Functions: inverse and composite - Edexcel GCSE Maths (Non-calculator) Exam Questions

Practise 28 exam-style questions on Functions: inverse and composite for Edexcel GCSE Maths (Non-calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

28 questions
1–5 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

1 mark

\(f(x) = \dfrac{6}{x - 2}\)

Explain why \(x = 2\) cannot be used as an input for this function.

Model answer

When \(x = 2\), \(x - 2 = 0\), so \(f(2) = \dfrac{6}{0}\). Division by 0 is not possible, so \(f(2)\) is undefined.

Explanation

A function can only take inputs that give a defined output. Any value that makes the denominator zero must be excluded, because dividing by zero has no answer.

Mark scheme

1 mark
C1 for explaining that when \(x = 2\) the denominator is 0 and you cannot divide by 0
oe, e.g. "6 ÷ 0 is undefined"

Question 2

2 marks

\(f(x) = x + 1\)
\(g(x) = 2x\)

Sara says,
"\(fg(3) = f(3) \times g(3) = 4 \times 6 = 24\)"

Sara is wrong. Explain why, and work out the correct value of \(fg(3)\).

Model answer

\(fg(3)\) is a composite function: it means \(f(g(3))\), not \(f(3) \times g(3)\).
\(g(3) = 6\)
\(fg(3) = f(6) = 6 + 1 = 7\)

Explanation

Writing two functions next to each other means applying one after the other, starting with the one on the right. The output of \(g\) becomes the input of \(f\). Multiplying the two outputs is a common misunderstanding of the notation.

Mark scheme

1 mark
C1 for explaining that \(fg(3)\) means apply \(g\) first and then \(f\) to the result (it is a composite function, not a product)
oe
1 mark
B1 for 7
cao

Question 3

3 marks

\(f(x) = \dfrac{3}{x + 1}\)

Find \(f^{-1}(x)\)

Model answer

Let \(y = \dfrac{3}{x + 1}\)
\(y(x + 1) = 3\)
\(x + 1 = \dfrac{3}{y}\)
\(x = \dfrac{3}{y} - 1\)
\(f^{-1}(x) = \dfrac{3}{x} - 1\)

Explanation

Multiply up to clear the fraction, then divide by \(y\) and subtract 1 to make \(x\) the subject. Finally write the inverse in terms of \(x\). Check: \(f(2) = 1\) and \(f^{-1}(1) = 3 - 1 = 2\).

Mark scheme

1 mark
M1 for \(y(x + 1) = 3\) or \(x + 1 = \dfrac{3}{y}\)
correct answer with no working scores full marks
1 mark
M1 for \(x = \dfrac{3}{y} - 1\) or \(x = \dfrac{3 - y}{y}\)
1 mark
A1 for \(f^{-1}(x) = \dfrac{3}{x} - 1\)
oe, e.g. \(\dfrac{3 - x}{x}\)

Question 4

4 marks

\(f(x) = 2x - 1\)
\(g(x) = x^2 + 3\)

Solve \(gf(x) = 3f(x) + 7\)

Model answer

\(gf(x) = (2x - 1)^2 + 3 = 4x^2 - 4x + 1 + 3 = 4x^2 - 4x + 4\)
\(3f(x) + 7 = 3(2x - 1) + 7 = 6x + 4\)
\(4x^2 - 4x + 4 = 6x + 4\)
\(4x^2 - 10x = 0\)
\(2x(2x - 5) = 0\)
\(x = 0\) or \(x = \dfrac{5}{2}\)

Explanation

Write both sides in terms of \(x\): \(gf(x)\) puts \(2x - 1\) into \(g\), and \(3f(x)\) means three times \(f(x)\). The resulting quadratic has no constant term, so factorise out \(2x\) - do not divide by \(x\), or you lose the solution \(x = 0\).

Mark scheme

1 mark
M1 for \(gf(x) = (2x - 1)^2 + 3\)
correct answer with no working scores full marks
1 mark
M1 for expanding to \(4x^2 - 4x + 4\)
or \(4x^2 - 4x + 1 + 3\)
1 mark
M1 for forming and simplifying the equation to \(4x^2 - 10x = 0\)
from \(4x^2 - 4x + 4 = 6x + 4\); oe, e.g. \(2x^2 - 5x = 0\)
1 mark
A1 for \(x = 0\) and \(x = \dfrac{5}{2}\)
accept 2.5; both values needed

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