Composite and inverse functions - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 18 exam-style questions on Composite and inverse functions for Edexcel A-Level Maths (Paper 2). 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.

18 questions
2–10 marks each
Instant marking
No sign-up needed

18 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 by
\[f(x) = 7 - \sqrt{2x + a}, \quad x \geqslant -\dfrac{a}{2}\]
where a is a positive constant.

Given that \(f^{-1}(4) = 2\), find the value of a.

State the range of f.

Model answer

\(f^{-1}(4) = 2\) means that \(f(2) = 4\).
\(7 - \sqrt{4 + a} = 4\)
\(\sqrt{4 + a} = 3\)
\(4 + a = 9\), so \(a = 5\)
\(\sqrt{2x + 5}\) is 0 when \(x = -\dfrac{5}{2}\) and takes every value \(\geqslant 0\), so \(7 - \sqrt{2x + 5}\) takes every value \(\leqslant 7\).
Range: \(f(x) \leqslant 7\)

Explanation

If \(f^{-1}(4) = 2\) then f maps 2 to 4, so substituting \(x = 2\) into f and setting it equal to 4 gives a without finding the inverse function. A square root is never negative, so f(x) is never more than 7. Common errors are using \(f(4) = 2\), which gives \(a = 17\), and writing the range in terms of x.

Mark scheme

1 mark
M1 Uses \(f^{-1}(4) = 2 \Rightarrow f(2) = 4\), forming \(7 - \sqrt{4 + a} = 4\); or a correct final answer of \(a = 5\) with no working
An answer of \(a = 5\) and \(f(x) \leqslant 7\) with no working scores 3 marks. Alternative: finds \(f^{-1}(x) = \dfrac{(7 - x)^2 - a}{2}\) and sets \(\dfrac{9 - a}{2} = 2\), for M1. Using \(f(4) = 2\) instead, \(7 - \sqrt{8 + a} = 2\), gives the common wrong answer \(a = 17\): M0 A0.
1 mark
A1 \(a = 5\)
From \(\sqrt{4 + a} = 3\).
1 mark
B1 \(f(x) \leqslant 7\)
Independent of the value of a: the square root is 0 at the start of the domain and takes every non-negative value. Accept \(y \leqslant 7\), \(f \leqslant 7\), \((-\infty, 7]\), \(\{f(x) : f(x) \leqslant 7\}\). Do not accept \(x \leqslant 7\), \(f(x) < 7\) or \(f(x) \geqslant 7\).

Question 2

5 marks

The functions f and g are defined by
\[f(x) = \sqrt{x - 3}, \quad x \geqslant 3\]
\[g(x) = 10 - x^2, \quad x \in \mathbb{R}\]
Find fg(x) in its simplest form.

Using set notation, find the set of values of x for which fg(x) is defined, and state the range of fg.

Find gf(x) in its simplest form, stating the set of values of x for which it is defined.

Model answer

\(fg(x) = f(10 - x^2) = \sqrt{10 - x^2 - 3} = \sqrt{7 - x^2}\)
fg(x) needs \(g(x) = 10 - x^2\) to lie in the domain of f: \(10 - x^2 \geqslant 3\)
\(x^2 \leqslant 7\), so \(-\sqrt{7} \leqslant x \leqslant \sqrt{7}\)
fg(x) is defined for \(\{x : -\sqrt{7} \leqslant x \leqslant \sqrt{7}\}\).
\(\sqrt{7 - x^2}\) is 0 at \(x = \pm\sqrt{7}\) and greatest, \(\sqrt{7}\), at \(x = 0\).
Range of fg: \(0 \leqslant fg(x) \leqslant \sqrt{7}\)
\(gf(x) = g\left(\sqrt{x - 3}\right) = 10 - \left(\sqrt{x - 3}\right)^2 = 10 - (x - 3) = 13 - x\)
gf(x) = g(f(x)) needs x to lie in the domain of f, so \(gf(x) = 13 - x\) for \(x \geqslant 3\).

Explanation

A composite is defined only where the inner function gives values in the domain of the outer one: fg needs \(10 - x^2 \geqslant 3\), which gives the set \(\{x : -\sqrt{7} \leqslant x \leqslant \sqrt{7}\}\), and the range of fg runs from 0 at the ends to \(\sqrt{7}\) at x = 0. For gf, f is applied first, so gf(x) simplifies to 13 - x but is still defined only for \(x \geqslant 3\). Common errors are using \(10 - x^2 \geqslant 0\), giving the set of values without set notation, and forgetting that gf keeps the domain of f.

