Function notation, domain, range, composite & inverse functions – IGCSE Maths exam questions with mark scheme

1Grade 4–5HigherCalculator4 marks

$\mathrm{f}(x) = 3x - 5 \qquad \mathrm{g}(x) = x^2 + 1$

(a) Find $\mathrm{f}(4)$. [1]

(b) Find $\mathrm{g}(-3)$. [1]

(a) $7$B1
(b) $10$B1
(c) for $\mathrm{g}(2) = 5$ or $3(x^2 + 1) - 5$ oeM1
$10$A1
Answer: (a) 7 (b) 10 (c) 10
Examiner tip: fg(2) means 'do g first, then f' — work out g(2) and feed the answer into f.

2Grade 6–7HigherCalculator4 marks

$\mathrm{f}(x) = \dfrac{2x + 1}{x - 3}$

(a) State the value of $x$ that must be excluded from the domain of f. [1]

(b) Find $\mathrm{f}^{-1}(x)$. [3]

(a) $3$ (accept $x = 3$ or $x \neq 3$)B1
(b) for $y(x - 3) = 2x + 1$ or $x(y - 3) = 2y + 1$ (correct first step to clear the fraction)M1
for collecting the terms in the new subject on one side and factorising, e.g. $x(y - 2) = 3y + 1$ or $y(x - 2) = 3x + 1$M1
$\mathrm{f}^{-1}(x) = \dfrac{3x + 1}{x - 2}$ oeA1
Answer: (a) $x = 3$ (b) $\mathrm{f}^{-1}(x) = \dfrac{3x + 1}{x - 2}$
Examiner tip: When the subject appears twice, gather both terms on one side and factorise — otherwise x is left on both sides.

3Grade 8–9HigherCalculator5 marks

$\mathrm{f}(x) = 2x + 3 \qquad \mathrm{g}(x) = x^2$

Solve $\mathrm{fg}(x) = \mathrm{gf}(x)$.

Give your solutions correct to 3 significant figures. [5]

for $\mathrm{fg}(x) = 2x^2 + 3$M1
for $\mathrm{gf}(x) = (2x + 3)^2$ or $4x^2 + 12x + 9$M1
for forming a correct 3-term quadratic, e.g. $2x^2 + 12x + 6 = 0$ or $x^2 + 6x + 3 = 0$M1
(dep on previous M1) for correct use of the formula or completing the square, e.g. $x = -3 \pm \sqrt{6}$M1
$-0.551$ and $-5.45$A1
Answer: $x = -0.551$ or $x = -5.45$
Examiner tip: gf(x) means put f(x) into g, so it is (2x + 3)², not 2x² + 3 — the order of the letters matters.

4Grade 6–7HigherCalculator3 marks

$\mathrm{h}(x) = \sqrt{x - 4}$

(a) Find $\mathrm{h}(20)$. [1]

(b) State which values of $x$ cannot be included in the domain of h. [1]

(a) $4$B1
(b) $x < 4$B1
(c) $\mathrm{h}(x) \geqslant 0$ (accept $y \geqslant 0$)B1
Answer: (a) 4 (b) $x < 4$ (c) $\mathrm{h}(x) \geqslant 0$
Examiner tip: You cannot take the square root of a negative number, so x − 4 must not be negative; x = 4 itself is allowed.

5Grade 8–9HigherCalculator5 marks

The function h is defined by

$\mathrm{h}(x) = x^2 - 6x + 11 \qquad x \geqslant 3$

(a) By writing $\mathrm{h}(x)$ in the form $(x - a)^2 + b$, find the range of h. [2]

(b) Find $\mathrm{h}^{-1}(x)$. [3]

(a) for $(x - 3)^2 + 2$M1
$\mathrm{h}(x) \geqslant 2$ (accept $\mathrm{h} \geqslant 2$ or $y \geqslant 2$; reject $x \geqslant 2$ or $\mathrm{h}(x) > 2$)A1
(b) for $(x - 3)^2 = y - 2$ or $(y - 3)^2 = x - 2$ oe (rearranging their completed square), or for a correct quadratic-formula start on $x^2 - 6x + 11 - y = 0$M1
for $x - 3 = \pm\sqrt{y - 2}$ or $y - 3 = \pm\sqrt{x - 2}$ oe (e.g. $x = 3 \pm \sqrt{y - 2}$; the $\pm$ may be omitted)M1
$\mathrm{h}^{-1}(x) = 3 + \sqrt{x - 2}$ (the $\pm$ must be resolved to $+$; accept $\mathrm{h}^{-1}(x) = \sqrt{x - 2} + 3$)A1
Answer: (a) $\mathrm{h}(x) \geqslant 2$ (b) $\mathrm{h}^{-1}(x) = 3 + \sqrt{x - 2}$
Examiner tip: Because the domain is x ≥ 3, only the positive square root is valid — leaving ± loses the final mark.

6Grade 4–5HigherCalculator3 marks

$\mathrm{f}(x) = 4x - 1$

(a) Find $\mathrm{f}^{-1}(x)$. [2]

(b) Solve $\mathrm{f}(x) = 19$. [1]

(a) for $y = 4x - 1 \Rightarrow x = \dfrac{y + 1}{4}$ or $x = 4y - 1$ rearranged, or a correct flow diagram reversedM1
$\mathrm{f}^{-1}(x) = \dfrac{x + 1}{4}$ oeA1
(b) $x = 5$B1
Answer: (a) $\mathrm{f}^{-1}(x) = \dfrac{x + 1}{4}$ (b) $x = 5$
Examiner tip: Your answer to (a) must be in terms of x, written as f⁻¹(x) = …, not left in terms of y.

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