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).