(a)     Show that \(f:\mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}\) defined by \(f(x,{\text{ }}y) = (2x + y,{\text{ }}x - y)\) is a bijection.

(b)     Find the inverse of f .

(a)     we need to show that the function is both injective and surjective to be a bijection     R1

suppose \(f(x,{\text{ }}y) = f(u,{\text{ }}v)\)     M1

\((2x + y,{\text{ }}x - y) = (2u + v,{\text{ }}u - v)\)

forming a pair of simultaneous equations     M1

\(2x + y = 2u + v\)     (i)

\(x - y = u - v\)     (ii)

\((i) + (ii) \Rightarrow 3x = 3u \Rightarrow x = u\)     A1

\((i) - 2(ii) \Rightarrow 3y = 3v \Rightarrow y = v\)     A1

hence function is injective     R1

let \(2x + y = s\) and \(x - y = t\)     M1

\( \Rightarrow 3x = s + t\)

\( \Rightarrow x = \frac{{s + t}}{3}\)     A1

also \(3y = s - 2t\)

\( \Rightarrow y = \frac{{s - 2t}}{3}\)     A1

for any \((s,{\text{ }}t) \in \mathbb{R} \times \mathbb{R}\) there exists \((x,{\text{ }}y) \in \mathbb{R} \times \mathbb{R}\) and the function is surjective     R1

[10 marks]

(b)     the inverse is \({f^{ - 1}}(x,{\text{ }}y) = \left( {\frac{{x + y}}{3},{\text{ }}\frac{{x - 2y}}{3}} \right)\)     A1

[1 mark]

Total [11 marks]

Many students were able to show that the expression was injective, but found more difficulty in showing it was subjective. As with question 1 part (e), a number of candidates did not realise that the answer to part (b) came directly from part (a), hence the reason for it being worth only one mark.