## 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

**for the open interval.**

*A1** *

*[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

Below are the graphs of the two functions \(F:P \to Q{\text{ and }}g:A \to B\) .

Determine, with reference to features of the graphs, whether the functions are injective and/or surjective.

Given two functions \(h:X \to Y{\text{ and }}k:Y \to Z\) .

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 \(k \circ h\) .

**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

Let \(f:\mathbb{Z} \times \mathbb{R} \to \mathbb{R},{\text{ }}f(m,{\text{ }}x) = {( – 1)^m}x\). Determine whether *f* is

(i) surjective;

(ii) injective.

*P* is the set of all polynomials such that \(P = \left\{ {\sum\limits_{i = 0}^n {{a_i}{x^i}|n \in \mathbb{N}} } \right\}\).

Let \(g:P \to P,{\text{ }}g(p) = xp\). Determine whether *g* is

(i) surjective;

(ii) injective.

Let \(h:\mathbb{Z} \to {\mathbb{Z}^ + }\), \(h(x) = \left\{ {\begin{array}{*{20}{c}}

{2x,}&{x > 0} \\

{1 – 2x,}&{x \leqslant 0}

\end{array}} \right\}\). Determine whether *h* is

(i) surjective;

(ii) injective.

**Answer/Explanation**

## Markscheme

(i) let \(x \in \mathbb{R}\)

for example, \(f(0,{\text{ }}x) = x\), *M1*

hence *f* is surjective *A1*

* *

(ii) for example, \(f(2,{\text{ }}3) = f(4,{\text{ }}3) = 3,{\text{ but }}(2,{\text{ }}3) \ne (4,{\text{ }}3)\) *M1*

hence *f* is not injective *A1*

*[4 marks]*

(i) there is no element of *P* such that \(g(p) = 7\), for example *R1*

hence *g* is not surjective *A1*

* *

(ii) \(g(p) = g(q) \Rightarrow xp = xq \Rightarrow p = q\), hence *g* is injective *M1A1*

*[4 marks]*

(i) for \(x > 0,{\text{ }}h(x) = 2,{\text{ }}4,{\text{ }}6,{\text{ }}8 \ldots \) *A1*

for \(x \leqslant 0,{\text{ }}h(x) = 1,{\text{ }}3,{\text{ }}5,{\text{ }}7 \ldots \) *A1*

therefore *h* is surjective *A1*

* *

(ii) for \(h(x) = h(y)\), since an odd number cannot equal an even number, there are only two possibilities: *R1*

\(x,{\text{ }}y > 0,{\text{ }}2x = 2y \Rightarrow x = y;\) *A1*

\(x,{\text{ }}y \leqslant 0,{\text{ }}1 – 2x = 1 – 2y \Rightarrow x = y\) *A1*

therefore *h* is injective *A1*

**Note:** This can be demonstrated in a variety of ways.

*[7 marks]*

## Examiners report

This was the least successfully answered question on the paper. Candidates often could quote the definitions of surjective and injective, but often could not apply the definitions in the examples.

a) Some candidates failed to show convincingly that the function was surjective, and not injective.

This was the least successfully answered question on the paper. Candidates often could quote the definitions of surjective and injective, but often could not apply the definitions in the examples.

b) Some candidates had trouble interpreting the notation used in the question, hence could not answer the question successfully.

This was the least successfully answered question on the paper. Candidates often could quote the definitions of surjective and injective, but often could not apply the definitions in the examples.

c) Many candidates failed to appreciate that the function is discrete, and hence erroneously attempted to differentiate the function to show that it is monotonic increasing, hence injective. Others who provided a graph again showed a continuous rather than discrete function.

## Question

Let \(f:\mathbb{Z} \times \mathbb{R} \to \mathbb{R},{\text{ }}f(m,{\text{ }}x) = {( – 1)^m}x\). Determine whether *f* is

(i) surjective;

(ii) injective.

*P* is the set of all polynomials such that \(P = \left\{ {\sum\limits_{i = 0}^n {{a_i}{x^i}|n \in \mathbb{N}} } \right\}\).

Let \(g:P \to P,{\text{ }}g(p) = xp\). Determine whether *g* is

(i) surjective;

(ii) injective.

Let \(h:\mathbb{Z} \to {\mathbb{Z}^ + }\), \(h(x) = \left\{ {\begin{array}{*{20}{c}}

{2x,}&{x > 0} \\

{1 – 2x,}&{x \leqslant 0}

\end{array}} \right\}\). Determine whether *h* is

(i) surjective;

(ii) injective.

