(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.