Mark scheme

1 mark
B1 \(fg(x) = \sqrt{7 - x^2}\); correct final answers of \(fg(x) = \sqrt{7 - x^2}\), \(\{x : -\sqrt{7} \leqslant x \leqslant \sqrt{7}\}\), \(0 \leqslant fg(x) \leqslant \sqrt{7}\) and \(gf(x) = 13 - x\) for \(x \geqslant 3\) with no working score 5 marks
An answer of \(fg(x) = \sqrt{7 - x^2}\), \(\{x : -\sqrt{7} \leqslant x \leqslant \sqrt{7}\}\), \(0 \leqslant fg(x) \leqslant \sqrt{7}\) and \(gf(x) = 13 - x\) for \(x \geqslant 3\) with no working scores 5 marks. From \(\sqrt{(10 - x^2) - 3}\); accept a typed form such as sqrt(7 - x²). \(\sqrt{10 - x^2 - 3}\) left unsimplified is B0 (simplest form asked). Do not accept \(13 - x\) (that is gf) or \(\sqrt{10 - x^2} - 3\).
1 mark
M1 Requires g(x) to lie in the domain of f: \(10 - x^2 \geqslant 3\) (or \(7 - x^2 \geqslant 0\)), and finds the critical values \(x = \pm\sqrt{7}\)
Using \(10 - x^2 \geqslant 0\) (giving \(\pm\sqrt{10}\)) is M0.
1 mark
A1 \(\{x : -\sqrt{7} \leqslant x \leqslant \sqrt{7}\}\)
Set notation is required: accept \(\{x : -\sqrt{7} \leqslant x \leqslant \sqrt{7}\}\), \(\{x \in \mathbb{R} : |x| \leqslant \sqrt{7}\}\), \(\{x : x \geqslant -\sqrt{7}\} \cap \{x : x \leqslant \sqrt{7}\}\). Do not accept \(-\sqrt{7} \leqslant x \leqslant \sqrt{7}\) without set brackets, interval notation \([-\sqrt{7}, \sqrt{7}]\), strict inequalities, or decimals (2.65).
1 mark
B1 \(0 \leqslant fg(x) \leqslant \sqrt{7}\)
\(fg(x) = \sqrt{7 - x^2}\) is 0 at \(x = \pm\sqrt{7}\) and greatest, \(\sqrt{7}\), when \(x = 0\). Accept y in place of fg(x), \([0, \sqrt{7}]\). \(0 \leqslant fg(x) \leqslant \sqrt{10}\) or \(fg(x) \geqslant 0\) is B0.
1 mark
B1 \(gf(x) = 13 - x\), defined for \(x \geqslant 3\)
Both the expression and the set of values are needed for the mark. From \(gf(x) = 10 - \left(\sqrt{x - 3}\right)^2 = 10 - (x - 3)\). gf keeps the domain of f, the function applied first, even though 13 - x looks defined for every x: 13 - x alone, or 'for all real x', is B0. Accept \(\{x : x \geqslant 3\}\) or \([3, \infty)\) here, since set notation is not demanded for this part.

Question 3

6 marks

The functions p and q are defined by
\[p(x) = \ln\left(x^2\right), \quad x \in \mathbb{R}, \ x ≠ 0\]
\[q(x) = \mathrm{e}^{\frac{1}{2}x}, \quad x \in \mathbb{R}\]
Show that \(pq(x) = x\) for all real values of x.

Find qp(x) in its simplest form.

Hence explain why q is not the inverse function of p.

The function r is defined by \(r(x) = \ln\left(x^2\right)\), \(x > 0\). Write down \(r^{-1}(x)\), stating the domain of \(r^{-1}\).

Model answer

\(pq(x) = p\left(\mathrm{e}^{\frac{1}{2}x}\right) = \ln\left(\left(\mathrm{e}^{\frac{1}{2}x}\right)^2\right) = \ln\left(\mathrm{e}^{x}\right) = x\)
This holds for every real x, since \(\mathrm{e}^{\frac{1}{2}x} > 0\) is never 0.
\(qp(x) = q\left(\ln\left(x^2\right)\right) = \mathrm{e}^{\frac{1}{2}\ln\left(x^2\right)} = \mathrm{e}^{\ln\sqrt{x^2}} = \sqrt{x^2} = |x|\)
For x < 0, qp(x) = -x ≠ x (for example, qp(-2) = 2), so q does not undo p for every x in the domain of p. In fact p is many-one, since p(-2) = p(2) = ln 4, so p has no inverse.
For x > 0, \(r(x) = 2\ln x\), which takes every real value.
\(y = \ln\left(x^2\right) \Rightarrow x^2 = \mathrm{e}^{y} \Rightarrow x = \mathrm{e}^{\frac{1}{2}y}\) (x > 0)
\(r^{-1}(x) = \mathrm{e}^{\frac{1}{2}x}, \quad x \in \mathbb{R}\)