**Answer/Explanation**

## Markscheme

(i) let \(x \in \mathbb{R}\)

for example, \(f(0,{\text{ }}x) = x\), *M1*

hence *f* is surjective *A1*

* *

(ii) for example, \(f(2,{\text{ }}3) = f(4,{\text{ }}3) = 3,{\text{ but }}(2,{\text{ }}3) \ne (4,{\text{ }}3)\) *M1*

hence *f* is not injective *A1*

*[4 marks]*

(i) there is no element of *P* such that \(g(p) = 7\), for example *R1*

hence *g* is not surjective *A1*

* *

(ii) \(g(p) = g(q) \Rightarrow xp = xq \Rightarrow p = q\), hence *g* is injective *M1A1*

*[4 marks]*

(i) for \(x > 0,{\text{ }}h(x) = 2,{\text{ }}4,{\text{ }}6,{\text{ }}8 \ldots \) *A1*

for \(x \leqslant 0,{\text{ }}h(x) = 1,{\text{ }}3,{\text{ }}5,{\text{ }}7 \ldots \) *A1*

therefore *h* is surjective *A1*

* *

(ii) for \(h(x) = h(y)\), since an odd number cannot equal an even number, there are only two possibilities: *R1*

\(x,{\text{ }}y > 0,{\text{ }}2x = 2y \Rightarrow x = y;\) *A1*

\(x,{\text{ }}y \leqslant 0,{\text{ }}1 – 2x = 1 – 2y \Rightarrow x = y\) *A1*

therefore *h* is injective *A1*

**Note:** This can be demonstrated in a variety of ways.

*[7 marks]*

## Examiners report

a) Some candidates failed to show convincingly that the function was surjective, and not injective.

b) Some candidates had trouble interpreting the notation used in the question, hence could not answer the question successfully.

c) Many candidates failed to appreciate that the function is discrete, and hence erroneously attempted to differentiate the function to show that it is monotonic increasing, hence injective. Others who provided a graph again showed a continuous rather than discrete function.

## Question

Let \(f:\mathbb{Z} \times \mathbb{R} \to \mathbb{R},{\text{ }}f(m,{\text{ }}x) = {( – 1)^m}x\). Determine whether *f* is

(i) surjective;

(ii) injective.

*P* is the set of all polynomials such that \(P = \left\{ {\sum\limits_{i = 0}^n {{a_i}{x^i}|n \in \mathbb{N}} } \right\}\).

Let \(g:P \to P,{\text{ }}g(p) = xp\). Determine whether *g* is

(i) surjective;

(ii) injective.

Let \(h:\mathbb{Z} \to {\mathbb{Z}^ + }\), \(h(x) = \left\{ {\begin{array}{*{20}{c}}

{2x,}&{x > 0} \\

{1 – 2x,}&{x \leqslant 0}

\end{array}} \right\}\). Determine whether *h* is

(i) surjective;

(ii) injective.

**Answer/Explanation**

## Markscheme

(i) let \(x \in \mathbb{R}\)

for example, \(f(0,{\text{ }}x) = x\), *M1*

hence *f* is surjective *A1*

* *

(ii) for example, \(f(2,{\text{ }}3) = f(4,{\text{ }}3) = 3,{\text{ but }}(2,{\text{ }}3) \ne (4,{\text{ }}3)\) *M1*

hence *f* is not injective *A1*

*[4 marks]*

(i) there is no element of *P* such that \(g(p) = 7\), for example *R1*

hence *g* is not surjective *A1*

* *

(ii) \(g(p) = g(q) \Rightarrow xp = xq \Rightarrow p = q\), hence *g* is injective *M1A1*

*[4 marks]*

(i) for \(x > 0,{\text{ }}h(x) = 2,{\text{ }}4,{\text{ }}6,{\text{ }}8 \ldots \) *A1*

for \(x \leqslant 0,{\text{ }}h(x) = 1,{\text{ }}3,{\text{ }}5,{\text{ }}7 \ldots \) *A1*

therefore *h* is surjective *A1*

* *

(ii) for \(h(x) = h(y)\), since an odd number cannot equal an even number, there are only two possibilities: *R1*

\(x,{\text{ }}y > 0,{\text{ }}2x = 2y \Rightarrow x = y;\) *A1*

\(x,{\text{ }}y \leqslant 0,{\text{ }}1 – 2x = 1 – 2y \Rightarrow x = y\) *A1*

therefore *h* is injective *A1*

**Note:** This can be demonstrated in a variety of ways.

*[7 marks]*

## Examiners report

a) Some candidates failed to show convincingly that the function was surjective, and not injective.

b) Some candidates had trouble interpreting the notation used in the question, hence could not answer the question successfully.

c) Many candidates failed to appreciate that the function is discrete, and hence erroneously attempted to differentiate the function to show that it is monotonic increasing, hence injective. Others who provided a graph again showed a continuous rather than discrete function.

## 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 \(f(a{\text{, }}b) = f(c,{\text{ }}d) \Rightarrow (a{\text{, }}b) = (c,{\text{ }}d)\).

## Question

Consider the functions \(f:A \to B\) and \(g:B \to C\).

Show that if both *f* and *g* are injective, then \(g \circ f\) is also injective.

Show that if both *f* and *g* are surjective, then \(g \circ f\) is also surjective.

Show, using a single counter example, that both of the converses to the results in part (a) and part (b) are false.

