	A1
(c)
$\dfrac{1}{4}(1+1)^{2}+c=3$	M1
$G(x)=\dfrac{1}{4}(x^{2}+1)^{2}+2$	A1	Accept $\dfrac{x^{4}}{4}+\dfrac{x^{2}}{2}+\dfrac{9}{4}$.

Answers: (a) $4x^{3}+4x$ · (b) $\frac14(x^{2}+1)^{2}+c$ · (c) $G(x)=\frac14(x^{2}+1)^{2}+2$

+ Add to test✓ In your test12AASL5-q00415.6 Differentiation rules Paper 1Easy6 marks

Differentiate each of the following with respect to $x$.

(a)

$f(x)=8\sqrt{x}-\dfrac{2}{x^{3}}+3$

(2)

(b)

$g(x)=\sin(3x^{2}-1)$

(2)

(c)

$h(x)=\ln(2x+5)+\mathrm{e}^{-4x}$

(2)

Show mark scheme

Working	Mark	Notes
(a)
$4x^{-\frac12}$	A1	Accept $\dfrac{4}{\sqrt{x}}$.
$+6x^{-4}$, so $f'(x)=4x^{-\frac12}+6x^{-4}$	A1	Award A0 if a term from the constant 3 is present.
(b)
attempt to use the chain rule, e.g. $\cos(3x^{2}-1)\times\ldots$	M1
$g'(x)=6x\cos(3x^{2}-1)$	A1
(c)
$\dfrac{2}{2x+5}$	A1
$-4\mathrm{e}^{-4x}$	A1

Answers: (a) $4x^{-\frac12}+6x^{-4}$ · (b) $6x\cos(3x^{2}-1)$ · (c) $\dfrac{2}{2x+5}-4\mathrm{e}^{-4x}$

+ Add to test✓ In your test13AASL5-q00425.6 Differentiation rules Paper 1Easy5 marks

Let $f(x)=x^{3}\ln x$, for $x\gt 0$.

(a)

Find $f'(x)$.

(2)

(b)

Find the exact $x$-coordinate of the stationary point on the graph of $f$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
attempt to use the product rule, e.g. $3x^{2}\ln x+x^{3}\times\ldots$	M1
$f'(x)=3x^{2}\ln x+x^{2}$	A1
(b)
$x^{2}(3\ln x+1)=0$	M1	Setting their $f'(x)=0$ and factorizing.
$\ln x=-\tfrac{1}{3}$	A1	Must reject or ignore $x=0$, as $x\gt 0$.
$x=\mathrm{e}^{-\frac{1}{3}}$	A1	Accept $\dfrac{1}{\sqrt[3]{\mathrm{e}}}$.

Answers: (a) $3x^{2}\ln x+x^{2}$ · (b) $x=\mathrm{e}^{-1/3}$

+ Add to test✓ In your test14AASL5-q00455.6 Differentiation rules Paper 1Easy7 marks

Let $f(x)=\dfrac{x^{2}+3}{\sqrt{x}}$, for $x\gt 0$.

(a)

Write $f(x)$ in the form $x^{p}+3x^{q}$, where $p,\,q\in\mathbb{Q}$.

(2)

(b)

Find $f'(x)$.

(2)

(c)

Find the equation of the tangent to the graph of $f$ at the point where $x=4$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$p=\tfrac32$	A1	Accept $x^{\frac32}$.
$q=-\tfrac12$	A1	Accept $3x^{-\frac12}$.
(b)
$\tfrac32x^{\frac12}$	A1	FT from their (a).
$-\tfrac32x^{-\frac32}$, so $f'(x)=\tfrac32x^{\frac12}-\tfrac32x^{-\frac32}$	A1	Accept $\dfrac{3\sqrt{x}}{2}-\dfrac{3}{2x\sqrt{x}}$.
(c)
$f(4)=\dfrac{19}{2}$	A1
$f'(4)=\tfrac32\times2-\tfrac32\times\tfrac18=\dfrac{45}{16}$	A1	FT from their (b).
$y-\dfrac{19}{2}=\dfrac{45}{16}(x-4)$	A1	Accept $y=\dfrac{45}{16}x-\dfrac{7}{4}$.

Answers: (a) $p=\tfrac32,\ q=-\tfrac12$ · (b) $\tfrac32x^{1/2}-\tfrac32x^{-3/2}$ · (c) $y=\dfrac{45}{16}x-\dfrac74$

+ Add to test✓ In your test15AASL5-q00465.6 Differentiation rules Paper 1Easy6 marks

Let $f(x)=\ln(x^{2}+5)$, for $x\in\mathbb{R}$.

(a)

Find $f'(x)$.

(2)

(b)

Find the gradient of the graph of $f$ at $x=1$.

(1)

(c)

Let $h(x)=\ln\left(\dfrac{(x^{2}+5)^{3}}{x}\right)$, for $x\gt 0$. By first writing $h(x)$ in terms of $f(x)$, show that the graph of $h$ has a stationary point at $x=1$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
attempt to use the chain rule, e.g. $\dfrac{1}{x^{2}+5}\times\ldots$	M1
$f'(x)=\dfrac{2x}{x^{2}+5}$	A1
(b)
$f'(1)=\dfrac{2}{6}=\dfrac13$	A1
(c)
$h(x)=3\ln(x^{2}+5)-\ln x=3f(x)-\ln x$	M1	Correct use of the laws of logarithms.
$h'(x)=3f'(x)-\dfrac{1}{x}\left(=\dfrac{6x}{x^{2}+5}-\dfrac1x\right)$	A1
$h'(1)=3\times\dfrac13-1=0$	A1	Must see substitution leading to 0.
stationary point at $x=1$	AG

Answers: (a) $\dfrac{2x}{x^{2}+5}$ · (b) $\tfrac13$ · (c) $h'(1)=3\times\tfrac13-1=0$

+ Add to test✓ In your test16AASL5-q00475.6 Differentiation rules Paper 1Easy5 marks

A curve has equation $y=\sin^{2}(3x)$, for $x\in\mathbb{R}$.

(a)

Find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$.

(2)

(b)

Find the gradient of the curve at the point where $x=\dfrac{\pi}{12}$.

(2)

(c)

Find the least positive value of $x$ for which the gradient of the curve is zero.

(1)

Show mark scheme

Working	Mark	Notes
(a)
attempt to use the chain rule twice, e.g. $2\sin(3x)\times\cos(3x)\times\ldots$	M1
$\dfrac{\mathrm{d}y}{\mathrm{d}x}=6\sin(3x)\cos(3x)$	A1	Accept $3\sin(6x)$.
(b)
$6\sin\left(\tfrac{\pi}{4}\right)\cos\left(\tfrac{\pi}{4}\right)=6\times\tfrac{\sqrt2}{2}\times\tfrac{\sqrt2}{2}$	(M1)	Or $3\sin\left(\tfrac{\pi}{2}\right)$.
$3$	A1
(c)
$x=\dfrac{\pi}{6}$	A1	From $3\sin(6x)=0$, $6x=\pi$.

Answers: (a) $6\sin3x\cos3x\ (=3\sin6x)$ · (b) $3$ · (c) $x=\dfrac{\pi}{6}$

+ Add to test✓ In your test17AASL5-q00565.7 The second derivative and graphs Paper 1Easy4 marks

The diagram shows part of the graph of $y=f(x)$. The points P, Q, R and S lie on the graph. Q is a local maximum point and R is a point of inflexion.

Copy and complete the table below, writing “positive”, “negative” or “zero” in each cell.

	P	Q	R	S
$f'(x)$
$f''(x)$

(a)

Complete the row for $f'(x)$.

(2)

(b)

Complete the row for $f''(x)$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
at least two of: P positive, Q zero, R negative, S negative	A1
all four correct: positive, zero, negative, negative	A1
(b)
at least two of: P negative, Q negative, R zero, S positive	A1
all four correct: negative, negative, zero, positive	A1

Answers: $f'$: positive, zero, negative, negative · $f''$: negative, negative, zero, positive

+ Add to test✓ In your test18AASL5-q00575.7 The second derivative and graphs Paper 1Easy5 marks

Let $g(x)=(x-1)^{4}+2x$, for $x\in\mathbb{R}$.

(a)

Find $g'(x)$ and $\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}$, where $y=g(x)$.

(2)

(b)

Show that $g''(1)=0$ but that the graph of $g$ has no point of inflexion at $x=1$.

(2)

(c)

Write down the set of values of $x$ for which the graph of $g$ is concave-up.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$g'(x)=4(x-1)^{3}+2$	A1
$\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=12(x-1)^{2}$	A1
(b)
$g''(1)=12(0)^{2}=0$	A1
$g''(x)=12(x-1)^{2}\gt 0$ for both $x\lt 1$ and $x\gt 1$, so $g''$ does not change sign (no change of concavity)	R1	Award R1 for a reason referring to the sign of $g''$ on both sides of $x=1$.
(c)
$x\in\mathbb{R},\ x\neq1$	A1	Accept $x\lt 1$ or $x\gt 1$.

Answers: (a) $g'(x)=4(x-1)^{3}+2$, $g''(x)=12(x-1)^{2}$ · (c) $x\neq1$

+ Add to test✓ In your test19AASL5-q00585.7 The second derivative and graphs Paper 1Easy6 marks

A curve has equation $y=4\sqrt{x}+\dfrac{1}{x}$, for $x\gt 0$.

(a)

Find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$.

(2)

(b)

Find $\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}$.

(2)

(c)

State, with a reason, whether the curve is concave-up or concave-down at the point where $x=4$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$2x^{-\frac12}$	A1
$-x^{-2}$, so $\dfrac{\mathrm{d}y}{\mathrm{d}x}=2x^{-\frac12}-x^{-2}$	A1
(b)
$-x^{-\frac32}$	A1	FT from their (a).
$+2x^{-3}$	A1
(c)
$\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\Big\|_{x=4}=-\dfrac18+\dfrac{2}{64}=-\dfrac{3}{32}$	A1
$\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\lt 0$, so concave-down	R1	Do not award R1 without a correct value or sign of the second derivative.

Answers: (a) $2x^{-1/2}-x^{-2}$ · (b) $-x^{-3/2}+2x^{-3}$ · (c) concave-down ($-\tfrac{3}{32}\lt 0$)

+ Add to test✓ In your test20AASL5-q00605.7 The second derivative and graphs Paper 1Easy6 marks

The diagram shows the graphs labelled A, B and C. One of them is the graph of $y=f(x)$, one is the graph of $y=f'(x)$ and one is the graph of $y=f''(x)$.

(a)

State which graph is the graph of $f$, which is the graph of $f'$ and which is the graph of $f''$.

(2)

(b)

Write down the number of points of inflexion on the graph of $f$. Give a reason for your answer.

(2)

(c)

Given that $f(x)=\dfrac{x^{4}}{4}-2x^{2}$, find the exact $x$-coordinates of the points of inflexion on the graph of $f$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
C is the graph of $f$	A1
A is the graph of $f'$ and B is the graph of $f''$	A1	Reasoning: the zeros of A are at the turning points of C; the zeros of B are at the turning points of A.
(b)
$2$	A1
graph B ($f''$) crosses the $x$-axis twice, changing sign each time	R1	Accept: graph A ($f'$) has two turning points. Do not award A1R0.
(c)
$f''(x)=3x^{2}-4=0$	M1
$x=\pm\dfrac{2}{\sqrt3}$	A1	Accept $\pm\dfrac{2\sqrt3}{3}$.

Answers: (a) $f$: C, $f'$: A, $f''$: B · (b) 2 · (c) $x=\pm\dfrac{2}{\sqrt3}$

+ Add to test✓ In your test21AASL5-q00615.6 Differentiation rules Paper 1Easy5 marks

Let $y=\mathrm{e}^{-2x}+\sin(2x)$, for $x\in\mathbb{R}$.

(a)

Find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$.

(2)

(b)

Find $\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}$.

(2)

(c)

Show that $\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}+4y=8\mathrm{e}^{-2x}$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$-2\mathrm{e}^{-2x}$	A1
$+2\cos(2x)$	A1
(b)
$4\mathrm{e}^{-2x}$	A1	FT from their (a).
$-4\sin(2x)$	A1
(c)
$4\mathrm{e}^{-2x}-4\sin(2x)+4\mathrm{e}^{-2x}+4\sin(2x)$	A1	Must see the substitution of both $y$ and their second derivative.
$=8\mathrm{e}^{-2x}$	AG

Answers: (a) $-2\mathrm{e}^{-2x}+2\cos2x$ · (b) $4\mathrm{e}^{-2x}-4\sin2x$

+ Add to test✓ In your test22AASL5-q00695.8 Stationary points, inflexion and optimization Paper 1Easy6 marks

Let $f(x)=x^{4}-8x^{2}+3$, for $x\in\mathbb{R}$.

(a)

Find $f'(x)$.

(1)

(b)

Find the $x$-coordinates of the stationary points on the graph of $f$.

(2)

(c)

By considering the sign of $f'(x)$, find the coordinates of the local maximum point and of the local minimum points.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$f'(x)=4x^{3}-16x$	A1
(b)
$4x(x^{2}-4)=0\Rightarrow4x(x-2)(x+2)=0$	M1
$x=-2,\ 0,\ 2$	A1
(c)
$f'\lt 0$ for $x\lt -2$, $f'\gt 0$ for $-2\lt x\lt0$, $f'\lt 0$ for $0\lt x\lt2$, $f'\gt 0$ for $x\gt2$	R1	e.g. $f'(-3)=-60$, $f'(-1)=12$, $f'(1)=-12$, $f'(3)=60$, or a correct sign diagram.
local maximum $(0,\,3)$	A1
local minima $(-2,\,-13)$ and $(2,\,-13)$	A1	Both points required.

Answers: (b) $x=-2,\ 0,\ 2$ · (c) max $(0,3)$, min $(-2,-13)$ and $(2,-13)$

+ Add to test✓ In your test23AASL5-q00705.8 Stationary points, inflexion and optimization Paper 1Easy7 marks

Let $f(x)=\mathrm{e}^{2x}-6x$, for $x\in\mathbb{R}$.

(a)

Find $f'(x)$.

(1)

(b)

Find the exact $x$-coordinate of the stationary point on the graph of $f$.

(2)

(c)

Use the second derivative to show that this stationary point is a minimum.

(2)

(d)

Find the exact minimum value of $f$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$f'(x)=2\mathrm{e}^{2x}-6$	A1
(b)
$2\mathrm{e}^{2x}=6\Rightarrow\mathrm{e}^{2x}=3$	M1
$x=\dfrac{\ln3}{2}$	A1	Accept $\ln\sqrt3$.
(c)
$f''(x)=4\mathrm{e}^{2x}$	A1
$f''\left(\tfrac{\ln3}{2}\right)=12\gt 0$ (or $4\mathrm{e}^{2x}\gt 0$ for all $x$), so a minimum	R1
(d)
$\mathrm{e}^{\ln3}-6\times\dfrac{\ln3}{2}$	(M1)
$3-3\ln3$	A1	Accept $3-\ln27$.

Answers: (a) $2\mathrm{e}^{2x}-6$ · (b) $x=\tfrac12\ln3$ · (d) $3-3\ln3$

+ Add to test✓ In your test24AASL5-q00735.8 Stationary points, inflexion and optimization Paper 1Easy8 marks

An open-topped water tank is a cuboid with a square base of side $x$ metres and height $h$ metres. Its volume is $32\ \text{m}^{3}$. Let $S$ be the area, in m², of the metal used to make the base and the four sides.

Diagram not to scale

(a)

Write down an expression for $h$ in terms of $x$.

(1)

(b)

Show that $S=x^{2}+\dfrac{128}{x}$.

(2)

(c)

Find $\dfrac{\mathrm{d}S}{\mathrm{d}x}$.

(1)

(d)

Find the value of $x$ for which $S$ is a minimum, and the minimum value of $S$.

(3)

(e)

Use the second derivative to justify that this value of $x$ gives a minimum.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$h=\dfrac{32}{x^{2}}$	A1
(b)
$S=x^{2}+4xh$	M1	Base plus four sides; no top.
$x^{2}+4x\times\dfrac{32}{x^{2}}$	A1
$S=x^{2}+\dfrac{128}{x}$	AG
(c)
$2x-\dfrac{128}{x^{2}}$	A1
(d)
$2x-\dfrac{128}{x^{2}}=0\Rightarrow x^{3}=64$	M1
$x=4$	A1
$S=16+32=48\ \text{m}^{2}$	A1
(e)
$\dfrac{\mathrm{d}^{2}S}{\mathrm{d}x^{2}}=2+\dfrac{256}{x^{3}}$, which is $6\gt 0$ at $x=4$, so a minimum	R1

Answers: (a) $h=\dfrac{32}{x^{2}}$ · (c) $2x-\dfrac{128}{x^{2}}$ · (d) $x=4$, $S=48\ \text{m}^{2}$

+ Add to test✓ In your test25AASL5-q00825.9 Kinematics Paper 1Easy8 marks

A particle moves in a straight line so that its displacement, $s$ metres, from a fixed point O after $t$ seconds is given by $s(t)=2\sin(3t)$, for $0\le t\le\pi$.

(a)

Find expressions for the velocity, $v$, and the acceleration, $a$, of the particle in terms of $t$.

(3)

(b)

Show that $a=-9s$.

(1)

(c)

Find the speed of the particle when $t=\dfrac{\pi}{9}$.

(2)

(d)

Find the first time $t\gt 0$ at which the particle is at rest.

(2)

Show mark scheme

Working	Mark	Notes
(a)
attempt to differentiate using the chain rule (factor of 3 seen)	(M1)
$v=6\cos(3t)$	A1
$a=-18\sin(3t)$	A1	FT from their $v$.
(b)
$a=-18\sin(3t)=-9\big(2\sin(3t)\big)$	M1	Must see the factorisation linked to $s$.
$a=-9s$	AG
(c)
$v\left(\tfrac{\pi}{9}\right)=6\cos\left(\tfrac{\pi}{3}\right)$	(M1)
$3\ \text{m s}^{-1}$	A1
(d)
$\cos(3t)=0\Rightarrow 3t=\tfrac{\pi}{2}$	(M1)
$t=\dfrac{\pi}{6}$	A1	Award A0 for $t=\tfrac{\pi}{2}$ (forgetting to divide by 3).

Answers: (a) $v=6\cos 3t,\ a=-18\sin 3t$ · (c) $3\ \text{m s}^{-1}$ · (d) $t=\tfrac{\pi}{6}$

+ Add to test✓ In your test26AASL5-q00855.9 Kinematics Paper 1Easy6 marks

A cyclist moves along a straight path. The graph shows her velocity, $v\ \text{m s}^{-1}$, after $t$ seconds, for $0\le t\le 8$. The graph consists of three straight line segments joining the points $(0,\,0)$, $(2,\,6)$, $(5,\,6)$ and $(8,\,-3)$.

(a)

Write down the acceleration of the cyclist for $0\lt t\lt 2$.

(1)

(b)

Write down the value of $t$ at which the cyclist changes direction.

(1)

(c)

Find the distance travelled by the cyclist in the first 7 seconds.

(2)

(d)

Find the total distance travelled by the cyclist in the 8 seconds.

(1)

(e)

Find the displacement of the cyclist from her starting point when $t=8$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$3\ \text{m s}^{-2}$	A1
(b)
$t=7$	A1
(c)
area under the graph from $t=0$ to $t=7$, e.g. $\tfrac12(7+3)\times6$, or $6+18+6$	(M1)
$30$ m	A1
(d)
$30+\tfrac12(1)(3)=31.5$ m	A1	FT from (c).
(e)
$30-1.5=28.5$ m	A1	FT from (c). Award A0 for 31.5.

Answers: (a) $3\ \text{m s}^{-2}$ · (b) $t=7$ · (c) 30 m · (d) 31.5 m · (e) 28.5 m

+ Add to test✓ In your test27AASL5-q00935.10 Indefinite integrals and reverse chain rule Paper 1Easy6 marks

Find each of the following integrals.

(a)

$\displaystyle\int\dfrac{6}{3x+1}\,\mathrm{d}x$

(2)

(b)

$\displaystyle\int\mathrm{e}^{4-2x}\,\mathrm{d}x$

(2)

(c)

$\displaystyle\int\sqrt{4x+1}\,\mathrm{d}x$

(2)

Show mark scheme

Working	Mark	Notes
(a)
$k\ln\|3x+1\|$	M1	Award for a log of the correct linear function.
$2\ln\|3x+1\|+C$	A1	Condone missing modulus. Do not penalize the absence of $+C$ more than once in the question.
(b)
$k\,\mathrm{e}^{4-2x}$	M1
$-\tfrac{1}{2}\mathrm{e}^{4-2x}+C$	A1
(c)
$k(4x+1)^{\frac32}$	M1
$\tfrac{1}{6}(4x+1)^{\frac32}+C$	A1

Answers: (a) $2\ln|3x+1|+C$ · (b) $-\tfrac12\mathrm{e}^{4-2x}+C$ · (c) $\tfrac16(4x+1)^{3/2}+C$

+ Add to test✓ In your test28AASL5-q00965.10 Indefinite integrals and reverse chain rule Paper 1Easy6 marks

In this question, $x\gt 0$.

(a)

Find $\displaystyle\int\left(6\sqrt{x}-\dfrac{4}{x}\right)\mathrm{d}x$.

(3)

(b)

Find $\displaystyle\int\dfrac{x^{2}-3}{\sqrt{x}}\,\mathrm{d}x$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$6x^{\frac12}$ integrated by increasing the power by one	(M1)
$4x^{\frac32}$	A1
$-4\ln x+C$	A1	Accept $-4\ln\|x\|$. Do not penalise the omission of $+C$ in both parts more than once.
(b)
$x^{\frac32}-3x^{-\frac12}$	M1	Splitting the fraction into powers of x.
$\dfrac{2}{5}x^{\frac52}$	A1
$-6x^{\frac12}+C$	A1

Answers: (a) $4x^{\frac32}-4\ln x+C$ · (b) $\tfrac25x^{\frac52}-6x^{\frac12}+C$

+ Add to test✓ In your test29AASL5-q00975.10 Indefinite integrals and reverse chain rule Paper 1Easy6 marks

Find each of the following integrals.

(a)

$\displaystyle\int\dfrac{1}{(2x-5)^{3}}\,\mathrm{d}x$, for $x\neq\tfrac52$

(2)

(b)

$\displaystyle\int\dfrac{3}{1-4x}\,\mathrm{d}x$, for $x\lt \tfrac14$

(2)

(c)

$\displaystyle\int\sin\left(\dfrac{\pi}{3}-2x\right)\mathrm{d}x$

(2)

Show mark scheme

Working	Mark	Notes
(a)
$(2x-5)^{-2}$ seen, or factor $\tfrac12$ seen	(M1)
$-\dfrac{1}{4(2x-5)^{2}}+C$	A1
(b)
$\ln(1-4x)$ seen	(M1)
$-\dfrac{3}{4}\ln(1-4x)+C$	A1	Accept $\ln\|1-4x\|$. Award A0 for $+\tfrac34\ln(1-4x)$.
(c)
$\cos\left(\tfrac{\pi}{3}-2x\right)$ seen	(M1)
$\dfrac{1}{2}\cos\left(\dfrac{\pi}{3}-2x\right)+C$	A1	Award A0 for $-\tfrac12\cos\left(\tfrac{\pi}{3}-2x\right)$.

Answers: (a) $-\dfrac{1}{4(2x-5)^{2}}+C$ · (b) $-\tfrac34\ln(1-4x)+C$ · (c) $\tfrac12\cos\left(\tfrac{\pi}{3}-2x\right)+C$

+ Add to test✓ In your test30AASL5-q00985.10 Indefinite integrals and reverse chain rule Paper 1Easy5 marks

A function $f$ has derivative $f'(x)=4\sin(2x)$, for $x\in\mathbb{R}$. The graph of $f$ passes through the point $\left(\dfrac{\pi}{2},\,5\right)$.

(a)

Find $f(x)$.

(4)

(b)

Write down the maximum value of $f$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
attempt to integrate $f'(x)$	M1
$-2\cos(2x)+C$	A1
$5=-2\cos\pi+C$	M1	Substituting the point into an expression containing C.
$f(x)=3-2\cos(2x)$	A1
(b)
$5$	A1	FT from their f.

Answers: (a) $f(x)=3-2\cos(2x)$ · (b) 5

+ Add to test✓ In your test31AASL5-q01045.10 Indefinite integrals and reverse chain rule Paper 1Easy5 marks

(a)

Expand $\left(\mathrm{e}^{x}+\mathrm{e}^{-x}\right)^{2}$.

(1)

(b)

Hence find $\displaystyle\int\left(\mathrm{e}^{x}+\mathrm{e}^{-x}\right)^{2}\mathrm{d}x$.

(2)

(c)

A student writes $\displaystyle\int\cos(5x)\,\mathrm{d}x=5\sin(5x)+C$. Explain the error and write down the correct answer.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\mathrm{e}^{2x}+2+\mathrm{e}^{-2x}$	A1
(b)
$\dfrac12\mathrm{e}^{2x}-\dfrac12\mathrm{e}^{-2x}$	A1	FT from (a).
$+2x+C$	A1
(c)
the student has multiplied by 5 (as when differentiating) instead of dividing by 5; differentiating $5\sin(5x)$ gives $25\cos(5x)$	R1
$\dfrac15\sin(5x)+C$	A1

Answers: (a) $\mathrm{e}^{2x}+2+\mathrm{e}^{-2x}$ · (b) $\tfrac12\mathrm{e}^{2x}+2x-\tfrac12\mathrm{e}^{-2x}+C$ · (c) $\tfrac15\sin(5x)+C$

+ Add to test✓ In your test32AASL5-q01085.11 Definite integrals and areas Paper 1Easy6 marks

Find the exact value of each of the following.

(a)

$\displaystyle\int_{1}^{4}\left(3\sqrt{x}-\dfrac{2}{x^{2}}\right)\mathrm{d}x$

(2)

(b)

$\displaystyle\int_{0}^{\frac{\pi}{6}}\cos(3x)\,\mathrm{d}x$

(2)

(c)

$\displaystyle\int_{0}^{\ln 2}\mathrm{e}^{2x}\,\mathrm{d}x$

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\left[2x^{\frac32}+\dfrac{2}{x}\right]_{1}^{4}=\left(16+\tfrac12\right)-(2+2)$	(M1)
$\dfrac{25}{2}$	A1
(b)
$\left[\tfrac13\sin(3x)\right]_{0}^{\pi/6}=\tfrac13\sin\tfrac{\pi}{2}-0$	(M1)
$\dfrac13$	A1
(c)
$\left[\tfrac12\mathrm{e}^{2x}\right]_{0}^{\ln 2}=\tfrac12(4)-\tfrac12(1)$	(M1)
$\dfrac32$	A1

Answers: (a) $\tfrac{25}{2}$ · (b) $\tfrac13$ · (c) $\tfrac32$

+ Add to test✓ In your test33AASL5-q01095.11 Definite integrals and areas Paper 1Easy6 marks

The functions $f$ and $g$ are continuous for all $x\in\mathbb{R}$. It is given that $\displaystyle\int_{1}^{5}f(x)\,\mathrm{d}x=8$, $\displaystyle\int_{3}^{5}f(x)\,\mathrm{d}x=10$ and $\displaystyle\int_{1}^{5}g(x)\,\mathrm{d}x=-3$.

(a)

Find $\displaystyle\int_{1}^{5}\big(2f(x)-3g(x)\big)\,\mathrm{d}x$.

(2)

(b)

Find $\displaystyle\int_{1}^{5}\big(f(x)+1\big)\,\mathrm{d}x$.

(2)

(c)

Write down the value of $\displaystyle\int_{1}^{3}f(x)\,\mathrm{d}x$.

(1)

(d)

Write down the value of $\displaystyle\int_{5}^{1}g(x)\,\mathrm{d}x$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$2(8)-3(-3)$	(M1)
$25$	A1
(b)
$8+\displaystyle\int_{1}^{5}1\,\mathrm{d}x=8+(5-1)$	(M1)
$12$	A1	Award A0 for 9.
(c)
$8-10=-2$	A1
(d)
$3$	A1

Answers: (a) 25 · (b) 12 · (c) −2 · (d) 3

+ Add to test✓ In your test34IBMA2S-052-00-015.2 Increasing and decreasing functions Paper 1Easy6 marks

No calculator

A delivery drone takes off vertically from a rooftop landing pad. For $0 \le t \le 5$, its height above the pad, $h$ metres, after $t$ seconds is modelled by $h(t) = 2t^3 - 15t^2 + 36t$.

(a)

Find $h'(t)$.

(2)

(b)

Find the values of $t$ at which the vertical velocity of the drone is zero.

(3)

(c)

Find the height of the drone at the later of these two times.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiates at least two terms correctly using the power rule A1 $h'(t) = 6t^2 - 30t + 36$	✓ 2
Part (b)
M1 sets $h'(t) = 0$ M1 factorises, e.g. $6(t-2)(t-3) = 0$, or uses the quadratic formula A1 $t = 2$ and $t = 3$	✓ 3
Part (c)
A1 $h(3) = 54 - 135 + 108 = 27$ m	✓ 1

Exam tip: Take out the common factor 6 before factorising; the quadratic $t^2 - 5t + 6$ is far easier to handle than $6t^2 - 30t + 36$.

+ Add to test✓ In your test35IBMA2S-052-00-025.3 Differentiating polynomials and integer powers Paper 1Easy4 marks

No calculator

Let $f(x) = \dfrac{6x^3 - 5}{2x^2}$, for $x \neq 0$.

(a)

Write $f(x)$ in the form $ax + bx^n$, where $a, b, n \in \mathbb{Q}$.

(2)

(b)

Hence find the gradient of the graph of $f$ at the point where $x = -1$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 splits the fraction into $\frac{6x^3}{2x^2} - \frac{5}{2x^2}$ A1 $f(x) = 3x - \frac{5}{2}x^{-2}$	✓ 2
Part (b)
A1 $f'(x) = 3 + 5x^{-3}$ A1 $f'(-1) = 3 - 5 = -2$	✓ 2

Common error: Differentiating the numerator and denominator separately is not valid. Rewrite as a sum of integer powers first, then differentiate term by term.

+ Add to test✓ In your test36IBMA2S-052-00-035.1 Limits and the derivative Paper 1Easy5 marks

No calculator

After a capsule's re-entry burn ends, the temperature of a sensor on its heat shield, $T$ °C, is modelled by $T(t) = 20 + 600(1 + 2t)^{-1}$, where $t$ is the time in minutes after the burn ends.

(a)

Find the temperature of the sensor at the moment the burn ends.

(1)

(b)

Find $T'(t)$.

(2)

(c)

Find the rate of change of the temperature when $t = 2$, and interpret its sign.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $T(0) = 620$ °C	✓ 1
Part (b)
M1 uses the chain rule, with power reduced to $-2$ and a factor of 2 from the inner function A1 $T'(t) = -1200(1 + 2t)^{-2}$	✓ 2
Part (c)
M1 substitutes $t = 2$: $-\frac{1200}{25}$ A1 $-48$ °C per minute, so the temperature is decreasing by 48 °C per minute	✓ 2

Common error: Forgetting to multiply by the derivative of the inner function $1 + 2t$ gives $-600(1+2t)^{-2}$, which is out by a factor of 2.

+ Add to test✓ In your test37IBMA2S-052-04-015.5 Introduction to integration Paper 1Easy5 marks

No calculator

A curve $y=f(x)$, defined for $x>0$, has gradient function $f'(x)=3\sqrt{x}-\frac{2}{x^{2}}$. The curve passes through the point $(4,\,10)$.

(a)

Find an expression for $f(x)$.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate using the power rule (e.g. at least one term of $3x^{1/2}-2x^{-2}$ with its power increased by 1)	✓ 1
A1 $2x^{3/2}$	✓ 1
A1 $+\frac{2}{x}$	✓ 1
M1 substitute $x=4$, $y=10$ into an expression including $+c$: $16+\frac{1}{2}+c=10$	✓ 1
A1 $f(x)=2x^{3/2}+\frac{2}{x}-\frac{13}{2}$	✓ 1

Common error: Integrating $-2x^{-2}$ gives $+2x^{-1}$, not $-2x^{-1}$: dividing by the new power $-1$ changes the sign.

+ Add to test✓ In your test38IBMA2S-052-04-025.10 Indefinite integrals and reverse chain rule Paper 1Easy4 marks

No calculator

Find each of the following indefinite integrals.

(a)

$\int (2x+5)^{4}\,dx$

(2)

(b)

$\int \frac{1}{4x-3}\,dx$

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 integral of the form $k(2x+5)^{5}$	✓ 1
A1 $\frac{1}{10}(2x+5)^{5}+c$	✓ 1
Part (b)
M1 integral of the form $k\ln\|4x-3\|$	✓ 1
A1 $\frac{1}{4}\ln\|4x-3\|+c$	✓ 1

Exam tip: For a linear inner function $ax+b$, integrate as normal and then divide by $a$. Check by differentiating your answer mentally.

+ Add to test✓ In your test39IBMA2S-052-04-035.9 Kinematics Paper 1Easy6 marks

No calculator

A track cyclist starts from rest at the start line. Her velocity, $v$ m s⁻¹, at time $t$ seconds is modelled by $v(t)=8-8e^{-0.5t}$, $t\geq 0$. Her displacement from the start line at time $t$ is $s$ metres.

(a)

Find an expression for $s(t)$.

(4)

(b)

Show that $s(2)=\frac{16}{e}$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate $v(t)$	✓ 1
A1 $8t+16e^{-0.5t}+c$	✓ 1
M1 use $s(0)=0$: $16+c=0$	✓ 1
A1 $s(t)=8t+16e^{-0.5t}-16$	✓ 1
Part (b)
M1 substitute $t=2$: $16+16e^{-1}-16$	✓ 1
A1 $=16e^{-1}=\frac{16}{e}$ (AG)	✓ 1

Common error: Assuming $c=0$ because the cyclist starts at the origin. At $t=0$ the term $16e^{0}=16$ is not zero, so $c=-16$.

+ Add to test✓ In your test40IBMA2S-052-04-045.10 Indefinite integrals and reverse chain rule Paper 1Easy4 marks

No calculator

This question uses the result of part (a) to find an integral by inspection.

(a)

Find $\frac{d}{dx}\left(\sin(x^{3})\right)$.

(2)

(b)

Hence find $\int x^{2}\cos(x^{3})\,dx$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the chain rule	✓ 1
A1 $3x^{2}\cos(x^{3})$	✓ 1
Part (b)
M1 use of part (a), recognising the integrand is $\frac{1}{3}$ of the derivative	✓ 1
A1 $\frac{1}{3}\sin(x^{3})+c$	✓ 1

Exam tip: "Hence" means use the previous result: adjust the constant factor so that differentiating your answer gives exactly the integrand.

+ Add to test✓ In your test41IBMA2S-052-05-015.9 Kinematics Paper 1Easy5 marks

No calculator

A delivery drone takes off from the roof of a building. Its height above the ground, h metres, t seconds after take-off is modelled by h(t) = t³ − 9t² + 24t + 10, for 0 ≤ t ≤ 6.

(a)

Find an expression for h''(t).

(2)

(b)

Find the value of t at which the graph of h has a point of inflection, justifying your answer.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate twice, e.g. h'(t) = 3t² − 18t + 24	✓ 1
A1 h''(t) = 6t − 18	✓ 1
Part (b)
M1 sets h''(t) = 0, i.e. 6t − 18 = 0	✓ 1
A1 t = 3 s	✓ 1
R1 h''(t) < 0 for t < 3 and h''(t) > 0 for t > 3, so the concavity changes at t = 3	✓ 1

Exam tip: Solving h''(t) = 0 only gives a candidate. The reasoning mark comes from showing the second derivative changes sign either side of the value.

+ Add to test✓ In your test42IBMA2S-052-05-035.3 Differentiating polynomials and integer powers Paper 1Easy6 marks

No calculator

Let f(x) = x⁴ − 6x² + 3, for x ∈ ℝ.

(a)

Find f''(x).

(2)

(b)

Find the set of values of x for which the graph of f is concave down.

(2)

(c)

Find the coordinates of the points of inflection on the graph of f.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 f'(x) = 4x³ − 12x	✓ 1
A1 f''(x) = 12x² − 12	✓ 1
Part (b)
M1 solves 12x² − 12 < 0, or finds the roots x = ±1 of f''(x) = 0	✓ 1
A1 −1 < x < 1	✓ 1
Part (c)
M1 substitutes x = ±1 into f(x), stating that f''(x) changes sign at x = ±1 (for example, using the concavity found in the previous part)	✓ 1
A1 (−1, −2) and (1, −2)	✓ 1

Exam tip: f''(x) = 0 alone does not guarantee a point of inflection. State that f'' changes sign there. Because f is even, both points must share the same y-coordinate, which is a quick check.

+ Add to test✓ In your test43IBMA2S-052-05-045.8 Stationary points, inflexion and optimization Paper 1Easy6 marks

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An e-bike company models its weekly profit, P thousand dollars, by P(x) = −2x³ + 15x² + 36x − 20, where x is the number of bikes produced, in hundreds, and 0 ≤ x ≤ 8.

(a)

Find the value of x for which P'(x) = 0.

(3)

(b)

Use the second derivative to show that this value of x gives a maximum profit.

(2)

(c)

Find the maximum weekly profit.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 P'(x) = −6x² + 30x + 36	✓ 1
M1 sets equal to zero and factorises: −6(x − 6)(x + 1) = 0	✓ 1
A1 x = 6 (rejecting x = −1)	✓ 1
Part (b)
M1 P''(x) = −12x + 30	✓ 1
R1 P''(6) = −42 < 0, so a local maximum	✓ 1
Part (c)
A1 P(6) = 304 thousand dollars ($304 000)	✓ 1

Exam tip: In a "show that" part, state the sign of P'' explicitly and draw the conclusion in words. A number alone does not earn the reasoning mark.

+ Add to test✓ In your test44IBMA2S-052-08-025.1 Limits and the derivative Paper 1Easy5 marks

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A 3D printer builds a solid cube layer by layer. When the side length of the cube is x cm, its volume is V(x) = x³ cm³.

(a)

Show that the average rate of change of volume with respect to side length, as the side increases from 2 cm to (2 + h) cm, is 12 + 6h + h².

(3)

(b)

Hence find the instantaneous rate of change of volume with respect to side length when x = 2, stating its units.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 writes $\frac{V(2+h)-V(2)}{h}=\frac{(2+h)^3-8}{h}$	✓ 1
A1 correct expansion (2 + h)³ = 8 + 12h + 6h² + h³	✓ 1
A1 $\frac{12h+6h^2+h^3}{h}=12+6h+h^2$ correctly simplified (AG)	✓ 1
Part (b)
M1 lets h → 0 in 12 + 6h + h²	✓ 1
A1 12 cm³ per cm (accept 12 cm²)	✓ 1

Common error: Substituting h = 0 before cancelling the h in the denominator gives 0/0. Simplify the quotient first, then let h tend to 0.

+ Add to test✓ In your test45IBMA2S-052-08-045.1 Limits and the derivative Paper 1Easy5 marks

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Oil leaking from a stationary tanker spreads out on calm water as a circular slick. When the radius of the slick is r metres, its area is A = πr² m².

(a)

Write down $\frac{dA}{dr}$.

(1)

(b)

Find the rate of change of area with respect to radius when r = 5, giving an exact answer with units.

(2)

(c)

Find the radius of the slick at the moment when the area is changing at 18π m² per metre of radius.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $\frac{dA}{dr}=2\pi r$	✓ 1
Part (b)
M1 substitutes r = 5 into 2πr	✓ 1
A1 10π m² per metre	✓ 1
Part (c)
M1 sets 2πr = 18π	✓ 1
A1 r = 9 m	✓ 1

Common error: dA/dr is a rate with respect to radius, not time, so its units are m² per metre, not m² per second.

+ Add to test✓ In your test46IBMA2S-052-09-015.1 Limits and the derivative Paper 1Easy6 marks

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During a storm, rainwater flows into a garden water tank. The volume of water in the tank, $V$ litres, $t$ hours after the storm begins, changes at a rate given by
$\frac{dV}{dt} = 6t^2 - 4t + 5$, for $0 \le t \le 4$.
When $t = 2$, the tank contains 40 litres of water.

(a)

Find an expression for $V$ in terms of $t$.

(5)

(b)

Write down the volume of water in the tank at the moment the storm begins.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate (power increased by one on at least one term)	✓ 1
A1 $2t^3$	✓ 1
A1 $-2t^2 + 5t + c$	✓ 1
M1 substitute $t = 2$, $V = 40$: $16 - 8 + 10 + c = 40$	✓ 1
A1 $V = 2t^3 - 2t^2 + 5t + 22$	✓ 1
Part (b)
A1 22 litres	✓ 1

Common error: Leaving out the constant of integration means the given condition cannot be used, and the method marks for finding it are lost. Write + c straight after integrating.

+ Add to test✓ In your test47IBMA2S-052-09-025.5 Introduction to integration Paper 1Easy6 marks

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The curved edge of a laptop cooling fin is modelled by the curve $y = f(x)$, $x > 0$, where
$f'(x) = 3 - \frac{8}{x^3}$.
The curve passes through the point $(2, 9)$.

(a)

Find $f(x)$.

(5)

(b)

Find the value of $f(4)$.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate, e.g. writing $-8x^{-3}$ and increasing a power by 1	✓ 1
A1 $3x$	✓ 1
A1 $+4x^{-2}$ (accept $\frac{4}{x^2}$) $+\,c$	✓ 1
M1 substitute $(2, 9)$: $6 + 1 + c = 9$	✓ 1
A1 $f(x) = 3x + \frac{4}{x^2} + 2$	✓ 1
Part (b)
A1 $12 + \frac{1}{4} + 2 = \frac{57}{4}$	✓ 1

Common error: Integrating $-8x^{-3}$ gives $\frac{-8x^{-2}}{-2} = +4x^{-2}$. Dropping one of the two minus signs gives the wrong sign on this term.

+ Add to test✓ In your test48IBMA2S-052-01-015.5 Introduction to integration Paper 1Easy5 marks

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An electric car is plugged into a smart charger. While it is charging, the rate at which energy is delivered is modelled by $r(t) = 2t + 1$ kilowatts, where $t$ is the time in hours after the charger is switched on. The car is charged from $t = 1$ to $t = k$, where $k > 1$. During this time it receives a total of 10 kWh of energy.

(a)

Show that $\displaystyle\int_1^k (2t+1)\,dt = k^2 + k - 2$.

(3)

(b)

Hence find the value of $k$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts to integrate $2t+1$	✓ 1
A1 obtains $[t^2 + t]$	✓ 1
M1 substitutes the limits: $(k^2 + k) - (1 + 1) = k^2 + k - 2$ (AG)	✓ 1
Part (b)
M1 sets $k^2 + k - 2 = 10$ and solves $k^2 + k - 12 = 0$, i.e. $(k+4)(k-3) = 0$	✓ 1
A1 $k = 3$, rejecting $k = -4$ because $k > 1$	✓ 1

Exam tip: When a definite integral produces a quadratic in the unknown limit, always check each root against the given condition (here $k > 1$) and state why you reject one.

+ Add to test✓ In your test49IBMA2S-052-01-025.5 Introduction to integration Paper 1Easy6 marks

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A graphic designer builds a company logo from the region enclosed between the curve $y = x^2 - 2x$ and the line $y = x$.

Diagram not to scale

(a)

Find the $x$-coordinates of the points where the curve and the line intersect.

(2)

(b)

Find the exact area of the region enclosed by the curve and the line.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 equates $x^2 - 2x = x$, giving $x^2 - 3x = 0$	✓ 1
A1 $x = 0$ and $x = 3$	✓ 1
Part (b)
M1 uses the integral of (upper − lower), with the line above the curve on $0 < x < 3$	✓ 1
A1 $\displaystyle\int_0^3 (3x - x^2)\,dx$	✓ 1
A1 $\left[\frac{3x^2}{2} - \frac{x^3}{3}\right]_0^3$	✓ 1
A1 $\frac{27}{2} - 9 = \frac{9}{2}$	✓ 1

Common error: Using (curve − line) gives $-\frac{9}{2}$. An area is positive, so check which graph is on top, for example by testing $x = 1$.

+ Add to test✓ In your test50IBMA2S-052-01-045.5 Introduction to integration Paper 1Easy5 marks

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The profile of a decorative wave on a phone case is modelled by $f(x) = x^3 - 4x$ for $-2 \le x \le 2$. The curve meets the $x$-axis at $x = -2$, $0$ and $2$.

Diagram not to scale

(a)

Show that $\displaystyle\int_{-2}^{2} f(x)\,dx = 0$.

(2)

(b)

Find the total area enclosed between the curve and the $x$-axis.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 integrates correctly to $\left[\frac{x^4}{4} - 2x^2\right]_{-2}^{2}$	✓ 1
A1 substitutes the limits: $(4 - 8) - (4 - 8) = -4 - (-4) = 0$ (AG)	✓ 1
Alternative: M1 shows $f(-x) = -x^3 + 4x = -f(x)$, so $f$ is odd; A1 states that the integral of an odd function over an interval symmetric about $0$ is $0$ (AG).	—
Part (b)
M1 splits the integral at $x = 0$, recognising that the curve is above the axis for $-2 < x < 0$ and below the axis for $0 < x < 2$	✓ 1
A1 $\displaystyle\int_{-2}^0 f(x)\,dx = 4$ (or $\displaystyle\int_0^2 f(x)\,dx = -4$)	✓ 1
A1 total area $= 4 + 4 = 8$	✓ 1
Accept a symmetry argument giving the total area as $2 \times 4$.	—

Common error: A definite integral of zero does not mean zero area. Regions below the $x$-axis contribute negative values, so split the integral at every root.

+ Add to test✓ In your test51IBMA2S-052-02-015.2 Increasing and decreasing functions Paper 1Easy5 marks

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The function $f$ is defined by $f(x)=x^3+kx^2+5$, where $k$ is a constant. The graph of $f$ has a stationary point at $x=4$.

(a)

Find the value of $k$.

(3)

(b)

Determine whether this stationary point is a local maximum or a local minimum. Justify your answer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiates: $f'(x)=3x^2+2kx$	✓ 1
M1 substitutes $x=4$ and sets $f'(4)=0$: $48+8k=0$	✓ 1
A1 $k=-6$	✓ 1
Part (b)
A1 $f''(x)=6x-12$, so $f''(4)=12$	✓ 1
R1 $f''(4)>0$, therefore the stationary point is a local minimum	✓ 1

Exam tip: A nature test earns its reasoning mark only when you state the sign of $f''$ at the point AND link it to the conclusion, e.g. '$f''(4)>0$ so minimum'.

+ Add to test✓ In your test52IBMA2S-052-02-045.1 Limits and the derivative Paper 1Easy5 marks

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A delivery drone flies at a constant speed of $v$ km h⁻¹. The energy it uses per kilometre, $E$ watt-hours, is modelled by
$E(v)=\frac{1600}{v}+\frac{v}{4}$, for $v>0$.

(a)

Find the speed that minimises the energy used per kilometre.

(3)

(b)

Verify that this speed gives a minimum.

(1)

(c)

Find the minimum energy used per kilometre.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiates: $E'(v)=-1600v^{-2}+\frac{1}{4}$	✓ 1
M1 sets $E'(v)=0$ to obtain $v^2=6400$	✓ 1
A1 $v=80$ km h⁻¹; reject $v=-80$ because $v>0$	✓ 1
Part (b)
R1 Either second-derivative test: $E''(v)=\frac{3200}{v^3}$, so $E''(80)=\frac{1}{160}>0$, hence a minimum; or first-derivative sign change: $E'(40)=-\frac{3}{4}<0$ and $E'(100)=\frac{9}{100}>0$, hence a minimum	✓ 1
Part (c)
A1 $E(80)=\frac{1600}{80}+\frac{80}{4}=20+20=40$ Wh per km	✓ 1

Common error: Writing the derivative of $\frac{1600}{v}$ as $1600\ln v$, which confuses differentiation with integration. Rewrite the term as $1600v^{-1}$ first, which gives $-1600v^{-2}$.

+ Add to test✓ In your test53IBMA2S-052-07-015.4 Tangents and normals Paper 1Easy6 marks

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In a scale model, a section of a roller-coaster track is modelled by $y = x^3 - 3x^2 + 4$, for $0 \le x \le 4$, where $x$ and $y$ are measured in metres.

(a)

Find the gradient of the track at the point where $x = 3$.

(2)

(b)

Find the equation of the tangent to the track at the point where $x = 3$. Give your answer in the form $y = mx + c$.

(3)

(c)

Write down the $y$-intercept of this tangent.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate: $\frac{dy}{dx} = 3x^2 - 6x$	✓ 1
A1 gradient = 9	✓ 1
Part (b)
A1 $y = 4$ when $x = 3$	✓ 1
M1 substitution of their point and gradient into a line equation, e.g. $y - 4 = 9(x - 3)$ or $4 = 9(3) + c$	✓ 1
A1 $y = 9x - 23$	✓ 1
Part (c)
A1 $-23$ (m), follow through from their tangent	✓ 1

Exam tip: Always find the $y$-coordinate from the original function, not from the derivative. Substituting $x = 3$ into $\frac{dy}{dx}$ gives the gradient, not the point.

+ Add to test✓ In your test54IBMA2S-052-07-085.6 Differentiation rules Paper 1Easy5 marks

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Part of the profile of a water-park slide is modelled by $y = x\ln x$, for $x \ge 1$.

(a)

Find $\frac{dy}{dx}$.

(2)

(b)

Find the coordinates of the point on the slide at which the tangent is parallel to the line $y = 2x + 5$.

(2)

(c)

Find the $y$-intercept of this tangent.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule	✓ 1
A1 $\ln x + 1$	✓ 1
Part (b)
M1 setting $\ln x + 1 = 2$	✓ 1
A1 $(e, e)$	✓ 1
Part (c)
A1 $-e$, from $y - e = 2(x - e)$	✓ 1

Exam tip: Parallel lines have equal gradients, so you only need the gradient of $y = 2x + 5$, not its intercept.

+ Add to test✓ In your test55IBMA2S-052-07-095.9 Kinematics Paper 1Easy7 marks

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During a sprint, the distance travelled by a cyclist is modelled by $s(t) = 0.6t^2 - 0.01t^3$, $0 \le t \le 40$, where $s$ is in metres and $t$ is in seconds.

(a)

Find the equation of the tangent to the graph of $s$ at $t = 30$.

(3)

(b)

Show that this tangent passes through the origin, and interpret this result in context.

(2)

(c)

Find the other value of $t$ at which the tangent to the graph of $s$ is parallel to the tangent at $t = 30$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $s'(t) = 1.2t - 0.03t^2$	✓ 1
A1 $s'(30) = 9$ and $s = 270$ at $t = 30$	✓ 1
A1 $s = 9t$	✓ 1
Part (b)
R1 the tangent $s = 9t$ has zero intercept, so passes through $(0, 0)$	✓ 1
R1 the average speed over the first 30 s equals the instantaneous speed at $t = 30$, namely 9 m s⁻¹	✓ 1
Part (c)
M1 solving $1.2t - 0.03t^2 = 9$, i.e. $t^2 - 40t + 300 = 0$	✓ 1
A1 $t = 10$ s	✓ 1

Exam tip: On a distance–time graph, the gradient of a chord from the origin is the average speed and the gradient of the tangent is the instantaneous speed.

+ Add to test✓ In your test56IBMA2S-052-06-015.2 Increasing and decreasing functions Paper 1Easy5 marks

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A robotic welding arm rotates about a fixed joint. Its angle of rotation, θ degrees, measured from its starting position, is modelled by θ(t) = t³ − 9t² + 24t, where t is the time in seconds and 0 ≤ t ≤ 5.

(a)

Find θ′(t).

(1)

(b)

Find the values of t at which the arm is momentarily stationary.

(2)

(c)

Hence find the interval of time during which θ is decreasing, justifying your answer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 θ′(t) = 3t² − 18t + 24	✓ 1
Part (b)
M1 sets 3t² − 18t + 24 = 0, e.g. 3(t − 2)(t − 4) = 0	✓ 1
A1 t = 2 and t = 4	✓ 1
Part (c)
R1 θ′(t) < 0 between the roots. Justify this either by testing a value, e.g. θ′(3) = 27 − 54 + 24 = −3 < 0, or by noting that the quadratic has a positive leading coefficient and so is negative between its roots	✓ 1
A1 2 < t < 4 (accept 2 ≤ t ≤ 4)	✓ 1

Exam tip: Always justify the sign of the derivative between stationary points. You can do this by testing a value or by referring to the shape of the quadratic.

+ Add to test✓ In your test57IBMA2S-052-06-045.8 Stationary points, inflexion and optimization Paper 1Easy5 marks

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A delivery company models the running cost, C dollars per kilometre, of a van travelling at a constant speed v km h⁻¹ by C(v) = v/20 + 80/v, for v > 0.

(a)

Find C′(v).

(2)

(b)

Find the set of speeds for which the running cost per kilometre is decreasing.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 derivative of v/20 is 1/20	✓ 1
A1 derivative of 80/v is −80v⁻², giving C′(v) = 1/20 − 80/v²	✓ 1
Part (b)
M1 sets C′(v) < 0, i.e. 1/20 − 80/v² < 0	✓ 1
A1 v² < 1600	✓ 1
A1 0 < v < 40	✓ 1

Common error: Writing −40 < v < 40 ignores the domain; the model only applies for v > 0.

+ Add to test✓ In your test58IBMA2S-052-03-015.9 Kinematics Paper 1Easy6 marks

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The gripper of a robotic arm slides along a straight rail. Its displacement from a fixed sensor, $s$ cm, at time $t$ seconds is given by $s(t)=2t^3-15t^2+36t$, for $t\ge 0$.

(a)

Find an expression for the velocity $v(t)$ of the gripper.

(2)

(b)

Find the values of $t$ at which the gripper is at rest.

(2)

(c)

Find the acceleration of the gripper the first time it is at rest.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate $s(t)$	✓ 1
A1 $v(t)=6t^2-30t+36$	✓ 1
Part (b)
M1 sets $v(t)=0$, e.g. $6(t-2)(t-3)=0$	✓ 1
A1 $t=2$ and $t=3$	✓ 1
Part (c)
M1 $a(t)=12t-30$, substitutes $t=2$	✓ 1
A1 $a=-6$ cm s⁻²	✓ 1

Exam tip: "At rest" means velocity is zero, not acceleration. Factorise the quadratic fully to find both times.

+ Add to test✓ In your test59IBMA2S-052-03-035.9 Kinematics Paper 1Easy5 marks

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A delivery drone takes off vertically from the ground. Its upward velocity is $v(t)=4t-t^2$ m s⁻¹ for $0\le t\le 4$, where $t$ is in seconds.

(a)

Find the maximum upward velocity of the drone.

(2)

(b)

Find the height of the drone at $t=4$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sets $v'(t)=4-2t=0$, giving $t=2$ (or completes the square: $v=4-(t-2)^2$, maximum when $t=2$)	✓ 1
A1 maximum velocity $v(2)=4$ m s⁻¹	✓ 1
Part (b)
M1 integrates velocity: $\int_0^4(4t-t^2)\,dt$	✓ 1
A1 $\left[2t^2-\frac{t^3}{3}\right]_0^4$ (accept $32-\frac{64}{3}$)	✓ 1
A1 height $=\frac{32}{3}$ m	✓ 1

Common error: Substituting $t=4$ into $v(t)$ gives 0. That is the velocity at $t=4$, not the height. To find the height, integrate the velocity from the take-off time.

+ Add to test✓ In your test60IBMA3S-5-17r5-015.2 Increasing and decreasing functions Paper 1Easy5 marks

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A space probe moves away from a planet. Its distance from the planet's surface is modelled by s(t) = 2t³ − 15t² + 24t + 10, where s is in thousands of kilometres and t is the time in hours after a course correction, 0 ≤ t ≤ 6.

(a)

Find ds/dt.

(2)

(b)

Find the values of t for which the probe is getting closer to the planet.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate (at least one power reduced correctly)	✓ 1
A1 6t² − 30t + 24	✓ 1
Part (b)
M1 set ds/dt < 0 (or = 0) and attempt to solve, e.g. 6(t − 1)(t − 4)	✓ 1
A1 critical values t = 1 and t = 4	✓ 1
A1 1 < t < 4	✓ 1

Exam tip: "Getting closer" means the distance is decreasing, so look for where the derivative is negative, not zero.

+ Add to test✓ In your test61IBMA3S-5-17r5-025.4 Tangents and normals Paper 1Easy4 marks

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The upper edge of a footbridge arch is modelled by y = 8x − x², 0 ≤ x ≤ 8, where x and y are in metres and the origin is at the left support.

Diagram not to scale

(a)

Find the gradient of the arch at the point where x = 3.

(2)

(b)

Find the equation of the tangent to the arch at x = 3. Give your answer in the form y = mx + c.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 dy/dx = 8 − 2x	✓ 1
A1 gradient = 2	✓ 1
Part (b)
M1 point (3, 15) found and used with their gradient, e.g. y − 15 = 2(x − 3)	✓ 1
A1 y = 2x + 9	✓ 1

Common error: Forgetting to find the y-coordinate: the gradient alone does not fix a tangent, you also need a point on the curve.

+ Add to test✓ In your test62IBMA3S-5-17r5-035.6 Differentiation rules Paper 1Easy5 marks

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The drum temperature T °C of a coffee roaster, t minutes after the beans are loaded, satisfies dT/dt = 3t² − 6t + 8, 0 ≤ t ≤ 12. When t = 0, T = 180.

(a)

Find an expression for T in terms of t.

(3)

(b)

Find the drum temperature when t = 5.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate (at least one term correct)	✓ 1
A1 T = t³ − 3t² + 8t + c	✓ 1
A1 c = 180	✓ 1
Part (b)
M1 substitute t = 5 into their expression (ECF from (a))	✓ 1
A1 270 °C	✓ 1

Common error: Leaving out the constant of integration. The initial condition is what fixes it, so use it straight after integrating.

+ Add to test✓ In your test63IBMA3S-5-17r5-045.9 Kinematics Paper 1Easy5 marks

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A model for the strain in a bridge cable at distance x metres from a pylon is f(x) = 5x³ − 4/x² + 6√x, for x > 0.

(a)

Find f′(x).

(3)

(b)

Find the exact value of f′(4).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rewrite as powers, e.g. −4x⁻² and 6x^1/2	✓ 1
A1 15x² + 8x⁻³	✓ 1
A1 + 3x^−1/2	✓ 1
Part (b)
M1 substitute x = 4: 15(16) + 8/64 + 3/2 (ECF from (a))	✓ 1
A1 241.625 (that is 1933/8)	✓ 1

Common error: Differentiating a term like 4/x² without first writing it as 4x⁻² often leads to a wrong power or sign.

+ Add to test✓ In your test64IBMA3S-5-16r7-015.3 Differentiating polynomials and integer powers Paper 1Easy4 marks

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Let $f(x)=2x^5-\frac{3}{x^2}+7x$, for $x\neq 0$.

(a)

Find $f'(x)$.

(3)

(b)

Find $f'(1)$.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate term by term	✓ 1
A1 $10x^4$	✓ 1
A1 $+6x^{-3}+7$, giving $f'(x)=10x^4+6x^{-3}+7$	✓ 1
Part (b)
A1 $f'$ evaluated at $x=1$: $10+6+7=23$	✓ 1

Common error: Write $\frac{3}{x^2}$ as $3x^{-2}$ before differentiating; the sign of the term changes when the power is brought down.

+ Add to test✓ In your test65IBMA3S-5-16r7-025.10 Indefinite integrals and reverse chain rule Paper 1Easy5 marks

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(a)

Find $\int 6x(x^2+3)^4\,dx$.

(3)

(b)

Hence find $\int_0^1 6x(x^2+3)^4\,dx$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to reverse the chain rule, e.g. $(x^2+3)^5$ seen	✓ 1
A1 coefficient $\frac{6}{5\times 2}=\frac35$, giving $\frac35(x^2+3)^5$	✓ 1
A1 $+C$	✓ 1
Part (b)
M1 substitute limits 1 and 0 into the antiderivative, $\frac35(4^5-3^5)$	✓ 1
A1 $\frac{2343}{5}$	✓ 1

Exam tip: Check an antiderivative by differentiating it; the result should return the original integrand.

+ Add to test✓ In your test66IBMA3S-5-16r7-035.9 Kinematics Paper 1Easy7 marks

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A train leaves a station and moves along a straight horizontal track. Its velocity, in m s⁻¹, at time t seconds after leaving is $v(t)=12t-3t^2$, for $0\le t\le 6$.

(a)

Find the acceleration of the train when $t=1$.

(2)

(b)

Find the value of t, for $t>0$, at which the train is momentarily at rest.

(2)

(c)

Find the distance travelled by the train in the first 4 seconds.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate $v$, $a=12-6t$	✓ 1
A1 $6\text{ m s}^{-2}$	✓ 1
Part (b)
M1 set $12t-3t^2=0$ and factorise	✓ 1
A1 $t=4$	✓ 1
Part (c)
M1 attempt to integrate $v$ from 0 to 4	✓ 1
A1 $6t^2-t^3$ seen	✓ 1
A1 $96-64=32$ m	✓ 1

Common error: Distance is found by integrating the velocity only while the velocity does not change sign; here it stays positive on $0<t<4$.

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The profile of a ski-jump ramp is modelled by the curve $y=2x^3-5x^2+4$, where x and y are in metres.

(a)

Find $\frac{dy}{dx}$.

(2)

(b)

Find the equation of the tangent to the curve at the point where $x=2$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate	✓ 1
A1 $6x^2-10x$	✓ 1
Part (b)
A1 $y=0$ and gradient $=4$ at $x=2$	✓ 1
A1 $y=4x-8$	✓ 1

Exam tip: Find both the gradient and the y-value at the given x before writing the tangent equation.

+ Add to test✓ In your test68IBMA3S-5-17r7-015.10 Indefinite integrals and reverse chain rule Paper 1Easy5 marks

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A function f has derivative $f'(x)=12x(3x^2-5)^3$.

(a)

Find $\int 12x(3x^2-5)^3\,dx$.

(3)

(b)

The graph of $y=f(x)$ passes through the point $(1,\,2)$. Find the value of the constant of integration.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the reverse chain rule, giving an expression of the form $k(3x^2-5)^4$	✓ 1
A1 correct coefficient, $\frac{1}{2}(3x^2-5)^4$	✓ 1
A1 includes the constant: $\frac{1}{2}(3x^2-5)^4+C$	✓ 1
Part (b)
M1 substitutes $x=1$, $y=2$ into their expression: $2=\frac{1}{2}(3-5)^4+C$	✓ 1
A1 $C=2-8=-6$ (ECF from (a))	✓ 1

Common error: Forgetting that the inner derivative 6x must be matched exactly: here 12x is twice 6x, so the integral needs a factor of 2 outside before dividing by 4.

+ Add to test✓ In your test69IBMA3S-5-17r7-035.4 Tangents and normals Paper 1Easy6 marks

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The curve C has equation y = x³ − 4x + 1. The point P on C has x-coordinate 2.

(a)

Find the gradient of the tangent to C at P.

(2)

(b)

Find the equation of the normal to C at P, giving your answer in the form ax + by = c, where a, b, c are integers.

(3)

(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 attempt to differentiate: $\frac{dy}{dx}=3x^2-4$	✓ 1
A1 substitutes $x=2$ to obtain gradient 8	✓ 1
Part (b)
A1 $y=8-8+1=1$, so P is $(2,\,1)$	✓ 1
M1 normal gradient $-\frac{1}{8}$ used in $y-1=-\frac{1}{8}(x-2)$	✓ 1
A1 $x+8y=10$	✓ 1
Part (c)
A1 sets $y=0$: $x=10$	✓ 1

Common error: The normal gradient is the negative reciprocal of the tangent gradient; using −8 or 1/8 loses the accuracy marks.

+ Add to test✓ In your test70IBMA3S-5-17r7-045.2 Increasing and decreasing functions Paper 1Easy5 marks

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The function f is defined by f(x) = 2x³ − 9x² − 24x + 5, for x ∈ ℝ.

(a)

Find f′(x).

(2)

(b)

Find the set of values of x for which f is decreasing.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate (at least two terms correct)	✓ 1
A1 $f'(x)=6x^2-18x-24$	✓ 1
Part (b)
M1 sets $f'(x)<0$ (or $=0$) and factorises: $6(x-4)(x+1)$	✓ 1
A1 critical values $x=-1$ and $x=4$	✓ 1
R1 the quadratic has positive leading coefficient so $f'(x)<0$ between the roots, giving $-1<x<4$	✓ 1

Exam tip: A sign diagram or a quick sketch of the gradient function justifies which side of the roots is negative.

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A cycling route is surveyed and the height of the road is modelled by $h(x)=2x^3-9x^2+12x$ for $0\le x\le 3$, where $x$ is the horizontal distance in hundreds of metres from the start and $h$ is the height in tens of metres above the start.

(a)

Find $h'(x)$.

(2)

(b)

Find the values of $x$ at the stationary points of the road profile.

(2)

(c)

Find $h''(x)$ and hence determine the nature of each stationary point.

(3)

(d)

Find the difference in height, in metres, between the local maximum and the local minimum.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate (at least one term correct)	✓ 1
A1 $h'(x)=6x^2-18x+12$	✓ 1
Part (b)
M1 sets $h'(x)=0$ and factorises: $6(x-1)(x-2)=0$	✓ 1
A1 $x=1$ and $x=2$	✓ 1
Part (c)
A1 $h''(x)=12x-18$	✓ 1
M1 evaluates $h''\left(1\right)=-6$ and $h''(2)=6$	✓ 1
A1 local maximum at $x=1$ ($h''<0$) and local minimum at $x=2$ ($h''>0$)	✓ 1
Part (d)
A1 $h\left(1\right)=5$ and $h(2)=4$	✓ 1
A1 difference $=1$ ten of metres $=10$ m	✓ 1

Common error: Remember the units of h here are tens of metres, so the final height difference needs converting.

+ Add to test✓ In your test72IBMA3S-5-17r9-025.4 Tangents and normals Paper 1Easy6 marks

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The side wall of an excavated ditch is modelled by the curve $y=x^2-4x+7$ for $0\le x\le 5$, where $x$ and $y$ are measured in metres from a reference peg.

(a)

Find the equation of the tangent to the curve at the point where $x=3$. Give your answer in the form $y=mx+c$.

(3)

(b)

The normal to the curve at $x=3$ is extended until it meets the $x$-axis. Find the $x$-coordinate of this point.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $\frac{dy}{dx}=2x-4$ and gradient $=2$ at $x=3$	✓ 1
A1 $y=4$ when $x=3$	✓ 1
A1 tangent $y=2x-2$	✓ 1
Part (b)
A1 gradient of the normal $=-\frac12$	✓ 1
M1 equation $y-4=-\frac12(x-3)$	✓ 1
A1 setting $y=0$ gives $x=11$	✓ 1

Exam tip: The normal gradient is the negative reciprocal of the tangent gradient; check it by confirming that the product of the two gradients is -1.

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After start-up, the extra electrical load of a data-centre cooling unit is modelled by $L(t)=\dfrac{3t}{t^2+1}$ kW, where $t\ge 0$ is the time in hours.

(a)

Use the quotient rule to show that $L'(t)=\dfrac{3(1-t^2)}{(t^2+1)^2}$.

(3)

(b)

Find the greatest load, in kW, for $t\ge0$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 quotient rule with $u=3t$, $v=t^2+1$	✓ 1
A1 $L'(t)=\frac{3(t^2+1)-3t(2t)}{(t^2+1)^2}$	✓ 1
A1 simplifies to $\frac{3-3t^2}{(t^2+1)^2}=\frac{3(1-t^2)}{(t^2+1)^2}$ (AG)	✓ 1
Part (b)
M1 sets $L'(t)=0$, so $t=1$ (since $t\ge 0$)	✓ 1
A1 $L\left(1\right)=\frac{3}{2}=1.5$ kW	✓ 1

Common error: When a fraction is zero, only its numerator needs to be zero; do not expand the denominator.

+ Add to test✓ In your test74IBMA3S-5-17r10-015.1 Limits and the derivative Paper 1Easy4 marks

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The gas supply to a coffee roaster has flow rate $r(t)=6t^2+4$ litres per minute, where $t$ is the time in minutes since the burner was lit.

(a)

Write down an integral for the total volume $V$ of gas used between $t=1$ and $t=3$.

(1)

(b)

Hence find $V$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$V=\int_1^3 (6t^2+4)\,dt$	✓ 1
Part (b)
antiderivative $2t^3+4t$	✓ 1
$(54+12)-(2+4)$	✓ 1
$V=60$ litres	✓ 1

Exam tip: Write the limits on the integral before integrating, and substitute upper limit minus lower limit.

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A curved footbridge ramp has height $y$ metres at horizontal distance $x$ metres, given by $y=x^3-3x+2$ for $0\le x\le 3$.

(a)

Find $\frac{dy}{dx}$ and hence the gradient of the ramp at $x=2$.

(2)

(b)

Find the equation of the tangent to the ramp profile at $x=2$. Give your answer in the form $y=mx+c$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$\frac{dy}{dx}=3x^2-3$	✓ 1
gradient $=3(4)-3=9$	✓ 1
Part (b)
$y=8-6+2=4$, so the point is $(2,4)$	✓ 1
attempt $y-4=9(x-2)$	✓ 1
$y=9x-14$	✓ 1

Common error: Substitute $x$ into the original function to find the $y$-coordinate, not into the derivative.

+ Add to test✓ In your test76IBMA3S-5-17r10-035.6 Differentiation rules Paper 1Easy5 marks

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The strength of a signal from a space probe is modelled by $S(t)=t\,e^{-t/2}$ millivolts, for $t\ge 0$, where $t$ is the time in hours after a manoeuvre.

(a)

Show that $S'(t)=e^{-t/2}\left(1-\frac{t}{2}\right)$.

(3)

(b)

Find the maximum signal strength, giving your answer in exact form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
attempt product rule with $u=t$, $v=e^{-t/2}$	✓ 1
$\frac{d}{dt}e^{-t/2}=-\frac12 e^{-t/2}$	✓ 1
$S'(t)=e^{-t/2}-\frac t2e^{-t/2}=e^{-t/2}\left(1-\frac t2\right)$	✓ 1
Part (b)
$S'(t)=0$ and $e^{-t/2}>0$ so $t=2$	✓ 1
$S(2)=2e^{-1}$ (≈ 0.7358 mV)	✓ 1

Exam tip: An exponential is never zero, so solve the remaining factor equal to zero.

+ Add to test✓ In your test77IBMA3S-5-17r10-045.2 Increasing and decreasing functions Paper 1Easy6 marks

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The height $h$ (in tens of metres) of a bridge deck above the river at horizontal position $x$ (in hundreds of metres) is $h(x)=x^3-6x^2+9x+1$ for $0\le x\le 4$.

(a)

Find the values of $x$ at which $\frac{dh}{dx}=0$.

(3)

(b)

Show that the deck has a point of inflexion and find its coordinates.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$h'(x)=3x^2-12x+9$	✓ 1
$3(x-1)(x-3)=0$	✓ 1
$x=1$ and $x=3$	✓ 1
Part (b)
$h''(x)=6x-12$ and $h''(x)=0$ gives $x=2$	✓ 1
$h''$ changes sign (negative for $x<2$, positive for $x>2$)	✓ 1
$h(2)=8-24+18+1=3$, so the point is $(2,3)$	✓ 1

Common error: Showing $f''(x)=0$ is not enough for an inflexion; also show that $f''$ changes sign.

+ Add to test✓ In your test78IBMA3S-5-16r12-015.10 Indefinite integrals and reverse chain rule Paper 1Easy5 marks

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The function f satisfies $f'(x) = 12x(3x^2-1)^3$, and the graph of y = f(x) passes through the point (1, 10). Find f(x).

(a)

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate by inspection or substitution (reverse chain rule), e.g. $u=3x^2-1$	✓ 1
A1 $(3x^2-1)^4$ seen	✓ 1
A1 $\frac{1}{2}(3x^2-1)^4 + c$	✓ 1
M1 substitutes (1, 10) into their expression for $f(x)$	✓ 1
A1 $c=2$, so $f(x)=\frac{1}{2}(3x^2-1)^4+2$	✓ 1

Exam tip: Differentiate your answer to check it: the derivative of $(3x^2-1)^4$ is $24x(3x^2-1)^3$, so a factor of $\frac{1}{2}$ is needed.

+ Add to test✓ In your test79IBMA3S-5-16r12-025.9 Kinematics Paper 1Easy6 marks

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A rover moves along a straight track. Its velocity, v m s⁻¹, at time t seconds is given by $v = 3t^2-12t+9$, for $0 \le t \le 4$.

(a)

Find the values of t at which the rover is at rest.

(3)

(b)

Find the displacement of the rover from its position at t = 0 to its position at t = 4.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sets $v=0$ (or $3t^2-12t+9=0$)	✓ 1
A1 factorises or solves correctly, e.g. $3(t-1)(t-3)=0$	✓ 1
A1 $t=1$ and $t=3$	✓ 1
Part (b)
M1 attempt to integrate $v$ with limits 0 and 4	✓ 1
A1 $t^3-6t^2+9t$ (condone missing limits)	✓ 1
A1 $64-96+36=4$ m	✓ 1

Common error: Displacement is the integral of velocity and can be positive or negative. It is not the total distance travelled when the direction changes.

+ Add to test✓ In your test80IBMA3S-5-16r12-095.10 Indefinite integrals and reverse chain rule Paper 1Easy5 marks

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Given that $k>0$ and $\int_0^k \dfrac{6}{(3x+1)^2}\,dx=\dfrac{3}{2}$, find the value of k.

(a)

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate by inspection or substitution, e.g. $u=3x+1$	✓ 1
A1 $-\frac{2}{3x+1}$	✓ 1
M1 substitutes limits $k$ and 0 and sets equal to $\frac{3}{2}$, e.g. $-\frac{2}{3k+1}+2=\frac{3}{2}$	✓ 1
A1 $3k+1=4$	✓ 1
A1 $k=1$	✓ 1

Common error: The integral of $(ax+b)^{-2}$ is $-\frac{1}{a}(ax+b)^{-1}$. Do not forget the $\frac{1}{a}$, here $\frac{1}{3}$.

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An F1 team models the lap-time penalty, $P$ seconds, caused by the rear-wing setting $x$ ($1\le x\le 4$) by $P(x)=2x^4-3\sqrt{x}+\frac{5}{x^2}$.

(a)

Find $\frac{dP}{dx}$.

(3)

(b)

Find the rate of change of $P$ with respect to $x$ when $x=1$.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rewriting as $2x^4-3x^{1/2}+5x^{-2}$ and attempting to differentiate	✓ 1
A1 $8x^3-\frac{3}{2}x^{-1/2}$	✓ 1
A1 $-10x^{-3}$, so $\frac{dP}{dx}=8x^3-\frac{3}{2\sqrt{x}}-\frac{10}{x^3}$	✓ 1
Part (b)
A1 $8-\frac32-10=-3.5$ seconds per unit of setting (ECF from (a))	✓ 1

Common error: Rewrite roots and reciprocals as powers of $x$ before differentiating; differentiating $\frac{5}{x^2}$ as $\frac{5}{2x}$ is a frequent slip.

+ Add to test✓ In your test82IBMA3S-5-17r14-025.9 Kinematics Paper 1Easy6 marks

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A syringe pump delivers a drug solution. The volume $V$ mL of solution left in the syringe $t$ minutes after the pump starts ($t\ge 0$) changes at the rate $\frac{dV}{dt}=-\frac{12}{(2t+1)^2}$. Initially the syringe holds 10 mL.

(a)

Find $V$ in terms of $t$.

(5)

(b)

State the volume that the model predicts the syringe approaches as $t$ becomes very large.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate, obtaining an expression of the form $k(2t+1)^{-1}$	✓ 1
A1 $V=\frac{-12(2t+1)^{-1}}{(-1)(2)}+C=\frac{6}{2t+1}+C$	✓ 1
M1 substituting $t=0$, $V=10$	✓ 1
A1 $6+C=10$	✓ 1
A1 $V=\frac{6}{2t+1}+4$	✓ 1
Part (b)
A1 4 mL, since $\frac{6}{2t+1}\to 0$ (ECF from (a))	✓ 1

Common error: When integrating a function of $ax+b$, divide by the coefficient $a$ (here 2) as well as applying the power rule.

+ Add to test✓ In your test83IBMA3S-5-17r14-035.5 Introduction to integration Paper 1Easy4 marks

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The mass $M$ kg of honey in a collecting tank $t$ hours after extraction begins is modelled by $M=t^3\ln t$ for $t\ge 1$.

(a)

Find $\frac{dM}{dt}$.

(3)

(b)

Find the exact rate at which the mass is increasing when $t=e$.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule with $u=t^3$, $v=\ln t$	✓ 1
A1 $3t^2\ln t$	✓ 1
A1 $+t^2$ (from $t^3\times\frac1t$)	✓ 1
Part (b)
A1 $3e^2\ln e+e^2=4e^2$ kg per hour ($\approx 29.556$) (ECF from (a))	✓ 1

Exam tip: Remember $\ln e = 1$; this makes an exact answer in terms of $e$ easy to simplify.

+ Add to test✓ In your test84IBMA3S-5-17r14-045.2 Increasing and decreasing functions Paper 1Easy6 marks

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During an outbreak, the number of patients $N$ on a hospital ward $d$ days after the first case is modelled by $N=d^3-12d^2+36d+20$ for $0\le d\le 8$.

(a)

Find $\frac{dN}{dd}$.

(2)

(b)

Find the values of $d$ at which $\frac{dN}{dd}=0$.

(2)

(c)

Use the second derivative to show that $N$ has a local maximum when $d=2$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate (at least two terms correct)	✓ 1
A1 $3d^2-24d+36$	✓ 1
Part (b)
M1 setting $3d^2-24d+36=0$ and attempting to factorise or solve	✓ 1
A1 $d=2$ and $d=6$	✓ 1
Part (c)
A1 $\frac{d^2N}{dd^2}=6d-24$	✓ 1
R1 at $d=2$ the value is $-12<0$, hence a local maximum	✓ 1

Exam tip: A negative second derivative at a stationary point means a local maximum; always state the value and the conclusion.

+ Add to test✓ In your test85IBMA3S-5-16r16-015.5 Introduction to integration Paper 1Easy5 marks

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The gradient of a curve is given by $\frac{dy}{dx} = 8x(x^2-3)^3$.

(a)

Find $\int 8x(x^2-3)^3\,dx$.

(2)

(b)

The curve passes through the point $(2, 5)$. Find an expression for $y$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at reverse chain rule, e.g. $(x^2-3)^4$ seen	✓ 1
A1 $(x^2-3)^4 + C$	✓ 1
Part (b)
M1 substitutes $x=2$, $y=5$ into their integral	✓ 1
A1 $1^4 + C = 5$, so $C = 4$	✓ 1
A1 $y = (x^2-3)^4 + 4$	✓ 1

Common error: After a reverse chain rule integral, check by differentiating: the factor 8x must reappear exactly.

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Let $f(x) = x^3 - 6x^2 + 9x + 2$.

(a)

Find $f'(x)$ and hence find the $x$-coordinates of the stationary points of the graph of $f$.

(3)

(b)

Find the coordinates of the stationary points and determine their nature.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $f'(x) = 3x^2 - 12x + 9$	✓ 1
M1 sets $f'(x)=0$	✓ 1
A1 $x=1$ and $x=3$	✓ 1
Part (b)
A1 $(1, 6)$ and $(3, 2)$	✓ 1
M1 $f''(x) = 6x-12$ found and evaluated at a stationary point	✓ 1
R1 at $x=1$, $f'' = -6 < 0$ so maximum; at $x=3$, $f'' = 6 > 0$ so minimum	✓ 1

Common error: A stationary point is found from f′(x) = 0, but its y-coordinate comes from substituting into f, not f′.

+ Add to test✓ In your test87IBMA3S-5-16r16-055.10 Indefinite integrals and reverse chain rule Paper 1Easy4 marks

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(a)

Find $\int 12x\,e^{x^2}\,dx$.

(2)

(b)

Hence find the exact value of $\int_0^1 12x\,e^{x^2}\,dx$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at reverse chain rule, e.g. $e^{x^2}$ seen	✓ 1
A1 $6e^{x^2} + C$	✓ 1
Part (b)
M1 $6e^{1} - 6e^{0}$	✓ 1
A1 $6e - 6$ (accept $6(e-1)$)	✓ 1

Common error: Remember the derivative of x² is 2x, so 12x needs a factor of 6, not 12, in front of ex².

