IB DP Maths Topic 8.3 Composition of functions and inverse functions HL Paper 3


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The function f is defined by

\[f(x) = \frac{{1 – {{\text{e}}^{ – x}}}}{{1 + {{\text{e}}^{ – x}}}},{\text{ }}x \in \mathbb{R}{\text{ .}}\]

(a)     Find the range of f .

(b)     Prove that f is an injection.

(c)     Taking the codomain of f to be equal to the range of f , find an expression for \({f^{ – 1}}(x)\) .



(a)     \(\left] { – 1,1} \right[\)     A1A1

Note: Award A1 for the values –1, 1 and A1 for the open interval.


[2 marks]


(b)     EITHER

Let \(\frac{{1 – {{\text{e}}^{ – x}}}}{{1 + {{\text{e}}^{ – x}}}} = \frac{{1 – {{\text{e}}^{ – y}}}}{{1 + {{\text{e}}^{ – y}}}}\)     M1

\(1 – {{\text{e}}^{ – x}} + {{\text{e}}^{ – y}} – {{\text{e}}^{ – (x + y)}} = 1 + {{\text{e}}^{ – x}} – {{\text{e}}^{ – y}} – {{\text{e}}^{ – (x + y)}}\)     A1

\({{\text{e}}^{ – x}} = {{\text{e}}^{ – y}}\)

\(x = y\)     A1

Therefore f is an injection     AG



\(f'(x) = \frac{{{{\text{e}}^{ – x}}(1 + {{\text{e}}^{ – x}}) + {{\text{e}}^{ – x}}(1 – {{\text{e}}^{ – x}})}}{{{{(1 + {{\text{e}}^{ – x}})}^2}}}\)     M1

\( = \frac{{2{{\text{e}}^{ – x}}}}{{{{(1 + {{\text{e}}^{ – x}})}^2}}}\)     A1

\( > 0\) for all x.     A1

Therefore f is an injection.     AG

Note: Award M1A1A0 for a graphical solution.


[3 marks]


(c)     Let \(y = \frac{{1 – {{\text{e}}^{ – x}}}}{{1 + {{\text{e}}^{ – x}}}}\)     M1

\(y(1 + {{\text{e}}^{ – x}}) = 1 – {{\text{e}}^{ – x}}\)     A1

\({{\text{e}}^{ – x}}(1 + y) = 1 – y\)     A1

\({{\text{e}}^{ – x}} = \frac{{1 – y}}{{1 + y}}\)

\(x = \ln \left( {\frac{{1 + y}}{{1 – y}}} \right)\)     A1

\({f^{ – 1}}(x) = \ln \left( {\frac{{1 + x}}{{1 – x}}} \right)\)     A1

[5 marks]

Total [10 marks]

Examiners report

Most candidates found the range of f correctly. Two algebraic methods were seen for solving (b), either showing that the derivative of f is everywhere positive or showing that \(f(a) = f(b) \Rightarrow a = b\) . Candidates who based their ‘proof’ on a graph produced on their graphical calculators were given only partial credit on the grounds that the whole domain could not be shown and, in any case, it was not clear from the graph that f was an injection.


Two functions, F and G , are defined on \(A = \mathbb{R}\backslash \{ 0,{\text{ }}1\} \) by

\[F(x) = \frac{1}{x},{\text{ }}G(x) = 1 – x,{\text{ for all }}x \in A.\]

(a)     Show that under the operation of composition of functions each function is its own inverse.

(b)     F and G together with four other functions form a closed set under the operation of composition of functions.

Find these four functions.



(a)     the following two calculations show the required result

\(F \circ F(x) = \frac{1}{{\frac{1}{x}}} = x\)

\(G \circ G(x) = 1 – (1 – x) = x\)     M1A1A1

[3 marks]


(b)     part (a) shows that the identity function defined by I(x) = x belongs to S     A1

the two compositions of F and G are:

\(F \circ G(x) = \frac{1}{{1 – x}};\)     (M1)A1

\(G \circ F(x) = 1 – \frac{1}{x}\left( { = \frac{{x – 1}}{x}} \right)\)     (M1)A1

the final element is

\(G \circ F \circ G(x) = 1 – \frac{1}{{1 – x}}\left( { = \frac{x}{{x – 1}}} \right)\)     (M1)A1

[7 marks]

Total [10 marks]

Examiners report

This question was generally well done. In part(a), the quickest answer involved showing that squaring the function gave the identity. Some candidates went through the more elaborate method of finding the inverse function in each case.


Three functions mapping \(\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}\) are defined by

\[{f_1}(m,{\text{ }}n) = m – n + 4;\,\,\,{f_2}(m,{\text{ }}n) = \left| m \right|;\,\,\,{f_3}(m,{\text{ }}n) = {m^2} – {n^2}.\]

Two functions mapping \(\mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}\) are defined by

\[{g_1}(k) = (2k,{\text{ }}k);\,\,\,{g_2}(k) = \left( {k,{\text{ }}\left| k \right|} \right).\]

(a)     Find the range of

(i)     \({f_1} \circ {g_1}\) ;

(ii)     \({f_3} \circ {g_2}\) .

