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IB Maths Analysis & Approaches · Standard Level · Topic 2

Functions

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Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 14 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →

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+ Add to test✓ In your test1AASL2-q00012.1 Equations of straight lines Paper 1Easy8 marks

The line $L_{1}$ passes through the points $\mathrm{A}(-1,\,7)$ and $\mathrm{B}(3,\,-1)$.

(a)

Find the gradient of $L_{1}$.

(2)

(b)

Find the equation of $L_{1}$, giving your answer in the form $ax+by+d=0$, where $a,b,d\in\mathbb{Z}$.

(2)

(c)

The line $L_{2}$ is perpendicular to $L_{1}$ and passes through $\mathrm{C}(6,\,4)$. Find the equation of $L_{2}$, giving your answer in the form $y=mx+c$.

(2)

(d)

Find the coordinates of the point of intersection of $L_{1}$ and $L_{2}$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\dfrac{-1-7}{3-(-1)}$	(M1)
$-2$	A1
(b)
$y-7=-2(x+1)$ or $y=-2x+5$	(M1)	FT from their gradient.
$2x+y-5=0$	A1	Accept any integer multiple, e.g. $-2x-y+5=0$.
(c)
gradient $\tfrac{1}{2}$ and $y-4=\tfrac{1}{2}(x-6)$	(M1)	FT from (a): gradient must be $-\dfrac{1}{m_{1}}$.
$y=\tfrac{1}{2}x+1$	A1
(d)
$2x+\left(\tfrac{1}{2}x+1\right)-5=0$, i.e. $\tfrac{5}{2}x=4$	(M1)	Attempt to solve their equations simultaneously.
$\left(\dfrac{8}{5},\,\dfrac{9}{5}\right)$	A1	Accept $(1.6,\,1.8)$.

Answers: (a) $-2$ · (b) $2x+y-5=0$ · (c) $y=\tfrac{1}{2}x+1$ · (d) $\left(\tfrac{8}{5},\tfrac{9}{5}\right)$

+ Add to test✓ In your test2AASL2-q00022.1 Equations of straight lines Paper 1Easy6 marks

The line $L_{1}$ has equation $3x-4y+12=0$.

(a)

Write the equation of $L_{1}$ in the form $y=mx+c$.

(2)

(b)

Find the coordinates of the points where $L_{1}$ meets the $x$-axis and the $y$-axis.

(2)

(c)

The line $L_{2}$ is parallel to $L_{1}$ and passes through the point $(2,\,-5)$. Write down the equation of $L_{2}$ in the form $y-y_{1}=m(x-x_{1})$.

(1)

(d)

Write the equation of $L_{2}$ in the form $ax+by+d=0$, where $a,b,d\in\mathbb{Z}$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$4y=3x+12$	(M1)	Rearranging to make y the subject.
$y=\tfrac{3}{4}x+3$	A1
(b)
$x$-axis: $(-4,\,0)$	A1
$y$-axis: $(0,\,3)$	A1	Award A1A0 if both are given as single numbers −4 and 3 without coordinates.
(c)
$y+5=\tfrac{3}{4}(x-2)$	A1	FT from their gradient in (a). Accept $y-(-5)=\tfrac34(x-2)$.
(d)
$3x-4y-26=0$	A1	Accept any integer multiple, e.g. $-3x+4y+26=0$.

Answers: (a) $y=\tfrac34x+3$ · (b) $(-4,0)$, $(0,3)$ · (c) $y+5=\tfrac34(x-2)$ · (d) $3x-4y-26=0$

+ Add to test✓ In your test3AASL2-q00042.1 Equations of straight lines Paper 1Easy7 marks

A candle burns at a constant rate. Its height is 21 cm two hours after it is lit and 15 cm five hours after it is lit. The height of the candle, $h$ cm, $t$ hours after it is lit is modelled by $h=mt+c$.

(a)

Find the value of $m$ and the value of $c$.

(2)

(b)

Interpret, in context, the value of $m$ and the value of $c$.

(2)

(c)

A second candle is lit at the same time. Its height is modelled by $h=18-1.2t$. Find the time when the two candles have the same height, and this height.

(2)

(d)

Write the model for the first candle in the form $at+bh+d=0$, where $a,b,d\in\mathbb{Z}$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$m=\dfrac{15-21}{5-2}$	(M1)
$m=-2,\ c=25$	A1	$21=-2(2)+c$.
(b)
$m$: the candle’s height decreases by 2 cm every hour	A1	Must refer to rate and units.
$c$: the height of the candle when it is lit is 25 cm	A1
(c)
$25-2t=18-1.2t$	(M1)	FT from (a).
$t=8.75$ hours, height $7.5$ cm	A1	Both values required.
(d)
$2t+h-25=0$	A1	Accept any integer multiple.

Answers: (a) $m=-2,\ c=25$ · (c) 8.75 hours, 7.5 cm · (d) $2t+h-25=0$

+ Add to test✓ In your test4AASL2-q00082.2 Functions, domain, range and inverse Paper 1Easy7 marks

Let $f(x)=\sqrt{6-2x}+1$, for the largest possible domain.

(a)

Find the domain of $f$.

(2)

(b)

Write down the range of $f$.

(1)

(c)

Solve $f(x)=4$.

(2)

(d)

Hence write down the value of $f^{-1}(4)$.

(1)

(e)

Explain why $f^{-1}(0)$ does not exist.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$6-2x\ge0$	(M1)
$x\le3$	A1
(b)
$f(x)\ge1$	A1	Accept $y\ge1$ or $[1,\infty)$.
(c)
$\sqrt{6-2x}=3$, so $6-2x=9$	(M1)
$x=-\dfrac{3}{2}$	A1
(d)
$-\dfrac{3}{2}$	A1	FT from (c).
(e)
$0$ is not in the range of $f$ (as $f(x)\ge1$), so it is not in the domain of $f^{-1}$	R1

Answers: (a) $x\le3$ · (b) $f(x)\ge1$ · (c) $x=-\tfrac{3}{2}$ · (d) $-\tfrac{3}{2}$

+ Add to test✓ In your test5AASL2-q00102.2 Functions, domain, range and inverse Paper 1Easy5 marks

Let $f(x)=7-2x$, for $-2\le x\lt 4$.

(a)

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

(1)

(b)

Find the range of $f$.

(2)

(c)

Solve $f(x)=0$.

(1)

(d)

Find the value of $f^{-1}(5)$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$11$	A1
(b)
end values $-1$ and $11$	A1
$-1\lt f(x)\le11$	A1	Award A1A0 for $-1\le f(x)\le11$ (the value $-1$ is not attained since $x\lt4$).
(c)
$x=3.5$	A1
(d)
$7-2x=5\Rightarrow f^{-1}(5)=1$	A1

Answers: (a) 11 · (b) $-1\lt f(x)\le11$ · (c) 3.5 · (d) 1

+ Add to test✓ In your test6AASL2-q00122.2 Functions, domain, range and inverse Paper 1Easy7 marks

A temperature of $C$ degrees Celsius is equal to $F(C)$ degrees Fahrenheit, where $F(C)=1.8C+32$.

(a)

Find $F(37)$.

(1)

(b)

Find the value of $F^{-1}(212)$.

(2)

(c)

Find an expression for $F^{-1}(x)$.

(2)

(d)

Find the temperature at which the reading in degrees Fahrenheit is equal to the reading in degrees Celsius.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$98.6$	A1
(b)
$1.8C+32=212$	(M1)
$100$	A1
(c)
$x=1.8y+32$ or $y=1.8x+32$ rearranged for $x$	(M1)	Interchanging x and y at any stage.
$F^{-1}(x)=\dfrac{x-32}{1.8}$	A1	Accept $\dfrac{5(x-32)}{9}$.
(d)
$1.8C+32=C$	(M1)
$-40$ (degrees)	A1

Answers: (a) 98.6 · (b) 100 · (c) $\tfrac{5(x-32)}{9}$ · (d) −40

+ Add to test✓ In your test7AASL2-q00132.2 Functions, domain, range and inverse Paper 1Easy7 marks

For each relation below, state whether $y$ is a function of $x$, for the given values of $x$. Give a reason for each answer.

(i) $y=x^{2}-1$, for $x\in\mathbb{R}$
(ii) $y^{2}=x$, for $x\ge0$
(iii) the set of ordered pairs $\{(1,\,2),\ (2,\,3),\ (3,\,2)\}$
(iv) the set of ordered pairs $\{(2,\,1),\ (2,\,3),\ (4,\,5)\}$

(a)(i)

Relation (i).

(1)

(a)(ii)

Relation (ii).

(1)

(a)(iii)

Relation (iii).

(1)

(a)(iv)

Relation (iv).

(1)

(b)

For the relation in (iii), write down the domain and the range.

(2)

(c)

Explain why the relation in (iii) does not have an inverse function.

(1)

Show mark scheme

Working	Mark	Notes
(a)(i)
function: each $x$ gives exactly one value of $y$	A1
(a)(ii)
not a function: e.g. $x=4$ gives $y=2$ and $y=-2$	A1	Must include a reason.
(a)(iii)
function: each first element appears exactly once	A1
(a)(iv)
not a function: $x=2$ is paired with both 1 and 3	A1	Must include a reason.
(b)
domain $\{1,\,2,\,3\}$	A1
range $\{2,\,3\}$	A1	Do not accept {2, 3, 2}.
(c)
not one-to-one: the output 2 comes from both $x=1$ and $x=3$	R1

Answers: (a) (i) function (ii) not (iii) function (iv) not · (b) domain $\{1,2,3\}$, range $\{2,3\}$

+ Add to test✓ In your test8AASL2-q00172.3 Graphs of functions Paper 1Easy8 marks

Let $f(x)=\tfrac{1}{2}\left(x^{3}-4x\right)$, for $-2\le x\le2$. The following table shows some values of $f(x)$.

$x$	−2	−1.5	−1	−0.5	0	0.5	1	1.5	2
$f(x)$	0	1.3125	$p$	0.9375	0	−0.9375	−1.5	$q$	0

The following grid is provided.

(a)

Find the value of $p$ and the value of $q$.

(2)

(b)

On the grid, draw the graph of $y=f(x)$.

(3)

(c)

Describe the symmetry of the graph.

(1)

(d)

Use your graph to estimate the solutions of the equation $x^{3}-4x+2=0$, for $-2\le x\le2$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$p=\tfrac12(-1+4)=1.5$	A1
$q=\tfrac12(3.375-6)=-1.3125$	A1
(b)
all nine points plotted correctly (within half a small square)	A1	FT their p and q.
smooth curve (not line segments) through the points	A1
curve drawn only for $-2\le x\le2$, with turning points near $(-1.15,\,1.54)$ and $(1.15,\,-1.54)$	A1	“Draw” requires an accurate curve on the grid.
(c)
rotational symmetry of order 2 about the origin	A1	Accept “symmetric under a half-turn about O” or “f(−x) = −f(x)”.
(d)
rewriting as $f(x)=-1$ and drawing the line $y=-1$	(M1)
$x\approx0.54$ and $x\approx1.68$	A1	Accept answers in the ranges $0.50$ to $0.60$ and $1.60$ to $1.75$. The third root, $-2.21$, is outside the domain.

Answers: (a) $p=1.5,\ q=-1.3125$ · (c) rotational symmetry of order 2 about O · (d) $x\approx0.54,\ 1.68$

+ Add to test✓ In your test9AASL2-q00292.5 Composite and inverse functions Paper 1Easy8 marks

Let $f(x)=3x-1$ and $g(x)=x^{2}+2$, for $x\in\mathbb{R}$.

(a)

Find $(f\circ g)(2)$.

(2)

(b)

Find $(g\circ f)(x)$, giving your answer in the form $ax^{2}+bx+c$.

(2)

(c)

Solve $(f\circ g)(x)=17$.

(2)

(d)

Find $f^{-1}(x)$ and verify that $(f\circ f^{-1})(x)=x$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$g(2)=6$, then $f(6)$	(M1)
$17$	A1
(b)
$(3x-1)^{2}+2$	(M1)
$9x^{2}-6x+3$	A1
(c)
$3(x^{2}+2)-1=17$, so $x^{2}=4$	(M1)
$x=\pm2$	A1	Both values required.
(d)
$f^{-1}(x)=\dfrac{x+1}{3}$	A1
$3\left(\dfrac{x+1}{3}\right)-1=x$	A1	Must show the substitution.

Answers: (a) 17 · (b) $9x^{2}-6x+3$ · (c) $x=\pm2$ · (d) $f^{-1}(x)=\tfrac{x+1}{3}$

+ Add to test✓ In your test10AASL2-q00312.5 Composite and inverse functions Paper 1Easy6 marks

The functions $f$ and $g$ are defined on the set $\{1,\,2,\,3,\,4,\,5\}$ by the following table.

$x$	1	2	3	4	5
$f(x)$	3	5	1	2	4
$g(x)$	4	3	5	1	2

(a)

Find $(f\circ g)(2)$.

(2)

(b)

Find $(g\circ f)(4)$.

(1)

(c)

Write down $f^{-1}(5)$.

(1)

(d)

Solve $(g\circ f)(x)=5$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$g(2)=3$	(M1)
$f(3)=1$	A1
(b)
$g(f(4))=g(2)=3$	A1
(c)
$2$	A1
(d)
$g(f(x))=5\Rightarrow f(x)=3$	(M1)
$x=1$	A1

Answers: (a) 1 · (b) 3 · (c) 2 · (d) $x=1$

+ Add to test✓ In your test11AASL2-q00382.6 Quadratic functions Paper 1Easy8 marks

The following diagram shows part of the graph of a quadratic function $f$. The vertex of the graph is $(3,\,-2)$ and the graph passes through the point $(1,\,6)$.

(a)

The function can be written in the form $f(x)=a(x-h)^{2}+k$. Find the value of $a$, of $h$ and of $k$.

(3)

(b)

Express $f(x)$ in the form $f(x)=ax^{2}+bx+c$ and hence write down the $y$-intercept of the graph.

(2)

(c)

Express $f(x)$ in the form $f(x)=a(x-p)(x-q)$ and hence write down the $x$-intercepts of the graph.

(2)

(d)

Write down the equation of the axis of symmetry.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$h=3,\ k=-2$	A1
$a(1-3)^{2}-2=6$	(M1)
$a=2$	A1
(b)
$f(x)=2x^{2}-12x+16$	A1	FT from (a).
$(0,\,16)$	A1	Accept 16.
(c)
$f(x)=2(x-2)(x-4)$	A1
$(2,\,0)$ and $(4,\,0)$	A1
(d)
$x=3$	A1

Answers: (a) $a=2,\ h=3,\ k=-2$ · (b) $2x^{2}-12x+16$; $(0,16)$ · (c) $2(x-2)(x-4)$; $(2,0),(4,0)$ · (d) $x=3$

+ Add to test✓ In your test12AASL2-q00392.6 Quadratic functions Paper 1Easy7 marks

Let $f(x)=-2(x+1)(x-5)$, for $x\in\mathbb{R}$.

(a)

Write down the $x$-intercepts of the graph of $f$.

(1)

(b)

Write down the equation of the axis of symmetry.

(1)

(c)

Find the coordinates of the vertex.

(2)

(d)

Write down the $y$-intercept.

(1)

(e)

Write $f(x)$ in the form $a(x-h)^{2}+k$.

(1)

(f)

Write $f(x)$ in the form $ax^{2}+bx+c$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$(-1,\,0)$ and $(5,\,0)$	A1	Accept −1 and 5.
(b)
$x=2$	A1
(c)
$f(2)=-2(3)(-3)$	(M1)
$(2,\,18)$	A1
(d)
$10$	A1	Accept $(0,\,10)$.
(e)
$-2(x-2)^{2}+18$	A1	FT from (c).
(f)
$-2x^{2}+8x+10$	A1

Answers: (a) $(-1,0)$, $(5,0)$ · (b) $x=2$ · (c) $(2,18)$ · (d) 10 · (e) $-2(x-2)^{2}+18$ · (f) $-2x^{2}+8x+10$

+ Add to test✓ In your test13AASL2-q00452.7 Quadratic equations and the discriminant Paper 1Easy7 marks

(a)

Solve $2x^{2}-5x-3=0$ by factorizing.

(2)

(b)

By completing the square, solve $x^{2}+6x+4=0$. Give your answers in the form $p\pm\sqrt{q}$, where $p,q\in\mathbb{Z}$.

(3)

(c)

Hence, or otherwise, solve the inequality $2x^{2}-5x-3\le0$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$(2x+1)(x-3)=0$	(M1)
$x=-\dfrac{1}{2}$ or $x=3$	A1
(b)
$(x+3)^{2}-9+4=0$	M1
$(x+3)^{2}=5$	A1
$x=-3\pm\sqrt{5}$	A1
(c)
recognizing the parabola opens upwards, so the expression is non-positive between the roots	(M1)	A sketch or sign diagram is acceptable.
$-\dfrac{1}{2}\le x\le3$	A1	Award A0 for strict inequalities.

Answers: (a) $x=-\tfrac{1}{2},\ 3$ · (b) $x=-3\pm\sqrt{5}$ · (c) $-\tfrac{1}{2}\le x\le3$

+ Add to test✓ In your test14AASL2-q00482.7 Quadratic equations and the discriminant Paper 1Easy6 marks

(a)

Solve the inequality $x(x-2)\gt 15$.

(3)

(b)

Solve the inequality $2x^{2}+3x-2\lt 0$.

(2)

(c)

Hence write down all the integers that satisfy the inequality in part (b).

(1)

Show mark scheme

Working	Mark	Notes
(a)
$x^{2}-2x-15\gt 0$	M1	Rearranging to compare with zero. Award M0 for $x\gt 15$ or $x-2\gt 15$.
$(x-5)(x+3)\gt 0$, critical values $-3$ and $5$	A1
$x\lt -3$ or $x\gt 5$	A1	Award A0 for $-3\lt x\lt 5$ or for non-strict inequalities.
(b)
$(2x-1)(x+2)\lt 0$, critical values $-2$ and $\tfrac{1}{2}$	(M1)	A sketch or sign diagram may be seen.
$-2\lt x\lt \dfrac{1}{2}$	A1
(c)
$-1,\ 0$	A1	FT from their (b). Both values and no others.

Answers: (a) $x\lt -3$ or $x\gt 5$ · (b) $-2\lt x\lt \tfrac{1}{2}$ · (c) $-1,\ 0$

+ Add to test✓ In your test15AASL2-q00492.7 Quadratic equations and the discriminant Paper 1Easy7 marks

(a)

Solve $3x^{2}-4x-2=0$, giving your answers in the form $\dfrac{a\pm\sqrt{b}}{c}$, where $a,b,c\in\mathbb{Z}$.

(3)

(b)

Hence solve $3(x-1)^{2}-4(x-1)-2=0$.

(2)

(c)

Use the discriminant to determine the number of real roots of $3x^{2}-4x+2=0$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$x=\dfrac{4\pm\sqrt{(-4)^{2}-4(3)(-2)}}{2(3)}$	(M1)	Correct substitution into the quadratic formula.
$x=\dfrac{4\pm\sqrt{40}}{6}$	A1
$x=\dfrac{2\pm\sqrt{10}}{3}$	A1	Accept $\dfrac{4\pm2\sqrt{10}}{6}$.
(b)
recognizing $x-1=\dfrac{2\pm\sqrt{10}}{3}$	M1	Must use (a). Award M0 for expanding and starting again (not ‘hence’).
$x=\dfrac{5\pm\sqrt{10}}{3}$	A1
(c)
$\Delta=(-4)^{2}-4(3)(2)=-8$	M1
$\Delta\lt 0$, so there are no real roots	A1

Answers: (a) $\dfrac{2\pm\sqrt{10}}{3}$ · (b) $\dfrac{5\pm\sqrt{10}}{3}$ · (c) no real roots

+ Add to test✓ In your test16AASL2-q00502.7 Quadratic equations and the discriminant Paper 1Easy8 marks

(a)

Solve $x^{4}-5x^{2}+4=0$.

(3)

(b)

Solve $x-5\sqrt{x}+6=0$, for $x\ge0$.