+ Add to test✓ In your test88IBMA3S-5-17r18-015.9 Kinematics Paper 1Easy5 marks

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The angle, $\theta$ degrees, turned through by a robot's elbow joint $t$ seconds after it starts to move is modelled by $\theta(t)=2t^3-9t^2+12t$, for $0\le t\le 3$.

(a)

Find $\dfrac{d\theta}{dt}$.

(2)

(b)

Find the values of $t$ for which the angle is decreasing.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate (at least one term correct)	✓ 1
A1 $\dfrac{d\theta}{dt}=6t^2-18t+12$	✓ 1
Part (b)
M1 attempt to solve $6t^2-18t+12<0$ or $6t^2-18t+12=0$ (ECF from (a))	✓ 1
A1 critical values $t=1$ and $t=2$	✓ 1
A1 $1<t<2$	✓ 1

Common error: Writing $t2$ for 'decreasing': that is where the quadratic derivative is positive. Sketch the parabola $\frac{d\theta}{dt}$ to see which side is negative.

+ Add to test✓ In your test89IBMA3S-5-17r18-025.4 Tangents and normals Paper 1Easy7 marks

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The curve $C$ has equation $y=x^3-2x^2+5$. The point $A(2,5)$ lies on $C$.

(a)

Find the gradient of $C$ at $A$.

(3)

(b)

Find the equation of the normal to $C$ at $A$. Give your answer in the form $ax+by=c$, where $a,b,c\in\mathbb{Z}$.

(3)

(c)

The normal meets the $x$-axis at the point $P$. Write down the $x$-coordinate of $P$.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate	✓ 1
A1 $\dfrac{dy}{dx}=3x^2-4x$	✓ 1
A1 gradient $=3(4)-4(2)=4$	✓ 1
Part (b)
M1 normal gradient $-\frac14$ (ECF from (a))	✓ 1
M1 line through $(2,5)$ with their normal gradient, e.g. $y-5=-\frac14(x-2)$	✓ 1
A1 $x+4y=22$ (accept any integer multiple)	✓ 1
Part (c)
A1 $22$ (ECF from (b))	✓ 1

Exam tip: Always write the normal gradient as the negative reciprocal of the tangent gradient, and check that your final line passes through the given point.

+ Add to test✓ In your test90IBMA3S-5-17r18-045.5 Introduction to integration Paper 1Easy4 marks

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A tailor draws a decorative panel on a grid where one unit represents 1 dm. The upper edge of the panel follows the curve $y=6x-x^2$ and the lower edge is the $x$-axis. Find the area of the panel, in dm².

Shaded panel

(a)

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 limits found from $6x-x^2=0$, i.e. $x=0$ and $x=6$	✓ 1
M1 $\int_0^6(6x-x^2)\,dx$	✓ 1
A1 $\left[3x^2-\dfrac{x^3}{3}\right]_0^6$	✓ 1
A1 $36$ dm²	✓ 1

Common error: Forgetting to divide by the new power when integrating, giving $x^3$ instead of $\frac{x^3}{3}$. Check by differentiating your answer.

+ Add to test✓ In your test91IBMA3S-5-17r19-015.4 Tangents and normals Paper 1Easy5 marks

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A bridge cable hangs in the curve $y = 0.5x^2 - 2x + 5$, where $x$ and $y$ are measured in metres from a reference point. P is the point on the cable where $x = 4$.

Diagram not to scale

(a)

Find the gradient of the cable at P.

(2)

(b)

Find the equation of the normal to the cable at P. Give your answer in the form $y = mx + c$.

(2)

(c)

The normal meets the x-axis at the point Q. Find the x-coordinate of Q.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate, obtaining $\frac{dy}{dx} = x - 2$	✓ 1
A1 gradient $= 2$	✓ 1
Part (b)
M1 use of perpendicular gradient $-\frac{1}{2}$ through P$(4, 5)$	✓ 1
A1 $y = -\frac{1}{2}x + 7$	✓ 1
Part (c)
A1 $x = 14$ (ECF from (b))	✓ 1

Common error: Using the gradient of the tangent in place of the perpendicular gradient. Always work out the value of $y$ at P before writing a line through it.

+ Add to test✓ In your test92IBMA3S-5-17r19-025.9 Kinematics Paper 1Easy4 marks

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A small space probe moves along a straight rail. Its displacement from a fixed marker is $s = t^3 - 6t^2 + 9t$ metres, where $t \ge 0$ is the time in seconds.

(a)

Find an expression for the velocity of the probe at time $t$.

(2)

(b)

The probe is momentarily at rest at two different times. Find the later of these times.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate (at least one term correct)	✓ 1
A1 $v = 3t^2 - 12t + 9$	✓ 1
Part (b)
M1 set $v = 0$ and attempt to factorise or solve, $3(t-1)(t-3) = 0$	✓ 1
A1 $t = 3$ s (accept $t = 1$ and $t = 3$ both found, with $t = 3$ identified as the later)	✓ 1

Exam tip: "At rest" means velocity is zero, not displacement. Differentiate displacement to get velocity before setting anything equal to zero.

+ Add to test✓ In your test93IBMA3S-5-17r19-035.1 Limits and the derivative Paper 1Easy5 marks

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During the first four minutes of a roast, the temperature $T$ °C of a batch of coffee beans changes at the rate $\frac{dT}{dt} = 20 + 15\sqrt{t}$, where $t$ is the time in minutes. At $t = 0$ the temperature is 20 °C.

(a)

Find an expression for $T$ in terms of $t$.

(4)

(b)

Find the temperature of the beans when $t = 4$.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $20t$	✓ 1
A1 $10t^{3/2}$	✓ 1
M1 use of $T(0) = 20$ to find the constant of integration	✓ 1
A1 $T = 20t + 10t^{3/2} + 20$	✓ 1
Part (b)
A1 $T = 180$ °C	✓ 1

Common error: Integrating $\sqrt{t}$ as $t^{1/2}$ with no change, or forgetting the constant. Write $\sqrt{t}$ as $t^{1/2}$ first, then raise the power by one and divide by the new power.

+ Add to test✓ In your test94IBMA3S-5-17r19-045.2 Increasing and decreasing functions Paper 1Easy5 marks

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The height of a bridge deck above the river is modelled by $h(x) = x^3 - 9x^2 + 24x + 3$ metres, for $0 \le x \le 6$, where $x$ is measured in units of 10 m along the bridge.

(a)

Find the values of $x$ at which the deck has a stationary point.

(2)

(b)

Use the second derivative to determine the nature of each stationary point.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $h'(x) = 3x^2 - 18x + 24$ set equal to zero	✓ 1
A1 $x = 2$ and $x = 4$	✓ 1
Part (b)
A1 $h''(x) = 6x - 18$	✓ 1
R1 $h''(2) = -6 < 0$, so a local maximum at $x = 2$	✓ 1
R1 $h''(4) = 6 > 0$, so a local minimum at $x = 4$	✓ 1

Exam tip: State the value of the second derivative and its sign, then give the conclusion. A bare statement of "maximum" or "minimum" earns nothing.

+ Add to test✓ In your test95IBMA3S-5-17r21-015.5 Introduction to integration Paper 1Easy6 marks

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A climber ascends a cliff face. Her height $h$ metres above base camp, $t$ hours after she sets off, satisfies $\frac{dh}{dt}=6t^2-12t+9$. At $t=0$ she is 50 m above base camp.

(a)

Find an expression for $h$ in terms of $t$.

(4)

(b)

Find her height above base camp after 3 hours.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate (at least one term with power increased by one)	✓ 1
A1 $2t^3-6t^2+9t$	✓ 1
M1 substitutes $t=0$, $h=50$ to find the constant	✓ 1
A1 $h=2t^3-6t^2+9t+50$	✓ 1
Part (b)
M1 substitutes $t=3$ into their $h$	✓ 1
A1 77 m (ECF from (a))	✓ 1

Common error: Write the constant of integration straight away and use the given condition to find it before doing anything else with the expression.

+ Add to test✓ In your test96IBMA3S-5-17r21-025.1 Limits and the derivative Paper 1Easy6 marks

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A telecoms company models the cost, $C(x)$ thousand euros, of installing $x$ km of fibre cable by $C(x)=3x^2-5x$.

(a)

Use differentiation from first principles to show that $C'(x)=6x-5$.

(4)

(b)

Find the rate of change of the cost when $x=2$, including units.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 writes $\frac{C(x+h)-C(x)}{h}$ with $C(x+h)=3(x+h)^2-5(x+h)$	✓ 1
A1 expands and simplifies to $\frac{6xh+3h^2-5h}{h}=6x+3h-5$	✓ 1
M1 takes the limit as $h\to0$	✓ 1
A1 $C'(x)=6x-5$ (AG, must show the limit step)	✓ 1
Part (b)
M1 substitutes $x=2$ into $6x-5$	✓ 1
A1 7 thousand euros per km	✓ 1

Exam tip: In a first-principles proof, keep $h$ in every term until you divide by $h$, and only then let $h\to0$.

+ Add to test✓ In your test97IBMA3S-5-17r21-035.4 Tangents and normals Paper 1Easy8 marks

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The height profile of a mountain ridge is modelled by $f(x)=x^3-3x^2+2$ for $0\leq x\leq 4$.

Diagram not to scale

(a)

Find $f'(x)$.

(2)

(b)

Find the equation of the tangent to the curve at the point where $x=3$. Give your answer in the form $y=mx+c$.

(3)

(c)

Find the equation of the normal to the curve at the point where $x=3$. Give your answer in the form $ax+by=c$ where $a$, $b$, $c$ are integers.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $3x^2$	✓ 1
A1 $-6x$	✓ 1
Part (b)
M1 $f(3)=2$ and $f'(3)=27-18=9$	✓ 1
M1 $y-2=9(x-3)$	✓ 1
A1 $y=9x-25$	✓ 1
Part (c)
M1 normal gradient $-\frac19$	✓ 1
M1 $y-2=-\frac19(x-3)$	✓ 1
A1 $x+9y=21$	✓ 1

Common error: The normal gradient is the negative reciprocal of the tangent gradient, so $-\frac{1}{9}$ here, not $-9$ and not $\frac19$.

+ Add to test✓ In your test98IBMA3S-5-16r23-015.10 Indefinite integrals and reverse chain rule Paper 1Easy5 marks

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Use differentiation from first principles to find the derivative of $f(x)=3x^2-5x$.

(a)

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $f(x+h)=3(x+h)^2-5(x+h)$	✓ 1
M1 substitutes into $\frac{f(x+h)-f(x)}{h}$	✓ 1
A1 numerator simplifies to $6xh+3h^2-5h$	✓ 1
M1 divides by $h$ to obtain $6x+3h-5$	✓ 1
A1 lets $h\to 0$ to obtain $f'(x)=6x-5$	✓ 1

Common error: Do not let $h\to 0$ before cancelling the factor $h$, because the quotient is then $\frac{0}{0}$.

+ Add to test✓ In your test99IBMA3S-5-16r23-025.2 Increasing and decreasing functions Paper 1Easy5 marks

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Let $f(x)=x^3-6x^2+9x+2$.

(a)

Find the set of values of $x$ for which $f$ is increasing.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $f'(x)=3x^2-12x+9$	✓ 1
M1 attempts to solve $f'(x)=0$, e.g. $3(x-1)(x-3)=0$	✓ 1
A1 $x=1$ and $x=3$	✓ 1
M1 considers the sign of $f'$, e.g. positive-$x^2$ quadratic is positive outside its roots, or tests a value	✓ 1
A1 $x<1$ or $x>3$	✓ 1

Exam tip: Increasing means $f'(x)>0$. Find where $f'=0$ first, then check the sign of $f'$ in each interval.

+ Add to test✓ In your test100IBMA3S-5-16r23-035.9 Kinematics Paper 1Easy6 marks

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A train leaves a station and moves along a straight track. Its displacement from the station is $s$ metres at time $t$ seconds, where $s=t^3-9t^2+24t$ for $0\le t\le 6$.

(a)

Find an expression for the velocity $v$ of the train at time $t$.

(2)

(b)

The train is momentarily at rest twice. Find the two times at which this happens.

(2)

(c)

Find the acceleration of the train when $t=4$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts to differentiate (at least one term correct)	✓ 1
A1 $v=3t^2-18t+24$	✓ 1
Part (b)
M1 sets $v=0$, i.e. $3(t-2)(t-4)=0$	✓ 1
A1 $t=2$ and $t=4$	✓ 1
Part (c)
M1 $a=\frac{dv}{dt}=6t-18$	✓ 1
A1 $a=6\ \text{m s}^{-2}$	✓ 1

Common error: Velocity is $\frac{ds}{dt}$ and acceleration is $\frac{dv}{dt}$. Do not substitute the time into $s$ when you are asked for $v$ or $a$.

+ Add to test✓ In your test101IBMA3S-5-17r23-015.2 Increasing and decreasing functions Paper 1Easy5 marks

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At an archaeological dig, the thickness of a sediment layer, $T$ cm, at a distance $x$ metres from the trench wall is modelled by $T(x) = 2x^3 - \frac{12}{x} + 5$, for $x > 0$.

(a)

Find $T'(x)$.

(3)

(b)

Find the value of $T'(2)$ and state what it tells you about the sediment layer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate (at least one term with the power reduced by one)	✓ 1
A1 $6x^2$	✓ 1
A1 $+\frac{12}{x^2}$ (accept $12x^{-2}$; the constant term differentiates to zero)	✓ 1
Part (b)
M1 substitute $x=2$: $6(4)+\frac{12}{4}$ (ECF from (a))	✓ 1
A1 $27$, so the thickness is increasing at 27 cm per metre at $x=2$	✓ 1

Common error: Write $-\frac{12}{x}$ as $-12x^{-1}$ before differentiating; the derivative is then $+12x^{-2}$, and a sign slip here is very common.

+ Add to test✓ In your test102IBMA3S-5-17r23-025.5 Introduction to integration Paper 1Easy7 marks

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The upper surface of a buried wall is modelled by the curve $y = f(x)$, $x>0$, where $x$ and $y$ are in metres. The gradient of the curve is $f'(x) = 8x^3 - \frac{3}{x^2} + 2$, and the curve passes through the point $(1, 5)$.

(a)

Find $f(x)$.

(5)

(b)

Find the height of the wall surface when $x=2$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate (at least one term with the power increased by one)	✓ 1
A1 $2x^4 + \frac{3}{x}$ (accept $3x^{-1}$)	✓ 1
A1 $+2x + c$	✓ 1
M1 substitute $(1,5)$: $2+3+2+c=5$	✓ 1
A1 $c=-2$, so $f(x)=2x^4+\frac{3}{x}+2x-2$	✓ 1
Part (b)
M1 substitute $x=2$ into their $f(x)$: $32+\frac{3}{2}+4-2$ (ECF from (a))	✓ 1
A1 $35.5$ m	✓ 1

Exam tip: Always include the constant of integration, then use the given point straight away to find it.

+ Add to test✓ In your test103IBMA3S-5-17r23-035.9 Kinematics Paper 1Easy6 marks

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During an acceleration test, a cyclist's distance from the start line is modelled by $s(t) = 2t^2 + 3t$, where $s$ is in metres and $t$ is in seconds.

(a)

Show that the average speed of the cyclist between $t=3$ and $t=3+h$ is $(15+2h)$ m s⁻¹.

(3)

(b)

Write down the limit of $15+2h$ as $h\to 0$, and state what this value represents.

(2)

(c)

Differentiate $s(t)$ to verify your answer to part (b).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at $\frac{s(3+h)-s(3)}{h}$	✓ 1
A1 $s(3)=27$ and $s(3+h)=27+15h+2h^2$	✓ 1
A1 $\frac{15h+2h^2}{h}=15+2h$ (AG)	✓ 1
Part (b)
A1 $15$	✓ 1
A1 the cyclist's instantaneous speed at $t=3$ s, in m s⁻¹	✓ 1
Part (c)
A1 $s'(t)=4t+3$, so $s'(3)=15$ m s⁻¹, agreeing with (b)	✓ 1

Exam tip: The limit of the average rate of change as the interval shrinks to zero is the derivative; quote units with the final value.

+ Add to test✓ In your test104IBMA3S-5-16r26-025.1 Limits and the derivative Paper 1Easy6 marks

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A shop's marginal cost of producing $x$ hundred gift sets is $C'(x) = 6x^2 - 4x + 1$ thousand euros per hundred sets. The fixed cost (the cost when $x=0$) is €40 thousand.

(a)

Find an expression for $C(x)$, the total cost in thousand euros.

(4)

(b)

Find the total cost of producing 300 gift sets.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate $C'(x)$	✓ 1
A1 $2x^3 - 2x^2 + x$ (condone missing $+c$)	✓ 1
M1 use of $C(0)=40$ to find the constant	✓ 1
A1 $C(x) = 2x^3 - 2x^2 + x + 40$	✓ 1
Part (b)
M1 substitute $x=3$ into $C(x)$	✓ 1
A1 $C(3) = 79$, so the cost is €79 thousand	✓ 1

Exam tip: Always include the constant of integration and use the given condition to find it before substituting any new values.

+ Add to test✓ In your test105IBMA3S-5-16r26-065.7 The second derivative and graphs Paper 1Easy5 marks

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A solar panel supplies power $P(t) = 3\sqrt{t} + 2$ kilowatts, where $t$ is the time in hours. The energy supplied between $t=1$ and $t=4$ is $\int_1^4 P(t)\,dt$ kilowatt-hours.

(a)

Find the energy supplied between $t=1$ and $t=4$.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate, writing $\sqrt{t}=t^{1/2}$	✓ 1
A1 $2t^{3/2}$	✓ 1
A1 $+2t$	✓ 1
M1 substitute the limits 4 and 1 into the antiderivative	✓ 1
A1 $(16+8)-(2+2) = 20$ kWh	✓ 1

Exam tip: Write roots as powers before integrating, and keep the limits in the substitution step in the order upper minus lower.

+ Add to test✓ In your test106IBMA3S-5-16r26-095.7 The second derivative and graphs Paper 1Easy6 marks

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Let $f(x) = x^3 + kx^2 - 9x + 4$, where $k$ is a constant. The graph of $f$ has a stationary point at $x=1$.

(a)

Find the value of $k$.

(4)

(b)

Use the second derivative to determine whether the stationary point at $x=1$ is a maximum or a minimum.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate $f$	✓ 1
A1 $f'(x) = 3x^2 + 2kx - 9$	✓ 1
M1 set $f'=0$ at $x=1$, so $3 + 2k - 9 = 0$	✓ 1
A1 $k=3$	✓ 1
Part (b)
M1 $f''(x) = 6x + 6$ and evaluate $f''$ at $x=1$, giving $12$	✓ 1
R1 $f''>0$ at $x=1$, so the stationary point is a local minimum	✓ 1

Common error: Substitute $x=1$ into $f'(x)$, not into $f(x)$, when told there is a stationary point.

+ Add to test✓ In your test107IBMA3S-5-18r30-015.4 Tangents and normals Paper 1Easy7 marks

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Let $f(x)=4x^3-\frac{6}{x^2}+5x-7$, for $x\neq 0$.

(a)

Find $f'(x)$.

(3)

(b)

Find the equation of the tangent to the curve $y=f(x)$ at the point where $x=1$. Give your answer in the form $y=mx+c$.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate, rewriting $\frac{6}{x^2}$ as $6x^{-2}$	✓ 1
A1 $12x^2+5$	✓ 1
A1 $+12x^{-3}$, so $f'(x)=12x^2+12x^{-3}+5$	✓ 1
Part (b)
A1 $f\left(1\right)=4-6+5-7=-4$	✓ 1
A1 $f'\left(1\right)=12+12+5=29$	✓ 1
M1 $y-(-4)=29(x-1)$	✓ 1
A1 $y=29x-33$	✓ 1

Common error: Rewrite $\frac{6}{x^2}$ as $6x^{-2}$ before differentiating; the derivative of $-6x^{-2}$ is $+12x^{-3}$, not $-12x^{-1}$.

+ Add to test✓ In your test108IBMA3S-5-18r30-025.3 Differentiating polynomials and integer powers Paper 1Easy4 marks

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Let $f(x)=(3x^2-1)^4$.

(a)

Find $f'(x)$.

(3)

(b)

Find the value of $f'(1)$.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the chain rule	✓ 1
A1 $4(3x^2-1)^3$	✓ 1
A1 multiplied by $6x$, giving $f'(x)=24x(3x^2-1)^3$	✓ 1
Part (b)
A1 $24\times 2^3=192$	✓ 1

Exam tip: In the chain rule, multiply the derivative of the outer function by the derivative of the inner function; do not forget the factor $6x$.

+ Add to test✓ In your test109IBMA3S-5-18r30-035.6 Differentiation rules Paper 1Easy5 marks

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Let $y=x^2e^{3x}$.

(a)

Find $\frac{dy}{dx}$.

(3)

(b)

Hence find the $x$-coordinates of the points on the curve where $\frac{dy}{dx}=0$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule	✓ 1
A1 $2xe^{3x}$	✓ 1
A1 $+3x^2e^{3x}$	✓ 1
Part (b)
M1 factorises $xe^{3x}(2+3x)=0$ and uses $e^{3x}\neq 0$	✓ 1
A1 $x=0$ and $x=-\frac{2}{3}$ ($-0.6667$)	✓ 1

Exam tip: An exponential factor is never zero, so it can be cancelled when you set a derivative equal to zero.

+ Add to test✓ In your test110IBMA3S-5-18r30-045.6 Differentiation rules Paper 1Easy5 marks

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Let $f(x)=\frac{2x+1}{x-3}$, for $x\neq 3$.

(a)

Show that $f'(x)=-\frac{7}{(x-3)^2}$.

(3)

(b)

Hence explain why $f$ is decreasing at every point of its domain.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the quotient rule	✓ 1
A1 $\frac{2(x-3)-(2x+1)\left(1\right)}{(x-3)^2}$	✓ 1
A1 simplifies to $-\frac{7}{(x-3)^2}$ (AG)	✓ 1
Part (b)
R1 the numerator $-7$ is negative	✓ 1
R1 $(x-3)^2>0$ for $x\neq 3$, so $f'(x)<0$ everywhere in the domain	✓ 1

Exam tip: In a quotient-rule “show that” question, write out $u$, $v$, $u'$ and $v'$ first, then substitute and simplify in clear steps.

+ Add to test✓ In your test111IBMA3S-5-17r34-015.2 Increasing and decreasing functions Paper 1Easy4 marks

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A beekeeper models the mass of honey, M kg, stored in a hive t days after a nectar flow begins by $M(t)=2t^3-9t^2+12t$, for $0\le t\le 3$.

(a)

Find $\frac{dM}{dt}$.

(2)

(b)

Find the values of t at which the mass of honey is neither increasing nor decreasing.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate (at least one term correct)	✓ 1
A1 $\frac{dM}{dt}=6t^2-18t+12$	✓ 1
Part (b)
M1 set $6t^2-18t+12=0$ and attempt to solve (for example $6(t-1)(t-2)=0$)	✓ 1
A1 $t=1$ and $t=2$	✓ 1

Exam tip: A mass that is neither increasing nor decreasing has a rate of change of zero, so set the derivative equal to zero. Dividing through by the common factor 6 first makes the quadratic easier to factorise.

+ Add to test✓ In your test112IBMA3S-5-17r34-035.7 The second derivative and graphs Paper 1Easy5 marks

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After an injection, the concentration of a drug in a patient's blood is modelled by $C(t)=t\,e^{-t/2}$ mg per litre, where t is the time in hours after the injection and $t\ge 0$.

(a)

Find $C'(t)$.

(3)

(b)

Find the value of t for which $C'(t)=0$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of the product rule with $u=t$ and $v=e^{-t/2}$	✓ 1
A1 $\frac{d}{dt}\left(e^{-t/2}\right)=-\frac12e^{-t/2}$	✓ 1
A1 $C'(t)=e^{-t/2}-\frac12te^{-t/2}$ (or $e^{-t/2}\left(1-\frac t2\right)$)	✓ 1
Part (b)
M1 set $C'(t)=0$; since $e^{-t/2}\ne0$, $1-\frac t2=0$	✓ 1
A1 $t=2$ hours	✓ 1

Exam tip: After the product rule, factorise out the common exponential. An exponential is never zero, so the other factor alone must be zero.

+ Add to test✓ In your test113IBMA3S-5-17r34-045.5 Introduction to integration Paper 1Easy5 marks

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Let $f(x)=\dfrac{4x-1}{x+2}$, for $x\ne-2$.

(a)

Show that $f'(x)=\dfrac{9}{(x+2)^2}$.

(3)

(b)

The gradient of the curve $y=f(x)$ is $\frac14$ at a point where $x>-2$. Find the x-coordinate of this point.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of the quotient rule with $u=4x-1$ and $v=x+2$	✓ 1
A1 $f'(x)=\frac{4(x+2)-(4x-1)}{(x+2)^2}$	✓ 1
A1 expanding the numerator: $4x+8-4x+1=9$, giving $\frac{9}{(x+2)^2}$	✓ 1
Part (b)
M1 $\frac{9}{(x+2)^2}=\frac14$, so $(x+2)^2=36$	✓ 1
A1 $x=4$ (reject $x=-8$ since $x>-2$)	✓ 1

Common error: Reversing the order of the numerator in the quotient rule gives the wrong sign. Write $v\,u'-u\,v'$ over $v^2$ and substitute carefully.

+ Add to test✓ In your test114IBMA3S-5-18r35-015.10 Indefinite integrals and reverse chain rule Paper 1Easy6 marks

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Let $f(x) = 2x^2 - 5x$.

(a)

Find and simplify $f(x+h) - f(x)$.

(2)

(b)

Hence use the definition of the derivative as a limit to show that $f'(x) = 4x - 5$.

(3)

(c)

Find the gradient of the graph of $y = f(x)$ at $x = 3$.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $2(x+h)^2 - 5(x+h) - (2x^2 - 5x)$	✓ 1
A1 $4xh + 2h^2 - 5h$	✓ 1
Part (b)
M1 $\frac{f(x+h)-f(x)}{h} = 4x + 2h - 5$	✓ 1
M1 $f'(x) = \lim_{h \to 0}(4x + 2h - 5)$	✓ 1
A1 $f'(x) = 4x - 5$	✓ 1
Part (c)
A1 $4(3) - 5 = 7$	✓ 1

Common error: Divide every term of $f(x+h)-f(x)$ by $h$ before letting $h \to 0$; substituting $h = 0$ too early gives the meaningless $\frac{0}{0}$.

+ Add to test✓ In your test115IBMA3S-5-18r35-025.10 Indefinite integrals and reverse chain rule Paper 1Easy4 marks

No calculator

Consider the expression $\dfrac{x^2 - 4}{x^2 - x - 2}$, defined for $x \neq 2$ and $x \neq -1$.

(a)

Explain why the limit as $x \to 2$ cannot be found by direct substitution.

(1)

(b)

Find $\displaystyle\lim_{x \to 2} \frac{x^2 - 4}{x^2 - x - 2}$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
R1 substituting $x = 2$ gives $\frac{0}{0}$, which is undefined	✓ 1
Part (b)
M1 factorise numerator $(x-2)(x+2)$	✓ 1
M1 factorise denominator $(x-2)(x+1)$ and cancel $(x-2)$	✓ 1
A1 limit $= \frac{4}{3}$	✓ 1

Exam tip: When a limit gives $\frac{0}{0}$, a common factor has usually cancelled out; factorise first, then substitute.

+ Add to test✓ In your test116IBMA3S-5-18r35-035.9 Kinematics Paper 1Easy5 marks

No calculator

A kelp bed in a marine reserve has mass $M$ kg at time $t$ months. Its mass increases at the rate $\dfrac{dM}{dt} = 12t^2 - 6t + 4$. At $t = 0$ the mass of the kelp bed is 50 kg.

(a)

Find an expression for $M$ in terms of $t$.

(4)

(b)

Find the mass of the kelp bed after 3 months.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate	✓ 1
A1 $4t^3 - 3t^2 + 4t$	✓ 1
M1 substitute $t=0$, $M=50$ to find the constant	✓ 1
A1 $M = 4t^3 - 3t^2 + 4t + 50$	✓ 1
Part (b)
A1 $M = 108 - 27 + 12 + 50 = 143$ kg	✓ 1

Common error: After integrating a rate, always add the constant and use the initial condition before substituting any other value of $t$.

+ Add to test✓ In your test117IBMA3S-5-18r35-045.8 Stationary points, inflexion and optimization Paper 1Easy5 marks

No calculator

The cross-section of a covered walkway roof is the region between the $x$-axis and the curve $y = 8x - x^2$, where $x$ and $y$ are in metres.

Cross-section of the roof (shaded)

(a)

Write down the width of the base of the cross-section.

(1)

(b)

Find the area of the cross-section.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 8 m	✓ 1
Part (b)
M1 $\int_0^8 (8x - x^2)\,dx$	✓ 1
A1 $\left[4x^2 - \frac{x^3}{3}\right]$	✓ 1
M1 substitute limits 8 and 0	✓ 1
A1 $256 - \frac{512}{3} = \frac{256}{3}$ m$^2$	✓ 1

Exam tip: Find where the curve meets the $x$-axis first; those values are the limits of the definite integral.

+ Add to test✓ In your test118IBMA3S-5-18r44-015.2 Increasing and decreasing functions Paper 1Easy4 marks

No calculator

Let $f(x)=3x^4-5x^2+7x-2$.

(a)

Find $f'(x)$.

(2)

(b)

Find the gradient of the graph of $f$ at the point where $x=-1$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate at least one term correctly	✓ 1
A1 $f'(x)=12x^3-10x+7$	✓ 1
Part (b)
M1 substitute $x=-1$ into $f'(x)$	✓ 1
A1 $-12+10+7=5$	✓ 1

Common error: Differentiating the constant term $-2$ gives 0, not $-2$; and a negative value of $x$ needs brackets when substituted into $x^3$.

+ Add to test✓ In your test119IBMA3S-5-18r44-025.3 Differentiating polynomials and integer powers Paper 1Easy6 marks

No calculator

Let $f(x)=2x^3+\frac{54}{x}$, for $x\neq 0$.

(a)

Find $f'(x)$.

(3)

(b)

Find the exact values of $x$ for which the gradient of the graph of $f$ is zero.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rewrite $\frac{54}{x}$ as $54x^{-1}$	✓ 1
A1 $6x^2$	✓ 1
A1 $-54x^{-2}$ (or $-\frac{54}{x^2}$)	✓ 1
Part (b)
M1 set $6x^2-\frac{54}{x^2}=0$	✓ 1
A1 $x^4=9$	✓ 1
A1 $x=\pm\sqrt{3}$	✓ 1

Exam tip: Rewrite $\frac{k}{x^n}$ as $kx^{-n}$ before differentiating; the quotient rule is not needed for a constant numerator.

+ Add to test✓ In your test120IBMA3S-5-18r44-085.10 Indefinite integrals and reverse chain rule Paper 1Easy4 marks

No calculator

Find the exact value of $\int_1^2\left(3x^2+\frac{2}{x^2}\right)dx$.

(a)

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate, rewriting $\frac{2}{x^2}$ as $2x^{-2}$	✓ 1
A1 $x^3-2x^{-1}$	✓ 1
M1 substitute limits 2 and 1 and subtract	✓ 1
A1 $8$	✓ 1

Exam tip: Write the integrand with negative powers first: $\frac{2}{x^2}=2x^{-2}$ integrates to $-2x^{-1}$.

+ Add to test✓ In your test121AASL5-q00045.1 Limits and the derivative Paper 1Medium7 marks

Water drains from a tank. The volume of water in the tank, $V$ litres, $t$ minutes after draining starts is given by $V(t)=1600-80t+t^{2}$, for $0\le t\le 40$.

(a)

Find $\dfrac{\mathrm{d}V}{\mathrm{d}t}$.

(1)

(b)

Find the value of $\dfrac{\mathrm{d}V}{\mathrm{d}t}$ when $t=10$, and interpret its sign in context.

(2)

(c)

Find the time at which the water is draining at a rate of 30 litres per minute.

(2)

(d)

Show that the tank is empty when $t=40$, and write down the rate at which the volume is changing at that instant.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$-80+2t$	A1
(b)
$-60$	A1
negative, so the volume is decreasing (by 60 litres per minute)	R1
(c)
$-80+2t=-30$	M1	Award M0 for $-80+2t=30$ (this gives $t=55$, outside the domain).
$t=25$ (minutes)	A1
(d)
$V(40)=1600-3200+1600=0$	A1
$\dfrac{\mathrm{d}V}{\mathrm{d}t}=-80+80=0$ litres per minute	A1

Answers: (a) $2t-80$ · (b) $-60$, decreasing · (c) $t=25$ · (d) $0$ litres per minute

+ Add to test✓ In your test122AASL5-q00105.2 Increasing and decreasing functions Paper 1Medium6 marks

The following diagram shows the graph of $y=f'(x)$, the derivative of a function $f$, for $-1.6\le x\le 3.2$. The graph of $f'$ crosses the $x$-axis at $(-1,\,0)$, has a local maximum point at $(0,\,4)$ and touches the $x$-axis at $(2,\,0)$.

(a)

Write down the $x$-coordinate of the local minimum point on the graph of $f$. Justify your answer.

(2)

(b)

State, with a reason, the nature of the point on the graph of $f$ where $x=2$.

(2)

(c)

Find the set of values of $x$ for which the graph of $f$ is concave-down.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$x=-1$	A1
$f'(x)$ changes sign from negative to positive at $x=-1$	R1	Award R0 for “$f'(x)=0$” alone.
(b)
$f'(2)=0$ but $f'(x)$ does not change sign (positive on both sides of $x=2$)	R1
a point of inflexion with zero gradient (stationary point of inflexion)	A1	Award A1 only if R1 is awarded. Also accept a statement that $f'$ has a local minimum at $x=2$ so $f''$ changes sign.
(c)
concave-down where $f''(x)\lt 0$, i.e. where the graph of $f'$ is decreasing	R1
$0\lt x\lt 2$	A1	Accept $0\le x\le 2$.

Answers: (a) $x=-1$ · (b) a point of inflexion with zero gradient · (c) $0\lt x\lt 2$

+ Add to test✓ In your test123AASL5-q00135.2 Increasing and decreasing functions Paper 1Medium6 marks

Let $f(x)=x^{3}+3x^{2}+5x-7$, for $x\in\mathbb{R}$.

(a)

Find $f'(x)$.

(1)

(b)

Show that $f'(x)=3(x+1)^{2}+2$.

(2)

(c)

Hence explain why $f$ is increasing for all values of $x$.

(1)

(d)

By considering $f(0)$ and $f(1)$, show that the equation $f(x)=0$ has exactly one real solution, and that this solution lies between $0$ and $1$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$3x^{2}+6x+5$	A1
(b)
$3(x^{2}+2x)+5$ or expanding $3(x+1)^{2}+2=3x^{2}+6x+3+2$	M1
$3\big((x+1)^{2}-1\big)+5$ or $3x^{2}+6x+5$	A1
$f'(x)=3(x+1)^{2}+2$	AG
(c)
$3(x+1)^{2}\ge0$, so $f'(x)\ge2\gt 0$ for all $x$	R1
(d)
$f(0)=-7\lt 0$ and $f(1)=2\gt 0$, so the graph crosses the $x$-axis between $0$ and $1$	A1
$f$ is increasing for all $x$, so the graph cannot cross the $x$-axis more than once	R1

Answers: (a) $3x^{2}+6x+5$ · (c) $f'(x)\ge2\gt 0$ · (d) one root, between 0 and 1

+ Add to test✓ In your test124AASL5-q00185.3 Differentiating polynomials and integer powers Paper 1Medium8 marks

Let $f(x)=\dfrac{x^{3}-4}{2x^{2}}$, for $x\neq0$.

(a)

Write $f(x)$ in the form $ax+bx^{-2}$, where $a,b\in\mathbb{Q}$.

(1)

(b)

Find $f'(x)$.

(2)

(c)

Find the coordinates of the point on the graph of $f$ where the gradient is zero.

(3)

(d)

Find the set of values of $x$ for which $f$ is increasing.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$f(x)=\tfrac{1}{2}x-2x^{-2}$	A1
(b)
$\tfrac{1}{2}$	A1	FT from (a).
$+4x^{-3}$, so $f'(x)=\tfrac{1}{2}+4x^{-3}$	A1	Accept $\tfrac12+\dfrac{4}{x^{3}}$.
(c)
$\tfrac{1}{2}+\dfrac{4}{x^{3}}=0\Rightarrow x^{3}=-8$	M1
$x=-2$	A1
$f(-2)=\dfrac{-8-4}{8}=-\tfrac{3}{2}$; point $\left(-2,\,-\tfrac{3}{2}\right)$	A1
(d)
$f'(x)\gt 0$: $\dfrac{4}{x^{3}}\gt -\tfrac12$, considering $x\gt 0$ and $x\lt 0$ separately (or a sign diagram for $f'$)	M1	For $x\gt 0$, $f'(x)\gt 0$ always; for $x\lt 0$, $x^{3}\lt -8$.
$x\lt -2$ or $x\gt 0$	A1	Both intervals required.

Answers: (a) $\tfrac12x-2x^{-2}$ · (b) $f'(x)=\tfrac12+4x^{-3}$ · (c) $\left(-2,\,-\tfrac32\right)$ · (d) $x\lt -2$ or $x\gt 0$

+ Add to test✓ In your test125AASL5-q00205.3 Differentiating polynomials and integer powers Paper 1Medium6 marks

Let $f(x)=ax^{3}+bx^{2}$, where $a,b\in\mathbb{R}$. The gradient of the graph of $f$ is $-3$ at $x=1$ and $0$ at $x=2$.

(a)

Find $f'(x)$.

(1)

(b)

Find the value of $a$ and the value of $b$.

(4)

(c)

Find the gradient of the graph of $f$ at $x=-1$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$3ax^{2}+2bx$	A1
(b)
substituting $x=1$ or $x=2$ into their $f'(x)$ and equating	M1
$3a+2b=-3$ and $12a+4b=0$	A1
attempting to solve simultaneously, e.g. $b=-3a$	M1
$a=1,\ b=-3$	A1
(c)
$f'(-1)=3+6=9$	A1	FT from (b).

Answers: (a) $3ax^{2}+2bx$ · (b) $a=1,\ b=-3$ · (c) $9$

+ Add to test✓ In your test126AASL5-q00215.3 Differentiating polynomials and integer powers Paper 1Medium8 marks

A curve has equation $y=x^{3}-6x^{2}+10x+1$.

(a)

Find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$.

(2)

(b)

Find the coordinates of the points on the curve at which the gradient is $1$.

(4)

(c)

Write $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ in the form $3(x-h)^{2}+k$ and hence write down the least possible value of the gradient of the curve.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$3x^{2}-12x$	A1	Award for any two correct terms.
$3x^{2}-12x+10$	A1
(b)
$3x^{2}-12x+10=1$	M1
$3(x-1)(x-3)=0\Rightarrow x=1,\ x=3$	A1
substituting their $x$-values into $y$	(M1)
$(1,\,6)$ and $(3,\,4)$	A1
(c)
$3(x-2)^{2}-2$	A1
$-2$	A1	FT from their $k$.

Answers: (a) $3x^{2}-12x+10$ · (b) $(1,6)$, $(3,4)$ · (c) $3(x-2)^{2}-2$; least gradient $-2$

+ Add to test✓ In your test127AASL5-q00225.3 Differentiating polynomials and integer powers Paper 1Medium7 marks

Let $f(x)=\left(x+\dfrac{2}{x}\right)^{2}$, for $x\neq0$.

(a)

Show that $f(x)=x^{2}+4+4x^{-2}$.

(2)

(b)

Find $f'(x)$.

(2)

(c)

Find the coordinates of the points on the graph of $f$ where the gradient is zero.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$x^{2}+2\cdot x\cdot\dfrac{2}{x}+\dfrac{4}{x^{2}}$	M1	Award for a correct expansion with at least the middle term attempted.
$x^{2}+4+\dfrac{4}{x^{2}}$	A1
$f(x)=x^{2}+4+4x^{-2}$	AG
(b)
$2x$	A1
$-8x^{-3}$	A1	Accept $f'(x)=2x-\dfrac{8}{x^{3}}$.
(c)
$2x-8x^{-3}=0\Rightarrow x^{4}=4$	M1
$x=\pm\sqrt{2}$	A1
$(\sqrt{2},\,8)$ and $(-\sqrt{2},\,8)$	A1	$f(\sqrt2)=2+4+2=8$

Answers: (b) $2x-8x^{-3}$ · (c) $(\sqrt2,\,8)$, $(-\sqrt2,\,8)$

+ Add to test✓ In your test128AASL5-q00245.4 Tangents and normals Paper 1Medium8 marks

Let $f(x)=\ln(2x-3)$, for $x\gt \tfrac{3}{2}$. The graph of $f$ meets the $x$-axis at the point $\text{P}(2,\,0)$.

(a)

Find $f'(x)$.

(2)

(b)

Find the equation of the tangent to the graph of $f$ at P.

(2)

(c)

Find the equation of the normal to the graph of $f$ at P.

(2)

(d)

The tangent and the normal at P meet the $y$-axis at the points A and B. Find the area of triangle ABP.

(2)

Show mark scheme

Working	Mark	Notes
(a)
attempt to use the chain rule, e.g. $\dfrac{1}{2x-3}\times\ldots$	M1
$f'(x)=\dfrac{2}{2x-3}$	A1
(b)
gradient $f'(2)=2$	A1
$y=2x-4$	A1	Accept $y-0=2(x-2)$.
(c)
gradient of normal $=-\dfrac{1}{2}$	M1	FT from their gradient in (b).
$y=-\tfrac{1}{2}x+1$	A1	Accept $y=-\tfrac12(x-2)$.
(d)
A$(0,\,-4)$, B$(0,\,1)$; base $\text{AB}=5$, height $=2$ (the $x$-coordinate of P)	(M1)	FT from their lines.
area $=\tfrac{1}{2}\times5\times2=5$	A1

Answers: (a) $\dfrac{2}{2x-3}$ · (b) $y=2x-4$ · (c) $y=-\tfrac12x+1$ · (d) $5$

+ Add to test✓ In your test129AASL5-q00265.4 Tangents and normals Paper 1Medium6 marks

Let $f(x)=x^{2}-4x+5$.

(a)

Find $f'(x)$.

(1)

(b)

Find the gradient of the normal to the graph of $f$ at the point where $x=1$.

(2)

(c)

The tangent to the graph of $f$ at the point Q is parallel to the normal in part (b). Find the coordinates of Q.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$2x-4$	A1
(b)
$f'(1)=-2$	(M1)
$\dfrac{1}{2}$	A1
(c)
$2x-4=\dfrac{1}{2}$	M1	FT from (b).
$x=\dfrac{9}{4}$	A1
$\text{Q}\left(\dfrac{9}{4},\,\dfrac{17}{16}\right)$	A1	$\dfrac{81}{16}-9+5=\dfrac{17}{16}$

Answers: (a) $2x-4$ · (b) $\tfrac12$ · (c) $\left(\tfrac94,\,\tfrac{17}{16}\right)$

+ Add to test✓ In your test130AASL5-q00285.4 Tangents and normals Paper 1Medium7 marks

Let $f(x)=x^{2}-3x$. The point $\text{P}(1,\,-2)$ lies on the graph of $f$.

Diagram not to scale

(a)

Find $f'(x)$.

(1)

(b)

Find the equation of the normal to the graph of $f$ at P.

(3)

(c)

The normal meets the graph of $f$ again at the point R. Find the coordinates of R.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$2x-3$	A1
(b)
$f'(1)=-1$	(M1)
gradient of normal $=1$	(A1)
$y=x-3$	A1	Accept $y+2=x-1$.
(c)
$x^{2}-3x=x-3$	M1	FT from (b).
$(x-1)(x-3)=0$	A1
$\text{R}(3,\,0)$	A1

Answers: (a) $2x-3$ · (b) $y=x-3$ · (c) $\text{R}(3,\,0)$

+ Add to test✓ In your test131AASL5-q00295.4 Tangents and normals Paper 1Medium6 marks

The line $y=6x+k$, where $k\in\mathbb{R}$, is a tangent to the curve $y=x^{3}-6x$.

(a)

Find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ for the curve.

(1)

(b)

Find the $x$-coordinates of the points on the curve where the gradient is $6$.

(2)

(c)

Hence find the possible values of $k$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$3x^{2}-6$	A1
(b)
$3x^{2}-6=6$	M1
$x=\pm2$	A1
(c)
substituting a point of contact into $y=6x+k$, e.g. $(2,\,-4)$: $-4=12+k$	M1
$k=-16$	A1
$k=16$	A1	From $(-2,\,4)$: $4=-12+k$.

Answers: (a) $3x^{2}-6$ · (b) $x=\pm2$ · (c) $k=-16$ or $k=16$

+ Add to test✓ In your test132AASL5-q00365.5 Introduction to integration Paper 1Medium7 marks

The gradient of a curve is given by $\dfrac{\mathrm{d}y}{\mathrm{d}x}=3x^{2}-8x+k$, where $k\in\mathbb{R}$. The curve has a horizontal tangent at the point where $x=2$, and passes through the point $(0,\,5)$.

(a)

Find the value of $k$.

(2)

(b)

Find the equation of the curve.

(3)

(c)

Find the equation of the tangent to the curve at the point where it meets the $y$-axis.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$3(2)^{2}-8(2)+k=0$	M1
$k=4$	A1
(b)
attempting to anti-differentiate	M1
$y=x^{3}-4x^{2}+4x+c$	A1	FT from their $k$.
$c=5$, so $y=x^{3}-4x^{2}+4x+5$	A1
(c)
gradient at $x=0$ is $k=4$	(M1)	FT from (a).
$y=4x+5$	A1	FT from (a) and (b).

Answers: (a) $k=4$ · (b) $y=x^{3}-4x^{2}+4x+5$ · (c) $y=4x+5$

+ Add to test✓ In your test133AASL5-q00445.6 Differentiation rules Paper 1Medium7 marks

The functions $f$ and $g$ are differentiable for all $x\in\mathbb{R}$. The table shows some values of $f$, $g$ and their derivatives.

$x$	$f(x)$	$f'(x)$	$g(x)$	$g'(x)$
1	3	$-2$	2	4
2	5	7	1	$-3$

(a)

Let $h(x)=f\big(g(x)\big)$. Find $h'(1)$.

(3)

(b)

Let $p(x)=f(x)\,g(x)$. Find $p'(2)$.

(2)

(c)

Let $q(x)=\dfrac{f(x)}{g(x)}$. Find $q'(1)$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$h'(1)=f'\big(g(1)\big)\times g'(1)$	M1
$f'(2)\times4=7\times4$	(A1)
$28$	A1
(b)
$p'(2)=f'(2)g(2)+f(2)g'(2)=7(1)+5(-3)$	M1
$-8$	A1
(c)
$q'(1)=\dfrac{g(1)f'(1)-f(1)g'(1)}{\big(g(1)\big)^{2}}=\dfrac{2(-2)-3(4)}{2^{2}}$	M1
$-4$	A1

Answers: (a) $28$ · (b) $-8$ · (c) $-4$

+ Add to test✓ In your test134AASL5-q00485.6 Differentiation rules Paper 1Medium6 marks

Let $f(x)=\mathrm{e}^{2x}\cos x$, for $-\dfrac{\pi}{2}\lt x\lt\dfrac{\pi}{2}$.

(a)

Find $f'(x)$.

(2)

(b)

Show that the $x$-coordinate of the stationary point on the graph of $f$ satisfies $\tan x=2$.

(2)

(c)

Find the equation of the tangent to the graph of $f$ at the point where $x=0$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
attempt to use the product rule, e.g. $2\mathrm{e}^{2x}\cos x+\mathrm{e}^{2x}\times\ldots$	M1
$f'(x)=2\mathrm{e}^{2x}\cos x-\mathrm{e}^{2x}\sin x$	A1	Accept $\mathrm{e}^{2x}(2\cos x-\sin x)$.
(b)
$\mathrm{e}^{2x}(2\cos x-\sin x)=0$ and $\mathrm{e}^{2x}\gt 0$, so $2\cos x-\sin x=0$	M1	Must see $\mathrm{e}^{2x}\neq0$ (or $\gt 0$) stated or implied by dividing.
$\sin x=2\cos x\Rightarrow\dfrac{\sin x}{\cos x}=2$	A1	Valid since $\cos x\neq0$ on the domain.
$\tan x=2$	AG
(c)
$f(0)=1$ and $f'(0)=2$	A1	Both values needed.
$y=2x+1$	A1	FT from their (a).

Answers: (a) $\mathrm{e}^{2x}(2\cos x-\sin x)$ · (c) $y=2x+1$

+ Add to test✓ In your test135AASL5-q00495.6 Differentiation rules Paper 1Medium8 marks

Let $f(x)=\dfrac{x}{\ln x}$, for $x\gt 1$.

(a)

Show that $f'(x)=\dfrac{\ln x-1}{(\ln x)^{2}}$.

(2)

(b)

Find the exact coordinates of the stationary point on the graph of $f$.

(3)

(c)

Justify that this point is a local minimum.

(1)

(d)

Find the equation of the tangent to the graph of $f$ at the point where $x=\mathrm{e}^{2}$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
attempt to use the quotient rule, with terms in the correct order	M1
$f'(x)=\dfrac{\ln x\times1-x\times\frac{1}{x}}{(\ln x)^{2}}$	A1
$f'(x)=\dfrac{\ln x-1}{(\ln x)^{2}}$	AG
(b)
$\ln x-1=0\Rightarrow\ln x=1$	M1
$x=\mathrm{e}$	A1
$y=\dfrac{\mathrm{e}}{\ln\mathrm{e}}=\mathrm{e}$	A1	Coordinates $(\mathrm{e},\,\mathrm{e})$.
(c)
$(\ln x)^{2}\gt 0$; for $1\lt x\lt\mathrm{e}$, $\ln x-1\lt 0$ so $f'(x)\lt 0$; for $x\gt\mathrm{e}$, $f'(x)\gt 0$; $f'$ changes from negative to positive	R1	Accept a correct sign diagram, or a valid second-derivative argument.
(d)
$f\left(\mathrm{e}^{2}\right)=\dfrac{\mathrm{e}^{2}}{2}$ and $f'\left(\mathrm{e}^{2}\right)=\dfrac{2-1}{4}=\dfrac14$	A1	Both values needed.
$y-\dfrac{\mathrm{e}^{2}}{2}=\dfrac14\left(x-\mathrm{e}^{2}\right)$	A1	Accept $y=\dfrac14x+\dfrac{\mathrm{e}^{2}}{4}$. FT from their values.

Answers: (b) $(\mathrm{e},\ \mathrm{e})$ · (d) $y=\dfrac14x+\dfrac{\mathrm{e}^{2}}{4}$

+ Add to test✓ In your test136AASL5-q00505.6 Differentiation rules Paper 1Medium6 marks

Let $f(x)=\sqrt{x}$, for $x\ge0$, and $g(x)=4x^{2}+9$, for $x\in\mathbb{R}$.

(a)

Write down an expression for $(f\circ g)(x)$.

(1)

(b)

Find $(f\circ g)'(x)$.

(2)

(c)

Find the gradient of the graph of $y=(f\circ g)(x)$ at $x=2$.

(1)

(d)

Show that the derivative of $(g\circ f)(x)$ is constant for $x\gt 0$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\sqrt{4x^{2}+9}$	A1
(b)
attempt to use the chain rule, e.g. $\tfrac12(4x^{2}+9)^{-\frac12}\times\ldots$	M1
$\dfrac{4x}{\sqrt{4x^{2}+9}}$	A1	Accept $\tfrac12(4x^{2}+9)^{-\frac12}(8x)$.
(c)
$\dfrac{8}{\sqrt{25}}=\dfrac85$	A1	FT from their (b).
(d)
$(g\circ f)(x)=4\left(\sqrt{x}\right)^{2}+9=4x+9$	A1
$(g\circ f)'(x)=4$, which is constant	A1	Accept a chain-rule approach: $8\sqrt{x}\times\dfrac{1}{2\sqrt{x}}=4$.

Answers: (a) $\sqrt{4x^{2}+9}$ · (b) $\dfrac{4x}{\sqrt{4x^{2}+9}}$ · (c) $\tfrac85$ · (d) $(g\circ f)'(x)=4$

+ Add to test✓ In your test137AASL5-q00515.6 Differentiation rules Paper 1Medium8 marks

Let $f(x)=\mathrm{e}^{x}(x^{2}+k)$, where $k\in\mathbb{R}$ and $x\in\mathbb{R}$.

(a)

Show that $f'(x)=\mathrm{e}^{x}(x^{2}+2x+k)$.

(2)

(b)

The graph of $f$ has a stationary point at $x=1$. Find the value of $k$.

(2)

(c)

Using your value of $k$, find the $x$-coordinate of the other stationary point.

(2)

(d)

Find the set of values of $k$ for which the graph of $f$ has no stationary points.

(2)

Show mark scheme

Working	Mark	Notes
(a)
attempt to use the product rule	M1
$\mathrm{e}^{x}(x^{2}+k)+\mathrm{e}^{x}(2x)$	A1
$f'(x)=\mathrm{e}^{x}(x^{2}+2x+k)$	AG
(b)
$f'(1)=0\Rightarrow\mathrm{e}(1+2+k)=0$	M1	Must see $\mathrm{e}\neq0$ used, or $1+2+k=0$.
$k=-3$	A1
(c)
$x^{2}+2x-3=0\Rightarrow(x+3)(x-1)=0$	M1
$x=-3$	A1
(d)
$x^{2}+2x+k=0$ has no real roots: discriminant $4-4k\lt 0$	M1	Since $\mathrm{e}^{x}\gt 0$.
$k\gt 1$	A1	Award A0 for $k\ge1$: when $k=1$, $f'(-1)=0$.

Answers: (b) $k=-3$ · (c) $x=-3$ · (d) $k\gt 1$

+ Add to test✓ In your test138AASL5-q00595.7 The second derivative and graphs Paper 1Medium6 marks

The function $f$ is defined for $x\in\mathbb{R}$. The diagram shows the graph of its second derivative, $y=f''(x)$. The graph of $f''$ crosses the $x$-axis at $x=-2$ and touches it at $x=1$.

(a)

Write down the $x$-coordinate of the point of inflexion on the graph of $f$.

(1)

(b)

Explain why the graph of $f$ does not have a point of inflexion at $x=1$.

(1)

(c)

Write down the set of values of $x$ for which the graph of $f$ is concave-down.

(1)

(d)

It is given that $f'(1)=0$. Determine the nature of the stationary point on the graph of $f$ at $x=1$. Justify your answer.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$x=-2$	A1
(b)
$f''(x)\ge0$ on both sides of $x=1$: $f''$ does not change sign, so there is no change of concavity	R1
(c)
$x\lt -2$	A1
(d)
$f''(x)\gt 0$ for $-2\lt x\lt1$ and $x\gt1$, so $f'$ is increasing on an interval containing $x=1$	R1
as $f'(1)=0$, $f'(x)\lt 0$ just to the left of $x=1$ and $f'(x)\gt 0$ just to the right	R1
local minimum	A1	Award A1 only if at least one R1 is awarded. The second-derivative test fails here, since $f''(1)=0$.

Answers: (a) $x=-2$ · (c) $x\lt -2$ · (d) local minimum

+ Add to test✓ In your test139AASL5-q00625.7 The second derivative and graphs Paper 1Medium7 marks

A function $f$ has derivative $f'(x)=x^{2}\mathrm{e}^{-x}$, for $x\in\mathbb{R}$.

(a)

Find $f''(x)$.

(2)

(b)

Find the set of values of $x$ for which the graph of $f$ is concave-up.

(2)

(c)

Explain why the graph of $f$ has no local maximum or local minimum points.

(1)

(d)

The graph of $f$ has two points of inflexion. State, with a reason, the $x$-coordinate of the point of inflexion with zero gradient.

(2)

Show mark scheme

Working	Mark	Notes
(a)
attempt to use the product rule	M1
$f''(x)=2x\mathrm{e}^{-x}-x^{2}\mathrm{e}^{-x}$	A1	Accept $x(2-x)\mathrm{e}^{-x}$.
(b)
$x(2-x)\mathrm{e}^{-x}\gt 0\Rightarrow x(2-x)\gt 0$	M1	Or sign diagram for $f''$ with zeros 0 and 2.
$0\lt x\lt2$	A1
(c)
$f'(x)=x^{2}\mathrm{e}^{-x}\ge0$ for all $x$, so $f'$ never changes sign	R1
(d)
$x=0$	A1
$f'(0)=0$, whereas $f'(2)=4\mathrm{e}^{-2}\neq0$	R1	Do not award A1R0.

Answers: (a) $x(2-x)\mathrm{e}^{-x}$ · (b) $0\lt x\lt2$ · (d) $x=0$

+ Add to test✓ In your test140AASL5-q00675.3 Differentiating polynomials and integer powers Paper 1Medium8 marks

Let $f(x)=3x^{4}-4x^{3}$, for $x\in\mathbb{R}$. The graph of $f$ has stationary points at $x=0$ and $x=1$.

(a)

Show that $f'(x)=12x^{2}(x-1)$.

(2)

(b)

Find $f''(x)$ and hence show that $(1,\,-1)$ is a local minimum point.

(2)

(c)

Given that $f''(0)=0$, determine the nature of the stationary point at $x=0$. Justify your answer.

(2)

(d)

Find the $x$-coordinate of the point of inflexion on the graph of $f$ that has a non-zero gradient.

(2)

Show mark scheme

Working	Mark	Notes
(a)
attempt to differentiate, e.g. one term correct	M1
$f'(x)=12x^{3}-12x^{2}$	A1	Must see this expanded form before the factorized answer.
$f'(x)=12x^{2}(x-1)$	AG
(b)
$f''(x)=36x^{2}-24x$	A1
$f''(1)=12\gt 0$ (and $f(1)=3-4=-1$), so a local minimum	R1
(c)
$f'(x)\lt 0$ on both sides of $x=0$, e.g. $f'(-1)=-24$, $f'(0.5)=-1.5$ (or $f''$ changes sign at $x=0$)	R1	Since $12x^{2}\ge0$ and $x-1\lt 0$ near $x=0$.
a point of inflexion with zero gradient	A1	Award A1 only if R1 is awarded.
(d)
$12x(3x-2)=0$	M1	Setting their $f''(x)=0$.
$x=\tfrac{2}{3}$	A1	$f''$ changes sign at $x=\tfrac23$ as it is a single root.

Answers: (b) $f''(x)=36x^{2}-24x$ · (c) point of inflexion with zero gradient · (d) $x=\tfrac23$

+ Add to test✓ In your test141AASL5-q00685.3 Differentiating polynomials and integer powers Paper 1Medium7 marks

Let $f(x)=x^{3}+bx^{2}+cx+d$, where $b,\,c,\,d\in\mathbb{R}$. The graph of $f$ has a point of inflexion at $(1,\,3)$, and the gradient of the graph at this point is $-2$.

(a)

Find $f'(x)$ and $f''(x)$.

(2)

(b)

Show that $b=-3$.

(1)

(c)

Find the value of $c$ and the value of $d$.

(3)

(d)

Write down the set of values of $x$ for which the graph of $f$ is concave-down.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$f'(x)=3x^{2}+2bx+c$	A1
$f''(x)=6x+2b$	A1
(b)
$f''(1)=0\Rightarrow6+2b=0$	A1
$b=-3$	AG
(c)
$f'(1)=-2\Rightarrow3-6+c=-2$	M1
$c=1$	A1
$f(1)=1-3+1+d=3\Rightarrow d=4$	A1	FT from their $c$.
(d)
$x\lt 1$	A1	$f''(x)=6x-6\lt 0$.

Answers: (c) $c=1,\ d=4$ · (d) $x\lt1$

+ Add to test✓ In your test142AASL5-q00715.8 Stationary points, inflexion and optimization Paper 1Medium9 marks

Let $f(x)=2x+4\cos x$, for $0\le x\le2\pi$.

(a)

Find $f'(x)$.

(1)

(b)

Find the $x$-coordinates of the stationary points on the graph of $f$.

(3)

(c)

Use the second derivative to determine the nature of each stationary point.

(3)

(d)

Find the greatest value of $f(x)$ for $0\le x\le2\pi$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$f'(x)=2-4\sin x$	A1
(b)
$2-4\sin x=0\Rightarrow\sin x=\dfrac12$	M1
$x=\dfrac{\pi}{6}$	A1
$x=\dfrac{5\pi}{6}$	A1
(c)
$f''(x)=-4\cos x$	A1
$f''\left(\tfrac{\pi}{6}\right)=-2\sqrt3\lt 0$, so a local maximum at $x=\tfrac{\pi}{6}$	R1
$f''\left(\tfrac{5\pi}{6}\right)=2\sqrt3\gt 0$, so a local minimum at $x=\tfrac{5\pi}{6}$	R1
(d)
comparing the local maximum $f\left(\tfrac{\pi}{6}\right)=\tfrac{\pi}{3}+2\sqrt3$ with the endpoint value $f(2\pi)=4\pi+4$	M1	$f$ is increasing for $\tfrac{5\pi}{6}\lt x\le2\pi$.
$4\pi+4$	A1	Award M0A0 for $\tfrac{\pi}{3}+2\sqrt3$.

Answers: (b) $x=\dfrac{\pi}{6},\ \dfrac{5\pi}{6}$ · (c) local max at $\tfrac{\pi}{6}$, local min at $\tfrac{5\pi}{6}$ · (d) $4\pi+4$

+ Add to test✓ In your test143AASL5-q00745.8 Stationary points, inflexion and optimization Paper 1Medium9 marks

Let $f(x)=x\mathrm{e}^{-x}$, for $x\in\mathbb{R}$.

(a)

Find $f'(x)$.

(2)

(b)

Show that $f''(x)=(x-2)\mathrm{e}^{-x}$.

(1)

(c)

Show that the graph of $f$ has a point of inflexion at $x=2$.

(2)

(d)

Find the equation of the tangent to the graph of $f$ at the point of inflexion.

(3)

(e)

Write down the $x$-coordinate of the point where this tangent meets the $x$-axis.

(1)

Show mark scheme

Working	Mark	Notes
(a)
attempt to use the product rule	M1
$f'(x)=\mathrm{e}^{-x}-x\mathrm{e}^{-x}$	A1	Accept $(1-x)\mathrm{e}^{-x}$.
(b)
$f''(x)=-\mathrm{e}^{-x}-\left(\mathrm{e}^{-x}-x\mathrm{e}^{-x}\right)$	A1
$f''(x)=(x-2)\mathrm{e}^{-x}$	AG
(c)
$f''(2)=0$	A1
$\mathrm{e}^{-x}\gt 0$ and $x-2$ changes sign at $x=2$, so $f''$ changes sign (concavity changes)	R1	e.g. $f''(1)=-\mathrm{e}^{-1}\lt 0$, $f''(3)=\mathrm{e}^{-3}\gt 0$.
(d)
point $\left(2,\ 2\mathrm{e}^{-2}\right)$	A1
gradient $f'(2)=-\mathrm{e}^{-2}$	A1
$y-2\mathrm{e}^{-2}=-\mathrm{e}^{-2}(x-2)$	A1	Accept $y=-\mathrm{e}^{-2}x+4\mathrm{e}^{-2}$.
(e)
$x=4$	A1

Answers: (a) $(1-x)\mathrm{e}^{-x}$ · (d) $y=-\mathrm{e}^{-2}x+4\mathrm{e}^{-2}$ · (e) $x=4$

+ Add to test✓ In your test144AASL5-q00805.9 Kinematics Paper 1Medium8 marks

A particle moves in a straight line. Its displacement, $s$ metres, from a fixed point O after $t$ seconds is given by $s(t)=t^{3}-6t^{2}+9t$, for $0\le t\le5$.

(a)

Find an expression for the velocity, $v$, of the particle.

(1)

(b)

Find the values of $t$ when the particle is at rest.

(2)

(c)

Find the total distance travelled by the particle in the 5 seconds.

(3)

(d)

Find the speed of the particle when its acceleration is zero.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$v=3t^{2}-12t+9$	A1
(b)
$3(t-1)(t-3)=0$	M1
$t=1,\ t=3$	A1
(c)
attempt to find $s$ at $t=0,1,3,5$: $0,\ 4,\ 0,\ 20$	(M1)	Or $\int_{0}^{5}\|v\|\,\mathrm{d}t$ split at $t=1$ and $t=3$.
$4+4+20$	(M1)	Adding distances for the three stages.
$28$ m	A1
(d)
$a=6t-12=0\Rightarrow t=2$	M1
$v(2)=-3$, so speed $=3\ \text{m s}^{-1}$	A1	Award A0 for $-3$.

Answers: (a) $v=3t^{2}-12t+9$ · (b) $t=1,\ 3$ · (c) 28 m · (d) $3\ \text{m s}^{-1}$

+ Add to test✓ In your test145AASL5-q00815.9 Kinematics Paper 1Medium8 marks

A particle moves in a straight line. Its acceleration, in m s⁻², after $t$ seconds is given by $a(t)=4-2t$, for $0\le t\le 6$. When $t=0$ its velocity is $5\ \text{m s}^{-1}$.

(a)

Find an expression for the velocity, $v(t)$.

(3)

(b)

Find the value of $t$ at which the particle changes direction.

(2)

(c)

Find the total distance travelled by the particle between $t=0$ and $t=6$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
attempt to integrate $a(t)$	M1
$4t-t^{2}+c$	A1
$v(t)=5+4t-t^{2}$	A1
(b)
$5+4t-t^{2}=0\Rightarrow(5-t)(1+t)=0$	M1
$t=5$	A1	Must reject $t=-1$. $v$ changes sign at $t=5$ since $v(6)=-7$.
(c)
$\displaystyle\int_{0}^{5}(5+4t-t^{2})\,\mathrm{d}t-\int_{5}^{6}(5+4t-t^{2})\,\mathrm{d}t$	M1	Or $\int_{0}^{6}\|v\|\,\mathrm{d}t$ split at their $t$.
$\dfrac{100}{3}$ and $\dfrac{10}{3}$	A1	Using $5t+2t^{2}-\tfrac{t^{3}}{3}$: values $\tfrac{100}{3}$ at $t=5$ and $30$ at $t=6$.
$\dfrac{110}{3}$ m	A1	Accept $36\tfrac23$ m.

Answers: (a) $v=5+4t-t^{2}$ · (b) $t=5$ · (c) $\tfrac{110}{3}$ m

+ Add to test✓ In your test146AASL5-q00835.9 Kinematics Paper 1Medium8 marks

A particle starts from a fixed point O and moves in a straight line. Its velocity, $v\ \text{m s}^{-1}$, after $t$ seconds is given by $v(t)=4-\mathrm{e}^{t}$, for $0\le t\le 2$.

(a)

Find the exact value of $t$ at which the particle is instantaneously at rest.

(2)

(b)

Show that the displacement of the particle from O at this time is $8\ln 2-3$ metres.

(3)

(c)

Find the exact total distance travelled by the particle in the 2 seconds.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$4-\mathrm{e}^{t}=0$	(M1)
$t=\ln 4$	A1	Accept $2\ln 2$.
(b)
$\displaystyle\int_{0}^{\ln 4}\left(4-\mathrm{e}^{t}\right)\mathrm{d}t$	M1
$\left[4t-\mathrm{e}^{t}\right]_{0}^{\ln 4}$	A1
$(4\ln 4-4)-(0-1)=4\ln 4-3$	A1	Must see $\mathrm{e}^{\ln 4}=4$ used and $4\ln 4=8\ln 2$.
$=8\ln 2-3$	AG
(c)
displacement at $t=2$: $\displaystyle\int_{0}^{2}\left(4-\mathrm{e}^{t}\right)\mathrm{d}t=9-\mathrm{e}^{2}$; or $\displaystyle\int_{\ln 4}^{2}\left(\mathrm{e}^{t}-4\right)\mathrm{d}t=\mathrm{e}^{2}-12+8\ln 2$	(M1)
distance $=2(8\ln 2-3)-(9-\mathrm{e}^{2})$, or $(8\ln 2-3)+(\mathrm{e}^{2}-12+8\ln 2)$	(M1)	Adding the two stages as positive quantities.
$\mathrm{e}^{2}+16\ln 2-15$ m	A1	$=3.479\ldots$ Accept equivalent exact forms, e.g. $\mathrm{e}^{2}+8\ln 4-15$.

Answers: (a) $t=\ln 4$ · (c) $\mathrm{e}^{2}+16\ln 2-15$ m

+ Add to test✓ In your test147AASL5-q00845.9 Kinematics Paper 1Medium9 marks

A particle moves in a straight line, starting from a fixed point O. Its velocity, $v\ \text{m s}^{-1}$, after $t$ seconds is given by $v(t)=3t^{2}-kt$, for $0\le t\le 4$, where $k$ is a positive constant. When $t=4$ the particle is back at O.

(a)

Show that $k=8$.

(2)

(b)

Find the value of $t$, for $0\lt t\lt 4$, at which the particle changes direction.

(2)

(c)

Find the total distance travelled by the particle in the 4 seconds.

(3)

(d)

Find the minimum velocity of the particle.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\displaystyle\int_{0}^{4}(3t^{2}-kt)\,\mathrm{d}t=0$	M1	Displacement after 4 seconds equated to zero.
$\left[t^{3}-\tfrac{k}{2}t^{2}\right]_{0}^{4}=64-8k$	A1
$64-8k=0\Rightarrow k=8$	AG
(b)
$3t^{2}-8t=0\Rightarrow t(3t-8)=0$	(M1)
$t=\dfrac{8}{3}$	A1	Must reject $t=0$.
(c)
$\displaystyle\int_{0}^{8/3}(8t-3t^{2})\,\mathrm{d}t+\int_{8/3}^{4}(3t^{2}-8t)\,\mathrm{d}t$	(M1)	Or $2\times\left\|\displaystyle\int_{0}^{8/3}(3t^{2}-8t)\,\mathrm{d}t\right\|$, since the net displacement is zero.
each part equals $\dfrac{256}{27}$	(A1)	$\left[4t^{2}-t^{3}\right]_{0}^{8/3}=\tfrac{256}{9}-\tfrac{512}{27}$
$\dfrac{512}{27}$ m	A1	Accept $18\tfrac{26}{27}$ m.
(d)
$a=6t-8=0\Rightarrow t=\tfrac{4}{3}$	(M1)
$v_{\min}=-\dfrac{16}{3}\ \text{m s}^{-1}$	A1	$3\left(\tfrac{16}{9}\right)-8\left(\tfrac{4}{3}\right)$. Award A0 for $\tfrac{16}{3}$ (that is the speed).

Answers: (b) $t=\tfrac{8}{3}$ · (c) $\tfrac{512}{27}$ m · (d) $-\tfrac{16}{3}\ \text{m s}^{-1}$

+ Add to test✓ In your test148AASL5-q00945.10 Indefinite integrals and reverse chain rule Paper 1Medium7 marks

(a)

Find $\displaystyle\int\dfrac{x}{x^{2}+4}\,\mathrm{d}x$.

(3)

(b)

Hence find the exact value of $\displaystyle\int_{0}^{2}\dfrac{x}{x^{2}+4}\,\mathrm{d}x$.

(2)

(c)

Find $\displaystyle\int\sin x\cos^{3}x\,\mathrm{d}x$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
recognizing $2x$ as the derivative of $x^{2}+4$, e.g. $u=x^{2}+4,\ \mathrm{d}u=2x\,\mathrm{d}x$	M1
$\displaystyle\int\dfrac{1}{2u}\,\mathrm{d}u$	A1	Or $k\ln(x^{2}+4)$ by inspection.
$\tfrac{1}{2}\ln(x^{2}+4)+C$	A1
(b)
$\tfrac12\ln8-\tfrac12\ln4$	M1	Correct substitution of both limits into their answer to (a).
$\tfrac12\ln2$	A1	Accept $\ln\sqrt2$.
(c)
$u=\cos x,\ \mathrm{d}u=-\sin x\,\mathrm{d}x$, or $k\cos^{4}x$ by inspection	M1
$-\tfrac14\cos^{4}x+C$	A1

Answers: (a) $\tfrac12\ln(x^{2}+4)+C$ · (b) $\tfrac12\ln2$ · (c) $-\tfrac14\cos^{4}x+C$

+ Add to test✓ In your test149AASL5-q00995.10 Indefinite integrals and reverse chain rule Paper 1Medium5 marks

(a)

Find $\displaystyle\int x^{2}\left(x^{3}+1\right)^{5}\mathrm{d}x$.

(3)

(b)

Hence find the value of $\displaystyle\int_{0}^{1}x^{2}\left(x^{3}+1\right)^{5}\mathrm{d}x$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
recognizing $x^{2}$ as a multiple of the derivative of $x^{3}+1$, or substitution $u=x^{3}+1$, $\mathrm{d}u=3x^{2}\,\mathrm{d}x$	M1
$(x^{3}+1)^{6}$ seen, or $\displaystyle\int\tfrac13u^{5}\,\mathrm{d}u$	A1
$\dfrac{1}{18}\left(x^{3}+1\right)^{6}+C$	A1
(b)
$\dfrac{2^{6}}{18}-\dfrac{1}{18}$	(M1)	Substituting both limits into their (a).
$\dfrac{7}{2}$	A1	Accept $\tfrac{63}{18}$.

Answers: (a) $\tfrac{1}{18}(x^{3}+1)^{6}+C$ · (b) $\tfrac72$

+ Add to test✓ In your test150AASL5-q01005.10 Indefinite integrals and reverse chain rule Paper 1Medium5 marks

(a)

By using the substitution $u=\cos x$, find $\displaystyle\int\sin x\,\mathrm{e}^{\cos x}\,\mathrm{d}x$.

(3)

(b)

Hence find the exact value of $\displaystyle\int_{0}^{\frac{\pi}{2}}\sin x\,\mathrm{e}^{\cos x}\,\mathrm{d}x$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\dfrac{\mathrm{d}u}{\mathrm{d}x}=-\sin x$, so the integral becomes $\displaystyle\int-\mathrm{e}^{u}\,\mathrm{d}u$	M1
$-\mathrm{e}^{u}$	A1
$-\mathrm{e}^{\cos x}+C$	A1	Must be in terms of x.
(b)
$-\mathrm{e}^{\cos(\pi/2)}-\left(-\mathrm{e}^{\cos 0}\right)=-\mathrm{e}^{0}+\mathrm{e}^{1}$	(M1)
$\mathrm{e}-1$	A1

Answers: (a) $-\mathrm{e}^{\cos x}+C$ · (b) $\mathrm{e}-1$

+ Add to test✓ In your test151AASL5-q01015.10 Indefinite integrals and reverse chain rule Paper 1Medium6 marks

(a)

Find $\displaystyle\int\dfrac{\cos x}{3+\sin x}\,\mathrm{d}x$.

(2)

(b)

By writing $\tan x=\dfrac{\sin x}{\cos x}$, find $\displaystyle\int\tan x\,\mathrm{d}x$, for $0\le x\lt\dfrac{\pi}{2}$.

(2)

(c)

Hence find the exact value of $\displaystyle\int_{0}^{\frac{\pi}{3}}\tan x\,\mathrm{d}x$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
recognizing the numerator as the derivative of the denominator, or $u=3+\sin x$	(M1)
$\ln(3+\sin x)+C$	A1	Accept $\ln\|3+\sin x\|$.
(b)
numerator is $-1\times$ the derivative of $\cos x$	(M1)
$-\ln(\cos x)+C$	A1	Accept $-\ln\|\cos x\|+C$ or $\ln(\sec x)+C$.
(c)
$-\ln\left(\cos\tfrac{\pi}{3}\right)+\ln(\cos 0)=-\ln\tfrac12+0$	(M1)
$\ln 2$	A1

Answers: (a) $\ln(3+\sin x)+C$ · (b) $-\ln(\cos x)+C$ · (c) $\ln 2$

+ Add to test✓ In your test152AASL5-q01025.10 Indefinite integrals and reverse chain rule Paper 1Medium9 marks

A function $f$ satisfies $f''(x)=8\mathrm{e}^{2x}$, for $x\in\mathbb{R}$. The gradient of the graph of $f$ at the point $(0,\,-1)$ is 3.

(a)

Find $f'(x)$.

(3)

(b)

Find $f(x)$.

(3)

(c)

Find the exact coordinates of the stationary point on the graph of $f$ and justify that it is a minimum.

(3)

Show mark scheme

Working	Mark	Notes
(a)
attempt to integrate $f''(x)$	M1
$4\mathrm{e}^{2x}+c$	A1
$4+c=3\Rightarrow f'(x)=4\mathrm{e}^{2x}-1$	A1
(b)
attempt to integrate their $f'(x)$	M1
$2\mathrm{e}^{2x}-x+d$	A1	FT from (a).
$2+d=-1\Rightarrow f(x)=2\mathrm{e}^{2x}-x-3$	A1
(c)
$4\mathrm{e}^{2x}=1\Rightarrow x=-\ln 2$	A1	Accept $\tfrac12\ln\tfrac14$.
$y=2\left(\tfrac14\right)+\ln 2-3=\ln 2-\dfrac52$	A1
$f''(x)=8\mathrm{e}^{2x}\gt 0$ for all $x$, so it is a minimum	R1

Answers: (a) $f'(x)=4\mathrm{e}^{2x}-1$ · (b) $f(x)=2\mathrm{e}^{2x}-x-3$ · (c) $\left(-\ln 2,\ \ln 2-\tfrac52\right)$

+ Add to test✓ In your test153AASL5-q01065.11 Definite integrals and areas Paper 1Medium8 marks

Let $f(x)=x^{3}-2x^{2}-3x$. The diagram shows part of the graph of $f$. The shaded region A lies above the $x$-axis and the shaded region B lies below the $x$-axis.

(a)

Find the $x$-coordinates of the points where the graph of $f$ meets the $x$-axis.

(2)

(b)

Find the area of region A.

(2)

(c)

Find the area of region B.

(3)

(d)

Hence write down the value of $\displaystyle\int_{-1}^{3}f(x)\,\mathrm{d}x$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$x(x^{2}-2x-3)=0\Rightarrow x(x+1)(x-3)=0$	M1
$x=-1,\ 0,\ 3$	A1
(b)
$\left[\tfrac{x^{4}}{4}-\tfrac{2x^{3}}{3}-\tfrac{3x^{2}}{2}\right]_{-1}^{0}=0-\left(\tfrac14+\tfrac23-\tfrac32\right)$	M1	Correct integration and substitution of limits.
$\dfrac{7}{12}$	A1
(c)
$\displaystyle\int_{0}^{3}f(x)\,\mathrm{d}x=\tfrac{81}{4}-18-\tfrac{27}{2}$	M1
$=-\dfrac{45}{4}$	A1
area $=\dfrac{45}{4}$	A1	Award A1 for the positive value.
(d)
$\dfrac{7}{12}-\dfrac{45}{4}=-\dfrac{32}{3}$	A1	FT from (b) and (c).

Answers: (a) $-1,\ 0,\ 3$ · (b) $\tfrac{7}{12}$ · (c) $\tfrac{45}{4}$ · (d) $-\tfrac{32}{3}$

+ Add to test✓ In your test154AASL5-q01105.11 Definite integrals and areas Paper 1Medium6 marks

The diagram shows the graphs of $y=5-x$ and $y=\dfrac{4}{x}$, for $x\gt 0$. The shaded region R is enclosed by the two graphs.

(a)

Show that the graphs intersect where $x=1$ and $x=4$.

(2)

(b)

Find the exact area of R. Give your answer in the form $p-q\ln 2$, where $p,q\in\mathbb{Q}$.

(4)

Show mark scheme

Working	Mark	Notes
(a)
$5-x=\dfrac{4}{x}\Rightarrow x^{2}-5x+4=0$	M1
$(x-1)(x-4)=0$	A1
$x=1,\ x=4$	AG
(b)
$\displaystyle\int_{1}^{4}\left(5-x-\dfrac{4}{x}\right)\mathrm{d}x$	M1	Correct expression, including limits, must be seen. Accept the integral of the difference in the opposite order with the modulus taken.
$\left[5x-\dfrac{x^{2}}{2}-4\ln x\right]_{1}^{4}$	A1
$(20-8-4\ln 4)-\left(5-\tfrac12-0\right)$	M1	Substituting both limits.
$\dfrac{15}{2}-8\ln 2$	A1	$=1.95\ldots$ Accept $p=\tfrac{15}{2},\ q=8$.

Answers: (b) $\tfrac{15}{2}-8\ln 2$

+ Add to test✓ In your test155AASL5-q01125.11 Definite integrals and areas Paper 1Medium7 marks

Let $f(x)=\sin(2x)$, for $0\le x\le\dfrac{2\pi}{3}$.