**Answer/Explanation**

## Markscheme

let *s* and *t* be in *A* and \(s \ne t\) *M1*

since *f* is injective \(f(s) \ne f(t)\) *A1*

since *g* is injective \(g \circ f(s) \ne g \circ f(t)\) *A1*

hence \(g \circ f\) is injective *AG*

*[3 marks]*

let *z* be an element of *C*

we must find *x* in *A* such that \(g \circ f(x) = z\) *M1*

since *g* is surjective, there is an element *y* in *B* such that \(g(y) = z\) *A1*

since *f* is surjective, there is an element *x* in *A* such that \(f(x) = y\) *A1*

thus \(g \circ f(x) = g(y) = z\) *R1*

hence \(g \circ f\) is surjective *AG*

*[4 marks]*

converses: if \(g \circ f\) is injective then *g* and *f* are injective

if \(g \circ f\) is surjective then *g* and *f* are surjective *(A1)*

*A2*

**Note:** There will be many alternative counter-examples.

*[3 marks]*

## Examiners report

This question was found difficult by a large number of candidates and no fully correct solutions were seen. A number of students made thought-through attempts to show it was surjective, but found more difficulty in showing it was injective. Very few were able to find a single counter example to show that the converses of the earlier results were false. Candidates struggled with the abstract nature of the question.

This question was found difficult by a large number of candidates and no fully correct solutions were seen. A number of students made thought-through attempts to show it was surjective, but found more difficulty in showing it was injective. Very few were able to find a single counter example to show that the converses of the earlier results were false. Candidates struggled with the abstract nature of the question.

This question was found difficult by a large number of candidates and no fully correct solutions were seen. A number of students made thought-through attempts to show it was surjective, but found more difficulty in showing it was injective. Very few were able to find a single counter example to show that the converses of the earlier results were false. Candidates struggled with the abstract nature of the question.

## Question

Consider the functions \(f:A \to B\) and \(g:B \to C\).

Show that if both *f* and *g* are injective, then \(g \circ f\) is also injective.

Show that if both *f* and *g* are surjective, then \(g \circ f\) is also surjective.

Show, using a single counter example, that both of the converses to the results in part (a) and part (b) are false.

**Answer/Explanation**

## Markscheme

let *s* and *t* be in *A* and \(s \ne t\) *M1*

since *f* is injective \(f(s) \ne f(t)\) *A1*

since *g* is injective \(g \circ f(s) \ne g \circ f(t)\) *A1*

hence \(g \circ f\) is injective *AG*

*[3 marks]*

let *z* be an element of *C*

we must find *x* in *A* such that \(g \circ f(x) = z\) *M1*

since *g* is surjective, there is an element *y* in *B* such that \(g(y) = z\) *A1*

since *f* is surjective, there is an element *x* in *A* such that \(f(x) = y\) *A1*

thus \(g \circ f(x) = g(y) = z\) *R1*

hence \(g \circ f\) is surjective *AG*

*[4 marks]*

converses: if \(g \circ f\) is injective then *g* and *f* are injective

if \(g \circ f\) is surjective then *g* and *f* are surjective *(A1)*

*A2*

**Note:** There will be many alternative counter-examples.

*[3 marks]*

## Examiners report

## Question

The function \(g:\mathbb{Z} \to \mathbb{Z}\) is defined by \(g(n) = \left| n \right| – 1{\text{ for }}n \in \mathbb{Z}\) . Show that *g *is neither surjective nor injective.

The set *S *is finite. If the function \(f:S \to S\) is injective, show that *f *is surjective.

Using the set \({\mathbb{Z}^ + }\) as both domain and codomain, give an example of an injective function that is not surjective.

**Answer/Explanation**

## Markscheme

non-S: for example –2 does not belong to the range of *g **R1*

non-I: for example \(g(1) = g( – 1) = 0\) *R1*** **

**Note**: Graphical arguments have to recognize that we are dealing with sets of integers and not all real numbers

*[2 marks]*

as *f *is injective \(n\left( {f(S)} \right) = n(S)\) *A1*

* R1*

** **

**Note: **Accept alternative explanations.

*f *is surjective *AG*

*[2 marks]*

for example, \(h(n) = n + 1\) *A1*** **

**Note: **Only award the ** A1 **if the function works.

I: \(n + 1 = m + 1 \Rightarrow n = m\) *R1*

non-S: 1 has no pre-image as \(0 \notin {\mathbb{Z}^ + }\) *R1*

*[3 marks]*

## Examiners report

Nearly all candidates were aware of the conditions for an injection and a surjection in part (a). However, many missed the fact that the function in question was mapping from the set of integers to the set of integers. This led some to lose marks by applying graphical tests that were relevant for functions on the real numbers but not appropriate in this case. However, many candidates were able to give two integer counter examples to prove that the function was neither injective or surjective. In part (b) candidates seemed to lack the communication skills to adequately demonstrate what they intuitively understood to be true. It was usually not stated that the number of elements in the sets of the image and pre – image was equal. Part (c) was well done by many candidates although a significant minority used functions that mapped the positive integers to non – integer values and thus not appropriate for the conditions required of the function.

