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 \(f(a{\text{, }}b) = f(c,{\text{ }}d) \Rightarrow (a{\text{, }}b) = (c,{\text{ }}d)\).