(a)

Find the value of $x$, for $0\lt x\lt\dfrac{2\pi}{3}$, at which the graph of $f$ meets the $x$-axis.

(1)

(b)

Find the value of $\displaystyle\int_{0}^{\frac{2\pi}{3}}f(x)\,\mathrm{d}x$.

(2)

(c)

Find the total area of the region enclosed by the graph of $f$, the $x$-axis and the line $x=\dfrac{2\pi}{3}$.

(3)

(d)

Explain why the answers to parts (b) and (c) are different.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$x=\dfrac{\pi}{2}$	A1
(b)
$\left[-\tfrac12\cos(2x)\right]_{0}^{2\pi/3}=-\tfrac12\cos\tfrac{4\pi}{3}+\tfrac12\cos 0=\tfrac14+\tfrac12$	(M1)
$\dfrac34$	A1
(c)
$\displaystyle\int_{0}^{\pi/2}\sin(2x)\,\mathrm{d}x+\left\|\displaystyle\int_{\pi/2}^{2\pi/3}\sin(2x)\,\mathrm{d}x\right\|$	M1	Correct expression for the area must be seen; FT from their (a).
$1$ and $\left\|-\tfrac14\right\|=\tfrac14$	A1
$\dfrac54$	A1
(d)
between $\tfrac{\pi}{2}$ and $\tfrac{2\pi}{3}$ the curve is below the $x$-axis, so that part contributes a negative value to the integral but a positive value to the area	R1

Answers: (a) $\tfrac{\pi}{2}$ · (b) $\tfrac34$ · (c) $\tfrac54$

+ Add to test✓ In your test156IBMA2S-052-00-055.6 Differentiation rules Paper 1Medium6 marks

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Let $f(x) = \dfrac{2x + 1}{x^2 + 2}$, for $x \in \mathbb{R}$.

(a)

Show that $f'(x) = \dfrac{-2(x + 2)(x - 1)}{(x^2 + 2)^2}$.

(3)

(b)

Find the coordinates of the stationary points on the graph of $f$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts the quotient rule A1 correct unsimplified derivative $\frac{2(x^2 + 2) - (2x + 1)(2x)}{(x^2 + 2)^2}$ A1 numerator simplified to $-2x^2 - 2x + 4 = -2(x + 2)(x - 1)$	✓ 3
Part (b)
M1 sets $f'(x) = 0$, giving $x = 1$ or $x = -2$ A1 $(1, 1)$ A1 $\left(-2, -\frac{1}{2}\right)$	✓ 3

Common error: The order in the quotient rule matters: it is $vu' - uv'$ over $v^2$. Reversing it flips the sign of every term in the numerator.

+ Add to test✓ In your test157IBMA2S-052-00-075.4 Tangents and normals Paper 1Medium6 marks

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A curve has equation $y = \ln(3x - 2)$, for $x > \frac{2}{3}$. The point P lies on the curve, and the gradient of the curve at P is $\frac{1}{2}$.

(a)

Find $\dfrac{dy}{dx}$.

(2)

(b)

Find the $x$-coordinate of P.

(2)

(c)

The tangent to the curve at P meets the $y$-axis at Q. Find the exact $y$-coordinate of Q.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses the chain rule with inner function $3x - 2$ A1 $\frac{dy}{dx} = \frac{3}{3x - 2}$	✓ 2
Part (b)
M1 sets $\frac{3}{3x - 2} = \frac{1}{2}$ A1 $x = \frac{8}{3}$	✓ 2
Part (c)
M1 uses the tangent $y - \ln 6 = \frac{1}{2}\left(x - \frac{8}{3}\right)$, with $y$-coordinate of P equal to $\ln 6$ A1 $y = \ln 6 - \frac{4}{3}$	✓ 2

Common error: The derivative of $\ln(3x - 2)$ is not $\frac{1}{3x - 2}$: the factor 3 from the inner function must be included.

+ Add to test✓ In your test158IBMA2S-052-00-095.6 Differentiation rules Paper 1Medium8 marks

No calculator

Let $f(x) = x^2e^{-x}$, for $x \in \mathbb{R}$.

(a)

Find $f'(x)$.

(2)

(b)

Find the $x$-coordinates of the stationary points of the graph of $f$.

(2)

(c)

Find $f''(x)$ and hence determine the nature of the stationary point where $x = 2$.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts the product rule A1 $f'(x) = e^{-x}(2x - x^2)$ or equivalent	✓ 2
Part (b)
M1 since $e^{-x} \neq 0$, solves $x(2 - x) = 0$ A1 $x = 0$ and $x = 2$	✓ 2
Part (c)
M1 applies the product rule to $f'(x)$ A1 $f''(x) = e^{-x}(x^2 - 4x + 2)$ A1 $f''(2) = -2e^{-2}$ R1 $f''(2) < 0$, so the point is a local maximum	✓ 4

Common error: When differentiating $e^{-x}$ the result is $-e^{-x}$; dropping this minus sign changes both the stationary points and the nature test.

+ Add to test✓ In your test159IBMA2S-052-00-105.7 The second derivative and graphs Paper 1Medium8 marks

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In a laboratory trial, the response of a cell culture, $R$ units, to a dose of $d$ milligrams of a hormone is modelled by $R(d) = \dfrac{kd}{d^2 + 4}$, for $d \ge 0$, where $k$ is a positive constant.

(a)

Show that $R'(d) = \dfrac{k(4 - d^2)}{(d^2 + 4)^2}$.

(3)

(b)

When the dose is 1 mg, the response is increasing at a rate of 0.6 units per mg. Find the value of $k$.

(2)

(c)

Find the maximum response, justifying that it is a maximum.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts the quotient rule (or the product rule on $kd(d^2+4)^{-1}$) A1 correct unsimplified derivative, e.g. $\frac{k(d^2 + 4) - kd(2d)}{(d^2 + 4)^2}$ A1 numerator simplified to $k(4 - d^2)$, leading to the given answer	✓ 3
Part (b)
M1 substitutes $d = 1$ into $R'(d)$ and sets it equal to 0.6: $\frac{3k}{25} = 0.6$ A1 $k = 5$	✓ 2
Part (c)
A1 $R'(d) = 0$ when $d = 2$ (rejecting $d = -2$ since $d \ge 0$) R1 valid justification: $R'(d) > 0$ for $0 \le d < 2$ and $R'(d) < 0$ for $d > 2$, so this is a maximum; or $R''(2) < 0$, so this is a maximum A1 $R(2) = \frac{10}{8} = \frac{5}{4} = 1.25$ units	✓ 3

Exam tip: A sign check of $R'(d)$ either side of the stationary point is a quick, complete justification and avoids finding a messy second derivative.

+ Add to test✓ In your test160IBMA2S-052-04-055.11 Definite integrals and areas Paper 1Medium6 marks

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A water tank in an industrial cooling system is filled from empty. The rate at which water enters the tank is $r(t)=\frac{12t}{t^{2}+4}$ litres per minute, where $t$ is the time in minutes. The volume of water in the tank at time $t$ is $V$ litres, and $V(0)=0$.

(a)

Find an expression for $V(t)$.

(4)

(b)

Find the time at which the volume of water in the tank is $6\ln 5$ litres.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 recognise the integrand as a multiple of $\frac{g'(t)}{g(t)}$ with $g(t)=t^{2}+4$	✓ 1
A1 $6\ln(t^{2}+4)+c$	✓ 1
M1 use $V(0)=0$: $6\ln 4+c=0$	✓ 1
A1 $V(t)=6\ln(t^{2}+4)-6\ln 4$ or $6\ln\left(\frac{t^{2}+4}{4}\right)$	✓ 1
Part (b)
M1 set $\frac{t^{2}+4}{4}=5$	✓ 1
A1 $t=4$ minutes	✓ 1

Exam tip: When the numerator is a constant multiple of the derivative of the denominator, the integral is a multiple of ln of the denominator. Here $t^2+4>0$, so no modulus is needed.

+ Add to test✓ In your test161IBMA2S-052-04-075.10 Indefinite integrals and reverse chain rule Paper 1Medium6 marks

No calculator

A function $f$, defined for $x>0$, has derivative $f'(x)=\frac{\ln x}{x}$. It is given that $f(e)=2$.

(a)

Find an expression for $f(x)$.

(4)

(b)

Find the exact value of $f(e^{3})$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 recognise $\frac{1}{x}$ as the derivative of $\ln x$, or use the substitution $u=\ln x$	✓ 1
A1 $\frac{1}{2}(\ln x)^{2}+c$	✓ 1
M1 use $f(e)=2$: $\frac{1}{2}+c=2$	✓ 1
A1 $f(x)=\frac{1}{2}(\ln x)^{2}+\frac{3}{2}$	✓ 1
Part (b)
M1 substitute, using $\ln e^{3}=3$: $\frac{9}{2}+\frac{3}{2}$	✓ 1
A1 $6$	✓ 1

Common error: Treating $\frac{\ln x}{x}$ as $\ln x\times\ln x$ or trying to integrate $\ln x$ and $\frac{1}{x}$ separately. Spot the pattern $g(x)\,g'(x)$ with $g(x)=\ln x$.

+ Add to test✓ In your test162IBMA2S-052-04-095.10 Indefinite integrals and reverse chain rule Paper 1Medium6 marks

No calculator

The profile of a ramp in a skate park is modelled by the curve $y=f(x)$, where $x$ and $y$ are in metres. The gradient of the profile is given by $f'(x)=\frac{x+1}{\sqrt{x^{2}+2x+5}}$. The curve passes through the point $(-1,\,1)$.

(a)

Find an expression for $f(x)$.

(5)

(b)

Find the exact height of the ramp profile when $x=1$.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitution $u=x^{2}+2x+5$, $\frac{du}{dx}=2x+2$, or equivalent recognition	✓ 1
A1 $\int\frac{1}{2}u^{-1/2}\,du$	✓ 1
A1 $\sqrt{x^{2}+2x+5}+c$	✓ 1
M1 use $(-1,\,1)$: $\sqrt{4}+c=1$	✓ 1
A1 $f(x)=\sqrt{x^{2}+2x+5}-1$	✓ 1
Part (b)
A1 $f=\sqrt{8}-1=2\sqrt{2}-1$ m	✓ 2

Common error: Forgetting that $x+1$ is only half of $\frac{du}{dx}=2x+2$; missing the factor $\frac{1}{2}$ doubles the answer.

+ Add to test✓ In your test163IBMA2S-052-05-055.6 Differentiation rules Paper 1Medium7 marks

No calculator

Engineers planning a hydro-electric dam model the cross-section of a valley floor by y = f(x), where f(x) = ln(x² + 4), for −5 ≤ x ≤ 5.

(a)

Show that $f''(x) = \frac{8 - 2x^2}{(x^2+4)^2}$.

(3)

(b)

Find the coordinates of the points of inflection on the graph of f. Give your answers in exact form.

(3)

(c)

Explain why both points found in the previous part are points of inflection.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 f'(x) = 2x/(x² + 4), using the chain rule	✓ 1
M1 attempts the quotient rule (or the product rule on 2x(x² + 4)⁻¹) on f'(x): [2(x² + 4) − 2x(2x)]/(x² + 4)²	✓ 1
A1 shows the numerator simplifies: 2x² + 8 − 4x² = 8 − 2x², giving f''(x) = (8 − 2x²)/(x² + 4)²	✓ 1
Part (b)
M1 sets f''(x) = 0, so 8 − 2x² = 0	✓ 1
A1 x = ±2	✓ 1
A1 (2, ln 8) and (−2, ln 8)	✓ 1
Part (c)
R1 the denominator (x² + 4)² is always positive, and 8 − 2x² is positive for \|x\| < 2 and negative for \|x\| > 2, so f''(x) changes sign at both x = 2 and x = −2	✓ 1

Common error: Writing the y-coordinate as ln 4 + ln 4 or (ln 4)² is wrong. Substitute directly: f(2) = ln(4 + 4) = ln 8.

+ Add to test✓ In your test164IBMA2S-052-05-095.7 The second derivative and graphs Paper 1Medium8 marks

No calculator

Let f(x) = x + 2cos x, for 0 ≤ x ≤ 2π.

(a)

Find f''(x).

(2)

(b)

Find the x-coordinates of the points of inflection on the graph of f, justifying your answer.

(3)

(c)

Find the x-coordinate of the local maximum point. Use the second derivative to justify your answer.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 f'(x) = 1 − 2sin x	✓ 1
A1 f''(x) = −2cos x	✓ 1
Part (b)
M1 sets f''(x) = 0, giving cos x = 0	✓ 1
A1 x = π/2 and x = 3π/2	✓ 1
R1 f''(x) = −2cos x changes sign at each value (negative to positive at π/2, positive to negative at 3π/2), so both are points of inflection	✓ 1
Part (c)
M1 sets f'(x) = 0, giving sin x = 1/2	✓ 1
A1 x = π/6 or x = 5π/6	✓ 1
R1 f''(π/6) = −√3 < 0, so the maximum is at x = π/6. Also f''(5π/6) = √3 > 0, so that point is a minimum	✓ 1

Common error: sin x = 1/2 has two solutions in 0 ≤ x ≤ 2π: π/6 and 5π/6. Find both, then test the sign of f'' at each. The negative value identifies the maximum.

+ Add to test✓ In your test165IBMA2S-052-08-065.1 Limits and the derivative Paper 1Medium7 marks

No calculator

Consider the limit $L=\lim_{h\to 0}\frac{\sqrt{9+h}-3}{h}$.

(a)

The limit L is equal to f′(a) for a function f and a number a. Write down a suitable f(x) and the value of a.

(2)

(b)

Hence find the exact value of L.

(3)

(c)

Find the coordinates of the point on the graph of y = √x at which the gradient of the tangent is ¼.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 f(x) = √x	✓ 1
A1 a = 9	✓ 1
(Accept equivalent pairs, e.g. f(x) = √(9 + x), a = 0.)	—
Part (b)
M1 differentiates: f′(x) = ½x^−1/2	✓ 1
M1 substitutes x = 9: ½ × 1/3	✓ 1
A1 L = 1/6	✓ 1
(Follow through from part 1: for f(x) = √(9 + x), award M1 for f′(x) = ½(9 + x)^−1/2 and M1 for substituting x = 0.)	—
Part (c)
M1 sets ½x^−1/2 = ¼	✓ 1
A1 (4, 2)	✓ 1

Exam tip: A limit of the form [f(a+h) − f(a)]/h is the gradient of f at x = a. Recognising it lets you use differentiation rules instead of algebraic manipulation.

+ Add to test✓ In your test166IBMA2S-052-08-075.7 The second derivative and graphs Paper 1Medium6 marks

No calculator

After an injection, the concentration of a drug in a patient's blood is modelled by C(t) = pt − qt² mg L⁻¹, where t is the time in hours after the injection and p, q are positive constants. Immediately after the injection the concentration is increasing at 12 mg L⁻¹ h⁻¹, and the concentration reaches its maximum when t = 4.

(a)

Find the value of p and the value of q.

(4)

(b)

Find C′(6) and interpret your answer in context.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 C′(t) = p − 2qt	✓ 1
A1 C′(0) = 12 gives p = 12	✓ 1
M1 sets C′(4) = 0: 12 − 8q = 0	✓ 1
A1 q = 1.5	✓ 1
Part (b)
A1 C′(6) = 12 − 18 = −6	✓ 1
R1 6 hours after injection the concentration is decreasing at 6 mg L⁻¹ per hour	✓ 1

Common error: Using C(0) = 12 instead of C′(0) = 12 confuses the value of the concentration with its rate of change; the word 'increasing at' signals a derivative.

+ Add to test✓ In your test167IBMA2S-052-09-055.1 Limits and the derivative Paper 1Medium7 marks

No calculator

A weather balloon is released from a platform. Its height, $h$ metres, $t$ minutes after release is such that
$h'(t) = kt^2 + 20t - 5$, for $0 \le t \le 8$,
where $k$ is a constant. The balloon is released from a height of 40 m, and after 3 minutes it is at a height of 160 m.

(a)

Show that $h(t) = \frac{k}{3}t^3 + 10t^2 - 5t + 40$.

(3)

(b)

Find the value of $k$.

(2)

(c)

Find the height of the balloon 6 minutes after release.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate $h'(t)$	✓ 1
A1 $\frac{k}{3}t^3 + 10t^2 - 5t + c$	✓ 1
A1 use $h(0) = 40$ to obtain $c = 40$	✓ 1
Part (b)
M1 substitute $t = 3$: $9k + 90 - 15 + 40 = 160$	✓ 1
A1 $k = 5$	✓ 1
Part (c)
M1 substitute $t = 6$: $\frac{5}{3}(216) + 360 - 30 + 40$	✓ 1
A1 730 m	✓ 1

Exam tip: An unknown constant inside the derivative stays as a letter when you integrate. Treat $k$ like any other coefficient: $\int kt^2\,dt = \frac{k}{3}t^3$.

+ Add to test✓ In your test168IBMA2S-052-09-075.2 Increasing and decreasing functions Paper 1Medium7 marks

No calculator

A section of a roller-coaster track is modelled by the curve $y = f(x)$, $0 \le x \le 3$, where $x$ and $y$ are in metres. It is given that
$f''(x) = 6x - 2$.
The track is horizontal at the point where $x = 1$, and the height of the track there is 4 m.

(a)

Find an expression for $f'(x)$.

(3)

(b)

Find an expression for $f(x)$.

(3)

(c)

Find the height of the track when $x = 2$.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate $f''(x)$	✓ 1
A1 $3x^2 - 2x + c$	✓ 1
A1 use $f'(x) = 0$ when $x = 1$, giving $3 - 2 + c = 0$, so $f'(x) = 3x^2 - 2x - 1$	✓ 1
Part (b)
M1 attempt to integrate $f'(x)$	✓ 1
A1 $x^3 - x^2 - x + d$	✓ 1
A1 use $f(x) = 4$ when $x = 1$, giving $1 - 1 - 1 + d = 4$, so $f(x) = x^3 - x^2 - x + 5$	✓ 1
Part (c)
A1 $8 - 4 - 2 + 5 = 7$ m	✓ 1

Common error: Each integration needs its own constant. Here "horizontal" means the gradient is zero at $x = 1$, and that condition fixes the first constant. The height of 4 m at $x = 1$ fixes the second. Mixing the two up gives the wrong curve.

+ Add to test✓ In your test169IBMA2S-052-09-105.3 Differentiating polynomials and integer powers Paper 1Medium7 marks

No calculator

A curve $y = f(x)$ has gradient function $f'(x) = ax + b$, where $a$ and $b$ are constants. The curve passes through the point $(0, 1)$ and has a stationary point at $(2, -7)$.

(a)

Show that $2a + b = 0$.

(2)

(b)

Find the value of $a$ and the value of $b$.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
R1 at a stationary point the gradient is zero, so $f'(2) = 0$	✓ 1
A1 $2a + b = 0$	✓ 1
Part (b)
M1 integrate to get $f(x) = \frac{a}{2}x^2 + bx + c$	✓ 1
A1 $c = 1$ from the point $(0, 1)$	✓ 1
M1 use $f(2) = -7$: $2a + 2b + 1 = -7$	✓ 1
M1 solve simultaneously with $2a + b = 0$	✓ 1
A1 $a = 4$, $b = -8$	✓ 1

Common error: The stationary point gives two conditions, not one: $f'(2) = 0$ and $f(2) = -7$. Using only one of them leaves too few equations to find both constants.

+ Add to test✓ In your test170IBMA2S-052-01-055.10 Indefinite integrals and reverse chain rule Paper 1Medium7 marks

No calculator

Let $f(x) = e^{2x}$. It is given that $\displaystyle\int_0^k f(x)\,dx = 4$, where $k > 0$.

Diagram not to scale

(a)

Find the exact value of $k$.

(4)

(b)

The region $R$ is bounded by the graph of $f$, the $y$-axis and the line $y = 9$. Find the exact area of $R$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts to integrate $e^{2x}$	✓ 1
A1 $\left[\frac{1}{2}e^{2x}\right]_0^k = \frac{1}{2}(e^{2k} - 1)$	✓ 1
M1 sets $\frac{1}{2}(e^{2k} - 1) = 4$, giving $e^{2k} = 9$	✓ 1
A1 $k = \frac{1}{2}\ln 9 = \ln 3$	✓ 1
Part (b)
M1 recognises that the curve meets $y = 9$ at $x = \ln 3$ and sets up $\displaystyle\int_0^{\ln 3} (9 - e^{2x})\,dx$, or writes a rectangle area minus the area under the curve	✓ 1
A1 correct intermediate value: either $\left[9x - \frac{1}{2}e^{2x}\right]_0^{\ln 3} = 9\ln 3 - \frac{9}{2} + \frac{1}{2}$, or rectangle area $9\ln 3$ with the area under the curve equal to 4	✓ 1
A1 area of $R = 9\ln 3 - 4$	✓ 1

Exam tip: Look for ways to reuse earlier results. The first part already gives the area under the curve from 0 to $\ln 3$ as 4, so the region is simply a $\ln 3$ by 9 rectangle minus 4.

+ Add to test✓ In your test171IBMA2S-052-01-065.11 Definite integrals and areas Paper 1Medium12 marks

No calculator

An irrigation channel has a parabolic cross-section. Using axes in metres, with the $y$-axis vertical and the lowest point of the channel at the origin, the channel wall is modelled by $y = \frac{1}{4}x^2$ for $-4 \le x \le 4$. When the channel is full, the water surface lies along the line $y = 4$.

Diagram not to scale

(a)

Show that the cross-sectional area of water in a full channel is $\frac{64}{3}$ m².

(3)

(b)

A straight section of the channel is 60 m long. Find the volume of water in this section when it is full.

(2)

(c)

The water is at depth $h$ metres, where $0 < h \le 4$. Show that the cross-sectional area of water is $\frac{8}{3}h^{3/2}$ m².

(4)

(d)

Find the exact depth of water when the channel holds half of its full cross-sectional area.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 forms a correct integral for the area, e.g. $\displaystyle\int_{-4}^{4} \left(4 - \frac{1}{4}x^2\right)dx$, or $2\displaystyle\int_{0}^{4} \left(4 - \frac{1}{4}x^2\right)dx$, or $\displaystyle\int_{0}^{4} 4\sqrt{y}\,dy$	✓ 1
A1 correct antiderivative, e.g. $\left[4x - \frac{x^3}{12}\right]$ or $\left[\frac{8}{3}y^{3/2}\right]$	✓ 1
A1 correct substitution of limits leading to $\frac{64}{3}$, e.g. $2\left(16 - \frac{16}{3}\right) = \frac{64}{3}$ or $\frac{8}{3}\times 8 = \frac{64}{3}$ (AG)	✓ 1
Part (b)
M1 multiplies the cross-sectional area by the length: $\frac{64}{3} \times 60$	✓ 1
A1 1280 m³	✓ 1
Part (c)
M1 finds the limits $x = \pm 2\sqrt{h}$ from $\frac{1}{4}x^2 = h$, or uses the width $4\sqrt{y}$ at height $y$	✓ 1
M1 forms a correct integral, e.g. $\displaystyle\int_{-2\sqrt{h}}^{2\sqrt{h}} \left(h - \frac{x^2}{4}\right)dx$ or $\displaystyle\int_{0}^{h} 4\sqrt{y}\,dy$	✓ 1
A1 correct antiderivative, e.g. $\left[hx - \frac{x^3}{12}\right]$ or $\left[\frac{8}{3}y^{3/2}\right]$	✓ 1
A1 correct substitution of limits leading to the result, e.g. $2\left(2h^{3/2} - \frac{8h^{3/2}}{12}\right) = \frac{8}{3}h^{3/2}$ (AG)	✓ 1
Part (d)
M1 sets $\frac{8}{3}h^{3/2} = \frac{32}{3}$	✓ 1
A1 $h^{3/2} = 4$	✓ 1
A1 $h = 4^{2/3} = 2\sqrt[3]{2}$ m (accept $2^{4/3}$ or $\sqrt[3]{16}$)	✓ 1

Common error: Half the area does not occur at half the depth. The channel is wider near the top, so the area grows like $h^{3/2}$, not like $h$.

+ Add to test✓ In your test172IBMA2S-052-01-085.5 Introduction to integration Paper 1Medium7 marks

No calculator

The functions $f$ and $g$ are continuous. It is given that $\displaystyle\int_1^4 f(x)\,dx = 7$ and $\displaystyle\int_1^4 g(x)\,dx = -2$.

(a)

Find $\displaystyle\int_1^4 \big(3f(x) - 2g(x) + 1\big)\,dx$.

(2)

(b)

Find $\displaystyle\int_4^1 \big(f(x) - 2\big)\,dx$.

(2)

(c)

Find the value of $k$ such that $\displaystyle\int_1^4 \big(f(x) + kx\big)\,dx = 0$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses linearity: $3(7) - 2(-2) + \displaystyle\int_1^4 1\,dx$	✓ 1
A1 $21 + 4 + 3 = 28$	✓ 1
Part (b)
M1 reverses the limits with a change of sign: $-\displaystyle\int_1^4 (f(x) - 2)\,dx = -(7 - 6)$	✓ 1
A1 $-1$	✓ 1
Part (c)
M1 evaluates $\displaystyle\int_1^4 kx\,dx = k\left[\frac{x^2}{2}\right]_1^4 = \frac{15k}{2}$	✓ 1
A1 $7 + \frac{15k}{2} = 0$	✓ 1
A1 $k = -\frac{14}{15}$	✓ 1

Common error: The constant 1 does not integrate to 1. It integrates to the length of the interval, here $4 - 1 = 3$.

+ Add to test✓ In your test173IBMA2S-052-01-105.10 Indefinite integrals and reverse chain rule Paper 1Medium7 marks

No calculator

Consider the function $f(x) = \sin 2x$ for $0 \le x \le \pi$. The region between the graph of $f$ and the $x$-axis from $x = 0$ to $x = k$, where $\frac{\pi}{2} < k < \pi$, has a total area of $\frac{3}{2}$.

(a)

Show that $\displaystyle\int_0^{\pi/2} \sin 2x\,dx = 1$.

(2)

(b)

Find an expression for $\displaystyle\int_{\pi/2}^{k} \sin 2x\,dx$ in terms of $k$.

(2)

(c)

Find the exact value of $k$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 integrates to $\left[-\frac{1}{2}\cos 2x\right]_0^{\pi/2}$	✓ 1
A1 $\frac{1}{2} + \frac{1}{2} = 1$ (AG)	✓ 1
Part (b)
M1 substitutes the limits into $-\frac{1}{2}\cos 2x$: $-\frac{1}{2}\cos 2k + \frac{1}{2}\cos\pi$	✓ 1
A1 $-\frac{1}{2}(1 + \cos 2k)$	✓ 1
Part (c)
M1 recognises that the region for $\frac{\pi}{2} < x < k$ is below the axis, and writes total area $= 1 + \frac{1}{2}(1 + \cos 2k) = \frac{3}{2}$	✓ 1
A1 $\cos 2k = 0$	✓ 1
A1 since $\pi < 2k < 2\pi$, $2k = \frac{3\pi}{2}$, so $k = \frac{3\pi}{4}$	✓ 1

Common error: Setting $\displaystyle\int_0^k \sin 2x\,dx = \frac{3}{2}$ is wrong: the part below the axis would be subtracted. Add the magnitude of the negative piece instead.

+ Add to test✓ In your test174IBMA2S-052-02-035.9 Kinematics Paper 1Medium7 marks

No calculator

The number of active users of a new fitness app, $N$ thousand, $t$ months after launch is modelled by
$N(t)=-t^3+9t^2+48t+200$, for $0\le t\le 12$.

(a)

Find the value of $t$ at which the number of active users is a maximum, and find this maximum number of users.

(4)

(b)

Find the time at which the number of users is increasing most rapidly, and find this rate of increase.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $N'(t)=-3t^2+18t+48$	✓ 1
M1 sets $N'(t)=0$: $t^2-6t-16=0$, $(t-8)(t+2)=0$	✓ 1
A1 $t=8$ (rejecting $t=-2$)	✓ 1
A1 $N(8)=648$, i.e. 648 000 users	✓ 1
Part (b)
M1 sets $N''(t)=-6t+18=0$	✓ 1
A1 $t=3$ months	✓ 1
A1 $N'(3)=-27+54+48=75$ thousand users per month	✓ 1

Exam tip: 'Increasing most rapidly' means the derivative $N'$ is a maximum, so you need $N''(t)=0$ — the point of inflexion of $N$, not its stationary point.

+ Add to test✓ In your test175IBMA2S-052-02-055.7 The second derivative and graphs Paper 1Medium6 marks

No calculator

Let $f(x)=x^4-4x^3$, for $x\in\mathbb{R}$.

(a)

Find the coordinates of the stationary points of the graph of $f$.

(3)

(b)

Show that the stationary point at the origin is a point of inflexion and not a local maximum or minimum.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $f'(x)=4x^3-12x^2$	✓ 1
M1 sets $f'(x)=0$ and factorises: $4x^2(x-3)=0$, so $x=0$ or $x=3$	✓ 1
A1 $(0,0)$ and $(3,-27)$	✓ 1
Part (b)
M1 valid method: either finds $f''(x)=12x^2-24x=12x(x-2)$ with $f''(0)=0$ and examines the sign of $f''$ either side of $x=0$, or examines the sign of $f'(x)=4x^2(x-3)$ either side of $x=0$	✓ 1
A1 correct signs, e.g. at $x=-1$, $f''=36>0$ and at $x=1$, $f''=-12<0$; or $f'(x)<0$ just either side of $x=0$ because $4x^2>0$ and $x-3<0$	✓ 1
R1 conclusion: $f''$ changes sign at $x=0$ (or $f'$ does not change sign), so the origin is a stationary point of inflexion and not a local maximum or minimum	✓ 1

Common error: Concluding 'inflexion' simply because $f''(0)=0$. A zero second derivative is inconclusive. You must show that $f''$ changes sign, or that $f'$ keeps the same sign, on either side of the point.

+ Add to test✓ In your test176IBMA2S-052-02-075.6 Differentiation rules Paper 1Medium7 marks

No calculator

Let $f(x)=2\sin x+\cos 2x$, for $0\le x\le \pi$.

(a)

Show that $f'(x)=2\cos x\,(1-2\sin x)$.

(2)

(b)

Find the $x$-coordinates of the stationary points of $f$.

(3)

(c)

Find the greatest value of $f(x)$ on the given domain.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiates to obtain $2\cos x-2\sin 2x$	✓ 1
A1 uses $\sin 2x=2\sin x\cos x$ to obtain $2\cos x-4\sin x\cos x=2\cos x(1-2\sin x)$	✓ 1
Part (b)
M1 sets $\cos x=0$ or $\sin x=\frac{1}{2}$	✓ 1
A1 $x=\frac{\pi}{2}$	✓ 1
A1 $x=\frac{\pi}{6}$ and $x=\frac{5\pi}{6}$	✓ 1
Part (c)
M1 evaluates $f$ at stationary points (and endpoints), e.g. $f(\frac{\pi}{6})=1+\frac{1}{2}$, $f(\frac{\pi}{2})=1$, $f(0)=f(\pi)=1$	✓ 1
A1 greatest value $\frac{3}{2}$	✓ 1

Common error: Dividing both sides of $2\cos x(1-2\sin x)=0$ by $\cos x$ loses the solution $x=\frac{\pi}{2}$. Set each factor equal to zero instead.

+ Add to test✓ In your test177IBMA2S-052-02-085.8 Stationary points, inflexion and optimization Paper 1Medium14 marks

No calculator

An offshore wind turbine T is 3 km from a straight coastline. The point on the coastline nearest to T is P, so PT = 3 km. A substation S is on the coastline, 8 km from P.
A power cable is laid along the coastline from S to a point Q, where Q lies between P and S and PQ = $x$ km, $0\le x\le 8$. The cable then runs in a straight line underwater from Q to T.
Laying cable along the coast costs 30 thousand dollars per km and laying cable underwater costs 50 thousand dollars per km. The total cost is $C$ thousand dollars.

Diagram not to scale

(a)

Show that $C(x)=30(8-x)+50\sqrt{9+x^2}$.

(2)

(b)

Find $C'(x)$.

(2)

(c)

Find the value of $x$ that minimises the cost.

(4)

(d)

Justify that this value of $x$ gives a minimum.

(2)

(e)

Find the minimum cost.

(2)

(f)

Without a calculator, show that laying the whole cable underwater directly from S to T would cost more than the minimum cost.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses Pythagoras in triangle PQT: QT $=\sqrt{9+x^2}$	✓ 1
A1 land length $8-x$; multiplies each length by its cost and adds to obtain the given expression	✓ 1
Part (b)
M1 attempts the chain rule on $\sqrt{9+x^2}$	✓ 1
A1 $C'(x)=-30+\frac{50x}{\sqrt{9+x^2}}$	✓ 1
Part (c)
M1 sets $C'(x)=0$: $50x=30\sqrt{9+x^2}$	✓ 1
M1 squares and rearranges: $25x^2=9(9+x^2)$	✓ 1
A1 $16x^2=81$	✓ 1
A1 $x=\frac{9}{4}$ (rejecting the negative root)	✓ 1
Part (d)
M1 tests the sign of $C'$ either side, e.g. $C'(0)=-30<0$ and $C'(4)=-30+\frac{200}{5}=10>0$	✓ 1
R1 $C'$ changes from negative to positive, so minimum	✓ 1
Part (e)
M1 substitutes $x=\frac{9}{4}$: $30\times\frac{23}{4}+50\sqrt{\frac{225}{16}}$	✓ 1
A1 $172.5+187.5=360$ thousand dollars	✓ 1
Part (f)
M1 direct cost $=50\sqrt{64+9}=50\sqrt{73}$	✓ 1
R1 $\sqrt{73}>\sqrt{64}=8$, so cost $>400>360$	✓ 1

Exam tip: When squaring an equation such as $50x=30\sqrt{9+x^2}$, check the root in the original equation and the context: only $x>0$ satisfies it here.

+ Add to test✓ In your test178IBMA2S-052-02-125.8 Stationary points, inflexion and optimization Paper 1Medium10 marks

No calculator

A conical lampshade is made from thin metal so that its slant height is 12 cm. The cone has base radius $r$ cm and vertical height $h$ cm, and its volume is $V$ cm³.

Diagram not to scale

(a)

Show that $V=\frac{\pi}{3}\left(144h-h^3\right)$.

(2)

(b)

Find $\frac{dV}{dh}$.

(1)

(c)

Find the exact value of $h$ that maximises the volume.

(3)

(d)

Justify that this value of $h$ gives a maximum.

(2)

(e)

Find the exact maximum volume.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses Pythagoras: $r^2+h^2=144$	✓ 1
A1 substitutes $r^2=144-h^2$ into $V=\frac{1}{3}\pi r^2h$ to obtain the given result	✓ 1
Part (b)
A1 $\frac{dV}{dh}=\frac{\pi}{3}(144-3h^2)=\pi(48-h^2)$	✓ 1
Part (c)
M1 sets $\frac{dV}{dh}=0$	✓ 1
A1 $h^2=48$	✓ 1
A1 $h=4\sqrt{3}$ (rejecting the negative root)	✓ 1
Part (d)
M1 $\frac{d^2V}{dh^2}=-2\pi h$	✓ 1
R1 at $h=4\sqrt{3}$, $\frac{d^2V}{dh^2}=-8\sqrt{3}\pi<0$, so maximum	✓ 1
Part (e)
M1 uses $r^2=144-48=96$ in $V=\frac{1}{3}\pi r^2h$	✓ 1
A1 $V=\frac{1}{3}\pi(96)(4\sqrt{3})=128\sqrt{3}\,\pi$ cm³	✓ 1

Exam tip: Choose the variable to eliminate wisely: replacing $r^2$ (not $h$) avoids square roots and gives a polynomial in $h$ that is easy to differentiate.

+ Add to test✓ In your test179IBMA2S-052-07-035.4 Tangents and normals Paper 1Medium7 marks

No calculator

The edge of a smartphone case is modelled by the curve $y = x^3 + kx$, where $k$ is a constant. The normal to the curve at the point where $x = 1$ is parallel to the line $x + 5y = 2$.

(a)

Find the value of $k$.

(4)

(b)

Find the $x$-coordinate of the point where this normal meets the $x$-axis.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{dy}{dx} = 3x^2 + k$	✓ 1
A1 gradient of the line is $-\frac{1}{5}$	✓ 1
M1 use of perpendicular gradients: tangent gradient $= 5$, so $3 + k = 5$	✓ 1
A1 $k = 2$	✓ 1
Part (b)
A1 point on curve is $(1, 3)$	✓ 1
M1 normal $y - 3 = -\frac{1}{5}(x - 1)$ with $y = 0$ substituted	✓ 1
A1 $x = 16$	✓ 1

Common error: The line given is parallel to the normal, not the tangent. Setting $3 + k = -\frac{1}{5}$ uses the wrong gradient.

+ Add to test✓ In your test180IBMA2S-052-07-065.4 Tangents and normals Paper 1Medium15 marks

No calculator

In a simplified model, the path of a spacecraft passing a planet is given by $y = \sqrt{x^2 + 9}$, where $x$ and $y$ are measured in thousands of kilometres.

(a)

Show that $\frac{dy}{dx} = \frac{x}{\sqrt{x^2 + 9}}$.

(2)

(b)

Find the equation of the tangent to the path at the point where $x = 4$.

(3)

(c)

Find the equation of the normal to the path at the point where $x = 4$.

(2)

(d)

Find the coordinates of the point on the path at which the tangent is parallel to the line $3x - 5y = 7$.

(4)

(e)

The tangent and the normal to the path at the point where $x = 4$ form a triangle with the $x$-axis. Find the area of this triangle.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the chain rule on $(x^2 + 9)^{\frac{1}{2}}$	✓ 1
A1 $\frac{1}{2}(x^2 + 9)^{-\frac{1}{2}} \times 2x$ leading to the given answer	✓ 1
Part (b)
A1 $y = 5$ when $x = 4$	✓ 1
M1 gradient $\frac{4}{5}$ substituted into $y - 5 = \frac{4}{5}(x - 4)$	✓ 1
A1 $y = \frac{4}{5}x + \frac{9}{5}$	✓ 1
Part (c)
M1 normal gradient $-\frac{5}{4}$ used in $y - 5 = -\frac{5}{4}(x - 4)$	✓ 1
A1 $y = -\frac{5}{4}x + 10$	✓ 1
Part (d)
M1 setting $\frac{x}{\sqrt{x^2 + 9}} = \frac{3}{5}$	✓ 1
A1 $25x^2 = 9x^2 + 81$, so $x^2 = \frac{81}{16}$	✓ 1
R1 rejecting $x = -\frac{9}{4}$ because the gradient must be positive, so $x > 0$	✓ 1
A1 $\left(\frac{9}{4}, \frac{15}{4}\right)$	✓ 1
Part (e)
A1 tangent meets the $x$-axis at $x = -\frac{9}{4}$	✓ 1
A1 normal meets the $x$-axis at $x = 8$	✓ 1
M1 area $= \frac{1}{2} \times \frac{41}{4} \times 5$, using base $8 - (-\frac{9}{4})$ and height 5 from the vertex $(4, 5)$	✓ 1
A1 $\frac{205}{8}$	✓ 1

Exam tip: Squaring an equation can introduce extra solutions. Check each root in the original equation, using the sign of the gradient, before giving a final answer.

+ Add to test✓ In your test181IBMA2S-052-06-025.2 Increasing and decreasing functions Paper 1Medium5 marks

No calculator

The function f is defined by f(x) = 2x³ − 3x² + kx, for x ∈ ℝ, where k is a constant.

(a)

Find f′(x).

(1)

(b)

Find the set of values of k for which f is increasing for all x ∈ ℝ.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 f′(x) = 6x² − 6x + k	✓ 1
Part (b)
R1 States that f is increasing for all x requires f′(x) ≥ 0 for all x	✓ 1
M1 Uses the condition that the discriminant of f′(x) is ≤ 0, i.e. b² − 4ac ≤ 0 for 6x² − 6x + k. Alternatively, finds the minimum of f′ at x = 1/2 (vertex or completing the square) and requires it to be ≥ 0	✓ 1
A1 (−6)² − 4(6)(k) ≤ 0, i.e. 36 − 24k ≤ 0. Alternatively, f′(1/2) = k − 3/2 ≥ 0	✓ 1
A1 k ≥ 3/2	✓ 1
Note: k = 3/2 is included. For this value f′(x) = 6(x − 1/2)², which is zero only at the single point x = 1/2, so f is still increasing. The answer k > 3/2 does not earn the final A1.	—

Common error: Requiring the discriminant to be greater than zero gives two roots, so f′ would change sign. For f′ never to be negative, the discriminant must be at most zero. Do not exclude the boundary value: when the discriminant is zero, f′ touches zero at one point only, and f is still increasing.

+ Add to test✓ In your test182IBMA2S-052-06-065.10 Indefinite integrals and reverse chain rule Paper 1Medium6 marks

No calculator

Let f(x) = ln(x² + 4) − x/2, for x ∈ ℝ.

(a)

Find f′(x).

(2)

(b)

Show that f′(x) = −(x − 2)² / (2(x² + 4)).

(2)

(c)

Hence justify that f is a decreasing function.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses the chain rule on ln(x² + 4)	✓ 1
A1 f′(x) = 2x/(x² + 4) − 1/2	✓ 1
Part (b)
M1 combines over the common denominator 2(x² + 4)	✓ 1
A1 numerator 4x − x² − 4 = −(x − 2)²	✓ 1
Part (c)
R1 (x − 2)² ≥ 0 and x² + 4 > 0, so f′(x) ≤ 0 for all x	✓ 1
R1 f′(x) = 0 only at the single point x = 2, so f is decreasing for all x	✓ 1

Exam tip: When a derivative is zero at an isolated point but negative everywhere else, the function is still decreasing; say so explicitly to secure the reasoning mark.

+ Add to test✓ In your test183IBMA2S-052-06-085.4 Tangents and normals Paper 1Medium7 marks

No calculator

The function f is defined by f(x) = x³ + ax² + bx, where a and b are constants. The function f is decreasing for −1 < x < 3 and increasing for x < −1 and for x > 3.

(a)

Find f′(x) in terms of a and b.

(1)

(b)

Find the value of a and the value of b.

(4)

(c)

Find the amount by which f(x) decreases as x increases from −1 to 3.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 f′(x) = 3x² + 2ax + b	✓ 1
Part (b)
R1 f′(x) = 0 at x = −1 and x = 3	✓ 1
M1 forms equations 3 − 2a + b = 0 and 27 + 6a + b = 0, or equates f′(x) with 3(x + 1)(x − 3)	✓ 1
A1 a = −3	✓ 1
A1 b = −9	✓ 1
Part (c)
M1 evaluates f at both ends, e.g. 5 and −27	✓ 1
A1 32	✓ 1

Exam tip: The endpoints of an interval where a polynomial switches from increasing to decreasing are its stationary points, so they are roots of the derivative.

+ Add to test✓ In your test184IBMA2S-052-03-065.9 Kinematics Paper 1Medium6 marks

No calculator

A camera dolly moves along a straight film-set track. Its acceleration is $a(t)=6t-4$ m s⁻² for $t\ge 0$ seconds. At $t=0$ the dolly is at its starting point and has velocity $v_0$ m s⁻¹. At $t=3$ it is 12 m from the starting point in the positive direction.

(a)

Find the value of $v_0$.

(4)

(b)

Find the minimum velocity of the dolly.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 integrates $a$: $v=3t^2-4t+c$	✓ 1
A1 $c=v_0$	✓ 1
M1 integrates again: $s=t^3-2t^2+v_0t$ and uses $s(3)=12$	✓ 1
A1 $9+3v_0=12$, so $v_0=1$	✓ 1
Part (b)
M1 $a=0$ at $t=\frac{2}{3}$	✓ 1
A1 $v=-\frac{1}{3}$ m s⁻¹	✓ 1

Exam tip: Keep the unknown $v_0$ as the constant through both integrations, and use $s(0)=0$ to remove the second constant.

+ Add to test✓ In your test185IBMA2S-052-03-085.9 Kinematics Paper 1Medium5 marks

No calculator

A lift in a mine shaft moves vertically. Its velocity, measured upwards, is $v(t)=t^2-6t+8$ m s⁻¹, for $0\le t\le 5$, where $t$ is in seconds.

(a)

Write down an expression for the acceleration of the lift.

(1)

(b)

Find the intervals of time during which the lift is speeding up. Justify your answer.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $a(t)=2t-6$	✓ 1
Part (b)
M1 recognises speeding up when $v$ and $a$ have the same sign	✓ 1
A1 $v=(t-2)(t-4)$: $v>0$ for $0\le t<2$ and $4<t\le5$; $v<0$ for $2<t<4$	✓ 1
A1 $a<0$ for $t<3$; $a>0$ for $t>3$	✓ 1
R1 so speeding up for $2<t<3$ and $4<t\le 5$	✓ 1

Common error: A positive acceleration does not always mean speeding up. If the velocity is negative, a positive acceleration slows the lift down.

+ Add to test✓ In your test186IBMA2S-052-03-105.9 Kinematics Paper 1Medium7 marks

No calculator

A test rig for an automatic sliding door moves the door along a straight track. The door's velocity is $v(t)=t^2-kt$ m s⁻¹, for $0\le t\le 6$, where $t$ is in seconds and $k$ is a positive constant. At $t=6$ the door is back at its starting position.

(a)

Find the value of $k$.

(3)

(b)

Find the total distance travelled by the door for $0\le t\le 6$.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sets $\int_0^6(t^2-kt)\,dt=0$	✓ 1
A1 $72-18k=0$	✓ 1
A1 $k=4$	✓ 1
Part (b)
M1 recognises $v=0$ at $t=4$ and splits the integral	✓ 1
A1 $\left\|\int_0^4(t^2-4t)\,dt\right\|=\frac{32}{3}$	✓ 1
A1 $\int_4^6(t^2-4t)\,dt=\frac{32}{3}$	✓ 1
A1 total $\frac{64}{3}$ m	✓ 1

Exam tip: "Back at its starting position" is a condition on displacement, so set the definite integral of velocity equal to zero.

+ Add to test✓ In your test187IBMA2S-052-03-125.9 Kinematics Paper 1Medium12 marks

No calculator

A float in a wave-energy device moves up and down a vertical guide rail. Its displacement above a fixed point is $s(t)=2\sin t+\cos 2t$ metres, for $0\le t\le\pi$, where $t$ is in seconds.

(a)

Find an expression for the velocity $v(t)$ of the float.

(2)

(b)

Find all the times in $0\le t\le\pi$ at which the float is at rest.

(4)

(c)

Find the acceleration of the float at $t=\frac{\pi}{6}$.

(2)

(d)

Find the total distance travelled by the float for $0\le t\le\pi$.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate, including the chain rule on $\cos2t$	✓ 1
A1 $v=2\cos t-2\sin2t$	✓ 1
Part (b)
M1 sets $v=0$	✓ 1
M1 uses $\sin2t=2\sin t\cos t$: $2\cos t(1-2\sin t)=0$	✓ 1
A1 $\cos t=0\Rightarrow t=\frac{\pi}{2}$	✓ 1
A1 $\sin t=\frac12\Rightarrow t=\frac{\pi}{6},\frac{5\pi}{6}$	✓ 1
Part (c)
M1 $a=-2\sin t-4\cos2t$	✓ 1
A1 $a=-1-2=-3$ m s⁻²	✓ 1
Part (d)
M1 evaluates $s$ at $0$ and at each time of rest	✓ 1
A1 $s=1,\ 1.5,\ 1,\ 1.5,\ 1$ at $t=0,\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\pi$	✓ 1
M1 adds the magnitudes of the changes in $s$	✓ 1
A1 $2$ m	✓ 1

Exam tip: Without a calculator, find distance by tracking position between turning points instead of integrating $|v|$.

+ Add to test✓ In your test188IBMA3S-5-17r5-065.4 Tangents and normals Paper 1Medium7 marks

No calculator

Part of a suspension cable is modelled by the curve y = x³ − 3x + 2, 0 ≤ x ≤ 3, where x and y are in metres. The point P on the curve has x-coordinate 2.

Diagram not to scale

(a)

Find the gradient of the curve at P.

(2)

(b)

Find the equation of the normal to the curve at P. Give your answer in the form ax + by + d = 0, where a, b and d are integers.

(3)

(c)

The normal meets the x-axis at the point Q. Find the x-coordinate of Q.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 dy/dx = 3x² − 3	✓ 1
A1 9	✓ 1
Part (b)
A1 gradient of normal = −1/9 (ECF from (a))	✓ 1
M1 use of P(2, 4) in the equation of a line, e.g. y − 4 = −(1/9)(x − 2)	✓ 1
A1 x + 9y − 38 = 0	✓ 1
Part (c)
M1 set y = 0 in the equation of the normal	✓ 1
A1 x = 38	✓ 1

Common error: Using the tangent gradient for the normal. The normal gradient is the negative reciprocal of the tangent gradient.

+ Add to test✓ In your test189IBMA3S-5-17r5-075.6 Differentiation rules Paper 1Medium6 marks

No calculator

The power P kW supplied to a coffee roaster, t minutes after it starts, is given by P(t) = 6√t + 4, 0 ≤ t ≤ 9. The energy supplied between t = 0 and t = 9 is E kW min, where E = ∫₀⁹ P(t) dt.

(a)

Find the value of E.

(4)

(b)

Hence find the mean power over the first 9 minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate, rewriting 6√t as 6t^1/2	✓ 1
A1 4t^3/2 + 4t	✓ 1
M1 substitute limits 9 and 0	✓ 1
A1 144 kW min	✓ 1
Part (b)
M1 divide their E by 9 (ECF from (a))	✓ 1
A1 16 kW	✓ 1

Exam tip: For a definite integral, show the antiderivative in square brackets with the limits, then substitute: it makes method marks easy to see.

+ Add to test✓ In your test190IBMA3S-5-17r5-085.2 Increasing and decreasing functions Paper 1Medium5 marks

No calculator

A bridge engineer models the deflection of a beam by f(x) = x³ + kx² + 12x + 5, where k is a constant.

(a)

Find f′(x).

(2)

(b)

Given that f is never decreasing, find the set of possible values of k.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate	✓ 1
A1 3x² + 2kx + 12	✓ 1
Part (b)
M1 f′(x) ≥ 0 for all x, so discriminant ≤ 0	✓ 1
A1 (2k)² − 4(3)(12) ≤ 0, that is 4k² − 144 ≤ 0	✓ 1
A1 −6 ≤ k ≤ 6	✓ 1

Common error: Using a strict inequality loses the end values. At k = ±6 the derivative is zero at a single point only, so f is still never decreasing.

+ Add to test✓ In your test191IBMA3S-5-17r5-095.2 Increasing and decreasing functions Paper 1Medium6 marks

No calculator

The profile of an access ramp on a bridge is modelled by f(x) = x³ − 6x² + 9x + 2, 0 ≤ x ≤ 4, where x and f(x) are in metres.

(a)

Find f′(x).

(2)

(b)

Show that there is exactly one point on the ramp where the gradient is −3, and find the height of the ramp at that point.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate	✓ 1
A1 3x² − 12x + 9	✓ 1
Part (b)
M1 set f′(x) = −3	✓ 1
A1 3x² − 12x + 12 = 0, that is 3(x − 2)² = 0	✓ 1
R1 repeated root, so only one value, x = 2	✓ 1
A1 f(2) = 4 m	✓ 1

Exam tip: When asked to show that there is exactly one solution, say why: a repeated root (or discriminant zero) is the reason.

+ Add to test✓ In your test192IBMA3S-5-17r5-115.3 Differentiating polynomials and integer powers Paper 1Medium6 marks

No calculator

The strength S (in arbitrary units) of a signal received from a probe, t hours after it passes behind a ridge, is modelled by S(t) = 5t/(t² + 4), t ≥ 0.

(a)

Show that S′(t) = 5(4 − t²)/(t² + 4)².

(3)

(b)

Hence find the values of t for which the signal strength is increasing.

(2)

(c)

Find the greatest signal strength.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the quotient rule	✓ 1
A1 [5(t² + 4) − 5t(2t)]/(t² + 4)²	✓ 1
A1 simplify numerator to 20 − 5t² to reach the given result (AG)	✓ 1
Part (b)
M1 S′(t) > 0 with denominator positive, so 4 − t² > 0	✓ 1
A1 0 ≤ t < 2	✓ 1
Part (c)
A1 S(2) = 10/8 = 1.25	✓ 1

Common error: Concluding from S′ = 0 alone that a point is a maximum. Check that the derivative changes from positive to negative.

+ Add to test✓ In your test193IBMA3S-5-16r7-045.6 Differentiation rules Paper 1Medium6 marks

No calculator

An astronomer models the brightness of a variable star using $f(x)=x^2\sin(3x)$, where x is in radians.

(a)

Find $f'(x)$.

(4)

(b)

Show that $f'\!\left(\frac{\pi}{6}\right)=\frac{\pi}{3}$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule	✓ 1
A1 $2x\sin(3x)$	✓ 1
M1 attempt to differentiate $\sin(3x)$ using the chain rule	✓ 1
A1 $+3x^2\cos(3x)$	✓ 1
Part (b)
M1 substitute $x=\frac{\pi}{6}$ with $\sin\frac{\pi}{2}=1$ and $\cos\frac{\pi}{2}=0$	✓ 1
A1 $2\times\frac{\pi}{6}\times 1+0=\frac{\pi}{3}$ (AG)	✓ 1

Common error: The derivative of $\sin(3x)$ is $3\cos(3x)$; forgetting the factor 3 is a frequent slip.

+ Add to test✓ In your test194IBMA3S-5-16r7-065.10 Indefinite integrals and reverse chain rule Paper 1Medium6 marks

No calculator

(a)

Find $\int \frac{3x}{x^2+4}\,dx$.

(3)

(b)

Hence find the exact value of $\int_0^2 \frac{3x}{x^2+4}\,dx$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 recognise the numerator is a multiple of the derivative of $x^2+4$	✓ 1
A1 $\frac32\ln(x^2+4)$	✓ 1
A1 $+C$	✓ 1
Part (b)
M1 substitute limits 2 and 0	✓ 1
A1 $\frac32(\ln 8-\ln 4)$	✓ 1
A1 $\frac32\ln 2$	✓ 1

Common error: Do not integrate the numerator and denominator separately; the form $\frac{f'(x)}{f(x)}$ integrates to $\ln|f(x)|$.

+ Add to test✓ In your test195IBMA3S-5-16r7-075.6 Differentiation rules Paper 1Medium6 marks

No calculator

The apparent brightness of a comet, in suitable units, t days after it is first observed is $B(t)=\frac{5t}{t^2+4}$, for $t\ge 0$.

(a)

Show that $B'(t)=\frac{5(4-t^2)}{(t^2+4)^2}$.

(3)

(b)

Find the maximum brightness, and the value of t at which it occurs.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the quotient rule	✓ 1
A1 $\frac{5(t^2+4)-5t(2t)}{(t^2+4)^2}$	✓ 1
A1 simplifying to $\frac{5(4-t^2)}{(t^2+4)^2}$ (AG)	✓ 1
Part (b)
M1 set $B'(t)=0$	✓ 1
A1 $t=2$	✓ 1
A1 $B(2)=\frac{10}{8}=1.25$	✓ 1

Exam tip: In a quotient-rule answer, look for a common factor before expanding; it often cancels a lot of work.

+ Add to test✓ In your test196IBMA3S-5-16r7-095.9 Kinematics Paper 1Medium7 marks

No calculator

A coach models an athlete's power output during a sprint test by $P(t)=t^3-6t^2+9t+2$, in hundreds of watts, where t is the time in seconds and $0\le t\le 4$.

(a)

Find $P'(t)$.

(2)

(b)

Find the values of t for which $P'(t)=0$.

(2)

(c)

Use the second derivative to show that $t=1$ gives a local maximum, and find the power output at that time.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate	✓ 1
A1 $3t^2-12t+9$	✓ 1
Part (b)
M1 factorise or solve $3t^2-12t+9=0$	✓ 1
A1 $t=1$ and $t=3$	✓ 1
Part (c)
A1 $P''(t)=6t-12$	✓ 1
R1 $P''$ at $t=1$ equals $-6<0$, so a local maximum	✓ 1
A1 $P$ at $t=1$ equals 6 hundred watts	✓ 1

Common error: State the sign of the second derivative as the reason for the nature of the point; stating only the value is not a full justification.

+ Add to test✓ In your test197IBMA3S-5-17r7-055.7 The second derivative and graphs Paper 1Medium5 marks

No calculator

A function f has derivative $f'(x)=(x+3)(x-2)^2$, for x ∈ ℝ.

(a)

Write down the values of x for which the graph of f has a horizontal tangent.

(1)

(b)

Find the set of values of x for which f is decreasing.

(2)

(c)

Explain why the graph of f has neither a local maximum nor a local minimum at x = 2.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $x=-3$ and $x=2$	✓ 1
Part (b)
M1 considers the sign of $f'(x)$: $(x-2)^2\ge 0$ for all $x$, so the sign is that of $(x+3)$	✓ 1
A1 $x<-3$	✓ 1
Part (c)
R1 $(x-2)^2$ is non-negative on both sides of $x=2$	✓ 1
R1 so $f'(x)$ does not change sign at $x=2$ (it is positive either side), hence not a turning point	✓ 1

Common error: A zero of f′ is not automatically a maximum or minimum: the sign of f′ must change.

+ Add to test✓ In your test198IBMA3S-5-17r7-085.10 Indefinite integrals and reverse chain rule Paper 1Medium7 marks

No calculator

Consider the function $g(x)=\frac{6x}{x^2+4}$, for x ≥ 0.

Shaded region

(a)

Show that $\frac{d}{dx}\left[\ln(x^2+4)\right]=\frac{2x}{x^2+4}$.

(2)

(b)

Hence find the exact area of the region enclosed by the graph of g, the x-axis and the line x = 2.

(3)

(c)

The area between the graph of g, the x-axis and the line x = k, where k > 0, equals 3 ln 5. Find the value of k.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the chain rule with inner function $x^2+4$	✓ 1
A1 $\frac{1}{x^2+4}\times 2x=\frac{2x}{x^2+4}$ (AG)	✓ 1
Part (b)
M1 writes $\int_0^2 g(x)\,dx=3\left[\ln(x^2+4)\right]_0^2$	✓ 1
M1 substitutes limits: $3(\ln 8-\ln 4)$	✓ 1
A1 $3\ln 2$	✓ 1
Part (c)
M1 $3\ln\left(\frac{k^2+4}{4}\right)=3\ln 5$, so $k^2+4=20$	✓ 1
A1 $k=4$	✓ 1

Exam tip: Use the part just proved: 6x/(x² + 4) is exactly 3 times the derivative of ln(x² + 4).

+ Add to test✓ In your test199IBMA3S-5-17r7-115.10 Indefinite integrals and reverse chain rule Paper 1Medium6 marks

No calculator

A solar installation stores energy in a battery. During the morning the rate of change of the stored energy is $E'(t)=40\sin t\cos^4 t$ kWh per hour, for 0 ≤ t ≤ $\frac{\pi}{2}$, where t is measured in hours (angles in radians).

(a)

Find $\int \sin t\cos^4 t\,dt$.

(3)

(b)

Find the total energy gained by the battery over the period 0 ≤ t ≤ $\frac{\pi}{2}$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the reverse chain rule with $\cos^5 t$	✓ 1
A1 $-\frac{1}{5}\cos^5 t$	✓ 1
A1 includes $+C$: $-\frac{1}{5}\cos^5 t+C$	✓ 1
Part (b)
M1 $40\left[-\frac{1}{5}\cos^5 t\right]_0^{\pi/2}$	✓ 1
A1 substitutes limits: $40\left(0+\frac{1}{5}\right)$	✓ 1
A1 8 kWh	✓ 1

Common error: When integrating a power of cos t times sin t, the sign changes: the derivative of cos t is −sin t.

+ Add to test✓ In your test200IBMA3S-5-17r9-065.1 Limits and the derivative Paper 1Medium7 marks

No calculator

A data centre operates $x$ identical server racks in one hall, where $x>0$. The running cost per hour is modelled by $C(x)=\dfrac{x}{4}+\dfrac{1600}{x}+30$, in euros (€).

(a)

Find $C'(x)$.

(2)

(b)

Find the number of racks that minimises the hourly cost.

(2)

(c)

Use $C''(x)$ to show that this gives a minimum.

(2)

(d)

Find the minimum hourly cost, in euros.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiates, writing $\frac{1600}{x}=1600x^{-1}$	✓ 1
A1 $C'(x)=\frac14-\frac{1600}{x^2}$	✓ 1
Part (b)
M1 sets $C'(x)=0$, so $x^2=6400$	✓ 1
A1 $x=80$ (rejecting $x=-80$)	✓ 1
Part (c)
A1 $C''(x)=\frac{3200}{x^3}$	✓ 1
R1 $C''(80)>0$, so a minimum	✓ 1
Part (d)
A1 $C(80)=20+20+30=70$, so €70	✓ 1

Common error: Reject any root that lies outside the stated domain (here x > 0) before substituting.

+ Add to test✓ In your test201IBMA3S-5-17r9-095.4 Tangents and normals Paper 1Medium8 marks

No calculator

The boundary of a burial mound, seen in cross-section, is modelled by the curve $y=x^3-3x^2+4$, where $x$ and $y$ are in metres.

(a)

Find the two values of $x$ at which the tangent to the curve is parallel to the line $y=9x$.

(3)

(b)

Find the equation of the tangent at the point where $x$ is positive, in the form $y=mx+c$.

(2)

(c)

The normal to the curve at the point where $x=-1$ meets the line $y=-1$ at the point Q. Find the $x$-coordinate of Q.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{dy}{dx}=3x^2-6x$	✓ 1
M1 $3x^2-6x=9$, so $x^2-2x-3=0$	✓ 1
A1 $x=3$ and $x=-1$	✓ 1
Part (b)
A1 point $(3,4)$	✓ 1
A1 $y=9x-23$	✓ 1
Part (c)
M1 point $(-1,0)$ and normal gradient $-\frac19$	✓ 1
A1 normal $y=-\frac19(x+1)$	✓ 1
A1 $-1=-\frac19(x+1)$ gives $x=8$	✓ 1

Common error: Substitute x into the original curve, not the derivative, to find the y-coordinate of the point of contact.

+ Add to test✓ In your test202IBMA3S-5-17r9-105.6 Differentiation rules Paper 1Medium10 marks

No calculator

A ground-penetrating radar signal has relative strength $f(x)=\dfrac{\ln x}{x}$ at a depth of $x$ metres, where $x>0$.

(a)

Use the quotient rule to show that $f'(x)=\dfrac{1-\ln x}{x^2}$.

(3)

(b)

Find the exact coordinates of the stationary point of the graph of $f$.

(2)

(c)

Show that $f''(x)=\dfrac{2\ln x-3}{x^3}$.

(3)

(d)

Hence find the value of $f''(e)$ in exact form and state the nature of the stationary point.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 quotient rule with $u=\ln x$, $v=x$	✓ 1
A1 $f'(x)=\frac{\frac1x\cdot x-\ln x}{x^2}$	✓ 1
A1 $=\frac{1-\ln x}{x^2}$ (AG)	✓ 1
Part (b)
M1 sets $1-\ln x=0$, so $x=e$	✓ 1
A1 $f(e)=\frac1e$, coordinates $\left(e,\frac1e\right)$	✓ 1
Part (c)
M1 quotient rule on $f'(x)$ with $u=1-\ln x$, $v=x^2$	✓ 1
A1 $f''(x)=\frac{-\frac1x\cdot x^2-2x(1-\ln x)}{x^4}$	✓ 1
A1 $=\frac{-3x+2x\ln x}{x^4}=\frac{2\ln x-3}{x^3}$ (AG)	✓ 1
Part (d)
A1 $f''(e)=\frac{2-3}{e^3}=-\frac{1}{e^3}$ ($\approx-0.0498$)	✓ 1
R1 $f''(e)<0$, so a maximum	✓ 1

Exam tip: Simplify numerators fully before using a second derivative: cancelling a common factor of x here turns a long fraction into a short one.

+ Add to test✓ In your test203IBMA3S-5-17r10-065.4 Tangents and normals Paper 1Medium7 marks

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A cable on a space probe lies along the curve $y=\sqrt{4x+1}$ for $x\ge 0$, with $x$ and $y$ in metres.

(a)

Use the chain rule to find $\frac{dy}{dx}$, and hence its value at $x=2$.

(2)

(b)

Find the equation of the normal to the curve at $x=2$.

(3)

(c)

The normal meets the axes at points $A$ and $B$. Find the area of triangle $OAB$, where $O$ is the origin.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$\frac{dy}{dx}=\frac12(4x+1)^{-1/2}\times 4=2(4x+1)^{-1/2}$	✓ 1
at $x=2$: $\frac{2}{3}$	✓ 1
Part (b)
$y=3$ at $x=2$	✓ 1
gradient of normal $=-\frac32$	✓ 1
$y=-\frac32x+6$	✓ 1
Part (c)
intercepts $x=4$ and $y=6$	✓ 1
area $=\frac12\times4\times6=12$ m$^2$	✓ 1

Common error: The normal's gradient is the negative reciprocal of the tangent's gradient, not just its negative.

+ Add to test✓ In your test204IBMA3S-5-17r10-075.8 Stationary points, inflexion and optimization Paper 1Medium7 marks

No calculator

An open-topped cylindrical canister stores roasted beans. It has radius $r$ decimetres, height $h$ decimetres and volume $216\pi$ dm$^3$. The metal used is to be minimised.

Diagram not to scale

(a)

Show that the surface area of the canister is $A=\pi r^2+\frac{432\pi}{r}$.

(2)

(b)

Find the value of $r$ that minimises $A$.

(3)

(c)

Justify that this value gives a minimum, and find the minimum surface area in exact form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$h=\frac{216}{r^2}$ from the volume	✓ 1
$A=\pi r^2+2\pi rh=\pi r^2+\frac{432\pi}{r}$	✓ 1
Part (b)
$A'=2\pi r-\frac{432\pi}{r^2}$	✓ 1
set $A'=0$ so $r^3=216$	✓ 1
$r=6$	✓ 1
Part (c)
$A''=2\pi+\frac{864\pi}{r^3}$ which is positive at $r=6$ (equals $6\pi$)	✓ 1
$A=36\pi+72\pi=108\pi$ dm$^2$ (≈ 339.3)	✓ 1

Exam tip: Express the quantity in one variable before differentiating, and justify the minimum with a second derivative or a sign change.

+ Add to test✓ In your test205IBMA3S-5-17r10-095.6 Differentiation rules Paper 1Medium5 marks

No calculator

A bridge cable carries a load of $w(x)=6x-4$ kN per metre at position $x$ metres from a tower. The total load on the section from $x=1$ to $x=k$, where $k>1$, is $16$ kN.

(a)

Find the value of $k$.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
integrate: $3x^2-4x$	✓ 1
substitute limits: $(3k^2-4k)-(3-4)$	✓ 1
$3k^2-4k+1=16$ so $3k^2-4k-15=0$	✓ 1
attempt to solve, e.g. $(3k+5)(k-3)=0$	✓ 1
$k=3$ (reject $-\frac53$ since $k>1$)	✓ 1

Exam tip: State why you reject a root when the question gives a condition such as $k>1$.

+ Add to test✓ In your test206IBMA3S-5-17r10-105.2 Increasing and decreasing functions Paper 1Medium6 marks

No calculator

The profile of a decorative roadway is the curve $y=x^3+ax^2+bx+c$, where $a$, $b$ and $c$ are constants. It has a local maximum at the point $(1,6)$ and a local minimum at $x=3$.

(a)

Find the values of $a$ and $b$.

(4)

(b)

Find the value of $c$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$y'=3x^2+2ax+b$	✓ 1
$y'=0$ at $x=1$ and at $x=3$ giving $3+2a+b=0$ and $27+6a+b=0$	✓ 1
$a=-6$	✓ 1
$b=9$	✓ 1
Part (b)
$1-6+9+c=6$	✓ 1
$c=2$	✓ 1

Common error: Use both pieces of information: the stationary condition on the derivative, and the point lying on the original curve.

+ Add to test✓ In your test207IBMA3S-5-17r10-115.5 Introduction to integration Paper 1Medium6 marks

No calculator

An antenna on a space probe is aligned along the tangent to the curve $y=\ln x$, for $x>0$, at the point $P(a,\ln a)$.

(a)

Show that the tangent at $P$ has equation $y=\frac{x}{a}+\ln a-1$.

(3)

(b)

The tangent passes through the origin. Find the exact value of $a$.

(2)

(c)

Write down the equation of this tangent.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$\frac{dy}{dx}=\frac1x$ so gradient $=\frac1a$	✓ 1
$y-\ln a=\frac1a(x-a)$	✓ 1
rearranged to $y=\frac xa+\ln a-1$	✓ 1
Part (b)
$0=\ln a-1$	✓ 1
$a=e$ (≈ 2.718)	✓ 1
Part (c)
$y=\frac{x}{e}$	✓ 1

Exam tip: If a line passes through the origin, substitute $(0,0)$ to get an equation for the unknown.

+ Add to test✓ In your test208IBMA3S-5-16r12-045.10 Indefinite integrals and reverse chain rule Paper 1Medium5 marks

No calculator

Find the exact value of $\int_0^{\sqrt{\pi}} x\sin(x^2)\,dx$.

(a)

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at reverse chain rule or substitution $u=x^2$	✓ 1
A1 $-\frac{1}{2}\cos(x^2)$	✓ 1
M1 substitutes both limits into their integral	✓ 1
A1 $-\frac{1}{2}\cos\pi+\frac{1}{2}\cos 0$	✓ 1
A1 $=1$	✓ 1

Exam tip: When the inner function has derivative $2x$, the factor $x$ in the integrand supplies half of it, which is why $\frac{1}{2}$ appears.

+ Add to test✓ In your test209IBMA3S-5-16r12-055.10 Indefinite integrals and reverse chain rule Paper 1Medium6 marks

No calculator

Let $f(x)=\dfrac{4x-2}{x^2-x+5}$.

(a)

Find $\int f(x)\,dx$.

(3)

(b)

Hence find the exact value of $\int_1^3 f(x)\,dx$, giving your answer in the form $a\ln b$ where $a, b \in \mathbb{Q}$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 recognises numerator as a multiple of the derivative of denominator, e.g. $4x-2=2(2x-1)$	✓ 1
A1 $2\ln(x^2-x+5)$	✓ 1
A1 $+c$ (denominator is always positive, so no modulus needed; accept $2\ln\|x^2-x+5\|+c$)	✓ 1
Part (b)
M1 substitutes limits 3 and 1 into their answer to (a)	✓ 1
A1 $2\ln 11-2\ln 5$	✓ 1
A1 $2\ln\frac{11}{5}$	✓ 1

Common error: Check the sign of the denominator before dropping a modulus in $\ln|g(x)|$. Here $x^2-x+5$ has no real roots, so it is always positive.

+ Add to test✓ In your test210IBMA3S-5-16r12-085.4 Tangents and normals Paper 1Medium7 marks

No calculator

A decorative steel panel on a bridge is bounded by the curves $y=6x-x^2$ and $y=x^2$, where x and y are in metres.

Shaded panel

(a)

Find the x-coordinates of the two points where the curves meet.

(3)

(b)

Find the area of the panel.

(3)

(c)

The panel is coated at a cost of €12 per square metre. Find the total cost of coating the panel.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 equates the two expressions, $6x-x^2=x^2$	✓ 1
A1 $2x(x-3)=0$ or equivalent	✓ 1
A1 $x=0$ and $x=3$	✓ 1
Part (b)
M1 integrates the difference of the functions, e.g. $\int_0^3(6x-2x^2)\,dx$	✓ 1
A1 $3x^2-\frac{2}{3}x^3$	✓ 1
A1 $27-18=9$ m²	✓ 1
Part (c)
A1 €108 (ECF from (b))	✓ 1

Exam tip: When finding the area between two curves, subtract the lower curve from the upper curve. Between the intersections, $6x-x^2$ lies above $x^2$.

+ Add to test✓ In your test211IBMA3S-5-16r12-135.5 Introduction to integration Paper 1Medium5 marks

No calculator

The curve $y=x(x-2)(x-4)$ crosses the x-axis at the points with x = 0, 2 and 4. The curve and the x-axis enclose two regions. Find the total area of these two regions.

The two enclosed regions are shaded

(a)

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 expands, $y=x^3-6x^2+8x$	✓ 1
A1 $\frac{x^4}{4}-2x^3+4x^2$	✓ 1
M1 evaluates the integral over $[0,2]$ and over $[2,4]$ separately	✓ 1
A1 $\int_0^2 = 4$ and $\int_2^4=-4$	✓ 1
A1 total area $=4+4=8$	✓ 1

Common error: Integrating from 0 to 4 in one step gives 0 because the regions cancel. Find each region's area separately and add the positive values.

+ Add to test✓ In your test212IBMA3S-5-17r14-065.10 Indefinite integrals and reverse chain rule Paper 1Medium7 marks

No calculator

(a)

Find $\int 6x^2(x^3+2)^4\,dx$.

(3)

(b)

The curve $y=f(x)$ has gradient function $6x^2(x^3+2)^4$ and passes through the point $\left(-1,\frac15\right)$. Find $f(0)$.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 recognising that $6x^2$ is a multiple of the derivative of $x^3+2$ (substitution $u=x^3+2$ or inspection)	✓ 1
A1 $\frac{2}{5}(x^3+2)^5$	✓ 1
A1 $+C$	✓ 1
Part (b)
M1 substituting $x=-1$, $y=\frac15$ into $\frac25(x^3+2)^5+C$	✓ 1
A1 $\frac25+C=\frac15$, so $C=-\frac15$	✓ 1
M1 substituting $x=0$: $\frac25\times 32-\frac15$	✓ 1
A1 $f(0)=\frac{63}{5}$ ($=12.6$)	✓ 1

Common error: Check that the factor outside the bracket really matches the derivative of the inner function before integrating by inspection; here $6x^2$ is twice $3x^2$, giving the factor $\frac{2}{5}$.

+ Add to test✓ In your test213IBMA3S-5-17r14-085.6 Differentiation rules Paper 1Medium7 marks

No calculator

The curve $y=\sqrt{4x^2+k}$, where $k$ is a positive constant, has gradient $\frac85$ at the point where $x=2$.

(a)

Find the value of $k$.

(4)

(b)

Find the equation of the tangent to the curve at $x=2$, giving your answer in the form $y=mx+c$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 using the chain rule: $\frac12(4x^2+k)^{-1/2}\times 8x$	✓ 1
A1 $\frac{dy}{dx}=\frac{4x}{\sqrt{4x^2+k}}$	✓ 1
M1 equating the gradient at $x=2$: $\frac{8}{\sqrt{16+k}}=\frac85$	✓ 1
A1 $\sqrt{16+k}=5$, so $k=9$	✓ 1
Part (b)
A1 $y=5$ when $x=2$	✓ 1
M1 $y-5=\frac85(x-2)$	✓ 1
A1 $y=\frac85x+\frac95$	✓ 1

Common error: When differentiating $\sqrt{4x^2+k}$, do not forget the factor $8x$ from the derivative of the inner function.

+ Add to test✓ In your test214IBMA3S-5-17r14-095.2 Increasing and decreasing functions Paper 1Medium11 marks

No calculator

A function is defined by $f(x)=x^4-8x^3+18x^2+5$.

(a)

Find $f'(x)$, giving your answer in fully factorised form.

(3)

(b)

Find the $x$-coordinates of the stationary points of the graph of $f$.

(2)

(c)

By considering the sign of $f'(x)$ on either side of each stationary point, determine the nature of each.

(4)

(d)

Find the $y$-coordinate of each stationary point.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate (at least two terms correct)	✓ 1
A1 $4x^3-24x^2+36x$	✓ 1
A1 $4x(x-3)^2$	✓ 1
Part (b)
M1 setting $f'(x)=0$	✓ 1
A1 $x=0$ and $x=3$	✓ 1
Part (c)
M1 considering the sign of $f'$ either side of $x=0$	✓ 1
A1 $f'$ changes from negative to positive, so a local minimum at $x=0$	✓ 1
M1 considering the sign of $f'$ either side of $x=3$	✓ 1
A1 $f'$ is positive on both sides, so a (horizontal) point of inflexion at $x=3$	✓ 1
Part (d)
A1 $f(0)=5$	✓ 1
A1 $f(3)=81-216+162+5=32$	✓ 1

Common error: At a stationary point where $f''(x)=0$ the second-derivative test is inconclusive, so check the sign of $f'$ on either side instead.

+ Add to test✓ In your test215IBMA3S-5-16r16-075.10 Indefinite integrals and reverse chain rule Paper 1Medium6 marks

No calculator

The gradient of a curve is given by $\frac{dy}{dx} = \frac{6x^2}{x^3+1}$, for $x > -1$. The curve passes through the point $(0, 3)$.

(a)

Find an expression for $y$ in terms of $x$.

(4)

(b)

Find the exact value of $y$ when $x=1$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 recognises the form $\frac{f'(x)}{f(x)}$, e.g. $\ln(x^3+1)$ seen	✓ 1
A1 $2\ln(x^3+1) + C$	✓ 1
M1 substitutes $(0,3)$ into their expression	✓ 1
A1 $y = 2\ln(x^3+1) + 3$	✓ 1
Part (b)
M1 substitutes $x=1$	✓ 1
A1 $y = 3 + 2\ln 2$	✓ 1

Exam tip: If the numerator is a multiple of the derivative of the denominator, the integral is a multiple of the natural logarithm of the denominator.

+ Add to test✓ In your test216IBMA3S-5-16r16-105.3 Differentiating polynomials and integer powers Paper 1Medium6 marks

No calculator

Let $f(x) = x^4 - 4x^3$.

(a)

Find $f'(x)$ and hence find the $x$-coordinates of the stationary points of the graph of $f$.

(3)

(b)

Show that the graph of $f$ has a point of inflexion at $x = 0$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $f'(x) = 4x^3 - 12x^2$	✓ 1
M1 sets $f'(x)=0$, e.g. $4x^2(x-3)=0$	✓ 1
A1 $x = 0$ and $x = 3$	✓ 1
Part (b)
A1 $f''(x) = 12x^2 - 24x$ and $f''(0) = 0$	✓ 1
M1 evaluates $f''$ either side of 0, e.g. $f''=36$ at $x=-1$ and $f''=-12$ at $x=1$	✓ 1
R1 $f''$ changes sign at $x=0$, so there is a point of inflexion	✓ 1

Common error: f″(x) = 0 alone does not prove an inflexion: the sign of f″ must change.

+ Add to test✓ In your test217IBMA3S-5-16r16-135.10 Indefinite integrals and reverse chain rule Paper 1Medium6 marks

No calculator

(a)

Differentiate $y = (2x^2+1)^{3/2}$.

(3)

(b)

Hence find the exact value of $\int_0^2 x\sqrt{2x^2+1}\,dx$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at the chain rule	✓ 1
A1 $\frac{3}{2}(2x^2+1)^{1/2} \times 4x$	✓ 1
A1 $\frac{dy}{dx} = 6x\sqrt{2x^2+1}$	✓ 1
Part (b)
M1 uses part (a): $\int x\sqrt{2x^2+1}\,dx = \frac{1}{6}(2x^2+1)^{3/2}$	✓ 1
M1 substitutes limits: $\frac{1}{6}\left(9^{3/2} - 1\right)$	✓ 1
A1 $\frac{13}{3}$	✓ 1

Common error: When using a 'hence' result, divide by the constant that appears (here 6) rather than integrating from scratch.

+ Add to test✓ In your test218IBMA3S-5-17r18-065.7 The second derivative and graphs Paper 1Medium7 marks

No calculator

The function $f$ is defined by $f(x)=x^4-4x^3$, for $x\in\mathbb{R}$.

(a)

Find the coordinates of the stationary points of the graph of $f$.

(4)

(b)

Determine the nature of each stationary point, justifying your answers.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 set $f'(x)=0$ and attempt to solve	✓ 1
A1 $f'(x)=4x^3-12x^2=4x^2(x-3)$	✓ 1
A1 $x=0$ and $x=3$	✓ 1
A1 $(0,0)$ and $(3,-27)$	✓ 1
Part (b)
M1 $f''(x)=12x^2-24x$	✓ 1
A1 $f''(3)=36>0$, so $(3,-27)$ is a local minimum	✓ 1
A1 $f''(0)=0$ so the test is inconclusive; $f'(x)=4x^2(x-3)$ is negative on both sides of $x=0$, so $(0,0)$ is a point of inflexion (not a maximum or minimum)	✓ 1

Common error: Concluding that a point is neither a maximum nor a minimum just because $f''=0$ is not enough: you must show the gradient (or $f''$) does or does not change sign.