Nearly all candidates were aware of the conditions for an injection and a surjection in part (a). However, many missed the fact that the function in question was mapping from the set of integers to the set of integers. This led some to lose marks by applying graphical tests that were relevant for functions on the real numbers but not appropriate in this case. However, many candidates were able to give two integer counter examples to prove that the function was neither injective or surjective. In part (b) candidates seemed to lack the communication skills to adequately demonstrate what they intuitively understood to be true. It was usually not stated that the number of elements in the sets of the image and pre – image was equal. Part (c) was well done by many candidates although a significant minority used functions that mapped the positive integers to non – integer values and thus not appropriate for the conditions required of the function.

Nearly all candidates were aware of the conditions for an injection and a surjection in part (a). However, many missed the fact that the function in question was mapping from the set of integers to the set of integers. This led some to lose marks by applying graphical tests that were relevant for functions on the real numbers but not appropriate in this case. However, many candidates were able to give two integer counter examples to prove that the function was neither injective or surjective. In part (b) candidates seemed to lack the communication skills to adequately demonstrate what they intuitively understood to be true. It was usually not stated that the number of elements in the sets of the image and pre – image was equal. Part (c) was well done by many candidates although a significant minority used functions that mapped the positive integers to non – integer values and thus not appropriate for the conditions required of the function.

## Question

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

\[f(x) = \left\{ {\begin{array}{*{20}{c}}

{2x + 1}&{{\text{for }}x \leqslant 2} \\

{{x^2} – 2x + 5}&{{\text{for }}x > 2.}

\end{array}} \right.\]

(i) Sketch the graph of *f*.

(ii) By referring to your graph, show that *f* is a bijection.

Find \({f^{ – 1}}(x)\).

**Answer/Explanation**

## Markscheme

(i)

*A1A1*

**Note:** Award ** A1** for each part of the piecewise function. Award

**if the two parts of the graph are of the correct shape but**

*A1A0**f*is not continuous at

*x*= 2. Do not penalise a discontinuity in the derivative at

*x*= 2.

(ii) demonstrating the need to show that *f* is both an injection and a surjection (seen anywhere) *(R1)*

*f* is an injection by any valid reason *eg* horizontal line test, strictly increasing function *R1*

the range of *f* is \(\mathbb{R}\) so that *f* is a surjection *R1*

*f* is therefore a bijection *AG*

*[5 marks]*

considering the linear section, put

\(y = 2x + 1\) or \(x = 2y + 1\) *(M1)*

\(x = \frac{{y – 1}}{2}\) or \(y = \frac{{x – 1}}{2}\) *A1*

so \({f^{ – 1}}(x) = \frac{{x – 1}}{2},{\text{ }}x \leqslant 5\) *A1*

**EITHER**

\(y = {(x – 1)^2} + 4\) *M1A1*

\({(x – 1)^2} = y – 4\)

\(x = 1 \pm \sqrt {y – 4} \) *A1*

\(x = 1 + \sqrt {y – 4} \)

taking the + sign to give the right hand half of the parabola *R1*

so \({f^{ – 1}}(x) = 1 + \sqrt {x – 4} ,{\text{ }}x > 5\) *A1*

**OR**

considering the quadratic section, put

\(y = {x^2} – 2x + 5\)

\({x^2} – 2x + 5 – y = 0\) *M1*

\(x = \frac{{2 \pm \sqrt {4 – 4(5 – y)} }}{2}{\text{ }}( = 1 \pm \sqrt {y – 4} )\) *M1A1*

taking the + sign to give the right hand half of the parabola *R1*

so \({f^{ – 1}}(x) = \frac{{2 + \sqrt {4 – 4(5 – x)} }}{2},{\text{ }}x > 5{\text{ }}({f^{ – 1}}(x) = 1 + \sqrt {x – 4} ,{\text{ }}x > 5)\) *A1*

**Note:** Award ** A0** for omission of \({f^{ – 1}}(x)\) or omission of the domain. Penalise the omission of the notation \({f^{ – 1}}(x)\) only once. The domain must be seen in both cases.

* *

*[8 marks]*

## Examiners report

For the most part the piecewise function was correctly graphed. Even though the majority of candidates knew that it is required to establish that the function is an injection and a surjection in order to prove it is a bijection, many just quoted the definition of injection or surjection and did not relate their reason to the graph.

The majority of candidates found the inverse of the first part of the piecewise function but some struggled with the algebra of the second part. In finding the inverse of the quadratic part of the function some candidates omitted the plus or minus sign in front of the square root. Others who had it often forgot to eliminate the negative sign and so did not gain the reasoning mark. Most did not state the correct domain for either part of the inverse function.

## Question

Consider the following functions

\(f:\left] {1,{\text{ }} + \infty } \right[ \to {\mathbb{R}^ + }\) where \(f(x) = (x – 1)(x + 2)\)

\(g:\mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}\) where \(g(x,{\text{ }}y) = \left( {\sin (x + y),{\text{ }}x + y} \right)\)

\(h:\mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}\) where \(h(x,{\text{ }}y) = (x + 3y,{\text{ }}2x + y)\)

(a) Show that \(f\) is bijective.