(3)

(c)

Solve $2x+1=\dfrac{15}{x}$, for $x\neq0$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$u^{2}-5u+4=0$, where $u=x^{2}$	M1
$x^{2}=1$ or $x^{2}=4$	A1
$x=\pm1,\ \pm2$	A1	All four values required.
(b)
$u^{2}-5u+6=0$, where $u=\sqrt{x}$	M1
$\sqrt{x}=2$ or $\sqrt{x}=3$	A1
$x=4$ or $x=9$	A1
(c)
$2x^{2}+x-15=0$	M1
$x=\dfrac{5}{2}$ or $x=-3$	A1	From $(2x-5)(x+3)=0$.

Answers: (a) $x=\pm1,\ \pm2$ · (b) $x=4,\ 9$ · (c) $x=\tfrac{5}{2},\ -3$

+ Add to test✓ In your test17AASL2-q00562.8 Reciprocal and rational functions Paper 1Easy7 marks

Let $f(x)=\dfrac{1}{x}$, for $x\neq0$, and $g(x)=\dfrac{1}{x-2}+3$, for $x\neq2$.

(a)

Write down $f^{-1}(x)$.

(1)

(b)

Describe fully the transformation that maps the graph of $y=f(x)$ onto the graph of $y=g(x)$.

(2)

(c)

Write down the equations of the asymptotes of the graph of $g$.

(2)

(d)

Find the coordinates of the points where the graph of $g$ meets the axes.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$f^{-1}(x)=\dfrac{1}{x}$ (the function is self-inverse)	A1
(b)
translation	A1	Accept horizontal and vertical translations described separately.
by $\begin{pmatrix}2\\3\end{pmatrix}$: 2 units right and 3 units up	A1
(c)
$x=2$	A1
$y=3$	A1
(d)
$\left(0,\,\dfrac{5}{2}\right)$	A1
$\left(\dfrac{5}{3},\,0\right)$	A1	From $\frac{1}{x-2}=-3$.

Answers: (a) $\tfrac{1}{x}$ · (b) translation by $\binom{2}{3}$ · (c) $x=2,\ y=3$ · (d) $\left(0,\tfrac{5}{2}\right),\ \left(\tfrac{5}{3},0\right)$

+ Add to test✓ In your test18AASL2-q00632.8 Reciprocal and rational functions Paper 1Easy7 marks

Let $f(x)=\dfrac{12}{x}$, for $x\neq0$.

(a)

Write down $f^{-1}(x)$.

(1)

(b)

Find the exact coordinates of the points where the graph of $f$ meets the line $y=x$.

(2)

(c)

Find the coordinates of the points where the graph of $f$ meets the line $y=8-x$.

(3)

(d)

State the geometrical relationship between the two points found in part (c), and explain how it follows from part (a).

(1)

Show mark scheme

Working	Mark	Notes
(a)
$f^{-1}(x)=\dfrac{12}{x}$	A1
(b)
$x^{2}=12$	(M1)
$\left(2\sqrt{3},\,2\sqrt{3}\right)$ and $\left(-2\sqrt{3},\,-2\sqrt{3}\right)$	A1	Accept $\sqrt{12}$.
(c)
$\dfrac{12}{x}=8-x$, i.e. $x^{2}-8x+12=0$	M1
$x=2$ or $x=6$	A1
$(2,\,6)$ and $(6,\,2)$	A1
(d)
each is the reflection of the other in $y=x$; since $f$ is self-inverse its graph is symmetric about $y=x$	R1

Answers: (a) $\tfrac{12}{x}$ · (b) $\left(\pm2\sqrt3,\ \pm2\sqrt3\right)$ · (c) $(2,6),\ (6,2)$

+ Add to test✓ In your test19AASL2-q00672.9 Exponential and logarithmic functions Paper 1Easy8 marks

Let $f(x)=4-2\mathrm{e}^{x}$, for $x\in\mathbb{R}$, and $g(x)=2+\ln(x-1)$, for $x\gt 1$.

(a)

Write down the $y$-intercept of the graph of $f$.

(1)

(b)

Write down the equation of the horizontal asymptote of the graph of $f$.

(1)

(c)

Find the exact $x$-intercept of the graph of $f$.

(2)

(d)

Write down the range of $f$.

(1)

(e)

Write down the equation of the vertical asymptote of the graph of $g$.

(1)

(f)

Find the exact $x$-intercept of the graph of $g$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$(0,\,2)$	A1	Accept 2.
(b)
$y=4$	A1
(c)
$2\mathrm{e}^{x}=4$, so $\mathrm{e}^{x}=2$	(M1)
$x=\ln2$	A1	Accept $(\ln2,\,0)$.
(d)
$f(x)\lt 4$	A1	Award A0 for $f(x)\le4$.
(e)
$x=1$	A1
(f)
$\ln(x-1)=-2$	(M1)
$x=1+\mathrm{e}^{-2}$	A1

Answers: (a) $(0,2)$ · (b) $y=4$ · (c) $x=\ln2$ · (d) $f(x)\lt 4$ · (e) $x=1$ · (f) $x=1+\mathrm{e}^{-2}$

+ Add to test✓ In your test20AASL2-q00692.9 Exponential and logarithmic functions Paper 1Easy7 marks

Let $f(x)=2^{x}$ and $g(x)=\left(\dfrac{1}{2}\right)^{x}$, for $x\in\mathbb{R}$, and let $h(x)=\log_{2}x$, for $x\gt 0$.

(a)

Show that $g(x)=f(-x)$, and hence describe the transformation that maps the graph of $f$ onto the graph of $g$.

(2)

(b)

Describe the transformation that maps the graph of $f$ onto the graph of $h$.

(1)

(c)

The graph of $y=8g(x)$ meets the graph of $y=f(x)$ at the point P. Find the exact coordinates of P.

(3)

(d)

Write down the coordinates of the point where the graph of $h$ meets the graph of $y=\log_{2}\left(\dfrac{8}{x}\right)$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$\left(\tfrac{1}{2}\right)^{x}=2^{-x}=f(-x)$	A1
reflection in the $y$-axis	A1
(b)
reflection in the line $y=x$	A1	They are inverse functions.
(c)
$2^{x}=8\times2^{-x}=2^{3-x}$	M1
$x=\dfrac{3}{2}$	A1
$\mathrm{P}\left(\dfrac{3}{2},\,2\sqrt{2}\right)$	A1	Accept $2^{3/2}$.
(d)
$\left(2\sqrt{2},\,\dfrac{3}{2}\right)$	A1	The reflection of P in $y=x$; or solve $\log_{2}x=3-\log_{2}x$.

Answers: (a) reflection in the $y$-axis · (b) reflection in $y=x$ · (c) $\left(\tfrac32,\,2\sqrt2\right)$ · (d) $\left(2\sqrt2,\,\tfrac32\right)$

+ Add to test✓ In your test21AASL2-q00752.10 Solving equations Paper 1Easy6 marks

Consider the equation $\sqrt{x+7}=x-5$, for $x\ge-7$.

(a)

Show that any solution of the equation satisfies $x^{2}-11x+18=0$.

(2)

(b)

Hence solve the equation $\sqrt{x+7}=x-5$.

(2)

(c)

By considering the graphs of $y=\sqrt{x+7}$ and $y=x-5$, solve the inequality $\sqrt{x+7}\gt x-5$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
squaring both sides: $x+7=(x-5)^{2}$	M1
$x+7=x^{2}-10x+25$	A1
$x^{2}-11x+18=0$	AG
(b)
$(x-2)(x-9)=0$, so $x=2$ or $x=9$	A1
$x=2$ is rejected, since $\sqrt{9}=3\neq-3$; so $x=9$ only	R1	Award R0 if both values are given as the final answer.
(c)
the graph of $y=\sqrt{x+7}$ lies above the line for $x$ less than the intersection at $x=9$	(M1)	A sketch may be seen.
$-7\le x\lt 9$	A1	FT from (b). Award A0 for $x\lt 9$ (domain omitted).

Answers: (b) $x=9$ · (c) $-7\le x\lt 9$

+ Add to test✓ In your test22AASL2-q00842.11 Transformations of graphs Paper 1Easy8 marks

The point $\mathrm{P}(4,\,-6)$ lies on the graph of $y=f(x)$.

(a)(i)

Write down the coordinates of the image of P on the graph of

$y=f(x)+5$;

(1)

(a)(ii)

$y=f(x-3)$;

(1)

(a)(iii)

$y=-f(x)$;

(1)

(a)(iv)

$y=f\left(\tfrac{1}{2}x\right)$;

(1)

(a)(v)

$y=3f(-x)$.

(1)

(b)

Describe a sequence of transformations that maps the graph of $y=x^{2}$ onto the graph of $y=4-2x^{2}$.

(3)

Show mark scheme

Working	Mark	Notes
(a)(i)
$(4,\,-1)$	A1
(a)(ii)
$(7,\,-6)$	A1
(a)(iii)
$(4,\,6)$	A1
(a)(iv)
$(8,\,-6)$	A1	Horizontal stretch with scale factor 2.
(a)(v)
$(-4,\,-18)$	A1	Both the reflection in the $y$-axis and the vertical stretch must be applied.
(b)
vertical stretch, scale factor 2	A1
reflection in the $x$-axis	A1
followed by a translation of 4 units up, $\begin{pmatrix}0\\4\end{pmatrix}$	A1	The translation must come after the stretch and reflection. Stretch and reflection may be combined as a stretch with scale factor −2.

Answers: (a) $(4,-1),\ (7,-6),\ (4,6),\ (8,-6),\ (-4,-18)$

+ Add to test✓ In your test23AASL2-q00872.11 Transformations of graphs Paper 1Easy8 marks

The following diagram shows the graph of $y=f(x)$. The graph has a local minimum point at $(-1,\,0)$, a local maximum point at $(1,\,4)$, meets the $y$-axis at $(0,\,2)$ and has a horizontal asymptote $y=2$.

(a)(i)

For each of the following graphs, write down the coordinates of the image of the point $(1,\,4)$ and the equation of the horizontal asymptote.

$y=f(x)-2$

(2)

(a)(ii)

$y=-f(x)$

(2)

(a)(iii)

$y=f\left(\tfrac{1}{2}x\right)$

(2)

(a)(iv)

$y=f(-x)+1$

(2)

Show mark scheme

Working	Mark	Notes
(a)(i)
$(1,\,2)$	A1
$y=0$	A1
(a)(ii)
$(1,\,-4)$	A1	The image is now a local minimum.
$y=-2$	A1
(a)(iii)
$(2,\,4)$	A1	Horizontal stretch with scale factor 2.
$y=2$	A1
(a)(iv)
$(-1,\,5)$	A1
$y=3$	A1

Answers: (i) $(1,2)$, $y=0$ · (ii) $(1,-4)$, $y=-2$ · (iii) $(2,4)$, $y=2$ · (iv) $(-1,5)$, $y=3$

+ Add to test✓ In your test24AASL2-q00882.11 Transformations of graphs Paper 1Easy7 marks

The graph of $y=\sqrt{x}$, for $x\ge0$, is transformed by the following sequence of transformations to give the graph of $y=g(x)$:

a horizontal stretch with scale factor $\tfrac{1}{4}$;
then a reflection in the $x$-axis;
then a translation of 3 units vertically upwards.

(a)

Show that $g(x)=3-2\sqrt{x}$.

(2)

(b)

Write down the domain and the range of $g$.

(2)

(c)

Find the $x$-intercept of the graph of $g$.

(2)

(d)

The horizontal stretch in the sequence above can be replaced by a vertical stretch, giving the same graph. Write down the scale factor of this vertical stretch.

(1)

Show mark scheme

Working	Mark	Notes
(a)
stretch: $y=\sqrt{4x}$	M1
reflection then translation: $y=3-\sqrt{4x}$	A1
$\sqrt{4x}=2\sqrt{x}$, so $g(x)=3-2\sqrt{x}$	AG
(b)
domain $x\ge0$	A1
range $g(x)\le3$	A1
(c)
$2\sqrt{x}=3$	(M1)
$\left(\dfrac{9}{4},\,0\right)$	A1	Accept $x=\tfrac{9}{4}$.
(d)
2	A1	Since $\sqrt{4x}=2\sqrt{x}$.

Answers: (b) $x\ge0$; $g(x)\le3$ · (c) $\left(\tfrac94,0\right)$ · (d) 2

+ Add to test✓ In your test25AASL2-q00032.1 Equations of straight lines Paper 1Medium8 marks

The points $\mathrm{P}(-3,\,2)$ and $\mathrm{Q}(5,\,-4)$ are shown on a coordinate plane. The line $L$ is the perpendicular bisector of $[\mathrm{PQ}]$.

(a)

Find the coordinates of the midpoint of $[\mathrm{PQ}]$.

(1)

(b)

Find the gradient of $[\mathrm{PQ}]$.

(1)

(c)

Find the equation of $L$, giving your answer in the form $ax+by+d=0$, where $a,b,d\in\mathbb{Z}$.

(3)

(d)

Find the $y$-intercept of $L$.

(1)

(e)

Show that the point $\mathrm{R}(4,\,3)$ lies on $L$, and hence explain why $\mathrm{RP}=\mathrm{RQ}$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$(1,\,-1)$	A1
(b)
$\dfrac{-4-2}{5-(-3)}=-\dfrac{3}{4}$	A1
(c)
perpendicular gradient $\dfrac{4}{3}$	(M1)	FT: $-\dfrac{1}{m}$ for their $m$ in (b).
$y+1=\tfrac{4}{3}(x-1)$	(M1)	Substituting their midpoint into a line equation with their perpendicular gradient.
$4x-3y-7=0$	A1	Accept any integer multiple.
(d)
$-\dfrac{7}{3}$	A1	Accept $\left(0,\,-\tfrac73\right)$.
(e)
$4(4)-3(3)-7=16-9-7=0$	A1
every point on the perpendicular bisector of $[\mathrm{PQ}]$ is the same distance from P and from Q	R1	Accept a calculation showing $\mathrm{RP}=\mathrm{RQ}=\sqrt{50}$.

Answers: (a) $(1,-1)$ · (b) $-\tfrac34$ · (c) $4x-3y-7=0$ · (d) $-\tfrac73$

+ Add to test✓ In your test26AASL2-q00092.2 Functions, domain, range and inverse Paper 1Medium8 marks

The following diagram shows the graph of $y=f(x)$, for $-2\le x\le3$. The graph passes through the points $(-2,\,-2.75)$, $(0,\,-2)$, $(2,\,1)$ and $(3,\,5)$. The line $y=x$ is also shown.

(a)

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

(1)

(b)

Write down the value of $f^{-1}(-2)$.

(1)

(c)

Explain why $f$ has an inverse.

(1)

(d)(i)

Write down the domain of $f^{-1}$.

(1)

(d)(ii)

Write down the range of $f^{-1}$.

(1)

(e)

On the same grid, sketch the graph of $y=f^{-1}(x)$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$1$	A1
(b)
$0$	A1
(c)
$f$ is one-to-one (each $y$-value comes from exactly one $x$-value; the graph is increasing)	R1	Accept a correct horizontal-line-test argument.
(d)(i)
$-2.75\le x\le5$	A1
(d)(ii)
$-2\le y\le3$	A1	Accept $-2\le f^{-1}(x)\le3$.
(e)
attempt to reflect the graph of $f$ in the line $y=x$	M1
curve passes through $(-2,\,0)$ and $(1,\,2)$, increasing and concave down	A1
endpoints at $(-2.75,\,-2)$ and $(5,\,3)$	A1

Answers: (a) 1 · (b) 0 · (d)(i) $-2.75\le x\le5$ · (d)(ii) $-2\le y\le3$

+ Add to test✓ In your test27AASL2-q00112.2 Functions, domain, range and inverse Paper 1Medium9 marks

The following diagram shows the graph of $y=f(x)$, for $-4\le x\le5$. The graph consists of three line segments joining the points $(-4,\,1)$, $(-2,\,-3)$, $(1,\,3)$ and $(5,\,1)$.

(a)

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

(1)

(b)

Write down the range of $f$.

(2)

(c)

Solve $f(x)=1$.

(2)

(d)

Explain why $f$ does not have an inverse.

(1)

(e)(i)

The function $g$ is defined by $g(x)=f(x)$, for $-2\le x\le1$.

Write down the domain of $g^{-1}$.

(1)

(e)(ii)

Write down the range of $g^{-1}$.

(1)

(f)

Find the value of $g^{-1}(2)$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$-3$	A1
(b)
end values $-3$ and $3$	A1
$-3\le f(x)\le3$	A1
(c)
drawing $y=1$, or using the middle segment $y=2x+1$	(M1)
$x=-4,\ x=0,\ x=5$	A1	All three values required.
(d)
$f$ is not one-to-one, e.g. $f(-4)=f(0)=f(5)=1$	R1	Accept a horizontal-line-test argument.
(e)(i)
$-3\le x\le3$	A1	The range of g.
(e)(ii)
$-2\le g^{-1}(x)\le1$	A1	Accept $-2\le y\le1$.
(f)
$2x+1=2\Rightarrow g^{-1}(2)=0.5$	A1

Answers: (a) −3 · (b) $-3\le f(x)\le3$ · (c) $x=-4,0,5$ · (e)(i) $-3\le x\le3$ (ii) $-2\le g^{-1}(x)\le1$ · (f) 0.5

+ Add to test✓ In your test28AASL2-q00142.2 Functions, domain, range and inverse Paper 1Medium8 marks

Let $g(x)=x^{2}-4x+1$, for $0\le x\le5$.

(a)

Write $g(x)$ in the form $(x-h)^{2}+k$.

(2)

(b)

Find the range of $g$.

(2)

(c)

Explain why $g$ does not have an inverse.

(1)

(d)

The function $h$ is defined by $h(x)=g(x)$, for $a\le x\le5$. Write down the least value of $a$ for which $h$ has an inverse.

(1)

(e)

Solve $g(x)=6$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$x^{2}-4x+4-4+1$	(M1)	Attempt to complete the square.
$(x-2)^{2}-3$	A1
(b)
least value $g(2)=-3$ (the vertex, not an end value) and greatest value $g(5)=6$	A1	$g(0)=1$ is not the least value.
$-3\le g(x)\le6$	A1	Award A1A0 for $1\le g(x)\le6$ (end values only).
(c)
$g$ is not one-to-one, e.g. $g(0)=g(4)=1$	R1	Accept a horizontal-line-test argument.
(d)
$a=2$	A1	FT from the vertex in (a).
(e)
$(x-2)^{2}=9\Rightarrow x=-1$ or $x=5$	(M1)	Or $x^{2}-4x-5=0\Rightarrow(x+1)(x-5)=0$.
$x=5$ only	A1	$x=-1$ must be rejected as it is not in the domain. Award M1A0 if $x=-1$ is also given.

Answers: (a) $(x-2)^{2}-3$ · (b) $-3\le g(x)\le6$ · (d) $a=2$ · (e) $x=5$

+ Add to test✓ In your test29AASL2-q00182.3 Graphs of functions Paper 1Medium6 marks

A function $f$ has domain $-3\le x\le5$. The following information is known about its graph.

$f(-3)=4$ and $f(5)=6$.
The only zeros of $f$ are $x=-1$ and $x=3$.
The $y$-intercept is $-1.5$.
$f$ is decreasing for $-3\lt x\lt1$ and increasing for $1\lt x\lt5$, with a minimum point at $(1,\,-2)$.

(a)

Sketch a possible graph of $y=f(x)$, labelling all the given points.

(3)

(b)

Write down the range of $f$.

(1)

(c)

Find the set of values of $k$ for which the equation $f(x)=k$ has exactly two solutions.

(2)

Show mark scheme

Working	Mark	Notes
(a)
smooth curve decreasing from $(-3,\,4)$ to the minimum $(1,\,-2)$, then increasing to $(5,\,6)$	A1
crossing the $x$-axis at $-1$ and $3$ only	A1
passing through $(0,\,-1.5)$, with all five points labelled and endpoints shown	A1
(b)
$-2\le f(x)\le6$	A1
(c)
considering horizontal lines $y=k$ meeting both the decreasing and the increasing parts	(M1)
$-2\lt k\le4$	A1	Award M1A0 for $-2\lt k\lt4$ or $-2\le k\le4$.

