18N.3.AHL.TZ0.HSRG_4

pestleMathematicsAAHLPaper 318N· sl-2-5-composite-functions-identity-finding-inversesource ↗

Consider the functions $f$ , $g$ : $\mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}$ defined by

$f\left( {\left( {x{\text{,}}\,\,y} \right)} \right) = \left( {x + y,\,\,x - y} \right)$ and $g\left( {\left( {x{\text{,}}\,\,y} \right)} \right) = \left( {xy,\,\,x + y} \right)$ .

Find $\left( {f \circ g} \right)\left( {\left( {x{\text{,}}\,\,y} \right)} \right)$ .

[3]

a.i.

Find $\left( {g \circ f} \right)\left( {\left( {x{\text{,}}\,\,y} \right)} \right)$ .

[2]

a.ii.

State with a reason whether or not $f$ and $g$ commute.

[1]

Find the inverse of $f$ .

[3]

Markscheme / solution

$\left( {f \circ g} \right)\left( {\left( {x{\text{,}}\,\,y} \right)} \right) = f\left( {g\left( {\left( {x{\text{,}}\,\,y} \right)} \right)} \right)$ ( $= f\left( {\left( {xy,\,\,x + y} \right)} \right)$ ) (M1)

$= \left( {xy + x + y,\,\,xy - x - y} \right)$ A1A1

[3 marks]

a.i.

$\left( {g \circ f} \right)\left( {\left( {x{\text{,}}\,\,y} \right)} \right) = g\left( {f\left( {\left( {x{\text{,}}\,\,y} \right)} \right)} \right)$

$= g\left( {\left( {x + y,\,\,x - y} \right)} \right)$

$= \left( {\left( {x + y} \right)\left( {x - y} \right),\,\,x + y + x - y} \right)$

$= \left( {{x^2} - {y^2}{\text{,}}\,\,2x} \right)$ A1A1

[2 marks]

a.ii.

no because $f \circ g \ne g \circ f$ R1

Note: Accept counter example.

[1 mark]

$f\left( {\left( {x{\text{,}}\,\,y} \right)} \right) = \left( {a{\text{,}}\,\,b} \right) \Rightarrow \left( {x + y,\,\,x - y} \right) = \left( {a{\text{,}}\,\,b} \right)$ (M1)

$\left\{ {\begin{array}{*{20}{c}} {x = \frac{{a + b}}{2}} \\ {y = \frac{{a - b}}{2}} \end{array}} \right.$ (M1)

${f^{ - 1}}\left( {\left( {x{\text{,}}\,\,y} \right)} \right) = \left( {\frac{{x + y}}{2},\,\,\frac{{x - y}}{2}} \right)$ A1

[3 marks]

Examiners’ report

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a.i.

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a.ii.

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