What is A7: Functions?

A function maps each allowed input to one output. To evaluate f(a), substitute the complete input a into every occurrence of the variable and then simplify.

How many marks are A7 questions worth?

In verified past-paper appearances, A7 questions on Edexcel GCSE Maths 1MA1 have been worth 2–7 marks.

Which past exam papers has A7 appeared in?

A7 has appeared in 13 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

Interpret simple expressions as functions with inputs and outputs; interpret the reverse as the 'inverse function' and two successive functions as a 'composite function' (formal notation expected)

Functions

Worked answers, methods and verified real exam appearances for A7 on Edexcel GCSE Maths 1MA1.

Explanation

A function maps each allowed input to one output. To evaluate $f(a)$ , substitute the complete input $a$ into every occurrence of the variable and then simplify.
A function can be represented by a rule, mapping diagram or input-output table.
Higher tier: an inverse function reverses a one-to-one function, while a composite applies two functions successively; in $fg(x)=f(g(x))$ , $g$ acts first.
Formal function notation is expected for those extensions.
Examiners require careful brackets when the input is an expression, because the whole input replaces $x$ .

Worked example

Given $f(x)=3x-4$ , work out $f(6)$ and $f(2a)$ .

1.Substitute $6$ : $f(6)=3(6)-4=14$ .
2.Substitute the complete input $2a$ : $f(2a)=3(2a)-4$ .
3.Simplify to $f(2a)=6a-4$ .

Answer: $f(6)=14$ and $f(2a)=6a-4$ .

Common mistakes

Don't treat $f(x)$ as $f\times x$ instead of function notation.
Don't substitute only part of an expression supplied as the input.
Don't read $f^{-1}(x)$ as $\dfrac{1}{f(x)}$ (Higher tier).

Exam tip

Put the complete input in brackets everywhere $x$ appears before simplifying.

Worked practice

Tier 1 · Easy

Given

f(x)=3x-4

, work out

f(6)

(1)

(Total for Question 1 is 1 mark)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$f(6)=14$	1	Substitute $x=6$ : $f(6)=3(6)-4=18-4=14$ .

Tier 2 · Standard

Given

f(x)=5x+2

, find

f^{-1}(x)

and work out

f^{-1}(27)

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$f^{-1}(x)=\frac{x-2}{5}$ $f^{-1}(27)=5$	3	Write $y=5x+2$ and rearrange: $x=(y-2)/5$ . Hence $f^{-1}(x)=(x-2)/5$ . Substituting $27$ gives $f^{-1}(27)=(27-2)/5=5$ .

Tier 3 · Hard

Let

f(x)=2x+3

and

g(x)=x^2-1

. Solve

fg(x)=19

, where

fg(x)=f(g(x))

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$x=-3$ or $x=3$	4	Form the composite by substituting $g(x)$ into $f$ : $fg(x)=2(x^2-1)+3=2x^2+1$ . Hence $2x^2+1=19$ , so $2x^2=18$ and $x^2=9$ . Therefore $x=-3$ or $x=3$ .

Tier 1 · Easy

A function machine adds

7

to its input. Write down the output when the input is

n

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$n+7$	1	Apply the function rule to the complete input $n$ . Adding $7$ gives the output $n+7$ .

Tier 2 · Standard

Given

g(x)=3x+1

, work out the value of

a

when

g(a)=16

(2)

(Total for Question 5 is 2 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$a=5$	2	Substitute $a$ as the input: $3a+1=16$ . Subtract $1$ and divide by $3$ to get $3a=15$ , so $a=5$ .

Tier 3 · Hard

The function

f(x)=3x+k

has

f(4)=17

. Work out the value of

k

and hence work out

f(-2)

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$k=5$ $f(-2)=-1$	3	Use $f(4)=17$ : $3(4)+k=17$ , so $12+k=17$ and $k=5$ . Therefore $f(-2)=3(-2)+5=-6+5=-1$ .

Tier 2 · Standard

The function

h

is given by

h(x)=x^2-2x

. Work out

h(-3)

and simplify

h(a+1)

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$h(-3)=15$ $h(a+1)=a^2-1$	3	Substitute the complete input each time. $h(-3)=(-3)^2-2(-3)=9+6=15$ . Also, $h(a+1)=(a+1)^2-2(a+1)=a^2+2a+1-2a-2=a^2-1$ .

Tier 3 · Hard

Let

f(x)=x^2+ax+b

. Given that

f(1)=8

and

f(-2)=2

, work out

a

and

b

. Hence work out

f(3)

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$a=3$ $b=4$ $f(3)=22$	5	From $f(1)=8$ , $1+a+b=8$ , so $a+b=7$ . From $f(-2)=2$ , $4-2a+b=2$ , so $-2a+b=-2$ . Subtracting the second equation from the first gives $3a=9$ , so $a=3$ and then $b=4$ . Therefore $f(3)=3^2+3(3)+4=22$ .

Tier 3 · Hard

Higher only: Let

f(t)=4t-3

and

g(x)=2x+7

. Find the function

h

such that

fh(x)=g(x)

for every value of

x

. Check your answer by forming

fh(x)

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$h(x)=\dfrac{x+5}{2}$	4	$fh(x)=f(h(x))=4h(x)-3$ . Setting this equal to $2x+7$ gives $4h(x)=2x+10$ , so $h(x)=\dfrac{x+5}{2}$ . Checking, $f(h(x))=4\left(\dfrac{x+5}{2}\right)-3=2x+10-3=2x+7=g(x)$ .

Q10

Tier 3 · Hard

Higher only: The functions are

f(x)=3x-5

and

g(x)=\dfrac{x+5}{3}

. Work out and simplify both

fg(x)

and

gf(x)

. Hence state the relationship between

f

and

g

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$fg(x)=x$ $gf(x)=x$ $g=f^{-1}$ (and $f=g^{-1}$ )	4	$fg(x)=3\left(\dfrac{x+5}{3}\right)-5=x+5-5=x$ . Also, $gf(x)=\dfrac{(3x-5)+5}{3}=x$ . Since each composite returns the original input, the functions reverse one another, so $g=f^{-1}$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2019-11	1H	Q18	6	Non-calculator	Higher	QP MS
2022-06	2F	Q12	3	Allowed	Foundation	QP MS
2022-11	2H	Q22	3	Allowed	Higher	QP MS
2024-11	1H	Q18	4	Non-calculator	Higher	QP MS
2024-11	1F	Q13	4	Non-calculator	Foundation	QP MS
2019-06	1H	Q21	7	Non-calculator	Higher	QP MS
2024-06	3H	Q16	5	Allowed	Higher	QP MS
2022-06	2H	Q19	4	Allowed	Higher	QP MS
2023-06	1H	Q20	5	Non-calculator	Higher	QP MS
2023-11	2H	Q19	4	Allowed	Higher	QP MS
2019-11	2F	Q8	2	Allowed	Foundation	QP MS
2024-06	1F	Q9	5	Non-calculator	Foundation	QP MS
2021-11	1H	Q21	6	Non-calculator	Higher	QP MS

Other points in A Algebra · equations and reasoning

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