Answers: (b) $-2\le f(x)\le6$ · (c) $-2\lt k\le4$

+ Add to test✓ In your test30AASL2-q00322.5 Composite and inverse functions Paper 1Medium6 marks

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

(a)

Find $(f\circ g)(x)$, giving your answer in the form $ax^{2}+bx+c$.

(2)

(b)

Find $(g\circ f)(x)$, giving your answer in the form $ax^{2}+bx+c$.

(2)

(c)

Solve $(f\circ g)(x)=(g\circ f)(x)$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$2(x^{2}-x+2)-2$	(M1)
$2x^{2}-2x+2$	A1
(b)
$(2x-2)^{2}-(2x-2)+2$	(M1)
$4x^{2}-10x+8$	A1
(c)
$2x^{2}-8x+6=0\Rightarrow x^{2}-4x+3=0$	(M1)	FT from (a) and (b).
$x=1,\ x=3$	A1

Answers: (a) $2x^{2}-2x+2$ · (b) $4x^{2}-10x+8$ · (c) $x=1,\ 3$

+ Add to test✓ In your test31AASL2-q00332.5 Composite and inverse functions Paper 1Medium7 marks

Let $f(x)=2x^{3}+1$, for $x\in\mathbb{R}$.

(a)

Explain why $f$ has an inverse.

(1)

(b)

Find $f^{-1}(x)$.

(3)

(c)

Find $f^{-1}(55)$.

(1)

(d)

Show that $(f\circ f^{-1})(x)=x$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$f$ is one-to-one (it is increasing for all $x$)	R1
(b)
$x=2y^{3}+1$	(M1)	Interchanging x and y at any stage.
$y^{3}=\dfrac{x-1}{2}$	A1
$f^{-1}(x)=\sqrt[3]{\dfrac{x-1}{2}}$	A1
(c)
$\sqrt[3]{27}=3$	A1
(d)
$f\left(\sqrt[3]{\dfrac{x-1}{2}}\right)=2\left(\sqrt[3]{\dfrac{x-1}{2}}\right)^{3}+1$	M1
$=2\times\dfrac{x-1}{2}+1$	A1
$=x$	AG

Answers: (b) $f^{-1}(x)=\sqrt[3]{\tfrac{x-1}{2}}$ · (c) 3

+ Add to test✓ In your test32AASL2-q00342.5 Composite and inverse functions Paper 1Medium6 marks

Let $g(x)=2x+5$, for $x\in\mathbb{R}$. The function $h$ is defined by $h(x)=(2x+5)^{3}$, for $x\in\mathbb{R}$.

(a)

Given that $h(x)=(f\circ g)(x)$, write down $f(x)$.

(1)

(b)

Let $k(x)=(g\circ f)(x)$. Write down $k(x)$.

(1)

(c)

Solve $k(x)=-11$.

(2)

(d)

Find $p(x)$ such that $(g\circ p)(x)=6x-1$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$f(x)=x^{3}$	A1
(b)
$k(x)=2x^{3}+5$	A1	FT from (a).
(c)
$2x^{3}=-16\Rightarrow x^{3}=-8$	(M1)
$x=-2$	A1
(d)
$2p(x)+5=6x-1$	(M1)
$p(x)=3x-3$	A1

Answers: (a) $x^{3}$ · (b) $2x^{3}+5$ · (c) $-2$ · (d) $3x-3$

+ Add to test✓ In your test33AASL2-q00402.6 Quadratic functions Paper 1Medium8 marks

The following diagram shows part of the graph of $f(x)=x^{2}+bx+c$. The graph crosses the $x$-axis at $\mathrm{A}(1,\,0)$ and $\mathrm{B}(5,\,0)$ and has vertex V. The line $L$ passes through A and V.

Diagram not to scale

(a)

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

(2)

(b)

Find the coordinates of V.

(2)

(c)

Find the equation of $L$.

(2)

(d)

The line $M$ is parallel to $L$ and passes through B. Find the $y$-intercept of $M$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$f(x)=(x-1)(x-5)$	(M1)	Or substituting both points into f(x) = x² + bx + c.
$b=-6,\ c=5$	A1
(b)
axis of symmetry $x=3$	(M1)
$\mathrm{V}(3,\,-4)$	A1
(c)
gradient $\dfrac{-4-0}{3-1}=-2$	(M1)	FT from (b).
$y=-2x+2$	A1	Accept equivalent forms, e.g. $y=-2(x-1)$.
(d)
$y=-2(x-5)$	(M1)	Gradient of L with point B.
$10$	A1

Answers: (a) $b=-6,\ c=5$ · (b) $(3,-4)$ · (c) $y=-2x+2$ · (d) 10

+ Add to test✓ In your test34AASL2-q00412.6 Quadratic functions Paper 1Medium7 marks

The graph of a quadratic function $f$ has axis of symmetry $x=3$ and passes through the points $(6,\,15)$ and $(4,\,-9)$. Let $f(x)=a(x-3)^{2}+k$.

(a)

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

(3)

(b)

Write $f(x)$ in the form $a(x-p)(x-q)$.

(2)

(c)

Write $f(x)$ in the form $ax^{2}+bx+c$.

(1)

(d)

Write down the $y$-intercept of the graph of $f$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$9a+k=15$ and $a+k=-9$	(M1)	Substituting both points.
$a=3$	A1
$k=-12$	A1
(b)
$3(x-3)^{2}=12\Rightarrow x-3=\pm2$	(M1)	Or expanding and factorizing $3x^{2}-18x+15$.
$3(x-1)(x-5)$	A1
(c)
$3x^{2}-18x+15$	A1
(d)
$15$	A1	Accept $(0,\,15)$.

Answers: (a) $a=3,\ k=-12$ · (b) $3(x-1)(x-5)$ · (c) $3x^{2}-18x+15$ · (d) 15

+ Add to test✓ In your test35AASL2-q00462.7 Quadratic equations and the discriminant Paper 1Medium7 marks

Consider the equation $kx^{2}-4x+(k-3)=0$, where $k\in\mathbb{R}$ and $k\neq0$.

(a)

Show that the discriminant is $16+12k-4k^{2}$.

(2)

(b)

Find the values of $k$ for which the equation has two equal real roots.

(2)

(c)

Find the set of values of $k$ for which the equation has two distinct real roots.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$\Delta=(-4)^{2}-4(k)(k-3)$	M1
$16-4k^{2}+12k$	A1
$\Delta=16+12k-4k^{2}$	AG
(b)
$16+12k-4k^{2}=0$, i.e. $k^{2}-3k-4=0$	(M1)
$k=-1$ or $k=4$	A1
(c)
$16+12k-4k^{2}\gt 0$, i.e. $(k-4)(k+1)\lt 0$	M1
$-1\lt k\lt 4$	A1
$k\neq0$	A1	Final answer $-1\lt k\lt 0$ or $0\lt k\lt 4$. For $k=0$ the equation is linear.

Answers: (b) $k=-1$ or $k=4$ · (c) $-1\lt k\lt 4,\ k\neq0$

+ Add to test✓ In your test36AASL2-q00512.7 Quadratic equations and the discriminant Paper 1Medium6 marks

Let $f(x)=x^{2}+(k-2)x-k$, where $k\in\mathbb{R}$.

(a)

Show that the discriminant of $f(x)$ is $k^{2}+4$.

(2)

(b)

Hence explain why the equation $f(x)=0$ has two distinct real roots for every value of $k$.

(1)

(c)

Show that the distance between the two $x$-intercepts of the graph of $f$ is $\sqrt{k^{2}+4}$.

(1)

(d)

Find the values of $k$ for which the $x$-intercepts of the graph of $f$ are $2\sqrt{5}$ units apart.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\Delta=(k-2)^{2}-4(1)(-k)$	M1
$k^{2}-4k+4+4k$	A1
$\Delta=k^{2}+4$	AG
(b)
$k^{2}\ge0$, so $\Delta\ge4\gt 0$ for all $k$	R1	Must refer to $\Delta\gt 0$.
(c)
roots $\dfrac{-(k-2)\pm\sqrt{k^{2}+4}}{2}$, difference $\dfrac{2\sqrt{k^{2}+4}}{2}$	M1
distance $=\sqrt{k^{2}+4}$	AG
(d)
$\sqrt{k^{2}+4}=2\sqrt{5}$, so $k^{2}+4=20$	M1
$k=\pm4$	A1	Both values required.

Answers: (d) $k=4$ or $k=-4$

+ Add to test✓ In your test37AASL2-q00572.8 Reciprocal and rational functions Paper 1Medium7 marks

Let $f(x)=\dfrac{2x+3}{x-1}$, for $x\neq1$.

(a)(i)

Write down the equation of

the vertical asymptote of the graph of $f$;

(1)

(a)(ii)

the horizontal asymptote of the graph of $f$.

(1)

(b)

Find the coordinates of the points where the graph of $f$ meets the axes.

(2)

(c)

Sketch the graph of $f$ for $-6\le x\le8$, showing the asymptotes and the intercepts found in parts (a) and (b).

(3)

Show mark scheme

Working	Mark	Notes
(a)(i)
$x=1$	A1
(a)(ii)
$y=2$	A1
(b)
$\left(-\dfrac{3}{2},\,0\right)$	A1
$(0,\,-3)$	A1
(c)
two branches of correct shape (decreasing on each side), approaching the asymptotes	A1	Branch in the top right and bottom left of the asymptotes.
asymptotes drawn and labelled $x=1$, $y=2$	A1	FT from (a).
intercepts $-\tfrac{3}{2}$ and $-3$ shown on the left branch	A1	FT from (b).

Answers: (a) $x=1,\ y=2$ · (b) $\left(-\tfrac{3}{2},0\right),\ (0,-3)$

+ Add to test✓ In your test38AASL2-q00592.8 Reciprocal and rational functions Paper 1Medium8 marks

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

(a)(i)

Write down the equation of

the vertical asymptote of the graph of $f$;

(1)

(a)(ii)

the horizontal asymptote of the graph of $f$.

(1)

(b)

Solve $f(x)=\dfrac{1}{2}$.

(2)

(c)

Explain, using algebra, why the equation $f(x)=-2$ has no solution.

(2)

(d)

The function $f$ is decreasing for $x\gt -3$. Find the range of $f$ for $x\ge0$.

(2)

Show mark scheme

Working	Mark	Notes
(a)(i)
$x=-3$	A1
(a)(ii)
$y=-2$	A1	Award A0 for $y=2$ (sign of the leading coefficient ignored).
(b)
$2(5-4x)=2x+6$	M1
$x=\dfrac{2}{5}$	A1
(c)
$5-4x=-2(2x+6)=-4x-12$	M1
this gives $5=-12$, which is impossible, so there is no solution	R1	Accept a correct reference to $y=-2$ being the horizontal asymptote only if supported by algebra, e.g. $f(x)=-2+\dfrac{17}{2(x+3)}\neq-2$.
(d)
$f(0)=\dfrac{5}{6}$	A1
$-2\lt f(x)\le\dfrac{5}{6}$	A1	FT from (a)(ii). Award A0 for $-2\le f(x)$ or $f(x)\lt\tfrac{5}{6}$.

Answers: (a) $x=-3$; $y=-2$ · (b) $x=\tfrac{2}{5}$ · (d) $-2\lt f(x)\le\tfrac{5}{6}$

+ Add to test✓ In your test39AASL2-q00602.8 Reciprocal and rational functions Paper 1Medium9 marks

The following diagram shows the graph of $f(x)=\dfrac{x+p}{q-x}$, for $x\neq q$, where $p,q\in\mathbb{R}$. The graph has asymptotes $x=2$ and $y=-1$, and passes through the points $(-2,\,0)$ and $(0,\,1)$.

(a)

Find the value of $p$ and the value of $q$.

(2)

(b)

Solve $f(x)=3$.

(2)

(c)

Hence, using the graph, solve $f(x)\gt 3$.

(2)

(d)

Find $f^{-1}(x)$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$q=2$	A1	From the vertical asymptote.
$p=2$	A1	From $f(-2)=0$; check $f(0)=\tfrac{2}{2}=1$.
(b)
$x+2=3(2-x)$	M1
$x=1$	A1
(c)
recognizing the graph lies above $y=3$ between $x=1$ and the vertical asymptote	M1
$1\lt x\lt 2$	A1	FT from (b). Award A0 for $x\gt 1$.
(d)
$x=\dfrac{y+2}{2-y}$	M1	Interchanging variables, at any stage.
$2x-xy=y+2$, so $y(x+1)=2x-2$	A1
$f^{-1}(x)=\dfrac{2x-2}{x+1}$	A1	Accept $\dfrac{2(x-1)}{x+1},\ x\neq-1$.

Answers: (a) $p=2,\ q=2$ · (b) $x=1$ · (c) $1\lt x\lt 2$ · (d) $f^{-1}(x)=\dfrac{2x-2}{x+1}$

+ Add to test✓ In your test40AASL2-q00612.8 Reciprocal and rational functions Paper 1Medium7 marks

Let $f(x)=\dfrac{2x+1}{x-3}$, for $x\neq3$.

(a)

Write down the equations of the asymptotes of the graph of $f$.

(2)

(b)

Show that $f(x)=2+\dfrac{7}{x-3}$.

(2)

(c)

Hence describe a sequence of transformations that maps the graph of $y=\dfrac{1}{x}$ onto the graph of $y=f(x)$.

(2)

(d)

Write down the value of $c$ for which the equation $f(x)=c$ has no solution.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$x=3$	A1
$y=2$	A1
(b)
$2+\dfrac{7}{x-3}=\dfrac{2(x-3)+7}{x-3}$	M1	Or $2x+1=2(x-3)+7$.
$=\dfrac{2x-6+7}{x-3}$	A1
$=\dfrac{2x+1}{x-3}$	AG
(c)
vertical stretch with scale factor 7	A1
translation by $\begin{pmatrix}3\\2\end{pmatrix}$	A1	The vertical translation must follow the stretch; the horizontal translation may come before or after it.
(d)
$c=2$	A1	The horizontal asymptote; FT from (a).

Answers: (a) $x=3,\ y=2$ · (c) vertical stretch, scale factor 7, then translation by $\binom{3}{2}$ · (d) $c=2$

+ Add to test✓ In your test41AASL2-q00652.9 Exponential and logarithmic functions Paper 1Medium9 marks

The following diagram shows the graph of $f(x)=a\times2^{x}+b$, where $a,b\in\mathbb{R}$. The graph passes through the points $(0,\,-2)$ and $(2,\,4)$ and has a horizontal asymptote $y=-4$.

(a)

Write down the value of $b$.

(1)

(b)

Find the value of $a$.

(2)

(c)

Find the $x$-intercept of the graph of $f$.

(2)

(d)

Find $f^{-1}(x)$, giving your answer in the form $\log_{2}(x+p)+q$.

(3)

(e)

Write down the equation of the vertical asymptote of the graph of $f^{-1}$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$b=-4$	A1
(b)
$a\times2^{0}-4=-2$ or $4a-4=4$	(M1)
$a=2$	A1
(c)
$2\times2^{x}=4$, so $2^{x}=2$	(M1)
$(1,\,0)$	A1	Accept $x=1$.
(d)
$x=2\times2^{y}-4$	(M1)	Interchanging variables, at any stage.
$2^{y}=\dfrac{x+4}{2}$, so $y=\log_{2}\left(\dfrac{x+4}{2}\right)$	A1
$f^{-1}(x)=\log_{2}(x+4)-1$	A1
(e)
$x=-4$	A1

Answers: (a) $b=-4$ · (b) $a=2$ · (c) $(1,0)$ · (d) $\log_{2}(x+4)-1$ · (e) $x=-4$

+ Add to test✓ In your test42AASL2-q00662.9 Exponential and logarithmic functions Paper 1Medium9 marks

Let $g(x)=2-\log_{3}(x+1)$, for $x\gt -1$.

(a)

Write down the equation of the vertical asymptote of the graph of $g$.

(1)

(b)

Find the coordinates of the points where the graph of $g$ meets the axes.

(3)

(c)

Find $g^{-1}(x)$.

(3)

(d)

Hence show that $g^{-1}(x)=9\mathrm{e}^{-x\ln3}-1$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$x=-1$	A1
(b)
$(0,\,2)$	A1	$\log_{3}1=0$
$\log_{3}(x+1)=2$	(M1)
$(8,\,0)$	A1
(c)
$x=2-\log_{3}(y+1)$	(M1)	Interchanging variables, at any stage.
$y+1=3^{2-x}$	A1
$g^{-1}(x)=3^{2-x}-1$	A1
(d)
$3^{2-x}=3^{2}\times3^{-x}=9\times3^{-x}$	M1
$3^{-x}=\mathrm{e}^{-x\ln3}$ (using $a^{x}=\mathrm{e}^{x\ln a}$)	A1
$g^{-1}(x)=9\mathrm{e}^{-x\ln3}-1$	AG

Answers: (a) $x=-1$ · (b) $(0,2),\ (8,0)$ · (c) $g^{-1}(x)=3^{2-x}-1$

+ Add to test✓ In your test43AASL2-q00682.9 Exponential and logarithmic functions Paper 1Medium8 marks

(a)

Write $4^{x}$ in the form $\mathrm{e}^{kx}$, giving the exact value of $k$.

(1)

(b)

Hence solve $\mathrm{e}^{x}\times4^{x}=\mathrm{e}^{3}$, giving your answer in the form $\dfrac{a}{b+\ln c}$, where $a,b,c\in\mathbb{Z}^{+}$.

(3)

(c)

Let $h(x)=\log_{4}x$, for $x\gt 0$. Show that $h(x)=\dfrac{\ln x}{2\ln2}$.

(2)

(d)

Find $h^{-1}(x)$, giving your answer in the form $\mathrm{e}^{px}$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$4^{x}=\mathrm{e}^{x\ln4}$, $k=\ln4$	A1	Accept $k=2\ln2$.
(b)
$\mathrm{e}^{x}\times\mathrm{e}^{x\ln4}=\mathrm{e}^{x(1+\ln4)}$	M1
$x(1+\ln4)=3$	A1
$x=\dfrac{3}{1+\ln4}$	A1
(c)
change of base: $\log_{4}x=\dfrac{\ln x}{\ln4}$	M1
$\ln4=\ln2^{2}=2\ln2$	A1
$h(x)=\dfrac{\ln x}{2\ln2}$	AG
(d)
$h^{-1}(x)=4^{x}$	(M1)	Exponential and logarithmic functions are inverses; or by interchanging variables.
$h^{-1}(x)=\mathrm{e}^{x\ln4}$	A1	Accept $\mathrm{e}^{2x\ln2}$. FT from (a).

Answers: (a) $\mathrm{e}^{x\ln4}$ · (b) $x=\dfrac{3}{1+\ln4}$ · (d) $h^{-1}(x)=\mathrm{e}^{x\ln4}$

+ Add to test✓ In your test44AASL2-q00742.10 Solving equations Paper 1Medium7 marks

(a)

Solve $\mathrm{e}^{2x}-3\mathrm{e}^{x}-10=0$.

(4)

(b)

Solve $\ln x+\ln(x-4)=\ln12$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
recognizing a quadratic in $\mathrm{e}^{x}$: $y^{2}-3y-10=0$, where $y=\mathrm{e}^{x}$	M1
$(y-5)(y+2)=0$, so $\mathrm{e}^{x}=5$ or $\mathrm{e}^{x}=-2$	A1
$\mathrm{e}^{x}=-2$ is rejected since $\mathrm{e}^{x}\gt 0$	R1
$x=\ln5$	A1
(b)
$\ln\big(x(x-4)\big)=\ln12$, so $x^{2}-4x-12=0$	M1
$(x-6)(x+2)=0$	A1
$x=6$ only	A1	Award only if $x=-2$ is rejected or not given.

Answers: (a) $x=\ln5$ · (b) $x=6$

+ Add to test✓ In your test45AASL2-q00762.10 Solving equations Paper 1Medium8 marks

(a)

Solve $2^{x+3}=7$, giving your answer in the form $\log_{2}a+b$, where $a,b\in\mathbb{Z}$.

(2)

(b)

Solve $3^{2x+1}=5^{x}$, giving your answer in the form $\dfrac{\ln p}{\ln q}$, where $p,q\in\mathbb{Q}$.

