Question
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 \({f^{ – 1}}(x)\) .
▶️Answer/Explanation
Markscheme
(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
OR
Consider
\(f'(x) = \frac{{{{\text{e}}^{ – x}}(1 + {{\text{e}}^{ – x}}) + {{\text{e}}^{ – x}}(1 – {{\text{e}}^{ – x}})}}{{{{(1 + {{\text{e}}^{ – x}})}^2}}}\) 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.
Question
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.
▶️Answer/Explanation
Markscheme
(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.
Question
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 .
▶️Answer/Explanation
Markscheme
(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.
Question
(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 .
▶️Answer/Explanation
Markscheme
(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.
Question
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 \({f^{ – 1}}(x)\) .
▶️Answer/Explanation
Markscheme
(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.
Question
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 \({f^{ – 1}}(x)\).
▶️Answer/Explanation
Markscheme
(a) EITHER
consider
\(f'(x) = 2{{\text{e}}^x} – {{\text{e}}^{ – x}} > 0\) for all x M1A1
so f is an injection A1
OR
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.
THEN
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.