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Date May 2016 Marks available 8 Reference code 16M.3srg.hl.TZ0.4
Level HL only Paper Paper 3 Sets, relations and groups Time zone TZ0
Command term Hence or otherwise and Prove Question number 4 Adapted from N/A

Question

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

Prove that \(f\) is an injection.

[5]
a.

(i)     Prove that \(f\) is a surjection.

(ii)     Hence, or otherwise, write down the inverse function \({f^{ - 1}}\).

[8]
b.

Markscheme

let \((a,{\text{ }}b)\) and \((c,{\text{ }}d) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + }\)

suppose that \(f(a,{\text{ }}b) = f(c,{\text{ }}d)\)     (M1)

so that \(\sqrt {ab}  = \sqrt {cd} \) and \(\frac{a}{b} = \frac{c}{d}\)     A1

leading to either \({a^2} = {c^2}\) or \({b^2} = {d^2}\) or equivalent     M1

state \(a = c\) and \(b = d\)     A1

this shows that \(f\) is an injection since \(f(a,{\text{ }}b) = f(c,{\text{ }}d) \Rightarrow (a,{\text{ }}b) = (c,{\text{ }}d)\)     R1AG

Note:     Accept final statement seen anywhere for R1.

[5 marks]

a.

(i)     now let \((u,{\text{ }}v) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + }\) and suppose that \(f(x,{\text{ }}y) = (u,{\text{ }}v)\)     (M1)

then, \(u = \sqrt {xy} ,{\text{ }}v = \frac{x}{y}\)     A1

attempt to eliminate \(x\) or \(y\)     M1

\( \Rightarrow x = u{v^{1/2}};{\text{ }}y = u{v^{ - 1/2}}\)    A1A1

this shows that \(f\) is a surjection since, given \((u,{\text{ }}v)\), there exists \((x,{\text{ }}y)\) such that \(f(x,{\text{ }}y) = (u,{\text{ }}v)\)     R1AG

Note: Accept final statement, seen anywhere, for R1.

(ii)     \({f^{ - 1}}(x,{\text{ }}y) = (x{y^{1/2}},{\text{ }}x{y^{ - 1/2}})\)     A1A1

[8 marks]

b.

Examiners report

Those candidates who formulated their responses in terms of the basic mathematical definitions of injectivity and surjectivity were usually successful. Otherwise, verbal attempts such as ‘\(f\) is one-to-one \( \Rightarrow f\) is injective’ or ‘\(g\) is surjective because its range equals its codomain’, received no credit.

a.

(i)     Those candidates who formulated their responses in terms of the basic mathematical definitions of injectivity and surjectivity were usually successful. Otherwise, verbal attempts such as ‘\(f\) is one-to-one \( \Rightarrow f\) is injective’ or ‘\(g\) is surjective because its range equals its codomain’, received no credit.

(ii)     It was surprising to see that some candidates were unable to relate what they had done in part (b)(i) to this part.

b.

Syllabus sections

Topic 8 - Option: Sets, relations and groups
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