+ Add to test✓ In your test219IBMA3S-5-17r18-085.4 Tangents and normals Paper 1Medium7 marks

No calculator

The curve $C$ has equation $y=x^2\ln x$, for $x>0$.

(a)

Find $\dfrac{dy}{dx}$.

(2)

(b)

Find the equation of the tangent to $C$ at the point where $x=e$. Give your answer in the form $y=mx+c$.

(3)

(c)

The tangent meets the $x$-axis at the point $P$. Find the $x$-coordinate of $P$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 product rule attempted	✓ 1
A1 $\dfrac{dy}{dx}=2x\ln x+x$	✓ 1
Part (b)
A1 $y=e^2$ and gradient $=2e\ln e+e=3e$ when $x=e$	✓ 1
M1 line through $(e,e^2)$ with gradient $3e$, e.g. $y-e^2=3e(x-e)$ (ECF from (a))	✓ 1
A1 $y=3ex-2e^2$	✓ 1
Part (c)
M1 put $y=0$: $3ex=2e^2$	✓ 1
A1 $x=\dfrac{2e}{3}$ ($\approx1.81$)	✓ 1

Exam tip: Remember that $\ln e=1$ when substituting exact values; leaving $\ln e$ unsimplified hides the gradient and invites arithmetic slips.

+ Add to test✓ In your test220IBMA3S-5-17r18-095.5 Introduction to integration Paper 1Medium6 marks

No calculator

The cross-section of a fishery's feeder channel is the region between the horizontal line $d=0$ and the curve $d=\dfrac{8}{x^2}$, from $x=1$ to $x=k$, where $k>1$. Here $x$ is the horizontal distance in metres and $d$ is a depth in metres.

Diagram not to scale

(a)

Find $\displaystyle\int_1^k\dfrac{8}{x^2}\,dx$ in terms of $k$.

(3)

(b)

The area of the cross-section is 6 m². Find the value of $k$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rewrite as $\int_1^k 8x^{-2}\,dx$	✓ 1
A1 $\left[-8x^{-1}\right]_1^k$	✓ 1
A1 $8-\dfrac{8}{k}$	✓ 1
Part (b)
M1 equate $8-\dfrac8k=6$ (ECF from (a))	✓ 1
A1 $\dfrac{8}{k}=2$	✓ 1
A1 $k=4$	✓ 1

Exam tip: Rewrite $\frac{1}{x^2}$ as $x^{-2}$ before integrating, and substitute the limits into the antiderivative in the order upper minus lower.

+ Add to test✓ In your test221IBMA3S-5-17r19-065.4 Tangents and normals Paper 1Medium6 marks

No calculator

In a scaled map, the path of a space probe passing an asteroid is modelled by the curve $y = x^3 - 5x + 1$. At two points on the path, the tangent is parallel to the line $y = 7x + 3$.

(a)

Find the coordinates of these two points.

(4)

(b)

Find the equation of the normal at the point where $x = 2$. Give your answer in the form $ax + by + c = 0$ where $a$, $b$, $c$ are integers.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{dy}{dx} = 3x^2 - 5$	✓ 1
M1 set equal to $7$	✓ 1
A1 $x = 2$ and $x = -2$	✓ 1
A1 $(2, -1)$ and $(-2, 3)$	✓ 1
Part (b)
M1 normal gradient $-\frac{1}{7}$ through $(2, -1)$, e.g. $y + 1 = -\frac{1}{7}(x - 2)$	✓ 1
A1 $x + 7y + 5 = 0$	✓ 1

Common error: Parallel lines share the same gradient, so set the derivative equal to 7. Setting $y = 7$ or using $-\frac{1}{7}$ in part (a) are frequent slips.

+ Add to test✓ In your test222IBMA3S-5-17r19-075.2 Increasing and decreasing functions Paper 1Medium7 marks

No calculator

The profile of a bridge deck is modelled by $f(x) = x^4 - 8x^3 + 18x^2$, for $x \ge 0$.

(a)

Find the values of $x$ at which the graph of $f$ has a stationary point.

(2)

(b)

Show that the stationary point at $x = 3$ is a point of inflection, and determine the nature of the stationary point at $x = 0$.

(3)

(c)

Find the coordinates of the other point of inflection of the graph.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $f'(x) = 4x^3 - 24x^2 + 36x$ set equal to zero	✓ 1
A1 $4x(x-3)^2 = 0$, so $x = 0$ and $x = 3$	✓ 1
Part (b)
A1 $f''(x) = 12x^2 - 48x + 36 = 12(x-1)(x-3)$	✓ 1
A1 $f''(0) = 36 > 0$, so a local minimum at $x = 0$	✓ 1
R1 $f''(3) = 0$ and $f''$ changes sign at $x = 3$ (negative for $1 < x < 3$, positive for $x > 3$), so a point of inflection	✓ 1
Part (c)
M1 $f''(x) = 0$ gives $x = 1$	✓ 1
A1 $(1, 11)$	✓ 1

Common error: $f''(x) = 0$ alone does not prove a point of inflection. You must show that the sign of $f''$ changes either side.

+ Add to test✓ In your test223IBMA3S-5-17r19-085.11 Definite integrals and areas Paper 1Medium6 marks

No calculator

A decorative panel on a bridge is the region enclosed between the curve $y = 6x - x^2$ and the straight line $y = x$, where $x$ and $y$ are in metres.

Shaded panel

(a)

Find the x-coordinates of the points where the curve and the line meet.

(2)

(b)

Find the exact area of the panel.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 equate $6x - x^2 = x$	✓ 1
A1 $x = 0$ and $x = 5$	✓ 1
Part (b)
M1 $\int_0^5 (6x - x^2 - x)\,dx = \int_0^5 (5x - x^2)\,dx$	✓ 1
A1 $\frac{5x^2}{2} - \frac{x^3}{3}$	✓ 1
M1 substitute limits $5$ and $0$	✓ 1
A1 $\frac{125}{2} - \frac{125}{3} = \frac{125}{6}$ m²	✓ 1

Exam tip: For an area between two curves, integrate (upper minus lower). Check which graph is on top by testing a value of $x$ between the limits.

+ Add to test✓ In your test224IBMA3S-5-17r19-115.5 Introduction to integration Paper 1Medium6 marks

No calculator

In a scale drawing, the gradient of a curve $y = f(x)$ is given by $f'(x) = \frac{6}{(3x+1)^2}$. The curve passes through the point $(0, 5)$.

(a)

Find $f(x)$.

(4)

(b)

Find the value of $f(1)$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate $6(3x+1)^{-2}$	✓ 1
A1 $-\frac{2}{3x+1}$	✓ 1
M1 use of the point $(0, 5)$ to find the constant	✓ 1
A1 $f(x) = 7 - \frac{2}{3x+1}$	✓ 1
Part (b)
M1 substitute $x = 1$ (ECF from (a))	✓ 1
A1 $\frac{13}{2}$	✓ 1

Common error: When integrating a linear function raised to a power, divide by the coefficient of $x$ (here 3) as well as by the new power. Differentiate your answer to check.

+ Add to test✓ In your test225IBMA3S-5-17r19-125.2 Increasing and decreasing functions Paper 1Medium7 marks

No calculator

The arch of a bridge is modelled by $f(x) = x^3 + ax^2 + bx + 5$ for $0 \le x \le 4$, where $a$ and $b$ are constants. The graph of $f$ has a point of inflection at $x = 2$ and a stationary point at $x = 1$.

(a)

Find the value of $a$ and the value of $b$.

(4)

(b)

Find the coordinates of the other stationary point and determine its nature.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $f''(x) = 6x + 2a$ and set $f''(2) = 0$	✓ 1
A1 $a = -6$	✓ 1
M1 $f'(x) = 3x^2 + 2ax + b$ and set $f'(x) = 0$ at $x = 1$	✓ 1
A1 $b = 9$	✓ 1
Part (b)
M1 $3x^2 - 12x + 9 = 0$ gives $x = 3$	✓ 1
A1 $(3, 5)$	✓ 1
R1 $f''(3) = 6 > 0$, so a local minimum	✓ 1

Exam tip: A point of inflection gives an equation from $f''$; a stationary point gives an equation from $f'$. Using each fact once gives two equations for two unknowns.

+ Add to test✓ In your test226IBMA3S-5-17r21-065.4 Tangents and normals Paper 1Medium9 marks

No calculator

Let $f(x)=\frac{x}{x^2+4}$ for $x\in\mathbb{R}$.

(a)

Use the quotient rule to show that $f'(x)=\frac{4-x^2}{(x^2+4)^2}$.

(3)

(b)

Find the coordinates of the points on the curve where the gradient is zero.

(3)

(c)

Find the equation of the normal to the curve at the origin.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 quotient rule $\frac{u'v-uv'}{v^2}$ with $u=x$, $v=x^2+4$	✓ 1
A1 $\frac{1\times(x^2+4)-x\times2x}{(x^2+4)^2}$	✓ 1
A1 simplifies to $\frac{4-x^2}{(x^2+4)^2}$ (AG)	✓ 1
Part (b)
M1 sets numerator $4-x^2=0$	✓ 1
A1 $x=\pm2$	✓ 1
A1 points $\left(2,\frac14\right)$ and $\left(-2,-\frac14\right)$	✓ 1
Part (c)
M1 $f'(0)=\frac{4}{16}=\frac14$	✓ 1
M1 normal gradient $-4$ through $(0,0)$	✓ 1
A1 $y=-4x$	✓ 1

Common error: Write the quotient rule out in full before substituting, and check the sign of the $uv'$ term as you simplify.

+ Add to test✓ In your test227IBMA3S-5-17r21-085.10 Indefinite integrals and reverse chain rule Paper 1Medium5 marks

No calculator

Consider the function $f(x)=6x(x^2+1)^3$.

(a)

Find $\int f(x)\,dx$.

(3)

(b)

Hence find the exact value of $\int_0^1 f(x)\,dx$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at substitution $u=x^2+1$ or inspection giving $(x^2+1)^4$	✓ 1
A1 correct factor $\frac34$ (since $\frac{d}{dx}(x^2+1)^4=8x(x^2+1)^3$)	✓ 1
A1 $\frac34(x^2+1)^4+c$	✓ 1
Part (b)
M1 $\frac34\left(2^4-1^4\right)$	✓ 1
A1 $\frac{45}{4}$ (= 11.25)	✓ 1

Exam tip: After integrating by inspection, differentiate your answer; if it does not return the original integrand, the constant factor is wrong.

+ Add to test✓ In your test228IBMA3S-5-17r21-095.4 Tangents and normals Paper 1Medium8 marks

No calculator

The cross-section of a snow slope is modelled by $f(x)=x^2+\frac{p}{x}$ for $x>0$, where $p$ is a constant. At the point where $x=2$ the tangent to the curve is parallel to the line $10x-2y=7$.

(a)

Find $f'(x)$ in terms of $p$.

(2)

(b)

Find the value of $p$.

(3)

(c)

Find the equation of the tangent to the curve at $x=2$, giving your answer in the form $y=mx+c$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $2x$	✓ 1
A1 $-\frac{p}{x^2}$	✓ 1
Part (b)
M1 rearranges line to $y=5x-\frac72$ so gradient is 5	✓ 1
M1 $f'(2)=4-\frac p4=5$	✓ 1
A1 $p=-4$	✓ 1
Part (c)
M1 $f(2)=4+\frac{-4}{2}=2$	✓ 1
M1 $y-2=5(x-2)$	✓ 1
A1 $y=5x-8$	✓ 1

Exam tip: Parallel lines have equal gradients, so rearrange the given line into $y=mx+c$ form first to read off the gradient.

+ Add to test✓ In your test229IBMA3S-5-17r21-145.1 Limits and the derivative Paper 1Medium6 marks

No calculator

A mobile network models the total data used by a customer as $D(t)=t^3$ gigabytes after $t$ months, $t\geq0$. The average rate of change of $D$ between $t=2$ and $t=2+h$ is $\frac{(2+h)^3-2^3}{h}$ gigabytes per month.

(a)

Expand and simplify $\frac{(2+h)^3-2^3}{h}$, for $h\neq0$.

(2)

(b)

Find the average rate of change between $t=2$ and $t=2.5$.

(2)

(c)

Find the limit of the expression as $h\to0$.

(1)

(d)

State what this limit represents in context.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 expands $(2+h)^3=8+12h+6h^2+h^3$	✓ 1
A1 $h^2+6h+12$	✓ 1
Part (b)
M1 substitutes $h=0.5$	✓ 1
A1 15.25 gigabytes per month	✓ 1
Part (c)
A1 12	✓ 1
Part (d)
R1 the instantaneous rate of change of data used at $t=2$ months, equal to $D'(2)$ (the gradient of the tangent)	✓ 1

Common error: A limit as $h\to0$ can only be taken after $h$ has been cancelled from the denominator.

+ Add to test✓ In your test230IBMA3S-5-16r23-055.2 Increasing and decreasing functions Paper 1Medium5 marks

No calculator

Let $f(x)=x^3+kx^2+12x-5$, where $k$ is a constant.

(a)

Find the set of values of $k$ for which $f'(x)\ge 0$ for all real $x$.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $f'(x)=3x^2+2kx+12$	✓ 1
R1 $f'(x)\ge0$ for all $x$ requires the quadratic to have at most one root, so discriminant $\le 0$	✓ 1
M1 discriminant $(2k)^2-4(3)(12)$	✓ 1
A1 $4k^2-144\le0$	✓ 1
A1 $-6\le k\le 6$	✓ 1

Common error: A quadratic that is never negative can touch the $x$-axis, so the condition is $\Delta\le0$, not $\Delta<0$.

+ Add to test✓ In your test231IBMA3S-5-16r23-075.3 Differentiating polynomials and integer powers Paper 1Medium6 marks

No calculator

The graph of $y=f(x)$ is given by $f(x)=x^4-4x^3+1$.

(a)

Find $f'(x)$ and $f''(x)$.

(2)

(b)

Find the $x$-coordinates of the points of inflection of the graph. Justify that each is a point of inflection.

(3)

(c)

Hence write down the values of $x$ for which the graph is concave up.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $f'(x)=4x^3-12x^2$	✓ 1
A1 $f''(x)=12x^2-24x$	✓ 1
Part (b)
M1 sets $f''(x)=0$, i.e. $12x(x-2)=0$	✓ 1
A1 $x=0$ and $x=2$	✓ 1
R1 $f''$ changes sign at each: positive for $x<0$, negative for $0<x<2$, positive for $x>2$	✓ 1
Part (c)
A1 $x<0$ or $x>2$	✓ 1

Common error: $f''(x)=0$ alone does not prove a point of inflection. Show that $f''$ changes sign.

+ Add to test✓ In your test232IBMA3S-5-16r23-095.10 Indefinite integrals and reverse chain rule Paper 1Medium5 marks

No calculator

Find $\lim_{x\to0}\dfrac{\sqrt{x+9}-3}{x}$.

(a)

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 multiplies numerator and denominator by the conjugate $\sqrt{x+9}+3$	✓ 1
A1 numerator becomes $(x+9)-9=x$	✓ 1
M1 cancels the factor $x$	✓ 1
A1 obtains $\dfrac{1}{\sqrt{x+9}+3}$	✓ 1
A1 limit $=\dfrac{1}{3+3}=\dfrac16$	✓ 1

Exam tip: When a limit gives $\frac00$, simplify first (factorise or multiply by a conjugate) and only then substitute.

+ Add to test✓ In your test233IBMA3S-5-16r23-115.9 Kinematics Paper 1Medium11 marks

No calculator

A sports scientist models the heart rate of an athlete during a training session. The heart rate is $H$ beats per minute above the resting rate, $t$ minutes after the session starts, where $H(t)=-t^3+9t^2$ for $0\le t\le 8$.

(a)

Find $H'(t)$.

(2)

(b)

Find the time at which the heart rate is greatest, and the greatest heart rate above the resting rate.

(4)

(c)

Find $H''(t)$ and hence show that the heart rate is rising fastest when $t=3$.

(3)

(d)

Find the greatest rate at which the heart rate is rising.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts to differentiate	✓ 1
A1 $H'(t)=-3t^2+18t$	✓ 1
Part (b)
M1 sets $H'(t)=0$, i.e. $-3t(t-6)=0$	✓ 1
A1 $t=6$ (reject $t=0$ or note $H(0)=0$)	✓ 1
R1 $H''(6)=-18<0$, or $H'$ changes from positive to negative, so a maximum	✓ 1
A1 $H(6)=-216+324=108$ beats per minute	✓ 1
Part (c)
A1 $H''(t)=-6t+18$	✓ 1
M1 sets $H''(t)=0$ to obtain $t=3$	✓ 1
R1 $H''$ changes from positive to negative at $t=3$, so $H'$ is greatest there	✓ 1
Part (d)
M1 substitutes $t=3$ into $H'(t)$	✓ 1
A1 $H'(3)=27$ beats per minute per minute	✓ 1

Exam tip: The greatest value of a rate of change $H'$ occurs where $H''=0$ and changes sign, not where $H'=0$.

+ Add to test✓ In your test234IBMA3S-5-17r23-065.4 Tangents and normals Paper 1Medium7 marks

No calculator

The curve $C$ has equation $y = f(x)$, where $f(x) = 2x^2 - \frac{8}{x}$, for $x \neq 0$.

(a)

Find the equation of the tangent to $C$ at the point where $x=2$. Give your answer in the form $y=mx+c$.

(4)

(b)

Find the equation of the normal to $C$ at the same point, in the form $ax+by+d=0$ where $a$, $b$ and $d$ are integers.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $f(2)=4$	✓ 1
M1 attempt to differentiate (at least one term correct)	✓ 1
A1 $f'(2)=4(2)+\frac{8}{4}=10$	✓ 1
A1 $y=10x-16$	✓ 1
Part (b)
A1 gradient of normal $-\frac{1}{10}$ (ECF from (a))	✓ 1
M1 $y-4=-\frac{1}{10}(x-2)$	✓ 1
A1 $x+10y-42=0$ (accept integer multiples)	✓ 1

Common error: The normal's gradient is the negative reciprocal of the tangent's gradient; do not just change the sign.

+ Add to test✓ In your test235IBMA3S-5-17r23-085.2 Increasing and decreasing functions Paper 1Medium7 marks

No calculator

The cross-section of an ancient ditch is the region between the ground surface (the $x$-axis) and the curve $y = x^2 - 6x + 5$, for $1 \le x \le 5$, where $x$ and $y$ are in metres.

Cross-section of the ditch (shaded)

(a)

Solve $x^2-6x+5=0$.

(2)

(b)

Find the area of the cross-section.

(4)

(c)

The ditch is straight and 15 m long with this constant cross-section. Find the volume of soil removed.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to factorise or use the quadratic formula	✓ 1
A1 $x=1$, $x=5$	✓ 1
Part (b)
M1 area $=\left\|\int_1^5 (x^2-6x+5)\,dx\right\|$ (accept the integral without the modulus)	✓ 1
A1 $\frac{x^3}{3}-3x^2+5x$	✓ 1
M1 substitute limits 5 and 1	✓ 1
A1 $\frac{32}{3}$ m² (area given as a positive value)	✓ 1
Part (c)
A1 $15\times\frac{32}{3}=160$ m³ (ECF from (b))	✓ 1

Common error: The curve lies below the axis here, so the definite integral is negative; an area must be quoted as a positive value.

+ Add to test✓ In your test236IBMA3S-5-17r23-105.2 Increasing and decreasing functions Paper 1Medium7 marks

No calculator

The function $g$ has derivative $g'(x) = x(x-4)^2$, for $x \in \mathbb{R}$.

(a)

Find the $x$-coordinates of the stationary points of $g$, and determine the nature of each.

(4)

(b)

Find the value of $g(4)-g(0)$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 set $g'(x)=0$	✓ 1
A1 $x=0$ and $x=4$	✓ 1
R1 $g'<0$ for $x<0$ and $g'>0$ for $0<x<4$, so a local minimum at $x=0$	✓ 1
R1 $g'>0$ on both sides of $x=4$ (no sign change), so not a maximum or minimum: a horizontal point of inflexion	✓ 1
Part (b)
M1 expand $g'(x)=x^3-8x^2+16x$ and attempt to integrate	✓ 1
A1 $\frac{x^4}{4}-\frac{8x^3}{3}+8x^2$	✓ 1
A1 $\frac{64}{3}$	✓ 1

Exam tip: When the derivative does not change sign at a stationary point, it is not a maximum or minimum; check values either side.

+ Add to test✓ In your test237IBMA3S-5-17r23-135.9 Kinematics Paper 1Medium12 marks

No calculator

A survey trolley moves along a straight rail in a trench. Its displacement from the start of the rail is $s$ metres at time $t$ seconds, where $s(t)=t^3-9t^2+24t$ for $0\le t\le 5$.

(a)

Find an expression for the velocity of the trolley.

(2)

(b)

Find the acceleration of the trolley when $t=3$, and describe what this tells you about its velocity at that instant.

(3)

(c)

Find the times at which the trolley is at rest.

(2)

(d)

Find the displacement of the trolley from the start at each of these times.

(2)

(e)

Find the total distance travelled by the trolley during the first 5 seconds.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate	✓ 1
A1 $v=3t^2-18t+24$	✓ 1
Part (b)
A1 $a=6t-18$	✓ 1
A1 $a(3)=0$	✓ 1
R1 the velocity is momentarily not changing; it has a minimum value of $v(3)=-3$ m s⁻¹ (the trolley is moving backwards at its greatest speed)	✓ 1
Part (c)
M1 set $v=0$: $3(t-2)(t-4)=0$	✓ 1
A1 $t=2$ and $t=4$	✓ 1
Part (d)
A1 $s(2)=20$ m	✓ 1
A1 $s(4)=16$ m	✓ 1
Part (e)
M1 split the motion into stages at $t=2$ and $t=4$ and add the distances	✓ 1
A1 $s(5)=20$ m, so the stages are 20 m, 4 m and 4 m (ECF from (d))	✓ 1
A1 $28$ m	✓ 1

Common error: Total distance is not the final displacement: when the velocity changes sign, add the distances of each stage separately.

+ Add to test✓ In your test238IBMA3S-5-16r26-045.3 Differentiating polynomials and integer powers Paper 1Medium6 marks

No calculator

Let $f(x) = x^4 - 4x^3 + 1$.

(a)

Find $f'(x)$ and $f''(x)$.

(2)

(b)

Find the coordinates of the points of inflexion of the graph of $f$, justifying that each is a point of inflexion.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $f'(x) = 4x^3 - 12x^2$	✓ 1
A1 $f''(x) = 12x^2 - 24x$	✓ 1
Part (b)
M1 set $f''(x)=0$, i.e. $12x(x-2)=0$	✓ 1
A1 $x=0$ and $x=2$	✓ 1
R1 $f''$ changes sign at each (e.g. $f''(-1)=36>0$, $f''=-12<0$ at $x=1$, $f''(3)=36>0$)	✓ 1
A1 points $(0,\,1)$ and $(2,\,-15)$	✓ 1

Common error: $f''(x)=0$ alone does not prove a point of inflexion; you must show that the sign of $f''$ changes.

+ Add to test✓ In your test239IBMA3S-5-16r26-055.8 Stationary points, inflexion and optimization Paper 1Medium7 marks

No calculator

A shop designs an open-topped storage box with a square base of side $x$ cm and height $h$ cm. The box is made from $1200\ \text{cm}^2$ of card, which covers the base and the four sides exactly.

Diagram not to scale

(a)

Show that the volume of the box is $V = 300x - \frac{x^3}{4}$.

(2)

(b)

Find the value of $x$ that maximises the volume.

(3)

(c)

Use the second derivative to show that this value of $x$ gives a maximum.

(1)

(d)

Find the maximum volume.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $x^2 + 4xh = 1200$ and attempt to make $h$ the subject, $h = \frac{1200 - x^2}{4x}$	✓ 1
A1 substitute into $V = x^2h$ and simplify to $V = 300x - \frac{x^3}{4}$ (AG)	✓ 1
Part (b)
M1 attempt to differentiate and set $\frac{dV}{dx}=0$	✓ 1
A1 $\frac{dV}{dx} = 300 - \frac{3x^2}{4}$	✓ 1
A1 $x^2 = 400$, so $x=20$ (reject $x=-20$)	✓ 1
Part (c)
R1 $\frac{d^2V}{dx^2} = -\frac{3x}{2} = -30 < 0$ when $x=20$, so a maximum	✓ 1
Part (d)
A1 $V = 300(20) - \frac{20^3}{4} = 4000\ \text{cm}^3$	✓ 1

Exam tip: When a question gives a constraint, use it to eliminate one variable before differentiating.

+ Add to test✓ In your test240IBMA3S-5-16r26-115.2 Increasing and decreasing functions Paper 1Medium10 marks

No calculator

The function $f(x) = x^3 - 9x^2 + 24x - 10$ has a graph with two stationary points.

(a)

Find the $x$-coordinates of the stationary points.

(3)

(b)

Use $f''(x)$ to determine the nature of each stationary point.

(4)

(c)

Show that the graph of $f$ has a point of inflexion at $(3,\,8)$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $f'(x) = 3x^2 - 18x + 24$	✓ 1
M1 set $f'(x)=0$	✓ 1
A1 $3(x-2)(x-4)=0$, so $x=2$ and $x=4$	✓ 1
Part (b)
M1 attempt to find $f''(x)$	✓ 1
A1 $f''(x) = 6x - 18$	✓ 1
M1 evaluate $f''(2)=-6$ and $f''(4)=6$	✓ 1
A1 maximum at $x=2$ and minimum at $x=4$	✓ 1
Part (c)
M1 set $f''(x)=0$	✓ 1
A1 $x=3$ and $f''$ changes sign (negative for $x<3$, positive for $x>3$)	✓ 1
A1 $f(3) = 27 - 81 + 72 - 10 = 8$, so the point is $(3,\,8)$	✓ 1

Common error: A positive second derivative means a minimum (concave up); do not mix up the two signs.

+ Add to test✓ In your test241IBMA3S-5-18r30-065.7 The second derivative and graphs Paper 1Medium11 marks

No calculator

Let $f(x)=2x^3-9x^2+12x-4$.

(a)

Find the coordinates of the stationary points of the graph of $f$.

(4)

(b)

Find $f''(x)$ and hence determine the nature of each stationary point.

(3)

(c)

Find the coordinates of the point of inflexion of the graph of $f$.

(3)

(d)

Write down the values of $x$ for which the graph of $f$ is concave up.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate	✓ 1
A1 $f'(x)=6x^2-18x+12$	✓ 1
M1 sets $f'(x)=0$ and attempts to solve, $6(x-1)(x-2)=0$	✓ 1
A1 stationary points at $(1,1)$ and $(2,0)$	✓ 1
Part (b)
A1 $f''(x)=12x-18$	✓ 1
A1 $f''\left(1\right)=-6<0$, so a local maximum at $(1,1)$	✓ 1
A1 $f''(2)=6>0$, so a local minimum at $(2,0)$	✓ 1
Part (c)
M1 sets $f''(x)=0$	✓ 1
A1 $x=1.5$	✓ 1
A1 $y=0.5$, so the point is $(1.5,0.5)$	✓ 1
Part (d)
A1 $x>1.5$ (accept $x>\frac{3}{2}$)	✓ 1

Exam tip: A point of inflexion needs $f''(x)=0$ and a change in the sign of $f''$ either side; the second derivative being zero is not enough on its own.

+ Add to test✓ In your test242IBMA3S-5-18r30-075.3 Differentiating polynomials and integer powers Paper 1Medium10 marks

No calculator

Let $f(x)=x^4-4x^3$.

(a)

Find the $x$-coordinates of the stationary points of the graph of $f$.

(4)

(b)

Show that the graph of $f$ has a point of inflexion at $x=0$, and not a local maximum or minimum.

(3)

(c)

The graph of $f$ has a second point of inflexion. Find its coordinates.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate	✓ 1
A1 $f'(x)=4x^3-12x^2$	✓ 1
M1 sets $f'(x)=0$ and factorises, $4x^2(x-3)=0$	✓ 1
A1 $x=0$ and $x=3$	✓ 1
Part (b)
A1 $f''(x)=12x^2-24x=12x(x-2)$	✓ 1
M1 considers the sign of $f''(x)$ on each side of $x=0$	✓ 1
R1 $f''(x)>0$ for $x<0$ and $f''(x)<0$ for $0<x<2$, so the concavity changes at $x=0$ and it is a (stationary) point of inflexion	✓ 1
Part (c)
M1 solves $f''(x)=0$	✓ 1
A1 $x=2$ (the other root is $x=0$)	✓ 1
A1 $y=2^4-4(2^3)=-16$, so the point is $(2,-16)$	✓ 1

Common error: When $f'(x)=0$ and $f''(x)=0$ at the same point, do not assume it is an inflexion; check that $f''$ changes sign.

+ Add to test✓ In your test243IBMA3S-5-18r30-085.8 Stationary points, inflexion and optimization Paper 1Medium7 marks

No calculator

A rail company is fencing a rectangular compound for storing track sections. One side of the compound lies along a long straight wall, so fencing is needed on only the other three sides. The company has 240 m of fencing. The two sides perpendicular to the wall each have length $x$ metres.

Plan view. Diagram not to scale

(a)

Show that the area of the compound, $A$ m², is given by $A=240x-2x^2$.

(2)

(b)

Find the value of $x$ that gives the maximum area.

(3)

(c)

Find the maximum area, and use the second derivative to justify that it is a maximum.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 length parallel to the wall is $240-2x$	✓ 1
A1 $A=x(240-2x)=240x-2x^2$ (AG)	✓ 1
Part (b)
M1 attempt to differentiate	✓ 1
A1 $\frac{dA}{dx}=240-4x$	✓ 1
A1 $x=60$	✓ 1
Part (c)
A1 $A=60\times 120=7200$ m²	✓ 1
R1 $\frac{d^2A}{dx^2}=-4<0$, so the area is a maximum	✓ 1

Exam tip: State the length of the third side in terms of $x$ before forming the area; the constraint is what lets you reduce the area to one variable.

+ Add to test✓ In your test244IBMA3S-5-18r30-105.4 Tangents and normals Paper 1Medium10 marks

No calculator

Let $f(x)=\frac{x^2}{x+1}$, for $x\neq -1$.

(a)

Show that $f'(x)=\frac{x^2+2x}{(x+1)^2}$.

(3)

(b)

Find the $x$-coordinates of the stationary points of the graph of $f$.

(2)

(c)

Find the equation of the normal to the graph of $f$ at the point where $x=1$. Give your answer in the form $ax+by+c=0$, where $a$, $b$ and $c$ are integers.

(3)

(d)

The normal meets the $x$-axis at the point P. Find the $x$-coordinate of P.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the quotient rule	✓ 1
A1 $\frac{2x(x+1)-x^2\left(1\right)}{(x+1)^2}$	✓ 1
A1 simplifies to $\frac{x^2+2x}{(x+1)^2}$ (AG)	✓ 1
Part (b)
M1 sets $x^2+2x=0$	✓ 1
A1 $x=0$ and $x=-2$	✓ 1
Part (c)
A1 $f\left(1\right)=\frac{1}{2}$ and $f'\left(1\right)=\frac{3}{4}$	✓ 1
M1 normal gradient $-\frac{4}{3}$ used in $y-\frac{1}{2}=-\frac{4}{3}(x-1)$	✓ 1
A1 $8x+6y-11=0$ (accept any integer multiple)	✓ 1
Part (d)
M1 sets $y=0$ in the equation of the normal	✓ 1
A1 $x=\frac{11}{8}$ ($1.375$)	✓ 1

Exam tip: The gradient of the normal is the negative reciprocal of the gradient of the tangent, so evaluate the derivative at the given $x$ first.

+ Add to test✓ In your test245IBMA3S-5-17r34-085.1 Limits and the derivative Paper 1Medium7 marks

No calculator

A Formula 1 car leaves its garage along a straight pit lane. Its distance from the garage is $s(t)=3t^2+2t$ metres, where t is the time in seconds, $t\ge0$.

(a)

Show that the average speed of the car between $t=2$ and $t=2+h$ is $(14+3h)$ m s⁻¹, where $h>0$.

(3)

(b)

Find the limit of $14+3h$ as $h\to0$, and state what this value represents.

(2)

(c)

Use differentiation to find the time at which the speed of the car is 32 m s⁻¹.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 average speed $=\frac{s(2+h)-s(2)}{h}$	✓ 1
A1 $s(2+h)=3(4+4h+h^2)+2(2+h)=16+14h+3h^2$ and $s(2)=16$	✓ 1
A1 $\frac{14h+3h^2}{h}=14+3h$	✓ 1
Part (b)
A1 limit $=14$	✓ 1
R1 the instantaneous speed of the car at $t=2$ s (the gradient of the tangent to the graph of s at $t=2$)	✓ 1
Part (c)
M1 $s'(t)=6t+2=32$	✓ 1
A1 $t=5$ s	✓ 1

Exam tip: The limit of the chord gradient is the derivative. Check it by differentiating directly: $s'(2)=6(2)+2=14$ agrees with part (b).

+ Add to test✓ In your test246IBMA3S-5-17r34-095.4 Tangents and normals Paper 1Medium7 marks

No calculator

A curve has equation $y=f(x)$, where $f(x)=x^2+\dfrac{16}{x}$, for $x>0$.

(a)

Find $f'(x)$.

(2)

(b)

The point P on the curve has x-coordinate 4. Find the equation of the normal to the curve at P, giving your answer in the form $y=mx+c$.

(3)

(c)

The normal at P meets the x-axis at the point Q. Find the x-coordinate of Q.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 rewrite $\frac{16}{x}=16x^{-1}$ and attempt to differentiate	✓ 1
A1 $f'(x)=2x-\frac{16}{x^2}$	✓ 1
Part (b)
A1 $f(4)=20$ and $f'(4)=8-1=7$	✓ 1
M1 normal gradient $=-\frac17$, and use of $y-20=-\frac17(x-4)$	✓ 1
A1 $y=-\frac17x+\frac{144}{7}$	✓ 1
Part (c)
M1 set $y=0$: $\frac17x=\frac{144}{7}$ (ECF from (b))	✓ 1
A1 $x=144$	✓ 1

Common error: The gradient of the normal is the negative reciprocal of the tangent gradient. Using $+7$ or $-7$ instead of $-\frac17$ gives the tangent or a wrong line.

+ Add to test✓ In your test247IBMA3S-5-17r34-105.9 Kinematics Paper 1Medium7 marks

No calculator

A hospital goods lift starts from rest at ground level and travels upwards in a vertical shaft. For $0\le t\le10$ its acceleration is $a(t)=0.6-0.12t$ m s⁻², where t is the time in seconds.

(a)

Find an expression for the velocity $v(t)$ of the lift.

(2)

(b)

Find the positive value of t at which the lift is next at rest.

(2)

(c)

Find the height of the lift above ground level when it is next at rest.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $v=\int(0.6-0.12t)\,dt=0.6t-0.06t^2+c$	✓ 1
A1 $c=0$ since $v=0$ when $t=0$, so $v=0.6t-0.06t^2$	✓ 1
Part (b)
M1 $0.6t-0.06t^2=0$, so $0.06t(10-t)=0$	✓ 1
A1 $t=10$ s	✓ 1
Part (c)
M1 $\int_0^{10}(0.6t-0.06t^2)\,dt$ (ECF from (a))	✓ 1
A1 $\left[0.3t^2-0.02t^3\right]_0^{10}$	✓ 1
A1 $30-20=10$ m	✓ 1

Exam tip: For displacement from a starting point, integrate velocity between the start and end times. Substitute the upper limit first and subtract the value at zero.

+ Add to test✓ In your test248IBMA3S-5-18r35-065.9 Kinematics Paper 1Medium7 marks

No calculator

A tagged sea turtle swims along a straight channel. Its displacement from a marker buoy is $s$ metres at time $t$ minutes, where $s = t^3 - 9t^2 + 24t$ for $0 \leq t \leq 5$.

(a)

Find an expression for the velocity of the turtle.

(1)

(b)

Find the times at which the turtle is momentarily at rest.

(3)

(c)

By considering the acceleration at each of these times, describe how the direction of motion of the turtle changes.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $v = 3t^2 - 18t + 24$	✓ 1
Part (b)
M1 set $v = 0$	✓ 1
M1 $t^2 - 6t + 8 = 0$, attempt to factorise or solve	✓ 1
A1 $t = 2$ and $t = 4$	✓ 1
Part (c)
A1 $a = 6t - 18$	✓ 1
A1 $a(2) = -6$ and $a(4) = 6$	✓ 1
R1 at $t=2$ velocity changes from positive to negative (away from buoy, then towards it); at $t=4$ it changes from negative back to positive	✓ 1

Exam tip: Zero velocity does not mean the object stays still; the sign of the acceleration at that instant tells you which way the velocity is changing.

+ Add to test✓ In your test249IBMA3S-5-18r35-075.2 Increasing and decreasing functions Paper 1Medium7 marks

No calculator

A designer models the height $y$ metres of a curved handrail above ground level by $y = x^3 - 6x^2 + 9x + 1$ for $0 \leq x \leq 5$, where $x$ is the horizontal distance in metres.

(a)

Find $\dfrac{dy}{dx}$ and hence find the values of $x$ at which $\dfrac{dy}{dx} = 0$.

(3)

(b)

Use the second derivative to determine the nature of each stationary point.

(2)

(c)

Find the coordinates of the point of inflection.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $3x^2 - 12x + 9$	✓ 1
M1 set equal to zero and attempt to factorise or solve	✓ 1
A1 $x = 1$ and $x = 3$	✓ 1
Part (b)
A1 $\frac{d^2y}{dx^2} = 6x - 12$	✓ 1
R1 at $x=1$ it is $-6 < 0$ so a local maximum; at $x=3$ it is $6 > 0$ so a local minimum	✓ 1
Part (c)
M1 set $6x - 12 = 0$, so $x = 2$	✓ 1
A1 $(2, 3)$	✓ 1

Common error: A second derivative of zero does not by itself prove a point of inflection; check that the sign of $\frac{d^2y}{dx^2}$ changes either side.

+ Add to test✓ In your test250IBMA3S-5-18r35-115.5 Introduction to integration Paper 1Medium5 marks

No calculator

The gradient of a curved steel rib in a pavilion roof is given by $f'(x) = (2x - 1)^3$, where $x$ is the horizontal distance in metres. The rib passes through the point $(1, 1)$.

(a)

Find $f(x)$.

(4)

(b)

Find $f(2)$.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate, e.g. $(2x-1)^4$	✓ 1
A1 $\frac{(2x-1)^4}{8}$	✓ 1
M1 use $x = 1$, $y = 1$ to find the constant	✓ 1
A1 $f(x) = \frac{(2x-1)^4}{8} + \frac{7}{8}$	✓ 1
Part (b)
A1 $\frac{81}{8} + \frac{7}{8} = 11$	✓ 1

Common error: When integrating a power of a linear expression $(ax+b)^n$, divide by $a$ as well as adding one to the power.

+ Add to test✓ In your test251IBMA3S-5-18r44-035.6 Differentiation rules Paper 1Medium6 marks

No calculator

Let $f(x)=x^2e^{3x}$.

(a)

Show that $f'(x)=xe^{3x}(3x+2)$.

(3)

(b)

Find the exact coordinates of the stationary points of the graph of $f$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule	✓ 1
A1 $f'(x)=2xe^{3x}+3x^2e^{3x}$	✓ 1
A1 factorise $xe^{3x}$ to reach the given form, with the intermediate line shown	✓ 1
Part (b)
M1 set $f'(x)=0$ and note $e^{3x}\neq0$	✓ 1
A1 $(0,0)$	✓ 1
A1 $\left(-\frac{2}{3},\frac{4}{9}e^{-2}\right)$	✓ 1

Exam tip: The factor $e^{3x}$ is never zero, so only the other factors can produce stationary points.

+ Add to test✓ In your test252IBMA3S-5-18r44-055.4 Tangents and normals Paper 1Medium6 marks

No calculator

A curve has equation $y=e^{2x}-3x$.

(a)

Find $\frac{dy}{dx}$.

(2)

(b)

Find the equation of the tangent to the curve at the point where $x=0$.

(2)

(c)

The normal to the curve at the point where $x=0$ crosses the $x$-axis at the point P. Find the $x$-coordinate of P.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate $e^{2x}$ using the chain rule	✓ 1
A1 $2e^{2x}-3$	✓ 1
Part (b)
A1 $y=1$ and gradient $-1$ at $x=0$	✓ 1
A1 $y=1-x$	✓ 1
Part (c)
M1 gradient of normal is $1$, so normal is $y=x+1$	✓ 1
A1 $x=-1$	✓ 1

Common error: The gradient of the normal is the negative reciprocal of the tangent gradient, so a tangent gradient of $-1$ gives a normal gradient of $1$, not $-1$.

+ Add to test✓ In your test253IBMA3S-5-18r44-075.3 Differentiating polynomials and integer powers Paper 1Medium5 marks

No calculator

A curve has gradient function $f'(x)=6x^2-4x+3$ and passes through the point $(2,10)$. Find $f(x)$.

(a)

Find $f(x)$.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate	✓ 1
A1 $2x^3$	✓ 1
A1 $-2x^2+3x$	✓ 1
M1 substitute $(2,10)$ into $f(x)=2x^3-2x^2+3x+C$	✓ 1
A1 $f(x)=2x^3-2x^2+3x-4$	✓ 1

Common error: Do not forget the constant of integration $+C$; the given point is needed to find its value.

+ Add to test✓ In your test254IBMA3S-5-18r44-095.2 Increasing and decreasing functions Paper 1Medium5 marks

No calculator

Let $k>0$ and suppose $\int_0^k(3x^2+2)\,dx=12$.

(a)

Find $k$ and show that there is no other real solution of the equation formed.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate	✓ 1
A1 $k^3+2k$	✓ 1
M1 equate to 12, $k^3+2k-12=0$	✓ 1
A1 $k=2$	✓ 1
R1 $(k-2)(k^2+2k+6)=0$ and the quadratic has discriminant $4-24<0$, so no other real solution	✓ 1

Common error: After finding one root, check whether the cubic has any others; here the discriminant of the remaining quadratic is negative.

+ Add to test✓ In your test255IBMA3S-5-18r44-115.4 Tangents and normals Paper 1Medium10 marks

No calculator

Let $f(x)=x^3-6x^2+9x$.

(a)

Find $f'(x)$.

(2)

(b)

Find the $x$-coordinates of the stationary points of the graph of $f$.

(2)

(c)

Find the equation of the tangent to the graph of $f$ at the point where $x=4$.

(3)

(d)

The normal to the graph at the point where $x=4$ meets the $x$-axis at the point Q. Find the $x$-coordinate of Q.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate	✓ 1
A1 $3x^2-12x+9$	✓ 1
Part (b)
M1 set $3x^2-12x+9=0$	✓ 1
A1 $x=1$ and $x=3$	✓ 1
Part (c)
A1 $f(4)=4$	✓ 1
A1 $f'(4)=9$	✓ 1
A1 $y=9x-32$	✓ 1
Part (d)
M1 gradient of normal $-\frac{1}{9}$	✓ 1
M1 $y-4=-\frac{1}{9}(x-4)$ with $y=0$	✓ 1
A1 $x=40$	✓ 1

Common error: Find the value of $f(4)$ from the original function, not from $f'(x)$.

+ Add to test✓ In your test256AASL5-q00145.2 Increasing and decreasing functions Paper 1Hard10 marks

Let $f(x)=x^{3}-kx^{2}+12x+1$, where $k\in\mathbb{R}$.

(a)

Find $f'(x)$.

(1)

(b)

Find the set of values of $k$ for which $f'(x)\gt 0$ for all $x\in\mathbb{R}$.

(4)

(c)

Let $k=6$. Find the coordinates of the point on the graph of $f$ where the gradient is zero.

(2)

(d)

Let $k=7$. Find the interval on which $f$ is decreasing. Give the endpoints in the form $\dfrac{a\pm\sqrt{b}}{3}$, where $a,b\in\mathbb{Z}$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$3x^{2}-2kx+12$	A1
(b)
recognizing that the quadratic $f'(x)$ has no real roots: discriminant $\lt 0$	M1
$(-2k)^{2}-4(3)(12)\lt 0$, i.e. $4k^{2}-144\lt 0$	A1
$k^{2}\lt 36$ or critical values $k=\pm6$	M1
$-6\lt k\lt 6$	A1	Award A0 for $-6\le k\le 6$ (when $k=\pm6$, $f'$ has a zero).
(c)
$3x^{2}-12x+12=3(x-2)^{2}=0\Rightarrow x=2$	(M1)
$(2,\,9)$	A1	$f(2)=8-24+24+1=9$
(d)
$3x^{2}-14x+12\lt 0$, attempting the quadratic formula	M1
$x=\dfrac{14\pm\sqrt{52}}{6}=\dfrac{7\pm\sqrt{13}}{3}$	A1
$\dfrac{7-\sqrt{13}}{3}\lt x\lt \dfrac{7+\sqrt{13}}{3}$	A1	Accept non-strict inequalities.

Answers: (a) $3x^{2}-2kx+12$ · (b) $-6\lt k\lt 6$ · (c) $(2,\,9)$ · (d) $\frac{7-\sqrt{13}}{3}\lt x\lt\frac{7+\sqrt{13}}{3}$

+ Add to test✓ In your test257AASL5-q00155.2 Increasing and decreasing functions Paper 1Hard9 marks

A function $f$ has derivative $f'(x)=(x+2)(x-1)(x-5)$, for $x\in\mathbb{R}$.

(a)

Find the intervals on which $f$ is decreasing.

(3)

(b)

Show that $f'(x)=x^{3}-4x^{2}-7x+10$, and hence find an expression for $f(x)$ in terms of $x$ and a constant of integration $c$.

(3)

(c)

The graph of $f$ has two local minimum points. Determine which of these is the lower point.

(3)

Show mark scheme

Working	Mark	Notes
(a)
considering the sign of $f'(x)$ in each interval (sign diagram or graph of the cubic)	M1
$x\lt -2$	A1
$1\lt x\lt 5$	A1	Accept non-strict inequalities.
(b)
$(x^{2}+x-2)(x-5)=x^{3}-5x^{2}+x^{2}-5x-2x+10$, so $f'(x)=x^{3}-4x^{2}-7x+10$	A1
attempting to anti-differentiate (power increased by one in at least two terms)	M1
$f(x)=\dfrac{x^{4}}{4}-\dfrac{4x^{3}}{3}-\dfrac{7x^{2}}{2}+10x+c$	A1
(c)
local minima at $x=-2$ and $x=5$ ($f'$ changes from negative to positive)	R1
$f(-2)=-\dfrac{58}{3}+c$ and $f(5)=-\dfrac{575}{12}+c$	A1	Or $f(5)-f(-2)=-\dfrac{343}{12}$.
$f(5)\lt f(-2)$, so the local minimum at $x=5$ is lower	A1	Award A1 only if correct values (or difference) are seen.

Answers: (a) $x\lt -2$, $1\lt x\lt 5$ · (b) $\frac{x^{4}}{4}-\frac{4x^{3}}{3}-\frac{7x^{2}}{2}+10x+c$ · (c) the minimum at $x=5$

+ Add to test✓ In your test258AASL5-q00235.3 Differentiating polynomials and integer powers Paper 1Hard9 marks

Let $f(x)=ax^{n}$, where $a\neq0$, $n\in\mathbb{Z}$ and $x\gt 0$. It is given that $f(2)=24$ and $f'(2)=-36$.

(a)

Show that $x\,f'(x)=n\,f(x)$ for all $x\gt 0$.

(2)

(b)

Hence find the value of $n$.

(2)

(c)

Find the value of $a$.

(2)

(d)

Find the value of $x$ for which $f'(x)=-\dfrac{9}{4}$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$f'(x)=anx^{n-1}$	M1
$x\,f'(x)=anx^{n}=n\,(ax^{n})$	A1
$=n\,f(x)$	AG
(b)
$2(-36)=24n$	M1	Must use the result from (a) with $x=2$.
$n=-3$	A1
(c)
$a(2)^{-3}=24$	(M1)
$a=192$	A1
(d)
$f'(x)=-576x^{-4}=-\dfrac{9}{4}$	M1	FT from their $a$ and $n$.
$x^{4}=256$	A1
$x=4$	A1	Award A0 if $x=-4$ is also given ($x\gt 0$).

Answers: (b) $n=-3$ · (c) $a=192$ · (d) $x=4$

+ Add to test✓ In your test259AASL5-q00275.4 Tangents and normals Paper 1Hard7 marks

The point $\text{A}(a,\,a^{2})$ lies on the curve $y=x^{2}$.

Diagram not to scale

(a)

Show that the equation of the tangent to the curve at A is $y=2ax-a^{2}$.

(2)

(b)

The tangent at A passes through the point $(1,\,-3)$. Show that $a^{2}-2a-3=0$.

(1)

(c)

Hence find the equations of the two tangents to $y=x^{2}$ that pass through the point $(1,\,-3)$.

(4)

Show mark scheme

Working	Mark	Notes
(a)
gradient $=2a$ and $y-a^{2}=2a(x-a)$	M1
$y=2ax-2a^{2}+a^{2}$	A1
$y=2ax-a^{2}$	AG
(b)
$-3=2a(1)-a^{2}$, so $a^{2}-2a-3=0$	A1
(c)
$(a-3)(a+1)=0$	M1
$a=3,\ a=-1$	A1
$y=6x-9$	A1
$y=-2x-1$	A1

Answers: (c) $y=6x-9$ and $y=-2x-1$

+ Add to test✓ In your test260AASL5-q00325.4 Tangents and normals Paper 1Hard14 marks

Let $f(x)=x^{3}-6x^{2}+9x+1$, for $x\in\mathbb{R}$.

(a)

Find $f'(x)$.

(2)

(b)

Find the intervals on which $f$ is increasing and the interval on which $f$ is decreasing.

(4)

(c)(i)

The line $L$ is the tangent to the graph of $f$ at the point where $x=0$.

Find the equation of $L$.

(2)

(c)(ii)

$L$ meets the graph of $f$ again at the point B. Find the coordinates of B.

(3)

(d)

Find the equation of the normal to the graph of $f$ at the point where $x=2$. Give your answer in the form $y=mx+c$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$3x^{2}-12x$	A1	Award for any two correct terms.
$f'(x)=3x^{2}-12x+9$	A1
(b)
$3x^{2}-12x+9=0$	M1
$3(x-1)(x-3)=0\Rightarrow x=1,\ x=3$	A1
increasing for $x\lt 1$ and $x\gt 3$	A1
decreasing for $1\lt x\lt 3$	A1	Accept non-strict inequalities.
(c)(i)
$f(0)=1$ and $f'(0)=9$	(A1)
$y=9x+1$	A1
(c)(ii)
$x^{3}-6x^{2}+9x+1=9x+1$	M1
$x^{2}(x-6)=0\Rightarrow x=6$	A1
$\text{B}(6,\,55)$	A1
(d)
$f(2)=3$ and $f'(2)=-3$	(A1)
gradient of normal $=\dfrac{1}{3}$: $y-3=\dfrac{1}{3}(x-2)$	(M1)
$y=\dfrac{1}{3}x+\dfrac{7}{3}$	A1

Answers: (a) $3x^{2}-12x+9$ · (b) increasing $x\lt1$, $x\gt3$; decreasing $1\lt x\lt3$ · (c)(i) $y=9x+1$ (ii) $\text{B}(6,\,55)$ · (d) $y=\frac13x+\frac73$

+ Add to test✓ In your test261AASL5-q00435.6 Differentiation rules Paper 1Hard8 marks

Let $g(x)=\dfrac{\sin x}{2+\cos x}$, for $0\le x\le 2\pi$.

(a)

Show that $g'(x)=\dfrac{2\cos x+1}{(2+\cos x)^{2}}$.

(3)

(b)

Find the $x$-coordinates of the stationary points on the graph of $g$.

(3)

(c)

Find the maximum value of $g(x)$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
attempt to use the quotient rule: $\dfrac{(2+\cos x)\cos x-\sin x(-\sin x)}{(2+\cos x)^{2}}$	M1	Award M1 for an attempt with the terms in the correct order in the numerator.
$\dfrac{2\cos x+\cos^{2}x+\sin^{2}x}{(2+\cos x)^{2}}$	A1
using $\cos^{2}x+\sin^{2}x=1$	A1
$g'(x)=\dfrac{2\cos x+1}{(2+\cos x)^{2}}$	AG
(b)
$2\cos x+1=0\Rightarrow\cos x=-\tfrac12$	M1
$x=\dfrac{2\pi}{3}$	A1
$x=\dfrac{4\pi}{3}$	A1	Award A1A0 for answers in degrees.
(c)
$g\left(\tfrac{2\pi}{3}\right)=\dfrac{\frac{\sqrt3}{2}}{2-\frac12}$	(M1)	Award for substituting their $x$ at which $\sin x\gt 0$. Note $g(0)=g(2\pi)=0$ and $g\left(\tfrac{4\pi}{3}\right)\lt 0$.
$\dfrac{\sqrt3}{3}$	A1	Accept $\dfrac{1}{\sqrt3}$.

Answers: (b) $x=\dfrac{2\pi}{3},\ \dfrac{4\pi}{3}$ · (c) $\dfrac{\sqrt3}{3}$

+ Add to test✓ In your test262AASL5-q00525.6 Differentiation rules Paper 1Hard10 marks

A curve has equation $y=\dfrac{1+\sin x}{\cos x}$, for $-\dfrac{\pi}{2}\lt x\lt\dfrac{\pi}{2}$.

(a)

Show that $\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1}{1-\sin x}$.

(4)

(b)

Find the gradient of the curve at $x=\dfrac{\pi}{6}$.

(1)

(c)

Find the value of $x$ at which the gradient of the curve is $\dfrac23$.

(3)

(d)

Explain why the gradient of the curve is always greater than $\dfrac12$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
attempt to use the quotient rule: $\dfrac{\cos x\cos x-(1+\sin x)(-\sin x)}{\cos^{2}x}$	M1
$\dfrac{\cos^{2}x+\sin x+\sin^{2}x}{\cos^{2}x}=\dfrac{1+\sin x}{\cos^{2}x}$	A1	Using $\sin^{2}x+\cos^{2}x=1$.
$\cos^{2}x=1-\sin^{2}x=(1-\sin x)(1+\sin x)$	M1
$\dfrac{1+\sin x}{(1-\sin x)(1+\sin x)}$, with $1+\sin x\neq0$ cancelled	A1
$\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{1}{1-\sin x}$	AG
(b)
$\dfrac{1}{1-\frac12}=2$	A1
(c)
$\dfrac{1}{1-\sin x}=\dfrac23\Rightarrow1-\sin x=\dfrac32$	M1
$\sin x=-\dfrac12$	A1
$x=-\dfrac{\pi}{6}$	A1	Award A0 if additional values outside the domain are given.
(d)
on the domain $-1\lt\sin x\lt1$, so $0\lt1-\sin x\lt2$	R1
so $\dfrac{1}{1-\sin x}\gt\dfrac12$	R1	Must refer to taking the reciprocal of a positive quantity less than 2.

Answers: (b) $2$ · (c) $x=-\dfrac{\pi}{6}$

+ Add to test✓ In your test263AASL5-q00555.6 Differentiation rules Paper 1Hard14 marks

Let $f(x)=\mathrm{e}^{\sin x}$, for $0\le x\le2\pi$.

(a)

Find $f'(x)$.

(2)

(b)

Find the coordinates of the stationary points on the graph of $f$.

(4)

(c)

Show that $f''(x)=\mathrm{e}^{\sin x}\left(\cos^{2}x-\sin x\right)$.

(2)

(d)

Hence show that, at a point of inflexion, $\sin^{2}x+\sin x-1=0$.

(2)

(e)

Hence find the exact value of $\sin x$ at the points of inflexion, justifying your answer.

(3)

(f)

Write down the number of points of inflexion on the graph of $f$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
attempt to use the chain rule	M1
$f'(x)=\cos x\,\mathrm{e}^{\sin x}$	A1
(b)
$\cos x\,\mathrm{e}^{\sin x}=0\Rightarrow\cos x=0$, since $\mathrm{e}^{\sin x}\gt 0$	M1
$x=\dfrac{\pi}{2},\ \dfrac{3\pi}{2}$	A1
$\left(\dfrac{\pi}{2},\ \mathrm{e}\right)$	A1
$\left(\dfrac{3\pi}{2},\ \mathrm{e}^{-1}\right)$	A1	Accept $\dfrac{1}{\mathrm{e}}$.
(c)
attempt to use the product rule on $\cos x\,\mathrm{e}^{\sin x}$	M1
$-\sin x\,\mathrm{e}^{\sin x}+\cos x\times\cos x\,\mathrm{e}^{\sin x}$	A1
$f''(x)=\mathrm{e}^{\sin x}\left(\cos^{2}x-\sin x\right)$	AG
(d)
$f''(x)=0\Rightarrow\cos^{2}x-\sin x=0$ and use of $\cos^{2}x=1-\sin^{2}x$	M1
$1-\sin^{2}x-\sin x=0$	A1
$\sin^{2}x+\sin x-1=0$	AG
(e)
attempt to solve the quadratic in $\sin x$, e.g. quadratic formula	M1
$\sin x=\dfrac{-1\pm\sqrt5}{2}$	A1
$\dfrac{-1-\sqrt5}{2}\lt -1$ is impossible, so $\sin x=\dfrac{\sqrt5-1}{2}$	R1
(f)
$2$	A1	$\sin x=0.618\ldots$ has two solutions in $[0,\,2\pi]$, both in $\left(0,\,\pi\right)$, and $f''$ changes sign at each.

Answers: (a) $\cos x\,\mathrm{e}^{\sin x}$ · (b) $\left(\tfrac{\pi}{2},\mathrm{e}\right)$, $\left(\tfrac{3\pi}{2},\mathrm{e}^{-1}\right)$ · (e) $\sin x=\dfrac{\sqrt5-1}{2}$ · (f) 2

+ Add to test✓ In your test264AASL5-q00635.7 The second derivative and graphs Paper 1Hard8 marks

Let $f(x)=\mathrm{e}^{x}+\mathrm{e}^{-2x}$, for $x\in\mathbb{R}$.

(a)

Find $f'(x)$ and $f''(x)$.

(2)

(b)

Show that the graph of $f$ is concave-up for all values of $x$.

(1)

(c)

Hence explain why the graph of $f$ has at most one stationary point.

(1)

(d)

Find the exact $x$-coordinate of the stationary point.

(2)

(e)

Show that the minimum value of $f$ is $\dfrac{3}{\sqrt[3]{4}}$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$f'(x)=\mathrm{e}^{x}-2\mathrm{e}^{-2x}$	A1
$f''(x)=\mathrm{e}^{x}+4\mathrm{e}^{-2x}$	A1
(b)
$\mathrm{e}^{x}\gt 0$ and $4\mathrm{e}^{-2x}\gt 0$, so $f''(x)\gt 0$ for all $x$	R1
concave-up for all $x$	AG
(c)
$f''\gt 0$ so $f'$ is (strictly) increasing; hence $f'(x)=0$ for at most one value of $x$	R1
(d)
$\mathrm{e}^{x}=2\mathrm{e}^{-2x}\Rightarrow\mathrm{e}^{3x}=2$	M1
$x=\dfrac{\ln2}{3}$	A1	Accept $\ln\sqrt[3]{2}$.
(e)
$f\left(\tfrac{\ln2}{3}\right)=2^{\frac13}+2^{-\frac23}$	M1
$=2^{-\frac23}(2+1)=\dfrac{3}{2^{\frac23}}$	A1	Minimum since $f''\gt 0$.
$\dfrac{3}{\sqrt[3]{4}}$	AG

Answers: (a) $f'=\mathrm{e}^{x}-2\mathrm{e}^{-2x}$, $f''=\mathrm{e}^{x}+4\mathrm{e}^{-2x}$ · (d) $x=\dfrac{\ln2}{3}$

+ Add to test✓ In your test265AASL5-q00645.6 Differentiation rules Paper 1Hard10 marks

Let $f(x)=\sin x+kx^{2}$, where $k\in\mathbb{R}$ and $x\in\mathbb{R}$.

(a)

Find $f'(x)$ and $f''(x)$.

(2)

(b)

Find the set of values of $k$ for which $f''(x)\gt 0$ for all values of $x$.

(3)

(c)

Let $k=\dfrac14$. Find the $x$-coordinates of the points of inflexion on the graph of $f$ for $0\le x\le2\pi$.

(3)

(d)

Let $k=\dfrac14$. Find the exact gradient of the graph of $f$ at the point of inflexion with the larger $x$-coordinate.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$f'(x)=\cos x+2kx$	A1
$f''(x)=-\sin x+2k$	A1
(b)
$2k-\sin x\gt 0$ for all $x$	M1
the greatest value of $\sin x$ is 1, so need $2k\gt 1$	R1
$k\gt\dfrac12$	A1	Award A0 for $k\ge\tfrac12$: when $k=\tfrac12$, $f''\left(\tfrac{\pi}{2}\right)=0$.
(c)
$\tfrac12-\sin x=0\Rightarrow\sin x=\tfrac12$	M1
$x=\dfrac{\pi}{6}$	A1	$f''$ changes sign at each value since $\cos x\neq0$ there.
$x=\dfrac{5\pi}{6}$	A1
(d)
$f'\left(\tfrac{5\pi}{6}\right)=\cos\left(\tfrac{5\pi}{6}\right)+\tfrac12\times\tfrac{5\pi}{6}$	(M1)
$\dfrac{5\pi}{12}-\dfrac{\sqrt3}{2}$	A1

Answers: (b) $k\gt\tfrac12$ · (c) $x=\dfrac{\pi}{6},\ \dfrac{5\pi}{6}$ · (d) $\dfrac{5\pi}{12}-\dfrac{\sqrt3}{2}$

+ Add to test✓ In your test266AASL5-q00725.8 Stationary points, inflexion and optimization Paper 1Hard11 marks

The point $\text{P}\left(a,\,\mathrm{e}^{-a}\right)$, where $a\ge0$, lies on the curve $y=\mathrm{e}^{-x}$. The line L is the tangent to the curve at P. The triangle formed by L and the coordinate axes is shaded in the diagram.

Diagram not to scale

(a)

Show that the equation of L is $y=-\mathrm{e}^{-a}x+(a+1)\mathrm{e}^{-a}$.

(2)

(b)

Find, in terms of $a$, the coordinates of the points where L meets the $x$-axis and the $y$-axis.

(2)

(c)

Hence show that the area of the shaded triangle is $T=\dfrac12(a+1)^{2}\mathrm{e}^{-a}$.

(1)

(d)

Find $\dfrac{\mathrm{d}T}{\mathrm{d}a}$.

(2)

(e)

Find the exact maximum value of $T$, justifying that it is a maximum.

(4)

Show mark scheme

Working	Mark	Notes
(a)
gradient of L $=-\mathrm{e}^{-a}$	A1
$y-\mathrm{e}^{-a}=-\mathrm{e}^{-a}(x-a)$	M1
$y=-\mathrm{e}^{-a}x+(a+1)\mathrm{e}^{-a}$	AG
(b)
$\big(a+1,\ 0\big)$	A1
$\big(0,\ (a+1)\mathrm{e}^{-a}\big)$	A1
(c)
$T=\tfrac12\times(a+1)\times(a+1)\mathrm{e}^{-a}$	A1
$T=\dfrac12(a+1)^{2}\mathrm{e}^{-a}$	AG
(d)
attempt to use the product rule	M1
$\dfrac{\mathrm{d}T}{\mathrm{d}a}=(a+1)\mathrm{e}^{-a}-\tfrac12(a+1)^{2}\mathrm{e}^{-a}$	A1	Accept $\tfrac12(a+1)(1-a)\mathrm{e}^{-a}$.
(e)
$\tfrac12(a+1)(1-a)\mathrm{e}^{-a}=0$	M1
$a=1$	A1	Reject $a=-1$ since $a\ge0$.
$\dfrac{\mathrm{d}T}{\mathrm{d}a}\gt 0$ for $0\le a\lt1$ and $\dfrac{\mathrm{d}T}{\mathrm{d}a}\lt 0$ for $a\gt1$, so a maximum	R1
$T=\dfrac{2}{\mathrm{e}}$	A1	$\tfrac12(2)^{2}\mathrm{e}^{-1}$.

Answers: (b) $(a+1,0)$, $\left(0,(a+1)\mathrm{e}^{-a}\right)$ · (d) $\tfrac12(a+1)(1-a)\mathrm{e}^{-a}$ · (e) $\dfrac{2}{\mathrm{e}}$ at $a=1$

+ Add to test✓ In your test267AASL5-q00955.11 Definite integrals and areas Paper 1Hard8 marks

Let $f(x)=\dfrac{\ln x}{x}$, for $x\gt 0$.

Diagram not to scale

(a)

Show that $\dfrac{\mathrm{d}}{\mathrm{d}x}\big((\ln x)^{2}\big)=2f(x)$.

(2)

(b)

Hence find $\displaystyle\int_{1}^{\mathrm{e}^{2}}f(x)\,\mathrm{d}x$.

(2)

(c)

Find the area of the region enclosed by the graph of $f$, the $x$-axis and the lines $x=\mathrm{e}^{-1}$ and $x=\mathrm{e}$.

(4)

Show mark scheme

Working	Mark	Notes
(a)
attempt to use the chain rule: $2\ln x\times\dfrac{\mathrm{d}}{\mathrm{d}x}(\ln x)$	M1
$2\ln x\times\dfrac{1}{x}=\dfrac{2\ln x}{x}$	A1
$=2f(x)$	AG
(b)
$\left[\tfrac12(\ln x)^{2}\right]_{1}^{\mathrm{e}^{2}}=\tfrac12(2)^{2}-\tfrac12(0)^{2}$	M1
$2$	A1
(c)
$f(x)\lt 0$ for $\mathrm{e}^{-1}\lt x\lt 1$ and $f(x)\gt 0$ for $1\lt x\lt\mathrm{e}$, so the integral is split at $x=1$	R1
$\displaystyle\int_{\mathrm{e}^{-1}}^{1}f(x)\,\mathrm{d}x=0-\tfrac12(-1)^{2}=-\tfrac12$	A1
$\displaystyle\int_{1}^{\mathrm{e}}f(x)\,\mathrm{d}x=\tfrac12$	A1
area $=\tfrac12+\tfrac12=1$	A1	Award A0 for an answer of 0 (from $\int_{\mathrm{e}^{-1}}^{\mathrm{e}}f$).

Answers: (b) $2$ · (c) $1$

+ Add to test✓ In your test268AASL5-q01035.10 Indefinite integrals and reverse chain rule Paper 1Hard8 marks

(a)

Show that $\sin^{3}x=\sin x-\sin x\cos^{2}x$.

(1)

(b)

Hence find $\displaystyle\int\sin^{3}x\,\mathrm{d}x$.

(3)

(c)

Find the exact value of $\displaystyle\int_{0}^{\pi}\sin^{3}x\,\mathrm{d}x$.

(2)

(d)

Without further integration, write down the value of $\displaystyle\int_{0}^{2\pi}\sin^{3}x\,\mathrm{d}x$. Justify your answer.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\sin^{3}x=\sin x\left(1-\cos^{2}x\right)$	M1	Use of $\sin^{2}x=1-\cos^{2}x$.
$=\sin x-\sin x\cos^{2}x$	AG
(b)
$\displaystyle\int\sin x\,\mathrm{d}x=-\cos x$	A1
recognizing $\sin x\cos^{2}x$ as a multiple of $f'(x)f(x)^{2}$ with $f(x)=\cos x$, or $u=\cos x$	M1
$-\cos x+\dfrac{1}{3}\cos^{3}x+C$	A1
(c)
$\left(1-\tfrac13\right)-\left(-1+\tfrac13\right)$	(M1)
$\dfrac{4}{3}$	A1
(d)
$0$	A1
$\sin^{3}(x+\pi)=-\sin^{3}x$, so the part of the graph from $\pi$ to $2\pi$ is below the axis with the same area as the part from $0$ to $\pi$	R1	Accept $\int_{\pi}^{2\pi}\sin^{3}x\,\mathrm{d}x=-\tfrac43$ found by symmetry or by their (b). Do not award A1R0.

Answers: (b) $-\cos x+\tfrac13\cos^{3}x+C$ · (c) $\tfrac43$ · (d) 0

+ Add to test✓ In your test269AASL5-q01055.10 Indefinite integrals and reverse chain rule Paper 1Hard15 marks

Let $f(x)=2x\,\mathrm{e}^{-x^{2}}$, for $x\in\mathbb{R}$.

(a)

Show that $f'(x)=2\left(1-2x^{2}\right)\mathrm{e}^{-x^{2}}$.

(2)

(b)

Find the exact coordinates of the local maximum point on the graph of $f$.

(3)

(c)

Find $\displaystyle\int f(x)\,\mathrm{d}x$.

(2)

(d)

Let $A(k)$ be the area of the region enclosed by the graph of $f$, the $x$-axis and the line $x=k$, where $k\gt 0$. Show that $A(k)=1-\mathrm{e}^{-k^{2}}$.

(2)

(e)

Find the exact value of $k$ for which $A(k)=\dfrac12$.

(2)

(f)

Explain why $A(k)\lt 1$ for all $k\gt 0$.

(1)

(g)

Hence find the exact value of $\displaystyle\int_{-1}^{1}|f(x)|\,\mathrm{d}x$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
attempt to use the product rule, with $\dfrac{\mathrm{d}}{\mathrm{d}x}\mathrm{e}^{-x^{2}}=-2x\,\mathrm{e}^{-x^{2}}$	M1
$2\mathrm{e}^{-x^{2}}-4x^{2}\mathrm{e}^{-x^{2}}$	A1
$=2\left(1-2x^{2}\right)\mathrm{e}^{-x^{2}}$	AG
(b)
$1-2x^{2}=0$	M1	Since $\mathrm{e}^{-x^{2}}\gt 0$.
$x=\dfrac{1}{\sqrt2}$	A1	Must reject $x=-\tfrac{1}{\sqrt2}$ (a minimum), or justify the maximum, e.g. sign of $f'$.
$y=\sqrt2\,\mathrm{e}^{-\frac12}$	A1	Accept $\dfrac{2}{\sqrt2}\mathrm{e}^{-\frac12}$ or $\sqrt{\dfrac{2}{\mathrm{e}}}$.
(c)
recognizing $-2x$ as the derivative of $-x^{2}$, or $u=-x^{2}$	(M1)
$-\mathrm{e}^{-x^{2}}+C$	A1
(d)
$\displaystyle\int_{0}^{k}2x\,\mathrm{e}^{-x^{2}}\,\mathrm{d}x$	M1	Must note or use $f(x)\ge0$ for $x\ge0$.
$\left[-\mathrm{e}^{-x^{2}}\right]_{0}^{k}=-\mathrm{e}^{-k^{2}}+1$	A1
$A(k)=1-\mathrm{e}^{-k^{2}}$	AG
(e)
$\mathrm{e}^{-k^{2}}=\tfrac12\Rightarrow k^{2}=\ln 2$	(M1)
$k=\sqrt{\ln 2}$	A1	Must reject the negative root.
(f)
$\mathrm{e}^{-k^{2}}\gt 0$ for all $k$, so $1-\mathrm{e}^{-k^{2}}\lt 1$	R1
(g)
$f(-x)=-f(x)$, so the area from $-1$ to $0$ equals the area from $0$ to $1$: $2A(1)$	M1
$A(1)=1-\mathrm{e}^{-1}$	A1
$2-2\mathrm{e}^{-1}$	A1	Accept $2-\tfrac{2}{\mathrm{e}}$. Award M0A0A0 for 0 (the signed integral).

Answers: (b) $\left(\tfrac{1}{\sqrt2},\ \sqrt2\,\mathrm{e}^{-\frac12}\right)$ · (c) $-\mathrm{e}^{-x^{2}}+C$ · (e) $k=\sqrt{\ln 2}$ · (g) $2-2\mathrm{e}^{-1}$

+ Add to test✓ In your test270AASL5-q01075.11 Definite integrals and areas Paper 1Hard7 marks

The diagram shows the graphs of $y=\sin2x$ and $y=\cos x$, for $0\le x\le\dfrac{\pi}{2}$. The shaded region is enclosed by the two graphs.

(a)

Show that the graphs intersect where $x=\dfrac{\pi}{6}$ and $x=\dfrac{\pi}{2}$.

(3)

(b)

Find the exact area of the shaded region.

(4)

Show mark scheme

Working	Mark	Notes
(a)
$2\sin x\cos x=\cos x$	M1	Use of the double-angle identity.
$\cos x(2\sin x-1)=0$	M1	Award M0 for dividing by $\cos x$ without considering $\cos x=0$.
$\cos x=0$ or $\sin x=\tfrac12$	A1
$x=\dfrac{\pi}{2}$ or $x=\dfrac{\pi}{6}$	AG
(b)
$\displaystyle\int_{\pi/6}^{\pi/2}(\sin2x-\cos x)\,\mathrm{d}x$	M1	Accept the reverse order of subtraction with an absolute value taken at the end.
$\left[-\tfrac12\cos2x-\sin x\right]$	A1	One mark for both terms correct.
$\left(-\tfrac12\cos\pi-1\right)-\left(-\tfrac12\cos\tfrac{\pi}{3}-\tfrac12\right)=-\tfrac12-\left(-\tfrac34\right)$	M1	Substituting both limits.
$\dfrac14$	A1

Answers: (b) $\tfrac14$

+ Add to test✓ In your test271AASL5-q01115.11 Definite integrals and areas Paper 1Hard10 marks

Let $f'(x)=\tfrac12x(x-3)(x-5)=\tfrac12\left(x^{3}-8x^{2}+15x\right)$, for $0\le x\le5$. The diagram shows the graph of $y=f'(x)$. Region A is enclosed by the graph of $f'$ and the $x$-axis for $0\le x\le3$, and region B is enclosed by the graph of $f'$ and the $x$-axis for $3\le x\le5$.

It is given that $f(0)=2$ and that the area of region B is $\dfrac{8}{3}$.

(a)

Find the $x$-coordinate of the local maximum point on the graph of $f$. Justify your answer.

(2)

(b)

Show that the area of region A is $\dfrac{63}{8}$.

(3)

(c)

Find the value of $f(3)$ and the value of $f(5)$.

(3)

(d)

Find the least value of $f(x)$ for $0\le x\le5$. Justify your answer.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$x=3$	A1
$f'$ changes sign from positive to negative at $x=3$	R1	Do not award A1R0.
(b)
$\displaystyle\int_{0}^{3}\tfrac12\left(x^{3}-8x^{2}+15x\right)\mathrm{d}x$	M1
$\tfrac12\left[\dfrac{x^{4}}{4}-\dfrac{8x^{3}}{3}+\dfrac{15x^{2}}{2}\right]_{0}^{3}$	A1
$\tfrac12\left(\dfrac{81}{4}-72+\dfrac{135}{2}\right)=\tfrac12\left(\dfrac{63}{4}\right)$	A1
$=\dfrac{63}{8}$	AG
(c)
$f(3)=f(0)+\displaystyle\int_{0}^{3}f'(x)\,\mathrm{d}x$, and $\displaystyle\int_{3}^{5}f'(x)\,\mathrm{d}x=-\tfrac83$	M1	Recognizing that the area of B gives a negative integral.
$f(3)=2+\dfrac{63}{8}=\dfrac{79}{8}$	A1
$f(5)=\dfrac{79}{8}-\dfrac{8}{3}=\dfrac{173}{24}$	A1	Award A0 for $\tfrac{301}{24}$ (adding the area of B).
(d)
$f$ increases on $0\lt x\lt 3$ and decreases on $3\lt x\lt 5$, so the least value is at an end point; $f(5)=\tfrac{173}{24}\gt 2=f(0)$	R1	FT from (c).
$2$	A1	Do not award A1R0.

Answers: (a) $x=3$ · (c) $f(3)=\tfrac{79}{8},\ f(5)=\tfrac{173}{24}$ · (d) 2

+ Add to test✓ In your test272AASL5-q01135.11 Definite integrals and areas Paper 1Hard9 marks

The line $y=kx$, where $k\gt 0$, and the curve $y=x^{2}$ enclose a region R.

Diagram not to scale

(a)

Find, in terms of $k$, the $x$-coordinates of the points of intersection of the line and the curve.

(2)

(b)

Show that the area of R is $\dfrac{k^{3}}{6}$.

(3)

(c)

Given that the area of R is 36, find the value of $k$.

(2)

(d)

For this value of $k$, show that the line $x=3$ divides R into two regions of equal area.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$x^{2}=kx\Rightarrow x(x-k)=0$	(M1)
$x=0,\ x=k$	A1
(b)
$\displaystyle\int_{0}^{k}\left(kx-x^{2}\right)\mathrm{d}x$	M1	Correct expression with limits.
$\left[\dfrac{kx^{2}}{2}-\dfrac{x^{3}}{3}\right]_{0}^{k}$	A1
$\dfrac{k^{3}}{2}-\dfrac{k^{3}}{3}$	A1
$=\dfrac{k^{3}}{6}$	AG
(c)
$k^{3}=216$	(M1)
$k=6$	A1
(d)
$\displaystyle\int_{0}^{3}\left(6x-x^{2}\right)\mathrm{d}x=\left[3x^{2}-\dfrac{x^{3}}{3}\right]_{0}^{3}$	M1	Or a symmetry argument: $6x-x^{2}=9-(x-3)^{2}$ is symmetric about $x=3$.
$27-9=18=\tfrac12(36)$	A1
so $x=3$ divides R into two regions of equal area	AG

Answers: (a) $x=0,\ x=k$ · (c) $k=6$

+ Add to test✓ In your test273AASL5-q01185.11 Definite integrals and areas Paper 1Hard16 marks

Let $f(x)=\dfrac{4x}{x^{2}+1}$, for $x\in\mathbb{R}$. The diagram shows part of the graph of $f$ and the line L, which is the tangent to the graph of $f$ at the origin.

(a)

Show that $f'(x)=\dfrac{4(1-x^{2})}{(x^{2}+1)^{2}}$.

(3)

(b)

Find the coordinates of the local maximum point and of the local minimum point on the graph of $f$.

(3)

(c)

The second derivative is $f''(x)=\dfrac{8x(x^{2}-3)}{(x^{2}+1)^{3}}$. Find the $x$-coordinates of the points of inflexion on the graph of $f$. Justify your answer.

(3)

(d)

Find the equation of L.

(2)

(e)

The shaded region is enclosed by L, the graph of $f$ and the line $x=1$. Find the exact area of the shaded region.

(5)

Show mark scheme

Working	Mark	Notes
(a)
attempt to use the quotient rule	M1
$\dfrac{4(x^{2}+1)-4x(2x)}{(x^{2}+1)^{2}}$	A1
$\dfrac{4x^{2}+4-8x^{2}}{(x^{2}+1)^{2}}=\dfrac{4-4x^{2}}{(x^{2}+1)^{2}}$	A1
$f'(x)=\dfrac{4(1-x^{2})}{(x^{2}+1)^{2}}$	AG
(b)
$1-x^{2}=0\Rightarrow x=\pm1$	M1
local maximum $(1,\,2)$	A1
local minimum $(-1,\,-2)$	A1
(c)
$8x(x^{2}-3)=0$	M1
$x=0,\ x=\pm\sqrt3$	A1	All three values required.
each is a single root of the numerator and $(x^{2}+1)^{3}\gt 0$, so $f''$ changes sign at each value	R1	Or a sign diagram for $f''$.
(d)
gradient $f'(0)=4$	A1
$y=4x$	A1
(e)
$\displaystyle\int_{0}^{1}\left(4x-\dfrac{4x}{x^{2}+1}\right)\mathrm{d}x$	M1	Award for the difference of their line and $f$, with limits 0 and 1.
$\int4x\,\mathrm{d}x=2x^{2}$	A1
recognizing $2x$ as the derivative of $x^{2}+1$ (inspection or substitution $u=x^{2}+1$)	M1
$2\ln(x^{2}+1)$	A1
$\left[2x^{2}-2\ln(x^{2}+1)\right]_{0}^{1}=2-2\ln2$	A1	Accept $2-\ln4$.

Answers: (b) max $(1,2)$, min $(-1,-2)$ · (c) $x=0,\ \pm\sqrt3$ · (d) $y=4x$ · (e) $2-2\ln2$

+ Add to test✓ In your test274AASL5-q01195.11 Definite integrals and areas Paper 1Hard16 marks

Let $f(x)=x\sqrt{9-x^{2}}$, for $0\le x\le3$. The diagram shows the graph of $f$ and the line $y=x$. The graph of $f$ has a maximum point at M. The line meets the graph of $f$ at the origin O and at the point P. The shaded region S is enclosed by the graph of $f$ and the line.

