Given that A is the interval $\left\{ {x\,{\text{:}}\,0 \leqslant x \leqslant 3} \right\}$ and B is the interval $\left\{ {y\,{\text{:}}\,0 \leqslant x \leqslant 4} \right\}$ then describe A × B in geometric form.

Show that the function $f$ is a bijection.

Hence find the inverse function ${f^{ - 1}}$.

A × B is a rectangle       A1

vertices at (0, 0), (3, 0), (0, 4) and (3, 4) or equivalent description      A1

and its interior      A1

Note: Accept diagrammatic answers.

[3 marks]

need to prove it is injective and surjective        R1

need to show if $f\left( {x,\,y} \right) = f\left( {u,\,v} \right)$ then $\left( {x,\,y} \right) = \left( {u,\,v} \right)$      M1

$\Rightarrow x + 3y = u + 3v$

$2x - y = 2u - v$      A1

Equation 2 – 2 Equation 1 $\Rightarrow y = v$

Equation 1 + 3 Equation 2 $\Rightarrow x = u$    A1

thus $\left( {x,\,y} \right) = \left( {u,\,v} \right) \Rightarrow f$ is injective

let $\left( {s,\,t} \right)$ be any value in the co-domain $\mathbb{R} \times \mathbb{R}$

we must find $\left( {x,\,y} \right)$ such that $f\left( {x,\,y} \right) = \left( {s,\,t} \right)$    M1

$s = x + 3y$ and $t = 2x - y$    M1

$\Rightarrow y = \frac{{2s - t}}{7}$      A1

and $x = \frac{{s + 3t}}{7}$      A1

hence $f\left( {x,\,y} \right) = \left( {s,\,t} \right)$ and is therefore surjective

[8 marks]

${f^{ - 1}}\left( {x,\,y} \right) = \left( {\frac{{x + 3y}}{7},\,\frac{{2x - y}}{7}} \right)$      A1A1

[2 marks]