| Date | None Specimen | Marks available | 3 | Reference code | SPNone.2.hl.TZ0.1 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Determine and Interpret | Question number | 1 | Adapted from | N/A |
Question
The relation \(R\) is defined on \({\mathbb{R}^ + } \times {\mathbb{R}^ + }\) such that \(({x_1},{y_1})R({x_2},{y_2})\) if and only if \(\frac{{{x_1}}}{{{x_2}}} = \frac{{{y_2}}}{{{y_1}}}\) .
Show that \(R\) is an equivalence relation.
Determine the equivalence class containing \(({x_1},{y_1})\) and interpret it geometrically.
Markscheme
\(\frac{{{x_1}}}{{{x_1}}} = \frac{{{y_1}}}{{{y_1}}} \Rightarrow ({x_1},{y_1})R({x_1},{y_1})\) so \(R\) is reflexive R1
\(({x_1},{y_1})R({x_2},{y_2}) \Rightarrow \frac{{{x_1}}}{{{x_2}}} = \frac{{{y_2}}}{{{y_1}}} \Rightarrow \frac{{{x_2}}}{{{x_1}}} = \frac{{{y_1}}}{{{y_2}}} \Rightarrow ({x_2},{y_2})R({x_1},{y_1})\) M1A1
so \(R\) is symmetric
\(({x_1},{y_1})R({x_2},{y_2})\) and \(({x_2},{y_2})R({x_3},{y_3}) \Rightarrow \frac{{{x_1}}}{{{x_2}}} = \frac{{{y_2}}}{{{y_1}}}\) and \(\frac{{{x_2}}}{{{x_3}}} = \frac{{{y_3}}}{{{y_2}}}\) M1
multiplying the two equations, M1
\( \Rightarrow \frac{{{x_1}}}{{{x_3}}} = \frac{{{y_3}}}{{{y_1}}} \Rightarrow ({x_1},{y_1})R({x_3},{y_3})\) so \(R\) is transitive A1
thus \(R\) is an equivalence relation AG
[6 marks]
\((x,y)R({x_1},{y_1}) \Rightarrow \frac{{{x_{}}}}{{{x_1}}} = \frac{{{y_1}}}{{{y_{}}}} \Rightarrow xy = {x_1}{y_1}\) (M1)
the equivalence class is therefore \(\left\{ {(x,y)|xy = {x_1}{y_1}} \right\}\) A1
geometrically, the equivalence class is (one branch of) a (rectangular) hyperbola A1
[3 marks]