(a)

Show that $f'(x)=\dfrac{9-2x^{2}}{\sqrt{9-x^{2}}}$, for $0\le x\lt 3$.

(3)

(b)

Find the exact coordinates of M.

(3)

(c)

Find $\displaystyle\int x\sqrt{9-x^{2}}\,\mathrm{d}x$.

(3)

(d)

Hence find the area of the region enclosed by the graph of $f$ and the $x$-axis.

(2)

(e)

Find the exact $x$-coordinate of P.

(2)

(f)

Find the area of S.

(3)

Show mark scheme

Working	Mark	Notes
(a)
attempt to use the product rule	M1
$\dfrac{\mathrm{d}}{\mathrm{d}x}\sqrt{9-x^{2}}=\dfrac{-x}{\sqrt{9-x^{2}}}$	A1	Correct use of the chain rule.
$\sqrt{9-x^{2}}-\dfrac{x^{2}}{\sqrt{9-x^{2}}}=\dfrac{(9-x^{2})-x^{2}}{\sqrt{9-x^{2}}}$	A1	Common denominator must be seen.
$f'(x)=\dfrac{9-2x^{2}}{\sqrt{9-x^{2}}}$	AG
(b)
$9-2x^{2}=0$	M1
$x=\dfrac{3}{\sqrt2}$	A1	Accept $\tfrac{3\sqrt2}{2}$. Must reject the negative root.
$y=\dfrac{3}{\sqrt2}\sqrt{\dfrac92}=\dfrac92$	A1
(c)
recognizing $x$ as a multiple of the derivative of $9-x^{2}$, or $u=9-x^{2}$, $\mathrm{d}u=-2x\,\mathrm{d}x$	M1
$\left(9-x^{2}\right)^{\frac32}$ seen, or $\displaystyle\int-\tfrac12u^{\frac12}\,\mathrm{d}u$	A1
$-\dfrac13\left(9-x^{2}\right)^{\frac32}+C$	A1
(d)
$\left[-\tfrac13\left(9-x^{2}\right)^{\frac32}\right]_{0}^{3}=0-\left(-\tfrac13\times27\right)$	(M1)	FT from (c).
$9$	A1
(e)
$x\sqrt{9-x^{2}}=x\Rightarrow\sqrt{9-x^{2}}=1$ (for $x\neq0$)	(M1)
$x=2\sqrt2$	A1	Accept $\sqrt8$.
(f)
$\displaystyle\int_{0}^{2\sqrt2}\left(x\sqrt{9-x^{2}}-x\right)\mathrm{d}x$	M1	Correct expression with limits; FT their (e).
$\left[-\tfrac13\left(9-x^{2}\right)^{\frac32}-\tfrac12x^{2}\right]_{0}^{2\sqrt2}=\left(-\tfrac13-4\right)-(-9)$	A1
$\dfrac{14}{3}$	A1

Answers: (b) $\mathrm{M}\left(\tfrac{3}{\sqrt2},\ \tfrac92\right)$ · (c) $-\tfrac13(9-x^{2})^{\frac32}+C$ · (d) 9 · (e) $2\sqrt2$ · (f) $\tfrac{14}{3}$

+ Add to test✓ In your test275IBMA2S-052-00-115.6 Differentiation rules Paper 1Hard15 marks

No calculator

Let $f(x) = \dfrac{\ln x}{x^2}$, for $x > 0$.

(a)

Show that $f'(x) = \dfrac{1 - 2\ln x}{x^3}$.

(3)

(b)

Find the exact coordinates of the local maximum point of the graph of $f$.

(3)

(c)

The graph of $f$ crosses the $x$-axis at the point A. Find the equation of the tangent to the graph at A.

(2)

(d)

Show that $f''(x) = \dfrac{6\ln x - 5}{x^4}$.

(4)

(e)

Find the exact $x$-coordinate of the point of inflection of the graph of $f$, justifying your answer.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts the quotient rule (or product rule on $x^{-2}\ln x$) A1 $\frac{x^2 \cdot \frac{1}{x} - 2x\ln x}{x^4}$ A1 divides through by $x$ to obtain the given result	✓ 3
Part (b)
M1 sets $1 - 2\ln x = 0$ A1 $x = e^{1/2} = \sqrt{e}$ A1 $y = \frac{1/2}{e} = \frac{1}{2e}$	✓ 3
Part (c)
A1 A is $(1, 0)$ and $f' = 1$ A1 $y = x - 1$	✓ 3
Part (d)
M1 attempts the quotient rule on $f'(x)$ A1 derivative of numerator is $-\frac{2}{x}$ A1 $\frac{-\frac{2}{x} \cdot x^3 - (1 - 2\ln x)(3x^2)}{x^6}$ A1 simplifies $\frac{-2x^2 - 3x^2 + 6x^2\ln x}{x^6}$ to the given result	✓ 4
Part (e)
M1 sets $6\ln x - 5 = 0$ A1 $x = e^{5/6}$ R1 since $x^4 > 0$ and $\ln x$ is increasing, $f''(x) < 0$ for $x < e^{5/6}$ and $f''(x) > 0$ for $x > e^{5/6}$, so the concavity changes	✓ 3

Exam tip: For a point of inflection, $f''(x) = 0$ alone is not enough: you must show that $f''(x)$ changes sign there to earn the reasoning mark.

+ Add to test✓ In your test276IBMA2S-052-04-115.10 Indefinite integrals and reverse chain rule Paper 1Hard13 marks

No calculator

This question builds up to integrating $\sin^{3}x$ without a formula booklet result.

(a)

Find $\int \sin x\cos^{2}x\,dx$.

(3)

(b)

By writing $\sin^{3}x=\sin x\,(1-\cos^{2}x)$, show that $\int\sin^{3}x\,dx=-\cos x+\frac{1}{3}\cos^{3}x+c$.

(3)

(c)

A function $g$ has derivative $g'(x)=\sin^{3}x$ and $g\left(\frac{\pi}{2}\right)=1$. Find $g(x)$ and hence find the exact value of $g(\pi)$.

(4)

(d)

Find the minimum value of $g(x)$ for $0\leq x\leq\pi$, justifying your answer.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitution $u=\cos x$, $du=-\sin x\,dx$, or equivalent recognition	✓ 1
A1 $-\int u^{2}\,du$	✓ 1
A1 $-\frac{1}{3}\cos^{3}x+c$	✓ 1
Part (b)
M1 use of $\sin^{2}x=1-\cos^{2}x$ to write $\sin x-\sin x\cos^{2}x$	✓ 1
M1 split into $\int\sin x\,dx-\int\sin x\cos^{2}x\,dx$ and use part (a)	✓ 1
A1 $-\cos x-\left(-\frac{1}{3}\cos^{3}x\right)+c=-\cos x+\frac{1}{3}\cos^{3}x+c$ (AG)	✓ 1
Part (c)
M1 substitute $x=\frac{\pi}{2}$: $-0+0+c=1$	✓ 1
A1 $c=1$, so $g(x)=-\cos x+\frac{1}{3}\cos^{3}x+1$	✓ 1
M1 substitute $x=\pi$: $1-\frac{1}{3}+1$	✓ 1
A1 $g(\pi)=\frac{5}{3}$	✓ 1
Part (d)
R1 $g'(x)=\sin^{3}x\geq 0$ for $0\leq x\leq\pi$, so $g$ is increasing on this interval	✓ 1
M1 minimum occurs at $x=0$: $g(0)=-1+\frac{1}{3}+1$	✓ 1
A1 minimum value $\frac{1}{3}$	✓ 1

Common error: Losing the sign: $\int\sin x\,dx=-\cos x$, and subtracting $-\frac{1}{3}\cos^{3}x$ gives $+\frac{1}{3}\cos^{3}x$. Track each minus sign separately.

+ Add to test✓ In your test277IBMA2S-052-05-085.4 Tangents and normals Paper 1Hard14 marks

No calculator

A roller-coaster designer models a section of track by y = f(x), where f(x) = x³ + ax² + bx + c. Here y is the vertical displacement in metres from station level and x is the horizontal distance in tens of metres. The graph of f has a point of inflection at (1, −2) and a stationary point where x = 3.

(a)

Find the value of a.

(2)

(b)

Find the value of b.

(2)

(c)

Find the value of c.

(2)

(d)

Find the coordinates of the other stationary point. Use the second derivative to classify both stationary points.

(4)

(e)

Find the equation of the tangent to the track at the point of inflection. Show that this tangent meets the curve at only one point.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 f''(x) = 6x + 2a and sets f'' = 0 at x = 1	✓ 1
A1 a = −3	✓ 1
Part (b)
M1 f'(x) = 3x² − 6x + b and sets f'(3) = 0	✓ 1
A1 b = −9	✓ 1
Part (c)
M1 substitutes the point of inflection: 1 − 3 − 9 + c = −2	✓ 1
A1 c = 9	✓ 1
Part (d)
M1 f'(x) = 3(x − 3)(x + 1) = 0	✓ 1
A1 (−1, 14)	✓ 1
R1 f''(−1) = −12 < 0, so a local maximum at (−1, 14)	✓ 1
R1 f''(3) = 12 > 0, so a local minimum at (3, −18)	✓ 1
Part (e)
A1 gradient at x = 1 is 3 − 6 − 9 = −12	✓ 1
A1 y = −12x + 10	✓ 1
M1 equates: x³ − 3x² − 9x + 9 = −12x + 10, giving x³ − 3x² + 3x − 1 = 0	✓ 1
R1 (x − 1)³ = 0 has the single solution x = 1, so there is only one meeting point	✓ 1

Exam tip: The point of inflection of a cubic lies exactly halfway between its turning points. Here (14 + (−18))/2 = −2, which is a quick check that a, b and c are right.

+ Add to test✓ In your test278IBMA2S-052-08-115.4 Tangents and normals Paper 1Hard14 marks

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The curve y = f(x), where f(x) = x³ − 3x, passes through the point P(1, −2). The point Q has x-coordinate 1 + h, where h ≠ 0, and also lies on the curve.

Diagram not to scale

(a)

Show that the gradient of the chord PQ is 3h + h².

(3)

(b)

By considering the limit as h → 0, find the gradient of the tangent at P and describe this tangent.

(2)

(c)

Find the possible coordinates of Q when the chord PQ has gradient 4.

(3)

(d)

For one value of h, the chord PQ is also the tangent to the curve at Q. Find this value of h and the coordinates of Q.

(4)

(e)

Find the equation of this tangent in the form y = mx + c.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 gradient = [f(1 + h) − (−2)] ÷ h	✓ 1
A1 f(1 + h) = 1 + 3h + 3h² + h³ − 3 − 3h = −2 + 3h² + h³	✓ 1
A1 (3h² + h³) ÷ h = 3h + h² (AG)	✓ 1
Part (b)
A1 limit = 0, so the gradient of the curve at x = 1 is 0	✓ 1
R1 the tangent at P is horizontal, the line y = −2	✓ 1
Part (c)
M1 sets h² + 3h − 4 = 0	✓ 1
A1 (h + 4)(h − 1) = 0, so h = 1 or h = −4	✓ 1
A1 Q is (2, 2) or (−3, −18)	✓ 1
Part (d)
M1 f′(1 + h) = 3(1 + h)² − 3 = 6h + 3h²	✓ 1
M1 equates this to the chord gradient: 3h + h² = 6h + 3h², giving h(2h + 3) = 0	✓ 1
A1 h = −3/2 (since h ≠ 0)	✓ 1
A1 Q(−1/2, 11/8)	✓ 1
Part (e)
M1 gradient m = 3(−3/2) + 9/4 = −9/4 and uses a point, e.g. y − 11/8 = −9/4(x + 1/2)	✓ 1
A1 y = −9/4 x + 1/4	✓ 1

Common error: In 3h + h² = 6h + 3h², factorise rather than cancelling h straight away. Then state that h = 0 is rejected because h ≠ 0, instead of losing that root without comment.

+ Add to test✓ In your test279IBMA2S-052-08-125.4 Tangents and normals Paper 1Hard7 marks

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A function is defined by f(x) = ax² + bx, where a and b are constants. For all h ≠ 0, the gradient of the chord joining the points on the graph of f with x-coordinates 2 and 2 + h is 3h + 7.

(a)

Find the value of a and the value of b.

(4)

(b)

Find the exact x-coordinate of the point on the graph of f at which the tangent is perpendicular to the tangent at x = 2.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 forms [f(2 + h) − f(2)] ÷ h = [a(4h + h²) + bh] ÷ h	✓ 1
A1 simplifies to (4a + b) + ah	✓ 1
A1 comparing coefficients, a = 3	✓ 1
A1 4a + b = 7, so b = −5	✓ 1
Part (b)
M1 gradient at x = 2 is the limit of 3h + 7 as h → 0, i.e. 7, so the perpendicular gradient is −1/7	✓ 1
M1 sets f′(x) = 6x − 5 = −1/7	✓ 1
A1 x = 17/21	✓ 1

Exam tip: An identity in h holds for every h, so you can compare the constant terms and the coefficients of h on both sides separately.

+ Add to test✓ In your test280IBMA2S-052-09-115.4 Tangents and normals Paper 1Hard15 marks

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A concrete "wave" in a skate park has a cross-section modelled by the region between the curve $y = f(x)$ and the $x$-axis, for $0 \le x \le 3$, where $x$ and $y$ are in metres. It is given that
$f'(x) = 3x^2 - 12x + 9$
and that the curve passes through the point $(1, 4)$.

(a)

Find $f(x)$.

(4)

(b)

Show that $f(x) = x(x - 3)^2$.

(2)

(c)

Find the coordinates of the stationary points of the curve and determine their nature.

(4)

(d)

Find the area of the cross-section.

(3)

(e)

The wave is 2.4 m wide and has this same cross-section throughout. Find the volume of concrete needed.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate $f'(x)$	✓ 1
A1 $x^3 - 6x^2$	✓ 1
A1 $+\,9x + c$	✓ 1
A1 substitute the point $x = 1$, $y = 4$: $1 - 6 + 9 + c = 4$, so $c = 0$ and $f(x) = x^3 - 6x^2 + 9x$	✓ 1
Part (b)
M1 take out the common factor: $x(x^2 - 6x + 9)$	✓ 1
A1 $x^2 - 6x + 9 = (x - 3)^2$, so $f(x) = x(x - 3)^2$	✓ 1
Part (c)
M1 solve $3x^2 - 12x + 9 = 0$, i.e. $3(x - 1)(x - 3) = 0$	✓ 1
A1 stationary points at $x = 1$, $y = 4$ and at $x = 3$, $y = 0$	✓ 1
M1 use $f''(x) = 6x - 12$, or a sign test on $f'$	✓ 1
A1 at $x = 1$, $f'' = -6 < 0$, so this point is a maximum; at $x = 3$, $f'' = 6 > 0$, so this point is a minimum	✓ 1
Part (d)
M1 $\int_0^3 (x^3 - 6x^2 + 9x)\,dx$	✓ 1
A1 $\left[\frac{x^4}{4} - 2x^3 + \frac{9x^2}{2}\right]_0^3$	✓ 1
A1 $\frac{81}{4} - 54 + \frac{81}{2} = \frac{27}{4}$ m²	✓ 1
Part (e)
M1 $\frac{27}{4} \times 2.4$	✓ 1
A1 16.2 m³	✓ 1

Exam tip: Writing $f(x) = x(x - 3)^2$ shows that the curve touches the $x$-axis at $x = 3$. Since $x \ge 0$ and $(x - 3)^2 \ge 0$, $f(x) \ge 0$ on $[0, 3]$, so the definite integral gives the area directly.

+ Add to test✓ In your test281IBMA2S-052-01-115.10 Indefinite integrals and reverse chain rule Paper 1Hard14 marks

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A new urban park is bounded by a straight road along the $x$-axis and a curved riverbank modelled by $y = 3x - \frac{1}{2}x^2$, where $x$ and $y$ are measured in hundreds of metres. A straight footpath runs from the origin along the line $y = mx$, where $0 < m < 3$. The path divides the park into two parts of equal area.

Diagram not to scale

(a)

Find the $x$-coordinates of the points where the riverbank meets the road.

(2)

(b)

Find the area of the park.

(3)

(c)

Show that the footpath meets the riverbank again at $x = 2(3 - m)$.

(2)

(d)

Show that the area between the footpath and the riverbank is $\frac{2}{3}(3 - m)^3$.

(4)

(e)

Find the exact value of $m$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 solves $x\left(3 - \frac{1}{2}x\right) = 0$	✓ 1
A1 $x = 0$ and $x = 6$	✓ 1
Part (b)
M1 writes $\displaystyle\int_0^6 \left(3x - \frac{1}{2}x^2\right)dx$	✓ 1
A1 $\left[\frac{3x^2}{2} - \frac{x^3}{6}\right]_0^6$	✓ 1
A1 $54 - 36 = 18$ square units (180 000 m²)	✓ 1
Part (c)
M1 equates $3x - \frac{1}{2}x^2 = mx$	✓ 1
A1 $x\left(3 - m - \frac{1}{2}x\right) = 0$, so $x \ne 0$ gives $x = 2(3 - m)$ (AG)	✓ 1
Part (d)
M1 writes $\displaystyle\int_0^{2(3-m)} \left((3 - m)x - \frac{1}{2}x^2\right)dx$	✓ 1
A1 $\left[\frac{(3 - m)x^2}{2} - \frac{x^3}{6}\right]$	✓ 1
M1 substitutes the upper limit: $\frac{(3 - m)\cdot 4(3 - m)^2}{2} - \frac{8(3 - m)^3}{6}$	✓ 1
A1 $2(3 - m)^3 - \frac{4}{3}(3 - m)^3 = \frac{2}{3}(3 - m)^3$ (AG)	✓ 1
Part (e)
M1 sets $\frac{2}{3}(3 - m)^3 = 9$ (half of 18)	✓ 1
A1 $(3 - m)^3 = \frac{27}{2}$	✓ 1
A1 $m = 3 - \frac{3}{\sqrt[3]{2}}$ (equivalently $3 - \frac{3\sqrt[3]{4}}{2}$)	✓ 1

Exam tip: Substituting $c = 3 - m$ before integrating keeps the algebra short. Check your formula with $m = 0$: it should give the whole park area of 18.

+ Add to test✓ In your test282IBMA2S-052-02-105.2 Increasing and decreasing functions Paper 1Hard15 marks

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Let $f(x)=x^3-3px^2+4$, where $p$ is a positive constant.

(a)

Show that the graph of $f$ has stationary points at $x=0$ and $x=2p$.

(3)

(b)

Determine the nature of each stationary point.

(3)

(c)

Find the set of values of $p$ for which the equation $f(x)=0$ has three distinct real roots.

(3)

(d)

Show that the point of inflexion of the graph of $f$ is the midpoint of the line segment joining the two stationary points.

(3)

(e)

The tangent to the graph of $f$ at the point of inflexion meets the $y$-axis at $(0,12)$. Find the value of $p$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $f'(x)=3x^2-6px$	✓ 1
M1 sets $f'(x)=0$ and factorises: $3x(x-2p)=0$	✓ 1
A1 $x=0$ or $x=2p$	✓ 1
Part (b)
M1 $f''(x)=6x-6p$	✓ 1
R1 $f''(0)=-6p<0$ since $p>0$, so local maximum at $x=0$	✓ 1
R1 $f''(2p)=6p>0$, so local minimum at $x=2p$	✓ 1
Part (c)
R1 local maximum value $f(0)=4>0$, so three roots require the local minimum value to be negative	✓ 1
M1 $f(2p)=8p^3-12p^3+4=4-4p^3$	✓ 1
A1 $4-4p^3<0\Rightarrow p>1$	✓ 1
Part (d)
M1 solves $f''(x)=0$ to obtain $x=p$, with $f''$ changing sign there	✓ 1
A1 $f(p)=p^3-3p^3+4=4-2p^3$	✓ 1
R1 midpoint of $(0,4)$ and $(2p,4-4p^3)$ is $(p,4-2p^3)$, the point of inflexion	✓ 1
Part (e)
M1 gradient at inflexion $f'(p)=3p^2-6p^2=-3p^2$	✓ 1
M1 tangent $y-(4-2p^3)=-3p^2(x-p)$; at $x=0$, $y=4+p^3$	✓ 1
A1 $4+p^3=12\Rightarrow p=2$	✓ 1

Exam tip: With a parameter, keep track of its sign: the nature of each stationary point in (b) depends on $p>0$, and you should say so explicitly.

+ Add to test✓ In your test283IBMA2S-052-07-055.4 Tangents and normals Paper 1Hard7 marks

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The cross-section of a grass mound is modelled by $y = \sqrt{x}$, $x \ge 0$. A spotlight on the ground at $L(-4, 0)$ shines a straight beam that just touches the surface of the mound at the point $P(a, \sqrt{a})$, so that the beam lies along the tangent at $P$.

(a)

Show that the tangent to the curve at $P$ has equation $y = \frac{x}{2\sqrt{a}} + \frac{\sqrt{a}}{2}$.

(2)

(b)

Find the coordinates of $P$.

(3)

(c)

The normal to the curve at $P$ meets the ground ($x$-axis) at $N$. Find the $x$-coordinate of $N$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$ substituted into $y - \sqrt{a} = \frac{1}{2\sqrt{a}}(x - a)$	✓ 1
A1 correct simplification $y = \frac{x}{2\sqrt{a}} - \frac{\sqrt{a}}{2} + \sqrt{a}$ leading to the given answer	✓ 1
Part (b)
M1 substitution of $(-4, 0)$: $0 = \frac{-4}{2\sqrt{a}} + \frac{\sqrt{a}}{2}$	✓ 1
M1 multiplying by $2\sqrt{a}$ to obtain $a - 4 = 0$	✓ 1
A1 $P(4, 2)$	✓ 1
Part (c)
M1 normal gradient $-4$: $y - 2 = -4(x - 4)$ with $y = 0$	✓ 1
A1 $x = \frac{9}{2}$	✓ 1

Common error: $L$ is not on the curve, so it cannot be used to find the gradient directly. Write the tangent at a general point first, then make it pass through $L$.

+ Add to test✓ In your test284IBMA2S-052-07-105.4 Tangents and normals Paper 1Hard14 marks

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The cross-section of a dome-shaped greenhouse is modelled by $y = 12 - x^2$, for $0 \le x \le \sqrt{12}$, where $x$ and $y$ are in metres. A vertical wall lies along the $y$-axis and the ground lies along the $x$-axis. A straight plank rests against the wall and the ground, touching the greenhouse at exactly one point $P(a, 12 - a^2)$, where $0 < a < \sqrt{12}$.

Diagram not to scale

(a)

Show that the plank lies along the line $y = -2ax + a^2 + 12$.

(3)

(b)

Find, in terms of $a$, the coordinates of the points where the plank meets the ground and the wall.

(2)

(c)

Show that the area $A$ of the triangle enclosed by the plank, the wall and the ground is given by $A = \frac{(a^2 + 12)^2}{4a}$.

(2)

(d)

Find the value of $a$ that minimises $A$, justifying that it gives a minimum.

(5)

(e)

Find the minimum area.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{dy}{dx} = -2x$, so gradient at $P$ is $-2a$	✓ 1
M1 substitution into $y - (12 - a^2) = -2a(x - a)$	✓ 1
A1 correct expansion $y = -2ax + 2a^2 + 12 - a^2$ leading to the given answer	✓ 1
Part (b)
A1 ground: $\left(\frac{a^2 + 12}{2a}, 0\right)$	✓ 1
A1 wall: $(0, a^2 + 12)$	✓ 1
Part (c)
M1 $A = \frac{1}{2} \times \frac{a^2 + 12}{2a} \times (a^2 + 12)$	✓ 1
A1 simplification to the given answer	✓ 1
Part (d)
M1 attempt to differentiate, e.g. writing $A = \frac{a^3}{4} + 6a + \frac{36}{a}$	✓ 1
A1 $\frac{dA}{da} = \frac{3a^2}{4} + 6 - \frac{36}{a^2}$	✓ 1
M1 setting $\frac{dA}{da} = 0$ to obtain $a^4 + 8a^2 - 48 = 0$, i.e. $(a^2 + 12)(a^2 - 4) = 0$	✓ 1
A1 $a = 2$	✓ 1
R1 $\frac{d^2A}{da^2} = \frac{3a}{2} + \frac{72}{a^3} > 0$ for $a > 0$, so minimum	✓ 1
Part (e)
M1 substituting $a = 2$ into $A$	✓ 1
A1 32 m²	✓ 1

Exam tip: Expanding $(a^2 + 12)^2$ and dividing each term by $4a$ before differentiating avoids a messy quotient rule and makes the factorisation of the quartic easier to spot.

+ Add to test✓ In your test285IBMA2S-052-06-105.7 The second derivative and graphs Paper 1Hard12 marks

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Let f(x) = x²ln x, for x > 0.

(a)

Show that f′(x) = x(2ln x + 1).

(2)

(b)

Find the interval on which f is decreasing, giving the endpoint in exact form.

(3)

(c)

Find the exact minimum value of f.

(2)

(d)

The function h is defined by h(x) = x²ln x − kx², for x > 0, where k is a constant.
(d) Find h′(x), giving your answer in factorised form.

(2)

(e)

Find the largest value of k for which h is increasing for all x ≥ e.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts the product rule	✓ 1
A1 2xln x + x, leading to given answer	✓ 1
Part (b)
R1 x > 0, so the sign of f′ depends on 2ln x + 1	✓ 1
M1 solves 2ln x + 1 < 0	✓ 1
A1 0 < x < e^−1/2	✓ 1
Part (c)
M1 substitutes x = e^−1/2 into f	✓ 1
A1 −1/(2e)	✓ 1
Part (d)
M1 uses part (a) and differentiates −kx²	✓ 1
A1 h′(x) = x(2ln x + 1 − 2k)	✓ 1
Part (e)
R1 requires 2ln x + 1 − 2k ≥ 0 for all x ≥ e	✓ 1
M1 ln x is increasing, so the condition is tightest at x = e, giving 3 − 2k ≥ 0	✓ 1
A1 k = 3/2	✓ 1

Exam tip: When a condition must hold for every x in an interval, find where the expression is smallest and impose the condition there.

+ Add to test✓ In your test286IBMA3S-5-17r5-125.4 Tangents and normals Paper 1Hard7 marks

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The path of a probe is modelled by the curve y = ln x, x > 0, with both axes in thousands of kilometres. A ground station at the origin O sends a beam along a straight line that just touches the path.

(a)

Show that the tangent to the curve at the point where x = a has equation y = x/a + ln a − 1.

(3)

(b)

The tangent passes through O. Find the value of a.

(3)

(c)

Hence write down the equation of the beam.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 dy/dx = 1/x, so the gradient at x = a is 1/a	✓ 1
M1 y − ln a = (1/a)(x − a)	✓ 1
A1 rearrange to reach y = x/a + ln a − 1 (AG)	✓ 1
Part (b)
M1 substitute (0, 0) into the tangent equation	✓ 1
A1 0 = ln a − 1, so ln a = 1	✓ 1
A1 a = e	✓ 1
Part (c)
A1 y = x/e	✓ 1

Exam tip: When a tangent must pass through a given point that is not on the curve, write the general tangent first and substitute the point into it.

+ Add to test✓ In your test287IBMA3S-5-16r7-115.10 Indefinite integrals and reverse chain rule Paper 1Hard8 marks

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(a)

Show that $\frac{d}{dx}\left(\cos^4 x\right)=-4\sin x\cos^3 x$.

(2)

(b)

Hence find the exact value of $\int_0^{\pi/2}\sin x\cos^3 x\,dx$.

(3)

(c)

A curve has gradient function $f'(x)=8\sin x\cos^3 x$ and passes through $(0,1)$. Find $f(x)$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the chain rule on $(\cos x)^4$	✓ 1
A1 $4\cos^3 x\times(-\sin x)=-4\sin x\cos^3 x$ (AG)	✓ 1
Part (b)
M1 use of part (a), antiderivative $-\frac14\cos^4 x$	✓ 1
A1 $-\frac14\left(\cos^4\frac{\pi}{2}-\cos^4 0\right)$	✓ 1
A1 $\frac14$	✓ 1
Part (c)
A1 $f(x)=-2\cos^4 x+C$	✓ 1
M1 substitute $(0,1)$: $1=-2+C$	✓ 1
A1 $f(x)=3-2\cos^4 x$	✓ 1

Exam tip: When a question says 'hence', the earlier result is the key to the antiderivative; use it rather than starting again.

+ Add to test✓ In your test288IBMA3S-5-16r7-125.10 Indefinite integrals and reverse chain rule Paper 1Hard11 marks

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The intensity profile of a star's image in a telescope is modelled by $g(x)=x\,e^{-x^2}$, for $x\ge 0$, where x is the distance from the centre of the image.

(a)

Show that $g'(x)=e^{-x^2}\left(1-2x^2\right)$.

(3)

(b)

Find the value of x, for $x>0$, at which $g'(x)=0$.

(2)

(c)

Find the exact value of $g\!\left(\frac{1}{\sqrt2}\right)$.

(2)

(d)

Show that $g''(x)=2x\,e^{-x^2}\left(2x^2-3\right)$.

(3)

(e)

Hence justify that the value found in part (c) is a maximum.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule	✓ 1
A1 derivative of $e^{-x^2}$ is $-2xe^{-x^2}$	✓ 1
A1 $e^{-x^2}-2x^2e^{-x^2}=e^{-x^2}(1-2x^2)$ (AG)	✓ 1
Part (b)
M1 set $1-2x^2=0$ (since $e^{-x^2}\neq 0$)	✓ 1
A1 $x=\frac{1}{\sqrt2}$	✓ 1
Part (c)
M1 substitute into $g$	✓ 1
A1 $\frac{1}{\sqrt2}e^{-1/2}=\frac{1}{\sqrt{2e}}$	✓ 1
Part (d)
M1 differentiate $e^{-x^2}(1-2x^2)$ using the product rule	✓ 1
A1 $-2xe^{-x^2}(1-2x^2)-4xe^{-x^2}$	✓ 1
A1 $2xe^{-x^2}(2x^2-3)$ (AG)	✓ 1
Part (e)
R1 at $x=\frac{1}{\sqrt2}$, $2x^2-3=-2<0$, so $g''<0$ and the point is a local maximum	✓ 1

Common error: Factorising out $e^{-x^2}$ early makes both the stationary-point equation and the second derivative far easier.

+ Add to test✓ In your test289IBMA3S-5-17r7-145.7 The second derivative and graphs Paper 1Hard12 marks

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A curve has gradient function $f'(x)=6x(x^2-1)^2$, for x ∈ ℝ. The curve passes through the point (2, 30).

(a)

Find f(x).

(4)

(b)

Find the coordinates of the stationary points of the curve and determine their nature.

(4)

(c)

Write down the set of values of x for which f is decreasing.

(1)

(d)

Find the x-coordinate of the point where the tangent at (2, 30) meets the x-axis. Give your answer as an exact fraction.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the reverse chain rule, giving $k(x^2-1)^3$	✓ 1
A1 $f(x)=(x^2-1)^3+C$	✓ 1
M1 substitutes $(2,\,30)$: $30=27+C$	✓ 1
A1 $f(x)=(x^2-1)^3+3$	✓ 1
Part (b)
A1 solves $6x(x^2-1)^2=0$, giving $x=0$ and $x=\pm1$	✓ 1
A1 $y$-values: $(0,\,2)$, $(1,\,3)$ and $(-1,\,3)$	✓ 1
R1 $f'$ changes from negative to positive at $x=0$, so a local minimum	✓ 1
R1 $(x^2-1)^2\ge0$, so $f'$ does not change sign at $x=\pm1$: both are points of inflection	✓ 1
Part (c)
A1 $x<0$	✓ 1
Part (d)
A1 gradient $f'(2)=12\times 9=108$	✓ 1
M1 tangent $y-30=108(x-2)$, setting $y=0$	✓ 1
A1 $x=2-\frac{30}{108}=\frac{31}{18}$ (=1.7222 to 4 d.p.)	✓ 1

Common error: At x = ±1 the gradient is zero but f′ keeps the same sign, so these are not maxima or minima.

+ Add to test✓ In your test290IBMA3S-5-17r9-135.4 Tangents and normals Paper 1Hard11 marks

No calculator

The profile of a cycling-track transition ramp is modelled by $y=ax^2+\dfrac{b}{x}$ for $x>0$, where $a$ and $b$ are constants and $x$ and $y$ are in metres. The graph has a stationary point at $(2,12)$.

(a)

Find the values of $a$ and $b$.

(5)

(b)

Show that the stationary point is a minimum.

(2)

(c)

The normal to the curve at the point where $x=1$ meets the $x$-axis at the point R. Find the $x$-coordinate of R.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{dy}{dx}=2ax-\frac{b}{x^2}$	✓ 1
M1 $\frac{dy}{dx}=0$ at $x=2$: $4a-\frac b4=0$, so $b=16a$	✓ 1
M1 $y(2)=4a+\frac b2=12$	✓ 1
A1 $a=1$	✓ 1
A1 $b=16$	✓ 1
Part (b)
M1 $\frac{d^2y}{dx^2}=2+\frac{32}{x^3}$	✓ 1
R1 at $x=2$ this is $6>0$, so a minimum	✓ 1
Part (c)
A1 $y\left(1\right)=17$	✓ 1
M1 gradient of curve at $x=1$ is $2-16=-14$	✓ 1
A1 normal gradient $\frac1{14}$ and equation $y-17=\frac{1}{14}(x-1)$	✓ 1
A1 setting $y=0$: $x=-237$	✓ 1

Common error: Use the stationary point twice: once for the zero gradient and once for the coordinates. Both conditions are needed to find two constants.

+ Add to test✓ In your test291IBMA3S-5-17r9-145.2 Increasing and decreasing functions Paper 1Hard12 marks

No calculator

A data-centre engineer models the response-time deviation, in milliseconds, by the cubic $f(x)=x^3+ax^2+bx+c$, where $x$ is the server load in arbitrary units. The graph of $f$ has a point of inflexion at $x=1$ and a local maximum at the point $(-1,4)$.

(a)

Find the values of $a$, $b$ and $c$.

(6)

(b)

Find the coordinates of the local minimum of the graph.

(3)

(c)

The equation $f(x)=k$ has exactly three distinct real solutions, where $k$ is an integer. Find the number of possible values of $k$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $f''(x)=6x+2a$ and sets $f''\left(1\right)=0$	✓ 1
A1 $a=-3$	✓ 1
M1 $f'(x)=3x^2+2ax+b$ with $f'(-1)=0$	✓ 1
A1 $b=-9$	✓ 1
M1 $f(-1)=4$: $-1+a-b+c=4$	✓ 1
A1 $c=-1$	✓ 1
Part (b)
M1 $f'(x)=3(x-3)(x+1)=0$	✓ 1
A1 $x=3$	✓ 1
A1 $f(3)=-28$, so $(3,-28)$	✓ 1
Part (c)
A1 $-28<k<4$	✓ 1
M1 lists the integers from $-27$ to $3$	✓ 1
A1 31 values	✓ 1

Exam tip: A cubic line y = k crosses the graph three times only when k lies strictly between the local minimum and local maximum values.

+ Add to test✓ In your test292IBMA3S-5-17r10-145.6 Differentiation rules Paper 1Hard11 marks

No calculator

The transmission efficiency of a space probe link at distance $x$ (in suitable units, $x\ge1$) is modelled by $f(x)=\dfrac{\ln x}{x}$.

(a)

Show that $f'(x)=\dfrac{1-\ln x}{x^2}$.

(3)

(b)

Find the exact coordinates of the maximum point of the graph of $f$.

(2)

(c)

It is given that $f''(x)=\dfrac{2\ln x-3}{x^3}$. Find the exact $x$-coordinate of the point of inflexion, and show it is a point of inflexion.

(3)

(d)

Find $\int_1^e f(x)\,dx$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
attempt quotient (or product) rule	✓ 1
numerator $\frac1x\cdot x-\ln x\cdot1$ (or equivalent)	✓ 1
$f'(x)=\frac{1-\ln x}{x^2}$	✓ 1
Part (b)
$f'(x)=0$ gives $\ln x=1$ so $x=e$	✓ 1
maximum at $\left(e,\frac1e\right)$	✓ 1
Part (c)
$2\ln x-3=0$	✓ 1
$x=e^{3/2}$	✓ 1
$f''$ changes sign (negative then positive) because $x^3>0$	✓ 1
Part (d)
substitution $u=\ln x$ with $du=\frac1x\,dx$	✓ 1
$\int_0^1 u\,du$	✓ 1
$\frac12$	✓ 1

Exam tip: When the integrand is a function of $\ln x$ multiplied by $\frac1x$, substitute $u=\ln x$ and change the limits.

+ Add to test✓ In your test293IBMA3S-5-16r12-125.9 Kinematics Paper 1Hard11 marks

No calculator

A robotic inspection crawler moves along a straight girder of a bridge. Its displacement from the starting point, s metres, after t minutes is given by $s=t^3-9t^2+24t$, for $0\le t\le 5$.

(a)

Find the times at which the crawler is instantaneously at rest.

(3)

(b)

Find the acceleration of the crawler at each of the times found in part (a).

(2)

(c)

Find the minimum velocity of the crawler for $0\le t\le 5$.

(2)

(d)

Find the total distance travelled by the crawler in the first 5 minutes.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiates, $v=3t^2-18t+24$, and sets $v=0$	✓ 1
A1 $3(t-2)(t-4)=0$ or equivalent	✓ 1
A1 $t=2$ and $t=4$	✓ 1
Part (b)
A1 $a=6t-18$	✓ 1
A1 $a=-6$ m min⁻² at $t=2$ and $a=6$ m min⁻² at $t=4$	✓ 1
Part (c)
M1 sets $a=0$ (or completes the square for $v$), giving $t=3$	✓ 1
A1 $v(3)=27-54+24=-3$ m min⁻¹	✓ 1
Part (d)
M1 evaluates s at the turning points and end points, e.g. $s(2)$, $s(4)$, $s(5)$	✓ 1
A1 $s(2)=20$, $s(4)=16$ and $s(5)=20$	✓ 1
M1 sums the three magnitudes of the changes, e.g. $20+\|16-20\|+\|20-16\|$	✓ 1
A1 28 m	✓ 1

Common error: Total distance needs the turning points. Using $s(5)-s(0)=20$ gives the displacement only, which is not the distance when the direction reverses.

+ Add to test✓ In your test294IBMA3S-5-17r14-135.10 Indefinite integrals and reverse chain rule Paper 1Hard8 marks

No calculator

(a)

Show that $\frac{d}{dx}\left(x^2\ln x\right)=2x\ln x+x$.

(2)

(b)

Hence find $\int x\ln x\,dx$.

(3)

(c)

Hence find the exact value of $\int_1^e x\ln x\,dx$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule	✓ 1
A1 $2x\ln x+x^2\times\frac1x=2x\ln x+x$ (AG)	✓ 1
Part (b)
M1 integrating both sides of the result in (a): $\int(2x\ln x+x)\,dx=x^2\ln x+C$	✓ 1
A1 $2\int x\ln x\,dx+\frac{x^2}{2}=x^2\ln x+C$	✓ 1
A1 $\int x\ln x\,dx=\frac12x^2\ln x-\frac14x^2+C$	✓ 1
Part (c)
M1 substituting the limits $e$ and 1 into the answer to (b)	✓ 1
A1 $\left(\frac{e^2}{2}-\frac{e^2}{4}\right)-\left(0-\frac14\right)$	✓ 1
A1 $\frac{e^2+1}{4}$ ($\approx 2.097$)	✓ 1

Exam tip: When a question says 'hence', rearrange the result you have just shown instead of starting a new method.

+ Add to test✓ In your test295IBMA3S-5-17r18-125.4 Tangents and normals Paper 1Hard15 marks

No calculator

The curve $C$ models the profile of a cam that guides a robot gripper. It has equation $y=x^3-6x^2+9x$, for $x\ge0$, with all lengths in cm.

(a)

Find the coordinates of the stationary points of $C$ and determine their nature.

(6)

(b)

Find the equation of the tangent to $C$ at the origin $O$.

(2)

(c)

The tangent in part (b) meets $C$ again at the point $B$. Find the coordinates of $B$.

(3)

(d)

Find the area of the region enclosed by $C$ and the tangent in part (b).

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate	✓ 1
A1 $\dfrac{dy}{dx}=3x^2-12x+9$	✓ 1
M1 set $\dfrac{dy}{dx}=0$ and solve	✓ 1
A1 $x=1$ and $x=3$	✓ 1
A1 $(1,4)$ and $(3,0)$	✓ 1
A1 $\dfrac{d^2y}{dx^2}=6x-12$; at $x=1$ it is $-6<0$ (maximum) and at $x=3$ it is $6>0$ (minimum)	✓ 1
Part (b)
M1 gradient at $x=0$ is $9$	✓ 1
A1 $y=9x$	✓ 1
Part (c)
M1 equate $x^3-6x^2+9x=9x$	✓ 1
A1 $x^2(x-6)=0$	✓ 1
A1 $B=(6,54)$	✓ 1
Part (d)
M1 $\int_0^6\left(9x-(x^3-6x^2+9x)\right)dx$ (line minus curve, limits $0$ and $6$)	✓ 1
A1 integrand simplified to $6x^2-x^3$	✓ 1
A1 $\left[2x^3-\dfrac{x^4}{4}\right]_0^6$	✓ 1
A1 $432-324=108$ cm²	✓ 1

Exam tip: When a line meets a cubic at a tangent point, the equation you solve has a repeated root there (here $x^2$): use it as a check that the tangent equation is correct.

+ Add to test✓ In your test296IBMA3S-5-17r18-145.5 Introduction to integration Paper 1Hard14 marks

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The curve $C$ has equation $y=f(x)$ where $f(x)=x+\dfrac{4}{x}$, for $x>0$.

(a)

Find the coordinates of the minimum point of $C$, justifying that it is a minimum.

(5)

(b)

Find the equation of the tangent to $C$ at the point where $x=1$.

(3)

(c)

Show that $f(x)-(8-3x)=\dfrac{4(x-1)^2}{x}$, and hence explain why the tangent never lies above $C$ for $x>0$.

(2)

(d)

Find the exact area of the region enclosed by $C$, the tangent at $x=1$, and the line $x=4$.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate ($x^{-1}\to -x^{-2}$)	✓ 1
A1 $f'(x)=1-\dfrac{4}{x^2}$	✓ 1
M1 set $f'(x)=0$	✓ 1
A1 $x=2$ (since $x>0$) and $y=4$, so the point is $(2,4)$	✓ 1
R1 $f''(x)=\dfrac{8}{x^3}$, so $f''(2)=1>0$, a minimum	✓ 1
Part (b)
A1 when $x=1$: $y=5$ and gradient $=1-4=-3$	✓ 1
M1 line through $(1,5)$ with gradient $-3$	✓ 1
A1 $y=8-3x$	✓ 1
Part (c)
M1 $x+\dfrac4x-8+3x$ combined over a common denominator: $\dfrac{4x^2-8x+4}{x}$	✓ 1
R1 $=\dfrac{4(x-1)^2}{x}\ge0$ for $x>0$, so $C$ is never below the tangent	✓ 1
Part (d)
M1 $\int_1^4\left(4x+\dfrac4x-8\right)dx$ (curve minus tangent)	✓ 1
A1 antiderivative $2x^2+4\ln x-8x$	✓ 1
M1 substitute limits: $(32+4\ln4-32)-(2-8)$	✓ 1
A1 $6+8\ln2$ ($\approx 11.545$)	✓ 1

Common error: Integrating $\frac{4}{x}$ as $\frac{4}{x^2}$ or $4x^{-1}\to 4x^0$. The integral of $\frac1x$ is $\ln x$, so $\int\frac4x\,dx=4\ln x$.

+ Add to test✓ In your test297IBMA3S-5-17r21-115.4 Tangents and normals Paper 1Hard8 marks

No calculator

A ridge has profile $y=x^2$, with units of 100 m. A straight cable starts at the point $A(2,-5)$ and just touches the ridge at a point $P$ with coordinates $(t,t^2)$.

Units of 100 m. Diagram not to scale

(a)

Show that the equation of the tangent to the ridge at $P$ is $y=2tx-t^2$.

(3)

(b)

The cable lies along the tangent. Hence show that $t^2-4t-5=0$.

(2)

(c)

Find the coordinates of the two possible points of contact.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $\frac{dy}{dx}=2x$, so gradient at $P$ is $2t$	✓ 1
M1 $y-t^2=2t(x-t)$	✓ 1
A1 $y=2tx-t^2$ (AG)	✓ 1
Part (b)
M1 substitutes $x=2$, $y=-5$: $-5=4t-t^2$	✓ 1
A1 $t^2-4t-5=0$ (AG)	✓ 1
Part (c)
M1 factorises $(t-5)(t+1)=0$ or equivalent	✓ 1
A1 $t=5$ and $t=-1$	✓ 1
A1 $(5,25)$ and $(-1,1)$	✓ 1

Exam tip: When a line is tangent to a curve at an unknown point, let the point of contact be $(t, f(t))$ and use the condition that the line passes through the known point.

+ Add to test✓ In your test298IBMA3S-5-17r21-125.4 Tangents and normals Paper 1Hard8 marks

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A relay beam is sent in a straight line from the origin $O$ and just touches the curve $y=\ln x$ at a point $P$ with coordinates $(a,\ln a)$, where $a>0$.

(a)

Find the gradient of the curve at $P$.

(2)

(b)

Write down the equation of the tangent at $P$ in the form $y=mx+c$.

(2)

(c)

The tangent passes through $O$. Find the exact value of $a$.

(3)

(d)

Hence write down the equation of the beam.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{d}{dx}\ln x=\frac1x$	✓ 1
A1 gradient $=\frac1a$	✓ 1
Part (b)
M1 $y-\ln a=\frac1a(x-a)$	✓ 1
A1 $y=\frac xa+\ln a-1$	✓ 1
Part (c)
M1 substitutes $(0,0)$: $0=\ln a-1$	✓ 1
A1 $\ln a=1$	✓ 1
A1 $a=e$	✓ 1
Part (d)
A1 $y=\frac xe$	✓ 1

Exam tip: A tangent through a given point gives an equation in the unknown point of contact; substitute the given point only after you have written the general tangent.

+ Add to test✓ In your test299IBMA3S-5-17r23-125.4 Tangents and normals Paper 1Hard11 marks

No calculator

The curve $C$ has equation $y=x^2$. Two tangents to $C$ pass through the point $P(1,-3)$.

Diagram not to scale

(a)

Show that the tangent to $C$ at the point $(a, a^2)$ has equation $y=2ax-a^2$.

(3)

(b)

Hence find the two possible values of $a$.

(3)

(c)

Write down the equations of the two tangents.

(2)

(d)

The two tangents cross the $y$-axis at $A$ and $B$. Find the area of triangle $PAB$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $\frac{dy}{dx}=2x$, so the gradient at $x=a$ is $2a$	✓ 1
M1 $y-a^2=2a(x-a)$	✓ 1
A1 $y=2ax-a^2$ (AG)	✓ 1
Part (b)
M1 substitute $(1,-3)$: $-3=2a-a^2$	✓ 1
A1 $a^2-2a-3=0$	✓ 1
A1 $a=3$ or $a=-1$	✓ 1
Part (c)
A1 $y=6x-9$	✓ 1
A1 $y=-2x-1$	✓ 1
Part (d)
A1 $A(0,-9)$ and $B(0,-1)$	✓ 1
M1 area $=\frac12\times AB\times$ (perpendicular distance of $P$ from the $y$-axis) $=\frac12\times 8\times 1$	✓ 1
A1 $4$ square units	✓ 1

Common error: A tangent through an outside point must be found by first writing the tangent at a general point, then forcing it through the given point.

+ Add to test✓ In your test300IBMA3S-5-17r23-145.2 Increasing and decreasing functions Paper 1Hard12 marks

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The function $f(x)=x^3+ax^2+bx+c$, where $a$, $b$ and $c$ are constants, has a point of inflexion at $x=1$. The graph of $f$ has a local maximum at the point where $x=-1$, and the value of $f$ there is 12.

(a)

Find the value of $a$.

(3)

(b)

Find the value of $b$.

(3)

(c)

Find the value of $c$.

(2)

(d)

Find the coordinates of the local minimum of the graph of $f$, justifying that it is a minimum.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $f''(x)=6x+2a$	✓ 1
M1 set $f''=0$ at $x=1$: $6+2a=0$	✓ 1
A1 $a=-3$	✓ 1
Part (b)
A1 $f'(x)=3x^2+2ax+b$	✓ 1
M1 $f'(-1)=0$: $3-2a+b=0$ (using their $a$)	✓ 1
A1 $b=-9$	✓ 1
Part (c)
M1 $f(-1)=12$: $-1+a-b+c=12$	✓ 1
A1 $c=7$	✓ 1
Part (d)
M1 $3x^2-6x-9=0$ (accept use of the factor $(x+1)$)	✓ 1
A1 $x=3$	✓ 1
A1 $f(3)=-20$	✓ 1
R1 $f''(3)=12>0$, so a minimum at $(3,-20)$	✓ 1

Exam tip: Each condition in the stem gives one equation: a point of inflexion gives $f''=0$, a stationary point gives $f'=0$, and a given value gives $f=$ that value.

+ Add to test✓ In your test301IBMA3S-5-16r26-075.7 The second derivative and graphs Paper 1Hard13 marks

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The derivative of a function $f$ is $f'(x) = (x-2)(x+1)^2$. The graph of $f$ passes through the point $(2,\,-5)$.

(a)

Find the $x$-coordinates of the stationary points of $f$ and determine the nature of each.

(5)

(b)

Find an expression for $f(x)$.

(6)

(c)

Find the $y$-coordinate of the point of inflexion at $x=-1$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 set $f'(x)=0$	✓ 1
A1 $x=-1$ and $x=2$	✓ 1
M1 consider the sign of $f'$ either side of each value (e.g. $f'(-2)=-4$, $f'(0)=-2$, $f'(3)=16$)	✓ 1
A1 $x=2$: $f'$ changes from negative to positive, so a local minimum	✓ 1
R1 $x=-1$: $f'$ is negative on both sides, so there is no sign change and it is a point of inflexion, not a maximum or minimum	✓ 1
Part (b)
M1 expand $f'(x)$	✓ 1
A1 $f'(x) = x^3 - 3x - 2$	✓ 1
M1 attempt to integrate	✓ 1
A1 $f(x) = \frac{x^4}{4} - \frac{3x^2}{2} - 2x + c$	✓ 1
M1 substitute $(2,-5)$ to find $c$	✓ 1
A1 $c=1$, so $f(x) = \frac{x^4}{4} - \frac{3x^2}{2} - 2x + 1$	✓ 1
Part (c)
M1 substitute $x=-1$ into $f$ (ECF from (b))	✓ 1
A1 $f(-1) = \frac{7}{4}$	✓ 1

Common error: A repeated factor in $f'(x)$ means $f'$ does not change sign there, so the stationary point is not a maximum or minimum.

+ Add to test✓ In your test302IBMA3S-5-16r26-145.2 Increasing and decreasing functions Paper 1Hard11 marks

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A function $f$ has second derivative $f''(x) = 6x - 12$. The graph of $f$ has a stationary point at $x=3$ and passes through the point $(1,\,7)$.

(a)

Find $f'(x)$.

(4)

(b)

Find $f(x)$.

(4)

(c)

Find the coordinates of the local minimum point of the graph of $f$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate $f''(x)$	✓ 1
A1 $f'(x) = 3x^2 - 12x + c$	✓ 1
M1 use $f'(3)=0$, so $27 - 36 + c = 0$	✓ 1
A1 $f'(x) = 3x^2 - 12x + 9$	✓ 1
Part (b)
M1 attempt to integrate $f'(x)$	✓ 1
A1 $f(x) = x^3 - 6x^2 + 9x + d$	✓ 1
M1 use the point $(1,7)$, so $1 - 6 + 9 + d = 7$	✓ 1
A1 $f(x) = x^3 - 6x^2 + 9x + 3$	✓ 1
Part (c)
R1 $f''(3) = 6 > 0$, so $x=3$ is a local minimum	✓ 1
M1 substitute $x=3$ into $f$	✓ 1
A1 $(3,\,3)$	✓ 1

Common error: Each integration needs its own constant; find the first constant using the stationary-point condition and the second using the point on the curve.

+ Add to test✓ In your test303IBMA3S-5-18r30-135.6 Differentiation rules Paper 1Hard13 marks

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Let $f(x)=xe^{-x^2}$.

(a)

Show that $f'(x)=(1-2x^2)e^{-x^2}$.

(3)

(b)

Find the exact $x$-coordinates of the stationary points of the graph of $f$.

(2)

(c)

Show that $f''(x)=(4x^3-6x)e^{-x^2}$.

(3)

(d)

Hence show that the graph of $f$ has a local maximum at $x=\frac{1}{\sqrt{2}}$, and find the exact $y$-coordinate of this point.

(3)

(e)

Find the exact $x$-coordinates of all the points of inflexion of the graph of $f$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule	✓ 1
A1 derivative of $e^{-x^2}$ is $-2xe^{-x^2}$ (chain rule)	✓ 1
A1 $f'(x)=e^{-x^2}-2x^2e^{-x^2}=(1-2x^2)e^{-x^2}$ (AG)	✓ 1
Part (b)
M1 sets $1-2x^2=0$, using $e^{-x^2}>0$	✓ 1
A1 $x=\pm\frac{1}{\sqrt{2}}$	✓ 1
Part (c)
M1 attempt to use the product rule on $f'(x)$	✓ 1
A1 $-4xe^{-x^2}+(1-2x^2)(-2x)e^{-x^2}$	✓ 1
A1 $(-4x-2x+4x^3)e^{-x^2}=(4x^3-6x)e^{-x^2}$ (AG)	✓ 1
Part (d)
A1 $f''\left(\frac{1}{\sqrt{2}}\right)=(\sqrt{2}-3\sqrt{2})e^{-1/2}=-2\sqrt{2}\,e^{-1/2}$	✓ 1
R1 $f''<0$, so a local maximum	✓ 1
A1 $y=\frac{1}{\sqrt{2}}e^{-1/2}$	✓ 1
Part (e)
M1 sets $f''(x)=0$, giving $x(2x^2-3)=0$	✓ 1
A1 $x=0$ and $x=\pm\sqrt{\frac{3}{2}}$ ($\pm 1.2247$), with $f''$ changing sign at each	✓ 1

Exam tip: Take out the common factor $e^{-x^2}$ after each differentiation; it is never zero, so it makes solving $f'(x)=0$ and $f''(x)=0$ much quicker.

+ Add to test✓ In your test304IBMA3S-5-17r34-135.9 Kinematics Paper 1Hard11 marks

No calculator

A hospital's automated medicine trolley moves along a straight rail. Its velocity is $v(t)=t^2-6t+8$ m s⁻¹ for $0\le t\le5$, where t is the time in seconds. Positive velocity is towards the pharmacy.

(a)

Find the times at which the trolley is momentarily at rest.

(2)

(b)

Find the time at which the acceleration of the trolley is zero.

(2)

(c)

At $t=0$ the trolley is at the origin. Find its displacement from the origin when $t=5$.

(3)

(d)

Find the total distance travelled by the trolley during the first 5 seconds.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $t^2-6t+8=0$, so $(t-2)(t-4)=0$	✓ 1
A1 $t=2$ and $t=4$	✓ 1
Part (b)
M1 $a=\frac{dv}{dt}=2t-6$ and set $a=0$	✓ 1
A1 $t=3$ s	✓ 1
Part (c)
M1 $\int_0^5\left(t^2-6t+8\right)dt$	✓ 1
A1 $\left[\frac{t^3}{3}-3t^2+8t\right]_0^5$	✓ 1
A1 $\frac{125}{3}-75+40=\frac{20}{3}$ m	✓ 1
Part (d)
M1 the direction changes at $t=2$ and $t=4$, so integrate over $[0,2]$, $[2,4]$ and $[4,5]$ separately, taking positive values	✓ 1
A1 $\int_0^2v\,dt=\frac{20}{3}$	✓ 1
A1 $\int_2^4v\,dt=-\frac43$ and $\int_4^5v\,dt=\frac43$	✓ 1
A1 total distance $=\frac{20}{3}+\frac43+\frac43=\frac{28}{3}$ m	✓ 1

Common error: Total distance is not the same as displacement. When the velocity changes sign, integrate each section separately and add the sizes of the results.

+ Add to test✓ In your test305IBMA3S-5-17r34-145.10 Indefinite integrals and reverse chain rule Paper 1Hard10 marks

No calculator

Let $f(x)=x^2\ln x$, for $x>0$.

(a)

Find $f'(x)$.

(3)

(b)

The graph of $y=f(x)$ has exactly one stationary point. Find its exact coordinates.

(4)

(c)

Find $f''(x)$, and hence determine the nature of the stationary point.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of the product rule with $u=x^2$ and $v=\ln x$	✓ 1
A1 $2x\ln x$	✓ 1
A1 $+\,x^2\cdot\frac1x=x$, so $f'(x)=2x\ln x+x$	✓ 1
Part (b)
M1 $f'(x)=0$: $x(2\ln x+1)=0$	✓ 1
A1 $\ln x=-\frac12$ (since $x\ne0$), so $x=e^{-1/2}$	✓ 1
M1 substitute into $f$: $e^{-1}\ln e^{-1/2}$	✓ 1
A1 $\left(e^{-1/2},-\frac{1}{2e}\right)$	✓ 1
Part (c)
M1 differentiate $2x\ln x+x$ using the product rule	✓ 1
A1 $f''(x)=2\ln x+3$	✓ 1
R1 at $x=e^{-1/2}$, $f''=-1+3=2>0$, so it is a minimum	✓ 1

Exam tip: For an exact answer, keep logarithms and exponentials in exact form: $\ln e^{-1/2}=-\frac12$ and $e^{-1}=\frac1e$. A decimal answer will not earn the accuracy mark.

+ Add to test✓ In your test306IBMA3S-5-18r35-085.3 Differentiating polynomials and integer powers Paper 1Hard7 marks

No calculator

The curve $C$ has equation $y = x^3 + ax^2 + bx + 4$, where $a$ and $b$ are constants. $C$ has a point of inflection at $x = 2$. The gradient of $C$ at the point where it crosses the $y$-axis is 3.

(a)

Show that $a = -6$.

(2)

(b)

Find the value of $b$.

(1)

(c)

Find the exact $x$-coordinates of the stationary points of $C$.

(2)

(d)

Use the second derivative to show that the stationary point with the larger $x$-coordinate is a local minimum.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{d^2y}{dx^2} = 6x + 2a$ and set equal to 0 at $x = 2$	✓ 1
A1 $12 + 2a = 0$, so $a = -6$	✓ 1
Part (b)
A1 $\frac{dy}{dx} = 3x^2 - 12x + b$, at $x=0$ equals $b = 3$	✓ 1
Part (c)
M1 solve $3x^2 - 12x + 3 = 0$, i.e. $x^2 - 4x + 1 = 0$	✓ 1
A1 $x = 2 \pm \sqrt{3}$	✓ 1
Part (d)
M1 $\frac{d^2y}{dx^2} = 6x - 12$ evaluated at $x = 2 + \sqrt{3}$	✓ 1
R1 $6\sqrt{3} > 0$, so a local minimum	✓ 1

Exam tip: Use each given condition once: one gives an equation in $a$, another an equation in $b$, and solving the quadratic in exact form keeps the surds for the next part.

+ Add to test✓ In your test307IBMA3S-5-18r44-135.4 Tangents and normals Paper 1Hard13 marks

No calculator

Let $f(x)=\frac{x^2}{x+1}$, for $x\neq-1$.

(a)

Show that $f'(x)=\frac{x(x+2)}{(x+1)^2}$.

(3)

(b)

Find the equation of the tangent to the graph of $f$ at the point where $x=1$.

(3)

(c)

Find the equation of the normal to the graph of $f$ at the point where $x=1$.

(2)

(d)

The tangent and the normal in (b) and (c) meet the $y$-axis at two points. Together with the point where $x=1$ on the graph, these form the vertices of a triangle. Find the exact area of this triangle.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the quotient rule	✓ 1
A1 $\frac{2x(x+1)-x^2}{(x+1)^2}$	✓ 1
A1 expand numerator to $x^2+2x$ and factorise to reach the given form	✓ 1
Part (b)
A1 $y=\frac{1}{2}$ when $x=1$	✓ 1
A1 gradient $\frac{3}{4}$ when $x=1$	✓ 1
A1 $y=\frac{3}{4}x-\frac{1}{4}$	✓ 1
Part (c)
M1 gradient of normal $-\frac{4}{3}$	✓ 1
A1 $y=-\frac{4}{3}x+\frac{11}{6}$	✓ 1
Part (d)
M1 set $x=0$ in both equations	✓ 1
A1 $-\frac{1}{4}$	✓ 1
A1 $\frac{11}{6}$	✓ 1
M1 area $=\frac{1}{2}\times\text{base}\times\text{height}$ with base on the $y$-axis and height 1	✓ 1
A1 $\frac{25}{24}$	✓ 1

Exam tip: Take the base of the triangle along the $y$-axis; the perpendicular height is then the $x$-coordinate of the third vertex.

+ Add to test✓ In your test308AASL5-q00055.1 Limits and the derivative Paper 2Easy5 marks

Let $g(h)=\dfrac{3^{h}-1}{h}$, where $h\neq0$. The table shows some values of $g(h)$, correct to four decimal places.

$h$	$-0.1$	$-0.01$	$-0.001$	$0.001$	$0.01$	$0.1$
$g(h)$	1.0404	$p$	1.0980	1.0992	$q$	1.1612

(a)

Find the value of $p$ and the value of $q$.

(2)

(b)

Use the table to write down an estimate, correct to three significant figures, for $\displaystyle\lim_{h\to0}\dfrac{3^{h}-1}{h}$.

(1)

(c)

The value of $\dfrac{3^{h}-1}{h}$ is the gradient of the chord joining the points $(0,\,1)$ and $(h,\,3^{h})$ on the curve $y=3^{x}$. State what the limit in part (b) represents for the curve $y=3^{x}$.

(1)

(d)

Hence write down the equation of the tangent to $y=3^{x}$ at the point $(0,\,1)$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$p=1.0926$	A1	$1.092599\ldots$
$q=1.1047$	A1	$1.104669\ldots$ Award A1A0 for both values correct but not to four decimal places.
(b)
$1.10$	A1	Do not accept $1.099$ or $1.1$ (not to three significant figures). The exact value is $\ln 3=1.0986\ldots$, which is not required.
(c)
the gradient of the tangent to the curve at $(0,\,1)$	A1	Accept “the gradient of the curve at $x=0$” or “the instantaneous rate of change of $3^{x}$ at $x=0$”.
(d)
$y=1.10x+1$	A1	FT from (b). Accept $y=1.1x+1$ or $y=(\ln3)x+1$.

Answers: (a) $p=1.0926$, $q=1.1047$ · (b) $1.10$ · (c) the gradient of the tangent at $(0,1)$ · (d) $y=1.10x+1$

+ Add to test✓ In your test309AASL5-q00075.1 Limits and the derivative Paper 2Easy6 marks

A ball is thrown vertically upwards. Its height, $s$ metres, above the point of release after $t$ seconds is given by $s(t)=20t-4.9t^{2}$. The average velocity of the ball between $t=1$ and $t=1+h$ is $\dfrac{s(1+h)-s(1)}{h}$. The table shows some values of this average velocity.

$h$	$0.1$	$0.01$	$0.001$
Average velocity (m s⁻¹)	9.71	$p$	$q$

(a)

Show that the average velocity between $t=1$ and $t=1.1$ is $9.71$ m s⁻¹.

(2)

(b)

Find the value of $p$ and the value of $q$.

(2)

(c)

Use the table to estimate the velocity of the ball at the instant $t=1$.

(1)

(d)

Write down, using derivative notation, the quantity that you have estimated in part (c).

(1)

Show mark scheme

Working	Mark	Notes
(a)
$s(1.1)=16.071$ and $s(1)=15.1$	M1	Award for both values seen.
$\dfrac{16.071-15.1}{0.1}=\dfrac{0.971}{0.1}$	A1
$=9.71$	AG
(b)
$p=10.151$	A1	Accept $10.2$.
$q=10.1951$	A1	Accept $10.195$ or $10.2$.
(c)
$10.2$ m s⁻¹	A1	Accept any value in $[10.19,\ 10.2]$.
(d)
$\dfrac{\mathrm{d}s}{\mathrm{d}t}$ at $t=1$	A1	Accept $s'(1)$.

Answers: (b) $p=10.151$, $q=10.1951$ · (c) $10.2$ m s⁻¹ · (d) $s'(1)$

+ Add to test✓ In your test310AASL5-q00385.5 Introduction to integration Paper 2Easy5 marks

Let $f(x)=2+3x-x^{2}$. The graph of $f$ meets the $x$-axis at the points where $x=p$ and $x=q$, where $p\lt q$.

Diagram not to scale

(a)

Find the value of $p$ and the value of $q$.

(2)

(b)

Write down an integral that represents the area of the region enclosed by the graph of $f$ and the $x$-axis.

(1)

(c)

Find the area of this region.

(1)

(d)

Explain why $\displaystyle\int_{0}^{5}f(x)\,\mathrm{d}x$ does not give the area of the region enclosed by the graph of $f$, the $x$-axis and the lines $x=0$ and $x=5$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$p=-0.562$	A1	$-0.561552\ldots$
$q=3.56$	A1	$3.56155\ldots$
(b)
$\displaystyle\int_{-0.562}^{3.56}(2+3x-x^{2})\,\mathrm{d}x$	A1	FT from (a). Accept $\int_{p}^{q}f(x)\,\mathrm{d}x$ with their values.
(c)
$11.7$	A1	$11.6821\ldots$ (exactly $\tfrac{17\sqrt{17}}{6}$)
(d)
$f(x)\lt 0$ for $3.56\lt x\le 5$, so part of the region is below the $x$-axis and that part contributes a negative value to the integral	R1

Answers: (a) $p=-0.562$, $q=3.56$ · (c) $11.7$ · (d) $f(x)\lt 0$ for part of the interval

+ Add to test✓ In your test311AASL5-q00865.9 Kinematics Paper 2Easy7 marks

A particle moves in a straight line. Its velocity, $v\ \text{m s}^{-1}$, after $t$ seconds is given by $v(t)=\mathrm{e}^{0.5t}-3$, for $0\le t\le 4$.

(a)

Write down the initial velocity of the particle.

(1)

(b)

Find the value of $t$ at which the particle changes direction.

(2)

(c)

Find the value of $t$, for $t\gt 0$, at which the speed of the particle is again equal to its initial speed.

(2)

(d)

Find the total distance travelled by the particle in the 4 seconds.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$-2\ \text{m s}^{-1}$	A1
(b)
$\mathrm{e}^{0.5t}-3=0$, or graph of $v$ crossing the $t$-axis	(M1)
$t=2.20$	A1	$2\ln 3=2.19722\ldots$
(c)
$\mathrm{e}^{0.5t}-3=2$	(M1)	Recognizing that the velocity must now be $+2$; $v=-2$ only at $t=0$.
$t=3.22$	A1	$2\ln 5=3.21887\ldots$ Award A0 for $t=0$.
(d)
$\displaystyle\int_{0}^{4}\left\|\mathrm{e}^{0.5t}-3\right\|\mathrm{d}t$	(M1)	Or the two separate integrals, split at their (b), added as positive values: $2.59167\ldots+3.36978\ldots$
$5.96$ m	A1	$5.96145\ldots$ Award A0 for $0.778$ (the displacement).

Answers: (a) $-2\ \text{m s}^{-1}$ · (b) $t=2.20$ · (c) $t=3.22$ · (d) 5.96 m

+ Add to test✓ In your test312IBMA2S-052-05-115.2 Increasing and decreasing functions Paper 2Easy6 marks

Calculator

The curved roof of a sports stadium is modelled by y = −0.0005x³ + 0.045x² + 10, for 0 ≤ x ≤ 80. Here y is the height in metres and x is the horizontal distance in metres from one edge of the roof.

(a)

Find expressions for $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$.

(2)

(b)

Find the coordinates of the point of inflection on the roof.

(3)

(c)

State the interval in which the roof is concave up.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 dy/dx = −0.0015x² + 0.09x	✓ 1
A1 d²y/dx² = −0.003x + 0.09	✓ 1
Part (b)
M1 sets −0.003x + 0.09 = 0. Because d²y/dx² is linear, it changes sign at this value.	✓ 1
A1 x = 30	✓ 1
A1 y = 37, giving the point (30, 37)	✓ 1
Part (c)
A1 0 ≤ x < 30. Accept 0 ≤ x ≤ 30 or 0 < x < 30.	✓ 1

Exam tip: Give both coordinates when asked for a point. Finding x = 30 but not substituting back to get the height loses a mark.

+ Add to test✓ In your test313IBMA2S-052-08-015.2 Increasing and decreasing functions Paper 2Easy6 marks

Calculator

A delivery drone rises vertically from a rooftop landing pad. Its height above the pad t seconds after take-off is modelled by h(t) = 2t² + 3t metres, for 0 ≤ t ≤ 8.

(a)

Find the average velocity of the drone over the interval 2 ≤ t ≤ 2.1 and over the interval 2 ≤ t ≤ 2.01.

(3)

(b)

Find the average velocity of the drone over the interval 1.99 ≤ t ≤ 2.

(1)

(c)

Hence estimate the instantaneous velocity of the drone at t = 2, justifying your answer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses average velocity = [h(b) − h(a)] ÷ (b − a), with h(2) = 14, h(2.1) = 15.12 or h(2.01) = 14.1102	✓ 1
A1 (15.12 − 14) ÷ 0.1 = 11.2 m s⁻¹	✓ 1
A1 (14.1102 − 14) ÷ 0.01 = 11.02 m s⁻¹	✓ 1
Part (b)
A1 h(1.99) = 13.8902, so (14 − 13.8902) ÷ 0.01 = 10.98 m s⁻¹	✓ 1
Part (c)
R1 average velocities over intervals shrinking towards t = 2 from both sides (11.2, 11.02 and 10.98) approach the same value	✓ 1
A1 instantaneous velocity ≈ 11 m s⁻¹	✓ 1

Exam tip: When estimating an instantaneous rate from chords, use intervals on both sides of the point: values that close in from above and below give a convincing limit.

+ Add to test✓ In your test314IBMA2S-052-08-035.1 Limits and the derivative Paper 2Easy6 marks

Calculator

A mobile phone is plugged into a fast charger. The battery charge, as a percentage of full capacity, t minutes after being plugged in is Q(t), for 0 ≤ t ≤ 90. It is known that Q(30) = 64 and Q′(30) = 0.8.

(a)

Interpret Q′(30) = 0.8 in the context of the question.

(2)

(b)

Use this information to estimate the charge 32.5 minutes after the phone is plugged in.

(2)

(c)

Charging slows down as the battery fills, so Q′(t) decreases for t > 30. State whether your estimate of the charge at 32.5 minutes is an overestimate or an underestimate, giving a reason.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
R1 it is the instantaneous rate of change of charge at the moment 30 minutes after plugging in	✓ 1
A1 the charge is increasing at 0.8 percentage points per minute at that moment	✓ 1
Part (b)
M1 Q(32.5) ≈ 64 + 2.5 × 0.8	✓ 1
A1 66 %	✓ 1
Part (c)
A1 overestimate	✓ 1
R1 the rate of charging between t = 30 and t = 32.5 is less than 0.8, so the actual increase is less than 2 percentage points	✓ 1

Exam tip: A full interpretation of a derivative mentions three things: the rate, the units of both quantities and the specific moment, here t = 30.

+ Add to test✓ In your test315IBMA2S-052-09-035.5 Introduction to integration Paper 2Easy4 marks

Calculator

After an injection, the rate at which insulin is absorbed into a patient's bloodstream is modelled by
$r(t) = 4t\,e^{-0.5t}$ units per hour,
where $t$ is the time in hours after the injection.

(a)

Find the rate of absorption 2 hours after the injection.

(1)

(b)

Find the total amount of insulin absorbed during the first 6 hours.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $r(2) = 8e^{-1} = 2.94$ units per hour	✓ 1
Part (b)
M1 recognise that the total is the integral of the rate	✓ 1
A1 $\int_0^6 4t\,e^{-0.5t}\,dt$ with correct limits and integrand	✓ 1
A1 12.8 units	✓ 1

Exam tip: When a question gives a rate and asks for a total, write the definite integral with its limits before you use the GDC. The written integral can earn method marks even if the final value is wrong.

+ Add to test✓ In your test316IBMA2S-052-09-045.11 Definite integrals and areas Paper 2Easy5 marks

Calculator

The front face of a pedestrian footbridge arch is the region between the curve $y = 6 - 0.15x^2$ and the $x$-axis, where $x$ and $y$ are measured in metres.

Diagram not to scale

(a)

Find the exact $x$-coordinates of the points where the arch meets the ground.

(2)

(b)

Find the area of the front face of the arch.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 set $6 - 0.15x^2 = 0$, giving $x^2 = 40$	✓ 1
A1 $x = \pm 2\sqrt{10}$ (accept $\pm\sqrt{40}$)	✓ 1
Part (b)
M1 an integral of $6 - 0.15x^2$ used for the area	✓ 1
A1 correct limits, e.g. $\int_{-2\sqrt{10}}^{2\sqrt{10}} (6 - 0.15x^2)\,dx$ or $2\int_{0}^{2\sqrt{10}} (6 - 0.15x^2)\,dx$	✓ 1
A1 50.6 m²	✓ 1

Exam tip: Write the integral with its limits before you evaluate it on the GDC. That line earns method marks even if the final value is keyed in wrongly. If you integrate from 0 to $2\sqrt{10}$ using symmetry, remember to double the result.

+ Add to test✓ In your test317IBMA2S-052-01-035.7 The second derivative and graphs Paper 2Easy4 marks

Calculator

After a patient takes a tablet, the concentration of the drug in the blood is modelled by $C(t) = 20t\,e^{-0.5t}$ mg L⁻¹, where $t$ is the time in hours, $0 \le t \le 8$. In pharmacology, the area under the concentration–time curve (AUC) measures the patient's total exposure to the drug.

(a)

Find the time at which the concentration is greatest, and write down this maximum concentration.

(2)

(b)

Find the AUC for $0 \le t \le 8$, giving its units.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $t = 2$ hours	✓ 1
A1 $C_{\max} = \frac{40}{e} \approx 14.7$ mg L⁻¹	✓ 1
Part (b)
M1 writes $\displaystyle\int_0^8 20t\,e^{-0.5t}\,dt$	✓ 1
A1 72.7 mg h L⁻¹	✓ 1

Exam tip: The units of an area under a graph are (units of $y$) × (units of $x$). Here that is mg L⁻¹ × h.

+ Add to test✓ In your test318IBMA2S-052-02-025.8 Stationary points, inflexion and optimization Paper 2Easy5 marks

Calculator

The fuel efficiency of a hybrid car, in kilometres per litre, when travelling at a constant speed of $v$ km h⁻¹ is modelled by
$F(v)=4+6\ln v-0.08v$, for $20\le v\le 130$.

(a)

Find $F'(v)$.

(1)

(b)

Find the speed at which the fuel efficiency is greatest.

(2)

(c)

Use the second derivative to justify that this speed gives a maximum.

(1)

(d)

Find the maximum fuel efficiency.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $F'(v)=\frac{6}{v}-0.08$	✓ 1
Part (b)
M1 sets $\frac{6}{v}-0.08=0$	✓ 1
A1 $v=75$ km h⁻¹	✓ 1
Part (c)
R1 $F''(v)=-\frac{6}{v^2}<0$ for all $v$ in the domain (in particular at $v=75$), so maximum	✓ 1
Part (d)
A1 $F(75)=4+6\ln 75-6=23.9$ km per litre	✓ 1

Common error: Giving only the speed when asked for the maximum efficiency: the maximum value of a function is $F(75)$, not the value $v=75$.

+ Add to test✓ In your test319IBMA2S-052-02-065.2 Increasing and decreasing functions Paper 2Easy6 marks

Calculator

A section of a roller-coaster track is modelled by
$h(x)=0.002x^3-0.18x^2+4.05x+10$, for $0\le x\le 65$,
where $h$ is the height in metres above the ground and $x$ is the horizontal distance in metres from the start of the section.

(a)

Find $h'(x)$.

(1)

(b)

Find the values of $x$ at which the track has stationary points.

(2)

(c)

Find the greatest height of the track in this section, stating where it occurs.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $h'(x)=0.006x^2-0.36x+4.05$	✓ 1
Part (b)
M1 sets $h'(x)=0$	✓ 1
A1 $x=15$ and $x=45$	✓ 1
Part (c)
M1 evaluates $h$ at the stationary points and at both endpoints	✓ 1
A1 $h(0)=10$, $h(15)=37$, $h(45)=10$, $h(65)=62$	✓ 1
A1 greatest height 62 m, at the end of the section, $x=65$	✓ 1

Exam tip: On a closed interval the greatest value can occur at an endpoint. Always compare the local maximum with the function values at both ends of the domain.

+ Add to test✓ In your test320IBMA2S-052-07-025.7 The second derivative and graphs Paper 2Easy6 marks

Calculator

After an injection, the concentration of a drug in a patient's bloodstream is modelled by $C(t) = 20t\,e^{-0.5t}$, where $C$ is measured in mg L⁻¹ and $t$ is the time in hours after the injection. Give answers correct to three significant figures.

(a)

Find the concentration when $t = 1$.

(1)

(b)

Find the gradient of the tangent to the graph of $C$ at $t = 1$.

(2)

(c)

Hence find the equation of the tangent to the graph of $C$ at $t = 1$.

(2)

(d)

Use your tangent to estimate the concentration when $t = 1.2$.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 12.1 mg L⁻¹ (accept $20e^{-0.5}$ or 12.1306…)	✓ 1
Part (b)
M1 attempt to find $C'(t)$, e.g. product rule $C'(t) = 20e^{-0.5t} - 10te^{-0.5t}$, or use of GDC to evaluate the derivative at $t = 1$	✓ 1
A1 6.07 (accept $10e^{-0.5}$ or 6.0653…)	✓ 1
Part (c)
M1 substitution of their values into $C - 12.13 = 6.065(t - 1)$ or equivalent, using values from parts 1 and 2	✓ 1
A1 $C = 6.07t + 6.07$ (accept $C = 6.07t + 6.03$ from using rounded values 12.1 and 6.07; accept $C = 10e^{-0.5}(t + 1)$)	✓ 1
Part (d)
A1 13.3 mg L⁻¹ (accept 13.4 from using the rounded coefficients $C = 6.07t + 6.07$; follow through from their tangent in part 3)	✓ 1

Exam tip: Keep the unrounded gradient and intercept stored in your GDC when making the estimate; rounding early can change the third significant figure.

+ Add to test✓ In your test321IBMA2S-052-06-035.7 The second derivative and graphs Paper 2Easy7 marks

Calculator

After an injection, the concentration of a painkiller in a patient's blood is modelled by C(t) = 12te^−0.4t mg L⁻¹, where t is the time in hours, t ≥ 0.

(a)

Show that C′(t) = 12e^−0.4t(1 − 0.4t).

(3)

(b)

Hence find the interval during which the concentration is increasing.

(2)

(c)

Find the rate of change of the concentration when t = 5.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts the product rule with u = 12t and v = e^−0.4t	✓ 1
A1 correct unsimplified derivative 12e^−0.4t + 12t(−0.4e^−0.4t)	✓ 1
A1 12e^−0.4t − 4.8te^−0.4t = 12e^−0.4t(1 − 0.4t)	✓ 1
Part (b)
R1 12e^−0.4t > 0, so C′(t) > 0 when 1 − 0.4t > 0	✓ 1
A1 0 ≤ t < 2.5. Also accept 0 < t < 2.5 or 0 ≤ t ≤ 2.5	✓ 1
Part (c)
M1 substitutes t = 5 into C′(t), giving 12e⁻²(1 − 2)	✓ 1
A1 −1.62 mg L⁻¹ per hour. Also accept −12e⁻²	✓ 1

Exam tip: An exponential factor is always positive, so state this explicitly: it is the reason the sign of the derivative depends only on the linear factor.

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The height, h metres, of a weather balloon above a launch platform is modelled by h(t) = −0.4t³ + 5t² + 8t + 100, where t is the time in minutes and 0 ≤ t ≤ 14.

(a)

Find h′(t).

(2)

(b)

Find the interval of time during which the balloon is descending.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 −1.2t²	✓ 1
A1 + 10t + 8	✓ 1
Part (b)
M1 solves h′(t) = 0	✓ 1
A1 t = 9.07 (rejecting t = −0.735)	✓ 1
A1 9.07 < t ≤ 14	✓ 1

Exam tip: Descending means h is decreasing, so you need h′(t) < 0; check which side of the root this occurs by substituting a value such as t = 10.

