Composite and inverse functions - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 17 exam-style questions on Composite and inverse functions for Edexcel A-Level Maths (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

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2–9 marks each
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17 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

3 marks

The function f is defined as
\[f(x) = x^3 + 2x - 1, \quad x \in \mathbb{R}\]
Explain why f has an inverse function.

Find the value of \(f^{-1}(11)\).

Model answer

\(f'(x) = 3x^2 + 2 > 0\) for all real x, so f is an increasing function. So f is one-one, and a one-one function has an inverse.
\(f^{-1}(11)\) is the value of x for which \(f(x) = 11\):
\(x^3 + 2x - 1 = 11 \Rightarrow x^3 + 2x - 12 = 0\)
\(f(2) = 8 + 4 - 1 = 11\), so \(f^{-1}(11) = 2\)

Explanation

A function has an inverse only if it is one-one, and f is increasing for all x, so it is one-one. To find \(f^{-1}(11)\) there is no need to find \(f^{-1}(x)\): solve \(f(x) = 11\), which gives \(x = 2\). A common error is to substitute 11 into f, or into 1/f, instead of solving \(f(x) = 11\).

Mark scheme

1 mark
B1 Explains that f is one-one, with a reason, e.g. \(x^3\) and \(2x\) are both increasing so f is an increasing function, or \(f'(x) = 3x^2 + 2 > 0\) for all x; a correct final answer of \(f^{-1}(11) = 2\) with no working scores the M1 and A1
An answer of \(f^{-1}(11) = 2\) with no working scores 2 marks (M1 A1); the B1 needs its reason. Accept 'each value of f(x) comes from only one value of x because f is increasing'. Do not accept 'f is one-one' with no reason, 'it is a cubic' (not every cubic is one-one) or 'it is one-many'.
1 mark
M1 Sets \(f(x) = 11\), i.e. \(x^3 + 2x - 1 = 11\), and attempts to solve \(x^3 + 2x - 12 = 0\) (e.g. by trial, factor theorem or calculator)
There is no need to find an expression for \(f^{-1}(x)\). Substituting 11 into f, into f' or into \(\dfrac{1}{f(x)}\) scores M0.
1 mark
A1 \(f^{-1}(11) = 2\)
cao. Allow x = 2. Ignore a check such as f(2) = 8 + 4 - 1 = 11.

Question 2

4 marks

The functions f and g are given by
\[f(x) = 6x - x^2, \quad x \in \mathbb{R}\]
\[g(x) = \mathrm{e}^{x - 9}, \quad x \in \mathbb{R}\]
Find gf(x), and hence find the range of gf.

Model answer

\(gf(x) = g(6x - x^2) = \mathrm{e}^{6x - x^2 - 9}\)
\(6x - x^2 - 9 = -(x^2 - 6x + 9) = -(x - 3)^2\), which takes every value \(\leqslant 0\)
So \(0 < \mathrm{e}^{6x - x^2 - 9} \leqslant \mathrm{e}^0 = 1\), the greatest value being at \(x = 3\).
Range: \(0 < gf(x) \leqslant 1\)

Explanation

gf(x) means apply f first, so the exponent is \(6x - x^2 - 9\), which equals \(-(x - 3)^2\) and is never positive. Since \(\mathrm{e}^t\) is positive and increasing, gf(x) is greater than 0 and at most \(\mathrm{e}^0 = 1\). Common errors are including 0 in the range and giving only one end of it.

Mark scheme

1 mark
M1 Applies f first and then g: \(gf(x) = \mathrm{e}^{(6x - x^2) - 9}\); or a correct final answer of \(gf(x) = \mathrm{e}^{6x - x^2 - 9}\) with range \(0 < gf(x) \leqslant 1\) and no working
An answer of \(gf(x) = \mathrm{e}^{6x - x^2 - 9}\) and range \(0 < gf(x) \leqslant 1\) with no working scores 4 marks. Finding \(fg(x) = 6\mathrm{e}^{x - 9} - \mathrm{e}^{2x - 18}\), or the product \(g(x)f(x)\), scores M0 A0.
1 mark
A1 \(gf(x) = \mathrm{e}^{6x - x^2 - 9}\) oe
Accept \(\mathrm{e}^{-(x - 3)^2}\), \(\dfrac{\mathrm{e}^{6x - x^2}}{\mathrm{e}^9}\).
1 mark
M1 Finds the greatest value of the exponent or of f: \(6x - x^2 - 9 = -(x - 3)^2 \leqslant 0\), or \(f(x) = 9 - (x - 3)^2 \leqslant 9\) (by completing the square or by differentiation)
This mark is independent of the first two marks.
1 mark
A1 \(0 < gf(x) \leqslant 1\)
Both ends are needed. Accept \(0 < y \leqslant 1\), (0, 1], \(\{gf(x) : 0 < gf(x) \leqslant 1\}\). Do not accept \(0 \leqslant gf(x) \leqslant 1\) (e to any power is never 0), \(gf(x) \leqslant 1\) alone, or the range written in terms of x.

