Functions: composite and inverse – GCSE Maths exam questions with mark scheme

1Grade 4–5HigherNon-calculator3 marks

The function $f$ is defined by $f(x) = 3x - 5$

(a) Find $f(4)$ [1]

(b) Solve $f(x) = 13$ [2]

(a) $7$B1
(b) for $3x - 5 = 13$M1
(b) $x = 6$A1
Answer: (a) $7$ (b) $x = 6$
Examiner tip: $f(x) = 13$ means the OUTPUT is 13 – set up an equation, don't substitute 13 for $x$.

2Grade 6–7HigherNon-calculator4 marks

$f(x) = 2x + 1$ and $g(x) = x^2$

(a) Find $fg(3)$ [2]

(b) Find $gf(x)$. Give your answer in its simplest form. [2]

(a) for $g(3) = 9$ or $fg(x) = 2x^2 + 1$M1
(a) $19$A1
(b) for $(2x + 1)^2$M1
(b) $4x^2 + 4x + 1$ (accept $(2x+1)^2$ as a final answer; if expanded, the expansion must be correct)A1
Answer: (a) $19$ (b) $4x^2 + 4x + 1$
Examiner tip: $fg(3)$ means do $g$ FIRST (the function nearest the $x$), then $f$.

3Grade 6–7HigherNon-calculator3 marks

$f(x) = \dfrac{x + 4}{3}$

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

(b) Find the value of $f^{-1}(5)$ [1]

(a) for $x = \dfrac{y + 4}{3}$ or $3y = x + 4$ or a correct first step of rearrangingM1
(a) $f^{-1}(x) = 3x - 4$A1
(b) $11$ (ft their $f^{-1}$)B1
Answer: (a) $f^{-1}(x) = 3x - 4$ (b) $11$
Examiner tip: Write the answer as $f^{-1}(x) = \ldots$ in terms of $x$, not $y$.

4Grade 8–9HigherNon-calculator3 marks

$f(x) = \dfrac{2x}{x - 3}$, $x \ne 3$

Find $f^{-1}(x)$ [3]

for $x = \dfrac{2y}{y - 3}$ and multiplying out, e.g. $xy - 3x = 2y$ (or $y = \dfrac{2x}{x - 3}$ and $xy - 3y = 2x$)M1
for collecting the $y$ terms and factorising, e.g. $y(x - 2) = 3x$ (or $x(y - 2) = 3y$)M1
$f^{-1}(x) = \dfrac{3x}{x - 2}$ (oe, e.g. $\dfrac{-3x}{2 - x}$; must be in terms of $x$)A1
Answer: $f^{-1}(x) = \dfrac{3x}{x - 2}$
Examiner tip: When $y$ appears twice, collect both $y$ terms on one side and factorise – that is the key step.

5Grade 8–9HigherNon-calculator4 marks

$f(x) = x^2 - 3$ and $g(x) = x + 2$

Solve $fg(x) = gf(x)$ [4]

for $fg(x) = (x + 2)^2 - 3$M1
for $gf(x) = x^2 - 3 + 2$M1
for equating their expressions and simplifying to a linear equation, e.g. $4x + 1 = -1$ or $4x = -2$M1
$x = -\tfrac{1}{2}$ (oe, e.g. $-0.5$)A1
Answer: $x = -\tfrac12$
Examiner tip: $fg(x)$ and $gf(x)$ are usually different – set up each carefully before equating.

6Grade 8–9HigherNon-calculator4 marks

$f(x) = 5 - 2x$

Solve the equation $f(x) = f^{-1}(x)$ [4]

for a correct first step to find the inverse, e.g. $x = 5 - 2y$M1
$f^{-1}(x) = \dfrac{5 - x}{2}$A1
for $5 - 2x = \dfrac{5 - x}{2}$ and a correct first step to solve, e.g. $10 - 4x = 5 - x$M1
$x = \tfrac{5}{3}$ (oe)A1
Answer: $x = \tfrac{5}{3}$
Examiner tip: $f^{-1}(x)$ is the inverse function, NOT $\dfrac{1}{f(x)}$.