(3)

(c)

Solve $\log_{2}x+\log_{4}x=6$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$x+3=\log_{2}7$	(M1)
$x=\log_{2}7-3$	A1
(b)
$(2x+1)\ln3=x\ln5$	M1	Taking logarithms of both sides (any base).
$x(\ln5-2\ln3)=\ln3$	A1
$x=\dfrac{\ln3}{\ln\frac{5}{9}}$	A1	Accept equivalent forms, e.g. $\dfrac{\ln\frac{1}{3}}{\ln\frac{9}{5}}$. $x\approx-1.87$ is not required.
(c)
$\log_{4}x=\dfrac{\log_{2}x}{\log_{2}4}=\dfrac{1}{2}\log_{2}x$	M1
$\dfrac{3}{2}\log_{2}x=6$, so $\log_{2}x=4$	A1
$x=16$	A1

Answers: (a) $\log_{2}7-3$ · (b) $x=\dfrac{\ln3}{\ln(5/9)}$ · (c) $x=16$

+ Add to test✓ In your test46AASL2-q00852.11 Transformations of graphs Paper 1Medium7 marks

The following diagram shows the graph of $y=f(x)$, for $-3\le x\le3$ (solid), which consists of three line segments joining $(-3,\,0)$, $(-1,\,2)$, $(1,\,-1)$ and $(3,\,1)$. The graph of $y=g(x)$ (dashed) is the image of the graph of $f$ under two transformations.

(a)(i)

Describe fully the two transformations.

(2)

(a)(ii)

Hence write down $g(x)$ in the form $pf(x-q)$, where $p,q\in\mathbb{Z}$.

(1)

(b)

Find $(g\circ f)(-1)$.

(2)

(c)

Find all solutions of the equation $f(x)=\dfrac{1}{2}$.

(2)

Show mark scheme

Working	Mark	Notes
(a)(i)
vertical stretch, scale factor 2	A1
translation of 1 unit to the right, $\begin{pmatrix}1\\0\end{pmatrix}$	A1	Either order is acceptable here.
(a)(ii)
$g(x)=2f(x-1)$	A1
(b)
$f(-1)=2$, then $g(2)$	(M1)
$-2$	A1	$g(2)=2f(1)=-2$, also readable from the graph of $g$.
(c)
line $y=\frac{1}{2}$ meets the graph of $f$ three times, or equation of at least one segment	(M1)
$x=-2.5,\ 0,\ 2.5$	A1	All three values required.

Answers: (a)(ii) $g(x)=2f(x-1)$ · (b) $-2$ · (c) $x=-2.5,\ 0,\ 2.5$

+ Add to test✓ In your test47AASL2-q00892.11 Transformations of graphs Paper 1Medium8 marks

(a)

Show that $\log_{2}(8x)=3+\log_{2}x$, for $x\gt 0$.

(1)

(b)

Hence describe two different single transformations, each of which maps the graph of $y=\log_{2}x$ onto the graph of $y=\log_{2}(8x)$.

(2)

(c)

Describe two different single transformations, each of which maps the graph of $y=2^{x}$ onto the graph of $y=2^{x+3}$.

(3)

(d)

The graph of $y=4^{x}$ is the image of the graph of $y=2^{x}$ under a single transformation. Describe this transformation fully.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\log_{2}8+\log_{2}x$ and $\log_{2}8=3$	M1
$3+\log_{2}x$	AG
(b)
translation of 3 units vertically upwards	A1
horizontal stretch with scale factor $\tfrac{1}{8}$	A1	$y=f(8x)$.
(c)
translation of 3 units in the negative $x$-direction	A1
$2^{x+3}=2^{3}\times2^{x}=8\times2^{x}$	M1
vertical stretch with scale factor 8	A1
(d)
$4^{x}=2^{2x}$	M1
horizontal stretch with scale factor $\tfrac{1}{2}$	A1

Answers: (b) translation 3 up; horizontal stretch, scale factor $\tfrac18$ · (c) translation 3 left; vertical stretch, scale factor 8 · (d) horizontal stretch, scale factor $\tfrac12$

+ Add to test✓ In your test48AASL2-q00902.11 Transformations of graphs Paper 1Medium8 marks

The points $\mathrm{P}(2,\,5)$ and $\mathrm{Q}(-1,\,1)$ lie on the graph of $y=f(x)$.

(a)

The graph of $y=f(x)$ is stretched vertically with scale factor 3 and then translated 4 units vertically downwards. Write down the equation of the new graph and the coordinates of the image of P.

(2)

(b)

The same two transformations are applied in the reverse order. Write down the equation of the new graph and the coordinates of the image of P.

(2)

(c)

The graph of $y=pf(x)+q$ passes through the images of P and Q, which are $(2,\,7)$ and $(-1,\,-5)$ respectively. Find the value of $p$ and the value of $q$.

(3)

(d)

Hence describe a sequence of two transformations that maps the graph of $y=f(x)$ onto the graph of $y=pf(x)+q$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$y=3f(x)-4$	A1
$(2,\,11)$	A1
(b)
$y=3\big(f(x)-4\big)=3f(x)-12$	A1
$(2,\,3)$	A1
(c)
$5p+q=7$ and $p+q=-5$	M1
$p=3$	A1
$q=-8$	A1
(d)
vertical stretch with scale factor 3, followed by a translation of 8 units downwards	A1	FT from (c). Accept a translation of $\tfrac{8}{3}$ units down followed by the stretch.

Answers: (a) $y=3f(x)-4$, $(2,11)$ · (b) $y=3f(x)-12$, $(2,3)$ · (c) $p=3,\ q=-8$

+ Add to test✓ In your test49AASL2-q00052.1 Equations of straight lines Paper 1Hard9 marks

The points $\mathrm{A}(-2,\,1)$ and $\mathrm{B}(4,\,4)$ lie on the line $L_{1}$. The line $L_{2}$ passes through B and is perpendicular to $L_{1}$. The line $L_{2}$ meets the $x$-axis at C.

(a)

Find the gradient of $L_{1}$.

(1)

(b)

Find the equation of $L_{2}$, giving your answer in the form $y=mx+c$.

(2)

(c)

Find the coordinates of C.

(1)

(d)

Find the area of triangle ABC.

(3)

(e)

The point D is such that ABCD is a rectangle. Find the coordinates of D.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\dfrac{4-1}{4-(-2)}=\dfrac{1}{2}$	A1
(b)
gradient $-2$ and $y-4=-2(x-4)$	(M1)	FT from (a).
$y=-2x+12$	A1
(c)
$(6,\,0)$	A1	FT from (b).
(d)
recognizing the right angle at B and attempting $\mathrm{AB}$ and $\mathrm{BC}$	(M1)
$\mathrm{AB}=\sqrt{45}=3\sqrt5,\ \mathrm{BC}=\sqrt{20}=2\sqrt5$	A1
area $=\tfrac12\times3\sqrt5\times2\sqrt5=15$	A1
(e)
$\overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{BC}}$, i.e. moving from A by $+2$ in $x$ and $-4$ in $y$; or intersecting the line through A parallel to $L_{2}$ with the line through C parallel to $L_{1}$	(M1)
$\mathrm{D}(0,\,-3)$	A1

Answers: (a) $\tfrac12$ · (b) $y=-2x+12$ · (c) $(6,0)$ · (d) 15 · (e) $(0,-3)$

+ Add to test✓ In your test50AASL2-q00152.2 Functions, domain, range and inverse Paper 1Hard8 marks

Let $f(x)=\dfrac{x^{2}+3}{4}$, for $x\ge0$, and $g(x)=\dfrac{x^{2}+3}{4}$, for $x\le0$. Both $f$ and $g$ have inverse functions.

(a)

Write down the range of $f$.

(1)

(b)

Find the value of $f^{-1}(7)$.

(2)

(c)(i)

Write down the domain of $f^{-1}$.

(1)

(c)(ii)

Write down the range of $f^{-1}$.

(1)

(d)

Find the value of $g^{-1}(7)$, justifying your choice.

(2)

(e)

Explain why the function $h(x)=\dfrac{x^{2}+3}{4}$, for $x\in\mathbb{R}$, does not have an inverse.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$f(x)\ge\dfrac{3}{4}$	A1
(b)
$\dfrac{x^{2}+3}{4}=7\Rightarrow x^{2}=25$	(M1)	Solving f(x) = 7.
$f^{-1}(7)=5$	A1	Award A0 if $-5$ is also given.
(c)(i)
$x\ge\dfrac34$	A1	FT from (a): the range of f.
(c)(ii)
$f^{-1}(x)\ge0$	A1	The domain of f.
(d)
$g^{-1}(7)=-5$	A1
the range of $g^{-1}$ is the domain of $g$, $x\le0$, so the negative root is taken	R1	Award R1 only with the value −5.
(e)
$h$ is not one-to-one, e.g. $h(5)=h(-5)=7$	R1

Answers: (a) $f(x)\ge\tfrac34$ · (b) 5 · (c)(i) $x\ge\tfrac34$ (ii) $f^{-1}(x)\ge0$ · (d) $-5$

+ Add to test✓ In your test51AASL2-q00302.5 Composite and inverse functions Paper 1Hard8 marks

Let $g(x)=\sqrt{x+4}$, for $x\ge-4$.

(a)

Write down the range of $g$.

(1)

(b)

Find $g^{-1}(x)$, stating its domain.

(3)

(c)

Explain why any point of intersection of the graphs of $y=g(x)$ and $y=g^{-1}(x)$ lies on the line $y=x$.

(1)

(d)

Hence solve $g(x)=g^{-1}(x)$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$g(x)\ge0$	A1
(b)
$x=\sqrt{y+4}$ (interchanging $x$ and $y$)	(M1)
$g^{-1}(x)=x^{2}-4$	A1
domain $x\ge0$	A1	FT from (a): the domain of the inverse is the range of g.
(c)
the graphs are reflections of each other in $y=x$ and $g$ is increasing, so they can only meet on $y=x$	R1	Must refer to both the reflection and the increasing nature of g.
(d)
$x^{2}-4=x$ (or $\sqrt{x+4}=x$)	M1
$x=\dfrac{1\pm\sqrt{17}}{2}$	A1
$x=\dfrac{1+\sqrt{17}}{2}$ only, since $x\ge0$ in the domain of $g^{-1}$	R1	Award only if $\frac{1-\sqrt{17}}{2}$ is rejected with a reason.

Answers: (a) $g(x)\ge0$ · (b) $g^{-1}(x)=x^{2}-4,\ x\ge0$ · (d) $x=\tfrac{1+\sqrt{17}}{2}$

+ Add to test✓ In your test52AASL2-q00352.5 Composite and inverse functions Paper 1Hard8 marks

Let $f(x)=\sqrt{x-2}$, for $x\ge2$, and $g(x)=x^{2}-7$, for $x\in\mathbb{R}$.

(a)

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

(1)

(b)

Find the largest possible domain of $f\circ g$.

(2)

(c)

Find $(g\circ f)(x)$ in its simplest form, and state its domain and range.

(3)

(d)

Solve $(f\circ g)(x)=4$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\sqrt{x^{2}-9}$	A1
(b)
$x^{2}-7\ge2$, i.e. $x^{2}\ge9$	(M1)	The output of g must lie in the domain of f.
$x\le-3$ or $x\ge3$	A1
(c)
$(g\circ f)(x)=x-9$	A1
domain $x\ge2$	A1	Must be the domain of f, not the reals.
range $(g\circ f)(x)\ge-7$	A1	FT from their domain.
(d)
$x^{2}-9=16$	(M1)
$x=\pm5$	A1	Both values required; both lie in the domain from (b).

Answers: (a) $\sqrt{x^{2}-9}$ · (b) $x\le-3$ or $x\ge3$ · (c) $x-9$, $x\ge2$, range $\ge-7$ · (d) $\pm5$

+ Add to test✓ In your test53AASL2-q00372.5 Composite and inverse functions Paper 1Hard14 marks

Let $f(x)=3x+4$, for $x\in\mathbb{R}$, and $g(x)=(x-1)^{2}$, for $x\ge1$.

(a)

Find $f^{-1}(x)$.

(2)

(b)

Find $(g\circ f)(x)$, giving your answer in the form $p(x+q)^{2}$, where $p,q\in\mathbb{Z}$.

(2)

(c)

Find the domain of $g\circ f$.

(1)

(d)

Write down the range of $g$.

(1)

(e)

Find $g^{-1}(x)$, stating its domain.

(3)

(f)

Solve $(f^{-1}\circ g)(x)=4$.

(3)

(g)

Find the value of $x$ for which $(f\circ g^{-1})(x)=13$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$x=3y+4$	(M1)	Interchanging x and y at any stage.
$f^{-1}(x)=\dfrac{x-4}{3}$	A1
(b)
$\big((3x+4)-1\big)^{2}$	(M1)
$9(x+1)^{2}$	A1
(c)
$3x+4\ge1\Rightarrow x\ge-1$	A1	The output of f must lie in the domain of g.
(d)
$g(x)\ge0$	A1
(e)
$x=(y-1)^{2}$	(M1)
$g^{-1}(x)=1+\sqrt{x}$	A1	The positive root is required because $g^{-1}(x)\ge1$. Award A0 for $1\pm\sqrt{x}$.
domain $x\ge0$	A1	FT from (d).
(f)
$\dfrac{g(x)-4}{3}=4\Rightarrow g(x)=16$	(M1)
$x-1=\pm4$	A1
$x=5$ only	A1	$x=-3$ must be rejected as it is not in the domain of $g$.
(g)
$3g^{-1}(x)+4=13\Rightarrow g^{-1}(x)=3$	(M1)
$x=g(3)=4$	A1

Answers: (a) $\tfrac{x-4}{3}$ · (b) $9(x+1)^{2}$ · (c) $x\ge-1$ · (d) $g(x)\ge0$ · (e) $1+\sqrt x,\ x\ge0$ · (f) $x=5$ · (g) $x=4$

+ Add to test✓ In your test54AASL2-q00432.6 Quadratic functions Paper 1Hard16 marks

Let $f(x)=2x^{2}-8x+5$, for $x\in\mathbb{R}$.

(a)

Express $f(x)$ in the form $a(x-h)^{2}+k$.

(3)

(b)(i)

Write down the coordinates of the vertex of the graph of $f$.

(1)

(b)(ii)

Write down the equation of the axis of symmetry.

(1)

(c)

Write down the $y$-intercept of the graph of $f$.

(1)

(d)

Solve $f(x)=0$, giving your answers in the form $p\pm\dfrac{\sqrt{q}}{r}$, where $p,q,r\in\mathbb{Z}^{+}$.

(3)

(e)

Describe the transformation that maps the graph of $y=2x^{2}$ onto the graph of $y=f(x)$.

(2)

(f)

Explain why $f$ does not have an inverse.

(1)

(g)

The function $h$ is defined by $h(x)=f(x)$, for $x\ge2$. Find $h^{-1}(x)$, stating its domain.

(4)

Show mark scheme

Working	Mark	Notes
(a)
$2\left(x^{2}-4x\right)+5$ and attempt to complete the square	M1
$2\big((x-2)^{2}-4\big)+5$	A1
$2(x-2)^{2}-3$	A1
(b)(i)
$(2,\,-3)$	A1	FT from (a).
(b)(ii)
$x=2$	A1
(c)
$(0,\,5)$	A1	Accept 5.
(d)
$(x-2)^{2}=\dfrac{3}{2}$	M1	Or correct substitution into the quadratic formula.
$x-2=\pm\sqrt{\dfrac{3}{2}}$	A1
$x=2\pm\dfrac{\sqrt{6}}{2}$	A1	$\sqrt{\tfrac{3}{2}}=\tfrac{\sqrt{6}}{2}$ must be rationalized.
(e)
translation	A1
by $\begin{pmatrix}2\\-3\end{pmatrix}$: 2 units right and 3 units down	A1
(f)
$f$ is not one-to-one (e.g. $f(0)=f(4)=5$; the parabola fails the horizontal-line test)	R1
(g)
$x=2(y-2)^{2}-3$	M1	Interchanging variables, at any stage.
$(y-2)^{2}=\dfrac{x+3}{2}$	A1
$h^{-1}(x)=2+\sqrt{\dfrac{x+3}{2}}$	A1	Must choose the positive root, since $h^{-1}(x)\ge2$.
domain $x\ge-3$	A1	The range of h.

Answers: (a) $2(x-2)^{2}-3$ · (b) $(2,-3)$; $x=2$ · (c) $(0,5)$ · (d) $2\pm\tfrac{\sqrt6}{2}$ · (e) translation by $\binom{2}{-3}$ · (g) $2+\sqrt{\tfrac{x+3}{2}},\ x\ge-3$

+ Add to test✓ In your test55AASL2-q00472.7 Quadratic equations and the discriminant Paper 1Hard8 marks

The line $y=mx+1$, where $m\in\mathbb{R}$, and the curve $y=x^{2}-x+5$ are drawn on the same axes.

(a)

Show that the $x$-coordinates of any points of intersection satisfy $x^{2}-(m+1)x+4=0$.

(1)

(b)

Find the values of $m$ for which the line is a tangent to the curve.

(3)

(c)

For the positive value of $m$ found in part (b), find the coordinates of the point where the line touches the curve.

(2)

(d)

Find the set of values of $m$ for which the line does not meet the curve.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$x^{2}-x+5=mx+1$, so $x^{2}-x-mx+4=0$	M1
$x^{2}-(m+1)x+4=0$	AG
(b)
tangent $\Leftrightarrow\Delta=0$: $(m+1)^{2}-16=0$	M1
$m+1=\pm4$	A1
$m=3$ or $m=-5$	A1
(c)
$x^{2}-4x+4=0$, so $(x-2)^{2}=0$	(M1)
$(2,\,7)$	A1
(d)
$(m+1)^{2}-16\lt 0$	(M1)
$-5\lt m\lt 3$	A1	FT from their values in (b).

Answers: (b) $m=3$ or $m=-5$ · (c) $(2,\,7)$ · (d) $-5\lt m\lt 3$

+ Add to test✓ In your test56AASL2-q00522.7 Quadratic equations and the discriminant Paper 1Hard8 marks

Consider the quadratic expression $(m-1)x^{2}+4x+m+2$, where $m\in\mathbb{R}$, $m\neq1$. The expression is positive for all real values of $x$.

(a)

Explain why $m\gt 1$.

(1)

(b)

Show that $m^{2}+m-6\gt 0$.

(2)

(c)

Hence find the set of possible values of $m$.

(3)

(d)

For a different value of $m$, the graph of $y=(m-1)x^{2}+4x+m+2$ touches the $x$-axis at exactly one point and lies on or above it. Find this value of $m$ and the coordinates of the point.

(2)

Show mark scheme

Working	Mark	Notes
(a)
if $m-1\lt 0$ the parabola opens downwards, so the expression is negative for large $\|x\|$; hence $m-1\gt 0$	R1
(b)
no real roots: $\Delta=16-4(m-1)(m+2)\lt 0$	M1
$16-4\left(m^{2}+m-2\right)\lt 0$, so $24-4m^{2}-4m\lt 0$	A1
$m^{2}+m-6\gt 0$	AG
(c)
$(m+3)(m-2)\gt 0$	M1
$m\lt -3$ or $m\gt 2$	A1
combining with $m\gt 1$: $m\gt 2$	A1
(d)
$m=2$	A1	$\Delta=0$ gives $m=2$ or $m=-3$; $m=-3$ is rejected as the graph would lie on or below the axis.
$x^{2}+4x+4=(x+2)^{2}$, point $(-2,\,0)$	A1

Answers: (c) $m\gt 2$ · (d) $m=2$, $(-2,0)$

+ Add to test✓ In your test57AASL2-q00552.7 Quadratic equations and the discriminant Paper 1Hard15 marks

A gardener has 60 m of fencing to enclose a rectangular vegetable plot against a straight wall. The wall forms one side of the rectangle and is not fenced. The two sides perpendicular to the wall each have length $x$ metres, where $0\lt x\lt 30$.

(a)

Show that the area of the plot, $A$ m², is given by $A=60x-2x^{2}$.

(1)

(b)

Find the values of $x$ for which the area is 400 m².