Question 3

6 marks

The function f is defined as
\[f(x) = 3x - 4, \quad x \in \mathbb{R}\]
The function g is defined for all real values of x and is such that
\[gf(x) = 9x^2 - 24x + 17, \quad x \in \mathbb{R}\]
Find g(x).

Hence solve the equation \(fg(x) = gf(x)\).

Model answer

\((3x - 4)^2 = 9x^2 - 24x + 16\), so \(9x^2 - 24x + 17 = (3x - 4)^2 + 1\)
So \(g(3x - 4) = (3x - 4)^2 + 1\), giving \(g(x) = x^2 + 1\)
\(fg(x) = f(x^2 + 1) = 3(x^2 + 1) - 4 = 3x^2 - 1\)
\(3x^2 - 1 = 9x^2 - 24x + 17\)
\(6x^2 - 24x + 18 = 0\)
\(x^2 - 4x + 3 = 0\)
\((x - 1)(x - 3) = 0\)
\(x = 1\) or \(x = 3\)

Explanation

Since \(gf(x) = g(3x - 4)\), writing \(9x^2 - 24x + 17\) in terms of \(3x - 4\) shows that g squares its input and adds 1; equivalently, substitute \(x = f^{-1}(t) = \dfrac{t + 4}{3}\). Then \(fg(x) = 3(x^2 + 1) - 4 = 3x^2 - 1\), and setting this equal to gf(x) gives a quadratic with roots 1 and 3. A common error is to treat gf(x) as the product g(x)f(x) and divide by \(3x - 4\).

Mark scheme

1 mark
M1 Writes \(9x^2 - 24x + 17\) in terms of \(3x - 4\), e.g. \((3x - 4)^2 + 1\), or substitutes \(x = \dfrac{t + 4}{3}\) (that is, uses \(f^{-1}\)) into \(9x^2 - 24x + 17\); a correct final answer of \(g(x) = x^2 + 1\) with \(x = 1\) and \(x = 3\) and no working scores 3 marks (special case)
An answer of \(g(x) = x^2 + 1\), \(x = 1\) and \(x = 3\) with no working scores 3 marks, as a special case: M1 A1 for g(x), which can be found by inspection, and the final A1 for \(x = 1\) and \(x = 3\); the M1 A1 M1 for forming fg(x) and solving the equation need working (a correct \(fg(x) = 3x^2 - 1\) stated earns its M1 A1). Alternative: writes \(g(x) = ax^2 + bx + c\), expands \(a(3x - 4)^2 + b(3x - 4) + c\) and compares coefficients (M1 for a correct expansion and at least one equation). Dividing \(9x^2 - 24x + 17\) by \(3x - 4\) (treating gf as a product) is M0.
1 mark
A1 \(g(x) = x^2 + 1\)
Must be in terms of x: \(g(t) = t^2 + 1\) is fine if then used correctly.
1 mark
M1 Forms fg(x) by substituting their g(x) into f: \(fg(x) = 3(x^2 + 1) - 4\)
ft their g(x). This mark does not depend on the first two marks.
1 mark
A1 \(fg(x) = 3x^2 - 1\)
May be implied by a correct equation \(3x^2 - 1 = 9x^2 - 24x + 17\).
1 mark
M1 Sets their fg(x) equal to \(9x^2 - 24x + 17\) and solves the resulting three-term quadratic, e.g. \(x^2 - 4x + 3 = 0\)
Usual rules for solving a quadratic; a calculator may be used. Their fg(x) must be a quadratic.
1 mark
A1 \(x = 1\) and \(x = 3\)
Both values are needed, and no others.