Explanation

In pq the exponential is applied first and the logarithm then undoes it, so pq(x) = x for every real x. In qp, \(\mathrm{e}^{\frac{1}{2}\ln\left(x^2\right)} = \sqrt{x^2} = |x|\), which is not x when x is negative, because squaring loses the sign of x: p is many-one and has no inverse. Restricting the domain to x > 0 makes it one-one, and then q is its inverse; a common error is to simplify \(\dfrac{1}{2}\ln\left(x^2\right)\) to \(\ln x\) and conclude that qp(x) = x.

Mark scheme

1 mark
M1 Applies q first and substitutes it into p: \(pq(x) = \ln\left(\left(\mathrm{e}^{\frac{1}{2}x}\right)^2\right)\). An answer with no working scores 0 marks for the show-that; for the rest, correct answers of \(qp(x) = |x|\) and \(r^{-1}(x) = \mathrm{e}^{\frac{1}{2}x}\), \(x \in \mathbb{R}\), with no working score 3 marks
An answer with no working scores 0 marks for the show-that; bare answers of \(qp(x) = |x|\) and \(r^{-1}(x) = \mathrm{e}^{\frac{1}{2}x}\), \(x \in \mathbb{R}\), with no working score 3 marks (M1 A1 B1), and the explanation earns its B1 on its own. Working with qp(x) instead of pq(x) for the show-that is M0.
1 mark
A1* \(pq(x) = \ln\left(\mathrm{e}^{x}\right) = x\) cso
\(\left(\mathrm{e}^{\frac{1}{2}x}\right)^2 = \mathrm{e}^{x}\) (or \(2 \ln\left(\mathrm{e}^{\frac{1}{2}x}\right) = 2 \times \dfrac{1}{2}x\)) must be seen, with no errors. This holds for every real x because \(q(x) = \mathrm{e}^{\frac{1}{2}x} > 0\), so q(x) is never 0 and always lies in the domain of p; this observation is not required for the mark.
1 mark
M1 Applies p first: \(qp(x) = \mathrm{e}^{\frac{1}{2}\ln\left(x^2\right)}\), and uses the power law of logarithms to write it as \(\mathrm{e}^{\ln\sqrt{x^2}}\) or \(\left(x^2\right)^{\frac{1}{2}}\)
Independent of the show-that. Writing \(\dfrac{1}{2}\ln\left(x^2\right) = \ln x\) and reaching \(qp(x) = x\) is a common error: M1 A0.
1 mark
A1 \(qp(x) = |x|\)
Accept \(\sqrt{x^2}\), typed |x| or abs(x), or the two-case form: x for x > 0 and -x for x < 0. qp(x) = x is A0. Condone the omission of x ≠ 0.
1 mark
B1 Explains, e.g. for x < 0, qp(x) = -x ≠ x (for example qp(-2) = 2), so q does not undo p for every x in the domain of p; or p is many-one (p(-2) = p(2) = ln 4), so p has no inverse
The reason must refer to negative values of x (or to a pair of values such as -2 and 2), or to p being many-one. 'Because qp(x) ≠ pq(x)' with no reference to negative x is B0.
1 mark
B1 \(r^{-1}(x) = \mathrm{e}^{\frac{1}{2}x}\), \(x \in \mathbb{R}\)
Both the expression and the domain are needed. For x > 0, \(r(x) = 2\ln x\) takes every real value, so the domain of \(r^{-1}\) is all real numbers. Accept \(\sqrt{\mathrm{e}^{x}}\) or any equivalent typed form, and 'for all real x'. A domain of x > 0 is B0.

Question 4

8 marks

Figure 1 shows the graph of \(y = f(x)\), where
\[f(x) = \dfrac{12}{x + 4}, \quad 0 \leqslant x \leqslant 8\]
The end points of the graph are \(A(0, 3)\) and \(B(8, 1)\).

State the range of f.

Find an expression for \(f^{-1}(x)\) and state the domain of \(f^{-1}\).

Write down the coordinates of the end points of the graph of \(y = f^{-1}(x)\).

