Question
Below are the graphs of the two functions \(F:P \to Q{\text{ and }}g:A \to B\) .
a.Determine, with reference to features of the graphs, whether the functions are injective and/or surjective.[4]
Show that
(i) if both h and k are injective then so is the composite function \(k \circ h\) ;
(ii) if both h and k are surjective then so is the composite function \(k \circ h\) .[9]
▶️Answer/Explanation
Markscheme
f is surjective because every horizontal line through Q meets the graph somewhere R1
f is not injective because it is a many-to-one function R1
g is injective because it always has a positive gradient R1
(accept horizontal line test reasoning)
g is not surjective because a horizontal line through the negative part of B would not meet the graph at all R1
[4 marks]
(i) EITHER
Let \({x_1},{\text{ }}{x_2} \in X{\text{ and }}{y_1} = h({x_1}){\text{ and }}{y_2} = h({x_2})\) M1
Then
\(k \circ \left( {h({x_1})} \right) = k \circ \left( {h({x_2})} \right)\)
\( \Rightarrow k({y_1}) = k({y_2})\) A1
\( \Rightarrow {y_1} = {y_2}\,\,\,\,\,{\text{(}}k{\text{ is injective)}}\) A1
\( \Rightarrow h({x_1}) = h({x_2})\,\,\,\,\,\left( {h({x_1}) = {y_1}{\text{ and }}h({x_2}) = {y_2}} \right)\) A1
\( \Rightarrow {x_1} \equiv {x_2}\,\,\,\,\,(h{\text{ is injective)}}\) A1
Hence \(k \circ h\) is injective AG
OR
\({{\text{x}}_1},{\text{ }}{x_2} \in X,{\text{ }}{x_1} \ne {x_2}\) M1
since h is an injection \( \Rightarrow h({x_1}) \ne h({x_2})\) A1
\(h({x_1}),{\text{ }}h({x_2}) \in Y\) A1
since k is an injection \( \Rightarrow k\left( {h({x_1})} \right) \ne k\left( {h({x_2})} \right)\) A1
\(k\left( {h({x_1})} \right),{\text{ }}k\left( {h({x_2})} \right) \in \mathbb{Z}\) A1
so \(k \circ h\) is an injection. AG
(ii) h and k are surjections and let \(z \in \mathbb{Z}\)
Since k is surjective there exists \(y \in Y\) such that k(y) = z R1
Since h is surjective there exists \(x \in X\) such that h(x) = y R1
Therefore there exists \(x \in X\) such that
\(k \circ h(x) = k\left( {h(x)} \right)\)
\( = k(y)\) R1
\( = z\) A1
So \(k \circ h\) is surjective AG
[9 marks]
Examiners report
‘Using features of the graph’ should have been a fairly open hint but too many candidates contented themselves with describing what injective and surjective meant rather than explaining which graph had which properties. Candidates found considerable difficulty with presenting a convincing argument in part (b).
‘Using features of the graph’ should have been a fairly open hint but too many candidates contented themselves with describing what injective and surjective meant rather than explaining which graph had which properties. Candidates found considerable difficulty with presenting a convincing argument in part (b).
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.
Question
The function \(f:{\mathbb{R}^ + } \times {\mathbb{R}^ + } \to {\mathbb{R}^ + } \times {\mathbb{R}^ + }\) is defined by \(f(x,{\text{ }}y) = \left( {x{y^2},\frac{x}{y}} \right)\).
Show that f is a bijection.
▶️Answer/Explanation
Markscheme
for f to be a bijection it must be both an injection and a surjection R1
Note: Award this R1 for stating this anywhere.
injection:
let \(f(a{\text{, }}b) = f(c,{\text{ }}d)\) so that (M1)
\(a{b^2} = c{d^2}\) and \(\frac{a}{b} = \frac{c}{d}\) A1
dividing the equations,
\({b^3} = {d^3}\) so \(b = d\) A1
substituting,
a = c A1
it follows that f is an injection because \(f(a{\text{, }}b) = f(c,{\text{ }}d) \Rightarrow (a{\text{, }}b) = (c,{\text{ }}d)\) R1
surjection:
let \(f(a{\text{, }}b) = (c,{\text{ }}d)\) where \((c,{\text{ }}d) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + }\) (M1)
then \(c = a{b^2}\) and \(d = \frac{a}{b}\) A1
dividing,
\({b^3} = \frac{c}{d}\) so \(b = \sqrt[3]{{\frac{c}{d}}}\) A1
substituting,
\(a = d \times \sqrt[3]{{\frac{c}{d}}}\) A1
it follows that f is a surjection because
given \((c,{\text{ }}d) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + }\) , there exists \((a,{\text{ }}b) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + }\) such that \(f(a{\text{, }}b) = (c,{\text{ }}d)\) R1
therefore f is a bijection AG
[11 marks]
Examiners report
Candidates who knew that they were required to give a rigorous demonstration that f was injective and surjective were generally successful, although the formality that is needed in this style of demonstration was often lacking. Some candidates, however, tried unsuccessfully to give a verbal explanation or even a 2-D version of the horizontal line test. In 2-D, the only reliable method for showing that a function f is injective is to show that \(f(a{\text{, }}b) = f(c,{\text{ }}d) \Rightarrow (a{\text{, }}b) = (c,{\text{ }}d)\).