The function $f:{\mathbb{R}^ + } \times {\mathbb{R}^ + } \to {\mathbb{R}^ + } \times {\mathbb{R}^ + }$ is defined by $f(x,{\text{ }}y) = \left( {x{y^2},\frac{x}{y}} \right)$.

Show that f is a bijection.

for f to be a bijection it must be both an injection and a surjection     R1

Note: Award this R1 for stating this anywhere.

injection:

let $f(a{\text{, }}b) = f(c,{\text{ }}d)$ so that     (M1)

$a{b^2} = c{d^2}$ and $\frac{a}{b} = \frac{c}{d}$     A1

dividing the equations,

${b^3} = {d^3}$ so $b = d$     A1

substituting,

a = c     A1

it follows that f is an injection because $f(a{\text{, }}b) = f(c,{\text{ }}d) \Rightarrow (a{\text{, }}b) = (c,{\text{ }}d)$     R1

surjection:

let $f(a{\text{, }}b) = (c,{\text{ }}d)$ where $(c,{\text{ }}d) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + }$     (M1)

then $c = a{b^2}$ and $d = \frac{a}{b}$     A1

dividing,

${b^3} = \frac{c}{d}$ so $b = \sqrt[3]{{\frac{c}{d}}}$     A1

substituting,

$a = d \times \sqrt[3]{{\frac{c}{d}}}$     A1

it follows that f is a surjection because

given $(c,{\text{ }}d) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + }$ , there exists $(a,{\text{ }}b) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + }$ such that $f(a{\text{, }}b) = (c,{\text{ }}d)$     R1

therefore f is a bijection     AG

[11 marks]

Candidates who knew that they were required to give a rigorous demonstration that f was injective and surjective were generally successful, although the formality that is needed in this style of demonstration was often lacking. Some candidates, however, tried unsuccessfully to give a verbal explanation or even a 2-D version of the horizontal line test. In 2-D, the only reliable method for showing that a function f is injective is to show that $f(a{\text{, }}b) = f(c,{\text{ }}d) \Rightarrow (a{\text{, }}b) = (c,{\text{ }}d)$.