Edexcel A level Maths (9MA0) · Pure mathematics › Algebra and functions
Practise Composite and inverse functions. 3 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.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy8 marks
The functions \(\mathrm{f}\) and \(\mathrm{g}\) are defined by \(\mathrm{f}(x) = 2x - 2, \quad x \in \mathbb{R}\) \(\mathrm{g}(x) = \frac{6}{x - 1}, \quad x \in \mathbb{R}, \ x \ne 1\)
(a) Find \(\mathrm{fg}(3)\).[2]
(b) Find \(\mathrm{gf}(x)\), giving your answer in its simplest form.[2]
(c) Find \(\mathrm{f}^{-1}(x)\).[2]
(d) Solve the equation \(\mathrm{f}(x) = \mathrm{f}^{-1}(x)\)[2]
Show the answer and mark scheme
(a)Answer: \(\mathrm{fg}(3) = 4\)
M1 for finding \(\mathrm{g}(3) = 3\) and substituting into \(\mathrm{f}\)
A1 for \(4\)
Worked solution: \(\mathrm{g}(3) = 3\), so \(\mathrm{fg}(3) = \mathrm{f}(3) = 2 \times 3 - 2 = 4\).
(b)Answer: \(\mathrm{gf}(x) = \frac{6}{2x - 3}\)
M1 for substituting \(\mathrm{f}(x)\) into \(\mathrm{g}\): \(\frac{6}{(2x - 2) - 1}\)
M1 for an attempt to make \(x\) the subject of \(y = 2x - 2\) (or to reverse the operations)
A1 for \(\mathrm{f}^{-1}(x) = \frac{x + 2}{2}\) oe
Worked solution: Let \(y = 2x - 2\). Then \(x = \frac{y + 2}{2}\), so \(\mathrm{f}^{-1}(x) = \frac{x + 2}{2}\), \(x \in \mathbb{R}\).
(d)Answer: \(x = 2\)
M1 for \(2x - 2 = \frac{x + 2}{2}\) (or \(\mathrm{f}(x) = x\)) and an attempt to solve
A1 for \(x = 2\)
Worked solution: \(2x - 2 = \frac{x + 2}{2}\) gives \(4x - 4 = x + 2\), so \(3x = 6\) and \(x = 2\). (Since \(\mathrm{f}\) is linear, this is also the solution of \(\mathrm{f}(x) = x\).)
Question 2Medium8 marks
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. The function \(\mathrm{f}\) is defined by \(\mathrm{f}(x) = \frac{5x - 1}{x + 1}, \quad x \gt -1\)
(a) Show that \(\mathrm{f}(x) = 5 - \frac{6}{x + 1}\)[2]
(b) Find the range of \(\mathrm{f}\).[2]
(c) Find \(\mathrm{f}^{-1}(x)\), stating its domain.[4]
Show the answer and mark scheme
(a)Answer: \(5x - 1 = 5(x + 1) - 6\)
M1 for writing the numerator as \(5(x + 1) - 6\) (or dividing)
A1* for the given result
Worked solution: \(5x - 1 = 5(x + 1) - 6\), so \(\mathrm{f}(x) = 5 - \frac{6}{x + 1}\).
(b)Answer: \(\mathrm{f}(x) \lt 5\)
M1 for using \(x + 1 \gt 0\) so that \(\frac{6}{x + 1}\) takes all positive values
A1 for \(\mathrm{f}(x) \lt 5\)
Worked solution: For \(x \gt -1\), \(x + 1 \gt 0\), so \(\frac{6}{x + 1}\) takes every positive value (tending to 0 as \(x\) increases). So \(\mathrm{f}(x)\) takes every value less than \(5\): the range is \(\mathrm{f}(x) \lt 5\).
M1 for \(y = \frac{5x - 1}{x + 1}\) and multiplying through by \((x + 1)\)
M1 for collecting the \(x\) terms and factorising
A1 for \(\mathrm{f}^{-1}(x) = \frac{-x - 1}{x - 5}\) oe
B1ft for the domain \(x \lt 5\) (their range of \(\mathrm{f}\))
Worked solution: Let \(y = \frac{5x - 1}{x + 1}\): \(y(x + 1) = 5x - 1\), so \(x(y - 5) = -y - 1\) and \(x = \frac{-y - 1}{y - 5}\). So \(\mathrm{f}^{-1}(x) = \frac{-x - 1}{x - 5}\), with domain equal to the range of \(\mathrm{f}\): \(x \lt 5\).
Question 3Hard9 marks
The functions \(\mathrm{f}\) and \(\mathrm{g}\) are defined by \(\mathrm{f}(x) = 3 + 2\mathrm{e}^{-x}, \quad x \in \mathbb{R}\) \(\mathrm{g}(x) = \ln(x - 1), \quad x \gt 1\)
(a) State the range of \(\mathrm{f}\).[1]
(b) Find \(\mathrm{f}^{-1}(x)\), stating its domain.[3]
(c) Show that \(\mathrm{fg}(x) = \frac{3x - 1}{x - 1}\).[3]
(d) Solve the equation \(\mathrm{fg}(x) = 7\).[2]
Show the answer and mark scheme
(a)Answer: \(\mathrm{f}(x) \gt 3\)
B1 for \(\mathrm{f}(x) \gt 3\) (accept \(y \gt 3\))
Worked solution: \(\mathrm{e}^{-x}\) takes every positive value, so \(2\mathrm{e}^{-x} \gt 0\) and \(\mathrm{f}(x) \gt 3\).
(b)Answer: \(\mathrm{f}^{-1}(x) = \ln\left(\frac{2}{x - 3}\right), \quad x \gt 3\)
M1 for making \(\mathrm{e}^{-x}\) (or \(\mathrm{e}^{-y}\)) the subject: \(\mathrm{e}^{-x} = \frac{y - 3}{2}\)
A1 for \(\mathrm{f}^{-1}(x) = -\ln\left(\frac{x - 3}{2}\right)\) or \(\ln\left(\frac{2}{x - 3}\right)\) oe
B1ft for the domain \(x \gt 3\)
Worked solution: Let \(y = 3 + 2\mathrm{e}^{-x}\). Then \(\mathrm{e}^{-x} = \frac{y - 3}{2}\), so \(-x = \ln\left(\frac{y - 3}{2}\right)\) and \(x = \ln\left(\frac{2}{y - 3}\right)\). So \(\mathrm{f}^{-1}(x) = \ln\left(\frac{2}{x - 3}\right)\), with domain \(x \gt 3\) (the range of \(\mathrm{f}\)).
M1 for \(\frac{3x - 1}{x - 1} = 7\) (or \(\frac{2}{x - 1} = 4\)) and an attempt to solve
A1 for \(x = \frac{3}{2}\)
Worked solution: \(3 + \frac{2}{x - 1} = 7\) gives \(\frac{2}{x - 1} = 4\), so \(x - 1 = \frac{1}{2}\) and \(x = \frac{3}{2}\), which is in the domain \(x \gt 1\).