(b)     Find all the solutions of \({f_1} \circ {g_2}(k) = {f_2} \circ {g_1}(k)\) .

(c)     Find all the solutions of \({f_3}(m,{\text{ }}n) = p\) in each of the cases p =1 and p = 2 .



(a)     (i)     \({f_1} \circ {g_1}(k) = k + 4\)     M1

Range \(({f_1} \circ {g_1}) = \mathbb{Z}\)     A1


(ii)     \({f_3} \circ {g_2}(k) = 0\)     M1

Range \(({f_3} \circ {g_2}) = \{ 0\} \)     A1

[4 marks]

(b)     the equation to solve is

\(k – \left| k \right| + 4 = \left| {2k} \right|\)     M1A1

the positive solution is k = 2     A1

the negative solution is k = –1     A1

[4 marks]

(c)     the equation factorizes: \((m + n)(m – n) = p\)     (M1)

for p = 1 , the possible factors over \(\mathbb{Z}\) are \(m + n =  \pm 1,{\text{ }}m – n =  \pm 1\)     (M1)(A1)

with solutions (1, 0) and (–1, 0)     A1

for p = 2 , the possible factors over \(\mathbb{Z}\) are \(m + n = \pm 1,{\text{ }} \pm 2;{\text{  }}m – n = \pm 2,{\text{ }} \pm 1\)     M1A1

there are no solutions over \(\mathbb{Z} \times \mathbb{Z}\)     A1

[7 marks]

Total [15 marks]

Examiners report

The majority of candidates were able to compute the composite functions involved in parts (a) and (b). Part(c) was satisfactorily tackled by a minority of candidates. There were more GDC solutions than the more obvious approach of factorizing a difference of squares. Some candidates seemed to forget that m and n belonged to the set of integers.


(a)     Show that \(f:\mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}\) defined by \(f(x,{\text{ }}y) = (2x + y,{\text{ }}x – y)\) is a bijection.

(b)     Find the inverse of f .



(a)     we need to show that the function is both injective and surjective to be a bijection     R1

suppose \(f(x,{\text{ }}y) = f(u,{\text{ }}v)\)     M1

\((2x + y,{\text{ }}x – y) = (2u + v,{\text{ }}u – v)\)

forming a pair of simultaneous equations     M1

\(2x + y = 2u + v\)     (i)

\(x – y = u – v\)     (ii)

\((i) + (ii) \Rightarrow 3x = 3u \Rightarrow x = u\)     A1

\((i) – 2(ii) \Rightarrow 3y = 3v \Rightarrow y = v\)     A1

hence function is injective     R1

let \(2x + y = s\) and \(x – y = t\)     M1

\( \Rightarrow 3x = s + t\)

\( \Rightarrow x = \frac{{s + t}}{3}\)     A1

also \(3y = s – 2t\)

\( \Rightarrow y = \frac{{s – 2t}}{3}\)     A1

for any \((s,{\text{ }}t) \in \mathbb{R} \times \mathbb{R}\) there exists \((x,{\text{ }}y) \in \mathbb{R} \times \mathbb{R}\) and the function is surjective     R1

[10 marks]


(b)     the inverse is \({f^{ – 1}}(x,{\text{ }}y) = \left( {\frac{{x + y}}{3},{\text{ }}\frac{{x – 2y}}{3}} \right)\)     A1

[1 mark]

Total [11 marks]

Examiners report

Many students were able to show that the expression was injective, but found more difficulty in showing it was subjective. As with question 1 part (e), a number of candidates did not realise that the answer to part (b) came directly from part (a), hence the reason for it being worth only one mark.


The function \(f:[0,{\text{ }}\infty [ \to [0,{\text{ }}\infty [\) is defined by \(f(x) = 2{{\text{e}}^x} + {{\text{e}}^{ – x}} – 3\) .

(a)     Find \(f'(x)\) .

(b)     Show that f is a bijection.

(c)     Find an expression for \({f^{ – 1}}(x)\) .



(a)     \(f'(x) = 2{{\text{e}}^x} – {{\text{e}}^{ – x}}\)     A1

[1 mark]


(b)     f is an injection because \(f'(x) > 0\) for \(x \in [0,{\text{ }}\infty [\)     R2

(accept GDC solution backed up by a correct graph)

since \(f(0) = 0\) and \(f(x) \to \infty \) as \(x \to \infty \) , (and f is continuous) it is a surjection     R1

hence it is a bijection     AG

[3 marks]


(c)     let \(y = 2{{\text{e}}^x} + {{\text{e}}^{ – x}} – 3\)     M1

so \(2{{\text{e}}^{2x}} – (y + 3){{\text{e}}^x} + 1 = 0\)     A1

\({{\text{e}}^x} = \frac{{y + 3 \pm \sqrt {{{(y + 3)}^2} – 8} }}{4}\)     A1

\(x = \ln \left( {\frac{{y + 3 \pm \sqrt {{{(y + 3)}^2} – 8} }}{4}} \right)\)     A1

since \(x \geqslant 0\) we must take the positive square root     (R1)

\({f^{ – 1}}(x) = \ln \left( {\frac{{x + 3 + \sqrt {{{(x + 3)}^2} – 8} }}{4}} \right)\)     A1

[6 marks]

Total [10 marks]

Examiners report

In many cases the attempts at showing that f is a bijection were unconvincing. The candidates were guided towards showing that f is an injection by noting that \(f'(x) > 0\) for all x, but some candidates attempted to show that \(f(x) = f(y) \Rightarrow x = y\) which is much more difficult. Solutions to (c) were often disappointing, with the algebra defeating many candidates.