(3)

(c)

Express $A$ in the form $-2(x-h)^{2}+k$.

(3)

(d)

Hence write down the maximum area and the dimensions of the plot that give this area.

(2)

(e)

Find the set of values of $x$ for which the area is at least 352 m².

(3)

(f)

Using the discriminant, find the set of values of $c$ for which there are two different plots each with area $c$ m².

(3)

Show mark scheme

Working	Mark	Notes
(a)
side parallel to the wall $=60-2x$, so $A=x(60-2x)$	A1
$A=60x-2x^{2}$	AG
(b)
$2x^{2}-60x+400=0$, i.e. $x^{2}-30x+200=0$	M1
$(x-10)(x-20)=0$	A1
$x=10$ or $x=20$	A1
(c)
$-2\left(x^{2}-30x\right)$ and attempt to complete the square	M1
$-2\big((x-15)^{2}-225\big)$	A1
$A=-2(x-15)^{2}+450$	A1
(d)
maximum area $450$ m²	A1	FT from (c).
15 m by 30 m	A1	$x=15$; side parallel to wall $60-30=30$.
(e)
$60x-2x^{2}\ge352$, i.e. $x^{2}-30x+176\le0$	M1
$(x-8)(x-22)\le0$	A1
$8\le x\le22$	A1
(f)
$2x^{2}-60x+c=0$ with $\Delta=3600-8c\gt 0$	M1
$c\lt 450$	A1
$0\lt c\lt 450$	A1	Area must be positive. Both roots $x=15\pm\sqrt{225-\tfrac{c}{2}}$ then lie in $0\lt x\lt 30$.

Answers: (b) $x=10,\ 20$ · (c) $-2(x-15)^{2}+450$ · (d) 450 m², 15 m by 30 m · (e) $8\le x\le22$ · (f) $0\lt c\lt 450$

+ Add to test✓ In your test58AASL2-q00582.8 Reciprocal and rational functions Paper 1Hard8 marks

Let $f(x)=\dfrac{3x+4}{x-3}$, for $x\neq3$.

(a)

Show that $(f\circ f)(x)=x$.

(3)

(b)

Hence write down $f^{-1}(x)$.

(1)

(c)

State what the result in part (b) shows about the graph of $f$.

(1)

(d)

Solve $f(x)=x$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$\dfrac{3\left(\frac{3x+4}{x-3}\right)+4}{\frac{3x+4}{x-3}-3}$	M1
numerator $\dfrac{9x+12+4x-12}{x-3}=\dfrac{13x}{x-3}$	A1
denominator $\dfrac{3x+4-3x+9}{x-3}=\dfrac{13}{x-3}$	A1
$(f\circ f)(x)=\dfrac{13x}{13}=x$	AG
(b)
$f^{-1}(x)=\dfrac{3x+4}{x-3}$	A1	Accept $f^{-1}(x)=f(x)$.
(c)
the graph is symmetric about (unchanged by a reflection in) the line $y=x$	A1
(d)
$3x+4=x(x-3)$, i.e. $x^{2}-6x-4=0$	M1
$x=\dfrac{6\pm\sqrt{52}}{2}$	A1
$x=3\pm\sqrt{13}$	A1

Answers: (b) $f^{-1}(x)=\dfrac{3x+4}{x-3}$ · (d) $x=3\pm\sqrt{13}$

+ Add to test✓ In your test59AASL2-q00622.8 Reciprocal and rational functions Paper 1Hard9 marks

Let $f(x)=\dfrac{x+3}{x-1}$, for $x\neq1$. The line $L$ has equation $y=x-3$.

(a)

Find the coordinates of the points where $L$ meets the graph of $f$.

(3)

(b)

Hence solve the inequality $\dfrac{x+3}{x-1}\gt x-3$.

(3)

(c)

The line with equation $y=x+c$, where $c\in\mathbb{R}$, meets the graph of $f$ at the points A and B. Show that the midpoint of AB lies on the line $x+y=2$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$x+3=(x-3)(x-1)$	M1
$x^{2}-5x=0$, so $x=0$ or $x=5$	A1
$(0,\,-3)$ and $(5,\,2)$	A1
(b)
considering the graphs: the curve is above $L$ for $x\lt 0$ and between the asymptote $x=1$ and $x=5$	M1	Or a sign diagram for $\dfrac{x(5-x)}{x-1}$ with critical values 0, 1, 5.
$x\lt 0$	A1
$1\lt x\lt 5$	A1	Award A0 if $x=1$ is included.
(c)
$x+3=(x+c)(x-1)$, i.e. $x^{2}+(c-2)x-(c+3)=0$	M1
$x=\dfrac{2-c\pm\sqrt{c^{2}+16}}{2}$, so the midpoint has $x=\dfrac{2-c}{2}$	A1	The discriminant $c^{2}+16\gt 0$, so A and B always exist.
$y=\dfrac{2-c}{2}+c=\dfrac{2+c}{2}$	A1
$x+y=\dfrac{2-c}{2}+\dfrac{2+c}{2}=2$	AG

Answers: (a) $(0,-3),\ (5,2)$ · (b) $x\lt 0$ or $1\lt x\lt 5$

+ Add to test✓ In your test60AASL2-q00772.10 Solving equations Paper 1Hard9 marks

Consider the equation $\left(x^{2}-3x\right)^{2}-2\left(x^{2}-3x\right)=m$, where $m\in\mathbb{R}$.

(a)

By completing the square, show that $x^{2}-3x\ge-\dfrac{9}{4}$ for all $x\in\mathbb{R}$.

(2)

(b)

Solve the equation when $m=8$.

(4)

(c)

Find the value of $m$ for which the equation has exactly three distinct real solutions.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$x^{2}-3x=\left(x-\dfrac{3}{2}\right)^{2}-\dfrac{9}{4}$	M1
$\left(x-\dfrac{3}{2}\right)^{2}\ge0$	A1
$x^{2}-3x\ge-\dfrac{9}{4}$	AG
(b)
$u^{2}-2u-8=0$, where $u=x^{2}-3x$	M1
$u=4$ or $u=-2$	A1
$x^{2}-3x-4=0\Rightarrow x=4,\ -1$	A1
$x^{2}-3x+2=0\Rightarrow x=1,\ 2$	A1
(c)
each value $u\gt -\tfrac{9}{4}$ gives two values of $x$ and $u=-\tfrac{9}{4}$ gives one, so one root of $u^{2}-2u=m$ must equal $-\tfrac{9}{4}$	R1	The other root, $2-\left(-\tfrac{9}{4}\right)=\tfrac{17}{4}$, then gives two values of $x$.
$m=\left(-\dfrac{9}{4}\right)^{2}-2\left(-\dfrac{9}{4}\right)$	M1
$m=\dfrac{153}{16}$	A1	Solutions are then $x=\tfrac{3}{2}$ and $x=\tfrac{3\pm\sqrt{26}}{2}$.

Answers: (b) $x=-1,\ 1,\ 2,\ 4$ · (c) $m=\dfrac{153}{16}$

+ Add to test✓ In your test61AASL2-q00862.11 Transformations of graphs Paper 1Hard8 marks

The graph of $y=\ln x$ is transformed by the following sequence of transformations to give the graph of $y=g(x)$:

a reflection in the $y$-axis;
then a translation of 2 units vertically upwards;
then a vertical stretch with scale factor 3.

(a)

Show that $g(x)=3\ln(-x)+6$.

(2)

(b)

Write down the domain of $g$ and the equation of the vertical asymptote of its graph.

(2)

(c)

Find the exact $x$-intercept of the graph of $g$.

(2)

(d)

The same three transformations are applied to $y=\ln x$, but with the vertical stretch first, then the reflection, then the translation. Write down the equation of the resulting graph, and explain why it is different from $y=g(x)$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
after reflection $y=\ln(-x)$; after translation $y=\ln(-x)+2$	A1
stretch applied to the whole expression: $y=3\big(\ln(-x)+2\big)$	M1
$g(x)=3\ln(-x)+6$	AG
(b)
domain $x\lt 0$	A1
$x=0$	A1
(c)
$\ln(-x)=-2$	(M1)
$x=-\mathrm{e}^{-2}$	A1	Accept $\left(-\mathrm{e}^{-2},\,0\right)$.
(d)
$y=3\ln(-x)+2$	A1
in $g$ the translation is applied before the stretch, so the upward shift of 2 is also stretched (to 6); here it is not	R1

Answers: (b) $x\lt 0$; $x=0$ · (c) $x=-\mathrm{e}^{-2}$ · (d) $y=3\ln(-x)+2$

+ Add to test✓ In your test62AASL2-q00912.11 Transformations of graphs Paper 1Hard10 marks

Let $f(x)=\dfrac{x+4}{x-2}$, for $x\neq2$. The graph of $y=f(x)$ is stretched horizontally with scale factor $\tfrac{1}{2}$ and then reflected in the $y$-axis to give the graph of $y=g(x)$.

(a)

Show that $g(x)=\dfrac{x-2}{x+1}$.

(3)

(b)

Write down the equations of the asymptotes of the graph of $g$.

(2)

(c)

Explain why the same graph is obtained if the reflection is applied before the stretch.

(2)

(d)

Find the coordinates of the point where the graphs of $f$ and $g$ intersect, and explain why this point is unchanged by the two transformations.

(3)

Show mark scheme

Working	Mark	Notes
(a)
stretch: $f(2x)=\dfrac{2x+4}{2x-2}$	M1
reflection: replacing $x$ by $-x$: $\dfrac{-2x+4}{-2x-2}$	M1
$=\dfrac{-2(x-2)}{-2(x+1)}$	A1
$g(x)=\dfrac{x-2}{x+1}$	AG
(b)
$x=-1$	A1
$y=1$	A1
(c)
reflection first gives $f(-x)$; then the stretch replaces $x$ by $2x$: $f(-2x)$	M1
both orders replace $x$ by $-2x$ in $f$, so they give the same function	R1	Award R1 for a valid general reason, e.g. both transformations act only on $x$-coordinates, multiplying them by $\tfrac{1}{2}$ and by $-1$, and multiplication is commutative.
(d)
$\dfrac{x+4}{x-2}=\dfrac{x-2}{x+1}$, so $(x+4)(x+1)=(x-2)^{2}$	M1
$9x=0$, point $(0,\,-2)$	A1
both transformations change only $x$-coordinates, by multiplying them, so a point with $x=0$ (on the $y$-axis) is mapped onto itself	R1

Answers: (b) $x=-1,\ y=1$ · (d) $(0,-2)$

+ Add to test✓ In your test63AASL2-q00062.1 Equations of straight lines Paper 2Easy8 marks

A straight mountain road climbs from a village at a height of 312 m above sea level to a pass at a height of 396 m. The horizontal distance from the village to the pass is 1200 m. Let $x$ be the horizontal distance, in metres, from the village and $h$ the height, in metres, of the road above sea level.

(a)

Find the gradient of the road.

(2)

(b)

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

(1)

(c)

A viewpoint on the road is 350 m above sea level. Find the horizontal distance of the viewpoint from the village.

(2)

(d)

A straight cable-car line is parallel to the road and passes through the point $(0,\,340)$. Write down the equation of the cable-car line.

(1)

(e)

Building regulations state that an access ramp must have a gradient no greater than $\dfrac{1}{12}$. A ramp must rise 0.65 m. Find the least horizontal length of the ramp.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\dfrac{396-312}{1200}$	(M1)
$0.07$	A1	Accept $\tfrac{7}{100}$ or 7%.
(b)
$h=0.07x+312$	A1	FT from (a).
(c)
$0.07x+312=350$	(M1)
$543$ m	A1	$542.857\ldots$ FT from (b).
(d)
$h=0.07x+340$	A1	Parallel lines have equal gradients, $m_{1}=m_{2}$. FT from (a).
(e)
$\dfrac{0.65}{\ell}\le\dfrac{1}{12}$ or $0.65\times12$	(M1)
$7.8$ m	A1

Answers: (a) 0.07 · (b) $h=0.07x+312$ · (c) 543 m · (d) $h=0.07x+340$ · (e) 7.8 m

+ Add to test✓ In your test64AASL2-q00162.2 Functions, domain, range and inverse Paper 2Easy8 marks

The fare, $F$ dollars, for a taxi journey of $d$ kilometres is modelled by $F(d)=3.20+1.45d$, for $0\lt d\le40$.

(a)

Find the fare for a journey of 12 km.

(2)

(b)

Interpret the value 1.45 in the context of the model.

(1)

(c)

Find an expression for $F^{-1}(x)$.

(2)

(d)

Find $F^{-1}(30.75)$ and interpret your answer in context.

(2)

(e)

Write down the range of $F$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$3.20+1.45(12)$	(M1)
$20.60	A1
(b)
the cost, in dollars, per kilometre travelled	A1	Accept the fare increases by $1.45 for each extra km.
(c)
interchanging variables: $x=3.20+1.45d$	(M1)	Or attempt to make $d$ the subject.
$F^{-1}(x)=\dfrac{x-3.20}{1.45}$	A1
(d)
$19$	A1
a fare of $30.75 corresponds to a journey of 19 km	R1
(e)
$3.20\lt F\le61.20$	A1

Answers: (a) $20.60 · (c) $\dfrac{x-3.20}{1.45}$ · (d) 19 (km) · (e) $3.20\lt F\le61.20$

+ Add to test✓ In your test65AASL2-q00272.4 Key features and intersections Paper 2Easy6 marks

A point O lies at the foot of a straight hillside. A ball is thrown from a point 1.8 m vertically above O, up the hillside. In a vertical cross-section, the path of the ball is modelled by $y=1.8+0.6x-0.025x^{2}$ and the hillside is modelled by $y=0.1x$, for $x\ge0$, where $x$ metres is the horizontal distance from O and $y$ metres is the height above the level of O.

(a)

Find the horizontal distance from O to the point where the ball lands on the hillside.

(2)

(b)

Find the maximum height of the ball above the level of O.

(1)

(c)

Find the greatest vertical distance between the ball and the hillside, and the value of $x$ at which it occurs.

(2)

(d)

Explain why the answer to (c) occurs at a smaller value of $x$ than the answer to (b).

(1)

Show mark scheme

Working	Mark	Notes
(a)
intersecting $y=1.8+0.6x-0.025x^{2}$ with $y=0.1x$ on a GDC	(M1)
$23.1$ m	A1	$23.1148\ldots$ (the other intersection is negative and is rejected).
(b)
$5.4$ m	A1	At $x=12$.
(c)
graphing $d(x)=1.8+0.6x-0.025x^{2}-0.1x$, or using its vertex	(M1)
$4.3$ m at $x=10$	A1	Both values required.
(d)
the hillside rises as $x$ increases, so beyond $x=10$ the ground rises faster than the ball does; the gap is largest before the ball reaches its highest point	R1	Accept: $d(x)=1.8+0.5x-0.025x^{2}$ has its vertex at $x=10\lt12$.

Answers: (a) 23.1 m · (b) 5.4 m · (c) 4.3 m at $x=10$

+ Add to test✓ In your test66AASL2-q00072.1 Equations of straight lines Paper 2Medium6 marks

A funicular railway runs in a straight line up a hillside. The track is 1500 m long, measured along the slope, and the upper station is 270 m higher than the lower station.

(a)

Find the horizontal distance between the two stations.

(2)

(b)

Find the gradient of the track. Give your answer as a decimal and as a percentage.

(2)

(c)

A second funicular is to be built on another hillside. Its track must rise 300 m and, for safety, its gradient must not exceed 0.25. Find the least possible length of this track, measured along the slope.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\sqrt{1500^{2}-270^{2}}$	(M1)	Use of Pythagoras with the track as the hypotenuse.
$1480$ m	A1	$1475.49\ldots$ Accept 1475 or 1475.5.
(b)
$\dfrac{270}{1475.49\ldots}=0.183$	A1	$0.182988\ldots$ FT from (a). Award A0 for $\tfrac{270}{1500}=0.18$ (slope length used).
$18.3\%$	A1	FT from their decimal.
(c)
least horizontal distance $\dfrac{300}{0.25}=1200$ m, and attempt to use Pythagoras	(M1)
$\sqrt{1200^{2}+300^{2}}=1240$ m	A1	$1236.93\ldots$ Accept 1237. Award M1A0 for 1200 (horizontal distance only).

Answers: (a) 1480 m (1475.5) · (b) 0.183, 18.3% · (c) 1240 m (1236.9)

+ Add to test✓ In your test67AASL2-q00192.3 Graphs of functions Paper 2Medium7 marks

Let $h(x)=\mathrm{e}^{\frac{x}{2}}-x^{2}+3$, for $-3\le x\le6$.

(a)

Sketch the graph of $y=h(x)$ on the given domain, showing the coordinates of any intercepts with the axes, the local maximum point and the endpoints.

(3)

(b)

Find the range of $h$.

(2)

(c)

Find the set of values of $k$ for which the equation $h(x)=k$ has exactly two solutions.

(2)

Show mark scheme

Working	Mark	Notes
(a)
correct shape: increases to a single maximum, then decreases throughout to the right endpoint	A1
$x$-intercepts $(-1.84,\,0)$ and $(2.57,\,0)$; $y$-intercept $(0,\,4)$	A1
maximum $(0.289,\,4.07)$; endpoints $(-3,\,-5.78)$ and $(6,\,-12.9)$	A1	Award for at least the maximum and both endpoints in approximately correct positions.
(b)
recognizing the least value occurs at the endpoint $x=6$	(M1)	$h(6)=-12.914\ldots$
$-12.9\le h(x)\le4.07$	A1
(c)
horizontal line meets the graph twice between $y=h(-3)$ and the maximum	(M1)
$-5.78\le k\lt 4.07$	A1	Both ends must be correct, including the inclusive left end: $h(-3)=-5.7768\ldots$, $h_{\max}=4.0719\ldots$

Answers: (a) intercepts $-1.84,\ 2.57,\ (0,4)$; max $(0.289,\,4.07)$ · (b) $-12.9\le h(x)\le4.07$ · (c) $-5.78\le k\lt 4.07$

+ Add to test✓ In your test68AASL2-q00202.3 Graphs of functions Paper 2Medium8 marks

Let $f(x)=x$ and $g(x)=\mathrm{e}^{-x}$. The function $h$ is defined by $h(x)=f(x)+g(x)$, for $-2\le x\le3$.

(a)

On the same set of axes, sketch the graphs of $y=f(x)$ and $y=h(x)$, for $-2\le x\le3$.

(3)

(b)

Write down the coordinates of the minimum point of the graph of $h$.

(2)

(c)

Explain why the graph of $h$ lies above the graph of $f$ for all values of $x$.

(1)

(d)

Solve $h(x)=2$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
straight line $y=x$ through the origin	A1
graph of $h$ with a single minimum near $(0,\,1)$, decreasing then increasing	A1
endpoints of $h$ approximately $(-2,\,5.39)$ and $(3,\,3.05)$, with $h$ above $y=x$ and approaching it as $x$ increases	A1
(b)
$x=0$	A1
$y=1$	A1	Award A1A0 if only the x-coordinate is given.
(c)
$h(x)-f(x)=\mathrm{e}^{-x}\gt 0$ for all $x$	R1
(d)
$x=-1.15$	A1	$-1.14619\ldots$
$x=1.84$	A1	$1.84140\ldots$

Answers: (b) $(0,\,1)$ · (d) $x=-1.15,\ 1.84$

+ Add to test✓ In your test69AASL2-q00222.4 Key features and intersections Paper 2Medium8 marks

The concentration of a drug in a patient’s blood, $C$ mg l⁻¹, $t$ hours after an injection is modelled by $C(t)=12t\,\mathrm{e}^{-0.4t}$, for $t\ge0$.

(a)

Find the maximum concentration and the time at which it occurs.

(2)

(b)

The drug is effective while the concentration is at least 8 mg l⁻¹. Find the length of time for which the drug is effective.

(3)

(c)

Find the time after which the concentration remains below 0.5 mg l⁻¹.