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A model for a sprinter's velocity during a 10-second run is $v(t)=10\left(1-e^{-0.8t}\right)$ m s⁻¹, for $0\le t\le 10$, where $t$ is in seconds.

(a)

Find the sprinter's acceleration at $t=0$.

(2)

(b)

Find the sprinter's velocity at $t=2$.

(2)

(c)

Find the distance the sprinter covers in the 10 seconds.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $a(t)=8e^{-0.8t}$	✓ 1
A1 $a(0)=8$ m s⁻²	✓ 1
Part (b)
M1 substitutes $t=2$: $10(1-e^{-1.6})$	✓ 1
A1 $7.98$ m s⁻¹	✓ 1
Part (c)
M1 $\int_0^{10}10(1-e^{-0.8t})\,dt$	✓ 1
A1 $87.5$ m	✓ 1

Exam tip: Velocity is always positive here, so the distance is simply the integral of $v$. Your GDC can evaluate it directly.

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A rocket sled starts from rest on a straight test track. Its acceleration is $a(t)=\dfrac{12}{t+1}$ m s⁻², where $t\ge 0$ is the time in seconds.

(a)

Find an expression for the velocity $v(t)$ of the sled.

(3)

(b)

Find the time at which the sled reaches a velocity of 30 m s⁻¹.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate $a(t)$	✓ 1
A1 $v=12\ln(t+1)+c$	✓ 1
A1 uses $v(0)=0$ to get $c=0$, so $v=12\ln(t+1)$	✓ 1
Part (b)
M1 $12\ln(t+1)=30$	✓ 1
A1 $t=e^{2.5}-1=11.2$ s	✓ 1

Common error: Forgetting the constant of integration. Here it happens to be zero, but you must show it using the initial condition.

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During an outbreak in a small town, the number of active cases is modelled by A(t) = −t³ + 9t² + 48t + 100, for 0 ≤ t ≤ 12, where t is the number of days since the outbreak was first recorded.

(a)

Find A′(t).

(2)

(b)

Find the value of t at which the number of active cases is greatest.

(2)

(c)

Find the greatest number of active cases.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate (at least one term correct)	✓ 1
A1 $A'(t)=-3t^2+18t+48$	✓ 1
Part (b)
M1 sets $A'(t)=0$ and solves: $-3(t-8)(t+2)=0$	✓ 1
A1 $t=8$ (rejects $t=-2$ as outside the domain)	✓ 1
Part (c)
A1 $A(8)=-512+576+384+100=548$	✓ 1

Exam tip: State why a root such as t = −2 is rejected: it lies outside the stated domain.

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The number of radioactive atoms remaining in an archaeological sample is modelled by $N(t)=600e^{-0.000121t}$, where $N$ is in thousands of atoms and $t$ is the age of the sample in years.

(a)

Find $\dfrac{dN}{dt}$.

(2)

(b)

Find the rate of change of $N$ when $t=4000$. Give your answer to 3 significant figures.

(2)

(c)

At a certain age the number of atoms is falling at a rate of $0.03$ thousand atoms per year. Find this age, to the nearest year.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiates using $\frac{d}{dt}e^{kt}=ke^{kt}$	✓ 1
A1 $\frac{dN}{dt}=-0.0726e^{-0.000121t}$	✓ 1
Part (b)
M1 substitutes $t=4000$ into $-0.0726e^{-0.000121t}$	✓ 1
A1 $-0.0447$ thousand atoms per year	✓ 1
Part (c)
M1 sets $0.0726e^{-0.000121t}=0.03$	✓ 1
M1 $e^{-0.000121t}=0.4132\ldots$ and takes natural logarithms	✓ 1
A1 $t=7304$ years	✓ 1

Exam tip: Keep unrounded values in your calculator until the final line, then round to the accuracy asked.

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During roasting, the temperature of a batch of coffee beans rises at a rate of $R(t)=40e^{-0.2t}$ °C per minute, where t is the time in minutes after the beans are loaded, for $0 \le t \le 10$.

(a)

Find the total rise in temperature of the beans during the first 10 minutes. Give your answer to 3 significant figures.

(4)

(b)

The beans are loaded at 22 °C. Find the temperature of the beans at the end of the 10 minutes. Use the unrounded answer from part (a) and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate $R(t)$ over 0 to 10, e.g. $\int_0^{10}40e^{-0.2t}\,dt$	✓ 1
M1 $-200e^{-0.2t}$ (or $200(1-e^{-2})$) seen	✓ 1
A1 172.93… seen (or value from calculator)	✓ 1
A1 173 °C	✓ 1
Part (b)
M1 adds 22 to the temperature rise	✓ 1
A1 194.93… so 195 °C (ECF from (a))	✓ 1

Common error: The integral of $e^{kt}$ is $\frac{1}{k}e^{kt}$. For $k=-0.2$ the factor $\frac{1}{k}$ is $-5$, so do not forget the negative sign.

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A marine biologist tracks an underwater camera moving along a straight line. Its velocity at time $t$ seconds is $v = t^2 - 6t + 8$ m s⁻¹, for $0 \le t \le 5$. At $t=0$ the camera is at its starting position.

(a)

Find the acceleration of the camera when $t = 4$.

(2)

(b)

Find the times at which the camera is momentarily at rest.

(2)

(c)

Find the displacement of the camera from its starting position when $t = 3$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $a = \frac{dv}{dt} = 2t - 6$	✓ 1
A1 2 m s⁻²	✓ 1
Part (b)
M1 sets $v = 0$ and solves $t^2-6t+8=0$	✓ 1
A1 $t = 2$ and $t = 4$	✓ 1
Part (c)
M1 $\int_0^3 (t^2-6t+8)\,dt$	✓ 1
A1 6 m	✓ 1

Exam tip: Velocity is the derivative of displacement and the integral of acceleration: choose the direction of the calculus carefully.

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A trawler's winch brings fish aboard at a rate of $R(t)=60e^{-0.2t}$ kilograms per minute, where $t$ is the time in minutes since hauling began, for $0\le t\le 10$.

(a)

Find the total mass of fish hauled aboard in the first 10 minutes. Give your answer to the nearest kilogram.

(3)

(b)

At time $T$ the rate has fallen to 20 kg per minute. Find $T$, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\int_0^{10}60e^{-0.2t}\,dt$ (correct limits)	✓ 1
A1 $\left[-300e^{-0.2t}\right]_0^{10}$ or $259.40\ldots$ from a calculator	✓ 1
A1 $259$ kg	✓ 1
Part (b)
M1 $60e^{-0.2T}=20$ (or $e^{-0.2T}=\frac13$)	✓ 1
A1 $T=5.49$ minutes ($5\ln 3=5.4931\ldots$)	✓ 1

Exam tip: A rate integrated over time gives a total amount: write the integral with limits before using your calculator, so the method mark is secure.

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In food science, the concentration $C$ g/L of a volatile flavour compound in a simmering sauce is modelled by $C(t)=8+12e^{-0.5t}$, where $t$ is the time in hours after the sauce reaches simmering point, $t\geq0$.

(a)

Find $C'(t)$.

(2)

(b)

Find the rate of change of concentration when $t=3$. Give your answer to 3 significant figures.

(2)

(c)

Find the time at which the concentration first reaches 10 g/L. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate $e^{-0.5t}$ (chain rule, factor $-0.5$)	✓ 1
A1 $C'(t)=-6e^{-0.5t}$	✓ 1
Part (b)
M1 substitutes $t=3$: $-6e^{-1.5}$	✓ 1
A1 $-1.34$ g/L per hour	✓ 1
Part (c)
M1 $8+12e^{-0.5t}=10$	✓ 1
M1 $e^{-0.5t}=\frac16$ so $t=-2\ln\frac16=2\ln6$	✓ 1
A1 $t=3.58$ hours	✓ 1

Common error: When differentiating $e^{kt}$ remember the factor $k$ from the chain rule, which here is $-0.5$.

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A diver steps off a 10 m platform. Her height above the water is $h$ metres at $t$ seconds after leaving the platform, where $h(t)=10-4.9t^2$ for $0\le t\le 1.4$.

(a)

Find $h'(t)$.

(2)

(b)

Find the rate of change of her height when $t=1.2$, and state what the sign of your answer tells you about her motion.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts to differentiate	✓ 1
A1 $h'(t)=-9.8t$	✓ 1
Part (b)
A1 $h'(1.2)=-9.8\times1.2=-11.76$, so $-11.8\ \text{m s}^{-1}$ to 3 s.f.	✓ 1
R1 negative, so her height is decreasing (she is moving downwards at about $11.8\ \text{m s}^{-1}$)	✓ 1

Exam tip: A negative rate of change means the quantity is decreasing. State this in context, with units.

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The electrical power drawn by a rack of servers in a data centre is modelled by $P(t) = 0.5t^2 - 4t + 12$ kilowatts, where $t$ is the time in hours after 08:00, for $0 \le t \le 6$.

(a)

Write down an integral for the energy used between $t=0$ and $t=6$, and evaluate it. Give the units of your answer.

(4)

(b)

Find the mean power over these six hours.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $\int_0^6 (0.5t^2-4t+12)\,dt$	✓ 1
M1 attempt to integrate	✓ 1
A1 $\frac{t^3}{6}-2t^2+12t$	✓ 1
A1 $36$ kWh	✓ 1
Part (b)
M1 $\frac{36}{6}$ (ECF from (a))	✓ 1
A1 $6$ kW	✓ 1

Exam tip: Power in kW integrated over time in hours gives energy in kWh, so the units can be used to check the setup.

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A Formula 1 car starts from rest at the start line and accelerates in a straight line. For $0\le t\le 10$ its acceleration is $a=12-0.6t$ m s⁻², where t is the time in seconds.

(a)

Find an expression for the velocity, v m s⁻¹, of the car in terms of t.

(2)

(b)

Find the speed of the car when $t=10$.

(1)

(c)

Find the distance travelled by the car during the first 10 seconds.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 integrate: $\int(12-0.6t)\,dt=12t-0.3t^2+c$	✓ 1
A1 $c=0$ because $v=0$ when $t=0$, so $v=12t-0.3t^2$	✓ 1
Part (b)
A1 $v=12(10)-0.3(10)^2=90$ m s⁻¹ (ECF from (a))	✓ 1
Part (c)
M1 $\int_0^{10}(12t-0.3t^2)\,dt=\left[6t^2-0.1t^3\right]_0^{10}$ (ECF from (a))	✓ 1
A1 distance $=600-100=500$ m	✓ 1

Common error: Leaving out the constant of integration. Here the condition that the car starts from rest shows that the constant is zero, so state this rather than ignoring it.

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The curve with equation $y=x^3-3x^2+4x$ has a tangent that is parallel to the line $y=7x-2$ at two points.

(a)

Find $\frac{dy}{dx}$.

(2)

(b)

Find the $x$-coordinates of these two points, giving your answers to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate	✓ 1
A1 $3x^2-6x+4$	✓ 1
Part (b)
M1 equate $\frac{dy}{dx}$ to $7$	✓ 1
A1 $x=-0.414$	✓ 1
A1 $x=2.41$	✓ 1

Exam tip: Parallel lines have equal gradients, so set the derivative equal to the gradient of the given line.

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The number of bees, $N$, in a colony $t$ weeks after it is established is modelled by $N(t)=\dfrac{2000}{1+49\mathrm{e}^{-0.6t}}$, for $t\ge0$.

(a)

Write down the initial number of bees in the colony.

(1)

(b)

Find the rate at which the number of bees is increasing when $t=4$.

(2)

(c)

Find the value of $t$ at which the number of bees is increasing most rapidly.

(2)

(d)

Write down the number of bees in the colony at this time.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$40$	A1
(b)
$N'(4)$, e.g. GDC derivative at $t=4$	(M1)
$180$ bees per week	A1	$179.906\ldots$ Award A0 for $N(4)=367$.
(c)
maximum of the graph of $\dfrac{\mathrm{d}N}{\mathrm{d}t}$ on GDC	(M1)	Or zero of $N''(t)$.
$t=6.49$ weeks	A1	$6.48636\ldots\ \left(=\dfrac{\ln49}{0.6}\right)$
(d)
$1000$	A1

Answers: (a) 40 · (b) 180 bees per week · (c) $t=6.49$ · (d) 1000

+ Add to test✓ In your test336AASL5-q00085.1 Limits and the derivative Paper 2Medium7 marks

A furniture maker produces $x$ tables per week. The weekly cost, in euros, of producing $x$ tables is modelled by $C(x)=0.001x^{3}-0.3x^{2}+45x+2000$, for $0\le x\le 250$. The marginal cost is defined as $C'(x)$.

(a)

Find $C'(x)$.

(1)

(b)

Find $C'(150)$ and interpret this value in context.

(2)

(c)

Find the actual increase in the weekly cost when production increases from 150 to 151 tables, and comment on your answer in relation to part (b).

(2)

(d)

Find the values of $x$ for which the marginal cost is €20 per table.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$C'(x)=0.003x^{2}-0.6x+45$	A1
(b)
$C'(150)=22.5$	A1
when 150 tables are made, the cost is increasing at about €22.50 per extra table	R1	Must refer to rate of change of cost per table.
(c)
$C(151)-C(150)=5397.651-5375=$€$22.65$	A1	$22.651$ Accept $22.7$.
this is close to $C'(150)$, so the marginal cost is a good estimate of the cost of one extra table	R1
(d)
$0.003x^{2}-0.6x+45=20$, e.g. graph of $C'$ and $y=20$	(M1)
$x=59.2$ or $x=141$	A1	$59.1751\ldots,\ 140.824\ldots$ Accept $59$ and $141$. Award A0 for only one value.

Answers: (a) $0.003x^{2}-0.6x+45$ · (b) €22.50 per table · (c) €22.65 · (d) $x=59.2,\ 141$

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The height, $h$ metres, of a drone above the ground $t$ seconds after it is switched on is modelled by $h(t)=0.0004t^{3}-0.036t^{2}+0.8t+5$, for $0\le t\le 60$.

(a)

Find $h'(t)$.

(1)

(b)

Find $h'(10)$ and state what it tells you about the motion of the drone at that time.

(2)

(c)

Find the interval of time during which the drone is descending.

(3)

(d)

Find the height of the drone at the instant it starts to descend.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$h'(t)=0.0012t^{2}-0.072t+0.8$	A1
(b)
$0.2$	A1
positive, so the drone is rising (at 0.2 m s⁻¹)	R1
(c)
$h'(t)\lt 0$, e.g. solving $0.0012t^{2}-0.072t+0.8=0$	M1
$t=14.7$ and $t=45.3$	A1	$14.7247\ldots,\ 45.2752\ldots$
$14.7\lt t\lt 45.3$	A1	Accept non-strict inequalities.
(d)
$h(14.7247\ldots)$	(M1)
$10.3$ m	A1	$10.2513\ldots$ Accept $10.25$.

Answers: (a) $0.0012t^{2}-0.072t+0.8$ · (b) $0.2$, rising · (c) $14.7\lt t\lt 45.3$ · (d) $10.3$ m

+ Add to test✓ In your test338AASL5-q00305.4 Tangents and normals Paper 2Medium6 marks

Let $f(x)=3\sin x+x$, for $0\le x\le 2\pi$.

(a)

Find $f'(x)$.

(1)

(b)

Find the equation of the tangent to the graph of $f$ at the point where $x=2$. Give your answer in the form $y=mx+c$.

(3)

(c)

The tangent meets the graph of $f$ again at the point Q. Find the $x$-coordinate of Q.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$f'(x)=3\cos x+1$	A1
(b)
$f(2)=4.72789\ldots$	A1
gradient $=f'(2)=3\cos2+1=-0.248440\ldots$	(M1)	Award for substituting $x=2$ into their $f'$ (or using the GDC derivative).
$y=-0.248x+5.22$	A1	$c=5.22477\ldots$ Accept $y-4.73=-0.248(x-2)$.
(c)
$3\sin x+x=-0.248440x+5.22477$, e.g. graph of both or of their difference	(M1)
$x=5.64$	A1	$5.63528\ldots$ Accept $5.63$ or $5.64$ (premature rounding of the tangent gives $5.6347$). Award A0 for $x=2$.

Answers: (a) $f'(x)=3\cos x+1$ · (b) $y=-0.248x+5.22$ · (c) $x=5.64$

+ Add to test✓ In your test339AASL5-q00375.11 Definite integrals and areas Paper 2Medium5 marks

Let $f(x)=1+\ln(x^{2}+1)$, for $x\ge0$. The shaded region R is enclosed by the graph of $f$, the $x$-axis, the $y$-axis and the line $x=3$.

(a)

Write down an integral that represents the area of R.

(1)

(b)

Find the area of R.

(1)

(c)

The line $x=k$ divides R into two regions of equal area. Find the value of $k$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$\displaystyle\int_{0}^{3}\big(1+\ln(x^{2}+1)\big)\,\mathrm{d}x$	A1	Accept the absence of d$x$.
(b)
$6.41$	A1	$6.40584\ldots$
(c)
$\displaystyle\int_{0}^{k}\big(1+\ln(x^{2}+1)\big)\,\mathrm{d}x=\tfrac12\times6.40584\ldots$	(M1)	Or $\int_{0}^{k}f=\int_{k}^{3}f$.
$=3.20292\ldots$	(A1)
$k=1.91$	A1	$1.91052\ldots$ Award A1 only for a value with $0\lt k\lt 3$.

Answers: (b) $6.41$ · (c) $k=1.91$

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Let $f(x)=kx(4-x)^{2}$, where $k\gt 0$. The following diagram shows part of the graph of $f$. The shaded region R is enclosed by the graph of $f$ and the $x$-axis, for $0\le x\le 4$.

Diagram not to scale

(a)

Write down an integral, in terms of $k$, that represents the area of R.

(1)

(b)

The area of R is 32 square units. Find the value of $k$.

(3)

(c)

By expanding $f(x)$, find the equation of the tangent to the graph of $f$ at the origin.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$\displaystyle\int_{0}^{4}kx(4-x)^{2}\,\mathrm{d}x$	A1
(b)
$k\displaystyle\int_{0}^{4}x(4-x)^{2}\,\mathrm{d}x=32$	(M1)
$\displaystyle\int_{0}^{4}x(4-x)^{2}\,\mathrm{d}x=21.3$	(A1)	$21.333\ldots=\tfrac{64}{3}$
$k=1.5$	A1
(c)
$f(x)=k(16x-8x^{2}+x^{3})$, so $f'(x)=k(16-16x+3x^{2})$	(M1)
$f'(0)=16k=24$	(A1)	FT from their $k$.
$y=24x$	A1

Answers: (a) $\int_0^4 kx(4-x)^2\,\mathrm{d}x$ · (b) $k=1.5$ · (c) $y=24x$

+ Add to test✓ In your test341AASL5-q00405.5 Introduction to integration Paper 2Medium14 marks

During a storm, the rate of rainfall, $r$ millimetres per hour, $t$ hours after the storm begins is modelled by $r(t)=0.1t^{2}(9-t)$, for $0\le t\le 9$.

(a)

Find the rate of rainfall when $t=5$.

(1)

(b)

Find $r'(t)$.

(2)

(c)

Find the interval of time during which the rate of rainfall is increasing.

(3)

(d)

Find $r'(7)$ and interpret your answer in context.

(2)

(e)

Find the total rainfall during the storm.

(2)

(f)

Find the rainfall during the last three hours of the storm.

(2)

(g)

Find the time at which half of the total rainfall has fallen.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$10$ mm per hour	A1
(b)
$r(t)=0.9t^{2}-0.1t^{3}$ or $1.8t$	A1	Award for expanding correctly or for one correct term of $r'(t)$.
$r'(t)=1.8t-0.3t^{2}$	A1
(c)
$1.8t-0.3t^{2}=0$ or $r'(t)\gt 0$	M1
$0.3t(6-t)=0\Rightarrow t=0,\ t=6$	A1
$0\lt t\lt 6$	A1	Accept $0\le t\le 6$.
(d)
$r'(7)=-2.1$	A1
the rate of rainfall is decreasing at 2.1 mm per hour, per hour	R1	Must refer to the rate of rainfall decreasing; units not required.
(e)
$\displaystyle\int_{0}^{9}0.1t^{2}(9-t)\,\mathrm{d}t$	(M1)
$54.7$ mm	A1	$54.675$ (exactly $\tfrac{2187}{40}$)
(f)
$\displaystyle\int_{6}^{9}r(t)\,\mathrm{d}t$	(M1)
$22.3$ mm	A1	$22.275$
(g)
$\displaystyle\int_{0}^{T}r(t)\,\mathrm{d}t=27.3375$	(M1)	FT from (e).
$T=5.53$ hours	A1	$5.52845\ldots$

Answers: (a) 10 mm h⁻¹ · (b) $1.8t-0.3t^{2}$ · (c) $0\lt t\lt 6$ · (d) $-2.1$ · (e) 54.7 mm · (f) 22.3 mm · (g) $5.53$ h

+ Add to test✓ In your test342AASL5-q00535.6 Differentiation rules Paper 2Medium7 marks

The height, $h$ metres, of a seat on a Ferris wheel above the ground $t$ minutes after it passes the lowest point is modelled by $h(t)=36-32\cos\left(\dfrac{\pi t}{15}\right)$, for $0\le t\le30$.

(a)

Find $h'(t)$.

(2)

(b)

Find the rate at which the height of the seat is changing when $t=4$.

(1)

(c)

Find the maximum rate at which the height of the seat increases, and the value of $t$ at which it occurs.

(2)

(d)

Find the values of $t$ for which the height of the seat is increasing at a rate of 5 metres per minute.

(2)

Show mark scheme

Working	Mark	Notes
(a)
attempt to use the chain rule, e.g. $\sin\left(\tfrac{\pi t}{15}\right)\times\tfrac{\pi}{15}$	M1
$h'(t)=\dfrac{32\pi}{15}\sin\left(\dfrac{\pi t}{15}\right)$	A1
(b)
$4.98$ m min⁻¹	A1	$4.98060\ldots$
(c)
$6.70$ m min⁻¹	A1	$6.70206\ldots\ \left(=\tfrac{32\pi}{15}\right)$
$t=7.5$	A1
(d)
$\dfrac{32\pi}{15}\sin\left(\dfrac{\pi t}{15}\right)=5$, e.g. intersection on GDC	(M1)
$t=4.02$ and $t=11.0$	A1	$4.02070\ldots$, $10.9793\ldots$ Both values required.

Answers: (a) $\dfrac{32\pi}{15}\sin\left(\dfrac{\pi t}{15}\right)$ · (b) 4.98 m min⁻¹ · (c) 6.70 m min⁻¹ at $t=7.5$ · (d) $t=4.02,\ 11.0$

+ Add to test✓ In your test343AASL5-q00545.6 Differentiation rules Paper 2Medium9 marks

Let $f(x)=4\ln x$ and $g(x)=\mathrm{e}^{\frac{x}{2}}$, for $x\gt 0$. The graphs of $f$ and $g$ intersect at $x=a$ and $x=b$, where $a\lt b$.

Diagram not to scale

(a)

Find $f'(x)$ and $g'(x)$.

(2)

(b)

Find the value of $x$ at which the tangents to the two graphs are parallel.

(2)

(c)

Find the value of $a$ and the value of $b$.

(2)

(d)

For $a\le x\le b$, let $d(x)=f(x)-g(x)$. Explain why $d$ has its maximum value at your answer to part (b), and find this maximum value.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$f'(x)=\dfrac{4}{x}$	A1
$g'(x)=\dfrac12\mathrm{e}^{\frac{x}{2}}$	A1
(b)
$\dfrac{4}{x}=\dfrac12\mathrm{e}^{\frac{x}{2}}$, e.g. intersection on GDC	(M1)
$x=2.40$	A1	$2.40433\ldots$
(c)
$a=1.92$	A1	$1.92306\ldots$
$b=2.89$	A1	$2.89429\ldots$
(d)
$d'(x)=f'(x)-g'(x)=0$ where the tangents are parallel	R1	Accept: this is the only stationary point of $d$, and $d(a)=d(b)=0$ with $d\gt 0$ between.
$d(2.40433\ldots)=4\ln(2.40433\ldots)-\mathrm{e}^{1.20216\ldots}$	(M1)
$0.182$	A1	$0.181772\ldots$ Accept 0.181 or 0.182.

Answers: (b) $x=2.40$ · (c) $a=1.92,\ b=2.89$ · (d) $0.182$

+ Add to test✓ In your test344AASL5-q00655.7 The second derivative and graphs Paper 2Medium7 marks

Let $g(x)=\mathrm{e}^{-\frac{x}{2}}\sin(2x)$, for $0\le x\le\pi$.

(a)

Find the coordinates of the local maximum point on the graph of $g$.

(2)

(b)

Find the $x$-coordinates of the points of inflexion on the graph of $g$.

(2)

(c)

Write down the interval on which the graph of $g$ is concave-up.

(1)

(d)

Find the least value of $g'(x)$ for $0\le x\le\pi$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$x=0.663$	A1	$0.662908\ldots$
$y=0.696$	A1	$0.696444\ldots$
(b)
$x=1.33$	A1	$1.32581\ldots$ Use of GDC on the graph of $g'$ (turning points) or of $g''$ (zeros).
$x=2.90$	A1	$2.89661\ldots$
(c)
$1.33\lt x\lt 2.90$	A1	FT from their (b).
(d)
minimum of the graph of $y=g'(x)$ on GDC, at $x=1.33$	(M1)	Endpoint values $g'(0)=2$ and $g'(\pi)=0.416$ are greater.
$-1.03$	A1	$-1.03070\ldots$

Answers: (a) $(0.663,\ 0.696)$ · (b) $x=1.33,\ 2.90$ · (c) $1.33\lt x\lt2.90$ · (d) $-1.03$

+ Add to test✓ In your test345AASL5-q00755.8 Stationary points, inflexion and optimization Paper 2Medium8 marks

The diagram shows part of the graph of $y=6\cos\left(\dfrac{x}{2}\right)$, for $-\pi\le x\le\pi$. The rectangle ABCD has vertices $\text{A}(-x,\,0)$ and $\text{B}(x,\,0)$ on the $x$-axis, where $0\lt x\lt\pi$, and vertices C and D on the curve.

(a)

Show that the area of the rectangle is $A=12x\cos\left(\dfrac{x}{2}\right)$.

(2)

(b)

Find the maximum area of the rectangle.

(3)

(c)

Find the maximum perimeter of the rectangle.

(3)

Show mark scheme

Working	Mark	Notes
(a)
width $\text{AB}=2x$	A1
height $\text{BC}=6\cos\left(\tfrac{x}{2}\right)$, so $A=2x\times6\cos\left(\tfrac{x}{2}\right)$	A1
$A=12x\cos\left(\tfrac{x}{2}\right)$	AG
(b)
use of GDC to find the maximum of $A$, e.g. graph, or solving $\dfrac{\mathrm{d}A}{\mathrm{d}x}=0$	(M1)
$x=1.72066\ldots$	(A1)
maximum area $=13.5$	A1	$13.4663\ldots$
(c)
perimeter $P=4x+12\cos\left(\tfrac{x}{2}\right)$	A1
use of GDC to find the maximum of $P$ (at $x=1.45946\ldots$)	(M1)	Or $\dfrac{\mathrm{d}P}{\mathrm{d}x}=4-6\sin\left(\tfrac{x}{2}\right)=0$.
maximum perimeter $=14.8$	A1	$14.7821\ldots$

Answers: (b) $13.5$ · (c) $14.8$

+ Add to test✓ In your test346AASL5-q00775.8 Stationary points, inflexion and optimization Paper 2Medium8 marks

A small factory makes $x$ bicycles per month, where $0\le x\le100$. The monthly profit, in dollars, is modelled by $P(x)=-0.01x^{3}+0.6x^{2}+30x-400$.

(a)

Find $P'(x)$.

(1)

(b)

Find the value of $x$ for which $P'(x)=0$.

(2)

(c)

Use the second derivative to show that this value of $x$ gives a maximum.

(2)

(d)

The factory can only make a whole number of bicycles. Find the number of bicycles that maximizes the monthly profit, and the maximum monthly profit.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$P'(x)=-0.03x^{2}+1.2x+30$	A1
(b)
$-0.03x^{2}+1.2x+30=0$, e.g. GDC	(M1)	Equivalent to $x^{2}-40x-1000=0$.
$x=57.4$	A1	$57.4165\ldots\ \left(=20+\sqrt{1400}\right)$; reject the negative root.
(c)
$P''(x)=-0.06x+1.2$	A1
$P''(57.4)=-2.24\lt 0$, so a maximum	R1	$-2.24499\ldots$
(d)
comparing $P(57)$ and $P(58)$	(M1)	$P(57)=1407.47$, $P(58)=1407.28$; endpoints $P(0)=-400$, $P(100)=-1400$.
57 bicycles	A1
$1407.47	A1	Award A0 for $P(57.4)=1407.66$.

Answers: (a) $-0.03x^{2}+1.2x+30$ · (b) $x=57.4$ · (d) 57 bicycles, $1407.47

+ Add to test✓ In your test347AASL5-q00785.8 Stationary points, inflexion and optimization Paper 2Medium9 marks

A lorry travels 600 km at a constant speed of $v$ km h⁻¹. The driver is paid $24 per hour. The cost of fuel is $0.00006v^{2}$ dollars per kilometre. Let $C$ be the total cost, in dollars, of the journey.

(a)

Show that $C(v)=\dfrac{14\,400}{v}+0.036v^{2}$.

(2)

(b)

Find the speed for which the total cost is a minimum.

(3)

(c)

Justify that this speed gives a minimum.

(1)

(d)

Find the minimum total cost.

(1)

(e)

Regulations require the lorry to travel at a speed between 70 km h⁻¹ and 90 km h⁻¹. Find the minimum total cost under these regulations. Justify your answer.

(2)

Show mark scheme

Working	Mark	Notes
(a)
time $=\dfrac{600}{v}$ hours, so driver’s pay $=24\times\dfrac{600}{v}=\dfrac{14\,400}{v}$	A1
fuel $=600\times0.00006v^{2}=0.036v^{2}$	A1
$C(v)=\dfrac{14\,400}{v}+0.036v^{2}$	AG
(b)
$C'(v)=-\dfrac{14\,400}{v^{2}}+0.072v=0$	M1	Or minimum of $C$ on GDC.
$v^{3}=200\,000$	A1
$v=58.5$ km h⁻¹	A1	$58.4803\ldots$
(c)
$C''(v)=\dfrac{28\,800}{v^{3}}+0.072\gt 0$ for $v\gt 0$	R1	Or $C'$ changes from negative to positive.
(d)
$369.35	A1	$369.354\ldots$ Accept $369.
(e)
$C'(v)\gt 0$ for $v\gt 58.5$, so $C$ is increasing on $70\le v\le90$ and the minimum is at $v=70$	R1
$382.11	A1	$382.114\ldots$ Accept $382.

Answers: (b) 58.5 km h⁻¹ · (d) $369.35 · (e) $382.11 at 70 km h⁻¹

+ Add to test✓ In your test348AASL5-q00895.9 Kinematics Paper 2Medium8 marks

A particle starts from a fixed point O and moves in a straight line. Its velocity, $v\ \text{m s}^{-1}$, after $t$ seconds is given by $v(t)=1-3\sin(1.2t)$, for $0\le t\le5$.

(a)

Find the first time at which the particle is at rest.

(2)

(b)

Find the number of times the particle changes direction during the 5 seconds. Justify your answer.

(2)

(c)

Find the total distance travelled by the particle.

(2)

(d)

Find the distance of the particle from O when $t=5$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$1-3\sin(1.2t)=0$, or first zero of $v$ on the GDC	(M1)
$t=0.283$ s	A1	$0.283197\ldots$
(b)
$v$ changes sign at $t=0.283$ and at $t=2.33$ only; the next zero, $t=5.52$, is outside the interval	R1	$2.33479\ldots$ and $5.51918\ldots$ Accept a sketch of $v$ showing exactly two sign changes.
2 times	A1	Do not award A1R0.
(c)
$\displaystyle\int_{0}^{5}\|1-3\sin(1.2t)\|\,\mathrm{d}t$	(M1)
$10.2$ m	A1	$10.2253\ldots$
(d)
$\left\|\displaystyle\int_{0}^{5}(1-3\sin(1.2t))\,\mathrm{d}t\right\|$	(M1)
$4.90$ m	A1	$4.90042\ldots$ Award A0 for $10.2$.

Answers: (a) $t=0.283$ s · (b) 2 · (c) 10.2 m · (d) 4.90 m

+ Add to test✓ In your test349AASL5-q00905.9 Kinematics Paper 2Medium8 marks

A sprinter starts from rest at the start line of a 100 m race. Her velocity, $v\ \text{m s}^{-1}$, $t$ seconds after the start is modelled by $v(t)=11\left(1-\mathrm{e}^{-0.8t}\right)$, for $t\ge0$.

(a)

Find her initial acceleration.

(2)

(b)

Write down the velocity that the model predicts she approaches.

(1)

(c)

Find the distance she runs in the first 5 seconds.

(2)

(d)

Find the time she takes to complete the 100 m race.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$a(t)=8.8\,\mathrm{e}^{-0.8t}$, or GDC derivative at $t=0$	(M1)
$8.8\ \text{m s}^{-2}$	A1
(b)
$11\ \text{m s}^{-1}$	A1
(c)
$\displaystyle\int_{0}^{5}11\left(1-\mathrm{e}^{-0.8t}\right)\mathrm{d}t$	(M1)
$41.5$ m	A1	$41.5018\ldots$
(d)
$\displaystyle\int_{0}^{T}11\left(1-\mathrm{e}^{-0.8t}\right)\mathrm{d}t=100$	M1
$11T-13.75\left(1-\mathrm{e}^{-0.8T}\right)=100$, or graph/table of the integral	(A1)	Award for any correct method of solving their equation with technology.
$T=10.3$ s	A1	$10.3405\ldots$

Answers: (a) $8.8\ \text{m s}^{-2}$ · (b) $11\ \text{m s}^{-1}$ · (c) 41.5 m · (d) 10.3 s

+ Add to test✓ In your test350AASL5-q01145.11 Definite integrals and areas Paper 2Medium7 marks

Let $f(x)=6-0.5x^{2}$ and $g(x)=\mathrm{e}^{0.5x}$. The shaded region R is enclosed by the graphs of $f$ and $g$.

(a)

Find the $x$-coordinates of the points of intersection of the two graphs.

(2)

(b)

Find the area of R.

(2)

(c)

Find the maximum vertical distance between the graphs of $f$ and $g$ within R.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$6-0.5x^{2}=\mathrm{e}^{0.5x}$, e.g. intersections on GDC	(M1)
$x=-3.41$ and $x=2.35$	A1	$-3.41125\ldots$, $2.35015\ldots$
(b)
$\displaystyle\int_{-3.41125}^{2.35015}\big(6-0.5x^{2}-\mathrm{e}^{0.5x}\big)\,\mathrm{d}x$	(M1)	FT their limits from (a).
$19.7$	A1	$19.6756\ldots$
(c)
$d(x)=6-0.5x^{2}-\mathrm{e}^{0.5x}$	(M1)
maximum of $d$ on GDC, or $d'(x)=-x-0.5\mathrm{e}^{0.5x}=0\Rightarrow x=-0.408$	(M1)	$-0.407776\ldots$
$5.10$	A1	$5.10130\ldots$

Answers: (a) $x=-3.41,\ 2.35$ · (b) $19.7$ · (c) $5.10$

+ Add to test✓ In your test351AASL5-q01155.11 Definite integrals and areas Paper 2Medium7 marks

Let $f(x)=3\ln(x+1)$ and $g(x)=x^{2}-2x$, for $x\ge0$. The graphs of $f$ and $g$ meet at the origin and at a second point. The shaded region R is enclosed by the two graphs.

(a)

Find the $x$-coordinate of the second point of intersection.

(2)

(b)

Write down an integral that represents the area of R.

(1)

(c)

Find the area of R.

(1)

(d)

The line $x=a$ divides R into two regions of equal area. Find the value of $a$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$3\ln(x+1)=x^{2}-2x$, or intersection on the GDC	(M1)
$x=3.32$	A1	$3.32186\ldots$
(b)
$\displaystyle\int_{0}^{3.32186\ldots}\left(3\ln(x+1)-x^{2}+2x\right)\mathrm{d}x$	A1	FT from (a). Must include correct limits and the difference in the correct order (or with a modulus).
(c)
$7.83$	A1	$7.82807\ldots$
(d)
$\displaystyle\int_{0}^{a}\big(f(x)-g(x)\big)\,\mathrm{d}x=\tfrac12(7.82807\ldots)$	M1	Or equating the two parts of R.
$\displaystyle\int_{0}^{a}\big(f(x)-g(x)\big)\,\mathrm{d}x=3.91403\ldots$	(A1)
$a=1.62$	A1	$1.61893\ldots$

Answers: (a) $x=3.32$ · (c) 7.83 · (d) $a=1.62$

+ Add to test✓ In your test352AASL5-q01165.11 Definite integrals and areas Paper 2Medium6 marks

Let $h(x)=\sqrt{1+x^{3}}$, for $x\ge0$.

(a)

Find $\displaystyle\int_{0}^{2}h(x)\,\mathrm{d}x$.

(2)

(b)

Given that $\displaystyle\int_{0}^{k}h(x)\,\mathrm{d}x=5$, where $k\gt 0$, find the value of $k$.

(3)

(c)

Explain why there is only one value of $k\gt 0$ that satisfies the equation in part (b).

(1)

Show mark scheme

Working	Mark	Notes
(a)
use of GDC integration with the correct function and limits	(M1)
$3.24$	A1	$3.24130\ldots$
(b)
$\displaystyle\int_{0}^{k}\sqrt{1+x^{3}}\,\mathrm{d}x-5=0$, or a graph of $y=\displaystyle\int_{0}^{x}h(t)\,\mathrm{d}t$ and $y=5$	M1
$k\gt 2$ since $3.24\lt 5$	(A1)	Or a sensible table of values of the integral.
$k=2.50$	A1	$2.49891\ldots$
(c)
$h(x)\gt 0$, so $\displaystyle\int_{0}^{k}h(x)\,\mathrm{d}x$ increases as $k$ increases; it can equal 5 only once	R1

Answers: (a) 3.24 · (b) $k=2.50$

+ Add to test✓ In your test353AASL5-q01175.11 Definite integrals and areas Paper 2Medium7 marks

Let $f(x)=x\cos x$, for $0\le x\le4$.

(a)

Find the value of $x$, for $0\lt x\lt 4$, at which the graph of $f$ meets the $x$-axis.

(1)

(b)

Find $\displaystyle\int_{0}^{4}f(x)\,\mathrm{d}x$.

(2)

(c)

Find the total area of the region enclosed by the graph of $f$, the $x$-axis and the line $x=4$.

(2)

(d)

Find the value of $b$, where $0\lt b\lt 4$, for which $\displaystyle\int_{0}^{b}f(x)\,\mathrm{d}x=0$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$x=1.57$	A1	$\tfrac{\pi}{2}=1.57079\ldots$
(b)
use of GDC integration with correct limits	(M1)
$-4.68$	A1	$-4.68085\ldots$
(c)
$\displaystyle\int_{0}^{4}\|x\cos x\|\,\mathrm{d}x$, or $0.570796\ldots+\|-5.25164\ldots\|$	(M1)
$5.82$	A1	$5.82244\ldots$
(d)
$\displaystyle\int_{0}^{b}x\cos x\,\mathrm{d}x=0$, e.g. graph of $y=\displaystyle\int_{0}^{x}f(u)\,\mathrm{d}u$ and its zero, or equating the area below the axis from $1.57$ to $b$ with $0.571$	(M1)
$b=2.33$	A1	$2.33112\ldots$ Award A0 for $b=1.57$.

Answers: (a) $x=1.57$ · (b) $-4.68$ · (c) 5.82 · (d) $b=2.33$

+ Add to test✓ In your test354IBMA2S-052-00-065.6 Differentiation rules Paper 2Medium6 marks

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A marker is painted on the tip of one blade of a wind turbine. The height of the marker above the ground, $h$ metres, $t$ seconds after it passes its lowest point is modelled by $h(t) = 45 - 38\cos\left(\dfrac{2\pi t}{3}\right)$.

(a)

Find $h'(t)$.

(2)

(b)

Find the rate at which the height of the marker is changing when $t = 0.5$.

(2)

(c)

Find the greatest rate at which the height of the marker increases, and the first time at which this occurs.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses the chain rule on the cosine term A1 $h'(t) = \frac{76\pi}{3}\sin\left(\frac{2\pi t}{3}\right)$	✓ 2
Part (b)
M1 substitutes $t = 0.5$ into $h'(t)$ A1 $68.9$ m s⁻¹	✓ 2
Part (c)
A1 maximum rate $\frac{76\pi}{3} = 79.6$ m s⁻¹ A1 when $\frac{2\pi t}{3} = \frac{\pi}{2}$, i.e. $t = 0.75$ s	✓ 2

Exam tip: Check your calculator is in radian mode: the argument $\frac{2\pi t}{3}$ is measured in radians, and degree mode gives a completely wrong value in part (b).

+ Add to test✓ In your test355IBMA2S-052-00-085.9 Kinematics Paper 2Medium6 marks

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In the first few seconds of a 100 m race, the velocity of a sprinter, $v$ m s⁻¹, is modelled by $v(t) = 12\left(1 - e^{-0.8t}\right)$, where $t$ is the time in seconds after the start.

(a)

Find an expression for the acceleration of the sprinter at time $t$.

(2)

(b)

Write down the acceleration of the sprinter at the start of the race.

(1)

(c)

Find the time at which the acceleration of the sprinter has fallen to 2 m s⁻².

(2)

(d)

Find the velocity of the sprinter at this time.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiates, using the chain rule on $e^{-0.8t}$ A1 $a(t) = 9.6e^{-0.8t}$	✓ 2
Part (b)
A1 $9.6$ m s⁻²	✓ 1
Part (c)
M1 sets $9.6e^{-0.8t} = 2$ A1 $t = \frac{\ln 4.8}{0.8} = 1.96$ s	✓ 2
Part (d)
A1 $v = 12\left(1 - \frac{2}{9.6}\right) = 9.5$ m s⁻¹	✓ 1

Exam tip: In part (d), notice that $e^{-0.8t} = \frac{2}{9.6}$ at this time; substituting that exact value avoids rounding errors from using $t = 1.96$.

+ Add to test✓ In your test356IBMA2S-052-04-065.7 The second derivative and graphs Paper 2Medium5 marks

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A transdermal patch releases a painkiller into a patient's skin. The rate of release is modelled by $r(t)=\frac{5}{2t+1}$ mg per hour, where $t$ is the time in hours after the patch is applied. The total mass of painkiller released by time $t$ is $A(t)$ mg.

(a)

Find an expression for $A(t)$.

(3)

(b)

The patch is designed to be removed once 10 mg has been released. Find the time at which the patch should be removed.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate, obtaining a multiple of $\ln(2t+1)$	✓ 1
A1 $\frac{5}{2}\ln(2t+1)+c$	✓ 1
A1 $A(0)=0$ gives $c=0$, so $A(t)=\frac{5}{2}\ln(2t+1)$	✓ 1
Part (b)
M1 set $\frac{5}{2}\ln(2t+1)=10$, giving $2t+1=e^{4}$	✓ 1
A1 $t=26.8$ hours	✓ 1

Common error: Writing $5\ln(2t+1)$ or $10\ln(2t+1)$. Divide by the coefficient of $t$ inside the logarithm, not multiply.

+ Add to test✓ In your test357IBMA2S-052-04-085.11 Definite integrals and areas Paper 2Medium6 marks

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The temperature of the water in a shallow tidal pool changes at a rate modelled by $\frac{dT}{dt}=0.6\cos(0.5t)$ °C per hour, where $t$ is the time in hours after 06:00. At 06:00 the water temperature is 24 °C.

(a)

Find an expression for $T$ in terms of $t$.

(3)

(b)

Write down the maximum water temperature predicted by the model.

(1)

(c)

Find the first time after 06:00 at which the water temperature reaches 24.9 °C.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate, obtaining a multiple of $\sin(0.5t)$	✓ 1
A1 $1.2\sin(0.5t)+C$	✓ 1
A1 uses $T(0)=24$ to obtain $C=24$, so $T=1.2\sin(0.5t)+24$	✓ 1
Part (b)
A1 25.2 °C	✓ 1
Part (c)
M1 sets $1.2\sin(0.5t)+24=24.9$, giving $\sin(0.5t)=0.75$, or uses a graph or equation solver on their $T$	✓ 1
A1 $t=1.70$ hours after 06:00	✓ 1

Exam tip: Make sure your calculator is in radian mode whenever a trigonometric model uses $t$ directly, as here.

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An autonomous delivery robot moves along a straight hospital corridor. It starts from rest at a charging point. Its velocity, $v$ m s⁻¹, at time $t$ seconds is modelled by $v(t)=3te^{-0.1t^{2}}$, $0\leq t\leq 10$. Its displacement from the charging point at time $t$ is $s$ metres.

(a)

Find the maximum velocity of the robot and the time at which it occurs.

(3)

(b)

Use the reverse chain rule to find an expression for $s(t)$.

(4)

(c)

Find the time taken for the robot to travel 12 m from the charging point.

(3)

(d)

Explain why, according to the model, the robot never travels more than 15 m from the charging point.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to find the maximum, e.g. solving $v'(t)=0$ or using a GDC	✓ 1
A1 $t=2.24$ s (exactly $\sqrt{5}$)	✓ 1
A1 $v=4.07$ m s⁻¹	✓ 1
Part (b)
M1 attempt reverse chain rule with inner function $-0.1t^{2}$	✓ 1
A1 $-15e^{-0.1t^{2}}+c$	✓ 1
M1 use $s(0)=0$: $-15+c=0$	✓ 1
A1 $s(t)=15-15e^{-0.1t^{2}}$	✓ 1
Part (c)
M1 set $15-15e^{-0.1t^{2}}=12$	✓ 1
A1 $e^{-0.1t^{2}}=0.2$ or $t^{2}=10\ln 5$	✓ 1
A1 $t=4.01$ s	✓ 1
Part (d)
R1 $v(t)\geq 0$ for $0\leq t\leq 10$, so the robot never reverses and its distance from the charging point is $s(t)$	✓ 1
R1 $e^{-0.1t^{2}}>0$ for all $t$, so $s(t)=15-15e^{-0.1t^{2}}<15$	✓ 1

Exam tip: Check an integral like $\int 3te^{-0.1t^2}dt$ by differentiating: $\frac{d}{dt}(-15e^{-0.1t^2})=-15(-0.2t)e^{-0.1t^2}=3te^{-0.1t^2}$.

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After a heavy storm, the rate at which water flows into a reservoir is modelled by R(t) = 12te^−0.35t, where R is measured in m³ s⁻¹ and t is the time in hours after the storm ends, t ≥ 0.

(a)

Find R'(t).

(2)

(b)

Find the time at which the flow rate is decreasing most rapidly. Justify your answer.

(3)

(c)

Find the rate of change of the flow rate at this time.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule	✓ 1
A1 R'(t) = 12e^−0.35t − 4.2te^−0.35t = 12e^−0.35t(1 − 0.35t)	✓ 1
Part (b)
M1 recognises the need to minimise R'(t), e.g. sets R''(t) = 0 where R''(t) = −4.2e^−0.35t(2 − 0.35t), or uses a GDC to find the minimum of R'(t)	✓ 1
A1 t = 5.71 hours (= 2/0.35)	✓ 1
R1 valid justification that R' has a minimum there, e.g. R''(t) changes from negative to positive at t = 5.71, or R'''(5.71) > 0, or a GDC graph of R' showing a minimum at t = 5.71	✓ 1
Part (c)
A1 R'(5.71) = −12e⁻² = −1.62 m³ s⁻¹ per hour	✓ 1

Common error: "Decreasing most rapidly" refers to the minimum of R', not the maximum of R. Setting R'(t) = 0 gives the peak flow at t = 2.86 hours, which answers a different question.

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The total number of people who have downloaded a new fitness app t days after its launch is modelled by $N(t) = \frac{50000}{1 + 49e^{-0.6t}}$, t ≥ 0.

(a)

Find the number of downloads at launch.

(1)

(b)

Write down the value that N approaches as t becomes large.

(1)

(c)

Show that $N'(t) = \frac{1470000e^{-0.6t}}{(1 + 49e^{-0.6t})^2}$.

(2)

(d)

Find the time at which the rate of downloads is greatest.

(2)

(e)

Find this greatest rate of downloads.

(2)

(f)

Describe the concavity of the graph of N before and after the time at which the rate of downloads is greatest. Interpret this in context.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 N(0) = 50000/50 = 1000	✓ 1
Part (b)
A1 50 000	✓ 1
Part (c)
M1 chain rule on 50000(1 + 49e^−0.6t)⁻¹	✓ 1
A1 −50000(1 + 49e^−0.6t)⁻² × (−29.4e^−0.6t), leading to the given answer	✓ 1
Part (d)
M1 recognises N''(t) = 0 or finds the maximum of N'(t), e.g. using a GDC or 49e^−0.6t = 1	✓ 1
A1 t = ln49/0.6 = 6.49 days	✓ 1
Part (e)
M1 evaluates N' at t = 6.49	✓ 1
A1 7500 downloads per day	✓ 1
Part (f)
A1 concave up for t < 6.49 and concave down for t > 6.49	✓ 1
R1 the daily download rate increases until about 6.49 days and then decreases	✓ 1

Exam tip: For logistic models, the point of inflection occurs at half the limiting value. Check that N(6.49) ≈ 25 000 to confirm your answer.

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A maglev test vehicle starts from rest. Its velocity, v m s⁻¹, t seconds after starting is modelled by v(t) = 3t²e^−0.4t, for 0 ≤ t ≤ 15.

(a)

Find the time at which the velocity of the vehicle is greatest.

(2)

(b)

Find the maximum acceleration of the vehicle.

(3)

(c)

The graph of v is concave down for p < t < q. Find the value of q, and interpret this interval in context.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sets v'(t) = 3e^−0.4t(2t − 0.4t²) = 0, or uses GDC to find the maximum of v	✓ 1
A1 t = 5 s	✓ 1
Part (b)
M1 recognises v''(t) = 0 or finds the maximum of v'(t) (e.g. on GDC)	✓ 1
A1 t = 1.46 s	✓ 1
A1 maximum acceleration = 3.46 m s⁻²	✓ 1
Part (c)
A1 q = 8.54	✓ 1
R1 for 1.46 < t < 8.54 the acceleration is decreasing	✓ 1

Common error: Maximum acceleration is not the same as maximum velocity. At t = 5 s the acceleration is zero. Maximum acceleration needs the stationary point of v', which is where v'' = 0.

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A lander descends towards the surface of Mars under rocket control. Its altitude s metres, t seconds after a timer starts, is recorded below.
t (s): 10, 12, 14, 16
s (m): 1700, 1584, 1476, 1376

(a)

Find the average velocity of the lander between t = 10 and t = 16.

(2)

(b)

By finding the gradients of the chords from t = 10 to t = 12 and from t = 12 to t = 14, estimate s′(12).

(3)

(c)

Explain the meaning of the sign of s′(12) and state the speed of the lander at t = 12.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 (1376 − 1700) ÷ (16 − 10)	✓ 1
A1 −54 m s⁻¹	✓ 1
Part (b)
M1 attempts at least one chord gradient, e.g. (1584 − 1700) ÷ 2	✓ 1
A1 both gradients −58 and −54 m s⁻¹	✓ 1
A1 estimate s′(12) ≈ −56 m s⁻¹ (mean of the two)	✓ 1
Part (c)
R1 negative because the altitude is decreasing, i.e. the lander is moving downwards	✓ 1
A1 speed ≈ 56 m s⁻¹	✓ 1

Exam tip: A chord that straddles the point, or the average of a chord just before and just after, usually gives a better estimate of the gradient than a one-sided chord.

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A cup of coffee is left on a desk to cool. Its temperature, in °C, t minutes after being poured is modelled by T(t) = 22 + 68e^−0.05t, for t ≥ 0.

(a)

Write down the temperature of the coffee when it is poured.

(1)

(b)

Find $\lim_{t\to\infty}T(t)$ and interpret this value in context.

(2)

(c)

Find the average rate of change of temperature over the first 10 minutes.

(2)

(d)

Find T′(t) and hence find the instantaneous rate of change of temperature when t = 10.

(3)

(e)

Find the time at which the instantaneous rate of change of temperature is equal to the average rate of change of temperature over the first 10 minutes.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 T(0) = 90 °C	✓ 1
Part (b)
A1 the limit is 22 (since e^−0.05t → 0)	✓ 1
R1 in the long run the coffee's temperature approaches 22 °C (the temperature of its surroundings / the room) but never falls below it	✓ 1
Part (c)
M1 [T(10) − T(0)] ÷ 10 with T(10) = 63.244…	✓ 1
A1 −2.68 °C per minute (−2.6756…)	✓ 1
Part (d)
A1 T′(t) = −3.4e^−0.05t	✓ 1
M1 substitutes t = 10	✓ 1
A1 −2.06 °C per minute	✓ 1
Part (e)
M1 sets −3.4e^−0.05t = −2.6756…	✓ 1
M1 e^−0.05t = 0.78694…, so −0.05t = ln 0.78694… (or uses GDC to solve)	✓ 1
A1 t = 4.79 minutes	✓ 1

Exam tip: Keep the unrounded average rate (−2.6756…) stored on your calculator for the final part. Using −2.68 instead gives 4.76 minutes rather than 4.79.

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In a rugby match a player takes a conversion kick from the ground. The path of the ball is modelled by y = 0.8x − 0.02x², where x metres is the horizontal distance from the kicking point and y metres is the height of the ball, for 0 ≤ x ≤ 40. Air resistance is ignored.

Diagram not to scale

(a)

Find $\frac{dy}{dx}$.

(1)

(b)

Find the angle above the horizontal at which the ball is kicked.

(3)

(c)

Find the coordinates of the highest point of the ball's path.

(3)

(d)

The crossbar is 3.0 m above the ground at a horizontal distance of 35 m from the kicking point. Show that the ball passes over the crossbar, and find the angle below the horizontal at which the ball is travelling at that point.

(3)

(e)

A spectator claims that at some point in its flight the ball is descending at 45° to the horizontal. Determine whether this claim is correct.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $\frac{dy}{dx}=0.8-0.04x$	✓ 1
Part (b)
A1 gradient at x = 0 is 0.8	✓ 1
M1 angle = arctan 0.8	✓ 1
A1 38.7°	✓ 1
Part (c)
M1 sets 0.8 − 0.04x = 0	✓ 1
A1 x = 20	✓ 1
A1 y = 8, so (20, 8)	✓ 1
Part (d)
A1 y(35) = 28 − 24.5 = 3.5 > 3.0, so the ball clears the bar	✓ 1
M1 gradient at x = 35 is 0.8 − 1.4 = −0.6	✓ 1
A1 angle = arctan 0.6 = 31.0° below the horizontal	✓ 1
Part (e)
M1 sets 0.8 − 0.04x = −1	✓ 1
A1 x = 45	✓ 1
R1 the ball lands when y = 0, i.e. at x = 40 < 45, so the claim is incorrect	✓ 1

Common error: The gradient of the path is tan of the angle with the horizontal, not the angle itself; take arctan of the gradient's magnitude and state whether the ball is rising or falling.

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A factory makes solar panels. The total cost, in dollars, of producing x panels in a day is modelled by C(x) = 5000 + 40x − 0.02x², for 0 ≤ x ≤ 800.

(a)

Find C′(x) and hence find C′(300).

(2)

(b)

Interpret C′(300) in context.

(1)

(c)

Find the exact additional cost of producing the 301st panel, and compare your answer with the value of C′(300).

(3)

(d)

Find the daily production level at which the rate of change of cost is $22 per panel.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 C′(x) = 40 − 0.04x	✓ 1
A1 C′(300) = 28	✓ 1
Part (b)
R1 when 300 panels are being made per day, the cost is increasing at a rate of about $28 per additional panel	✓ 1
Part (c)
M1 uses C(301) − C(300) = 15227.98 − 15200	✓ 1
A1 $27.98	✓ 1
R1 very close to C′(300) = 28, so the derivative gives a good estimate of the cost of one more panel	✓ 1
Part (d)
M1 sets 40 − 0.04x = 22	✓ 1
A1 x = 450 panels	✓ 1

Exam tip: The derivative gives an instantaneous rate. The actual cost of one extra unit is the gradient of a chord over an interval of width 1, so the two values are close but rarely identical.

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A large software update is downloaded to a games console. The download speed, in megabytes per second (MB s⁻¹), $t$ seconds after the download starts is modelled by
$s(t) = 40 - 30e^{-0.2t}$.

(a)

Find the amount of data downloaded in the first 10 seconds.

(2)

(b)

The first 500 MB of the update are downloaded in $T$ seconds. Find the value of $T$.

(3)

(c)

Find the mean download speed during these $T$ seconds.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\int_0^{10} (40 - 30e^{-0.2t})\,dt$	✓ 1
A1 270 MB	✓ 1
Part (b)
M1 form the equation $\int_0^{T} s(t)\,dt = 500$	✓ 1
M1 attempt to solve using technology, e.g. graph or solver	✓ 1
A1 $T = 16.1$	✓ 1
Part (c)
M1 $\frac{500}{T}$	✓ 1
A1 31.1 MB s⁻¹	✓ 1

Common error: In part (b) the unknown is the upper limit, not the integrand. Setting $s(T) = 500$ confuses the speed with the total amount downloaded.

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In a pharmacology trial, the concentration of a drug in a patient's blood, $C$ mg L⁻¹, $t$ hours after an oral dose is modelled by
$C(t) = k\left(e^{-0.2t} - e^{-0.9t}\right)$, for $0 \le t \le 10$,
where $k$ is a positive constant. The area under the concentration–time curve for $0 \le t \le 10$ is 25 mg h L⁻¹.

(a)

Show that $C(0) = 0$.

(1)

(b)

Find the value of $k$.

(3)

(c)

The drug is effective while its concentration is at least 2 mg L⁻¹. Find the length of time for which the drug is effective.

(3)

(d)

Find the maximum concentration of the drug and the time at which it occurs.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Substitutes $t=0$ to get $C(0) = k(e^0 - e^0) = k(1 - 1) = 0$	✓ 1
Part (b)
M1 $\int_0^{10} k\left(e^{-0.2t} - e^{-0.9t}\right)dt = 25$	✓ 1
A1 $\int_0^{10} \left(e^{-0.2t} - e^{-0.9t}\right)dt = 3.2123\ldots$	✓ 1
A1 $k = 7.78$ (7.7824…)	✓ 1
Part (c)
M1 solve $C(t) = 2$ using technology	✓ 1
A1 $t = 0.474$ and $t = 6.75$ (6.749…)	✓ 1
A1 6.27 hours (accept 6.28 from subtracting rounded roots)	✓ 1
Part (d)
M1 use technology to find the maximum of $C$, or solve $C'(t) = 0$	✓ 1
A1 $t = 2.15$ hours	✓ 1
A1 3.94 mg L⁻¹	✓ 1

Exam tip: Store $k$ and the roots in your GDC unrounded and use the stored values in the later parts. Working with rounded values can shift the final answers.

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An artificial ski hill is to be built. Its cross-section is modelled by the region between the curve
$h(x) = \frac{60}{1 + 0.0004x^2}$, for $-150 \le x \le 150$,
and the $x$-axis, where $x$ is the horizontal distance from the centre of the hill and $h$ is the height, both in metres.

Diagram not to scale

(a)

Write down the height of the summit.

(1)

(b)

Find the width of the hill at a height of 30 m.

(3)

(c)

Find the greatest magnitude of the gradient of the hill.

(2)

(d)

Find the area of the cross-section.

(3)

(e)

The hill has this same cross-section along its full length of 200 m. Find the volume of material needed to build it.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 60 m	✓ 1
Part (b)
M1 set $\frac{60}{1 + 0.0004x^2} = 30$	✓ 1
A1 $x^2 = 2500$, so $x = \pm 50$	✓ 1
A1 width = 100 m	✓ 1
Part (c)
M1 use technology to find the extreme value of $h'(x)$	✓ 1
A1 0.779 (at $x = \pm 28.9$)	✓ 1
Part (d)
M1 use an integral of $h(x)$ for the area	✓ 1
A1 $\int_{-150}^{150} \frac{60}{1 + 0.0004x^2}\,dx$	✓ 1
A1 7490 m²	✓ 1
Part (e)
M1 multiply the area by 200	✓ 1
A1 1.50×10⁶ m³	✓ 1

Exam tip: For a solid with a uniform cross-section, volume = cross-sectional area × length. Use the unrounded area from part (d) when you multiply.

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A sports club badge is cut from the region enclosed between the curve $y = 3\sin x$ and the line $y = x$ for $0 \le x \le \pi$, where lengths are in centimetres. The curve and the line meet at the origin and at the point where $x = a$, $a > 0$.

Diagram not to scale

(a)

Find the value of $a$.

(2)

(b)

Write down an integral that gives the area of the badge.

(2)

(c)

Find the area of the badge.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts to solve $3\sin x = x$, e.g. by graphing both functions on the GDC	✓ 1
A1 $a = 2.28$ (2.27886…)	✓ 1
Part (b)
A1 correct integrand $3\sin x - x$	✓ 1
A1 correct limits 0 and $a$: $\displaystyle\int_0^{a}(3\sin x - x)\,dx$	✓ 1
Part (c)
A2 2.35 cm² (2.3545…)	✓ 2

Exam tip: Store the intersection value in your GDC at full accuracy and use the stored value as the limit. Rounding $a$ first can change the third significant figure of the area.

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The entrance to a road tunnel is modelled by the region between the curve $y = 6 - 0.15x^2$ and the ground, $y = 0$. Both $x$ and $y$ are measured in metres, and the origin is at ground level directly below the highest point of the entrance.

Diagram not to scale

(a)

Find the width of the tunnel entrance at ground level.

(2)

(b)

Find the area of the tunnel entrance.

(3)

(c)

The part of the entrance above a horizontal beam at height 3 m is fitted with glass. Find the area of glass.

(4)

(d)

Find the percentage of the entrance that is fitted with glass.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 solves $6 - 0.15x^2 = 0$, giving $x = \pm\sqrt{40}$	✓ 1
A1 width $= 2\sqrt{40} = 4\sqrt{10} \approx 12.6$ m	✓ 1
Part (b)
M1 writes an integral of $6 - 0.15x^2$	✓ 1
A1 uses the correct limits $\pm\sqrt{40}$	✓ 1
A1 50.6 m² ($16\sqrt{10}$)	✓ 1
Part (c)
M1 solves $6 - 0.15x^2 = 3$, giving $x^2 = 20$	✓ 1
A1 limits $x = \pm\sqrt{20}$ $(\pm 4.47)$	✓ 1
M1 writes $\displaystyle\int_{-\sqrt{20}}^{\sqrt{20}} (6 - 0.15x^2 - 3)\,dx$	✓ 1
A1 17.9 m² ($8\sqrt{5}$)	✓ 1
Part (d)
M1 calculates $\frac{17.9}{50.6} \times 100$	✓ 1
A1 35.4%	✓ 1

Exam tip: For the glass, integrate the curve minus the line $y = 3$, not the curve alone. The region between the curve and the $x$-axis includes the part below the beam as well.

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The power output, $P$ kW, of a small wind turbine when the wind speed is $v$ m s⁻¹ is modelled by
$P(v)=0.5v^3e^{-0.25v}$, for $0\le v\le 25$.

(a)

Show that $P'(v)=e^{-0.25v}(1.5v^2-0.125v^3)$.

(2)

(b)

Find the wind speed at which the power output is a maximum, and find the maximum power output.

(3)

(c)

By considering the sign of $P'(v)$, justify that this is a maximum.

(2)

(d)

It is given that $P''(v)=e^{-0.25v}(0.03125v^3-0.75v^2+3v)$. Find the wind speed, with $0<v<12$, at which the power output is increasing most rapidly.

(3)

(e)

Find the rate at which the power output is increasing at this wind speed.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts the product rule	✓ 1
A1 $1.5v^2e^{-0.25v}+0.5v^3(-0.25)e^{-0.25v}=e^{-0.25v}(1.5v^2-0.125v^3)$	✓ 1
Part (b)
M1 sets $P'(v)=0$: $v^2(1.5-0.125v)=0$, since $e^{-0.25v}\ne 0$ (or uses GDC to find the maximum of $P$)	✓ 1
A1 $v=12$ m s⁻¹	✓ 1
A1 $P(12)=864e^{-3}=43.0$ kW	✓ 1
Part (c)
M1 evaluates $P'$ either side, e.g. $P'(10)=25e^{-2.5}\approx 2.05>0$ and $P'(14)=-49e^{-3.5}\approx -1.48<0$	✓ 1
R1 $P'$ changes from positive to negative, so maximum	✓ 1
Part (d)
M1 recognises that the required point is where $P''(v)=0$, i.e. the maximum of $P'(v)$	✓ 1
M1 obtains $v^2-24v+96=0$ (rejecting $v=0$), or uses GDC to find the maximum of $P'(v)$ (or the root of $P''(v)$) on $0<v<12$	✓ 1
A1 $v=12-\sqrt{48}=5.07$ m s⁻¹	✓ 1
Part (e)
M1 substitutes $v=5.07\ldots$ into $P'(v)$	✓ 1
A1 6.27 kW per m s⁻¹	✓ 1

Common error: Using the rounded value $v=5.07$ in part (e) can shift the third significant figure. Keep the full calculator value of $v$ until the final answer.

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A basketball is thrown towards a hoop. Its path is modelled by $y = -0.08x^2 + 0.9x + 2.1$, where $x$ is the horizontal distance from the point of release and $y$ is the height above the floor, both in metres. The centre of the hoop is at a height of 3.05 m.

(a)

Find the angle above the horizontal at which the ball is released.

(2)

(b)

The ball reaches the height of the hoop while descending. Find the horizontal distance of the ball from the point of release at this moment.

(2)

(c)

Find the acute angle between the path of the ball and the horizontal at this moment.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{dy}{dx} = -0.16x + 0.9$, giving gradient 0.9 at $x = 0$	✓ 1
A1 $\arctan 0.9 = 42.0°$	✓ 1
Part (b)
M1 attempt to solve $-0.08x^2 + 0.9x + 2.1 = 3.05$	✓ 1
A1 $x = 10.1$ m (choosing the larger root)	✓ 1
Part (c)
M1 substitution of $x = 10.07...$ into $\frac{dy}{dx}$	✓ 1
A1 gradient $= -0.711$	✓ 1
A1 angle $= 35.4°$	✓ 1

Exam tip: The direction of motion at any instant is along the tangent, so the angle with the horizontal is $\arctan$ of the absolute value of the gradient. Check your GDC is in degree mode.

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The cross-section of a curved mirror in a light sensor is modelled by $y = e^{0.5x}$, where $x$ and $y$ are measured in centimetres. When a ray of light strikes the mirror, the angle of incidence is measured from the normal to the mirror at that point.

(a)

Find the gradient of the curve at the point where $x = 2$.

(2)

(b)

Find the equation of the normal to the curve at the point where $x = 2$.

(3)

(c)

The normal at the point $P(a, e^{0.5a})$ passes through the origin. Find the value of $a$.

(4)

(d)

A ray of light travelling vertically downwards strikes the mirror at the point where $x = 2$. Find the angle of incidence.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\frac{dy}{dx} = 0.5e^{0.5x}$	✓ 1
A1 $\frac{e}{2} = 1.36$	✓ 1
Part (b)
A1 $y = e$ (2.72) when $x = 2$	✓ 1
M1 normal gradient $-\frac{2}{e}$ ($-0.736$) used in point–gradient form	✓ 1
A1 $y = -0.736x + 4.19$	✓ 1
Part (c)
M1 normal at $P$: $y - e^{0.5a} = -\frac{2}{e^{0.5a}}(x - a)$	✓ 1
M1 substituting $(0, 0)$	✓ 1
A1 $e^{a} + 2a = 0$ or equivalent	✓ 1
A1 $a = -0.352$	✓ 1
Part (d)
M1 recognising the normal makes an angle $\arctan\frac{2}{e}$ (36.3°) with the horizontal, or that the required angle is $\arctan\frac{e}{2}$	✓ 1
A1 53.7°	✓ 1

Common error: The angle of incidence is between the ray and the normal, not between the ray and the mirror surface. Giving 36.3° measures the angle with the wrong line.

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A curved road is modelled by $y = 20\ln(0.1x)$, for $10 \le x \le 300$, where $x$ and $y$ are in metres. A car drives along the road in the direction of increasing $x$, and its headlight beam points along the tangent to the road in the direction of travel. A road sign is located at $S(80, 50)$.

(a)

Find $\frac{dy}{dx}$.

(2)

(b)

Find the equation of the tangent to the road at the point where $x = 20$.

(3)

(c)

The headlight beam points directly at $S$ when the car is at the point where $x = a$. Show that $\frac{1600}{a} + 20\ln(0.1a) - 70 = 0$.

(3)

(d)

Find the value of $a$.

(2)

(e)

A pedestrian crosses the road at the point where $x = 20$, walking along the normal to the road. Find the $x$-coordinate of the point where the pedestrian's path meets the $x$-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of the chain rule or $\ln(0.1x) = \ln 0.1 + \ln x$	✓ 1
A1 $\frac{20}{x}$	✓ 1
Part (b)
A1 $y = 20\ln 2 = 13.9$	✓ 1
A1 gradient $= 1$	✓ 1
A1 $y = x - 6.14$	✓ 1
Part (c)
M1 tangent at $x = a$: $y - 20\ln(0.1a) = \frac{20}{a}(x - a)$	✓ 1
M1 substituting $(80, 50)$	✓ 1
A1 $50 = \frac{1600}{a} - 20 + 20\ln(0.1a)$ leading to the given equation	✓ 1
Part (d)
A1 $a = 36.1$	✓ 1
R1 rejecting $a = 236$ because the car would have passed the sign ($a > 80$) and the beam points forwards	✓ 1
Part (e)
M1 normal $y - 20\ln 2 = -(x - 20)$ with $y = 0$	✓ 1
A1 $x = 33.9$	✓ 1

Common error: A GDC finds every root of the equation in the window. Always check each root against the context; here only a car position before the sign makes sense.

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In a design for a stadium roof, a steel arch follows the curve $y = 6 - \frac{x^2}{2}$ and a support cable follows the curve $y = \frac{x^2}{4}$, where $x$ and $y$ are in tens of metres. The angle between two curves at a point of intersection is defined as the angle between their tangents at that point.

Diagram not to scale

(a)

Find the coordinates of the point of intersection of the curves for which $x > 0$.

(2)

(b)

Find the gradient of each curve at this point.

(2)

(c)

Find the acute angle between the arch and the cable at this point.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 equating: $6 - \frac{x^2}{2} = \frac{x^2}{4}$, giving $x^2 = 8$	✓ 1
A1 $(2\sqrt{2}, 2)$ or $(2.83, 2)$	✓ 1
Part (b)
A1 cable: $\frac{dy}{dx} = \frac{x}{2} = \sqrt{2}$ (1.41)	✓ 1
A1 arch: $\frac{dy}{dx} = -x = -2\sqrt{2}$ ($-2.83$)	✓ 1
Part (c)
M1 a valid method for the angle between the tangents. For example, angles with the horizontal $\arctan\sqrt{2} = 54.7°$ and $\arctan 2\sqrt{2} = 70.5°$; or $\tan\theta = \left\|\frac{m_1 - m_2}{1 + m_1 m_2}\right\|$; or the dot product of the direction vectors $(1, \sqrt{2})$ and $(1, -2\sqrt{2})$	✓ 1
M1 correct combination or substitution. For example, $180° - 54.7° - 70.5°$; or $\tan\theta = \left\|\frac{3\sqrt{2}}{-3}\right\| = \sqrt{2}$; or $\cos\theta = \frac{\|1-4\|}{\sqrt{3}\cdot 3}$	✓ 1
A1 54.7° (accept 54.7° to 54.8°)	✓ 1

Common error: Subtracting the two angles with the horizontal ($70.5° - 54.7°$) only works when both tangents slope the same way. Here one gradient is positive and the other negative. Sketch the two lines before combining the angles, or use $\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$.

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The temperature, T °C, of a component on an orbiting satellite is modelled by T(t) = 20 + 15sin(0.5t) − 0.8t, where t is the time in minutes after the satellite leaves Earth's shadow and 0 ≤ t ≤ 12.

(a)

Find T′(t).

(2)

(b)

Find the values of t for which T′(t) = 0.

(2)

(c)

Find the total length of time, within 0 ≤ t ≤ 12, for which the temperature is decreasing.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 7.5cos(0.5t)	✓ 1
A1 −0.8	✓ 1
Part (b)
M1 attempts to solve 7.5cos(0.5t) − 0.8 = 0, e.g. by GDC	✓ 1
A1 t = 2.93 and t = 9.64	✓ 1
Part (c)
R1 T′(t) < 0 for 2.93 < t < 9.64	✓ 1
A1 6.71 minutes	✓ 1

Exam tip: Make sure your calculator is in radian mode for models such as sin(0.5t); degree mode gives completely different roots.

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During a 10-second sprint start, the power output of a track cyclist is modelled by P(t) = 400t²e^−0.5t watts, where t is the time in seconds and 0 ≤ t ≤ 10.

(a)

Show that P′(t) = 200te^−0.5t(4 − t).

(3)

(b)

Hence find the interval on which the power output is increasing, justifying your answer.

(3)

(c)

Find the maximum power output.

(2)

(d)

Find the time at which the power output is decreasing most rapidly, and the rate of change of power at that time.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts the product rule	✓ 1
A1 800te^−0.5t − 200t²e^−0.5t	✓ 1
A1 factorises to 200te^−0.5t(4 − t)	✓ 1
Part (b)
M1 sets P′(t) > 0	✓ 1
R1 200te^−0.5t > 0 for t > 0, so sign depends on 4 − t	✓ 1
A1 0 < t < 4	✓ 1
Part (c)
M1 substitutes t = 4 into P(t)	✓ 1
A1 866 W	✓ 1
Part (d)
M1 attempts to find the minimum of P′(t) for t > 4, e.g. P″(t) = 0 or GDC	✓ 1
A1 t = 6.83 s (exact value 4 + 2√2)	✓ 1
M1 substitutes this t into P′(t)	✓ 1
A1 P′ = −127 W s⁻¹	✓ 1

Common error: The time of most rapid decrease is where P′ is minimised, not where P is minimised; look at the graph of P′ rather than P.

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The displacement, in millimetres, of a faulty loudspeaker cone is modelled by f(x) = e^0.2xcos x, where x is the time in milliseconds and 0 ≤ x ≤ 2π.

(a)

Show that f′(x) = e^0.2x(0.2cos x − sin x).

(2)

(b)

Find the interval of x on which f is decreasing.

(3)

(c)

Find the minimum displacement of the cone.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts the product rule	✓ 1
A1 0.2e^0.2xcos x − e^0.2xsin x leading to given answer	✓ 1
Part (b)
M1 sets f′(x) = 0, giving tan x = 0.2	✓ 1
A1 x = 0.197 and x = 3.34	✓ 1
A1 0.197 < x < 3.34	✓ 1
Part (c)
M1 recognises minimum at x = 3.34 (end values are positive)	✓ 1
A1 −1.91 mm	✓ 1

Common error: Solving tan x = 0.2 gives more than one solution in the domain; adding π to the principal value finds the second stationary point.

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An autonomous delivery robot moves along a straight hospital corridor. Its velocity is $v(t)=(t-2)e^{0.4t}$ m s⁻¹, for $0\le t\le 4$, where $t$ is in seconds.

(a)

Find the value of $t$ at which the robot changes direction.

(2)

(b)

Find the displacement of the robot from its starting position at $t=4$.

(2)

(c)

Find the total distance travelled by the robot for $0\le t\le 4$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sets $v(t)=0$ and notes $v$ changes sign	✓ 1
A1 $t=2$	✓ 1
Part (b)
M1 $\int_0^4 (t-2)e^{0.4t}\,dt$	✓ 1
A1 $5.06$ m	✓ 1
Part (c)
M1 $\int_0^4\|v(t)\|\,dt$ or splits at $t=2$	✓ 1
A1 values $2.66$ and $7.72$	✓ 1
A1 $10.4$ m	✓ 1

Exam tip: Displacement and distance differ whenever the velocity changes sign. Enter $|v(t)|$ in your GDC to get the distance in one step.

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Two self-driving cars, P and Q, are tested on a straight track. At $t=0$, car P is at the origin and car Q is 20 m ahead of it. Car Q moves forward at a constant 4 m s⁻¹. Car P starts from rest with velocity $v_P(t)=8-8e^{-0.5t}$ m s⁻¹, where $t$ is in seconds.

(a)

Find the acceleration of car P at $t=0$.

(2)

(b)

Show that the displacement of car P from the origin is $s_P(t)=8t+16e^{-0.5t}-16$.

(3)

(c)

Find the time at which car P overtakes car Q.

(3)

(d)

Find the distance travelled by car P when it overtakes car Q.

(2)

(e)

Find the greatest distance between the cars before car P overtakes.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $a_P(t)=4e^{-0.5t}$	✓ 1
A1 $4$ m s⁻²	✓ 1
Part (b)
M1 attempt to integrate $v_P$	✓ 1
A1 $8t+16e^{-0.5t}+c$	✓ 1
A1 uses $s_P(0)=0$ to get $c=-16$	✓ 1
Part (c)
M1 equates $s_P=20+4t$	✓ 1
A1 $4t+16e^{-0.5t}-36=0$ or equivalent	✓ 1
A1 $t=8.95$ s	✓ 1
Part (d)
M1 substitutes $t=8.95$ into $s_P$ or $20+4t$	✓ 1
A1 $55.8$ m	✓ 1
Part (e)
M1 gap $D=36-4t-16e^{-0.5t}$	✓ 1
M1 $D'(t)=-4+8e^{-0.5t}=0$	✓ 1
A1 $t=2\ln2$ ($=1.39$)	✓ 1
A1 $D=28-8\ln2=22.5$ m	✓ 1

Exam tip: The gap is greatest when the two cars have equal velocity, which is a quick check on your answer: $8-8e^{-0.5t}=4$.

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The electrical power P watts from a probe's generator, t days after launch, is modelled by P(t) = 120e^−0.05t, t ≥ 0.

(a)

Find P′(t).

(2)

(b)

Find the rate of change of power when t = 10. Give your answer to 3 significant figures, with units.

(2)

(c)

Find the value of t at which the power is falling at exactly 2 W per day. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate an exponential with the chain rule (factor −0.05)	✓ 1
A1 −6e^−0.05t	✓ 1
Part (b)
M1 substitute t = 10: −6e^−0.5	✓ 1
A1 −3.64 W per day	✓ 1
Part (c)
M1 −6e^−0.05t = −2	✓ 1
M1 e^−0.05t = 1/3 and take natural logarithms	✓ 1
A1 t = 22.0	✓ 1

Exam tip: Write down the sign of a rate of change: a negative derivative means the quantity is decreasing, so state it as such in context.

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The concentration C mg per litre of a flavour compound in a bag of freshly roasted coffee, t hours after roasting, is modelled by C(t) = 8te^−0.4t, t ≥ 0.

(a)

Show that C′(t) = 8e^−0.4t(1 − 0.4t).

(3)

(b)

Find the values of t for which the concentration is increasing.

(2)

(c)

Find the maximum concentration. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule	✓ 1
A1 8e^−0.4t + 8t(−0.4e^−0.4t)	✓ 1
A1 factorise to reach 8e^−0.4t(1 − 0.4t) (AG)	✓ 1
Part (b)
M1 C′(t) > 0 and e^−0.4t > 0, so 1 − 0.4t > 0	✓ 1
A1 0 ≤ t < 2.5	✓ 1
Part (c)
M1 substitute t = 2.5	✓ 1
A1 7.36 mg per litre	✓ 1

Exam tip: The exponential factor is always positive, so the sign of the derivative comes only from the bracket.

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A sprinter's velocity, in m s⁻¹, t seconds after the start of a race is modelled by $v(t)=10\left(1-e^{-0.8t}\right)$, for $t\ge 0$.

(a)

Find the sprinter's acceleration when $t=2$. Give your answer to 3 significant figures.

(2)

(b)

Write down the value that v approaches as t becomes very large.

(1)

(c)

Find the distance the sprinter runs in the first 5 seconds. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate, $a=8e^{-0.8t}$	✓ 1
A1 $a(2)=1.62\text{ m s}^{-2}$	✓ 1
Part (b)
A1 $10\text{ m s}^{-1}$	✓ 1
Part (c)
M1 attempt to integrate $v$ from 0 to 5	✓ 1
A1 antiderivative $10t+12.5e^{-0.8t}$	✓ 1
A1 $37.7$ m	✓ 1

Exam tip: Keep your calculator in radians or degrees as the question requires; here no angles are involved, so only the exponential needs care.