Using algebra, find the coordinates of the point where the graph of \(y = f(x)\) meets the graph of \(y = f^{-1}(x)\).

Figure 1: a sketch of the graph of y = f(x) on axes with origin O and no scale. The graph starts at the point A(0, 3) on the y-axis, marked with a dot, and falls, steeply at first and then more and more slowly, to the point B(8, 1), also marked with a dot, where it ends.

Model answer

f is decreasing for \(0 \leqslant x \leqslant 8\), from \(f(0) = 3\) to \(f(8) = 1\).
Range: \(1 \leqslant f(x) \leqslant 3\)
\(y = \dfrac{12}{x + 4} \Rightarrow x + 4 = \dfrac{12}{y} \Rightarrow x = \dfrac{12}{y} - 4\)
\(f^{-1}(x) = \dfrac{12}{x} - 4, \quad 1 \leqslant x \leqslant 3\)
The graph of \(y = f^{-1}(x)\) is the reflection of the graph of \(y = f(x)\) in the line \(y = x\), so its end points are \((3, 0)\) and \((1, 8)\).
\(\dfrac{12}{x + 4} = \dfrac{12}{x} - 4\)
\(12x = 12(x + 4) - 4x(x + 4)\)
\(4x^2 + 16x - 48 = 0\)
\(x^2 + 4x - 12 = 0\)
\((x + 6)(x - 2) = 0\), so \(x = 2\) or \(x = -6\)
\(x = -6\) is outside the domains, so \(x = 2\) and \(y = f(2) = 2\).
The graphs meet at \((2, 2)\).

Explanation

The range of f becomes the domain of \(f^{-1}\), and reflecting in \(y = x\) swaps the coordinates of every point, so A and B become \((3, 0)\) and \((1, 8)\). Setting \(f(x) = f^{-1}(x)\) gives \(x^2 + 4x - 12 = 0\), and the root \(x = -6\) lies outside the domains and is rejected. Common errors are giving the domain of \(f^{-1}\) as \(0 \leqslant x \leqslant 8\) and keeping \(x = -6\).

Mark scheme

1 mark
B1 \(1 \leqslant f(x) \leqslant 3\); correct final answers for the range, the inverse with its domain and the end points with no working score 6 marks
An answer of \(1 \leqslant f(x) \leqslant 3\), \(f^{-1}(x) = \dfrac{12}{x} - 4\) with domain \(1 \leqslant x \leqslant 3\), and end points \((3, 0)\) and \((1, 8)\) with no working scores 6 marks; the point \((2, 2)\) needs the algebra ('Using algebra'), so the final M1 A1 need working. Accept \(1 \leqslant y \leqslant 3\), \([1, 3]\); not strict inequalities or a range in terms of x.
1 mark
M1 Makes x the subject of \(y = \dfrac{12}{x + 4}\): \(x + 4 = \dfrac{12}{y}\)
Swapping x and y first is fine.
1 mark
A1 \(f^{-1}(x) = \dfrac{12}{x} - 4\)
Accept \(\dfrac{12 - 4x}{x}\), \(\dfrac{4(3 - x)}{x}\), typed 12/x - 4. Must be labelled \(f^{-1}(x) =\).
1 mark
B1ft Domain \(1 \leqslant x \leqslant 3\)
Follow through their range of f, written in terms of x. \(0 \leqslant x \leqslant 8\) (the domain of f) is B0.
1 mark
B1 \((3, 0)\)
The reflection of A in the line \(y = x\).
1 mark
B1 \((1, 8)\)
The reflection of B in the line \(y = x\).
1 mark
M1 Forms an equation from \(f(x) = f^{-1}(x)\), such as \(\dfrac{12}{x + 4} = \dfrac{12}{x} - 4\) (or from \(f(x) = x\)), reaches a 3-term quadratic, \(x^2 + 4x - 12 = 0\), and solves it
Both equations lead to \(x^2 + 4x - 12 = 0\). The quadratic may be solved on a calculator once it has been formed, but a point given with no equation seen scores M0 A0.
1 mark
A1 \((2, 2)\) only, with \(x = -6\) rejected
\(x = -6\) is outside the domains of f and \(f^{-1}\) and must be rejected or not used.

Ready to try the rest? Every answer you write is marked against the mark scheme, so you can see exactly where the marks are.

Study Helper Hints first, answers after you try

Stuck on this one? Ask for a hint and I’ll point you in the right direction, without giving the answer away. Once you’ve checked your answer, I’ll show you how the marks are awarded.