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A7Composite and inverse functions

Edexcel GCSE Maths (1MA1), Higher tier · Algebra

Practise Composite and inverse functions. 2 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.

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Sample questions

Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.

Question 1Easy3 marks
The function \(f\) is such that \(f(x) = 4x - 9\).
(a) Find \(f(3)\).[1]
(b) Solve \(f(x) = 11\).[2]
Show the answer and mark scheme
(a) Answer: \(3\)
  • B1 for 3

Worked solution: \(f(3) = 4 \times 3 - 9 = 3\)

(b) Answer: \(x = 5\)
  • M1 for \(4x - 9 = 11\) and a correct first step
  • A1 for 5

Worked solution: \(4x - 9 = 11\), \(4x = 20\), \(x = 5\)

Question 2Medium4 marks
The function \(g\) is such that \(g(x) = \frac{15}{x - 2}\).
(a) Find \(g(7)\).[1]
(b) State the value of \(x\) that must be excluded from any domain of \(g\).[1]
(c) Solve \(g(x) = 5\).[2]
Show the answer and mark scheme
(a) Answer: \(3\)
  • B1 for 3

Worked solution: \(g(7) = \frac{15}{7 - 2} = \frac{15}{5} = 3\)

(b) Answer: \(x = 2\)
  • B1 for 2

Worked solution: When \(x = 2\) the denominator is 0, and you cannot divide by 0.

(c) Answer: \(x = 5\)
  • M1 for \(15 = 5(x - 2)\) oe
  • A1 for 5

Worked solution: \(15 = 5(x - 2)\) so \(x - 2 = 3\), \(x = 5\).

Question 3Hard6 marks
\(f(x) = 2x + 3\) and \(g(x) = x^2\).
(a) Show that \(gf(x) - fg(x) = 2x^2 + 12x + 6\).[3]
(b) Solve \(gf(x) = fg(x)\).
Give your answers in the form \(a \pm \sqrt{b}\), where \(a\) and \(b\) are integers.[3]
Show the answer and mark scheme
(a) Answer: \((2x + 3)^2 - (2x^2 + 3) = 4x^2 + 12x + 9 - 2x^2 - 3 = 2x^2 + 12x + 6\)
  • M1 for \(gf(x) = (2x + 3)^2\)
  • M1 for \(fg(x) = 2x^2 + 3\)
  • C1 for \(4x^2 + 12x + 9 - 2x^2 - 3 = 2x^2 + 12x + 6\)

Worked solution: \(gf(x) = g(2x + 3) = (2x + 3)^2 = 4x^2 + 12x + 9\)
\(fg(x) = f(x^2) = 2x^2 + 3\)
\(gf(x) - fg(x) = 2x^2 + 12x + 6\)

(b) Answer: \(x = -3 \pm \sqrt{6}\)
  • M1 for \(x^2 + 6x + 3 = 0\) (or \(2x^2 + 12x + 6 = 0\))
  • M1 for \((x + 3)^2 = 6\) or a correct substitution into the quadratic formula
  • A1 for \(-3 \pm \sqrt{6}\)

Worked solution: \(2x^2 + 12x + 6 = 0\), so \(x^2 + 6x + 3 = 0\).
\((x + 3)^2 - 9 + 3 = 0\), \((x + 3)^2 = 6\), \(x = -3 \pm \sqrt{6}\).

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