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.
Show that the function \(f\) is a bijection.
Hence find the inverse function \({f^{ - 1}}\).
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]
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]
\({f^{ - 1}}\left( {x,\,y} \right) = \left( {\frac{{x + 3y}}{7},\,\frac{{2x - y}}{7}} \right)\) A1A1
[2 marks]