7Grade 8–9HigherNon-calculator3 marks

$f(x) = x^2 + 2$ and $g(x) = 2x + a$, where $a$ is a constant.

Given that $fg(x) = 4x^2 - 12x + 11$, find the value of $a$. [3]

for $fg(x) = (2x + a)^2 + 2$M1
for expanding to $4x^2 + 4ax + a^2 + 2$ and comparing coefficients, e.g. $4a = -12$M1
$a = -3$ (answer $a = \pm 3$ from the constant term alone, without rejecting $+3$, scores A0)A1
Answer: $a = -3$
Examiner tip: Compare the $x$ terms first: $4a = -12$ gives $a$ directly; then check the constant term.

8Grade 1–3Non-calculator3 marks

Here is a function machine.

(a) The input is $6$. Work out the output. [1]

(b) The output is $2$. Work out the input. [2]

(a) $26$B1
(b) for reversing the machine, e.g. $(2 - 8) \div 3$ or $-6 \div 3$, or forming and starting to solve $3x + 8 = 2$ (e.g. $3x = -6$)M1
(b) $-2$A1
Answer: (a) $26$ (b) $-2$
Examiner tip: Working backwards you must undo the operations in reverse order with inverse operations: subtract 8 first, then divide by 3.

9Grade 4–5HigherNon-calculator4 marks

The functions $f$ and $g$ are defined only for $x = 1, 2, 3, 4, 5$. The table shows all their values.

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

(a) Find $fg(4)$ [1]

(b) Find $f^{-1}(1)$ [1]

(a) $5$B1
(b) $3$B1
(c) for correctly working out $gf(x)$ for at least two values of $x$, e.g. $gf(1) = g(4) = 2$, $gf(2) = g(5) = 4$, $gf(3) = g(1) = 3$M1
(c) $x = 3$ (with no other values given)A1
Answer: (a) $5$ (b) $3$ (c) $x = 3$
Examiner tip: In $fg(4)$ the function nearest the bracket acts first: find $g(4)$, then put that answer into $f$.

10Grade 6–7HigherNon-calculator4 marks

$f(x) = 10 - x$ and $g(x) = x^2 + 1$

(a) Find $gf(13)$ [2]

(b) Show that $ff(x) = x$ and explain what this tells you about $f^{-1}(x)$. [2]

(a) for $f(13) = -3$ seen or $(10 - 13)^2 + 1$M1
(a) $10$A1
(b) for $ff(x) = 10 - (10 - x)$ leading to $x$ (must show the bracket or $10 - 10 + x$)M1
(b) explanation that $f$ is its own inverse, e.g. $f^{-1}(x) = f(x)$ or $f^{-1}(x) = 10 - x$ (accept 'f undoes itself')C1
Answer: (a) $10$ (b) $ff(x) = 10 - (10 - x) = x$, so $f^{-1}(x) = f(x) = 10 - x$
Examiner tip: When substituting a whole expression into $f$, put it in a bracket — writing $10 - 10 - x$ loses the sign and the mark.

11Grade 6–7HigherCalculator3 marks

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

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

(b) Work out $f^{-1}(50)$. Give your answer correct to 3 significant figures. [1]

(a) for a correct first step in rearranging $y = \dfrac{x^3 - 1}{2}$ (or $x = \dfrac{y^3 - 1}{2}$), e.g. $2y + 1 = x^3$M1
(a) $f^{-1}(x) = \sqrt[3]{2x + 1}$ (oe; must be in terms of $x$)A1
(b) $4.66$ (ft their $f^{-1}(x)$ from (a), correctly evaluated and given to 3 s.f.)B1
Answer: (a) $f^{-1}(x) = \sqrt[3]{2x + 1}$ (b) $4.66$
Examiner tip: Undo the steps in reverse order: multiply by 2, add 1, then cube root — taking the cube root before dealing with the 2 is the usual slip.