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A train brakes along a straight track. Its velocity, in m s⁻¹, t seconds after braking begins is $v(t)=40e^{-0.15t}$, for $t\ge 0$.

(a)

Find the time at which the velocity is 10 m s⁻¹. Give your answer to 3 significant figures.

(3)

(b)

Find the distance travelled in the first 8 seconds. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 set $40e^{-0.15t}=10$	✓ 1
M1 attempt to take logarithms, $t=\frac{\ln 4}{0.15}$	✓ 1
A1 $t=9.24$ s	✓ 1
Part (b)
M1 attempt to integrate $v$ from 0 to 8	✓ 1
A1 $\frac{40}{0.15}\left(1-e^{-1.2}\right)$	✓ 1
A1 $186$ m	✓ 1

Exam tip: Distance travelled is the area under the velocity–time graph, so use a definite integral with the given time limits.

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A rocket test sled moves along a straight track. Its displacement from the start, in metres, after t seconds is $s(t)=2t^3-9t^2+12t$, for $0\le t\le 3$.

(a)

Find the values of t at which the sled is momentarily at rest.

(3)

(b)

Find the total distance travelled by the sled in the first 3 seconds.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate, $v=6t^2-18t+12$	✓ 1
M1 set $v=0$	✓ 1
A1 $t=1$ and $t=2$	✓ 1
Part (b)
M1 evaluate $s$ at $t=0,1,2,3$	✓ 1
A1 $s=5$ at $t=1$, $s=4$ at $t=2$, $s=9$ at $t=3$	✓ 1
M1 add the magnitudes of the separate movements: $5+1+5$	✓ 1
A1 $11$ m	✓ 1

Common error: Total distance is not $s(3)-s(0)$ when the direction of motion changes; split the motion at each time of rest.

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The curve C has equation y = 2x³ − 5x² + 3. The tangent to C at the point P is parallel to the line y = 4x − 7.

(a)

Find the possible x-coordinates of P.

(4)

(b)

Find the equation of the tangent at the point P for which x > 0, giving your answer in the form y = mx + c.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate: $\frac{dy}{dx}=6x^2-10x$	✓ 1
M1 sets $6x^2-10x=4$	✓ 1
A1 $3x^2-5x-2=0$, factorised as $(3x+1)(x-2)=0$	✓ 1
A1 $x=2$ or $x=-\frac{1}{3}$	✓ 1
Part (b)
A1 $y=16-20+3=-1$ at $x=2$	✓ 1
M1 uses $y-(-1)=4(x-2)$	✓ 1
A1 $y=4x-9$	✓ 1

Exam tip: Parallel lines share the same gradient, so set the derivative equal to 4 rather than to a tangent equation.

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In an epidemiological model the cumulative number of cases, in thousands, I(t), after t weeks satisfies $\frac{dI}{dt}=4t\,e^{-0.5t^2}$, with I(0) = 0.

(a)

Find $\int 4t\,e^{-0.5t^2}\,dt$.

(3)

(b)

Hence find an expression for I(t).

(2)

(c)

Find the cumulative number of cases, in thousands, after 2 weeks. Give your answer to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the reverse chain rule: an expression of the form $k\,e^{-0.5t^2}$	✓ 1
A1 $k=-4$	✓ 1
A1 $-4e^{-0.5t^2}+C$	✓ 1
Part (b)
M1 uses $I=0$ when $t=0$: $0=-4+C$	✓ 1
A1 $I(t)=4-4e^{-0.5t^2}$	✓ 1
Part (c)
A1 $I=4-4e^{-2}=3.46$ (thousand)	✓ 1

Exam tip: Check an antiderivative by differentiating it: the chain rule should return the original integrand exactly, including the factor 4t.

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The curve C has equation y = x²e^−x, for x ∈ ℝ.

(a)

Find $\frac{dy}{dx}$.

(3)

(b)

Show that the tangent to C at the point where x = 1 passes through the origin.

(3)

(c)

Write down the gradient of the normal to C at the point where x = 1, to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule	✓ 1
A1 $2xe^{-x}$ and $-x^2e^{-x}$ correct	✓ 1
A1 $\frac{dy}{dx}=(2x-x^2)e^{-x}$	✓ 1
Part (b)
A1 $y=e^{-1}$ and gradient $e^{-1}$ at $x=1$	✓ 1
M1 tangent: $y-e^{-1}=e^{-1}(x-1)$	✓ 1
A1 simplifies to $y=e^{-1}x$, which passes through $(0,\,0)$ (AG)	✓ 1
Part (c)
A1 $-e=-2.72$	✓ 1

Exam tip: In a "show that" question, present the equation of the tangent in a form where substituting (0, 0) is immediate.

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A heat-pump installer models the total annual cost, C, in hundreds of pounds (£100 units), of heating a house with insulation thickness x cm by C(x) = x² + 54x⁻¹, for 1 ≤ x ≤ 10.

(a)

Find C′(x).

(2)

(b)

Find the value of x that minimises the annual cost.

(2)

(c)

Show that this value of x gives a minimum.

(1)

(d)

Find the minimum annual cost, in pounds.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate, including a negative power	✓ 1
A1 $C'(x)=2x-54x^{-2}$	✓ 1
Part (b)
M1 sets $2x=\frac{54}{x^2}$, so $x^3=27$	✓ 1
A1 $x=3$	✓ 1
Part (c)
R1 $C''(x)=2+108x^{-3}$, which is $6>0$ at $x=3$ (or C′ changes from negative to positive)	✓ 1
Part (d)
A1 $C(3)=9+18=27$, so the cost is £2700 (27 hundred pounds)	✓ 1

Exam tip: Remember the units of the model: C is in hundreds of pounds, so the final answer must be converted.

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A cyclist's closed cylindrical drinks flask has volume $750\ \text{cm}^3$, radius $r$ cm and height $h$ cm. The manufacturer wants to minimise the total surface area $A\ \text{cm}^2$ of the flask. Use unrounded values in later parts and give answers as stated.

Diagram not to scale

(a)

Show that $A=2\pi r^2+\dfrac{1500}{r}$.

(2)

(b)

Find $\dfrac{dA}{dr}$ and hence find the value of $r$ that gives a stationary point. Give $r$ to 3 significant figures.

(3)

(c)

Use the second derivative to show that this value of $r$ gives the minimum surface area.

(2)

(d)

Find the minimum surface area, to the nearest $\text{cm}^2$.

(2)

(e)

Find the height $h$ of this flask, to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\pi r^2h=750$, so $h=\frac{750}{\pi r^2}$	✓ 1
A1 $A=2\pi r^2+2\pi rh=2\pi r^2+\frac{1500}{r}$ (AG)	✓ 1
Part (b)
A1 $\frac{dA}{dr}=4\pi r-\frac{1500}{r^2}$	✓ 1
M1 sets $\frac{dA}{dr}=0$, so $r^3=\frac{375}{\pi}$	✓ 1
A1 $r=4.92$ cm	✓ 1
Part (c)
A1 $\frac{d^2A}{dr^2}=4\pi+\frac{3000}{r^3}$	✓ 1
R1 positive for $r>0$, so a minimum	✓ 1
Part (d)
M1 substitutes the unrounded $r$ into $A$	✓ 1
A1 $A_{\min}=457\ \text{cm}^2$	✓ 1
Part (e)
A1 $h=\frac{750}{\pi r^2}=9.85$ cm	✓ 1

Exam tip: When asked to show a result, write the substitution step explicitly before reaching the given expression.

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During a sprint, a cyclist's speed is modelled by $v(t)=12te^{-0.5t}$ m s⁻¹ for $0\le t\le 10$, where $t$ is the time in seconds from the start of the sprint.

(a)

Use the product rule to find $v'(t)$, giving your answer in the form $12e^{-0.5t}(a+bt)$.

(3)

(b)

Find the time at which the speed is greatest, and the greatest speed in m s⁻¹, to 3 significant figures.

(3)

(c)

Find $v''(t)$ and show that it is zero when $t=4$.

(3)

(d)

The acceleration is $a(t)=v'(t)$. Explain what the result in part (c) shows about the acceleration at $t=4$, and find its value to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts the product rule	✓ 1
A1 $12e^{-0.5t}-6te^{-0.5t}$	✓ 1
A1 $v'(t)=12e^{-0.5t}(1-0.5t)$	✓ 1
Part (b)
M1 sets $v'(t)=0$	✓ 1
A1 $t=2$	✓ 1
A1 $v(2)=24e^{-1}=8.83$ m s⁻¹	✓ 1
Part (c)
M1 differentiates $v'(t)$ with the product rule	✓ 1
A1 $v''(t)=12e^{-0.5t}(0.25t-1)$	✓ 1
A1 $v''(t)=0\Rightarrow t=4$	✓ 1
Part (d)
R1 $v''$ changes from negative to positive at $t=4$, so the acceleration has a minimum (the cyclist is slowing down most rapidly)	✓ 1
A1 $a(4)=12e^{-2}(1-2)=-1.62$ m s⁻²	✓ 1

Exam tip: Acceleration is the first derivative of speed, so a turning point of the acceleration is a zero of the second derivative of speed.

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A cyclist rides 30 km at a constant speed of $v$ km h⁻¹. The energy used per kilometre is modelled by $\left(\dfrac{900}{v}+0.1v^2\right)$ kJ, so the total energy for the ride is $E(v)=\dfrac{27000}{v}+3v^2$ kJ, for $v>0$.

(a)

Find $E'(v)$.

(2)

(b)

Find the speed that minimises the total energy. Give your answer to 3 significant figures.

(2)

(c)

Show that this speed gives a minimum.

(2)

(d)

Find the minimum total energy, to the nearest kJ. Use the unrounded value of $v$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiates, writing $\frac{27000}{v}=27000v^{-1}$	✓ 1
A1 $E'(v)=-\frac{27000}{v^2}+6v$	✓ 1
Part (b)
M1 sets $E'(v)=0$, so $v^3=4500$	✓ 1
A1 $v=16.5$ km h⁻¹	✓ 1
Part (c)
A1 $E''(v)=\frac{54000}{v^3}+6$	✓ 1
R1 positive for all $v>0$, so a minimum	✓ 1
Part (d)
M1 substitutes $v=4500^{1/3}$ into $E$	✓ 1
A1 $E_{\min}=2453$ kJ	✓ 1

Exam tip: If the second derivative is positive for every value in the domain, you can state the minimum without testing the particular value.

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The cross-section of a bridge support is the region enclosed between the curve $y=6x-x^2$ and the straight line $y=x$, with $x$ and $y$ in metres.

Shaded cross-section

(a)

Find the $x$-coordinates of the points where the curve and the line meet.

(2)

(b)

Find the area of the cross-section, giving your answer to three significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
set $6x-x^2=x$, so $x^2-5x=0$	✓ 1
$x=0$ and $x=5$	✓ 1
Part (b)
area $=\int(\text{curve}-\text{line})\,dx$ with integrand $5x-x^2$	✓ 1
correct limits 0 and 5	✓ 1
$\left[\frac{5x^2}{2}-\frac{x^3}{3}\right]_0^5=\frac{125}{2}-\frac{125}{3}$	✓ 1
area $=20.8$ m$^2$	✓ 1

Exam tip: Subtract the lower graph from the upper graph; check which is higher between the intersections.

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A lander moves along a straight rail on an asteroid. Its velocity is $v(t)=3t^2-12t+9$ m s$^{-1}$ for $0\le t\le4$, where $t$ is in seconds. At $t=0$ the lander is at the base station, $s=0$.

(a)

Find the times at which the lander is momentarily at rest.

(2)

(b)

Find the displacement of the lander from the base station at $t=4$.

(2)

(c)

Find the total distance travelled by the lander between $t=0$ and $t=4$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$3t^2-12t+9=0$	✓ 1
$t=1$ and $t=3$	✓ 1
Part (b)
$s=\int v\,dt=t^3-6t^2+9t$	✓ 1
$s(4)=64-96+36=4$ m	✓ 1
Part (c)
position at $t=1$ is $s=4$ and at $t=3$ is $s=0$	✓ 1
distances $4$, $4$ and $4$ for the three stages	✓ 1
total distance $=12$ m	✓ 1

Common error: Total distance needs the motion split where $v$ changes sign; the displacement alone can cancel.

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The upper edge of a footbridge arch is modelled by $y=8\sin\left(\dfrac{\pi x}{20}\right)$ for $0\le x\le 20$, where x and y are in metres and y = 0 is the level of the ground.
Point P on the arch has x = 5.

Diagram not to scale

(a)

Find the gradient of the arch at P. Give your answer to 3 significant figures.

(3)

(b)

The tangent to the arch at P meets the line y = 0 at the point A. Find the x-coordinate of A. Use the exact (unrounded) gradient from part (a) and give your answer to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate using the chain rule, e.g. $\frac{8\pi}{20}\cos(\cdot)$	✓ 1
A1 $\frac{dy}{dx}=0.4\pi\cos\left(\frac{\pi x}{20}\right)$	✓ 1
A1 0.889 (exact value $0.2\sqrt{2}\pi$)	✓ 1
Part (b)
A1 $y$-coordinate of P is $4\sqrt{2}$ (5.657…)	✓ 1
M1 equation of tangent, e.g. $y-4\sqrt{2}=0.2\sqrt{2}\pi(x-5)$	✓ 1
M1 sets $y=0$ and solves for $x$	✓ 1
A1 $x=5-\frac{20}{\pi}=-1.366…$ so $x=-1.37$	✓ 1

Common error: Set your calculator to radians for any trigonometric function whose argument is a multiple of $\pi$. A gradient found in degree mode will be wrong.

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The power used by a coffee roaster during a six-hour shift is modelled by $P(t)=12+5\sin\left(\dfrac{\pi t}{12}\right)$ kW, where t is the time in hours after the start of the shift, $0\le t\le 6$.
Electricity costs €0.30 per kWh. Find the cost of electricity for the shift, giving your answer to the nearest cent. Use the unrounded energy value in your cost calculation.

(a)

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 energy $=\int_0^6 P(t)\,dt$	✓ 1
A1 $12t-\frac{60}{\pi}\cos\left(\frac{\pi t}{12}\right)$	✓ 1
A1 energy $=91.0986…$ kWh (accept 91.1)	✓ 1
M1 multiplies by 0.30	✓ 1
A1 €27.33	✓ 1

Common error: Using the average of the power at the start and end of the shift does not give the energy, because $P(t)$ is not linear. Integrate the power over the full time interval.

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A beekeeper builds an open-topped box from thin wood. The box has a square base of side $x$ cm and height $h$ cm, and its volume is 13 500 cm³. The outside surface area (the base and four sides, with no lid) is $S$ cm².

Diagram not to scale

(a)

Show that $S=x^2+\frac{54000}{x}$.

(2)

(b)

Find $\frac{dS}{dx}$.

(2)

(c)

Find the value of $x$ for which $S$ is a minimum, and find this minimum value of $S$.

(3)

(d)

Justify that this value of $S$ is a minimum.

(2)

(e)

The wood costs €0.0045 per cm². Find the cost of the wood for the box of minimum surface area.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $h=\frac{13500}{x^2}$ from $x^2h=13500$	✓ 1
A1 $S=x^2+4xh=x^2+4x\times\frac{13500}{x^2}=x^2+\frac{54000}{x}$ (AG)	✓ 1
Part (b)
A1 $2x$	✓ 1
A1 $-\frac{54000}{x^2}$	✓ 1
Part (c)
M1 setting $\frac{dS}{dx}=0$	✓ 1
A1 $x^3=27000$, so $x=30$ cm	✓ 1
A1 $S=900+1800=2700$ cm²	✓ 1
Part (d)
M1 $\frac{d^2S}{dx^2}=2+\frac{108000}{x^3}$	✓ 1
R1 at $x=30$ it equals $6>0$, so a minimum	✓ 1
Part (e)
M1 $2700\times0.0045$	✓ 1
A1 €12.15	✓ 1

Exam tip: In an optimisation question, write the quantity to be optimised in terms of one variable first, using the constraint to eliminate the other.

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After a fresh set of tyres is fitted, an F1 team models the grip index $G$ of the tyres on lap $t$ ($t\ge 0$) by $G=\frac{20t}{t^2+4}$.

(a)

Show that $\frac{dG}{dt}=\frac{20(4-t^2)}{(t^2+4)^2}$.

(3)

(b)

Hence find the lap on which the grip index is greatest, and state this greatest value.

(3)

(c)

Find the rate of change of the grip index on lap 5, giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the quotient rule with $u=20t$, $v=t^2+4$	✓ 1
A1 $\frac{20(t^2+4)-20t(2t)}{(t^2+4)^2}$	✓ 1
A1 $\frac{80-20t^2}{(t^2+4)^2}=\frac{20(4-t^2)}{(t^2+4)^2}$ (AG)	✓ 1
Part (b)
M1 setting $\frac{dG}{dt}=0$	✓ 1
A1 $t=2$	✓ 1
A1 $G=\frac{40}{8}=5$	✓ 1
Part (c)
M1 substituting $t=5$: $\frac{20(4-25)}{29^2}$	✓ 1
A1 $-0.499$ per lap	✓ 1

Exam tip: A quotient is zero only where its numerator is zero (and the denominator is not), so you can ignore the denominator when solving for stationary points.

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In the first 5 seconds after a standing start, the acceleration of an F1 car is modelled by $a=14-2t$ m s⁻², where $t$ is the time in seconds. The car starts at rest at the line, so at $t=0$ both its velocity and its displacement are zero.

(a)

Find an expression for the velocity $v$ of the car at time $t$.

(3)

(b)

Find an expression for the displacement $s$ of the car from the line at time $t$.

(3)

(c)

Find the distance travelled by the car in the first 5 seconds, giving your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate $14-2t$	✓ 1
A1 $v=14t-t^2+C$	✓ 1
A1 $C=0$ since $v(0)=0$, so $v=14t-t^2$	✓ 1
Part (b)
M1 attempt to integrate $v$ (ECF from (a))	✓ 1
A1 $s=7t^2-\frac{t^3}{3}+D$	✓ 1
A1 $D=0$ since $s(0)=0$	✓ 1
Part (c)
M1 substituting $t=5$ into $s$: $7(25)-\frac{125}{3}$	✓ 1
A1 133 m (exact value $\frac{400}{3}\approx133.33$)	✓ 1

Exam tip: Use the starting conditions (here $v=0$ and $s=0$ at $t=0$) to find each constant of integration straight after integrating.

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A honey producer's total cost, in €, of filling and packing $x$ jars in one day is $C(x)=0.02x^2+4x+1250$ for $x>0$. The average cost per jar is $A(x)=\frac{C(x)}{x}$.

(a)

Write down an expression for $A(x)$ as a sum of three terms.

(2)

(b)

Find $\frac{dA}{dx}$.

(2)

(c)

Find the number of jars that minimises the average cost per jar.

(2)

(d)

Find the minimum average cost per jar, and justify that it is a minimum.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 dividing each term of $C$ by $x$	✓ 1
A1 $A(x)=0.02x+4+\frac{1250}{x}$	✓ 1
Part (b)
M1 attempt to differentiate, with $\frac{1250}{x}$ written as $1250x^{-1}$	✓ 1
A1 $0.02-\frac{1250}{x^2}$	✓ 1
Part (c)
M1 setting $\frac{dA}{dx}=0$	✓ 1
A1 $x^2=62500$, so $x=250$	✓ 1
Part (d)
A1 $A(250)=5+4+5=$ €14 (per jar)	✓ 1
R1 $\frac{d^2A}{dx^2}=\frac{2500}{x^3}>0$ for $x>0$, so a minimum	✓ 1

Exam tip: Average cost is total cost divided by quantity, so divide every term of the cost function before you differentiate.

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Parcels arrive at a logistics depot at a rate of $R(t) = 50 + 6t$ parcels per hour, where $t$ is the time in hours after the depot opens, for $0 \le t \le 10$.

(a)

Find the number of parcels that arrive during the first 4 hours.

(3)

(b)

Find the time, in hours, at which the total number of parcels that have arrived first reaches 600. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\int_0^4 (50+6t)\,dt$	✓ 1
A1 $[50t + 3t^2]_0^4$ seen	✓ 1
A1 248 parcels	✓ 1
Part (b)
M1 forms equation $50T + 3T^2 = 600$	✓ 1
M1 solves for positive $T$ (e.g. using a GDC or the quadratic formula)	✓ 1
A1 $T = 8.08$ hours	✓ 1

Exam tip: The total from time 0 to T is a definite integral with upper limit T: set it equal to the target and solve.

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A sea lion swims along a straight line from a rock. Its displacement from the rock is $s$ metres after $t$ seconds, where $s = t^3 - 9t^2 + 24t$ for $0 \le t \le 5$.

(a)

Find an expression for the velocity of the sea lion, and hence find the times at which it changes direction.

(3)

(b)

Find the total distance the sea lion swims in the first 5 seconds.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate	✓ 1
A1 $v = 3t^2 - 18t + 24$	✓ 1
A1 $t = 2$ and $t = 4$	✓ 1
Part (b)
M1 finds $s$ at the times it changes direction and at $t=5$	✓ 1
A1 $s(2)=20$ and $s(4)=16$ (and $s(5)=20$)	✓ 1
M1 adds the distances $20 + 4 + 4$	✓ 1
A1 28 m	✓ 1

Common error: Displacement can decrease while distance never does: split the motion at each change of direction.

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The entrance to a tunnel under a museum is arch-shaped. In a model, with $x$ and $y$ in metres, the arch is the region between the $x$-axis and the curve $y = 4 - \frac{x^2}{4}$, for $-4 \le x \le 4$.

Cross-section of the arch

(a)

Find the width of the arch at a height of 3 m.

(2)

(b)

Find the area of the arch opening. Give your answer to 3 significant figures.

(3)

(c)

The tunnel has this cross-section along its whole length of 25 m. Find its volume, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 solves $4 - \frac{x^2}{4} = 3$ giving $x = \pm 2$	✓ 1
A1 width 4 m	✓ 1
Part (b)
M1 $\int_{-4}^{4}\left(4 - \frac{x^2}{4}\right)dx$ (or twice the integral from 0 to 4)	✓ 1
A1 antiderivative $4x - \frac{x^3}{12}$	✓ 1
A1 $21.3$ m² (exact value $\frac{64}{3}$)	✓ 1
Part (c)
M1 volume = cross-sectional area × length ($25 \times \frac{64}{3}$)	✓ 1
A1 533 m³	✓ 1

Exam tip: Use the unrounded (or exact) area when it feeds into a later part, and round only the final answer.

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A delivery robot moves along a straight aisle in a warehouse. Its acceleration at time $t$ seconds is $a = 6 - 2t$ m s⁻², for $0 \le t \le 6$. At $t=0$ its velocity is 3 m s⁻¹ and its displacement is 0 m.

(a)

Find an expression for the velocity of the robot at time $t$.

(3)

(b)

Find the maximum velocity of the robot.

(2)

(c)

Find the displacement of the robot from its starting point when $t = 3$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate $a$	✓ 1
A1 $v = 6t - t^2 + C$	✓ 1
A1 $C = 3$, so $v = 6t - t^2 + 3$	✓ 1
Part (b)
M1 sets $a = 0$, giving $t = 3$	✓ 1
A1 $v = 12$ m s⁻¹	✓ 1
Part (c)
M1 $\int_0^3 (6t - t^2 + 3)\,dt$	✓ 1
A1 27 m	✓ 1

Exam tip: Each integration from acceleration to velocity to displacement introduces a constant, found from the initial conditions.

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A fashion designer cuts a decorative panel from fabric. On a grid with units of 1 dm, the panel is the region enclosed between the curves $y=6-x^2$ (upper edge) and $y=e^x$ (lower edge).

Shaded panel (1 unit = 1 dm)

(a)

Find the $x$-coordinates of the two points where the curves intersect. Give each answer to 3 significant figures.

(2)

(b)

Find the area of the panel, in dm². Use the unrounded values of the intersection points from your calculator, and give your answer to 3 significant figures.

(3)

(c)

The fabric costs €0.90 per dm². Using the unrounded area from (b), find the cost of the fabric for one panel, to the nearest euro.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 equate $6-x^2=e^x$ and solve (e.g. graphically or with a solver)	✓ 1
A1 $x=-2.43$ and $x=1.40$	✓ 1
Part (b)
M1 $\int(6-x^2-e^x)\,dx$ between the two intersection points	✓ 1
A1 correct limits ($-2.4315\ldots$ and $1.3978\ldots$) and correct integrand (upper minus lower)	✓ 1
A1 $13.3$ dm² ($13.315\ldots$)	✓ 1
Part (c)
M1 $13.315\ldots\times0.90$	✓ 1
A1 €12 (cost $=11.98\ldots$)	✓ 1

Exam tip: For the area between two curves always integrate (upper curve) minus (lower curve); store the intersection values in your calculator rather than retyping rounded numbers.

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A closed cylindrical hatbox has base radius $r$ cm and height $h$ cm. Its volume is $2000$ cm³. The total outer surface area of the hatbox is $A$ cm².

Diagram not to scale

(a)

Show that $A=2\pi r^2+\dfrac{4000}{r}$.

(2)

(b)

Find the value of $r$ for which $A$ is a minimum, correct to 3 significant figures.

(3)

(c)

Show 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 $h=\dfrac{2000}{\pi r^2}$ from $\pi r^2h=2000$	✓ 1
A1 substitute into $A=2\pi r^2+2\pi rh$ to obtain $2\pi r^2+\dfrac{4000}{r}$ (AG)	✓ 1
Part (b)
A1 $\dfrac{dA}{dr}=4\pi r-\dfrac{4000}{r^2}$	✓ 1
M1 set $\dfrac{dA}{dr}=0$	✓ 1
A1 $r=6.83$ ($r^3=\dfrac{1000}{\pi}$, so $r=6.8278\ldots$)	✓ 1
Part (c)
R1 $\dfrac{d^2A}{dr^2}=4\pi+\dfrac{8000}{r^3}>0$ for $r>0$, so a minimum	✓ 1
A1 $A=879$ cm² ($878.75\ldots$)	✓ 1

Exam tip: In a 'show that' question, write the substitution step explicitly; the final expression is given, so the marks are for the working that leads to it.

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A warehouse robot uses energy $E(x)=x^2-6\ln x$ kilojoules per delivery cycle when its speed setting is $x$, where $x>0$.

(a)

Find $E'(x)$.

(2)

(b)

Find the values of $x$ for which $E$ is increasing.

(3)

(c)

Find the minimum value of $E$, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate (one term correct)	✓ 1
A1 $E'(x)=2x-\dfrac6x$	✓ 1
Part (b)
M1 $E'(x)>0$, i.e. $2x-\dfrac6x>0$ (ECF from (a))	✓ 1
A1 $x^2>3$ (multiplying by $x>0$)	✓ 1
A1 $x>\sqrt3$	✓ 1
Part (c)
M1 $E(\sqrt3)=3-6\ln\sqrt3=3-3\ln3$	✓ 1
A1 $-0.296$ kJ	✓ 1

Common error: Writing $x^2>3$ as $x>3$ or forgetting the condition $x>0$ when taking the square root. State the domain restriction explicitly.

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A robot moves along a straight rail. Its velocity, $v$ m s⁻¹, at time $t$ seconds is given by $v(t)=t(t-2)(t-5)$, for $0\le t\le 5$.

(a)

Find the robot's displacement from its starting point after 5 seconds. Give your answer to 3 significant figures.

(2)

(b)

Find the total distance travelled by the robot during the first 5 seconds. Give your answer to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\int_0^5 v(t)\,dt$	✓ 1
A1 $-10.4$ m ($-\frac{125}{12}$)	✓ 1
Part (b)
M1 recognise that $v$ changes sign at $t=2$ (roots $0,2,5$)	✓ 1
M1 attempt to find $\int_0^2 v\,dt$ and $\left\|\int_2^5 v\,dt\right\|$ separately and add	✓ 1
A1 $5.33$ m and $15.75$ m (or $\frac{16}{3}$ and $\frac{63}{4}$)	✓ 1
A1 $21.1$ m ($\frac{253}{12}=21.083\ldots$)	✓ 1

Common error: Integrating once from 0 to 5 gives displacement, not distance. Where the velocity changes sign, split the integral and take the magnitude of each part.

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A probe descends towards a moon in a straight line. Its velocity is $v = 5te^{-0.5t}$ m s⁻¹, where $t \ge 0$ is the time in seconds.

(a)

Find an expression for the acceleration of the probe.

(2)

(b)

Find the greatest velocity of the probe. Give your answer to 3 significant figures.

(3)

(c)

Find the distance travelled by the probe during the first 4 seconds. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule	✓ 1
A1 $a = 5e^{-0.5t}(1 - 0.5t)$ or equivalent	✓ 1
Part (b)
M1 set $a = 0$	✓ 1
A1 $t = 2$	✓ 1
A1 $v = 10e^{-1} = 3.68$ m s⁻¹	✓ 1
Part (c)
M1 $\int_0^4 5te^{-0.5t}\,dt$	✓ 1
A1 $11.9$ m	✓ 1

Exam tip: Distance is the area under the velocity-time graph only while the velocity does not change sign. Here $v \ge 0$, so a single definite integral is enough.

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The roof of a footbridge canopy is modelled by the curve $y = \ln(2x + 1)$, for $x \ge 0$, where $x$ and $y$ are in metres. The point A on the curve has $x = 1$.

Diagram not to scale

(a)

Find the gradient of the curve at A.

(2)

(b)

Find the equation of the normal to the curve at A.

(2)

(c)

Find the y-coordinate of the point where the normal meets the y-axis. Give your answer to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 chain rule, $\frac{dy}{dx} = \frac{2}{2x+1}$	✓ 1
A1 gradient $= \frac{2}{3}$	✓ 1
Part (b)
M1 normal gradient $-\frac{3}{2}$ through $(1, \ln 3)$	✓ 1
A1 $y - \ln 3 = -\frac{3}{2}(x - 1)$ or equivalent	✓ 1
Part (c)
A1 $y = \ln 3 + \frac{3}{2} = 2.60$ (ECF from (b))	✓ 1

Exam tip: The point of tangency has $y = \ln 3$, not 0. Keep $\ln 3$ exact in the equation and round only the final answer.

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The strength $S$ of a mobile signal, in suitable units, at a distance $d$ km from a mast is modelled by $S(d)=30de^{-0.4d}$ for $d\geq0$.

(a)

Use the product rule to show that $S'(d)=30e^{-0.4d}(1-0.4d)$.

(3)

(b)

Find the value of $d$ at which the signal strength is a maximum.

(2)

(c)

Find the maximum signal strength. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 product rule with $u=30d$, $v=e^{-0.4d}$	✓ 1
A1 $S'(d)=30e^{-0.4d}+30d(-0.4e^{-0.4d})$	✓ 1
A1 factorises to $30e^{-0.4d}(1-0.4d)$ (AG)	✓ 1
Part (b)
M1 sets $S'(d)=0$ and notes $e^{-0.4d}>0$ so $1-0.4d=0$	✓ 1
A1 $d=2.5$ km	✓ 1
Part (c)
M1 $S(2.5)=30\times2.5\times e^{-1}$	✓ 1
A1 27.6	✓ 1

Exam tip: Factorise a derivative that contains a common exponential factor; the exponential is never zero, so it is the bracket that gives the stationary point.

+ Add to test✓ In your test412IBMA3S-5-17r21-075.10 Indefinite integrals and reverse chain rule Paper 2Medium6 marks

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The rate at which a phone downloads data is modelled by $R(t)=5+4\sin(0.5t)$ megabits per second, where $t$ is the time in seconds after the download starts and $0\leq t\leq10$. Angles are in radians.

(a)

Find $\int\left(5+4\sin(0.5t)\right)dt$.

(3)

(b)

Hence find the total amount of data downloaded in the first 10 seconds. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate $\sin(0.5t)$ giving a $\cos$ term	✓ 1
A1 $5t$	✓ 1
A1 $-8\cos(0.5t)+c$	✓ 1
Part (b)
M1 substitutes limits 10 and 0 into their antiderivative	✓ 1
A1 $\left(50-8\cos5\right)-\left(0-8\cos0\right)=58-8\cos5$	✓ 1
A1 55.7 megabits	✓ 1

Common error: Make sure the calculator is in radian mode whenever the angle is a plain number of seconds, not degrees.

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A snowbank sits on rock. In a cross-section, with $x$ and $y$ in metres, the upper surface of the snow is modelled by $y=6x-x^2$ and the rock surface by $y=x$, for $0\leq x\leq 5$.

Cross-section of the snowbank (shaded)

(a)

Find the values of $x$ where the two surfaces meet.

(2)

(b)

Find the area of the cross-section of snow. Give your answer to 3 significant figures.

(4)

(c)

The snowbank has the same cross-section along a length of 40 m. Find the volume of snow, using the unrounded area from (b). Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sets $6x-x^2=x$	✓ 1
A1 $x=0$ and $x=5$	✓ 1
Part (b)
M1 $\int_0^5\left(\text{upper}-\text{lower}\right)dx$	✓ 1
A1 integrand $5x-x^2$	✓ 1
A1 $\left[\frac52x^2-\frac13x^3\right]_0^5=\frac{125}{6}$	✓ 1
A1 20.8 m$^2$	✓ 1
Part (c)
M1 $\frac{125}{6}\times40$	✓ 1
A1 833 m$^3$	✓ 1

Exam tip: The area between two curves uses upper minus lower; sketch quickly or test a point between the limits to decide which is which.

+ Add to test✓ In your test414IBMA3S-5-16r23-065.9 Kinematics Paper 2Medium7 marks

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In the first phase of a sprint, the velocity of an athlete $t$ seconds after leaving the blocks is modelled by $v(t)=10.4\left(1-e^{-0.8t}\right)$ m s⁻¹, for $t\ge0$. Give answers to 3 significant figures.

(a)

Find the athlete's initial acceleration.

(3)

(b)

Find the distance the athlete covers in the first 4 seconds.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts to differentiate using the chain rule	✓ 1
A1 $a=8.32e^{-0.8t}$	✓ 1
A1 $a(0)=8.32\ \text{m s}^{-2}$	✓ 1
Part (b)
M1 $\int_0^4 10.4\left(1-e^{-0.8t}\right)dt$	✓ 1
A1 antiderivative $10.4\left(t+\frac{e^{-0.8t}}{0.8}\right)$	✓ 1
M1 substitutes the limits 4 and 0	✓ 1
A1 $29.1$ m	✓ 1

Exam tip: Distance from velocity needs an integral, and the integral of $e^{kt}$ is $\frac{1}{k}e^{kt}$. Include the $\frac1k$ factor, which here is negative.

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The brightness $B$ (in suitable units) of a variable star, $t$ days after observations began, is modelled by $B(t)=5+2\sin(0.4t)$ for $0\le t\le 10$, where the angle is in radians. Give non-integer answers to 3 significant figures.

(a)

Find $B'(t)$.

(2)

(b)

Find the values of $t$ for which the brightness is decreasing.

(3)

(c)

Find the rate of change of brightness when $t=8$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts to differentiate using the chain rule	✓ 1
A1 $B'(t)=0.8\cos(0.4t)$	✓ 1
Part (b)
M1 attempts to solve $B'(t)=0$, i.e. $0.4t=\frac{\pi}{2}$	✓ 1
A1 $t=3.93$	✓ 1
A1 $3.93<t\le 10$ (accept $<10$)	✓ 1
Part (c)
M1 $B'(8)=0.8\cos(3.2)$	✓ 1
A1 $-0.799$ (units per day)	✓ 1

Common error: Put the calculator in radian mode when the angle is given in radians. Check by noting that $0.4t$ only reaches 4 radians on this interval.

+ Add to test✓ In your test416IBMA3S-5-17r23-055.2 Increasing and decreasing functions Paper 2Medium7 marks

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The elevation of a cycle route, $h$ metres above sea level, at a distance $x$ km along the route is modelled by $h(x) = x^3 - 12x^2 + 36x + 20$, for $0 \le x \le 7$.

(a)

Find the values of $x$ for which $h'(x)=0$.

(4)

(b)

Show that the route has a local maximum elevation at $x=2$, and find this elevation.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $h'(x)=3x^2-24x+36$	✓ 1
M1 set $h'(x)=0$	✓ 1
A1 $x=2$	✓ 1
A1 $x=6$	✓ 1
Part (b)
A1 $h''(x)=6x-24$	✓ 1
R1 $h''(2)=-12<0$, so a local maximum	✓ 1
A1 $h(2)=52$ m	✓ 1

Exam tip: To show a maximum, state the value of the second derivative and the conclusion it gives; a sign alone is not enough.

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A data centre uses a coolant tank in the shape of a cuboid with a square base of side $x$ metres and no lid. The tank must hold exactly 36 m³ of coolant. The base is made from steel costing €16 per m² and the four vertical sides from steel costing €6 per m². The total cost of the steel is €$C$.

Diagram not to scale

(a)

Show that $C = 16x^2 + \frac{864}{x}$.

(3)

(b)

Find the value of $x$ that minimises $C$.

(4)

(c)

Show that this value of $x$ gives a minimum, and find the minimum cost.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 height $=\frac{36}{x^2}$	✓ 1
A1 area of the four sides $=4x\cdot\frac{36}{x^2}=\frac{144}{x}$	✓ 1
A1 $C=16x^2+6\cdot\frac{144}{x}=16x^2+\frac{864}{x}$ (AG)	✓ 1
Part (b)
A1 $\frac{dC}{dx}=32x-864x^{-2}$	✓ 1
M1 set $\frac{dC}{dx}=0$	✓ 1
A1 $x^3=27$	✓ 1
A1 $x=3$	✓ 1
Part (c)
M1 $\frac{d^2C}{dx^2}=32+1728x^{-3}$	✓ 1
R1 positive at $x=3$ (value 96), so a minimum	✓ 1
A1 €432	✓ 1

Exam tip: After finding a stationary value, justify the minimum with the second derivative (or a sign change of the first), then answer the actual question asked, here a cost.

+ Add to test✓ In your test418IBMA3S-5-17r23-095.9 Kinematics Paper 2Medium6 marks

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A cyclist accelerates away from a junction and then brakes to rest at traffic lights. Her velocity, $v$ m s⁻¹, is modelled by $v(t) = 1.2t - 0.06t^2$ for $0 \le t \le 20$, where $t$ is the time in seconds after leaving the junction. She comes to rest when $t=20$.

(a)

Find her acceleration when $t=15$.

(2)

(b)

Find the distance between the junction and the traffic lights.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate: $a=1.2-0.12t$	✓ 1
A1 $-0.6$ m s⁻²	✓ 1
Part (b)
A1 $\int_0^{20}(1.2t-0.06t^2)\,dt$	✓ 1
M1 attempt to integrate	✓ 1
A1 $0.6t^2-0.02t^3$	✓ 1
A1 $80$ m	✓ 1

Exam tip: The distance travelled by a cyclist who does not reverse is the definite integral of velocity; the negative acceleration here just means she is slowing down.

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During an outbreak, the number of new infections per day is modelled by $I(t) = 120\,t\,e^{-0.25t}$, where $t$ is the number of days since the first case was recorded and $t \ge 0$.

(a)

Find $I'(t)$.

(3)

(b)

Find the value of $t$ at which the number of new infections per day is greatest.

(2)

(c)

Find the greatest number of new infections per day, giving your answer to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule	✓ 1
A1 $120e^{-0.25t}$ as one of the two terms	✓ 1
A1 $I'(t) = 120e^{-0.25t} - 30t\,e^{-0.25t}$ (or $30e^{-0.25t}(4-t)$)	✓ 1
Part (b)
M1 set $I'(t)=0$ (and note $e^{-0.25t} \ne 0$)	✓ 1
A1 $t = 4$	✓ 1
Part (c)
A1 $I(4) = 480e^{-1} = 176.58\ldots$, so $177$ new infections per day	✓ 1

Exam tip: Remember that $e^{kt}$ is never zero, so a stationary point of $te^{kt}$ comes from the other factor.

+ Add to test✓ In your test420IBMA3S-5-16r26-085.1 Limits and the derivative Paper 2Medium6 marks

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The number of new cases of a disease reported each day is modelled by $r(t) = 50e^{-0.2t}$, where $t$ is the number of days since a vaccination campaign began.

(a)

Find the total number of new cases predicted in the first 10 days. Give your answer to the nearest whole number.

(4)

(b)

According to the model, find the total number of cases that will ever be reported.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to write $\int_0^{10} 50e^{-0.2t}\,dt$ and integrate	✓ 1
A1 $-250e^{-0.2t}$	✓ 1
M1 substitute the limits 10 and 0	✓ 1
A1 $250(1-e^{-2}) = 216.17\ldots$, so $216$ cases	✓ 1
Part (b)
M1 as $t\to\infty$, $e^{-0.2t}\to 0$	✓ 1
A1 $250$ cases	✓ 1

Exam tip: For a model that decays to zero, the long-run total is found by letting the upper limit tend to infinity in the antiderivative.

+ Add to test✓ In your test421IBMA3S-5-16r26-105.1 Limits and the derivative Paper 2Medium7 marks

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A bookshop sells a popular novel. When the price is €$p$, the number of copies sold per week is $640 - 48p$. Each copy costs the shop €5, and the weekly profit is €$P$.

(a)

Show that $P = -48p^2 + 880p - 3200$.

(2)

(b)

Find the price that maximises the weekly profit. Give your answer to the nearest cent.

(3)

(c)

Find the maximum weekly profit, using the unrounded value of $p$. Give your answer to the nearest cent.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $P = (p-5)(640-48p)$, profit per copy multiplied by copies sold	✓ 1
A1 expand to $P = -48p^2 + 880p - 3200$ (AG)	✓ 1
Part (b)
M1 attempt to differentiate and set $P'(p)=0$	✓ 1
A1 $P'(p) = -96p + 880$	✓ 1
A1 $p = 9.17$ (€9.17)	✓ 1
Part (c)
M1 substitute $p=\frac{880}{96}$ into $P$	✓ 1
A1 $P = 833.33$ (€833.33)	✓ 1

Exam tip: When a later part depends on an earlier answer, keep the unrounded value on your calculator rather than re-typing a rounded one.

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The power output of a solar farm, in megawatts, is modelled by $P(t) = -t^2 + 12t - 20$ for $2 \le t \le 10$, where $t$ is the number of hours after 06:00.

(a)

Find the greatest power output and the time at which it occurs.

(3)

(b)

The energy generated between $t=2$ and $t=10$ is $\int_2^{10} P(t)\,dt$ megawatt-hours. Find this energy, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate and set $P'(t)=0$, with $P'(t) = -2t + 12$	✓ 1
A1 $t = 6$, that is 12:00	✓ 1
A1 greatest output $P(6) = 16$ MW	✓ 1
Part (b)
M1 attempt to integrate $P(t)$ with limits 2 and 10	✓ 1
A1 antiderivative $-\frac{t^3}{3} + 6t^2 - 20t$	✓ 1
A1 $\frac{256}{3} = 85.3$ MWh	✓ 1

Exam tip: The energy is the area under the power curve, so a definite integral gives it directly.

+ Add to test✓ In your test423IBMA3S-5-18r30-055.2 Increasing and decreasing functions Paper 2Medium10 marks

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A shot-putter releases a shot at a height of 1.2 m. The height of the shot above the ground, $h$ metres, $t$ seconds after release is modelled by $h(t)=1.2+14t-4.9t^2$, for $t\geq 0$ until the shot lands.

(a)

Find $\frac{dh}{dt}$.

(2)

(b)

Find the time at which the shot reaches its maximum height. Give your answer to 3 significant figures.

(2)

(c)

Find the maximum height reached by the shot.

(2)

(d)

Find the speed of the shot at the instant it hits the ground. Give your answer to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate	✓ 1
A1 $\frac{dh}{dt}=14-9.8t$	✓ 1
Part (b)
M1 sets $14-9.8t=0$	✓ 1
A1 $t=1.43$ s	✓ 1
Part (c)
M1 substitutes $t=1.4286\ldots$ into $h(t)$	✓ 1
A1 $h=11.2$ m	✓ 1
Part (d)
M1 sets $h(t)=0$	✓ 1
A1 $t=2.94$ s (positive root)	✓ 1
M1 substitutes into $\frac{dh}{dt}$	✓ 1
A1 speed $=14.8$ m s⁻¹	✓ 1

Common error: Speed is a magnitude: the velocity on landing is negative, so state the speed as a positive value.

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The concentration of a supplement in an athlete’s blood, $C$ mg per litre, $t$ hours after it is taken is modelled by $C(t)=12te^{-0.25t}$, for $t\geq 0$.

(a)

Find $C'(t)$.

(3)

(b)

Find the time at which the concentration is greatest.

(2)

(c)

Find the greatest concentration. Give your answer to 3 significant figures.

(2)

(d)

Find $C'(8)$ and interpret your answer in context. Give $C'(8)$ to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule	✓ 1
A1 $12e^{-0.25t}$	✓ 1
A1 $-3te^{-0.25t}$, so $C'(t)=12e^{-0.25t}-3te^{-0.25t}$	✓ 1
Part (b)
M1 sets $C'(t)=0$, giving $3e^{-0.25t}(4-t)=0$	✓ 1
A1 $t=4$ hours	✓ 1
Part (c)
M1 $C(4)=48e^{-1}$	✓ 1
A1 $17.7$ mg per litre	✓ 1
Part (d)
M1 substitutes $t=8$ into $C'(t)$	✓ 1
A1 $C'(8)=-12e^{-2}=-1.62$	✓ 1
R1 at $t=8$ the concentration is decreasing at a rate of 1.62 mg per litre per hour	✓ 1

Common error: The chain rule on $e^{-0.25t}$ gives a factor of $-0.25$; forgetting it changes both the maximum time and the maximum concentration.

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A shunting locomotive leaves a depot along a straight track. Its displacement from the depot, $s$ metres, $t$ seconds after leaving is modelled by $s(t)=0.6t^2-0.01t^3$, for $0\leq t\leq 40$.

(a)

Find an expression for the velocity, $v(t)$, of the locomotive.

(2)

(b)

Show that the locomotive is instantaneously at rest when $t=40$.

(2)

(c)

Find the maximum speed of the locomotive and the time at which it occurs.

(3)

(d)

Find the distance travelled by the locomotive before it comes to rest at $t=40$.

(1)

(e)

Find the acceleration of the locomotive when $t=10$, and describe how its speed changes for $t>20$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate	✓ 1
A1 $v(t)=1.2t-0.03t^2$	✓ 1
Part (b)
M1 substitutes $t=40$ into $v(t)$	✓ 1
A1 $1.2(40)-0.03(1600)=48-48=0$ (AG)	✓ 1
Part (c)
M1 $a(t)=1.2-0.06t=0$	✓ 1
A1 $t=20$ s	✓ 1
A1 maximum speed $=12$ m s⁻¹	✓ 1
Part (d)
A1 $s(40)=0.6(1600)-0.01(64\,000)=320$ m	✓ 1
(no reversal, since $v(t)=0.03t(40-t)\geq 0$ on the interval)	—
Part (e)
A1 $a(10)=1.2-0.06(10)=0.6$ m s⁻²	✓ 1
R1 for $t>20$ the acceleration is negative, so the speed is decreasing	✓ 1

Exam tip: Maximum speed occurs where the acceleration is zero; confirm that the velocity itself is positive there before you quote a speed.

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The number of bees in a hive, in thousands, is modelled by $N(t)=30+8\sin(0.5t)$, where t is the time in weeks since the start of spring, and $t\ge0$. Angles are in radians.

(a)

Find $N'(t)$.

(2)

(b)

Find the rate at which the number of bees is changing when $t=2$. Give your answer to 3 significant figures, with units.

(2)

(c)

Find the smallest positive value of t for which $N'(t)=0$. Give your answer in exact form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 chain rule: the derivative of $\sin(0.5t)$ is $0.5\cos(0.5t)$	✓ 1
A1 $N'(t)=4\cos(0.5t)$	✓ 1
Part (b)
M1 substitute $t=2$: $4\cos 1$ (calculator in radian mode)	✓ 1
A1 $2.16$ thousand bees per week	✓ 1
Part (c)
M1 $\cos(0.5t)=0$, so $0.5t=\frac{\pi}{2}$	✓ 1
A1 $t=\pi$ weeks	✓ 1

Common error: Calculators set to degrees give the wrong value for $\cos 1$ here. The model uses radians, so check the angle mode before evaluating.

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A Formula 1 car brakes for a corner. At time t seconds after braking begins, its velocity is $v=90e^{-0.5t}$ m s⁻¹, for $0\le t\le4$.

(a)

Find the acceleration of the car when $t=4$. Give your answer to 3 significant figures.

(3)

(b)

Find the distance travelled by the car in the first 4 seconds. Give your answer to the nearest metre.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiate $v$ with respect to t	✓ 1
A1 $a=-45e^{-0.5t}$	✓ 1
A1 $a(4)=-45e^{-2}=-6.09$ m s⁻²	✓ 1
Part (b)
M1 $\int_0^4 90e^{-0.5t}\,dt$	✓ 1
A1 $\left[-180e^{-0.5t}\right]_0^4=180\left(1-e^{-2}\right)$	✓ 1
A1 $155.6\ldots$, so distance $=156$ m	✓ 1

Exam tip: When integrating $e^{kt}$, divide by k, which here is −0.5, so the factor 90 becomes −180. A quick check is to differentiate your answer.

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During daylight hours, the rate at which a hive brings in nectar is modelled by $R(t)=0.4+0.3\sin\!\left(\dfrac{\pi t}{12}\right)$ kg per hour, where t is the number of hours after 06:00, for $0\le t\le12$. Angles are in radians.

(a)

Find the rate at which nectar is brought in at 12:00.

(1)

(b)

Find the total mass of nectar brought in between 06:00 and 18:00. Give your answer to 3 significant figures.

(4)

(c)

Find the mean rate at which nectar is brought in over these 12 hours. Give your answer to 3 significant figures, using the unrounded total from part (b).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $R(6)=0.4+0.3\sin\frac{\pi}{2}=0.7$ kg per hour	✓ 1
Part (b)
M1 $\int_0^{12}R(t)\,dt$	✓ 1
A1 antiderivative $0.4t-\frac{3.6}{\pi}\cos\!\left(\frac{\pi t}{12}\right)$	✓ 1
M1 substitute the limits 12 and 0	✓ 1
A1 $4.8+\frac{7.2}{\pi}=7.09$ kg	✓ 1
Part (c)
M1 mean rate $=\frac{\text{total}}{12}$	✓ 1
A1 $0.591$ kg per hour	✓ 1

Exam tip: When integrating $\sin(kt)$ you must divide by k. Here $k=\frac{\pi}{12}$, so the factor $\frac{12}{\pi}$ appears in the antiderivative.

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A warehouse shuttle moves along a straight track. Its acceleration at time $t$ seconds is $a = 4 - 2t$ m s$^{-2}$. At $t = 0$ the shuttle is moving towards the loading bay with velocity $-3$ m s$^{-1}$.

(a)

Find an expression for the velocity of the shuttle at time $t$.

(2)

(b)

Find the times at which the shuttle is instantaneously at rest.

(2)

(c)

Find the total distance travelled by the shuttle during the first 3 seconds. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate $a$	✓ 1
A1 $v = 4t - t^2 - 3$	✓ 1
Part (b)
M1 set $4t - t^2 - 3 = 0$	✓ 1
A1 $t = 1$ and $t = 3$	✓ 1
Part (c)
M1 recognise velocity changes sign at $t=1$, so split the integral	✓ 1
M1 $\left\|\int_0^1 v\,dt\right\| + \int_1^3 v\,dt = \frac{4}{3} + \frac{4}{3}$	✓ 1
A1 $2.67$ m	✓ 1

Common error: Distance is not the same as displacement: here the displacement over the first 3 seconds is zero, so you must split the integral where $v$ changes sign.

+ Add to test✓ In your test430IBMA3S-5-18r35-095.9 Kinematics Paper 2Medium12 marks

Calculator

A remotely operated vehicle (ROV) moves vertically in a deep trench. Its velocity is $v$ m s$^{-1}$, positive upwards, at time $t$ seconds after it leaves a platform, where $v = 6e^{-0.25t} - 2$ for $0 \leq t \leq 10$. Let $s$ be the height of the ROV above the platform, with $s = 0$ at $t = 0$. Use unrounded values in later parts and give final answers to 3 significant figures.

(a)

Find the initial velocity of the ROV.

(1)

(b)

Find the time at which the ROV changes from moving upwards to moving downwards.

(3)

(c)

Find the acceleration of the ROV when $t = 2$.

(2)

(d)

Find the greatest height of the ROV above the platform.

(3)

(e)

Find the total distance travelled by the ROV during the first 10 seconds.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $v(0) = 6 - 2 = 4$ m s$^{-1}$	✓ 1
Part (b)
M1 set $6e^{-0.25t} - 2 = 0$	✓ 1
M1 $e^{-0.25t} = \frac{1}{3}$, attempt to take logarithms	✓ 1
A1 $t = 4\ln 3 = 4.39$ s	✓ 1
Part (c)
M1 $a = \frac{dv}{dt} = -1.5e^{-0.25t}$	✓ 1
A1 $a(2) = -0.910$ m s$^{-2}$	✓ 1
Part (d)
M1 $s = \int v\,dt$, with at least one term correct: $-24e^{-0.25t} - 2t$	✓ 1
M1 evaluate from $t=0$ to $t = 4\ln 3$	✓ 1
A1 $16 - 8\ln 3 = 7.21$ m	✓ 1
Part (e)
M1 $s(10) = 4 - 24e^{-2.5} = 2.03$ m	✓ 1
M1 distance down $= 7.21 - 2.03 = 5.18$ m, added to the distance up	✓ 1
A1 $12.4$ m	✓ 1

Common error: The greatest height is found at the instant the velocity is zero, not at $t = 10$; the total distance counts the downward motion as well.

+ Add to test✓ In your test431IBMA3S-5-18r35-105.11 Definite integrals and areas Paper 2Medium7 marks

Calculator

Parcels arrive at a sorting depot at a rate of $R$ parcels per hour, where $R = 80 + 30\cos\left(\frac{\pi t}{4}\right)$ and $t$ is the number of hours after 06:00, for $0 \leq t \leq 8$. Angles are in radians.

(a)

Find the rate of arrival of parcels at 08:00.

(1)

(b)

Find the total number of parcels that arrive between 06:00 and 14:00.

(3)

(c)

Find the number of parcels that arrive between 06:00 and 09:00. Give your answer to the nearest whole number.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $R(2) = 80 + 30\cos\left(\frac{\pi}{2}\right) = 80$ parcels per hour	✓ 1
Part (b)
M1 $\int_0^8 R\,dt$	✓ 1
A1 $80t + \frac{120}{\pi}\sin\left(\frac{\pi t}{4}\right)$	✓ 1
A1 640 parcels	✓ 1
Part (c)
M1 $\int_0^3 R\,dt$	✓ 1
A1 $240 + \frac{120}{\pi}\sin\left(\frac{3\pi}{4}\right) = 267.01$	✓ 1
A1 267 parcels	✓ 1

Exam tip: When integrating $\cos(kt)$, divide by $k$; with your calculator in radian mode, the result for a whole period of a cosine term is zero.

+ Add to test✓ In your test432IBMA3S-5-18r35-135.5 Introduction to integration Paper 2Medium7 marks

Calculator

The value $V$ euros of a delivery van $t$ years after purchase is modelled by $V = 1200e^{-0.1t}$.

(a)

Find $\dfrac{dV}{dt}$.

(2)

(b)

Find the rate at which the value is decreasing when $t = 3$, giving your answer in euros per year to 3 significant figures.

(2)

(c)

Find the time at which the value is decreasing at a rate of €60 per year. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate the exponential	✓ 1
A1 $\frac{dV}{dt} = -120e^{-0.1t}$	✓ 1
Part (b)
M1 substitute $t = 3$	✓ 1
A1 €88.9 per year	✓ 1
Part (c)
M1 $120e^{-0.1t} = 60$	✓ 1
M1 $e^{-0.1t} = 0.5$, attempt to take logarithms	✓ 1
A1 $t = 10\ln 2 = 6.93$ years	✓ 1

Common error: A negative derivative means the value is falling; the rate of decrease is the positive size of the derivative, so state it as a positive amount.

+ Add to test✓ In your test433IBMA3S-5-18r44-045.7 The second derivative and graphs Paper 2Medium5 marks

Calculator

The brightness of a variable star is modelled by $B(t)=\frac{2t+1}{t^2+3}$, where $t\ge0$ is the time in days after it is first observed and $B$ is measured in suitable units.

(a)

Find $\frac{dB}{dt}$.

(3)

(b)

The brightness has a single maximum for $t>0$. Find the time at which this occurs, giving your answer in days to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the quotient rule	✓ 1
A1 $\frac{2(t^2+3)-2t(2t+1)}{(t^2+3)^2}$	✓ 1
A1 $\frac{-2t^2-2t+6}{(t^2+3)^2}$	✓ 1
Part (b)
M1 set $\frac{dB}{dt}=0$, i.e. $t^2+t-3=0$	✓ 1
A1 $t=1.30$ days	✓ 1

Common error: A fraction is zero only when its numerator is zero, so you only need to solve the numerator equal to zero; do not expand the denominator.

+ Add to test✓ In your test434IBMA3S-5-18r44-125.9 Kinematics Paper 2Medium7 marks

Calculator

A sprinter's speed over the first 4 seconds of a race is modelled by $v(t)=6t-0.75t^2$ for $0\le t\le4$, where $v$ is in m s⁻¹ and $t$ is the time in seconds from the start.

(a)

Find the acceleration of the sprinter when $t=2$.

(2)

(b)

Find the value of $t$ at which the acceleration is zero, and the sprinter's speed at that time.

(2)

(c)

Find the distance the sprinter covers in the first 4 seconds.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $v'(t)=6-1.5t$	✓ 1
A1 $3$ m s⁻²	✓ 1
Part (b)
A1 $t=4$	✓ 1
A1 $v(4)=12$ m s⁻¹	✓ 1
Part (c)
M1 attempt to integrate $v$ from 0 to 4	✓ 1
A1 $3t^2-0.25t^3$	✓ 1
A1 $32$ m	✓ 1

Exam tip: Acceleration is the derivative of speed, and distance travelled is the integral of speed with respect to time.

+ Add to test✓ In your test435AASL5-q00175.2 Increasing and decreasing functions Paper 2Hard16 marks

The cross-section of a garden mound is modelled by $d(x)=\dfrac{-x^{3}+8x^{2}+48x}{80}$, for $0\le x\le 12$, where $d$ is the height of the mound in metres at a horizontal distance $x$ metres from one end.

(a)

Find $d'(x)$.

(2)

(b)

Find the gradient of the mound at $x=4$, and state whether the mound is rising or falling there.

(2)

(c)

Find the set of values of $x$ for which the height of the mound is increasing.

(3)

(d)

Write down the maximum height of the mound.

(1)

(e)(i)

A straight wooden beam rests along the tangent to the mound at the point where $x=10$.

Find the equation of this tangent.

(3)

(e)(ii)

Find the value of $x$ at which the beam reaches the ground.

(1)

(f)

Find the area of the cross-section of the mound.

(2)

(g)

Find the horizontal width of the part of the mound that is more than 2 m high.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\dfrac{-3x^{2}+16x}{80}$	A1	Award for any two correct terms.
$d'(x)=\dfrac{-3x^{2}+16x+48}{80}$	A1	Accept $-0.0375x^{2}+0.2x+0.6$.
(b)
$d'(4)=0.8$	A1
positive, so rising	R1
(c)
$d'(x)\gt 0$, e.g. solving $-3x^{2}+16x+48=0$	M1
$x=7.47$	A1	$7.47406\ldots$ (the other root, $-2.14$, is outside the domain)
$0\le x\lt 7.47$	A1	Accept $0\lt x\lt 7.47$.
(d)
$4.85$ m	A1	$4.85168\ldots$
(e)(i)
$d(10)=3.5$	(A1)
$d'(10)=-1.15$	(A1)
$y=-1.15x+15$	A1	Accept $y-3.5=-1.15(x-10)$.
(e)(ii)
$x=13.0$	A1	$13.0434\ldots$ FT from (e)(i).
(f)
$\displaystyle\int_{0}^{12}d(x)\,\mathrm{d}x$	(M1)
$36\ \text{m}^{2}$	A1
(g)
$d(x)=2\Rightarrow x=2.58$ or $x=11.0$	(M1)	$2.58118\ldots,\ 11.0357\ldots$
$8.45$ m	A1	$8.45455\ldots$

Answers: (a) $\frac{-3x^{2}+16x+48}{80}$ · (b) $0.8$, rising · (c) $0\le x\lt 7.47$ · (d) $4.85$ m · (e)(i) $y=-1.15x+15$ (ii) $x=13.0$ · (f) $36$ m² · (g) $8.45$ m

+ Add to test✓ In your test436AASL5-q00315.4 Tangents and normals Paper 2Hard8 marks

Let $f(x)=\dfrac{12}{1+x^{2}}$, for $-5\le x\le 5$.

(a)

Use your graphic display calculator to find the gradient of the graph of $f$ at the point where $x=1$.

(1)

(b)

Find the equation of the normal to the graph of $f$ at the point where $x=1$.

(3)

(c)

The normal meets the graph of $f$ again at the point A. Find the $x$-coordinate of A.

(2)

(d)

There is one other point on the graph of $f$ at which the tangent is parallel to the tangent at $x=1$. Find the $x$-coordinate of this point.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$-6$	A1
(b)
$f(1)=6$	(A1)
gradient of normal $=\dfrac{1}{6}$	(M1)
$y-6=\dfrac{1}{6}(x-1)$	A1	Accept $y=\dfrac{1}{6}x+\dfrac{35}{6}$ or $y=0.167x+5.83$.
(c)
$\dfrac{12}{1+x^{2}}=\dfrac{1}{6}x+\dfrac{35}{6}$, e.g. intersection of graphs	(M1)
$x=-1.06$	A1	$-1.05888\ldots$
(d)
graph of $y=f'(x)$ and $y=-6$, or solving $f'(x)=-6$	(M1)
$x=0.296$	A1	$0.295597\ldots$ Award A0 for $x=1$.

Answers: (a) $-6$ · (b) $y=\frac16x+\frac{35}{6}$ · (c) $-1.06$ · (d) $0.296$

+ Add to test✓ In your test437AASL5-q00665.7 The second derivative and graphs Paper 2Hard13 marks

The concentration of a drug in a patient’s blood, $C$ milligrams per litre (mg L⁻¹), $t$ hours after a tablet is taken is modelled by $C(t)=12\left(\mathrm{e}^{-0.2t}-\mathrm{e}^{-0.8t}\right)$, for $t\ge0$.

(a)

Write down the value of $C(0)$.

(1)

(b)

Find $C'(t)$.

(2)

(c)

Show that the concentration is a maximum when $t=\dfrac{5}{3}\ln4$.

(3)

(d)

Find the maximum concentration.

(1)

(e)

Find the time at which the concentration is decreasing most rapidly, and the rate of decrease at this time.

(3)

(f)

The drug is effective while the concentration is at least 3 mg L⁻¹. Find the length of time for which the drug is effective.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$0$	A1
(b)
$-2.4\mathrm{e}^{-0.2t}$	A1
$+9.6\mathrm{e}^{-0.8t}$	A1	$C'(t)=-2.4\mathrm{e}^{-0.2t}+9.6\mathrm{e}^{-0.8t}$
(c)
$C'(t)=0\Rightarrow9.6\mathrm{e}^{-0.8t}=2.4\mathrm{e}^{-0.2t}$	M1
$\mathrm{e}^{0.6t}=4$	A1
$t=\dfrac{\ln4}{0.6}$, and $C'$ changes from positive to negative (or $C''\lt 0$ there)	A1	$C''\left(\tfrac53\ln4\right)=-0.907\ldots\lt 0$.
$t=\dfrac{5}{3}\ln4$	AG
(d)
$5.67$ mg L⁻¹	A1	$5.66964\ldots\ \left(=9\times4^{-\frac13}\right)$
(e)
$C''(t)=0$, or minimum of the graph of $C'$ on GDC	(M1)
$t=4.62$ hours	A1	$4.62098\ldots\ \left(=\tfrac53\ln16\right)$
decreasing at $0.714$ mg L⁻¹ h⁻¹	A1	$C'(4.62\ldots)=-0.714330\ldots$ Accept $-0.714$.
(f)
$C(t)=3$, e.g. intersection on GDC	(M1)
$t=0.545$ and $t=6.85$	A1	$0.544666\ldots$, $6.84868\ldots$
$6.30$ hours	A1	$6.30402\ldots$ Accept 6.30–6.31 (6.305 from rounded values).

Answers: (a) 0 · (b) $-2.4\mathrm{e}^{-0.2t}+9.6\mathrm{e}^{-0.8t}$ · (d) 5.67 mg L⁻¹ · (e) $t=4.62$ h, 0.714 mg L⁻¹ h⁻¹ · (f) 6.30 h

+ Add to test✓ In your test438AASL5-q00765.8 Stationary points, inflexion and optimization Paper 2Hard9 marks

A closed cylindrical container of radius $r$ cm and height $h$ cm has a volume of $500\ \text{cm}^{3}$. The material for the curved surface and the top costs 2 cents per cm². The material for the base costs 4 cents per cm². Let $C$ be the total cost of the material, in cents.

Diagram not to scale

(a)

Write down an expression for $h$ in terms of $r$.

(1)

(b)

Show that $C=6\pi r^{2}+\dfrac{2000}{r}$.

(2)

(c)

Find $\dfrac{\mathrm{d}C}{\mathrm{d}r}$.

(2)

(d)

Find the value of $r$ for which the cost is a minimum. Justify that this value gives a minimum.

(3)

(e)

Find the minimum cost, in dollars, correct to the nearest cent.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$h=\dfrac{500}{\pi r^{2}}$	A1
(b)
$C=2(2\pi rh)+2(\pi r^{2})+4(\pi r^{2})$	M1	Award for a cost expression with the three surfaces and correct prices.
$4\pi r\times\dfrac{500}{\pi r^{2}}+6\pi r^{2}$	A1	Substitution of their $h$.
$C=6\pi r^{2}+\dfrac{2000}{r}$	AG
(c)
$12\pi r$	A1
$-2000r^{-2}$	A1	$\dfrac{\mathrm{d}C}{\mathrm{d}r}=12\pi r-\dfrac{2000}{r^{2}}$
(d)
$12\pi r-\dfrac{2000}{r^{2}}=0\Rightarrow r^{3}=\dfrac{500}{3\pi}$	M1	Or use of GDC on their derivative or on $C$.
$r=3.76$	A1	$3.75750\ldots$
$\dfrac{\mathrm{d}^{2}C}{\mathrm{d}r^{2}}=12\pi+\dfrac{4000}{r^{3}}\gt 0$ for $r\gt 0$, so a minimum	R1	Accept a sign change of $\dfrac{\mathrm{d}C}{\mathrm{d}r}$ from negative to positive, with values shown.
(e)
$7.98	A1	$798.402\ldots$ cents. Award A0 for 798 or 7.984.

Answers: (a) $h=\dfrac{500}{\pi r^{2}}$ · (c) $12\pi r-\dfrac{2000}{r^{2}}$ · (d) $r=3.76$ · (e) $7.98

+ Add to test✓ In your test439AASL5-q00795.8 Stationary points, inflexion and optimization Paper 2Hard13 marks

An open gutter is made by bending a sheet of metal 50 cm wide. The cross-section is a trapezium with a horizontal base of 20 cm and two sloping sides of 15 cm, each making an angle $\theta$ with the horizontal, where $0\lt\theta\le\dfrac{\pi}{2}$, as shown in the diagram.

Diagram not to scale

(a)

Show that the area of the cross-section, $A\ \text{cm}^{2}$, is given by $A=300\sin\theta+225\sin\theta\cos\theta$.

(3)

(b)

Find $\dfrac{\mathrm{d}A}{\mathrm{d}\theta}$.

(3)

(c)

Show that $\dfrac{\mathrm{d}A}{\mathrm{d}\theta}=0$ when $6\cos^{2}\theta+4\cos\theta-3=0$.

(2)

(d)

Hence find the value of $\theta$ for which the area of the cross-section is a maximum, and find this maximum area.

(3)

(e)

Find the percentage by which this maximum area exceeds the area of the cross-section when the sides are vertical.

(2)

Show mark scheme

Working	Mark	Notes
(a)
height $=15\sin\theta$	A1
width of the top $=20+2(15\cos\theta)=20+30\cos\theta$	A1
$A=\tfrac12\big(20+(20+30\cos\theta)\big)\times15\sin\theta=(20+15\cos\theta)\,15\sin\theta$	M1	Use of the trapezium area formula with their lengths.
$A=300\sin\theta+225\sin\theta\cos\theta$	AG
(b)
$300\cos\theta$	A1
attempt to use the product rule on $\sin\theta\cos\theta$	M1
$225\left(\cos^{2}\theta-\sin^{2}\theta\right)$	A1	$\dfrac{\mathrm{d}A}{\mathrm{d}\theta}=300\cos\theta+225\left(\cos^{2}\theta-\sin^{2}\theta\right)$. Accept $225\cos2\theta$.
(c)
use of $\sin^{2}\theta=1-\cos^{2}\theta$: $300\cos\theta+225(2\cos^{2}\theta-1)=0$	M1
$450\cos^{2}\theta+300\cos\theta-225=0$, then divide by 75	A1
$6\cos^{2}\theta+4\cos\theta-3=0$	AG
(d)
$\cos\theta=0.448$, rejecting $\cos\theta=-1.11$	A1	$\cos\theta=\dfrac{-2+\sqrt{22}}{6}=0.448402\ldots$; the other root is less than $-1$.
$\theta=1.11$	A1	$1.10581\ldots$ radians. Accept $63.4^{\circ}$.
$A=358\ \text{cm}^{2}$	A1	$358.328\ldots$ Accept GDC maximum of $A$ directly.
(e)
area with vertical sides $=300\ \text{cm}^{2}$	A1	$\theta=\tfrac{\pi}{2}$.
$19.4\%$	A1	$19.4429\ldots\%$

Answers: (b) $300\cos\theta+225(\cos^{2}\theta-\sin^{2}\theta)$ · (d) $\theta=1.11$, $A=358\ \text{cm}^{2}$ · (e) 19.4%

+ Add to test✓ In your test440AASL5-q00875.9 Kinematics Paper 2Hard10 marks

Two particles, P and Q, leave a fixed point O at the same time, $t=0$, and move along the same straight line. The velocity of P, $v_{\text{P}}\ \text{m s}^{-1}$, after $t$ seconds is given by $v_{\text{P}}(t)=10\sin(0.8t)\,\mathrm{e}^{-0.2t}$, for $0\le t\le 8$. Particle Q moves with a constant velocity of $2\ \text{m s}^{-1}$.

(a)

Find the value of $t$ at which P first changes direction.

(2)

(b)

Find the total distance travelled by P in the 8 seconds.

(2)

(c)

Determine whether P is speeding up or slowing down when $t=5$. Justify your answer.

(3)

(d)

Find the values of $t$, for $0\lt t\le 8$, at which P and Q are at the same position.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$\sin(0.8t)=0$, or first zero of $v_{\text{P}}$ after $t=0$ on the GDC	(M1)
$t=3.93$	A1	$\tfrac{5\pi}{4}=3.92699\ldots$
(b)
$\displaystyle\int_{0}^{8}\|v_{\text{P}}(t)\|\,\mathrm{d}t$	(M1)	Or $17.1286\ldots+7.80962\ldots+0.0173679\ldots$, split at $t=\tfrac{5\pi}{4}$ and $t=\tfrac{5\pi}{2}$.
$25.0$ m	A1	$24.9556\ldots$ Accept $24.9$ to $25.0$. Award A0 for $9.34$ (the displacement).
(c)
attempt to find $v_{\text{P}}(5)$ and $a_{\text{P}}(5)=v_{\text{P}}'(5)$	M1
$v_{\text{P}}(5)=-2.78$ and $a_{\text{P}}(5)=-1.37$	A1	$-2.78412\ldots$ and $-1.36687\ldots$
velocity and acceleration have the same sign, so P is speeding up	R1	Do not award R1 without the A1 values (or their signs) seen.
(d)
$\displaystyle\int_{0}^{t}v_{\text{P}}(u)\,\mathrm{d}u=2t$	M1	Equating the displacement of P with the displacement of Q; e.g. graphs of both displacements on the GDC.
$t=0.546$	A1	$0.546089\ldots$
$t=6.07$	A1	$6.06781\ldots$

Answers: (a) $t=3.93$ · (b) 25.0 m · (c) speeding up · (d) $t=0.546,\ 6.07$

+ Add to test✓ In your test441AASL5-q00885.9 Kinematics Paper 2Hard9 marks

A particle starts from a fixed point O and moves in a straight line. Its velocity, $v\ \text{m s}^{-1}$, after $t$ seconds is given by $v(t)=5-t\,\mathrm{e}^{0.4t}$, for $t\ge0$. The particle first moves away from O, then turns and returns to O.

(a)

Find the value of $t$ at which the particle turns.

(2)

(b)

Find the greatest distance of the particle from O before it returns.

(2)

(c)

Find the time at which the particle returns to O.

(3)

(d)

Find the speed of the particle when it returns to O.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$5-t\,\mathrm{e}^{0.4t}=0$	(M1)
$t=2.13$	A1	$2.13151\ldots$
(b)
$\displaystyle\int_{0}^{2.13151\ldots}v(t)\,\mathrm{d}t$	(M1)	FT from (a).
$6.57$ m	A1	$6.56851\ldots$
(c)
$\displaystyle\int_{0}^{T}v(t)\,\mathrm{d}t=0$	M1	Displacement equal to zero; accept use of a graph of the displacement function.
$T\gt 2.13$ (rejects $T=0$)	(A1)
$T=3.61$ s	A1	$3.61393\ldots$
(d)
$\|v(3.61393\ldots)\|$	(M1)
$10.3\ \text{m s}^{-1}$	A1	$v=-10.3386\ldots$ Award A0 for $-10.3$.

Answers: (a) $t=2.13$ · (b) 6.57 m · (c) $t=3.61$ s · (d) $10.3\ \text{m s}^{-1}$

+ Add to test✓ In your test442AASL5-q00915.9 Kinematics Paper 2Hard15 marks

A particle P moves in a straight line, starting from a fixed point O. Its velocity, $v\ \text{m s}^{-1}$, after $t$ seconds is given by $v(t)=12t\,\mathrm{e}^{-0.5t}-3$, for $0\le t\le10$. The graph of $v$ is shown below.

(a)

Write down the initial speed of P.

(1)

(b)

Find the maximum velocity of P.

(2)

(c)

Find the values of $t$ at which P changes direction.

(2)

(d)

Find the acceleration of P when $t=1$.

(2)

(e)

Determine whether P is speeding up or slowing down when $t=7$. Justify your answer.

(2)

(f)

Find the total distance travelled by P in the 10 seconds.

(3)

(g)

Find the greatest distance of P from O during the 10 seconds.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$3\ \text{m s}^{-1}$	A1	Award A0 for $-3$.
(b)
use of GDC to find the maximum of $v$ (at $t=2$)	(M1)
$5.83\ \text{m s}^{-1}$	A1	$5.82910\ldots\ (=24\mathrm{e}^{-1}-3)$
(c)
$v(t)=0$, e.g. zeros on graph	(M1)
$t=0.289$ and $t=6.52$	A1	$0.288842\ldots$, $6.52337\ldots$
(d)
$a=v'(1)$, e.g. GDC derivative, or $a=12\mathrm{e}^{-0.5t}-6t\mathrm{e}^{-0.5t}$	(M1)
$3.64\ \text{m s}^{-2}$	A1	$3.63918\ldots$
(e)
$v(7)=-0.463\lt 0$ and $a(7)=-0.906\lt 0$; velocity and acceleration have the same sign	R1	$v(7)=-0.463419\ldots$, $a(7)=-0.905921\ldots$
speeding up	A1	Award A1 only if R1 is awarded.
(f)
$\displaystyle\int_{0}^{10}\left\|12t\,\mathrm{e}^{-0.5t}-3\right\|\mathrm{d}t$	(M1)	Must see an absolute value or the integral split at the times found in (c).
$0.41163+21.0020+4.53088$	(A1)	Areas of the three parts (seen or implied).
$25.9$ m	A1	$25.9444\ldots$
(g)
displacement $s(t)=\displaystyle\int_{0}^{t}v\,\mathrm{d}u$, considered at the turning times and at $t=10$	(M1)
$s(0.289)=-0.412$, $s(6.52)=20.6$, $s(10)=16.1$	(A1)	$20.5903\ldots$, $16.0595\ldots$
$20.6$ m	A1

Answers: (a) $3\ \text{m s}^{-1}$ · (b) $5.83\ \text{m s}^{-1}$ · (c) $t=0.289,\ 6.52$ · (d) $3.64\ \text{m s}^{-2}$ · (e) speeding up · (f) 25.9 m · (g) 20.6 m

+ Add to test✓ In your test443AASL5-q00925.9 Kinematics Paper 2Hard15 marks

A train leaves station A from rest at time $t=0$ and travels along a straight track to station B, where it stops. Its velocity, $v\ \text{m s}^{-1}$, $t$ seconds after leaving A is modelled by $v(t)=0.0005\,t^{2}(60-t)$, for $0\le t\le60$.

(a)

Find the maximum velocity of the train.

(2)

(b)

Find the distance between station A and station B.

(2)

(c)

Find the maximum acceleration of the train.

(2)

(d)

Find the time taken for the train to travel half of the distance from A to B.

(3)

(e)

At $t=0$, a cyclist passes station A riding at a constant velocity of $8\ \text{m s}^{-1}$ in the same direction as the train. Find the time at which the train catches up with the cyclist, and the distance from A at which this happens.

(3)

(f)

Find the greatest distance between the cyclist and the train before the train catches up with the cyclist.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$v'(t)=0$, or maximum on the graph of $v$	(M1)	$v'(t)=0.06t-0.0015t^{2}=0\Rightarrow t=40$
$16\ \text{m s}^{-1}$	A1
(b)
$\displaystyle\int_{0}^{60}0.0005\,t^{2}(60-t)\,\mathrm{d}t$	(M1)
$540$ m	A1
(c)
$a(t)=0.06t-0.0015t^{2}$, maximum where $a'(t)=0$, i.e. $t=20$; or maximum of the graph of $v'$	(M1)
$0.6\ \text{m s}^{-2}$	A1	Award A0 for $1.8$ or $-1.8$ (the deceleration at $t=60$).
(d)
$\displaystyle\int_{0}^{T}v(t)\,\mathrm{d}t=270$	M1	FT from (b).
$\dfrac{T^{3}}{100}-\dfrac{T^{4}}{8000}=270$, or graph of the integral	(A1)
$T=36.9$ s	A1	$36.8563\ldots$
(e)
$\displaystyle\int_{0}^{t}v(u)\,\mathrm{d}u=8t$	M1	Equating the distances travelled.
$t=40$ s	A1	Other solutions of $t^{3}-80t^{2}+64\,000=0$ are outside $0\lt t\le 60$.
$320$ m	A1
(f)
the gap $8t-s(t)$ is greatest when the velocities are equal, $v(t)=8$	R1	Or attempt to maximize the gap with technology.
$t=20$	A1	The other solution, $t=54.6$, is after the train has caught up.
$8(20)-60=100$ m	A1	$s(20)=60$.

Answers: (a) $16\ \text{m s}^{-1}$ · (b) 540 m · (c) $0.6\ \text{m s}^{-2}$ · (d) 36.9 s · (e) $t=40$ s, 320 m from A · (f) 100 m

+ Add to test✓ In your test444AASL5-q01205.11 Definite integrals and areas Paper 2Hard15 marks

At 00:00 a tank holds 2000 litres of water. For the next 24 hours, water flows into the tank at a rate of $R(t)=60t\,\mathrm{e}^{-0.25t}$ litres per hour, where $t$ is the time in hours after 00:00. At the same time, water is pumped out of the tank at a constant rate of 30 litres per hour.

(a)

Find the maximum rate at which water flows into the tank.

(2)

(b)

Find the total volume of water that flows into the tank during the 24 hours.

(2)

(c)

Find the values of $t$ at which the rate of inflow is equal to the rate of outflow.

(2)

(d)

Explain why the volume of water in the tank is greatest when $t$ is equal to the larger of your values in part (c).

(2)

(e)

Find the greatest volume of water in the tank during the 24 hours. Give your answer to the nearest litre.

(3)

(f)

Find the volume of water in the tank at the end of the 24 hours. Give your answer to the nearest litre.