The function \(f:\mathbb{R} \to \mathbb{R}\) is defined by

\[f(x) = 2{{\text{e}}^x} – {{\text{e}}^{ – x}}.\]

(a)     Show that f is a bijection.

(b)     Find an expression for \({f^{ – 1}}(x)\).



(a)     EITHER


\(f'(x) = 2{{\text{e}}^x} – {{\text{e}}^{ – x}} > 0\) for all x     M1A1

so f is an injection     A1


let \(2{{\text{e}}^x} – {{\text{e}}^{ – x}} = 2{{\text{e}}^y} – {{\text{e}}^{ – y}}\)     M1

\(2({{\text{e}}^x} – {{\text{e}}^y}) + {{\text{e}}^{ – y}} – {{\text{e}}^{ – x}} = 0\)

\(2({{\text{e}}^x} – {{\text{e}}^y}) + {{\text{e}}^{ – (x + y)}}({{\text{e}}^x} – {{\text{e}}^y}) = 0\)

\(\left( {2 + {{\text{e}}^{ – (x + y)}}} \right)({{\text{e}}^x} – {{\text{e}}^y}) = 0\)

\({{\text{e}}^x} = {{\text{e}}^y}\)

\(x = y\)     A1

Note: Sufficient working must be shown to gain the above A1.


so f is an injection     A1

Note: Accept a graphical justification i.e. horizontal line test.



it is also a surjection (accept any justification including graphical)     R1

therefore it is a bijection     AG

[4 marks]


(b)     let \(y = 2{{\text{e}}^x} – {{\text{e}}^{ – x}}\)     M1

\(2{{\text{e}}^{2x}} – y{{\text{e}}^x} – 1 = 0\)     A1

\({{\text{e}}^x} = \frac{{y \pm \sqrt {{y^2} + 8} }}{4}\)     M1A1

since \({{\text{e}}^x}\) is never negative, we take the + sign     R1

\({f^{ – 1}}(x) = \ln \left( {\frac{{x + \sqrt {{x^2} + 8} }}{4}} \right)\)     A1

[6 marks]

Total [10 marks]

Examiners report

Solutions to (a) were often disappointing. Many candidates tried to use the result that, for an injection, \(f(a) = f(b) \Rightarrow a = b\) – although this is the definition, it is often much easier to proceed by showing that the derivative is everywhere positive or everywhere negative or even to use a horizontal line test. Although (b) is based on core material, solutions were often disappointing with some very poor use of algebra seen.


Let \(X\) and \(Y\) be sets. The functions \(f:X \to Y\) and \(g:Y \to X\) are such that \(g \circ f\) is the identity function on \(X\).

Prove that: 

(i)     \(f\) is an injection,

(ii)     \(g\) is a surjection.


Given that \(X = {\mathbb{R}^ + } \cup \{ 0\} \) and \(Y = \mathbb{R}\), choose a suitable pair of functions \(f\) and \(g\) to show that \(g\) is not necessarily a bijection.



(i)     to test injectivity, suppose \(f({x_1}) = f({x_2})\)     M1

apply \(g\) to both sides \(g\left( {f({x_1})} \right) = g\left( {f({x_2})} \right)\)     M1

\( \Rightarrow {x_1} = {x_2}\)     A1

so \(f\) is injective     AG

Note:     Do not accept arguments based on “\(f\) has an inverse”.

(ii)     to test surjectivity, suppose \(x \in X\)     M1

define \(y = f(x)\)     M1

then \(g(y) = g\left( {f(x)} \right) = x\)     A1

so \(g\) is surjective     AG

[6 marks]


choose, for example, \(f(x) = \sqrt x \) and \(g(y) = {y^2}\)     A1

then \(g \circ f(x) = {\left( {\sqrt x } \right)^2} = x\)     A1

the function \(g\) is not injective as \(g(x) = g( – x)\)     R1

[3 marks]

Total [9 marks]


Examiners report

Those candidates who formulated the questions in terms of the basic definitions of injectivity and surjectivity were usually sucessful. Otherwise, verbal attempts such as ‘\(f{\text{ is one – to – one }} \Rightarrow f{\text{ is injective}}\)’ or ‘\(g\) is surjective because its range equals its codomain’, received no credit. Some candidates made the false assumption that \(f\) and \(g\) were mutual inverses.


Few candidates gave completely satisfactory answers. Some gave functions satisfying the mutual identity but either not defined on the given sets or for which \(g\) was actually a bijection.


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