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]