Date | May 2017 | Marks available | 12 | Reference code | 17M.3srg.hl.TZ0.3 |
Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ0 |
Command term | Show that | Question number | 3 | Adapted from | N/A |
Question
The function \(f:\mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}\) is defined by \(f(x,{\text{ }}y) = (2{x^3} + {y^3},{\text{ }}{x^3} + 2{y^3})\).
Show that \(f\) is a bijection.
Hence write down the inverse function \({f^{ - 1}}(x,{\text{ }}y)\).
Markscheme
for \(f\) to be a bijection it must be both an injection and a surjection R1
Note: Award this R1 for stating this anywhere.
suppose that \(f(a,{\text{ }}b) = f(c,{\text{ }}d)\) (M1)
it follows that
\(2{a^3} + {b^3} = 2{c^3} + {d^3}\) and \({a^3} + 2{b^3} = {c^3} + 2{d^3}\) A1
attempting to solve the two equations M1
to obtain \(3{a^3} = 3{c^3}\)
Note: Award M1 only if a good attempt is made to solve the system.
\( \Rightarrow a = c\) and therefore \(b = d\) A1
\(f\) is an injection because \(f(a,{\text{ }}b) = f(c,{\text{ }}d) \Rightarrow (a,{\text{ }}b) = (c,{\text{ }}d)\) R1
Note: Award this R1 for stating this anywhere providing that an attempt is made to prove injectivity.
let \((p,{\text{ }}q) \in \mathbb{R} \times \mathbb{R}\) and let \(f(r,{\text{ }}s) = (p,{\text{ }}q)\) (M1)
then, \(p = 2{r^3} + {s^3}\) and \(q = {r^3} + 2{s^3}\) A1
attempting to solve the two equations M1
\(r = \sqrt[3]{{\frac{{2p - q}}{3}}}\) and \(s = \sqrt[3]{{\frac{{2q - p}}{3}}}\) A1A1
\(f\) is a surjection because given \((p,{\text{ }}q) \in \mathbb{R} \times \mathbb{R}\), there exists \((r,{\text{ }}s) \in \mathbb{R} \times \mathbb{R}\) such that \(f(r,{\text{ }}s) = (p,{\text{ }}q)\) R1
Note: Award this R1 for stating this anywhere providing that an attempt is made to prove surjectivity.
[12 marks]
\(\left( {{f^{ - 1}}(x,{\text{ }}y) = } \right)\,\,\,\left( {\sqrt[3]{{\frac{{2x - y}}{3},}}{\text{ }}\sqrt[3]{{\frac{{2y - x}}{3}}}} \right)\) A1
Note: A1 for correct expressions in \(x\) and \(y\), allow FT only if the expression is deduced in part (a).
[1 mark]