(b) Determine, with reasons, whether

(i) \(g\) is injective;

(ii) \(g\)* *is surjective.

(c) Find an expression for \({h^{ – 1}}(x,{\text{ }}y)\) and hence justify that \(h\)* *has an inverse function.

**Answer/Explanation**

## Markscheme

(a) **Method 1**

sketch of the graph of \(f\) *(M1)*

range of \(f = \) co-domain, therefore \(f\) is surjective *R1*

graph of \(f\) passes the horizontal line test, therefore \(f\) is injective *R1*

therefore \(f\) is bijective *AG*

**Note: **Other explanations may be given (*eg *use of derivative or description of parabola).

**Method 2**

Injective: \(f(a) = f(b) \Rightarrow a = b\) *M1*

\((a – 1)(a + 2) = (b – 1)(b + 2)\)

\({a^2} + a = {b^2} + b\)

solving for \(a\) by completing the square, or the quadratic formula, *A1*

\(a = b\)

surjective: for all \(y \in {\mathbb{R}^ + }\) there exists \(x \in \left] {1,{\text{ }}\infty } \right[\) such that \(f(x) = y\)

solving \(y = {x^2} + x – 2\) for x, \(x = \frac{{\sqrt {4y + 9} – 1}}{2}\). For all positive real \(y\), the minimum value for \(\sqrt {4y + 9} \) is \(3\). Hence, \(x \geqslant 1\) *R1*

since \(f\) is both injective and surjective, \(f\) is bijective. *AG*

**Method 3**

\(f\) is bijective if and only if \(f\) has an inverse *(M1)*

solving \(y = {x^2} + x – 2\) for \(x,{\text{ }}x = \frac{{\sqrt {4y + 9} – 1}}{2}\). For all positive real \(y\), the minimum value for \(\sqrt {4y + 9} \) is \(3\). Hence, \(x \geqslant 1\) *R1*

\({f^{ – 1}}(x) = \frac{{\sqrt {4x + 9} – 1}}{2}\) *R1*

\(f\) has an inverse, hence \(f\) is bijective *AG*

*[3 marks]*

(b) (i) attempt to find counterexample *(M1)*

*eg* \(g(x,{\text{ }}y) = g(y,{\text{ }}x),{\text{ }}x \ne y\) *A1*

*g *is not injective *R1*

(ii) \( – 1 \leqslant \sin (x + y) \leqslant 1\) *(M1)*

range of \(g\) is \(\left[ { – 1,{\text{ 1}}} \right] \times \mathbb{R} \ne \mathbb{R} \times \mathbb{R}\) *A1*

\(g\) is not surjective *R1*

*[6 marks]*

(c) let \(h(x,{\text{ }}y) = (u,{\text{ }}v)\)

then \(u = x + 3y\)

\(v = 2x + y\) *(M1)*

solving simultaneous equations *(M1)*

*eg* \(\left( \begin{array}{l}x\\y\end{array} \right) = {\left( {\begin{array}{*{20}{c}}1&3\\2&1\end{array}} \right)^{ – 1}}\left( \begin{array}{l}u\\v\end{array} \right) \Rightarrow \left( \begin{array}{l}x\\y\end{array} \right) = – \frac{1}{5}\left( {\begin{array}{*{20}{c}}1&{ – 3}\\{ – 2}&1\end{array}} \right)\left( \begin{array}{l}u\\v\end{array} \right)\)

\(x = \frac{{ – u + 3v}}{5}y = \frac{{2u – v}}{5}\) *A1*

hence \({h^{ – 1}}(x,{\text{ }}y) = \left( {\frac{{ – x + 3y}}{5},{\text{ }}\frac{{2x – y}}{5}} \right)\) *A1*

as this expression is defined for any values of \((x,{\text{ }}y) \in \mathbb{R} \times \mathbb{R}\) *R1*

the inverse of \(h\) exists *AG*

*[5 marks] *

## Examiners report

For part (a), given the command term ‘show that’ and the number of marks for this part, the best approach is a graphical one, i.e., an informal approach. Many candidates chose an algebraic approach and generally made correct statements for injective and surjective. However, they often did not follow through with the necessary algebraic manipulation to make a valid conclusion. In part (b), many candidates were not able to provide valid counter-examples. In part (c) It was obvious that quite a few candidates had not seen this type of function before. Those that were able to find the inverse generally did not justify their result, and hence could not earn the final R mark.

## Question

Sets *X *and *Y *are defined by \({\text{ }}X = \left] {0,{\text{ }}1} \right[;{\text{ }}Y = \{ 0,{\text{ }}1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5\} \).

(i) Sketch the set \(X \times Y\) in the Cartesian plane.

(ii) Sketch the set \(Y \times X\) in the Cartesian plane.

(iii) State \((X \times Y) \cap (Y \times X)\).

