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 $k \circ h$ .

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]

‘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).