12Grade 6–7HigherNon-calculator4 marks

The graph of $y = f(x)$ is a straight line, shown on the grid.

(a) Use the graph to find $f^{-1}(3)$ [1]

(b) Find an expression for $f^{-1}(x)$ [2]

(c) The graph of $y = f^{-1}(x)$ is a reflection of the graph of $y = f(x)$. Write down the equation of the mirror line. [1]

(a) $4$B1
(b) for finding $f(x) = \tfrac12 x + 1$ (oe) from the graph, or for a correct first step in rearranging $y = \tfrac12 x + 1$, e.g. $2y = x + 2$M1
(b) $f^{-1}(x) = 2x - 2$ (oe, e.g. $2(x - 1)$)A1
(c) $y = x$B1
Answer: (a) $4$ (b) $f^{-1}(x) = 2x - 2$ (c) $y = x$
Examiner tip: $f^{-1}(3)$ means 'which input gives an output of 3', so read across from $y = 3$ to the line and then down, not up from $x = 3$ (which gives 2.5).

13Grade 6–7HigherNon-calculator5 marks

$f(x) = 2x + 3$ and $g(x) = x^2 - 5$

(a) Ali says, '$fg(2)$ and $gf(2)$ give the same answer.'
Show that Ali is wrong. [2]

(b) Solve $gf(x) = 20$ [3]

(a) for correctly working out one of $fg(2) = 1$ or $gf(2) = 44$M1
(a) both $fg(2) = 1$ and $gf(2) = 44$ with a conclusion that they are differentC1
(b) for $(2x + 3)^2 - 5 = 20$ (oe)M1
(b) for $2x + 3 = \pm 5$, or a correct method to solve $4x^2 + 12x - 16 = 0$, e.g. $4(x + 4)(x - 1) = 0$M1
(b) $x = 1$ and $x = -4$A1
Answer: (a) $fg(2) = 1$, $gf(2) = 44$, so not equal (b) $x = 1$ or $x = -4$
Examiner tip: Square-rooting $(2x + 3)^2 = 25$ gives $\pm 5$ — candidates who use only $+5$ lose the final mark for missing $x = -4$.

14Grade 8–9HigherNon-calculator3 marks

$f(x) = 4x - 3$ and $g(x) = \dfrac{x}{2} + c$, where $c$ is a constant.

Given that $fg(x) = gf(x)$ for all values of $x$, find the value of $c$. [3]

for $fg(x) = 4\left(\dfrac{x}{2} + c\right) - 3$ $(= 2x + 4c - 3)$ or $gf(x) = \dfrac{4x - 3}{2} + c$ $(= 2x - \tfrac32 + c)$M1
for both composites correct and equating the constant terms, e.g. $4c - 3 = c - \tfrac32$M1
$c = \tfrac12$ (oe, e.g. $0.5$)A1
Answer: $c = \tfrac12$
Examiner tip: The $x$ terms cancel because both composites contain $2x$ — the equation you need comes from the constant terms only.

15Grade 8–9HigherNon-calculator4 marks

$f(x) = x^2 + 6x + 1$ for $x \ge -3$

(a) Write $x^2 + 6x + 1$ in the form $(x + a)^2 + b$, where $a$ and $b$ are integers. [2]

(b) Hence find $f^{-1}(x)$ [2]

(a) for $(x + 3)^2$ seen or $a = 3$M1
(a) $(x + 3)^2 - 8$A1
(b) for $y + 8 = (x + 3)^2$ or $x + 8 = (y + 3)^2$ (ft their $a$, $b$), leading to taking a square rootM1
(b) $f^{-1}(x) = \sqrt{x + 8} - 3$ (oe; do not accept $\pm$ unless the negative root is rejected using $x \ge -3$)A1
Answer: (a) $(x + 3)^2 - 8$ (b) $f^{-1}(x) = \sqrt{x + 8} - 3$
Examiner tip: Because $f$ is only defined for $x \ge -3$, the inverse takes the positive square root only — leaving $\pm$ in loses the final mark.

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