Consider the function \(f:X \times Y \to \mathbb{R}\) defined by \(f(x,{\text{ }}y) = x + y\) and the function \(g:X \times Y \to \mathbb{R}\) defined by \(g(x,{\text{ }}y) = xy\).

(i) Find the range of the function *f*.

(ii) Find the range of the function *g*.

(iii) Show that \(f\) is an injection.

(iv) Find \({f^{ – 1}}(\pi )\), expressing your answer in exact form.

(v) Find all solutions to \(g(x,{\text{ }}y) = \frac{1}{2}\).

**Answer/Explanation**

## Markscheme

(i)

correct horizontal lines *A1*

correctly labelled axes *A1*

clear indication that the endpoints are not included *A1*

(ii)

fully correct diagram *A1*

**Note:** Do not penalize the inclusion of endpoints twice.

(iii) the intersection is empty *A1*

*[5 marks]*

(i) range \((f) = \left] {0,{\text{ 1}}} \right[ \cup \left] {1,{\text{ 2}}} \right[ \cup \rm{L} \cup \left] {5,{\text{ 6}}} \right[\) *A1A1*

**Note: A1 **for six intervals and

**for fully correct notation.**

*A1*Accept \(0 < x < 6,{\text{ }}x \ne 0{\text{, 1, 2, 3, 4, 5, 6}}\).

(ii) range \((g) = \left[ {0,{\text{ 5}}} \right[\) *A1*

(iii) Attempt at solving

\(f({x_1},{\text{ }}{y_1}) = f({x_2},{\text{ }}{y_2})\) **M1**

\(f(x,{\text{ }}y) \in \left] {y,{\text{ }}y + 1} \right[ \Rightarrow {y_1} = {y_2}\) **M1**

and then \({x_1} = {x_2}\) *A1*

so \(f\) is injective *AG*

(iv) \({f^{ – 1}}(\pi ) = (\pi – 3,{\text{ }}3)\) **A1A1**

(v) solutions: (0.5, 1), (0.25, 2), \(\left( {\frac{1}{6},{\text{ 3}}} \right)\), (0.125, 4), (0.1, 5) *A2*

**Note:** ** A2 **for all correct,

**for 2 correct.**

*A1*

*[10 marks]*

## Examiners report

[N/A]

[N/A]

## Question

Define \(f:\mathbb{R}\backslash \{ 0.5\} \to \mathbb{R}\) by \(f(x) = \frac{{4x + 1}}{{2x – 1}}\).

Prove that *\(f\) *is an injection.

Prove that *\(f\) *is not a surjection.

**Answer/Explanation**

## Markscheme

**METHOD 1**

\(f(x) = f(y) \Rightarrow \frac{{4x + 1}}{{2x – 1}} = \frac{{4y + 1}}{{2y – 1}}\) *M1A1*

for attempting to cross multiply and simplify *M1*

\((4x + 1)(2y – 1) = (2x – 1)(4y + 1)\)

\( \Rightarrow 8xy + 2y – 4x – 1 = 8xy + 2x – 4y – 1 \Rightarrow 6y = 6x\)

\( \Rightarrow x = y\) *A1*

hence an injection *AG*

**METHOD 2**

\(f'(x) = \frac{{4(2x – 1) – 2(4x + 1)}}{{{{(2x – 1)}^2}}} = \frac{{ – 6}}{{{{(2x – 1)}^2}}}\) *M1A1*

\( < 0\;\;\;{\text{(for all }}x \ne 0.5{\text{)}}\) *R1*

therefore the function is decreasing on either side of the discontinuity

and \(f(x) < 2\) and \(x < 0.5\) for \(f(x) > 0.5\) *R1*

hence an injection *AG*

**Note: **If a correct graph of the function is shown, and the candidate states this is decreasing in each part (or horizontal line test) and hence an injection, award ** M1A1R1**.

**[4 marks]**

**METHOD 1**

attempt to solve \(y = \frac{{4x + 1}}{{2x – 1}}\) *M1*

\(y(2x – 1) = 4x + 1 \Rightarrow 2xy – y = 4x + 1\) *A1*

\(2xy – 4x = 1 + y \Rightarrow x = \frac{{1 + y}}{{2y – 4}}\) *A1*

no value for \(y = 2\) *R1*

hence not a surjection *AG*

**METHOD 2**

consider \(y = 2\) *A1*

attempt to solve \(2 = \frac{{4x + 1}}{{2x – 1}}\) *M1*

\(4x – 2 = 4x + 1\) *A1*

which has no solution *R1*

hence not a surjection *AG*

**Note: **If a correct graph of the function is shown, and the candidate states that because there is a horizontal asymptote at \(y = 2\) then the function is not a surjection, award ** M1R1**.

**[4 marks]**

**Total [8 marks]**

## Examiners report

Most students indicated an understanding of the concepts of Injection and Surjection, but many did not give rigorous proofs. Even where graphs were used, it was very common for a sketch to be so imprecise with no asymptotes marked that it was difficult to award even partial credit. Some candidates mistakenly stated that the function was not surjective because 0.5 was not in the domain.

