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Date May 2018 Marks available 8 Reference code 18M.1.hl.TZ0.9
Level HL only Paper 1 Time zone TZ0
Command term Show that Question number 9 Adapted from N/A

Question

Let \(f\,{\text{:}}\,\mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}\) be defined by \(f\left( {x,\,y} \right) = \left( {x + 3y,\,2x - y} \right)\).

Given that A is the interval \(\left\{ {x\,{\text{:}}\,0 \leqslant x \leqslant 3} \right\}\) and B is the interval \(\left\{ {y\,{\text{:}}\,0 \leqslant x \leqslant 4} \right\}\) then describe A × B in geometric form.

[3]
a.

Show that the function \(f\) is a bijection.

[8]
b.i.

Hence find the inverse function \({f^{ - 1}}\).

[2]
b.ii.

Markscheme

A × B is a rectangle       A1

vertices at (0, 0), (3, 0), (0, 4) and (3, 4) or equivalent description      A1

and its interior      A1

Note: Accept diagrammatic answers.

[3 marks]

a.

need to prove it is injective and surjective        R1

need to show if \(f\left( {x,\,y} \right) = f\left( {u,\,v} \right)\) then \(\left( {x,\,y} \right) = \left( {u,\,v} \right)\)      M1

\( \Rightarrow x + 3y = u + 3v\)

\(2x - y = 2u - v\)      A1

Equation 2 – 2 Equation 1 \( \Rightarrow y = v\)

Equation 1 + 3 Equation 2 \( \Rightarrow x = u\)    A1

thus \(\left( {x,\,y} \right) = \left( {u,\,v} \right) \Rightarrow f\) is injective

let \(\left( {s,\,t} \right)\) be any value in the co-domain \(\mathbb{R} \times \mathbb{R}\)

we must find \(\left( {x,\,y} \right)\) such that \(f\left( {x,\,y} \right) = \left( {s,\,t} \right)\)    M1

\(s = x + 3y\) and \(t = 2x - y\)    M1

\( \Rightarrow y = \frac{{2s - t}}{7}\)      A1

and \(x = \frac{{s + 3t}}{7}\)      A1

hence \(f\left( {x,\,y} \right) = \left( {s,\,t} \right)\) and is therefore surjective

[8 marks]

b.i.

\({f^{ - 1}}\left( {x,\,y} \right) = \left( {\frac{{x + 3y}}{7},\,\frac{{2x - y}}{7}} \right)\)      A1A1

[2 marks]

b.ii.

Examiners report

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a.
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b.i.
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b.ii.

Syllabus sections

Topic 4 - Sets, relations and groups » 4.3 » Functions: injections; surjections; bijections.

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