Question 4

8 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

The function f is given by \(f(x) = (x - 1)^3 + 1\) for all real values of x. The function f is one-one, so it has an inverse \(f^{-1}\).

Obtain an expression for \(f^{-1}(x)\) and state the set of values of x for which it is defined.

The curve \(y = f(x)\) and the curve \(y = f^{-1}(x)\) cross at exactly three points. With a reason, find the coordinates of these three points.

Model answer

Let \(y = (x - 1)^3 + 1\). Then \((x - 1)^3 = y - 1\), so \(x = 1 + \sqrt[3]{y - 1}\).
\(f^{-1}(x) = 1 + \sqrt[3]{x - 1}\). The range of f is \(\mathbb{R}\), so the domain of \(f^{-1}\) is \(x \in \mathbb{R}\).
f is increasing. Suppose the curves meet at a point \((a, b)\), so \(b = f^{-1}(a)\), i.e. \(f(b) = a\), and also \(b = f(a)\). If \(a < b\), then \(f(a) < f(b)\), i.e. \(b < a\), which is impossible; similarly \(a > b\) is impossible. So \(a = b\): the points lie on \(y = x\) and satisfy \(f(x) = x\).
\((x - 1)^3 + 1 = x\)
\((x - 1)^3 - (x - 1) = 0\)
\((x - 1)\left[(x - 1)^2 - 1\right] = 0\)
\((x - 1)(x - 2)x = 0\)
\(x = 0\), 1 or 2
The curves meet at \((0, 0)\), \((1, 1)\) and \((2, 2)\).

Explanation

Making x the subject of \(y = (x - 1)^3 + 1\) gives \(f^{-1}(x) = 1 + \sqrt[3]{x - 1}\), defined for all real x because the range of f is \(\mathbb{R}\). Because f is increasing, its graph and the graph of its inverse can only meet on the line \(y = x\), so the points come from \(f(x) = x\), which factorises as \(x(x - 1)(x - 2) = 0\). Common errors are giving the domain as \(x \geqslant 1\), assuming without reason that the curves meet on \(y = x\), and dividing by \((x - 1)\), which loses the point \((1, 1)\).

Mark scheme

1 mark
M1 Rearranges \(y = (x - 1)^3 + 1\) to make x the subject: \((x - 1)^3 = y - 1\), \(x = 1 + \sqrt[3]{y - 1}\). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: the question demands all stages of working, and the coordinates must come from algebra. Condone one slip in the rearrangement.
1 mark
A1 \(f^{-1}(x) = 1 + \sqrt[3]{x - 1}\)
Accept \(1 + (x - 1)^{\frac{1}{3}}\) and typed 1 + cbrt(x - 1). Must be labelled \(f^{-1}(x)\) (or \(y = \ldots\) after "\(f^{-1}(x)\):"); \(1 \pm \sqrt[3]{x - 1}\) is A0.
1 mark
B1 Domain \(x \in \mathbb{R}\) (all real numbers)
The range of f is \(\mathbb{R}\), so \(f^{-1}\) is defined for every real x. \(x \geqslant 1\) is B0.
1 mark
B1 Explains that the points of intersection lie on the line \(y = x\), e.g. because f is increasing: if \(f(a) = b\) and \(f(b) = a\) with \(a < b\), then \(f(a) < f(b)\) gives \(b < a\), a contradiction; so it is enough to solve \(f(x) = x\)
Accept a clear argument that an increasing function meets its inverse only on \(y = x\), or solving \(f(x) = f^{-1}(x)\) directly and reducing it correctly to \(f(x) = x\). Simply assuming that the curves meet on \(y = x\) with no reason scores B0, but the later marks are available.
1 mark
M1 Sets \(f(x) = x\) (or \(f^{-1}(x) = x\)) and forms a cubic equation: \((x - 1)^3 + 1 = x\), i.e. \((x - 1)^3 - (x - 1) = 0\) or \(x^3 - 3x^2 + 2x = 0\)
1 mark
M1 (dM1) Factorises their cubic: \((x - 1)\left[(x - 1)^2 - 1\right] = 0\) or \(x(x - 1)(x - 2) = 0\)
Depends on the previous M mark. Roots written down from a calculator score M0.
1 mark
A1 \(x = 0\), \(x = 1\) and \(x = 2\)
All three needed.
1 mark
A1 \((0, 0)\), \((1, 1)\) and \((2, 2)\)
All three coordinate pairs needed.

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