Most students indicated an understanding of the concepts of Injection and Surjection, but many did not give rigorous proofs. Even where graphs were used, it was very common for a sketch to be so imprecise with no asymptotes marked that it was difficult to award even partial credit. Some candidates mistakenly stated that the function was not surjective because 0.5 was not in the domain.

## Question

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.

**Answer/Explanation**

## Markscheme

(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.

## Question

The function \(f\) is defined by \(f:{\mathbb{R}^ + } \times {\mathbb{R}^ + } \to {\mathbb{R}^ + } \times {\mathbb{R}^ + }\) where \(f(x,{\text{ }}y) = \left( {\sqrt {xy} ,{\text{ }}\frac{x}{y}} \right)\)

Prove that \(f\) is an injection.

(i) Prove that \(f\) is a surjection.

(ii) Hence, or otherwise, write down the inverse function \({f^{ – 1}}\).

**Answer/Explanation**

## Markscheme

let \((a,{\text{ }}b)\) and \((c,{\text{ }}d) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + }\)

suppose that \(f(a,{\text{ }}b) = f(c,{\text{ }}d)\) *(M1)*

so that \(\sqrt {ab} = \sqrt {cd} \) and \(\frac{a}{b} = \frac{c}{d}\) *A1*

leading to either \({a^2} = {c^2}\) or \({b^2} = {d^2}\) or equivalent *M1*

state \(a = c\) and \(b = d\) *A1*

this shows that \(f\) is an injection since \(f(a,{\text{ }}b) = f(c,{\text{ }}d) \Rightarrow (a,{\text{ }}b) = (c,{\text{ }}d)\) *R1AG*

**Note: **Accept final statement seen anywhere for ** R1**.

*[5 marks]*

(i) now let \((u,{\text{ }}v) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + }\) and suppose that \(f(x,{\text{ }}y) = (u,{\text{ }}v)\) *(M1)*

then, \(u = \sqrt {xy} ,{\text{ }}v = \frac{x}{y}\) *A1*

attempt to eliminate \(x\) or \(y\) *M1*

\( \Rightarrow x = u{v^{1/2}};{\text{ }}y = u{v^{ – 1/2}}\) *A1A1*

this shows that \(f\) is a surjection since, given \((u,{\text{ }}v)\), there exists \((x,{\text{ }}y)\) such that \(f(x,{\text{ }}y) = (u,{\text{ }}v)\) *R1AG*

**Note: **Accept final statement, seen anywhere, for ** R1**.

(ii) \({f^{ – 1}}(x,{\text{ }}y) = (x{y^{1/2}},{\text{ }}x{y^{ – 1/2}})\) *A1A1*

*[8 marks]*

## Examiners report

Those candidates who formulated their responses in terms of the basic mathematical definitions of injectivity and surjectivity were usually successful. Otherwise, verbal attempts such as ‘\(f\) is one-to-one \( \Rightarrow f\) is injective’ or ‘\(g\) is surjective because its range equals its codomain’, received no credit.

(i) Those candidates who formulated their responses in terms of the basic mathematical definitions of injectivity and surjectivity were usually successful. Otherwise, verbal attempts such as ‘\(f\) is one-to-one \( \Rightarrow f\) is injective’ or ‘\(g\) is surjective because its range equals its codomain’, received no credit.

(ii) It was surprising to see that some candidates were unable to relate what they had done in part (b)(i) to this part.

## Question

Let \(A\) be the set \(\{ x|x \in \mathbb{R},{\text{ }}x \ne 0\} \). Let \(B\) be the set \(\{ x|x \in ] – 1,{\text{ }} + 1[,{\text{ }}x \ne 0\} \).

A function \(f:A \to B\) is defined by \(f(x) = \frac{2}{\pi }\arctan (x)\).

Let \(D\) be the set \(\{ x|x \in \mathbb{R},{\text{ }}x > 0\} \).

A function \(g:\mathbb{R} \to D\) is defined by \(g(x) = {{\text{e}}^x}\).

(i) Sketch the graph of \(y = f(x)\) and hence justify whether or not \(f\) is a bijection.

(ii) Show that \(A\) is a group under the binary operation of multiplication.

(iii) Give a reason why \(B\) is not a group under the binary operation of multiplication.

(iv) Find an example to show that \(f(a \times b) = f(a) \times f(b)\) is not satisfied for all \(a,{\text{ }}b \in A\).

(i) Sketch the graph of \(y = g(x)\) and hence justify whether or not \(g\) is a bijection.

(ii) Show that \(g(a + b) = g(a) \times g(b)\) for all \(a,{\text{ }}b \in \mathbb{R}\).

(iii) Given that \(\{ \mathbb{R},{\text{ }} + \} \) and \(\{ D,{\text{ }} \times \} \) are both groups, explain whether or not they are isomorphic.