(2)

(d)

Write down the equation of the horizontal asymptote of the graph of $C$ and interpret it in context.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$t=2.5$ hours	A1	Using the maximum feature of the GDC.
$11.0$ mg l⁻¹	A1	$11.036\ldots$
(b)
intersecting $y=C(t)$ with $y=8$	(M1)	Or using the GDC equation solver twice.
$t=0.991$ and $t=5.07$	(A1)
$4.08$ hours	A1	$4.0830\ldots$ Accept 4.07–4.08.
(c)
$12t\,\mathrm{e}^{-0.4t}=0.5$, with $t\gt 2.5$	(M1)
$14.7$ hours	A1	$14.657\ldots$
(d)
$C=0$: in the long term the drug is eliminated from the blood	A1	Both equation and interpretation required.

Answers: (a) 11.0 mg l⁻¹ at 2.5 h · (b) 4.08 h · (c) 14.7 h · (d) $C=0$

+ Add to test✓ In your test70AASL2-q00232.4 Key features and intersections Paper 2Medium6 marks

Let $f(x)=4-x^{2}$ and $g(x)=2^{x}$, for $x\in\mathbb{R}$. The graphs of $f$ and $g$ intersect at two points, A and B, where A has the smaller $x$-coordinate.

(a)

Find the coordinates of A and of B.

(3)

(b)

Write down the set of values of $x$ for which $f(x)\gt g(x)$.

(1)

(c)

Let $d(x)=f(x)-g(x)$. Find the maximum value of $d(x)$ for values of $x$ between A and B, and the value of $x$ at which it occurs.

(2)

Show mark scheme

Working	Mark	Notes
(a)
intersecting the two graphs on a GDC, or solving $4-x^{2}=2^{x}$	(M1)
$\mathrm{A}(-1.93,\,0.262)$	A1
$\mathrm{B}(1.26,\,2.40)$	A1
(b)
$-1.93\lt x\lt 1.26$	A1	FT from (a).
(c)
graphing $y=4-x^{2}-2^{x}$ and finding its maximum	(M1)
maximum $3.10$ at $x=-0.285$	A1	$3.0980\ldots$ at $x=-0.28454\ldots$

Answers: (a) A$(-1.93,\,0.262)$, B$(1.26,\,2.40)$ · (b) $-1.93\lt x\lt 1.26$ · (c) 3.10 at $x=-0.285$

+ Add to test✓ In your test71AASL2-q00242.4 Key features and intersections Paper 2Medium7 marks

Let $f(x)=\dfrac{2x^{2}+1}{x^{2}-9}$, for $x\in\mathbb{R}$, $x\neq\pm3$.

(a)

Write down the equations of the vertical and horizontal asymptotes of the graph of $f$.

(3)

(b)

Write down the coordinates of the local maximum point of the graph of $f$.

(1)

(c)

Find the range of $f$.

(2)

(d)

Describe the symmetry of the graph of $f$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$x=3$	A1
$x=-3$	A1
$y=2$	A1
(b)
$(0,\,-0.111)$	A1	Accept $\left(0,\,-\tfrac19\right)$.
(c)
$f(x)\le-0.111$	A1	Accept $f(x)\le-\tfrac19$.
or $f(x)\gt2$	A1	The graph approaches y = 2 from above for \|x\| > 3.
(d)
symmetrical about the $y$-axis (the line $x=0$)	A1

Answers: (a) $x=\pm3$, $y=2$ · (b) $(0,-0.111)$ · (c) $f(x)\le-0.111$ or $f(x)\gt2$ · (d) about the $y$-axis

+ Add to test✓ In your test72AASL2-q00252.4 Key features and intersections Paper 2Medium6 marks

For a particular product, the price, $p$ dollars, at which consumers will buy $q$ thousand units per week is modelled by the demand function $p=50-0.4q$. The price at which producers will supply $q$ thousand units per week is modelled by the supply function $p=5+0.02q^{2}$. Both models apply for $0\le q\le120$.

(a)

Market equilibrium occurs where the graphs of the two functions intersect. Find the equilibrium quantity and price.

(2)

(b)

A tax raises the supply function to $p=11+0.02q^{2}$. Find the new equilibrium quantity and price.

(2)

(c)

Find the increase in the equilibrium price, and express it as a percentage of the tax of $6.

(2)

Show mark scheme

Working	Mark	Notes
(a)
intersecting the graphs on a GDC, or solving $50-0.4q=5+0.02q^{2}$	(M1)
$q=38.5$ (thousand), $p=\$34.61$	A1	$38.4768\ldots,\ 34.6093\ldots$ Accept $p=34.6$.
(b)
solving $50-0.4q=11+0.02q^{2}$	(M1)
$q=35.3$ (thousand), $p=\$35.89$	A1	$35.2769\ldots,\ 35.8892\ldots$ Accept $p=35.9$.
(c)
$\$1.28$	A1	$1.27995\ldots$ Accept 1.3 from 3 s.f. values.
$21.3\%$	A1	Accept 21.7% from 1.3. FT from their increase.

Answers: (a) $q=38.5$, $p=\$34.61$ · (b) $q=35.3$, $p=\$35.89$ · (c) $1.28, 21.3%

+ Add to test✓ In your test73AASL2-q00362.5 Composite and inverse functions Paper 2Medium9 marks

A bank converts euros (€) to US dollars ($). The function $f(x)=1.08x$ gives the number of dollars equivalent to $x$ euros. The bank then deducts a commission of 2% of the dollar amount and a fixed fee of $3, so that a dollar amount $x$ becomes $g(x)=0.98x-3$ dollars.

(a)

Find $(g\circ f)(500)$ and interpret your answer in context.

(2)

(b)

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

(2)

(c)

Find $(g\circ f)^{-1}(x)$.

(2)

(d)

Hence find the number of euros that a customer must exchange to receive exactly $1000. Give your answer correct to the nearest cent.

(1)

(e)

Find $(f\circ g)(x)$ and show that $(f\circ g)(x)-(g\circ f)(x)$ is constant.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\$526.20$	A1	$0.98\times540-3$
a customer exchanging €500 receives $526.20	A1
(b)
$0.98(1.08x)-3$	(M1)
$1.0584x-3$	A1
(c)
$x=1.0584y-3$	(M1)
$(g\circ f)^{-1}(x)=\dfrac{x+3}{1.0584}$	A1
(d)
$\dfrac{1003}{1.0584}=947.66$, i.e. €947.66	A1	$947.656\ldots$
(e)
$(f\circ g)(x)=1.08(0.98x-3)=1.0584x-3.24$	A1
difference $=-0.24$, which does not depend on $x$	A1

Answers: (a) $526.20 · (b) $1.0584x-3$ · (c) $\tfrac{x+3}{1.0584}$ · (d) €947.66 · (e) $-0.24$

+ Add to test✓ In your test74AASL2-q00422.6 Quadratic functions Paper 2Medium7 marks

The cross-section of a tunnel entrance is modelled by the arch $y=h(x)$, where $h(x)=ax(x-40)$, for $0\le x\le40$. Distances are in metres, the ground is the $x$-axis and the highest point of the arch is $(20,\,12)$.

Diagram not to scale

(a)

Show that $a=-0.03$.

(2)

(b)

Write down $h(x)$ in the form $a(x-p)^{2}+q$.

(1)

(c)

Find the height of the arch at a horizontal distance of 5 m from O.

(1)

(d)

A lorry with a rectangular cross-section is 10.5 m high. Find the maximum width of the lorry that can pass through the arch.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$12=a(20)(20-40)$	M1
$12=-400a$	A1
$a=-0.03$	AG
(b)
$h(x)=-0.03(x-20)^{2}+12$	A1
(c)
$5.25$ m	A1
(d)
$-0.03(x-20)^{2}+12=10.5$	(M1)
$x=12.93,\ 27.07$	(A1)	$20\pm\sqrt{50}$
$14.1$ m	A1	$2\sqrt{50}=14.142\ldots$

Answers: (b) $-0.03(x-20)^{2}+12$ · (c) 5.25 m · (d) 14.1 m

+ Add to test✓ In your test75AASL2-q00532.7 Quadratic equations and the discriminant Paper 2Medium8 marks

A ball is thrown upwards from the edge of a platform. Its height above the ground, $h$ metres, $t$ seconds after it is thrown is modelled by $h(t)=1.6+12t-4.9t^{2}$, for $t\ge0$, until the ball reaches the ground.

(a)

Write down the height from which the ball is thrown.

(1)

(b)

Find the maximum height of the ball and the time at which it occurs.

(3)

(c)

Find the time at which the ball reaches the ground.

(2)

(d)

Find the length of time for which the ball is more than 6 m above the ground.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$1.6$ m	A1
(b)
$t=-\dfrac{12}{2(-4.9)}$, or maximum on GDC	(M1)
$t=1.22$ s	A1	$1.2244\ldots$
$8.95$ m	A1	$8.9469\ldots$
(c)
$1.6+12t-4.9t^{2}=0$	(M1)
$t=2.58$ s	A1	$2.5757\ldots$ Award A0 if the negative root is also given as an answer.
(d)
$1.6+12t-4.9t^{2}=6$, giving $t=0.449$ and $t=2$	(M1)
$1.55$ s	A1	$1.5510\ldots$

Answers: (a) 1.6 m · (b) 8.95 m at 1.22 s · (c) 2.58 s · (d) 1.55 s

+ Add to test✓ In your test76AASL2-q00542.7 Quadratic equations and the discriminant Paper 2Medium8 marks

On a dry road, the stopping distance, $d$ metres, of a car travelling at $v$ km h⁻¹ is modelled by $d=0.2v+0.006v^{2}$, for $v\ge0$.

(a)

Find the stopping distance of the car when it travels at 80 km h⁻¹.

(1)

(b)

Find the greatest speed at which the car can stop within 60 m.

(3)

(c)

On a wet road the model becomes $d=0.2v+kv^{2}$. At 50 km h⁻¹ the stopping distance on the wet road is 32.5 m. Find the value of $k$.

(2)

(d)

Find the speed on the wet road that gives the same stopping distance as 80 km h⁻¹ on the dry road.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$54.4$ m	A1
(b)
$0.006v^{2}+0.2v-60=0$ or $0.006v^{2}+0.2v\le60$	(M1)
$v=\dfrac{-0.2+\sqrt{0.04+1.44}}{0.012}$, or GDC	(A1)
$84.7$ km h⁻¹	A1	$84.712\ldots$ Award A0 if the negative root is also given as a final answer.
(c)
$10+2500k=32.5$	(M1)
$k=0.009$	A1
(d)
$0.2v+0.009v^{2}=54.4$	(M1)	FT from their (a) and (c).
$67.4$ km h⁻¹	A1	$67.424\ldots$

Answers: (a) 54.4 m · (b) 84.7 km h⁻¹ · (c) $k=0.009$ · (d) 67.4 km h⁻¹

+ Add to test✓ In your test77AASL2-q00642.8 Reciprocal and rational functions Paper 2Medium9 marks

A tank initially contains 200 litres of pure water. Salt water with a concentration of 40 grams per litre is pumped into the tank at a rate of 5 litres per minute, and mixes instantly. No liquid leaves the tank. The tank can hold at most 1000 litres.

(a)

Show that the concentration of salt in the tank, $C$ grams per litre, $t$ minutes after pumping starts is $C(t)=\dfrac{40t}{t+40}$.

(2)

(b)

Find the concentration after 20 minutes.

(1)

(c)

Find the time at which the concentration reaches 30 grams per litre.

(2)

(d)

Write down the domain of $C$ in this context, and find the greatest concentration that can be reached.

(2)

(e)

If the tank had no maximum capacity, the graph of $C$ would have a horizontal asymptote. Write down its equation and interpret it in context.

(2)

Show mark scheme

Working	Mark	Notes
(a)
mass of salt $=200t$ g and volume $=200+5t$ L	M1
$C=\dfrac{200t}{200+5t}$	A1
$C(t)=\dfrac{40t}{t+40}$	AG	Dividing numerator and denominator by 5.
(b)
$13.3$ g L⁻¹	A1	$\tfrac{40}{3}=13.33\ldots$
(c)
$40t=30(t+40)$	(M1)
$t=120$ minutes	A1
(d)
$0\le t\le160$	A1	From $200+5t=1000$.
$C(160)=32$ g L⁻¹	A1
(e)
$C=40$	A1
the concentration approaches (but never reaches) that of the incoming salt water, 40 g L⁻¹	R1

Answers: (b) 13.3 g L⁻¹ · (c) 120 min · (d) $0\le t\le160$; 32 g L⁻¹ · (e) $C=40$

+ Add to test✓ In your test78AASL2-q00702.9 Exponential and logarithmic functions Paper 2Medium8 marks

A radioactive sample initially has a mass of 80 mg. Its mass, $m$ mg, after $t$ years is modelled by $m(t)=80\mathrm{e}^{-kt}$, where $k$ is a positive constant. The half-life of the substance is 5.3 years.

(a)

Show that $k=0.131$, correct to three significant figures.

(2)

(b)

Find the mass of the sample after 12 years.

(2)

(c)

Find the mass of the substance that decays during the third year.

(2)

(d)

Find the time it takes for the mass of the sample to fall below 1 mg.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$80\mathrm{e}^{-5.3k}=40$, so $\mathrm{e}^{-5.3k}=\dfrac{1}{2}$	M1
$k=\dfrac{\ln2}{5.3}=0.13078\ldots$	A1	Must see a value to more than 3 s.f., or the exact form.
$k=0.131$	AG
(b)
$80\mathrm{e}^{-12k}$	(M1)
$16.7$ mg	A1	$16.653\ldots$ Accept 16.6 from $k=0.131$.
(c)
$m(2)-m(3)$	(M1)	Award M0 for $m(3)-m(4)$ or for $80-m(3)$.
$7.55$ mg	A1	$7.5501\ldots$ Accept 7.56 from $k=0.131$.
(d)
$80\mathrm{e}^{-kt}=1$, or $t=\dfrac{\ln80}{k}$	(M1)
$33.5$ years	A1	$33.506\ldots$ Accept 33.4–33.6.

Answers: (b) 16.7 mg · (c) 7.55 mg · (d) 33.5 years

+ Add to test✓ In your test79AASL2-q00712.9 Exponential and logarithmic functions Paper 2Medium8 marks

The number of birds, $N$, in a protected wetland $t$ years after the start of a conservation programme is modelled by $N(t)=A\mathrm{e}^{kt}$, where $A$ and $k$ are positive constants. After 2 years there are 3000 birds and after 6 years there are 4800 birds.

(a)

Show that $k=\dfrac{1}{4}\ln1.6$.

(2)

(b)

Find the value of $A$.

(2)

(c)

The aim of the programme is to have at least 6000 birds after 8 years. Determine whether the model predicts that this aim is met.

(2)

(d)

Find the value of $t$ when the number of birds first reaches 10 000.

(2)

Show mark scheme

Working	Mark	Notes
(a)
dividing: $\dfrac{A\mathrm{e}^{6k}}{A\mathrm{e}^{2k}}=\dfrac{4800}{3000}$	M1
$\mathrm{e}^{4k}=1.6$, so $4k=\ln1.6$	A1
$k=\dfrac{1}{4}\ln1.6$	AG
(b)
$A\mathrm{e}^{2k}=3000$, $A=\dfrac{3000}{\mathrm{e}^{0.5\ln1.6}}=\dfrac{3000}{\sqrt{1.6}}$	(M1)
$A=2372$	A1	$2371.70\ldots$ Accept 2370.
(c)
$N(8)=A\mathrm{e}^{8k}$	(M1)	Or $4800\times1.6^{0.5}$.
$N(8)=6072\gt 6000$, so the aim is met	A1	$6071.57\ldots$ Accept 6070. Conclusion required.
(d)
$A\mathrm{e}^{kt}=10\,000$	(M1)
$t=12.2$	A1	$12.246\ldots$ Accept 12.2–12.3.

Answers: (b) $A=2372$ · (c) $N(8)=6072$, aim met · (d) $t=12.2$

+ Add to test✓ In your test80AASL2-q00782.10 Solving equations Paper 2Medium7 marks

(a)

Solve $x^{4}-5x+1=0$.

(2)

(b)

Solve $\ln x=3-x$.

(2)

(c)

Find the set of values of $k$ for which the equation $x^{4}-5x+k=0$ has no real solutions.

(3)

Show mark scheme

Working	Mark	Notes
(a)
graph of $y=x^{4}-5x+1$ and its zeros, or GDC solver	(M1)
$x=0.200$ or $x=1.64$	A1	Both values required.
(b)
intersection of $y=\ln x$ and $y=3-x$	(M1)
$x=2.21$	A1	$2.2079\ldots$
(c)
rearranging to $x^{4}-5x=-k$ and considering the graph of $y=x^{4}-5x$	M1	Or considering the minimum of $y=x^{4}-5x+k$.
minimum value of $x^{4}-5x$ is $-4.04$ (at $x=1.08$)	A1	$-4.0395\ldots$
$k\gt 4.04$	A1

Answers: (a) $x=0.200,\ 1.64$ · (b) $x=2.21$ · (c) $k\gt 4.04$

+ Add to test✓ In your test81AASL2-q00792.10 Solving equations Paper 2Medium6 marks

Consider the equation $\ln(x+2)=x^{2}-1$, for $x\gt -2$.

(a)

Show that $x=-1$ is a solution of the equation.

(1)

(b)

Find the other solution.

(2)

(c)

Let $k(x)=\ln(x+2)-x^{2}+1$. Find the maximum value of $k(x)$ and the value of $x$ at which it occurs.

(2)

(d)

Hence write down the set of values of $x$ for which $\ln(x+2)\gt x^{2}-1$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$\ln1=0$ and $(-1)^{2}-1=0$	A1	Both sides must be evaluated.
(b)
intersection of $y=\ln(x+2)$ and $y=x^{2}-1$, or zeros of $k$	(M1)
$x=1.50$	A1	$1.50101\ldots$
(c)
maximum value $1.75$	A1	$1.74913\ldots$
at $x=0.225$	A1	$0.224744\ldots$
(d)
$-1\lt x\lt 1.50$	A1	FT from (b).

Answers: (b) $x=1.50$ · (c) 1.75 at $x=0.225$ · (d) $-1\lt x\lt 1.50$

+ Add to test✓ In your test82AASL2-q00802.10 Solving equations Paper 2Medium7 marks

Car A is bought for $30 000 and its value decreases by 20% each year. Car B is bought for $24 000 at the same time and its value decreases by 15% each year. The value of each car after $t$ years is modelled by a continuous function of $t$.

(a)

Find the value of car A after 3 years.

(2)

(b)

Show that the two cars have the same value when $\left(\dfrac{16}{17}\right)^{t}=0.8$.

(2)

(c)

Hence find the value of $t$ when the two cars have the same value.

(1)

(d)

Find the number of complete years after which the value of car B is first less than one third of its purchase price.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$30\,000\times0.8^{3}$	(M1)
$15 360	A1
(b)
$30\,000(0.8)^{t}=24\,000(0.85)^{t}$	M1
$\left(\dfrac{0.8}{0.85}\right)^{t}=\dfrac{24\,000}{30\,000}$	A1
$\left(\dfrac{16}{17}\right)^{t}=0.8$	AG
(c)
$t=3.68$	A1	$3.6807\ldots$ $\left(=\dfrac{\ln0.8}{\ln(16/17)}\right)$
(d)
$0.85^{t}\lt \dfrac{1}{3}$, $t\gt 6.76$; or $24\,000(0.85)^{6}=9051.59$, $24\,000(0.85)^{7}=7693.85$	(M1)
7 years	A1

Answers: (a) $15 360 · (c) $t=3.68$ · (d) 7 years

+ Add to test✓ In your test83AASL2-q00812.10 Solving equations Paper 2Medium10 marks

Two students start a typing course at the same time. Student A’s typing speed, in words per minute, $t$ weeks after the start of the course is modelled by $W(t)=20+12\ln(t+1)$. Student B’s typing speed is modelled by $V(t)=60-45\mathrm{e}^{-0.15t}$. Both models apply for $t\ge0$.

(a)

Write down the typing speed of each student at the start of the course.

(1)

(b)

Find the typing speed of student A after 10 weeks.

(1)

(c)

Find the exact time at which student A’s typing speed reaches 50 words per minute.

(2)

(d)

Write down the equation of the horizontal asymptote of the graph of $V$, and interpret it in context.

(2)

(e)

Find the set of values of $t$ for which student B types faster than student A.