(2)

(g)

Find the least volume of water in the tank during the 24 hours. Give your answer to the nearest litre.

(2)

Show mark scheme

Working	Mark	Notes
(a)
maximum of $R$ on GDC, or $R'(t)=0\Rightarrow t=4$	(M1)
$88.3$ litres per hour	A1	$88.2910\ldots\ (=240\mathrm{e}^{-1})$
(b)
$\displaystyle\int_{0}^{24}60t\,\mathrm{e}^{-0.25t}\,\mathrm{d}t$	(M1)
$943$ litres	A1	$943.342\ldots$
(c)
$60t\,\mathrm{e}^{-0.25t}=30$	(M1)
$t=0.578$ and $t=13.0$	A1	$0.577685\ldots$, $13.0467\ldots$
(d)
for $0.578\lt t\lt 13.0$, $R(t)\gt 30$, so the volume is increasing	R1
for $13.0\lt t\le 24$, $R(t)\lt 30$, so the volume is decreasing	R1	Must refer to the comparison of rates on both sides of $t=13.0$. The volume also decreases for $0\lt t\lt 0.578$, but only by about 8 litres, so $V(13.0)\gt V(0)$.
(e)
$V=2000+\displaystyle\int_{0}^{13.0467}\big(R(t)-30\big)\,\mathrm{d}t$	(M1)
$\displaystyle\int_{0}^{13.0467}\big(R(t)-30\big)\,\mathrm{d}t=411.807\ldots$	(A1)
$2412$ litres	A1	$2411.81\ldots$ Award A0 for 2410 (not to the nearest litre).
(f)
$2000+943.342-24\times30$	(M1)	FT from (b).
$2223$ litres	A1	$2223.34\ldots$
(g)
$2000+\displaystyle\int_{0}^{0.577685}\big(R(t)-30\big)\,\mathrm{d}t$	(M1)	Must consider the smaller time found in (c); $V(24)\gt 2000$.
$1992$ litres	A1	$1991.77\ldots$

Answers: (a) 88.3 L h⁻¹ · (b) 943 L · (c) $t=0.578,\ 13.0$ · (e) 2412 L · (f) 2223 L · (g) 1992 L

+ Add to test✓ In your test445IBMA2S-052-04-125.9 Kinematics Paper 2Hard15 marks

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During a contact window with a ground station, an Earth-observation satellite uploads data. The rate of upload is modelled by $R(t)=\frac{6e^{0.5t}}{4+e^{0.5t}}$ gigabytes (GB) per minute, where $t$ is the time in minutes from the start of the window. The total data uploaded by time $t$ is $D(t)$ GB, with $D(0)=0$.

(a)

Find $R(0)$ and show that $R(t)=6-\frac{24}{4+e^{0.5t}}$.

(3)

(b)

Hence state the value that $R(t)$ approaches as $t$ increases, and explain why $R(t)$ never reaches this value.

(2)

(c)

Find an expression for $D(t)$.

(5)

(d)

Find the total data uploaded in the first 10 minutes.

(2)

(e)

The upload is complete when 60 GB has been transferred. Find the time at which the upload is complete.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $R(0)=\frac{6}{5}=1.2$ GB per minute	✓ 1
M1 write numerator as $6(4+e^{0.5t})-24$	✓ 1
A1 divide to obtain $6-\frac{24}{4+e^{0.5t}}$ (AG)	✓ 1
Part (b)
A1 6 GB per minute	✓ 1
R1 $\frac{24}{4+e^{0.5t}}>0$ for all $t$, so $R(t)<6$	✓ 1
Part (c)
M1 recognise integrand as a multiple of $\frac{g'(t)}{g(t)}$ with $g(t)=4+e^{0.5t}$	✓ 1
A1 coefficient 12, since $g'(t)=0.5e^{0.5t}$	✓ 1
A1 $12\ln(4+e^{0.5t})+c$	✓ 1
M1 use $D(0)=0$: $12\ln 5+c=0$	✓ 1
A1 $D(t)=12\ln\left(\frac{4+e^{0.5t}}{5}\right)$	✓ 1
Part (d)
M1 substitute $t=10$ into $D(t)$	✓ 1
A1 41.0 GB	✓ 1
Part (e)
M1 set $12\ln\left(\frac{4+e^{0.5t}}{5}\right)=60$	✓ 1
A1 $e^{0.5t}=5e^{5}-4$	✓ 1
A1 $t=13.2$ minutes	✓ 1

Exam tip: A quotient whose numerator is a multiple of the derivative of its denominator integrates to a logarithm: find the multiplier by comparing $6e^{0.5t}$ with $0.5e^{0.5t}$.

+ Add to test✓ In your test446IBMA2S-052-05-125.7 The second derivative and graphs Paper 2Hard12 marks

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The concentration of an anaesthetic in a patient's bloodstream, C mg L⁻¹, t hours after an injection is modelled by $C(t) = \frac{10t}{t^2 + k}$, where t ≥ 0 and k is a positive constant.

(a)

Show that $C'(t) = \frac{10(k - t^2)}{(t^2 + k)^2}$.

(2)

(b)

The maximum concentration occurs 2 hours after the injection. Find the value of k.

(2)

(c)

Find the maximum concentration.

(1)

(d)

It is given that $C''(t) = \frac{20t(t^2 - 12)}{(t^2 + 4)^3}$. Find the time at which the concentration is decreasing most rapidly, justifying your answer. Find the rate of decrease at this time.

(4)

(e)

The anaesthetic is effective while the concentration is at least 1.5 mg L⁻¹. Find the length of time for which it is effective.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the quotient rule	✓ 1
A1 [10(t² + k) − 10t(2t)]/(t² + k)², leading to the given answer	✓ 1
Part (b)
M1 sets C'(2) = 0, giving k − t² = 0 with t = 2	✓ 1
A1 k = 4	✓ 1
Part (c)
A1 C(2) = 20/8 = 2.5 mg L⁻¹	✓ 1
Part (d)
M1 sets C''(t) = 0	✓ 1
A1 t = 2√3 ≈ 3.46 hours	✓ 1
R1 C''(t) < 0 for 0 < t < 2√3 and C''(t) > 0 for t > 2√3, so C'(t) has a minimum there	✓ 1
A1 C'(2√3) = −80/256, so the rate of decrease is 0.3125 mg L⁻¹ per hour (accept C' = −0.3125)	✓ 1
Part (e)
M1 solves 10t/(t² + 4) = 1.5, e.g. 3t² − 20t + 12 = 0	✓ 1
A1 t = 2/3 and t = 6	✓ 1
A1 6 − 2/3 = 16/3 ≈ 5.33 hours	✓ 1

Exam tip: "Decreasing most rapidly" means C' takes its most negative value. Find where C'' = 0 and confirm that C'' changes from negative to positive there. Then quote the rate of decrease as a positive number.

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Two sprinters run a 100 m race, both starting from rest at $t = 0$. Their velocities, in m s⁻¹, $t$ seconds after the start are modelled by
Runner A: $v_A(t) = 12\left(1 - e^{-0.9t}\right)$
Runner B: $v_B(t) = 11.5\left(1 - e^{-1.2t}\right)$.

(a)

Find the velocity of Runner A when $t = 2$.

(2)

(b)

Find the distance Runner A covers in the first 4 seconds.

(3)

(c)

Find the time Runner A takes to finish the race.

(3)

(d)

Find how far Runner B is behind Runner A at the moment Runner A finishes. Use the unrounded value of $T$ from part (c).

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 substitute $t = 2$ into $v_A$: $12\left(1 - e^{-1.8}\right)$	✓ 1
A1 10.0 m s⁻¹	✓ 1
Part (b)
M1 use an integral of $v_A$ for the distance	✓ 1
A1 $\int_0^4 12\left(1 - e^{-0.9t}\right)dt$	✓ 1
A1 35.0 m	✓ 1
Part (c)
M1 form the equation $\int_0^{T} v_A(t)\,dt = 100$	✓ 1
M1 solve using technology, e.g. $12\left(T - \frac{1 - e^{-0.9T}}{0.9}\right) = 100$	✓ 1
A1 $T = 9.44$ s (9.4442…)	✓ 1
Part (d)
M1 Runner B's distance is given by $\int_0^{T} v_B(t)\,dt$	✓ 1
M1 use the unrounded value $T = 9.4442…$ from part (c)	✓ 1
A1 Runner B has run 99.03 m (99.025…)	✓ 1
A1 0.975 m behind (accept 0.97 to 0.98)	✓ 1

Common error: Using the rounded time 9.44 s in part (d) gives about 1.02 m, which is wrong. The gap is a small difference between two large distances, so keep the full stored value of $T$. Comparing top speeds also does not decide the race: compare the areas under the velocity–time curves.

+ Add to test✓ In your test448IBMA2S-052-01-125.9 Kinematics Paper 2Hard12 marks

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Two delivery drones, A and B, leave the same depot at time $t = 0$ and fly in the same direction along the same straight line. Their velocities, in m s⁻¹, are modelled by $v_A(t) = 12\left(1 - e^{-0.5t}\right)$ and $v_B(t) = 1.5t$, for $0 \le t \le 20$, where $t$ is in seconds. Initially A moves ahead of B.

(a)

Find the value of $T$, $T > 0$, for which $v_A(T) = v_B(T)$.

(2)

(b)

Explain why the distance by which A is ahead of B is greatest at $t = T$.

(2)

(c)

Find the greatest distance by which A is ahead of B.

(3)

(d)

Find the time at which drone B catches up with drone A.

(3)

(e)

Find the distance of the drones from the depot at this moment.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts to solve $12\left(1 - e^{-0.5t}\right) = 1.5t$, e.g. using the intersection of two graphs	✓ 1
A1 $T = 7.84$ s (7.8414…)	✓ 1
Part (b)
R1 for $0 < t < T$, $v_A > v_B$, so A's lead over B is increasing	✓ 1
R1 for $t > T$, $v_B > v_A$, so A's lead decreases; hence A's lead is greatest at $t = T$	✓ 1
Part (c)
M1 uses $\displaystyle\int (v_A - v_B)\,dt$	✓ 1
A1 $\displaystyle\int_0^{T} \left(12(1 - e^{-0.5t}) - 1.5t\right)dt$, with the correct limits	✓ 1
A1 24.5 m (24.457…)	✓ 1
Part (d)
M1 sets the displacements equal: $\displaystyle\int_0^{t} (v_A - v_B)\,dt = 0$	✓ 1
A1 $12t + 24e^{-0.5t} - 24 - 0.75t^2 = 0$, or an equivalent equation entered into the GDC	✓ 1
A1 $t = 13.7$ s (13.659…)	✓ 1
Part (e)
M1 evaluates $\displaystyle\int_0^{13.66} v_B\,dt$ (or $\displaystyle\int_0^{13.66} v_A\,dt$)	✓ 1
A1 140 m (139.94…)	✓ 1

Common error: Drone B catches up when the displacements are equal, not when the velocities are equal. Equal velocities ($t = T$) is the moment when A's lead is greatest. After B overtakes, the gap between the drones grows again.

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After a tablet is swallowed, the concentration of a drug in a patient's blood, $C$ mg L⁻¹, $t$ hours later is modelled by
$C(t)=10\left(e^{-0.2t}-e^{-t}\right)$, for $t\ge 0$.

(a)

Find $C'(t)$.

(2)

(b)

Show that the concentration has a stationary value when $t=\frac{5}{4}\ln 5$, and find this concentration.

(4)

(c)

Use the second derivative to show that this concentration is a maximum.

(2)

(d)

Show that the graph of $C$ has a point of inflexion at exactly twice the time at which the maximum concentration occurs.

(3)

(e)

Find the rate of change of the concentration at the point of inflexion, and interpret its sign.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 differentiates the exponential terms using the chain rule	✓ 1
A1 $C'(t)=-2e^{-0.2t}+10e^{-t}$	✓ 1
Part (b)
M1 sets $C'(t)=0$: $10e^{-t}=2e^{-0.2t}$	✓ 1
A1 $e^{0.8t}=5\Rightarrow t=\frac{\ln 5}{0.8}=\frac{5}{4}\ln 5$ (AG)	✓ 1
M1 substitutes $t=2.01\ldots$ into $C(t)$	✓ 1
A1 $C=5.35$ mg L⁻¹	✓ 1
Part (c)
M1 $C''(t)=0.4e^{-0.2t}-10e^{-t}$	✓ 1
R1 $C''(\frac{5}{4}\ln 5)\approx 0.267-1.34=-1.07<0$, so it is a maximum	✓ 1
Part (d)
M1 sets $C''(t)=0$: $0.4e^{-0.2t}=10e^{-t}$	✓ 1
A1 $e^{0.8t}=25\Rightarrow t=\frac{5}{4}\ln 25=\frac{5}{2}\ln 5=2\times\frac{5}{4}\ln 5$	✓ 1
R1 $C''$ changes sign there, e.g. $C''(3)\approx-0.278<0$ and $C''(5)\approx 0.080>0$, so it is a point of inflexion	✓ 1
Part (e)
A1 $C'(\frac{5}{2}\ln 5)=-0.716$ mg L⁻¹ h⁻¹	✓ 1
R1 negative, so the concentration is decreasing (at its fastest rate), by 0.716 mg L⁻¹ per hour	✓ 1

Common error: Forgetting the factor from the chain rule: the derivative of $e^{-0.2t}$ is $-0.2e^{-0.2t}$, not $e^{-0.2t}$.

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An antibiotic is taken orally. Its concentration in the bloodstream is modelled by C(t) = 10(e^−0.2t − e^−0.8t) mg L⁻¹, where t is the time in hours after the dose, t ≥ 0. The antibiotic is effective while C(t) ≥ 3.

(a)

Find C′(t).

(2)

(b)

Show that C is increasing for 0 < t < (10/3)ln 2.

(4)

(c)

Find the maximum concentration.

(2)

(d)

Find the length of time for which the concentration is decreasing and the antibiotic is still effective.

(3)

(e)

Find the value of C′(t) when the concentration is decreasing most rapidly.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 −2e^−0.2t	✓ 1
A1 + 8e^−0.8t	✓ 1
Part (b)
M1 sets C′(t) > 0, i.e. 8e^−0.8t > 2e^−0.2t	✓ 1
M1 rearranges to e^0.6t < 4	✓ 1
A1 t < ln 4 / 0.6	✓ 1
A1 ln 4 / 0.6 = (10/3)ln 2	✓ 1
Part (c)
M1 substitutes t = 2.31 into C(t)	✓ 1
A1 4.72 mg L⁻¹	✓ 1
Part (d)
M1 solves C(t) = 3 for t > 2.31, e.g. by GDC	✓ 1
A1 t = 5.87	✓ 1
A1 5.87 − 2.31 = 3.56 hours	✓ 1
Part (e)
M1 attempts to minimise C′(t), e.g. C″(t) = 0 or GDC	✓ 1
A1 t = 4.62	✓ 1
A1 C′ = −0.595 mg L⁻¹ h⁻¹	✓ 1

Common error: In part (d), the interval starts at the maximum, not at t = 0; the earlier solution of C(t) = 3 lies in the increasing phase and is not needed.

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A bungee jumper leaves a platform at $t=0$. Her velocity, measured downwards, is modelled by $v(t)=20e^{-0.1t}\sin(0.6t)$ m s⁻¹, for $0\le t\le 10$, where $t$ is in seconds.

(a)

Find the time at which the jumper reaches her lowest point.

(2)

(b)

Find the maximum downward speed of the jumper.

(3)

(c)

Find the distance between the platform and the jumper's lowest point.

(3)

(d)

Find the jumper's acceleration at $t=1$.

(2)

(e)

Find the total distance travelled by the jumper for $0\le t\le 10$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sets $v(t)=0$ for $t>0$	✓ 1
A1 $t=\frac{\pi}{0.6}=5.24$ s	✓ 1
Part (b)
M1 attempt to maximise $v$, e.g. GDC or $v'(t)=0$	✓ 1
A1 $t=2.34$ s	✓ 1
A1 $15.6$ m s⁻¹	✓ 1
Part (c)
M1 integral of $v$	✓ 1
A1 limits 0 and $5.24$	✓ 1
A1 $51.6$ m	✓ 1
Part (d)
M1 attempt to differentiate $v$ and substitute $t=1$	✓ 1
A1 $7.94$ m s⁻²	✓ 1
Part (e)
M1 $\int_0^{10}\|v(t)\|\,dt$ (or $51.64+30.11$)	✓ 1
A1 $81.8$ m	✓ 1

Common error: Using $\int_0^{10}v\,dt$ gives only the net displacement of about 21.5 m, because the jumper moves back up after $t=5.24$.

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A cable-suspended camera moves along a straight line above a stadium pitch. Its velocity is $v(t)=4\cos t-t$ m s⁻¹, for $0\le t\le 4$, where $t$ is in seconds. At $t=0$ the camera is at point O.

(a)

Find the initial acceleration of the camera.

(2)

(b)

Find the time at which the camera changes direction.

(2)

(c)

Find the greatest distance of the camera from O in the positive direction.

(3)

(d)

Find the displacement of the camera from O at $t=4$.

(2)

(e)

Find the total distance travelled by the camera for $0\le t\le 4$.

(2)

(f)

Find the maximum speed of the camera for $0\le t\le 4$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $a(t)=-4\sin t-1$	✓ 1
A1 $a(0)=-1$ m s⁻²	✓ 1
Part (b)
M1 solves $4\cos t-t=0$	✓ 1
A1 $t=1.25$ s (1.2524)	✓ 1
Part (c)
M1 $\int_0^{1.2524}(4\cos t-t)\,dt$	✓ 1
A1 $4\sin(1.2524)-\frac{1.2524^2}{2}$	✓ 1
A1 $3.01$ m	✓ 1
Part (d)
M1 $\int_0^4 v\,dt=4\sin4-8$	✓ 1
A1 $-11.0$ m	✓ 1
Part (e)
M1 $\int_0^4\|v\|\,dt$ or $3.01+(3.01+11.0)$	✓ 1
A1 $17.1$ m	✓ 1
Part (f)
M1 considers $a=0$ or graph of $\|v\|$	✓ 1
A1 $t=\pi+\arcsin(0.25)=3.39$ s	✓ 1
A1 maximum speed $7.27$ m s⁻¹	✓ 1

Common error: Maximum velocity and maximum speed are different. The largest value of $v$ is 4 at $t=0$, but the speed is larger at $t=3.39$, where $v$ is most negative.

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A coffee roaster burns gas at a rate of R(t) = 4 + 6e^−0.5t litres per minute, where t is the time in minutes after ignition, 0 ≤ t ≤ 15.

(a)

Find ∫ R(t) dt.

(3)

(b)

Find the volume of gas burned in the first 6 minutes. Give your answer to 3 significant figures.

(2)

(c)

The gas supply is shut off when exactly 30 litres have been burned. Find the time, in minutes, at which this happens. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate	✓ 1
A1 4t	✓ 1
A1 −12e^−0.5t (+ c)	✓ 1
Part (b)
M1 substitute limits 6 and 0 into their antiderivative	✓ 1
A1 35.4 litres	✓ 1
Part (c)
M1 4T + 12 − 12e^−0.5T = 30 (ECF from (a))	✓ 1
A1 T = 4.78 minutes	✓ 1

Exam tip: For an equation you cannot rearrange, write it down in full, then solve it with your calculator and state the accuracy asked for.

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A tram moves along a straight track. Its velocity, in m s⁻¹, t seconds after leaving a stop is $v(t)=6t\,e^{-0.25t^2}$, for $t\ge 0$. The tram's displacement from the stop is $s$ metres, with $s=0$ when $t=0$.

(a)

Show that $v'(t)=e^{-0.25t^2}\left(6-3t^2\right)$.

(3)

(b)

Find the maximum speed of the tram. Give your answer to 3 significant figures.

(3)

(c)

Show that $s(t)=12\left(1-e^{-0.25t^2}\right)$.

(3)

(d)

Find the displacement of the tram when $t=4$, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule	✓ 1
A1 $6e^{-0.25t^2}+6t\left(-0.5t\right)e^{-0.25t^2}$	✓ 1
A1 $e^{-0.25t^2}(6-3t^2)$ (AG)	✓ 1
Part (b)
M1 set $v'(t)=0$	✓ 1
A1 $t=\sqrt2$	✓ 1
A1 $v=5.15\text{ m s}^{-1}$	✓ 1
Part (c)
M1 attempt to reverse the chain rule, $s=-12e^{-0.25t^2}+C$	✓ 1
M1 use $s(0)=0$ to find $C$	✓ 1
A1 $C=12$, giving $s(t)=12(1-e^{-0.25t^2})$ (AG)	✓ 1
Part (d)
M1 substitute $t=4$	✓ 1
A1 $s=11.8$ m	✓ 1

Exam tip: Keep an exact form like $\sqrt2$ in your calculator until the final step to avoid rounding errors.

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Calculator

The curve C has equation y = x³ + ax + b, where a and b are constants. The curve passes through the point (1, 6) and the gradient of the tangent to C at this point is 7.

(a)

Find the value of a.

(2)

(b)

Find the value of b.

(2)

(c)

The tangent at (1, 6) meets C again at the point Q. Find the coordinates of Q.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate: $\frac{dy}{dx}=3x^2+a$	✓ 1
A1 $3+a=7$, so $a=4$	✓ 1
Part (b)
M1 substitutes $(1,\,6)$ into $y=x^3+4x+b$	✓ 1
A1 $b=1$	✓ 1
Part (c)
M1 tangent $y=7x-1$ and equates $x^3+4x+1=7x-1$	✓ 1
A1 $x^3-3x+2=0$, factorised as $(x-1)^2(x+2)=0$	✓ 1
A1 $Q=(-2,\,-15)$	✓ 1

Exam tip: At the point of tangency the tangent and curve meet with a repeated root, so (x − 1) appears squared in the equation.

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A cylindrical data-storage capsule of height $h$ cm and radius $r$ cm is placed inside a spherical protective casing of radius $10$ cm so that the rim of each circular end of the capsule touches the inside of the sphere. The axis of the capsule passes through the centre of the sphere.

Cross-section through the axis. Diagram not to scale

(a)

Show that the volume of the capsule is $V=\pi\left(100h-\dfrac{h^3}{4}\right)$.

(3)

(b)

Find $\dfrac{dV}{dh}$ and hence find the exact value of $h$ for which $V$ has a stationary point, with $0<h<20$.

(3)

(c)

Show that this value of $h$ gives the maximum volume.

(2)

(d)

Find the maximum volume of the capsule, to the nearest $\text{cm}^3$.

(2)

(e)

Find the maximum volume of the capsule as a fraction of the volume of the sphere, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 Pythagoras in a right-angled triangle: $r^2+\left(\frac h2\right)^2=10^2$	✓ 1
A1 $r^2=100-\frac{h^2}{4}$	✓ 1
A1 $V=\pi r^2h=\pi\left(100h-\frac{h^3}{4}\right)$ (AG)	✓ 1
Part (b)
A1 $\frac{dV}{dh}=\pi\left(100-\frac{3h^2}{4}\right)$	✓ 1
M1 sets $\frac{dV}{dh}=0$, so $h^2=\frac{400}{3}$	✓ 1
A1 $h=\frac{20}{\sqrt3}$	✓ 1
Part (c)
A1 $\frac{d^2V}{dh^2}=-\frac{3\pi h}{2}$	✓ 1
R1 negative for $h>0$, so a maximum	✓ 1
Part (d)
M1 substitutes $h=\frac{20}{\sqrt3}$ into $V$	✓ 1
A1 $V_{\max}=\frac{4000\pi}{3\sqrt3}=2418\ \text{cm}^3$	✓ 1
Part (e)
M1 divides by $\frac43\pi(10)^3$	✓ 1
A1 fraction $=\frac{1}{\sqrt3}=0.577$	✓ 1

Exam tip: Express the volume in one variable before differentiating; use the geometric constraint to eliminate the other variable.

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Calculator

A relay station is at the point $R(4,0)$. A probe moves along the path $y=\sqrt{x}$ for $x\ge0$, where distances are in thousands of kilometres. The probe is at the point $P(x,\sqrt x)$ and $D$ is the distance $PR$.

Distances in thousands of km. Diagram not to scale

(a)

Show that $D^2=(x-4)^2+x$.

(2)

(b)

Find the value of $x$ for which $D$ is a minimum, and justify that it is a minimum.

(4)

(c)

Find the minimum distance between the probe and the relay station, in thousands of kilometres, to three significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
distance formula $D^2=(x-4)^2+(\sqrt x-0)^2$	✓ 1
$(\sqrt x)^2=x$, giving the result	✓ 1
Part (b)
$\frac{d(D^2)}{dx}=2(x-4)+1$	✓ 1
set equal to zero	✓ 1
$x=3.5$	✓ 1
second derivative $=2>0$ (or sign change of the derivative)	✓ 1
Part (c)
$D^2=0.5^2+3.5=3.75$	✓ 1
$D=1.94$	✓ 1

Exam tip: Minimising $D^2$ gives the same position as minimising $D$ and avoids differentiating a square root.

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Calculator

The power drawn by an instrument on a space probe is $P(t)=2t\,e^{t^2}$ watts, where $t$ is the time in hours since switch-on. The energy used between $t=0$ and $t=T$ is $E$ watt-hours.

(a)

Show that $E=e^{T^2}-1$.

(3)

(b)

The instrument has used $15$ watt-hours. Find $T$ to three significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$E=\int_0^T 2t\,e^{t^2}\,dt$ and recognise the derivative of $t^2$ (or substitute $u=t^2$)	✓ 1
antiderivative $e^{t^2}$	✓ 1
$e^{T^2}-e^0=e^{T^2}-1$	✓ 1
Part (b)
$e^{T^2}=16$	✓ 1
$T^2=\ln16$	✓ 1
$T=1.67$ hours	✓ 1

Common error: With $u=t^2$, the factor $2t\,dt$ becomes $du$ exactly; do not introduce an extra factor of 2.

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A space probe moves along a straight line. Its velocity, v m s⁻¹, at time t seconds is given by $v=6e^{-0.5t}-2$ for $t\ge 0$. Give all non-exact answers to 3 significant figures.

(a)

Write down the initial velocity of the probe.

(1)

(b)

Find the acceleration of the probe when t = 2.

(2)

(c)

Find the time at which the probe is instantaneously at rest.

(2)

(d)

Find the total distance travelled by the probe in the first 5 seconds. Use the unrounded time from part (c).

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $v(0)=4$ m s⁻¹	✓ 1
Part (b)
M1 differentiates, $a=-3e^{-0.5t}$	✓ 1
A1 $-3e^{-1}=-1.10$ m s⁻²	✓ 1
Part (c)
M1 sets $6e^{-0.5t}-2=0$, so $e^{-0.5t}=\frac{1}{3}$	✓ 1
A1 $t=2\ln 3=2.20$ s	✓ 1
Part (d)
M1 recognises that the direction changes at $t=2\ln 3$ so the integral must be split, or $\|v\|$ is used	✓ 1
M1 attempt to integrate $v$ over $0$ to $2\ln 3$ and over $2\ln 3$ to 5, e.g. $s=-12e^{-0.5t}-2t$	✓ 1
A1 distance forward $8-4\ln 3=3.61$ m (3.6056…)	✓ 1
A1 distance backward $2.59$ m (2.5906…)	✓ 1
A1 total distance 6.20 m	✓ 1

Common error: Integrating velocity from 0 to 5 gives the displacement, which is about 1.01 m. The total distance is larger because the probe reverses.

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Calculator

A curve C has equation $y=xe^{-x}$ for $x\ge 0$.

(a)

Show that $\dfrac{dy}{dx}=(1-x)e^{-x}$.

(3)

(b)

Find the coordinates of the maximum point of C.

(3)

(c)

The normal to C at the point where x = 2 meets the x-axis at the point Q. Find the x-coordinate of Q, giving your answer to 3 significant figures.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt at the product rule	✓ 1
A1 $e^{-x}-xe^{-x}$	✓ 1
R1 factorises to $(1-x)e^{-x}$ to reach the printed result	✓ 1
Part (b)
M1 sets $\frac{dy}{dx}=0$	✓ 1
A1 $x=1$	✓ 1
A1 $y=e^{-1}$, so the point is $(1, 0.368)$	✓ 1
Part (c)
A1 gradient of tangent at $x=2$ is $-e^{-2}$ (−0.135…)	✓ 1
M1 gradient of normal is $e^{2}$ (negative reciprocal)	✓ 1
A1 $y$-coordinate at $x=2$ is $2e^{-2}$ (0.270…)	✓ 1
M1 equation of normal, e.g. $y-2e^{-2}=e^{2}(x-2)$, then sets $y=0$	✓ 1
A1 $x=2-2e^{-4}=1.96$	✓ 1

Exam tip: A normal's gradient is the negative reciprocal of the tangent's gradient. Find the tangent gradient accurately first and keep exact values until the final line.

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A hospital is to buy a closed cylindrical water-storage tank of volume 2 m³, with radius $r$ m and height $h$ m. The two circular ends cost €400 per m² and the curved surface costs €250 per m². The total cost of the material is €$C$.

Diagram not to scale

(a)

Show that $C=800\pi r^2+\frac{1000}{r}$.

(3)

(b)

Find $\frac{dC}{dr}$.

(2)

(c)

Find the value of $r$ that gives the minimum cost, to 3 significant figures.

(3)

(d)

Find the minimum cost to the nearest euro, using the unrounded value of $r$.

(2)

(e)

Show that this cost is a minimum.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $\pi r^2h=2$, so $h=\frac{2}{\pi r^2}$	✓ 1
M1 $C=400(2\pi r^2)+250(2\pi rh)$	✓ 1
A1 substituting $h$: $800\pi r^2+500\pi r\times\frac{2}{\pi r^2}=800\pi r^2+\frac{1000}{r}$ (AG)	✓ 1
Part (b)
A1 $1600\pi r$	✓ 1
A1 $-\frac{1000}{r^2}$	✓ 1
Part (c)
M1 setting $\frac{dC}{dr}=0$	✓ 1
A1 $r^3=\frac{1000}{1600\pi}\ (=0.1989\ldots)$	✓ 1
A1 $r=0.584$ m	✓ 1
Part (d)
M1 substituting the unrounded $r$ into $C$	✓ 1
A1 €2569	✓ 1
Part (e)
A1 $\frac{d^2C}{dr^2}=1600\pi+\frac{2000}{r^3}$	✓ 1
R1 this is positive for all $r>0$, so the cost is a minimum	✓ 1

Common error: Cost per unit area differs for the ends and the curved surface; apply each price to the correct area before adding.

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Calculator

In a hospital trial, the response $R$ of a patient to a drug, $t$ hours after it is given ($t\ge 0$), is modelled by $R(t)=At\,e^{-kt}$, where $A$ and $k$ are positive constants.

(a)

Show that $\frac{dR}{dt}=A(1-kt)e^{-kt}$.

(3)

(b)

The response is greatest when $t=2.5$. Find the value of $k$.

(2)

(c)

The greatest response is 20 units. Find the exact value of $A$, and its value to 3 significant figures.

(3)

(d)

Use the second derivative to confirm that $R$ is a maximum at $t=2.5$. Use the exact value of $A$.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the product rule	✓ 1
A1 $Ae^{-kt}+At(-ke^{-kt})$	✓ 1
A1 factorising to $A(1-kt)e^{-kt}$ (AG)	✓ 1
Part (b)
M1 setting $\frac{dR}{dt}=0$, noting $e^{-kt}\ne0$	✓ 1
A1 $1-2.5k=0$, so $k=0.4$	✓ 1
Part (c)
M1 $R(2.5)=A(2.5)e^{-1}=20$	✓ 1
A1 $A=8e$	✓ 1
A1 $A\approx21.7$	✓ 1
Part (d)
M1 attempt to differentiate $A(1-kt)e^{-kt}$ using the product rule	✓ 1
A1 $\frac{d^2R}{dt^2}=Ae^{-kt}(k^2t-2k)$	✓ 1
R1 at $t=2.5$: $8e\times e^{-1}\times(0.16\times2.5-0.8)=-3.2<0$, so a maximum	✓ 1

Common error: $e^{-kt}$ is never zero, so a stationary point of $Ate^{-kt}$ comes only from the bracket $(1-kt)$.

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An architect designs a window made from a rectangle of width $2r$ metres and height $h$ metres, with a semicircle of radius $r$ metres on top. The total length of the outer boundary (the base $2r$, the two vertical sides, and the semicircular arc) is 10 m. The area of the window is $A$ m².

Diagram not to scale

(a)

Show that $A = 10r - \left(2 + \frac{\pi}{2}\right)r^2$.

(3)

(b)

Find $\frac{dA}{dr}$.

(2)

(c)

Find the value of $r$ for which $A$ has a stationary point. Give your answer to 3 significant figures.

(2)

(d)

Justify that this value of $r$ gives the maximum area, and find the maximum area.

(3)

(e)

Planning rules mean that $r \le 1.2$. Find the greatest possible area of the window under this rule. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $2r + 2h + \pi r = 10$, so $h = \frac{10 - 2r - \pi r}{2}$	✓ 1
M1 $A = 2rh + \frac{1}{2}\pi r^2$ with $h$ substituted	✓ 1
A1 $A = 10r - 2r^2 - \pi r^2 + \frac{\pi}{2}r^2 = 10r - \left(2+\frac{\pi}{2}\right)r^2$ (AG)	✓ 1
Part (b)
M1 attempt to differentiate	✓ 1
A1 $\frac{dA}{dr} = 10 - (4+\pi)r$	✓ 1
Part (c)
M1 sets $\frac{dA}{dr} = 0$	✓ 1
A1 $r = \frac{10}{4+\pi} = 1.40$ m	✓ 1
Part (d)
R1 $\frac{d^2A}{dr^2} = -(4+\pi) < 0$, so a maximum	✓ 1
M1 substitutes $r$ into $A$	✓ 1
A1 $A = 7.00$ m²	✓ 1
Part (e)
R1 $A$ increases for $r < 1.40$ (since $\frac{dA}{dr} > 0$), so the maximum is at $r = 1.2$	✓ 1
M1 substitutes $r=1.2$ into $A$	✓ 1
A1 $A = 6.86$ m²	✓ 1

Exam tip: When a constraint stops you reaching the stationary point, check the sign of the derivative to decide which end of the allowed interval gives the optimum.

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A survey submersible moves in a straight line along a coral reef. Its velocity at time $t$ seconds is $v = 5\sin(0.4t)$ m s⁻¹, for $t \ge 0$ (angles in radians). At $t=0$ the submersible is at the start of the reef, with displacement $s = 0$.

(a)

Find the first time $t > 0$ at which the submersible is momentarily at rest. Give your answer to 3 significant figures.

(2)

(b)

Find the acceleration of the submersible when $t = 2$.

(2)

(c)

Show that the displacement is given by $s = 12.5 - 12.5\cos(0.4t)$.

(3)

(d)

Find the greatest displacement of the submersible from the start of the reef.

(2)

(e)

Find the first time at which the displacement is 15 m. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sets $5\sin(0.4t)=0$ with $0.4t = \pi$	✓ 1
A1 $t = 7.85$ s	✓ 1
Part (b)
M1 $a = \frac{dv}{dt} = 2\cos(0.4t)$	✓ 1
A1 $a = 1.39$ m s⁻²	✓ 1
Part (c)
M1 attempt to integrate $v$, giving a cosine term	✓ 1
A1 $s = -12.5\cos(0.4t) + C$	✓ 1
A1 uses $s(0)=0$ to find $C = 12.5$ (AG)	✓ 1
Part (d)
M1 uses the time of first rest, where $\cos(0.4t) = -1$	✓ 1
A1 25 m	✓ 1
Part (e)
M1 $12.5 - 12.5\cos(0.4t) = 15$	✓ 1
M1 $\cos(0.4t) = -0.2$, so $0.4t = \arccos(-0.2)$	✓ 1
A1 $t = 4.43$ s	✓ 1

Exam tip: Set the calculator to radians for any calculus with trigonometric functions, and take the smallest positive solution when asked for the first time.

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A logistics company designs an open-topped rectangular storage crate with a square base of side $x$ metres and height $h$ metres. The volume of the crate is 32 m³. The base material costs £30 per m² and the side material costs £20 per m². The total material cost is £$C$.

Diagram not to scale

(a)

Show that $C = 30x^2 + \frac{2560}{x}$.

(3)

(b)

Find $\frac{dC}{dx}$.

(2)

(c)

Find the value of $x$ that minimises the cost. Give your answer to 3 significant figures.

(3)

(d)

Show that this value of $x$ gives a minimum cost.

(2)

(e)

Find the minimum cost, to the nearest pound.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $h = \frac{32}{x^2}$	✓ 1
M1 $C = 30x^2 + 20(4xh)$	✓ 1
A1 $80x \times \frac{32}{x^2} = \frac{2560}{x}$ (AG)	✓ 1
Part (b)
M1 attempt to differentiate, at least one term correct	✓ 1
A1 $\frac{dC}{dx} = 60x - \frac{2560}{x^2}$	✓ 1
Part (c)
M1 sets $\frac{dC}{dx}=0$	✓ 1
M1 $x^3 = \frac{2560}{60}$	✓ 1
A1 $x = 3.49$ m	✓ 1
Part (d)
M1 $\frac{d^2C}{dx^2} = 60 + \frac{5120}{x^3}$	✓ 1
R1 this is positive for $x>0$, so the cost is a minimum	✓ 1
Part (e)
M1 substitutes the unrounded value of $x$ into $C$	✓ 1
A1 £1099	✓ 1

Exam tip: Substitute the unrounded optimum value into the cost function, otherwise the final answer can be out by a pound or more.

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Calculator

A fishing vessel makes a 180 nautical mile voyage at a constant speed of $v$ knots (nautical miles per hour), where $v>0$. Its running costs are $(0.05v^3+300)$ euros per hour. The total cost of the voyage is €$C$.

(a)

Show that $C=9v^2+\dfrac{54000}{v}$.

(3)

(b)

Find $\dfrac{dC}{dv}$.

(2)

(c)

Find the speed that minimises the total cost. Give your answer to 3 significant figures.

(3)

(d)

Show that this speed gives a minimum and find the minimum total cost, to the nearest euro. Use the unrounded speed.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 time taken $=\dfrac{180}{v}$ hours	✓ 1
M1 $C=\dfrac{180}{v}\left(0.05v^3+300\right)$	✓ 1
A1 $C=9v^2+\dfrac{54000}{v}$ (AG)	✓ 1
Part (b)
M1 attempt to differentiate (one term correct)	✓ 1
A1 $\dfrac{dC}{dv}=18v-\dfrac{54000}{v^2}$	✓ 1
Part (c)
M1 set $\dfrac{dC}{dv}=0$ (ECF from (b))	✓ 1
A1 $v^3=3000$	✓ 1
A1 $v=14.4$ knots	✓ 1
Part (d)
R1 $\dfrac{d^2C}{dv^2}=18+\dfrac{108000}{v^3}>0$ for $v>0$, so a minimum	✓ 1
A1 €5616 ($5616.2\ldots$)	✓ 1

Exam tip: In an optimisation problem, build the cost function in one variable first and keep the unrounded optimum in your calculator memory for the final value.

+ Add to test✓ In your test467IBMA3S-5-17r19-095.9 Kinematics Paper 2Hard10 marks

Calculator

A drill head on a probe lander moves along a straight rail. Its acceleration is $a = 6t - 18$ m s⁻² for $t \ge 0$, where $t$ is the time in seconds. At $t = 0$ its velocity is $24$ m s⁻¹.

(a)

Find an expression for the velocity of the drill head at time $t$.

(3)

(b)

Find the times at which the drill head is momentarily at rest.

(2)

(c)

Find the total distance travelled by the drill head during the first 5 seconds.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate $a$	✓ 1
A1 $3t^2 - 18t$	✓ 1
A1 use of $v(0) = 24$ gives $v = 3t^2 - 18t + 24$	✓ 1
Part (b)
M1 set $v = 0$, $3(t-2)(t-4) = 0$	✓ 1
A1 $t = 2$ and $t = 4$	✓ 1
Part (c)
M1 recognise that the direction changes at $t = 2$ and $t = 4$, so integrate over the three intervals separately	✓ 1
A1 $\int_0^2 v\,dt = 20$	✓ 1
A1 $\int_2^4 v\,dt = -4$	✓ 1
A1 $\int_4^5 v\,dt = 4$	✓ 1
A1 total distance $= 20 + 4 + 4 = 28$ m	✓ 1

Common error: Integrating velocity from 0 to 5 gives displacement (here 20 m), not distance. Split the integral where the velocity changes sign and add the magnitudes.

+ Add to test✓ In your test468IBMA3S-5-17r19-135.4 Tangents and normals Paper 2Hard12 marks

Calculator

In a design drawing, the curve $C$ has equation $y = f(x) = x^3 - 6x^2 + 8x$. The point P$(1, 3)$ lies on $C$.

Diagram not to scale

(a)

Find the equation of the normal to $C$ at P.

(4)

(b)

The normal meets $C$ again at two other points. Show that their x-coordinates satisfy $x^2 - 5x + 2 = 0$.

(3)

(c)

Solve $x^2 - 5x + 2 = 0$. Give both solutions to 3 significant figures.

(2)

(d)

Let $\alpha$ be the smaller solution of $x^2 - 5x + 2 = 0$. Between $x = \alpha$ and $x = 1$ the curve lies above the normal. Find the area of the region enclosed by $C$ and the normal between these values. Use the unrounded value of $\alpha$ and give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $f'(x) = 3x^2 - 12x + 8$	✓ 1
A1 gradient at P $= -1$	✓ 1
M1 normal gradient $= 1$	✓ 1
A1 $y = x + 2$	✓ 1
Part (b)
M1 equate $x^3 - 6x^2 + 8x = x + 2$ to obtain $x^3 - 6x^2 + 7x - 2 = 0$	✓ 1
M1 use $x = 1$ as a root, factorising $(x-1)(\ldots)$	✓ 1
A1 $(x-1)(x^2 - 5x + 2) = 0$, hence $x^2 - 5x + 2 = 0$ (AG)	✓ 1
Part (c)
M1 quadratic formula or GDC, $x = \frac{5 \pm \sqrt{17}}{2}$	✓ 1
A1 $x = 0.438$ and $x = 4.56$	✓ 1
Part (d)
M1 $\int_\alpha^1 (x^3 - 6x^2 + 7x - 2)\,dx$	✓ 1
A1 antiderivative $\frac{x^4}{4} - 2x^3 + \frac{7x^2}{2} - 2x$	✓ 1
A1 area $= 0.113$	✓ 1

Exam tip: When a line meets a cubic at a known point, that point's x-value is a root of the equated equation, so factorise it out rather than solving the cubic from scratch.

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A boom on a space probe moves along a straight line. Its displacement from its starting position is $s = 2t - 3\sin t$ metres, where $t$ is the time in seconds and $0 \le t \le 2\pi$. Use radians throughout.

(a)

Find an expression for the velocity of the boom.

(2)

(b)

The boom is momentarily at rest at two times, $t_1$ and $t_2$, with $t_1 < t_2$. Find $t_1$ and $t_2$ to 3 significant figures.

(3)

(c)

Find the total distance travelled by the boom for $0 \le t \le 2\pi$. Use unrounded values of $t_1$ and $t_2$ and give your answer to 3 significant figures.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to differentiate	✓ 1
A1 $v = 2 - 3\cos t$	✓ 1
Part (b)
M1 set $2 - 3\cos t = 0$, so $\cos t = \frac{2}{3}$	✓ 1
A1 $t_1 = 0.841$	✓ 1
A1 $t_2 = 2\pi - t_1 = 5.44$	✓ 1
Part (c)
M1 evaluate $s$ at $t_1$, $t_2$ and $2\pi$ (the velocity changes sign at $t_1$ and $t_2$)	✓ 1
A1 $s(t_1) = -0.554$	✓ 1
A1 $s(t_2) = 13.1$ and $s(2\pi) = 12.6$	✓ 1
M1 sum of the magnitudes of the three displacements	✓ 1
A1 $0.554 + 13.67 + 0.554 = 14.8$ m	✓ 1

Common error: Switch your calculator to radian mode. Also, $\cos t = \frac{2}{3}$ has a second solution in $[0, 2\pi]$ at $2\pi - t_1$, which is easy to miss.

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The cross-section of a snow drift beside a hut is modelled by $g(x)=xe^{-0.5x}$ for $x\geq0$, where $x$ and $g(x)$ are in metres.

Diagram not to scale

(a)

Show that $g'(x)=e^{-0.5x}(1-0.5x)$.

(3)

(b)

Find the equation of the tangent to the curve at the origin.

(2)

(c)

Find the equation of the normal to the curve at the point where $x=2$.

(3)

(d)

Find the area of the cross-section between $x=0$ and $x=4$. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 product rule with chain rule on $e^{-0.5x}$	✓ 1
A1 $g'(x)=e^{-0.5x}+x(-0.5e^{-0.5x})$	✓ 1
A1 factorises to $e^{-0.5x}(1-0.5x)$ (AG)	✓ 1
Part (b)
M1 $g'(0)=1$ and $g(0)=0$	✓ 1
A1 $y=x$	✓ 1
Part (c)
M1 substitutes $x=2$ into $g'$: $e^{-1}(1-1)$	✓ 1
R1 $g'(2)=0$ so the tangent is horizontal and the normal is vertical	✓ 1
A1 $x=2$	✓ 1
Part (d)
M1 $\int_0^4xe^{-0.5x}\,dx$	✓ 1
A1 2.38 m$^2$	✓ 1

Common error: If the tangent is horizontal, do not try to use a negative reciprocal: the normal is a vertical line $x=k$.

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A robotic rover moves along a straight line in a crater. Its velocity $v$ cm s⁻¹ at time $t$ seconds is $v(t)=t^2-7t+10$ for $0\le t\le 6$. The rover starts at the origin, so its displacement is $s=0$ when $t=0$.

(a)

Find the times at which the rover changes direction.

(2)

(b)

Find the minimum velocity of the rover.

(3)

(c)

Find an expression for the displacement $s$ of the rover at time $t$.

(3)

(d)

Find the total distance travelled by the rover in the interval $0\le t\le 6$.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 solves $v=0$, i.e. $(t-2)(t-5)=0$	✓ 1
A1 $t=2$ and $t=5$	✓ 1
Part (b)
M1 differentiates $v'(t)=2t-7$ and sets it to zero	✓ 1
A1 $t=3.5$	✓ 1
A1 $v(3.5)=-2.25\ \text{cm s}^{-1}$	✓ 1
Part (c)
M1 attempts to integrate $v$	✓ 1
A1 $\dfrac{t^3}{3}-\dfrac{7t^2}{2}+10t$	✓ 1
A1 uses $s(0)=0$ to show the constant is $0$	✓ 1
Part (d)
M1 splits the motion at $t=2$ and $t=5$ and uses absolute changes in displacement	✓ 1
A1 $s(2)=\dfrac{26}{3}$	✓ 1
A1 $s(5)=\dfrac{25}{6}$	✓ 1
A1 $s(6)=6$	✓ 1
A1 total distance $=\dfrac{26}{3}+\left(\dfrac{26}{3}-\dfrac{25}{6}\right)+\left(6-\dfrac{25}{6}\right)=15$ cm	✓ 1

Common error: Total distance is not the final displacement. Split the motion wherever the velocity changes sign and add the sizes of each part.

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A sports scientist models the concentration $C$ micrograms per millilitre ($\mu$g mL⁻¹) of a supplement in an athlete's blood plasma $t$ hours after it is taken by $C(t)=8te^{-0.5t}$ for $t\ge0$. Give non-integer answers to 3 significant figures.

(a)

Find $C'(t)$.

(3)

(b)

Find the maximum concentration and the time at which it occurs.

(4)

(c)

Show that $C''(t)=2e^{-0.5t}(t-4)$.

(3)

(d)

The graph of $C$ has a point of inflection. Find its coordinates.

(3)

(e)

Find the rate of change of concentration at the point of inflection, and explain what it shows about the concentration at this time.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempts the product rule	✓ 1
A1 $C'(t)=8e^{-0.5t}-4te^{-0.5t}$	✓ 1
A1 $C'(t)=8e^{-0.5t}(1-0.5t)$	✓ 1
Part (b)
M1 sets $C'(t)=0$	✓ 1
A1 $t=2$	✓ 1
R1 $C'$ changes from positive to negative at $t=2$, so a maximum	✓ 1
A1 $C(2)=16e^{-1}=5.89\ \mu\text{g mL}^{-1}$	✓ 1
Part (c)
M1 attempts the product rule on $-4te^{-0.5t}$ or on the factorised form of $C'$	✓ 1
A1 $C''(t)=-4e^{-0.5t}-4e^{-0.5t}+2te^{-0.5t}$	✓ 1
A1 $C''(t)=(2t-8)e^{-0.5t}=2e^{-0.5t}(t-4)$	✓ 1
Part (d)
M1 sets $C''(t)=0$	✓ 1
A1 $t=4$	✓ 1
A1 $C(4)=32e^{-2}=4.33$, so the point is $(4,\,4.33)$	✓ 1
Part (e)
A1 $C'(4)=-8e^{-2}=-1.08\ \mu\text{g mL}^{-1}\ \text{h}^{-1}$	✓ 1
R1 $C''$ changes from negative to positive at $t=4$, so $C'$ is at its minimum: the concentration is falling most rapidly at 4 hours	✓ 1

Exam tip: A point of inflection of $C$ is where the gradient $C'$ has a turning value. Use it to find the time of the greatest rate of increase or decrease.

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A museum designs a closed cylindrical tube, of radius $r$ cm and height $h$ cm, to store rolled maps. The tube must have volume 1500 cm³. The surface area of the tube is $S$ cm².

Diagram not to scale

(a)

Show that $S = 2\pi r^2 + \frac{3000}{r}$.

(2)

(b)

Find the value of $r$ that minimises $S$. Give your answer to 3 significant figures.

(4)

(c)

Use the second derivative to justify that this value of $r$ gives a minimum.

(2)

(d)

Find the minimum surface area, to 3 significant figures. Use the unrounded value of $r$ from (b).

(2)

(e)

Show that, for the tube of minimum surface area, $h=2r$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $h=\frac{1500}{\pi r^2}$	✓ 1
A1 $S=2\pi r^2+2\pi r\cdot\frac{1500}{\pi r^2}=2\pi r^2+\frac{3000}{r}$ (AG)	✓ 1
Part (b)
A1 $\frac{dS}{dr}=4\pi r-3000r^{-2}$	✓ 1
M1 set $\frac{dS}{dr}=0$	✓ 1
A1 $r^3=\frac{750}{\pi}$	✓ 1
A1 $r=6.20$ cm	✓ 1
Part (c)
M1 $\frac{d^2S}{dr^2}=4\pi+6000r^{-3}$	✓ 1
R1 positive for all $r>0$, so a minimum	✓ 1
Part (d)
M1 substitute $r=\left(\frac{750}{\pi}\right)^{1/3}\approx 6.2035$ into $S$	✓ 1
A1 $725$ cm²	✓ 1
Part (e)
M1 $\pi r^3=750$, so $h=\frac{1500}{\pi r^2}=\frac{2\pi r^3}{\pi r^2}$	✓ 1
A1 $h=2r$ (AG)	✓ 1

Exam tip: In 'show that' parts, finish with the exact target expression; use unrounded values for later calculations and round only at the end.

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A jeweller cuts a 48 cm length of wire into two pieces. One piece, of length $x$ cm, is bent into a square. The other piece is bent into a circle.

The wire is cut into two pieces. Diagram not to scale

(a)

Show that the total area enclosed by the two shapes is $A = \frac{x^2}{16} + \frac{(48-x)^2}{4\pi}$.

(3)

(b)

Find the value of $x$ for which $A$ is a minimum. Give your answer to 3 significant figures.

(4)

(c)

Justify that this value of $x$ gives a minimum.

(1)

(d)

Find the minimum total area, using the unrounded value of $x$. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 side of square is $\frac{x}{4}$, so its area is $\frac{x^2}{16}$	✓ 1
M1 circumference of circle is $48-x$, so radius $r = \frac{48-x}{2\pi}$	✓ 1
A1 area of circle $\pi r^2 = \frac{(48-x)^2}{4\pi}$, giving the result (AG)	✓ 1
Part (b)
M1 attempt to differentiate $A$	✓ 1
A1 $\frac{dA}{dx} = \frac{x}{8} - \frac{48-x}{2\pi}$	✓ 1
M1 set $\frac{dA}{dx}=0$ and solve, giving $\pi x = 4(48-x)$	✓ 1
A1 $x = \frac{192}{\pi+4} = 26.9$	✓ 1
Part (c)
R1 $\frac{d^2A}{dx^2} = \frac{1}{8} + \frac{1}{2\pi} > 0$ for all $x$, so a minimum	✓ 1
Part (d)
M1 substitute $x = \frac{192}{\pi+4}$ into $A$	✓ 1
A1 $A = 80.7\ \text{cm}^2$	✓ 1

Exam tip: Both pieces of wire share one length, so write the radius in terms of $48-x$ before forming the total area.

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A rail operator designs an open-topped storage crate for track fixings. The crate is a cuboid with a square base of side $x$ metres and height $h$ metres, and its volume is 4 m³. The base material costs €30 per m² and the four sides cost €20 per m². The total cost of materials is €$C$.

Diagram not to scale

(a)

Write down an expression for $h$ in terms of $x$.

(1)

(b)

Show that $C=30x^2+\frac{320}{x}$.

(3)

(c)

Find $\frac{dC}{dx}$.

(2)

(d)

Find the value of $x$ that minimises the cost. Give your answer to 3 significant figures.

(3)

(e)

Find the minimum cost, to the nearest cent. Use the unrounded value of $x$.

(2)

(f)

Use the second derivative to justify that this cost is a minimum.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 $h=\frac{4}{x^2}$	✓ 1
Part (b)
A1 base cost $30x^2$	✓ 1
M1 area of the four sides $4xh$, at €20 per m² gives $80xh$	✓ 1
A1 substitutes $h=\frac{4}{x^2}$ to get $\frac{320}{x}$ (AG)	✓ 1
Part (c)
M1 attempt to differentiate, writing $\frac{320}{x}$ as $320x^{-1}$	✓ 1
A1 $\frac{dC}{dx}=60x-320x^{-2}$	✓ 1
Part (d)
M1 sets $\frac{dC}{dx}=0$	✓ 1
A1 $60x^3=320$, so $x^3=\frac{16}{3}$	✓ 1
A1 $x=1.75$	✓ 1
Part (e)
M1 substitutes $x=\left(\frac{16}{3}\right)^{1/3}=1.7471\ldots$ into $C$	✓ 1
A1 €274.73	✓ 1
Part (f)
R1 $\frac{d^2C}{dx^2}=60+640x^{-3}>0$ for $x>0$, so a minimum	✓ 1

Common error: The four sides of the crate have total area $4xh$, not $2xh$; check the surface area before applying the cost per m².

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The brightness of a variable star, in arbitrary units, is modelled by $B(t)=30+8\sin(0.5t)+2\cos t$, where $t$ is the time in days and $0\leq t\leq 12$. Angles are in radians.

(a)

Find $B'(t)$.

(3)

(b)

Show that $B'(t)=4\cos(0.5t)\,(1-\sin(0.5t))$.

(2)

(c)

Find the values of $t$, for $0\leq t\leq 12$, at which $B'(t)=0$. Give your answers to 3 significant figures.

(3)

(d)

Determine the maximum and the minimum values of $B(t)$ on the interval $0\leq t\leq 12$, justifying your answer.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the chain rule	✓ 1
A1 $4\cos(0.5t)$	✓ 1
A1 $-2\sin t$	✓ 1
Part (b)
M1 uses $\sin t=2\sin(0.5t)\cos(0.5t)$	✓ 1
A1 $4\cos(0.5t)-4\sin(0.5t)\cos(0.5t)$ factorised to the given form (AG)	✓ 1
Part (c)
M1 sets $\cos(0.5t)=0$ or $\sin(0.5t)=1$	✓ 1
A1 $t=\pi=3.14$	✓ 1
A1 $t=3\pi=9.42$ (the next solution, $5\pi=15.7$, is outside the interval)	✓ 1
Part (d)
M1 evaluates $B$ at the stationary points and at both endpoints	✓ 1
A1 $B(\pi)=36$ and $B(3\pi)=20$	✓ 1
A1 $B(0)=32$ and $B(12)=29.5$	✓ 1
R1 comparing values, the maximum is 36 (at $t=\pi$) and the minimum is 20 (at $t=3\pi$)	✓ 1

Common error: On a closed interval, compare the stationary values with the endpoint values before naming the greatest and least values.

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A beekeeper builds a hive box in the shape of a cuboid with a square base of side x cm and height h cm. The box has a base and four vertical sides but no top, and its volume is 48 000 cm³. The base is made from wood costing €0.003 per cm²; each side is made from wood costing €0.002 per cm².

Diagram not to scale

(a)

Show that the total cost of the wood, €C, is given by $C=0.003x^2+\dfrac{384}{x}$.

(3)

(b)

Find $\dfrac{dC}{dx}$.

(2)

(c)

Find the value of x for which the cost is a minimum.

(3)

(d)

Justify that this value of x gives a minimum cost.

(2)

(e)

Find the height of the box and the minimum cost of the wood.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 $h=\frac{48000}{x^2}$	✓ 1
M1 $C=0.003x^2+0.002\times4xh$	✓ 1
A1 $C=0.003x^2+\frac{0.008\times48000}{x}=0.003x^2+\frac{384}{x}$	✓ 1
Part (b)
M1 attempt to differentiate (at least one term correct)	✓ 1
A1 $\frac{dC}{dx}=0.006x-\frac{384}{x^2}$	✓ 1
Part (c)
M1 set $\frac{dC}{dx}=0$	✓ 1
A1 $x^3=\frac{384}{0.006}=64000$	✓ 1
A1 $x=40$	✓ 1
Part (d)
M1 $\frac{d^2C}{dx^2}=0.006+\frac{768}{x^3}$	✓ 1
R1 at $x=40$ this is $0.006+0.012=0.018>0$, so the cost is a minimum	✓ 1
Part (e)
A1 $h=\frac{48000}{40^2}=30$ cm	✓ 1
A1 $C=0.003(1600)+\frac{384}{40}=$ €14.40	✓ 1

Common error: Forgetting that the box has no top, or costing the sides at the base rate. Write the cost of each face type separately before combining them.

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A depot truck moves along a straight lane. Its velocity at time $t$ seconds is $v = 0.3t^2 - 2.4t + 3.6$ m s$^{-1}$ for $t \geq 0$. At $t = 0$ the truck is at the point $O$, so its displacement from $O$ is $s = 0$.

(a)

Find the times at which the truck is momentarily at rest.

(3)

(b)

Find the acceleration of the truck when $t = 6$.

(2)

(c)

Find an expression for $s$ in terms of $t$.

(2)

(d)

Show that $s = 0.1t(t-6)^2$. Hence explain why the truck is never behind $O$.

(3)

(e)

Find the total distance travelled by the truck in the first 8 seconds.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 set $v = 0$	✓ 1
M1 $t^2 - 8t + 12 = 0$, attempt to solve	✓ 1
A1 $t = 2$ and $t = 6$	✓ 1
Part (b)
M1 $a = 0.6t - 2.4$	✓ 1
A1 $a = 1.2$ m s$^{-2}$	✓ 1
Part (c)
M1 attempt to integrate $v$	✓ 1
A1 $s = 0.1t^3 - 1.2t^2 + 3.6t$ (no constant, since $s=0$ at $t=0$)	✓ 1
Part (d)
M1 expand $0.1t(t^2 - 12t + 36)$	✓ 1
A1 $0.1t^3 - 1.2t^2 + 3.6t$, matching the result in (c)	✓ 1
R1 for $t \geq 0$, both $t$ and $(t-6)^2$ are non-negative, so $s \geq 0$	✓ 1
Part (e)
M1 $s(2) = 3.2$, $s(6) = 0$, $s(8) = 3.2$	✓ 1
M1 distance $= 3.2 + (3.2 - 0) + (3.2 - 0)$, splitting at $t=2$ and $t=6$	✓ 1
A1 $9.6$ m	✓ 1

Common error: Total distance needs the displacement between every pair of turning points, added together as positive amounts; $s(8)$ alone gives only the final displacement.

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Let $f(x)=e^{-x^2}$. The graph of $f$ has a normal at the point where $x=1$.

(a)

Find $f'(x)$.

(2)

(b)

Find the gradient of the normal at $x=1$.

(2)

(c)

Find the $y$-intercept of this normal, giving your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use the chain rule	✓ 1
A1 $f'(x)=-2xe^{-x^2}$	✓ 1
Part (b)
M1 gradient of tangent at $x=1$ is $-2e^{-1}$ and use of the negative reciprocal	✓ 1
A1 $\frac{e}{2}$	✓ 1
Part (c)
A1 point on curve $\left(1,e^{-1}\right)$	✓ 1
M1 $y-e^{-1}=\frac{e}{2}(x-1)$ and set $x=0$	✓ 1
A1 $y=e^{-1}-\frac{e}{2}=-0.991$	✓ 1

Exam tip: Keep the exact values $e^{-1}$ and $\frac{e}{2}$ until the final step, then round once.

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A curve $y=f(x)$ has gradient function $f'(x)=6\cos 2x+8e^{4x}$ and passes through the point $(0,7)$. Angles are in radians.

(a)

Find $f(x)$.

(5)

(b)

Find $f(0.5)$, giving your answer to 3 significant figures.

(2)

(c)

Find the equation of the tangent to the curve at the point where $x=0$.

(3)

(d)

The normal to the curve at the point where $x=0$ meets the $x$-axis at the point R. Find the $x$-coordinate of R.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to integrate	✓ 1
A1 $3\sin 2x$	✓ 1
A1 $2e^{4x}$	✓ 1
M1 substitute $(0,7)$ to find the constant	✓ 1
A1 $f(x)=3\sin 2x+2e^{4x}+5$	✓ 1
Part (b)
M1 substitute $x=0.5$	✓ 1
A1 $22.3$	✓ 1
Part (c)
A1 $f'(0)=14$	✓ 1
M1 $y-7=14(x-0)$	✓ 1
A1 $y=14x+7$	✓ 1
Part (d)
M1 gradient of normal $-\frac{1}{14}$	✓ 1
M1 $y=-\frac{1}{14}x+7$ with $y=0$	✓ 1
A1 $x=98$	✓ 1

Common error: Integrating $\cos 2x$ gives $\frac{1}{2}\sin 2x$, so the factor of 2 inside the function divides the result.

Working	Mark	Notes
(a)
\(f(1+h)=(1+h)^{2}+3(1+h)=4+5h+h^{2}\)	M1	Award for a correct expansion of \(f(1+h)\).
gradient \(=\dfrac{4+5h+h^{2}-4}{(1+h)-1}=\dfrac{5h+h^{2}}{h}\)	A1
\(=5+h\)	AG
(b)
\(5\)	A1
(c)
\(f'(x)=2x+3\)	A1
\(f'(1)=2(1)+3=5\), which equals the limit in (b)	A1	Must see the substitution.

Working	Mark	Notes
(a)
\(3\)	A1
(b)
substituting \(x=1\) gives \(\dfrac{0}{0}\) (division by zero is undefined)	R1	Accept “\(x=1\) is not in the domain” or “the denominator is zero when \(x=1\)”.
(c)
\((x-1)(x^{2}+x+1)=x^{3}+x^{2}+x-x^{2}-x-1\)	M1
\(=x^{3}-1\), so \(\dfrac{x^{3}-1}{x-1}=\dfrac{(x-1)(x^{2}+x+1)}{x-1}\) and the factor \(x-1\neq0\) cancels	A1
\(f(x)=x^{2}+x+1\) for \(x\neq1\)	AG
(d)
\(f'(x)=2x+1\)	(M1)
\(f'(2)=5\)	A1

Working	Mark	Notes
(a)
\(4\pi r^{2}\)	A1
(b)
\(36\pi\)	A1
(c)
when the radius is 3 cm, the volume is increasing at a rate of \(36\pi\ \text{cm}^{3}\) per cm increase in the radius	R1	Must refer to rate of change of volume with respect to radius and give units cm³ per cm (or cm²).
(d)
\(4\pi r^{2}=100\pi\Rightarrow r^{2}=25\)	(M1)
\(r=5\) cm	A1	Award A0 if \(r=-5\) is also given and not rejected.

Working	Mark	Notes
(a)
\(6x^{2}-18x\)	A1	Award for any two correct terms.
\(f'(x)=6x^{2}-18x+12\)	A1
(b)
\(6x^{2}-18x+12=0\Rightarrow 6(x-1)(x-2)=0\)	M1	Or correct use of the quadratic formula.
\(x=1,\ x=2\)	A1
(c)
\(1\lt x\lt 2\)	A1	FT from (b). Accept \(1\le x\le 2\) or \(]1,\,2[\). Award A0 for \(f'(x)\lt 0\) alone.

Working	Mark	Notes
(a)
\(-2\lt x\lt 3\)	A1	Award A0 for \(-2\le x\le 3\), since \(f'(-2)=f'(3)=0\).
(b)
\(0\)	A1
(c)
\(f'(x)=(x+2)(x-3)\)	(M1)	Or \(4-2b+c=0\) and \(9+3b+c=0\).
\(b=-1,\ c=-6\)	A1
(d)
concave-up parabola	A1
\(x\)-intercepts \(-2\) and \(3\)	A1
\(y\)-intercept \(-6\)	A1	FT from (c). The vertex \(\left(\tfrac12,\,-\tfrac{25}{4}\right)\) need not be labelled.

Working	Mark	Notes
(a)
\(f'(x)=2ax+b\)	A1
(b)
\(0\)	A1
(c)
\(f'(0)=b=-8\)	A1
\(4a-8=0\Rightarrow a=2\)	A1	FT from their \(b\).

Working	Mark	Notes
(a)
\(20x^{4}-6x\)	A1	Award for any two correct terms.
\(\dfrac{\mathrm{d}y}{\mathrm{d}x}=20x^{4}-6x+1\)	A1
(b)
\(y=3x^{-4}\)	(M1)
\(-12x^{-5}\)	A1	Accept \(-\dfrac{12}{x^{5}}\).
(c)
\(y=2x^{2}-x-15\)	(M1)
\(4x-1\)	A1

Working	Mark	Notes
(a)
\(\dfrac{\mathrm{d}y}{\mathrm{d}x}=3x^{2}-2\)	(M1)
\(3(2)^{2}-2=10\)	A1
(b)
\(y-5=10(x-2)\)	(M1)
\(y=10x-15\)	A1
(c)
gradient \(-\dfrac{1}{10}\): \(y-5=-\dfrac{1}{10}(x-2)\)	(M1)
\(x+10y-52=0\)	A1	Accept any integer multiple.

Working	Mark	Notes
(a)
attempt to integrate, e.g. one power increased by 1	M1
\(x^{4}-6x^{-1}+C\)	A1	Accept \(x^{4}-\dfrac{6}{x}\). Condone the absence of \(+C\) here.
substituting \((1,\,3)\): \(1-6+C=3\)	M1	Must include \(C\).
\(f(x)=x^{4}-\dfrac{6}{x}+8\)	A1
(b)
gradient \(f'(1)=4+6=10\)	A1
\(y=10x-7\)	A1	Accept \(y-3=10(x-1)\).

Working	Mark	Notes
(a)
\(2x^{3}-2x^{2}\)	A1
\(+3x+c\)	A1	Penalize a missing \(+c\) once only in this question.
(b)
\(\dfrac{x^{4}}{4}\)	A1
\(+2x^{-1}+c\)	A1	Accept \(\dfrac{2}{x}\).
(c)
\(\displaystyle\int(4x^{2}+4x+1)\,\mathrm{d}x\)	(M1)
\(\dfrac{4x^{3}}{3}+2x^{2}+x+c\)	A1	Accept \(\dfrac{(2x+1)^{3}}{6}+c\).

	P	Q	R	S
\(f'(x)\)
\(f''(x)\)

Working	Mark	Notes
(a)
\(F(x)=x^{4}+2x^{2}+1\)	(M1)
\(F'(x)=4x^{3}+4x\)	A1
(b)
\(\dfrac{1}{4}(x^{2}+1)^{2}\)	A1	Accept \(\dfrac{x^{4}}{4}+\dfrac{x^{2}}{2}\).
\(+c\)	A1
(c)
\(\dfrac{1}{4}(1+1)^{2}+c=3\)	M1
\(G(x)=\dfrac{1}{4}(x^{2}+1)^{2}+2\)	A1	Accept \(\dfrac{x^{4}}{4}+\dfrac{x^{2}}{2}+\dfrac{9}{4}\).

Working	Mark	Notes
(a)
\(4x^{-\frac12}\)	A1	Accept \(\dfrac{4}{\sqrt{x}}\).
\(+6x^{-4}\), so \(f'(x)=4x^{-\frac12}+6x^{-4}\)	A1	Award A0 if a term from the constant 3 is present.
(b)
attempt to use the chain rule, e.g. \(\cos(3x^{2}-1)\times\ldots\)	M1
\(g'(x)=6x\cos(3x^{2}-1)\)	A1
(c)
\(\dfrac{2}{2x+5}\)	A1
\(-4\mathrm{e}^{-4x}\)	A1

Working	Mark	Notes
(a)
attempt to use the product rule, e.g. \(3x^{2}\ln x+x^{3}\times\ldots\)	M1
\(f'(x)=3x^{2}\ln x+x^{2}\)	A1
(b)
\(x^{2}(3\ln x+1)=0\)	M1	Setting their \(f'(x)=0\) and factorizing.
\(\ln x=-\tfrac{1}{3}\)	A1	Must reject or ignore \(x=0\), as \(x\gt 0\).
\(x=\mathrm{e}^{-\frac{1}{3}}\)	A1	Accept \(\dfrac{1}{\sqrt[3]{\mathrm{e}}}\).

Working	Mark	Notes
(a)
\(p=\tfrac32\)	A1	Accept \(x^{\frac32}\).
\(q=-\tfrac12\)	A1	Accept \(3x^{-\frac12}\).
(b)
\(\tfrac32x^{\frac12}\)	A1	FT from their (a).
\(-\tfrac32x^{-\frac32}\), so \(f'(x)=\tfrac32x^{\frac12}-\tfrac32x^{-\frac32}\)	A1	Accept \(\dfrac{3\sqrt{x}}{2}-\dfrac{3}{2x\sqrt{x}}\).
(c)
\(f(4)=\dfrac{19}{2}\)	A1
\(f'(4)=\tfrac32\times2-\tfrac32\times\tfrac18=\dfrac{45}{16}\)	A1	FT from their (b).
\(y-\dfrac{19}{2}=\dfrac{45}{16}(x-4)\)	A1	Accept \(y=\dfrac{45}{16}x-\dfrac{7}{4}\).

Working	Mark	Notes
(a)
attempt to use the chain rule, e.g. \(\dfrac{1}{x^{2}+5}\times\ldots\)	M1
\(f'(x)=\dfrac{2x}{x^{2}+5}\)	A1
(b)
\(f'(1)=\dfrac{2}{6}=\dfrac13\)	A1
(c)
\(h(x)=3\ln(x^{2}+5)-\ln x=3f(x)-\ln x\)	M1	Correct use of the laws of logarithms.
\(h'(x)=3f'(x)-\dfrac{1}{x}\left(=\dfrac{6x}{x^{2}+5}-\dfrac1x\right)\)	A1
\(h'(1)=3\times\dfrac13-1=0\)	A1	Must see substitution leading to 0.
stationary point at \(x=1\)	AG

Working	Mark	Notes
(a)
attempt to use the chain rule twice, e.g. \(2\sin(3x)\times\cos(3x)\times\ldots\)	M1
\(\dfrac{\mathrm{d}y}{\mathrm{d}x}=6\sin(3x)\cos(3x)\)	A1	Accept \(3\sin(6x)\).
(b)
\(6\sin\left(\tfrac{\pi}{4}\right)\cos\left(\tfrac{\pi}{4}\right)=6\times\tfrac{\sqrt2}{2}\times\tfrac{\sqrt2}{2}\)	(M1)	Or \(3\sin\left(\tfrac{\pi}{2}\right)\).
\(3\)	A1
(c)
\(x=\dfrac{\pi}{6}\)	A1	From \(3\sin(6x)=0\), \(6x=\pi\).

Working	Mark	Notes
(a)
\(g'(x)=4(x-1)^{3}+2\)	A1
\(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}=12(x-1)^{2}\)	A1
(b)
\(g''(1)=12(0)^{2}=0\)	A1
\(g''(x)=12(x-1)^{2}\gt 0\) for both \(x\lt 1\) and \(x\gt 1\), so \(g''\) does not change sign (no change of concavity)	R1	Award R1 for a reason referring to the sign of \(g''\) on both sides of \(x=1\).
(c)
\(x\in\mathbb{R},\ x\neq1\)	A1	Accept \(x\lt 1\) or \(x\gt 1\).

Working	Mark	Notes
(a)
\(2x^{-\frac12}\)	A1
\(-x^{-2}\), so \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=2x^{-\frac12}-x^{-2}\)	A1
(b)
\(-x^{-\frac32}\)	A1	FT from their (a).
\(+2x^{-3}\)	A1
(c)
\(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\Big\|_{x=4}=-\dfrac18+\dfrac{2}{64}=-\dfrac{3}{32}\)	A1
\(\dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\lt 0\), so concave-down	R1	Do not award R1 without a correct value or sign of the second derivative.

Working	Mark	Notes
(a)
C is the graph of \(f\)	A1
A is the graph of \(f'\) and B is the graph of \(f''\)	A1	Reasoning: the zeros of A are at the turning points of C; the zeros of B are at the turning points of A.
(b)
\(2\)	A1
graph B (\(f''\)) crosses the \(x\)-axis twice, changing sign each time	R1	Accept: graph A (\(f'\)) has two turning points. Do not award A1R0.
(c)
\(f''(x)=3x^{2}-4=0\)	M1
\(x=\pm\dfrac{2}{\sqrt3}\)	A1	Accept \(\pm\dfrac{2\sqrt3}{3}\).

Working	Mark	Notes
(a)
\(-2\mathrm{e}^{-2x}\)	A1
\(+2\cos(2x)\)	A1
(b)
\(4\mathrm{e}^{-2x}\)	A1	FT from their (a).
\(-4\sin(2x)\)	A1
(c)
\(4\mathrm{e}^{-2x}-4\sin(2x)+4\mathrm{e}^{-2x}+4\sin(2x)\)	A1	Must see the substitution of both \(y\) and their second derivative.
\(=8\mathrm{e}^{-2x}\)	AG

Working	Mark	Notes
(a)
\(f'(x)=4x^{3}-16x\)	A1
(b)
\(4x(x^{2}-4)=0\Rightarrow4x(x-2)(x+2)=0\)	M1
\(x=-2,\ 0,\ 2\)	A1
(c)
\(f'\lt 0\) for \(x\lt -2\), \(f'\gt 0\) for \(-2\lt x\lt0\), \(f'\lt 0\) for \(0\lt x\lt2\), \(f'\gt 0\) for \(x\gt2\)	R1	e.g. \(f'(-3)=-60\), \(f'(-1)=12\), \(f'(1)=-12\), \(f'(3)=60\), or a correct sign diagram.
local maximum \((0,\,3)\)	A1
local minima \((-2,\,-13)\) and \((2,\,-13)\)	A1	Both points required.

Working	Mark	Notes
(a)
\(f'(x)=2\mathrm{e}^{2x}-6\)	A1
(b)
\(2\mathrm{e}^{2x}=6\Rightarrow\mathrm{e}^{2x}=3\)	M1
\(x=\dfrac{\ln3}{2}\)	A1	Accept \(\ln\sqrt3\).
(c)
\(f''(x)=4\mathrm{e}^{2x}\)	A1
\(f''\left(\tfrac{\ln3}{2}\right)=12\gt 0\) (or \(4\mathrm{e}^{2x}\gt 0\) for all \(x\)), so a minimum	R1
(d)
\(\mathrm{e}^{\ln3}-6\times\dfrac{\ln3}{2}\)	(M1)
\(3-3\ln3\)	A1	Accept \(3-\ln27\).

Working	Mark	Notes
(a)
\(h=\dfrac{32}{x^{2}}\)	A1
(b)
\(S=x^{2}+4xh\)	M1	Base plus four sides; no top.
\(x^{2}+4x\times\dfrac{32}{x^{2}}\)	A1
\(S=x^{2}+\dfrac{128}{x}\)	AG
(c)
\(2x-\dfrac{128}{x^{2}}\)	A1
(d)
\(2x-\dfrac{128}{x^{2}}=0\Rightarrow x^{3}=64\)	M1
\(x=4\)	A1
\(S=16+32=48\ \text{m}^{2}\)	A1
(e)
\(\dfrac{\mathrm{d}^{2}S}{\mathrm{d}x^{2}}=2+\dfrac{256}{x^{3}}\), which is \(6\gt 0\) at \(x=4\), so a minimum	R1

Working	Mark	Notes
(a)
\(3\ \text{m s}^{-2}\)	A1
(b)
\(t=7\)	A1
(c)
area under the graph from \(t=0\) to \(t=7\), e.g. \(\tfrac12(7+3)\times6\), or \(6+18+6\)	(M1)
\(30\) m	A1
(d)
\(30+\tfrac12(1)(3)=31.5\) m	A1	FT from (c).
(e)
\(30-1.5=28.5\) m	A1	FT from (c). Award A0 for 31.5.

Working	Mark	Notes
(a)
\(k\ln\|3x+1\|\)	M1	Award for a log of the correct linear function.
\(2\ln\|3x+1\|+C\)	A1	Condone missing modulus. Do not penalize the absence of \(+C\) more than once in the question.
(b)
\(k\,\mathrm{e}^{4-2x}\)	M1
\(-\tfrac{1}{2}\mathrm{e}^{4-2x}+C\)	A1
(c)
\(k(4x+1)^{\frac32}\)	M1
\(\tfrac{1}{6}(4x+1)^{\frac32}+C\)	A1

Working	Mark	Notes
(a)
\(6x^{\frac12}\) integrated by increasing the power by one	(M1)
\(4x^{\frac32}\)	A1
\(-4\ln x+C\)	A1	Accept \(-4\ln\|x\|\). Do not penalise the omission of \(+C\) in both parts more than once.
(b)
\(x^{\frac32}-3x^{-\frac12}\)	M1	Splitting the fraction into powers of x.
\(\dfrac{2}{5}x^{\frac52}\)	A1
\(-6x^{\frac12}+C\)	A1

Working	Mark	Notes
(a)
\((2x-5)^{-2}\) seen, or factor \(\tfrac12\) seen	(M1)
\(-\dfrac{1}{4(2x-5)^{2}}+C\)	A1
(b)
\(\ln(1-4x)\) seen	(M1)
\(-\dfrac{3}{4}\ln(1-4x)+C\)	A1	Accept \(\ln\|1-4x\|\). Award A0 for \(+\tfrac34\ln(1-4x)\).
(c)
\(\cos\left(\tfrac{\pi}{3}-2x\right)\) seen	(M1)
\(\dfrac{1}{2}\cos\left(\dfrac{\pi}{3}-2x\right)+C\)	A1	Award A0 for \(-\tfrac12\cos\left(\tfrac{\pi}{3}-2x\right)\).

Working	Mark	Notes
(a)
attempt to integrate \(f'(x)\)	M1
\(-2\cos(2x)+C\)	A1
\(5=-2\cos\pi+C\)	M1	Substituting the point into an expression containing C.
\(f(x)=3-2\cos(2x)\)	A1
(b)
\(5\)	A1	FT from their f.

Working	Mark	Notes
(a)
\(\mathrm{e}^{2x}+2+\mathrm{e}^{-2x}\)	A1
(b)
\(\dfrac12\mathrm{e}^{2x}-\dfrac12\mathrm{e}^{-2x}\)	A1	FT from (a).
\(+2x+C\)	A1
(c)
the student has multiplied by 5 (as when differentiating) instead of dividing by 5; differentiating \(5\sin(5x)\) gives \(25\cos(5x)\)	R1
\(\dfrac15\sin(5x)+C\)	A1

Working	Mark	Notes
(a)
\(\left[2x^{\frac32}+\dfrac{2}{x}\right]_{1}^{4}=\left(16+\tfrac12\right)-(2+2)\)	(M1)
\(\dfrac{25}{2}\)	A1
(b)
\(\left[\tfrac13\sin(3x)\right]_{0}^{\pi/6}=\tfrac13\sin\tfrac{\pi}{2}-0\)	(M1)
\(\dfrac13\)	A1
(c)
\(\left[\tfrac12\mathrm{e}^{2x}\right]_{0}^{\ln 2}=\tfrac12(4)-\tfrac12(1)\)	(M1)
\(\dfrac32\)	A1