CAIE A-Level Mathematics AS 1.2 Functions Questions

Practise domains and ranges, composite and inverse functions and graph transformations while checking restrictions, operation order and exact function notation.

Exam points

state domains and ranges with the correct variable, inequality and included endpoint
form fg or gf in the stated order and verify that the inner range fits the outer domain
find an inverse by restricting one-to-one branches and solve transformation or composite equations

Covered syllabus points

1.2.1—Function terminology 1.2.2—Range and composition 1.2.3—One-one and inverse functions 1.2.4—One-one and inverse functions 1.2.5—Graph transformations

[Maximum number: 8]

Functions f and g are defined by

\begin{aligned} & \mathrm{f}(x)=(x+a)^{2}-a \text { for } x \leqslant-a, \\ & \mathrm{~g}(x)=2 x-1 \text { for } x \in \mathbb{R}, \end{aligned}

where a is a positive constant.

(a)

Find an expression for $\mathrm{f}^{-1}(x)$ .

[ 3 ]

$y=(x+a)^{2}-a$ leading to $(x+a)^{2}=y \pm a$

*M1

x and y may be interchanged initially.

Marking guidance:

Allow ± errors for these method marks.

$x=[ \pm] \sqrt{y \pm a} \pm a$
Alternative method for first 2 marks of Question 8(a)

$x=(y+a)^{2}-a$ leading to $y^{2}+2 a y+a^{2}-a-x[=0]$

*M1

Allow ± errors for this method mark.

$y=\frac{-2 a \pm \sqrt{4 a^{2}-4\left(a^{2}-a-x\right)}}{2}$ $\left[y\right.$ or $\left.f^{-1}(x)=\right]-\sqrt{x+a}-a$
OE
Must choose negative root.

(b)

State the domain of the function $\mathrm{f}^{-1}$ .

[ 1 ]

$x \geqslant-a$
Marking guidance:

Ignore infinity limit if included.

(c)

State the range of the function $\mathrm{f}^{-1}$ .

[ 1 ]

y or $f^{-1}[(x)] \leqslant-a$
Marking guidance:

Ignore negative infinity limit if included.

(d)

Given that $a=\frac{7}{2}$ , solve the equation gf(x)=0.

[ 3 ]

$\left[\operatorname{gf}\left(\frac{7}{2}\right)=\right] 2\left(\left(x+\frac{7}{2}\right)^{2}-\frac{7}{2}\right)-1$ or $2 x^{2}+4\left(\frac{7}{2}\right) x+2\left(\frac{7}{2}\right)^{2}-2\left(\frac{7}{2}\right)-1[=0]$
OE

Alternatively, $[\mathrm{gf}(x)=0 \Rightarrow] \mathrm{f}(x)=\frac{1}{2}$ .

$\begin{aligned} {[x=]-\frac{7}{2} \pm 2 \text { or } \frac{-14 \pm \sqrt{14^{2}-4 \times 2 \times \frac{33}{2}}}{4}} {\left[\frac{-14 \pm \sqrt{64}}{4}\right] \text { or factorising }} \end{aligned}$
OE

Solving their three term quadratic equation as far as two solutions or correctly selecting the negative root only.

Alternatively, $\pm \sqrt{\frac{1}{2}+\frac{7}{2}}-\frac{7}{2}$ .

$[x=]-\frac{11}{2}$
If B1M0 scored then award SCB1 for the correct final answer.

[Maximum number: 7]

The function f is defined as follows:

f(x)=\sqrt{x}-1 \text { for } x>1 .

(a)

Find an expression for $\mathrm{f}^{-1}(x)$ .

[ 1 ]

$\left[\mathrm{f}^{-1}(x)=\right](x+1)^{2}$
ISW
Condone ' y= '.

(b)

Figure for Question (b) — CAIE A-Level Mathematics AS

The diagram shows the graph of y=g(x) where $\mathrm{g}(x)=\frac{1}{x^{2}+2}$ for $x \in \mathbb{R}$ .

State the range of g and explain whether $\mathrm{g}^{-1}$ exists.