(3)

(f)

Suggest one reason why the model for student A is not appropriate for large values of $t$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
A: 20 and B: 15 words per minute	A1	Both values required.
(b)
$48.8$ words per minute	A1	$48.774\ldots$
(c)
$12\ln(t+1)=30$, so $\ln(t+1)=2.5$	(M1)
$t=\mathrm{e}^{2.5}-1$ weeks	A1	$(=11.2)$. Award A0 for 11.2 only (exact answer required).
(d)
$V=60$	A1
student B’s speed approaches (but never reaches) 60 words per minute	R1
(e)
intersections of the graphs of $W$ and $V$	(M1)
$t=7.89$ and $t=24.5$	A1	$7.8877\ldots$, $24.4838\ldots$
$7.89\lt t\lt 24.5$	A1
(f)
$W(t)\to\infty$ as $t\to\infty$, but a typing speed cannot increase without limit	R1	Accept any sensible equivalent.

Answers: (a) 20, 15 · (b) 48.8 · (c) $\mathrm{e}^{2.5}-1$ weeks · (d) $V=60$ · (e) $7.89\lt t\lt 24.5$

+ Add to test✓ In your test84AASL2-q00212.3 Graphs of functions Paper 2Hard9 marks

Let $s(x)=\tfrac{1}{2}x^{2}+4\mathrm{e}^{-x^{2}}$, for $-3\le x\le3$.

(a)

Sketch the graph of $y=s(x)$, for $-3\le x\le3$.

(3)

(b)

Show that the graph is symmetrical about the $y$-axis.

(1)

(c)

Find the coordinates of the two minimum points of the graph.

(2)

(d)

Find the set of values of $k$ for which the equation $s(x)=k$ has exactly four solutions.

(2)

(e)

Write down the number of solutions of $s(x)=4$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
local maximum at $(0,\,4)$	A1
two minimum points, symmetrically placed either side of the $y$-axis	A1
endpoints approximately $(\pm3,\,4.50)$, above the level of the maximum	A1
(b)
$s(-x)=\tfrac12(-x)^{2}+4\mathrm{e}^{-(-x)^{2}}=\tfrac12x^{2}+4\mathrm{e}^{-x^{2}}=s(x)$	A1	An algebraic argument is required; a sketch alone is not sufficient for “show”.
(c)
$x=\pm1.44$	A1	$\pm1.44202\ldots$ (exactly $\pm\sqrt{\ln8}$)
$y=1.54$	A1	$1.53972\ldots$
(d)
horizontal line between the minimum and the local maximum	(M1)
$1.54\lt k\lt4$	A1	FT from (c).
(e)
$3$	A1

Answers: (c) $(\pm1.44,\ 1.54)$ · (d) $1.54\lt k\lt4$ · (e) 3

+ Add to test✓ In your test85AASL2-q00262.4 Key features and intersections Paper 2Hard7 marks

During a chemical process, the temperature of a liquid, $T\,{}^{\circ}\mathrm{C}$, $h$ minutes after the start is modelled by $T(h)=0.02h^{3}-0.75h^{2}+7.2h+8$, for $0\le h\le24$.

(a)

Find the maximum temperature of the liquid during the process, and the time at which it occurs.

(2)

(b)

Find the minimum temperature of the liquid during the process.

(2)

(c)

A cooling fan is switched off whenever the temperature is below $20\,{}^{\circ}\mathrm{C}$. Find the total length of time for which the fan is switched off.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$28.6\,{}^{\circ}\mathrm{C}$	A1	$28.6051\ldots$ The value at $h=24$ is $25.28$, which is less.
$h=6.48$ minutes	A1	$6.47920\ldots$
(b)
comparing the local minimum $T(18.5)=11.1$ with the endpoint values $T(0)=8$ and $T(24)=25.28$	(M1)	Evidence of considering the endpoints.
$8\,{}^{\circ}\mathrm{C}$ (at $h=0$)	A1	Award M0A0 for 11.1 (local minimum only).
(c)
solving $T(h)=20$ on a GDC	(M1)
$h=2.10,\ 12.44,\ 22.96$	A1	$2.10053\ldots,\ 12.4425\ldots,\ 22.9569\ldots$
$2.10+(22.96-12.44)=12.6$ minutes	A1	$12.6149\ldots$

Answers: (a) $28.6\,{}^{\circ}\mathrm{C}$ at $h=6.48$ · (b) $8\,{}^{\circ}\mathrm{C}$ · (c) 12.6 minutes

+ Add to test✓ In your test86AASL2-q00282.4 Key features and intersections Paper 2Hard14 marks

A cable hangs between the tops of two towers that stand 100 m apart on opposite banks of a river. The height of the cable above the river, $h$ metres, at a horizontal distance $x$ metres from the midpoint between the towers, is modelled by $h(x)=10\left(\mathrm{e}^{\frac{x}{40}}+\mathrm{e}^{-\frac{x}{40}}\right)-12$, for $-50\le x\le50$.

(a)

Write down the height of the cable at the midpoint between the towers.

(1)

(b)

Find the height of each tower.

(1)

(c)

Find the total horizontal distance for which the cable is less than 10 m above the river.

(2)

(d)

A straight walkway below the cable is modelled by $w(x)=4+0.04x$, for $-50\le x\le50$. Vertical hangers join the cable to the walkway. Find the length of the shortest hanger, and the value of $x$ at which it occurs.

(3)

(e)

Find the greatest vertical distance between the cable and the walkway.

(2)

(f)

Hangers longer than 5 m need extra bracing. Find the total horizontal length of walkway for which the hangers need extra bracing.

(3)

(g)

The walkway is to be replaced by a horizontal walkway, $y=c$. The shortest vertical distance between the cable and the new walkway must be 3 m. Find the value of $c$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$h(0)=8$ m	A1
(b)
$h(\pm50)=25.8$ m	A1	$25.7684\ldots$
(c)
solving $h(x)=10$, e.g. $x=\pm17.74$	(M1)	$\pm17.7427\ldots$
$35.5$ m	A1	$35.4854\ldots$
(d)
graphing $h(x)-w(x)$ and finding its minimum	(M1)
$3.94$ m	A1	$3.93603\ldots$
$x=3.20$	A1	$3.19659\ldots$ Accept 3.2.
(e)
considering the end values $h(-50)-w(-50)$ and $h(50)-w(50)$	(M1)	Evidence of checking the endpoints of the domain.
$23.8$ m (at $x=-50$)	A1	$23.7684\ldots$ Award M1A0 for 19.8 (at x = 50).
(f)
solving $h(x)-w(x)=5$	(M1)
$x=-9.83$ and $x=16.1$	A1	$-9.82892\ldots,\ 16.1116\ldots$
$(-9.83+50)+(50-16.11)=74.1$ m	A1	$74.0594\ldots$ FT from their roots.
(g)
the least value of $h$ is $8$, at $x=0$	(M1)	FT from (a).
$c=5$	A1

Answers: (a) 8 m · (b) 25.8 m · (c) 35.5 m · (d) 3.94 m at $x=3.20$ · (e) 23.8 m · (f) 74.1 m · (g) $c=5$

+ Add to test✓ In your test87AASL2-q00442.6 Quadratic functions Paper 2Hard13 marks

A workshop sells handmade lamps. When the selling price is $x$ dollars, the weekly profit, $P$ dollars, is modelled by $P(x)=ax^{2}+bx+c$. The workshop has fixed weekly costs of $200, so that $P(0)=-200$. It records a weekly profit of $150 when the price is $10 and a weekly profit of $400 when the price is $20.

(a)

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

(4)

(b)

Express $P(x)$ in the form $a(x-h)^{2}+k$.

(2)

(c)

Write down the maximum weekly profit and the price at which it occurs.

(2)

(d)

Find the range of selling prices for which the workshop makes a profit.

(3)

(e)

Find the range of selling prices for which the weekly profit is at least $450.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$c=-200$	A1
$100a+10b-200=150$ and $400a+20b-200=400$	(M1)	Substituting both points to form two equations in a and b.
$a=-0.5$	A1
$b=40$	A1
(b)
$-0.5\left(x^{2}-80x\right)-200$, or vertex at $x=-\dfrac{40}{2(-0.5)}=40$	(M1)
$P(x)=-0.5(x-40)^{2}+600$	A1
(c)
maximum profit $\$600$	A1	FT from (b).
at a price of $\$40$	A1
(d)
solving $P(x)=0$	(M1)
$x=5.36$ and $x=74.6$	A1	$40\pm\sqrt{1200}=5.3589\ldots,\ 74.641\ldots$
$5.36\lt x\lt74.6$	A1	Accept $\$5.36\lt x\lt\$74.64$.
(e)
$-0.5(x-40)^{2}+600\ge450$, i.e. $(x-40)^{2}\le300$	(M1)
$22.7\le x\le57.3$	A1	$40\pm\sqrt{300}=22.679\ldots,\ 57.320\ldots$

Answers: (a) $a=-0.5,\ b=40,\ c=-200$ · (b) $-0.5(x-40)^{2}+600$ · (c) $600 at $40 · (d) $5.36\lt x\lt74.6$ · (e) $22.7\le x\le57.3$

+ Add to test✓ In your test88AASL2-q00722.9 Exponential and logarithmic functions Paper 2Hard8 marks

Alex invests $5000 in an account in which the value after $t$ years is $A(t)=5000\mathrm{e}^{0.04t}$. Beth invests $6000 in an account that pays 3% per year, compounded annually, so the value after $t$ complete years is $B(t)=6000(1.03)^{t}$.

(a)

Find the value of Alex’s investment after 10 years. Give your answer correct to the nearest cent.

(2)

(b)

Show that Alex’s account is equivalent to an annual percentage increase of 4.08%, correct to three significant figures.

(2)

(c)

Show that $A(t)=B(t)$ when $t=\dfrac{\ln1.2}{0.04-\ln1.03}$.

(2)

(d)

Find the least number of complete years after which Alex’s investment is worth more than Beth’s.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$5000\mathrm{e}^{0.4}$	(M1)
$7459.12	A1
(b)
$\dfrac{A(t+1)}{A(t)}=\mathrm{e}^{0.04}$	M1
$\mathrm{e}^{0.04}=1.04081\ldots$	A1	Must see at least 5 s.f.
increase of 4.08% per year	AG
(c)
taking logarithms: $\ln5000+0.04t=\ln6000+t\ln1.03$	M1
$t(0.04-\ln1.03)=\ln6000-\ln5000=\ln1.2$	A1
$t=\dfrac{\ln1.2}{0.04-\ln1.03}$	AG
(d)
$t=17.46\ldots$, or comparing values: $A(17)=9869.39\lt B(17)=9917.09$, $A(18)=10\,272.17\gt B(18)=10\,214.60$	(M1)
18 years	A1

Answers: (a) $7459.12 · (d) 18 years

+ Add to test✓ In your test89AASL2-q00732.9 Exponential and logarithmic functions Paper 2Hard13 marks

Two colonies of bacteria are being studied. The size of colony A, $t$ hours after the start of the study, is modelled by $A(t)=500\mathrm{e}^{0.12t}$. The size of colony B is modelled by $B(t)=2000+150t$. Both models apply for $t\ge0$.

(a)

Write down the initial size of colony A.

(1)

(b)

Show that, according to the model, colony A increases by approximately 12.7% each hour.

(2)

(c)

Find the time taken for the size of colony A to quadruple.

(2)

(d)

Find the value of $t$ when the two colonies are the same size.

(2)

(e)

For values of $t$ between 0 and the value found in part (d), find the greatest value of $B(t)-A(t)$, and the time at which it occurs.

(3)

(f)

Find the time at which colony A is twice the size of colony B.

(2)

(g)

Suggest one reason why the model for colony A may not be valid for large values of $t$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
500	A1
(b)
$\dfrac{A(t+1)}{A(t)}=\mathrm{e}^{0.12}$	M1
$\mathrm{e}^{0.12}=1.1275\ldots$	A1	Must see a value to at least 4 s.f.
increase of approximately 12.7% per hour	AG
(c)
$\mathrm{e}^{0.12t}=4$, so $t=\dfrac{\ln4}{0.12}$	(M1)
$11.6$ hours	A1	$11.552\ldots$
(d)
intersection of $y=500\mathrm{e}^{0.12t}$ and $y=2000+150t$	(M1)
$t=18.9$	A1	$18.911\ldots$
(e)
graph of $y=2000+150t-500\mathrm{e}^{0.12t}$ and its maximum	(M1)
$1895$ ($1.90\times10^{3}$)	A1	$1894.6\ldots$ Accept 1890–1900.
at $t=7.64$ hours	A1	$7.6357\ldots$
(f)
$500\mathrm{e}^{0.12t}=2(2000+150t)$	(M1)
$t=26.4$ hours	A1	$26.435\ldots$
(g)
exponential growth would continue without limit, but in reality space and nutrients are limited	R1	Accept any reasonable limiting factor.

Answers: (a) 500 · (c) 11.6 h · (d) $t=18.9$ · (e) 1895 at $t=7.64$ · (f) 26.4 h

+ Add to test✓ In your test90AASL2-q00822.10 Solving equations Paper 2Hard14 marks

A cup of tea is left to cool in a room. Its temperature, $T\,{}^{\circ}\mathrm{C}$, $t$ minutes after it is poured is modelled by $T(t)=22+68\mathrm{e}^{-kt}$, for $t\ge0$, where $k$ is a positive constant.

(a)

Write down the temperature of the tea when it is poured.

(1)

(b)

After 10 minutes the temperature of the tea is $60\,{}^{\circ}\mathrm{C}$. Find the value of $k$.

(3)

(c)

Find the temperature of the tea after 25 minutes.

(2)

(d)

Find the time taken for the temperature of the tea to fall to $40\,{}^{\circ}\mathrm{C}$.

(2)

(e)

Write down the equation of the horizontal asymptote of the graph of $T$, and interpret it in context.

(2)

(f)

At $t=0$, a glass of cold juice is placed in the same room. Its temperature is modelled by $U(t)=22-18\mathrm{e}^{-0.08t}$. Write down the initial temperature of the juice.

(1)

(g)

Find the time at which the temperature of the tea is $5\,{}^{\circ}\mathrm{C}$ more than the temperature of the juice.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$90\,{}^{\circ}\mathrm{C}$	A1
(b)
$22+68\mathrm{e}^{-10k}=60$	(M1)
$\mathrm{e}^{-10k}=\dfrac{38}{68}$, or GDC solver	(A1)
$k=0.0582$	A1	$0.058192\ldots$ Exact $k=\frac{1}{10}\ln\frac{68}{38}$.
(c)
$22+68\mathrm{e}^{-25k}$	(M1)
$37.9\,{}^{\circ}\mathrm{C}$	A1	$37.874\ldots$
(d)
$22+68\mathrm{e}^{-kt}=40$	(M1)
$22.8$ minutes	A1	$22.840\ldots$ Accept 22.8 from $k=0.0582$.
(e)
$T=22$	A1
the tea approaches room temperature, $22\,{}^{\circ}\mathrm{C}$, but never falls below it	R1
(f)
$4\,{}^{\circ}\mathrm{C}$	A1
(g)
$T(t)-U(t)=5$, i.e. $68\mathrm{e}^{-kt}+18\mathrm{e}^{-0.08t}=5$	M1
graphing $y=T(t)-U(t)$ and $y=5$, or GDC solver	(M1)	Must use technology.
$46.4$ minutes	A1	$46.430\ldots$ (46.42 with $k=0.0582$).

Answers: (a) 90 °C · (b) $k=0.0582$ · (c) 37.9 °C · (d) 22.8 min · (e) $T=22$ · (f) 4 °C · (g) 46.4 min

+ Add to test✓ In your test91AASL2-q00832.10 Solving equations Paper 2Hard15 marks

The concentration of a drug in a patient’s blood, $C$ mg L⁻¹, $t$ hours after a single dose is taken is modelled by $C(t)=20\left(\mathrm{e}^{-0.2t}-\mathrm{e}^{-0.9t}\right)$, for $t\ge0$.

(a)

Write down the value of $C(0)$.

(1)

(b)

Find the maximum concentration and the time at which it occurs.

(3)

(c)

The drug is effective when the concentration is at least 8 mg L⁻¹. Find the length of time for which a single dose is effective.

(3)

(d)

Find the time after which the concentration remains below 0.5 mg L⁻¹.

(2)

(e)(i)

A second, identical dose is taken at $t=6$. For $t\ge6$, the concentration is modelled by $D(t)=C(t)+C(t-6)$.

Describe the transformation that maps the graph of $y=C(t)$ onto the graph of $y=C(t-6)$.

(2)

(e)(ii)

Find $D(6)$.

(1)

(f)

Find the maximum value of $D(t)$ for $t\ge6$, and the time at which it occurs.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$0$	A1
(b)
maximum on the graph of $C$	(M1)
$10.1$ mg L⁻¹	A1	$10.1216\ldots$
$t=2.15$ hours	A1	$2.14868\ldots$
(c)
intersections of $y=C(t)$ and $y=8$	(M1)
$t=0.942$ and $t=4.33$	A1	$0.94233\ldots$, $4.33498\ldots$
$3.39$ hours	A1	$3.39264\ldots$
(d)
$C(t)=0.5$ on the decreasing part of the graph	(M1)
$t=18.4$ hours	A1	$18.4443\ldots$
(e)(i)
translation	A1
6 units in the positive $t$-direction, $\begin{pmatrix}6\\0\end{pmatrix}$	A1
(e)(ii)
$D(6)=C(6)+C(0)=5.93$ mg L⁻¹	A1	$5.93355\ldots$
(f)
maximum on the graph of $y=D(t)$ for $t\ge6$	(M1)
$14.2$ mg L⁻¹	A1	$14.1810\ldots$
$t=7.78$ hours	A1	$7.77900\ldots$

Answers: (a) 0 · (b) 10.1 mg L⁻¹ at 2.15 h · (c) 3.39 h · (d) 18.4 h · (e)(i) translation by $\binom{6}{0}$; (ii) 5.93 mg L⁻¹ · (f) 14.2 mg L⁻¹ at 7.78 h

+ Add to test✓ In your test92AASL2-q00922.11 Transformations of graphs Paper 2Hard13 marks

The weekly demand, $D$ units, for a product sold at a price of $p$ dollars is modelled by $D(p)=800\mathrm{e}^{-0.05p}$, for $0\le p\le60$.

(a)

Find the demand when the price is $10.

(1)

(b)

Find the price at which the weekly demand is 300 units.

(2)

(c)(i)

After an advertising campaign, the graph of the demand function is translated 4 units in the positive $p$-direction. The new demand is $E(p)=D(p-4)$.

Show that $E(p)=800\mathrm{e}^{0.2}\mathrm{e}^{-0.05p}$.

(1)

(c)(ii)

Hence describe the same change as a single vertical stretch, giving the scale factor to three significant figures.

(2)

(d)

Without solving a further equation, find the price at which the weekly demand after the campaign is 300 units. Justify your answer.

(2)

(e)

The weekly revenue after the campaign is $R(p)=pE(p)$. Find the price that maximizes the revenue and the maximum revenue.

(3)

(f)

Find the horizontal translation of the original demand graph that would be equivalent to a vertical stretch with scale factor 1.6.