**Answer/Explanation**

## Markscheme

(i) *A1*

**Notes: **Award ** A1 **for general shape, labelled asymptotes, and showing that \(x \ne 0\).

graph shows that it is injective since it is increasing or by the horizontal line test *R1*

graph shows that it is surjective by the horizontal line test *R1*

**Note: **Allow any convincing reasoning.

so \(f\) is a bijection *A1*

(ii) closed since non-zero real times non-zero real equals non-zero real *A1R1*

we know multiplication is associative *R1*

identity is 1 *A1*

inverse of \(x\) is \(\frac{1}{x}(x \ne 0)\) *A1*

hence it is a group *AG*

(iii) \(B\) does not have an identity *A2*

hence it is not a group *AG*

(iv) \(f(1 \times 1) = f(1) = \frac{1}{2}\) whereas \(f(1) \times f(1) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\) is one counterexample *A2*

hence statement is not satisfied *AG*

*[13 marks]*

award ** A1 **for general shape going through (0, 1) and with domain \(\mathbb{R}\)

*A1*graph shows that it is injective since it is increasing or by the horizontal line test and graph shows that it is surjective by the horizontal line test *R1*

**Note: **Allow any convincing reasoning.

so \(g\) is a bijection *A1*

(ii) \(g(a + b) = {{\text{e}}^{a + b}}\) and \(g(a) \times g(b) = {{\text{e}}^a} \times {{\text{e}}^b} = {{\text{e}}^{a + b}}\) *M1A1*

hence \(g(a + b) = g(a) \times g(b)\) *AG*

(iii) since \(g\) is a bijection and the homomorphism rule is obeyed *R1R1*

the two groups are isomorphic *A1*

*[8 marks]*

## Examiners report

[N/A]

[N/A]

## Question

The function \(f\,{\text{: }}\mathbb{Z} \to \mathbb{Z}\) is defined by \(f\left( n \right) = n + {\left( { – 1} \right)^n}\).

Prove that \(f \circ f\) is the identity function.

Show that \(f\) is injective.

Show that \(f\) is surjective.

**Answer/Explanation**

## Markscheme

**METHOD 1**

\(\left( {f \circ f} \right)\left( n \right) = n + {\left( { – 1} \right)^n} + {\left( { – 1} \right)^{n + {{\left( { – 1} \right)}^n}}}\) **M1A1**

\( = n + {\left( { – 1} \right)^n} + {\left( { – 1} \right)^n} \times {\left( { – 1} \right)^{{{\left( { – 1} \right)}^n}}}\) **(A1)**

considering \({\left( { – 1} \right)^n}\) for even and odd \(n\) **M1**

if \(n\) is odd, \({\left( { – 1} \right)^n} = – 1\) and if \(n\) is even, \({\left( { – 1} \right)^n} = 1\) and so \({\left( { – 1} \right)^{ \pm 1}} = – 1\) **A1**

\( = n + {\left( { – 1} \right)^n} – {\left( { – 1} \right)^n}\) **A1**

= \(n\) and so \(f \circ f\) is the identity function **AG**

**METHOD 2**

\(\left( {f \circ f} \right)\left( n \right) = n + {\left( { – 1} \right)^n} + {\left( { – 1} \right)^{n + {{\left( { – 1} \right)}^n}}}\) **M1A1**

\( = n + {\left( { – 1} \right)^n} + {\left( { – 1} \right)^n} \times {\left( { – 1} \right)^{{{\left( { – 1} \right)}^n}}}\) **(A1)**

\( = n + {\left( { – 1} \right)^n} \times \left( {1 + {{\left( { – 1} \right)}^{{{\left( { – 1} \right)}^n}}}} \right)\) **M1**

\({\left( { – 1} \right)^{ \pm 1}} = – 1\) **R1**

\(1 + {\left( { – 1} \right)^{{{\left( { – 1} \right)}^n}}} = 0\) **A1**

\(\left( {f \circ f} \right)\left( n \right) = n\) and so \(f \circ f\) is the identity function **AG**

**METHOD 3**

\(\left( {f \circ f} \right)\left( n \right) = f\left( {n + {{\left( { – 1} \right)}^n}} \right)\) **M1**

considering even and odd \(n\) **M1**

if \(n\) is even, \(f\left( n \right) = n + 1\) which is odd **A1**

so \(\left( {f \circ f} \right)\left( n \right) = f\left( {n + 1} \right) = \left( {n + 1} \right) – 1 = n\) **A1**

if \(n\) is odd, \(f\left( n \right) = n – 1\) which is even **A1**

so \(\left( {f \circ f} \right)\left( n \right) = f\left( {n – 1} \right) = \left( {n – 1} \right) + 1 = n\) **A1**

\(\left( {f \circ f} \right)\left( n \right) = n\) in both cases

hence \(f \circ f\) is the identity function ** AG**

**[6 marks]**

suppose \(f\left( n \right) = f\left( m \right)\) **M1**

applying \(f\) to both sides \( \Rightarrow n = m\) **R1**

hence \(f\) is injective **AG**

**[2 marks]**

\(m = f\left( n \right)\) has solution \(n = f\left( m \right)\) **R1**

hence surjective **AG**

**[1 mark]**

## Examiners report

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