[ 2 ]

$0<\mathrm{g}(x) \leqslant \frac{1}{2}$ or g(x)>0 and $\mathrm{g}(x) \leqslant \frac{1}{2}$ or $\left(0, \frac{1}{2}\right]$
Marking guidance:

Do not allow $\mathrm{g}(x)>0, \mathrm{~g}(x) \leqslant \frac{1}{2}$ .

Do not allow g(x)>0 or $\mathrm{g}(x) \leqslant \frac{1}{2}$ .

Condone g or y in place of g(x).

$\mathrm{g}^{-1}$ does not exist because it is one to many or $\mathrm{g}^{-1}$ does not exist because it is not one to one.

Or

$\mathrm{g}^{-1}$ does not exist because g is not one to one or

$\mathrm{g}^{-1}$ does not exist because g is many to one or

$\mathrm{g}^{-1}$ does not exist because g fails the horizontal line test.
$\mathrm{g}^{-1}$ can be replaced by 'It' throughout.

A correct statement followed by any further incorrect explanation can be awarded B1.

(c)

The function h is defined by $\mathrm{h}(x)=\frac{1}{x^{2}+2}$ for $x \geqslant 0$ .

Solve the equation $\mathrm{hf}(x)=\mathrm{f}\left(\frac{25}{16}\right)$ . Give your answer in the form $a+b \sqrt{c}$ , where a, b and c are
integers.

[ 4 ]

$\mathrm{f}\left(\frac{25}{16}\right)=\frac{1}{4}$
SOI

$\frac{1}{(\sqrt{x}-1)^{2}+2}=\frac{1}{4}$
Equating $\frac{1}{(\sqrt{x}-1)^{2}+2}$ , or their 'simplified'
version, to their $\mathrm{f}\left(\frac{25}{16}\right)$ .

$\left[(\sqrt{x}-1)^{2}+2=4 \text { leading to }\right]$

Or
$[x-2 \sqrt{x}+1+2=4 \text { leading to }]$

Or
$[x-1=2 \sqrt{x} \text { leading to }]$
Simplification as far as $x=\ldots$

Marking guidance:

Allow just + in the results because - can be disregarded at this stage.

Can be implied by the final answer.

Note: $x=1 \pm \sqrt{2}$ scores A0.

$3+2 \sqrt{2}$
Must discount the solution $3-2 \sqrt{2}$ .

[Maximum number: 6]

The function f is defined by $\mathrm{f}(x)=10+6 x-x^{2}$ for $x \in \mathbb{R}$ .

Figure for Question 3 — CAIE A-Level Mathematics AS

The function g is defined by g(x)=4 x+k for $x \in \mathbb{R}$ where k is a constant.

It is given that the graph of $y=\mathrm{g}^{-1} \mathrm{f}(x)$ meets the graph of y=g(x) at a single point P.

Determine the coordinates of P.

$\mathrm{g}^{-1}(x)=\frac{1}{4}(x-k)$ $g^{-1} f(x)=\frac{1}{4}\left(10+6 x-x^{2}-k\right)=4 x+k$
OE
May use their completed square form for f(x).

Simplify the quadratic equation obtained from $\mathrm{g}^{-1} \mathrm{f}(x)=\mathrm{g}(x)$ provided k is
present and apply $b^{2}-4 a c=0$ to this quadratic equation

*M1

Expect $x^{2}+10 x-10+5 k=0$ .

Obtain 100-4(5 k-10)=0 and hence k=7
Use their k to form and solve a quadratic in x
Marking guidance:

Allow if their quadratic has two solutions.

(-5,-13) only
SC B1 if no method seen.

Alternative Method for first 4 marks

State $\mathrm{f}(x)=\operatorname{gg}(x)$ $\operatorname{gg}(x)=16 x+5 k$
Apply $b^{2}-4 a c=0$ to quadratic equation obtained from $\mathrm{f}(x)=\operatorname{gg}(x)$

(*M1)

Provided k is present.

100-4(5 k-10)=0 and hence k=7

All question bank results loaded

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question (d)

Question 2

Question (a)

Question (b)

Question (c)

Question 3