(2)

Show mark scheme

Working	Mark	Notes
(a)
485 units	A1	$485.22\ldots$
(b)
$800\mathrm{e}^{-0.05p}=300$	(M1)
$p=\$19.62$	A1	$19.616\ldots\ \left(=20\ln\tfrac{8}{3}\right)$. Accept 19.6.
(c)(i)
$800\mathrm{e}^{-0.05(p-4)}=800\mathrm{e}^{0.2-0.05p}$	M1
$=800\mathrm{e}^{0.2}\mathrm{e}^{-0.05p}$	AG
(c)(ii)
vertical stretch	A1
scale factor $\mathrm{e}^{0.2}=1.22$	A1	$1.2214\ldots$
(d)
every point on the graph of $D$ moves 4 units in the positive $p$-direction, so the demand $E(p)=300$ occurs at a price $4$ dollars higher	R1
$p=\$23.62$	A1	FT from (b). Accept 23.6.
(e)
maximum on the graph of $y=800p\,\mathrm{e}^{0.2}\mathrm{e}^{-0.05p}$	(M1)
$p=\$20$	A1
$\$7189$	A1	$7189.26\ldots$ $(=16\,000\mathrm{e}^{-0.8})$. Accept 7190.
(f)
$\mathrm{e}^{0.05a}=1.6$	M1	Translation of $a$ units in the positive $p$-direction.
$a=20\ln1.6=9.40$ units in the positive $p$-direction	A1	$9.4000\ldots$

Answers: (a) 485 · (b) $19.62 · (c)(ii) vertical stretch, scale factor 1.22 · (d) $23.62 · (e) $20, $7189 · (f) 9.40 to the right

Working	Mark	Notes
(a)
\(\dfrac{-1-7}{3-(-1)}\)	(M1)
\(-2\)	A1
(b)
\(y-7=-2(x+1)\) or \(y=-2x+5\)	(M1)	FT from their gradient.
\(2x+y-5=0\)	A1	Accept any integer multiple, e.g. \(-2x-y+5=0\).
(c)
gradient \(\tfrac{1}{2}\) and \(y-4=\tfrac{1}{2}(x-6)\)	(M1)	FT from (a): gradient must be \(-\dfrac{1}{m_{1}}\).
\(y=\tfrac{1}{2}x+1\)	A1
(d)
\(2x+\left(\tfrac{1}{2}x+1\right)-5=0\), i.e. \(\tfrac{5}{2}x=4\)	(M1)	Attempt to solve their equations simultaneously.
\(\left(\dfrac{8}{5},\,\dfrac{9}{5}\right)\)	A1	Accept \((1.6,\,1.8)\).

Working	Mark	Notes
(a)
\(4y=3x+12\)	(M1)	Rearranging to make y the subject.
\(y=\tfrac{3}{4}x+3\)	A1
(b)
\(x\)-axis: \((-4,\,0)\)	A1
\(y\)-axis: \((0,\,3)\)	A1	Award A1A0 if both are given as single numbers −4 and 3 without coordinates.
(c)
\(y+5=\tfrac{3}{4}(x-2)\)	A1	FT from their gradient in (a). Accept \(y-(-5)=\tfrac34(x-2)\).
(d)
\(3x-4y-26=0\)	A1	Accept any integer multiple, e.g. \(-3x+4y+26=0\).

Working	Mark	Notes
(a)
\(m=\dfrac{15-21}{5-2}\)	(M1)
\(m=-2,\ c=25\)	A1	\(21=-2(2)+c\).
(b)
\(m\): the candle’s height decreases by 2 cm every hour	A1	Must refer to rate and units.
\(c\): the height of the candle when it is lit is 25 cm	A1
(c)
\(25-2t=18-1.2t\)	(M1)	FT from (a).
\(t=8.75\) hours, height \(7.5\) cm	A1	Both values required.
(d)
\(2t+h-25=0\)	A1	Accept any integer multiple.

Working	Mark	Notes
(a)
\(6-2x\ge0\)	(M1)
\(x\le3\)	A1
(b)
\(f(x)\ge1\)	A1	Accept \(y\ge1\) or \([1,\infty)\).
(c)
\(\sqrt{6-2x}=3\), so \(6-2x=9\)	(M1)
\(x=-\dfrac{3}{2}\)	A1
(d)
\(-\dfrac{3}{2}\)	A1	FT from (c).
(e)
\(0\) is not in the range of \(f\) (as \(f(x)\ge1\)), so it is not in the domain of \(f^{-1}\)	R1

Working	Mark	Notes
(a)
\(11\)	A1
(b)
end values \(-1\) and \(11\)	A1
\(-1\lt f(x)\le11\)	A1	Award A1A0 for \(-1\le f(x)\le11\) (the value \(-1\) is not attained since \(x\lt4\)).
(c)
\(x=3.5\)	A1
(d)
\(7-2x=5\Rightarrow f^{-1}(5)=1\)	A1

Working	Mark	Notes
(a)
\(98.6\)	A1
(b)
\(1.8C+32=212\)	(M1)
\(100\)	A1
(c)
\(x=1.8y+32\) or \(y=1.8x+32\) rearranged for \(x\)	(M1)	Interchanging x and y at any stage.
\(F^{-1}(x)=\dfrac{x-32}{1.8}\)	A1	Accept \(\dfrac{5(x-32)}{9}\).
(d)
\(1.8C+32=C\)	(M1)
\(-40\) (degrees)	A1

Working	Mark	Notes
(a)(i)
function: each \(x\) gives exactly one value of \(y\)	A1
(a)(ii)
not a function: e.g. \(x=4\) gives \(y=2\) and \(y=-2\)	A1	Must include a reason.
(a)(iii)
function: each first element appears exactly once	A1
(a)(iv)
not a function: \(x=2\) is paired with both 1 and 3	A1	Must include a reason.
(b)
domain \(\{1,\,2,\,3\}\)	A1
range \(\{2,\,3\}\)	A1	Do not accept {2, 3, 2}.
(c)
not one-to-one: the output 2 comes from both \(x=1\) and \(x=3\)	R1

Working	Mark	Notes
(a)
\(p=\tfrac12(-1+4)=1.5\)	A1
\(q=\tfrac12(3.375-6)=-1.3125\)	A1
(b)
all nine points plotted correctly (within half a small square)	A1	FT their p and q.
smooth curve (not line segments) through the points	A1
curve drawn only for \(-2\le x\le2\), with turning points near \((-1.15,\,1.54)\) and \((1.15,\,-1.54)\)	A1	“Draw” requires an accurate curve on the grid.
(c)
rotational symmetry of order 2 about the origin	A1	Accept “symmetric under a half-turn about O” or “f(−x) = −f(x)”.
(d)
rewriting as \(f(x)=-1\) and drawing the line \(y=-1\)	(M1)
\(x\approx0.54\) and \(x\approx1.68\)	A1	Accept answers in the ranges \(0.50\) to \(0.60\) and \(1.60\) to \(1.75\). The third root, \(-2.21\), is outside the domain.

Working	Mark	Notes
(a)
\(g(2)=6\), then \(f(6)\)	(M1)
\(17\)	A1
(b)
\((3x-1)^{2}+2\)	(M1)
\(9x^{2}-6x+3\)	A1
(c)
\(3(x^{2}+2)-1=17\), so \(x^{2}=4\)	(M1)
\(x=\pm2\)	A1	Both values required.
(d)
\(f^{-1}(x)=\dfrac{x+1}{3}\)	A1
\(3\left(\dfrac{x+1}{3}\right)-1=x\)	A1	Must show the substitution.

Working	Mark	Notes
(a)
\(g(2)=3\)	(M1)
\(f(3)=1\)	A1
(b)
\(g(f(4))=g(2)=3\)	A1
(c)
\(2\)	A1
(d)
\(g(f(x))=5\Rightarrow f(x)=3\)	(M1)
\(x=1\)	A1

\(x\)	1	2	3	4	5
\(f(x)\)	3	5	1	2	4
\(g(x)\)	4	3	5	1	2

Working	Mark	Notes
(a)
\(h=3,\ k=-2\)	A1
\(a(1-3)^{2}-2=6\)	(M1)
\(a=2\)	A1
(b)
\(f(x)=2x^{2}-12x+16\)	A1	FT from (a).
\((0,\,16)\)	A1	Accept 16.
(c)
\(f(x)=2(x-2)(x-4)\)	A1
\((2,\,0)\) and \((4,\,0)\)	A1
(d)
\(x=3\)	A1

Working	Mark	Notes
(a)
\((-1,\,0)\) and \((5,\,0)\)	A1	Accept −1 and 5.
(b)
\(x=2\)	A1
(c)
\(f(2)=-2(3)(-3)\)	(M1)
\((2,\,18)\)	A1
(d)
\(10\)	A1	Accept \((0,\,10)\).
(e)
\(-2(x-2)^{2}+18\)	A1	FT from (c).
(f)
\(-2x^{2}+8x+10\)	A1

Working	Mark	Notes
(a)
\((2x+1)(x-3)=0\)	(M1)
\(x=-\dfrac{1}{2}\) or \(x=3\)	A1
(b)
\((x+3)^{2}-9+4=0\)	M1
\((x+3)^{2}=5\)	A1
\(x=-3\pm\sqrt{5}\)	A1
(c)
recognizing the parabola opens upwards, so the expression is non-positive between the roots	(M1)	A sketch or sign diagram is acceptable.
\(-\dfrac{1}{2}\le x\le3\)	A1	Award A0 for strict inequalities.

Working	Mark	Notes
(a)
\(x^{2}-2x-15\gt 0\)	M1	Rearranging to compare with zero. Award M0 for \(x\gt 15\) or \(x-2\gt 15\).
\((x-5)(x+3)\gt 0\), critical values \(-3\) and \(5\)	A1
\(x\lt -3\) or \(x\gt 5\)	A1	Award A0 for \(-3\lt x\lt 5\) or for non-strict inequalities.
(b)
\((2x-1)(x+2)\lt 0\), critical values \(-2\) and \(\tfrac{1}{2}\)	(M1)	A sketch or sign diagram may be seen.
\(-2\lt x\lt \dfrac{1}{2}\)	A1
(c)
\(-1,\ 0\)	A1	FT from their (b). Both values and no others.

Working	Mark	Notes
(a)
\(x=\dfrac{4\pm\sqrt{(-4)^{2}-4(3)(-2)}}{2(3)}\)	(M1)	Correct substitution into the quadratic formula.
\(x=\dfrac{4\pm\sqrt{40}}{6}\)	A1
\(x=\dfrac{2\pm\sqrt{10}}{3}\)	A1	Accept \(\dfrac{4\pm2\sqrt{10}}{6}\).
(b)
recognizing \(x-1=\dfrac{2\pm\sqrt{10}}{3}\)	M1	Must use (a). Award M0 for expanding and starting again (not ‘hence’).
\(x=\dfrac{5\pm\sqrt{10}}{3}\)	A1
(c)
\(\Delta=(-4)^{2}-4(3)(2)=-8\)	M1
\(\Delta\lt 0\), so there are no real roots	A1

Working	Mark	Notes
(a)
\(u^{2}-5u+4=0\), where \(u=x^{2}\)	M1
\(x^{2}=1\) or \(x^{2}=4\)	A1
\(x=\pm1,\ \pm2\)	A1	All four values required.
(b)
\(u^{2}-5u+6=0\), where \(u=\sqrt{x}\)	M1
\(\sqrt{x}=2\) or \(\sqrt{x}=3\)	A1
\(x=4\) or \(x=9\)	A1
(c)
\(2x^{2}+x-15=0\)	M1
\(x=\dfrac{5}{2}\) or \(x=-3\)	A1	From \((2x-5)(x+3)=0\).

Working	Mark	Notes
(a)
\(f^{-1}(x)=\dfrac{1}{x}\) (the function is self-inverse)	A1
(b)
translation	A1	Accept horizontal and vertical translations described separately.
by \(\begin{pmatrix}2\\3\end{pmatrix}\): 2 units right and 3 units up	A1
(c)
\(x=2\)	A1
\(y=3\)	A1
(d)
\(\left(0,\,\dfrac{5}{2}\right)\)	A1
\(\left(\dfrac{5}{3},\,0\right)\)	A1	From \(\frac{1}{x-2}=-3\).

Working	Mark	Notes
(a)
\(f^{-1}(x)=\dfrac{12}{x}\)	A1
(b)
\(x^{2}=12\)	(M1)
\(\left(2\sqrt{3},\,2\sqrt{3}\right)\) and \(\left(-2\sqrt{3},\,-2\sqrt{3}\right)\)	A1	Accept \(\sqrt{12}\).
(c)
\(\dfrac{12}{x}=8-x\), i.e. \(x^{2}-8x+12=0\)	M1
\(x=2\) or \(x=6\)	A1
\((2,\,6)\) and \((6,\,2)\)	A1
(d)
each is the reflection of the other in \(y=x\); since \(f\) is self-inverse its graph is symmetric about \(y=x\)	R1

Working	Mark	Notes
(a)
\((0,\,2)\)	A1	Accept 2.
(b)
\(y=4\)	A1
(c)
\(2\mathrm{e}^{x}=4\), so \(\mathrm{e}^{x}=2\)	(M1)
\(x=\ln2\)	A1	Accept \((\ln2,\,0)\).
(d)
\(f(x)\lt 4\)	A1	Award A0 for \(f(x)\le4\).
(e)
\(x=1\)	A1
(f)
\(\ln(x-1)=-2\)	(M1)
\(x=1+\mathrm{e}^{-2}\)	A1

Working	Mark	Notes
(a)
\(\left(\tfrac{1}{2}\right)^{x}=2^{-x}=f(-x)\)	A1
reflection in the \(y\)-axis	A1
(b)
reflection in the line \(y=x\)	A1	They are inverse functions.
(c)
\(2^{x}=8\times2^{-x}=2^{3-x}\)	M1
\(x=\dfrac{3}{2}\)	A1
\(\mathrm{P}\left(\dfrac{3}{2},\,2\sqrt{2}\right)\)	A1	Accept \(2^{3/2}\).
(d)
\(\left(2\sqrt{2},\,\dfrac{3}{2}\right)\)	A1	The reflection of P in \(y=x\); or solve \(\log_{2}x=3-\log_{2}x\).

Working	Mark	Notes
(a)
squaring both sides: \(x+7=(x-5)^{2}\)	M1
\(x+7=x^{2}-10x+25\)	A1
\(x^{2}-11x+18=0\)	AG
(b)
\((x-2)(x-9)=0\), so \(x=2\) or \(x=9\)	A1
\(x=2\) is rejected, since \(\sqrt{9}=3\neq-3\); so \(x=9\) only	R1	Award R0 if both values are given as the final answer.
(c)
the graph of \(y=\sqrt{x+7}\) lies above the line for \(x\) less than the intersection at \(x=9\)	(M1)	A sketch may be seen.
\(-7\le x\lt 9\)	A1	FT from (b). Award A0 for \(x\lt 9\) (domain omitted).

Working	Mark	Notes
(a)(i)
\((4,\,-1)\)	A1
(a)(ii)
\((7,\,-6)\)	A1
(a)(iii)
\((4,\,6)\)	A1
(a)(iv)
\((8,\,-6)\)	A1	Horizontal stretch with scale factor 2.
(a)(v)
\((-4,\,-18)\)	A1	Both the reflection in the \(y\)-axis and the vertical stretch must be applied.
(b)
vertical stretch, scale factor 2	A1
reflection in the \(x\)-axis	A1
followed by a translation of 4 units up, \(\begin{pmatrix}0\\4\end{pmatrix}\)	A1	The translation must come after the stretch and reflection. Stretch and reflection may be combined as a stretch with scale factor −2.

Working	Mark	Notes
(a)(i)
\((1,\,2)\)	A1
\(y=0\)	A1
(a)(ii)
\((1,\,-4)\)	A1	The image is now a local minimum.
\(y=-2\)	A1
(a)(iii)
\((2,\,4)\)	A1	Horizontal stretch with scale factor 2.
\(y=2\)	A1
(a)(iv)
\((-1,\,5)\)	A1
\(y=3\)	A1

Working	Mark	Notes
(a)
stretch: \(y=\sqrt{4x}\)	M1
reflection then translation: \(y=3-\sqrt{4x}\)	A1
\(\sqrt{4x}=2\sqrt{x}\), so \(g(x)=3-2\sqrt{x}\)	AG
(b)
domain \(x\ge0\)	A1
range \(g(x)\le3\)	A1
(c)
\(2\sqrt{x}=3\)	(M1)
\(\left(\dfrac{9}{4},\,0\right)\)	A1	Accept \(x=\tfrac{9}{4}\).
(d)
2	A1	Since \(\sqrt{4x}=2\sqrt{x}\).

Working	Mark	Notes
(a)
\((1,\,-1)\)	A1
(b)
\(\dfrac{-4-2}{5-(-3)}=-\dfrac{3}{4}\)	A1
(c)
perpendicular gradient \(\dfrac{4}{3}\)	(M1)	FT: \(-\dfrac{1}{m}\) for their \(m\) in (b).
\(y+1=\tfrac{4}{3}(x-1)\)	(M1)	Substituting their midpoint into a line equation with their perpendicular gradient.
\(4x-3y-7=0\)	A1	Accept any integer multiple.
(d)
\(-\dfrac{7}{3}\)	A1	Accept \(\left(0,\,-\tfrac73\right)\).
(e)
\(4(4)-3(3)-7=16-9-7=0\)	A1
every point on the perpendicular bisector of \([\mathrm{PQ}]\) is the same distance from P and from Q	R1	Accept a calculation showing \(\mathrm{RP}=\mathrm{RQ}=\sqrt{50}\).

Working	Mark	Notes
(a)
\(1\)	A1
(b)
\(0\)	A1
(c)
\(f\) is one-to-one (each \(y\)-value comes from exactly one \(x\)-value; the graph is increasing)	R1	Accept a correct horizontal-line-test argument.
(d)(i)
\(-2.75\le x\le5\)	A1
(d)(ii)
\(-2\le y\le3\)	A1	Accept \(-2\le f^{-1}(x)\le3\).
(e)
attempt to reflect the graph of \(f\) in the line \(y=x\)	M1
curve passes through \((-2,\,0)\) and \((1,\,2)\), increasing and concave down	A1
endpoints at \((-2.75,\,-2)\) and \((5,\,3)\)	A1

Working	Mark	Notes
(a)
\(-3\)	A1
(b)
end values \(-3\) and \(3\)	A1
\(-3\le f(x)\le3\)	A1
(c)
drawing \(y=1\), or using the middle segment \(y=2x+1\)	(M1)
\(x=-4,\ x=0,\ x=5\)	A1	All three values required.
(d)
\(f\) is not one-to-one, e.g. \(f(-4)=f(0)=f(5)=1\)	R1	Accept a horizontal-line-test argument.
(e)(i)
\(-3\le x\le3\)	A1	The range of g.
(e)(ii)
\(-2\le g^{-1}(x)\le1\)	A1	Accept \(-2\le y\le1\).
(f)
\(2x+1=2\Rightarrow g^{-1}(2)=0.5\)	A1

Working	Mark	Notes
(a)
\(x^{2}-4x+4-4+1\)	(M1)	Attempt to complete the square.
\((x-2)^{2}-3\)	A1
(b)
least value \(g(2)=-3\) (the vertex, not an end value) and greatest value \(g(5)=6\)	A1	\(g(0)=1\) is not the least value.
\(-3\le g(x)\le6\)	A1	Award A1A0 for \(1\le g(x)\le6\) (end values only).
(c)
\(g\) is not one-to-one, e.g. \(g(0)=g(4)=1\)	R1	Accept a horizontal-line-test argument.
(d)
\(a=2\)	A1	FT from the vertex in (a).
(e)
\((x-2)^{2}=9\Rightarrow x=-1\) or \(x=5\)	(M1)	Or \(x^{2}-4x-5=0\Rightarrow(x+1)(x-5)=0\).
\(x=5\) only	A1	\(x=-1\) must be rejected as it is not in the domain. Award M1A0 if \(x=-1\) is also given.

Working	Mark	Notes
(a)
smooth curve decreasing from \((-3,\,4)\) to the minimum \((1,\,-2)\), then increasing to \((5,\,6)\)	A1
crossing the \(x\)-axis at \(-1\) and \(3\) only	A1
passing through \((0,\,-1.5)\), with all five points labelled and endpoints shown	A1
(b)
\(-2\le f(x)\le6\)	A1
(c)
considering horizontal lines \(y=k\) meeting both the decreasing and the increasing parts	(M1)
\(-2\lt k\le4\)	A1	Award M1A0 for \(-2\lt k\lt4\) or \(-2\le k\le4\).

Working	Mark	Notes
(a)
\(2(x^{2}-x+2)-2\)	(M1)
\(2x^{2}-2x+2\)	A1
(b)
\((2x-2)^{2}-(2x-2)+2\)	(M1)
\(4x^{2}-10x+8\)	A1
(c)
\(2x^{2}-8x+6=0\Rightarrow x^{2}-4x+3=0\)	(M1)	FT from (a) and (b).
\(x=1,\ x=3\)	A1