| Date | May 2015 | Marks available | 11 | Reference code | 15M.1.hl.TZ0.15 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Determine | Question number | 15 | Adapted from | N/A |
Question
The relations \({\rho _1}\) and \({\rho _2}\) are defined on the Cartesian plane as follows
\(({x_1},{\text{ }}{y_1}){\rho _1}({x_2},{\text{ }}{y_2}) \Leftrightarrow x_1^2 - x_2^2 = y_1^2 - y_2^2\)
\(({x_1},{\text{ }}{y_1}){\rho _2}({x_2},{\text{ }}{y_2}) \Leftrightarrow \sqrt {x_1^2 + x_2^2} \leqslant \sqrt {y_1^2 + y_2^2} \).
For \({\rho _1}\) and \({\rho _2}\) determine whether or not each is reflexive, symmetric and transitive.
For each of \({\rho _1}\) and \({\rho _2}\) which is an equivalence relation, describe the equivalence classes.
Markscheme
\({\rho _1}\)
\(({x_1},{\text{ }}{y_1}){\rho _1}({x_1},{\text{ }}{y_1}) \Rightarrow 0 = 0\;\;\;\)hence reflexive. R1
\(({x_1},{\text{ }}{y_1}){\rho _1}({x_2},{\text{ }}{y_2}) \Rightarrow x_1^2 - x_2^2 = y_1^2 - y_2^2\)
\( \Rightarrow (x_1^2 - x_2^2) = - (y_1^2 - y_2^2)\)
\( \Rightarrow x_2^2 - x_1^2 = y_2^2 - y_1^2 \Rightarrow ({x_2},{\text{ }}{y_2}){\rho _1}({x_1},{\text{ }}{y_1})\;\;\;\)hence symmetric M1A1
\(({x_1},{\text{ }}{y_1}){\rho _1}({x_2},{\text{ }}{y_2}) \Rightarrow x_1^2 - x_2^2 = y_1^2 - y_2^2{\text{ - i}}\)
\(({x_2},{\text{ }}{y_2}){\rho _1}({x_3},{\text{ }}{y_3}) \Rightarrow x_2^2 - x_3^2 = y_2^2 - y_3^2{\text{ - ii}}\) M1
\({\text{i}} + {\text{ii}} \Rightarrow x_1^2 - x_3^2 = y_1^2 - y_3^2 \Rightarrow ({x_1},{\text{ }}{y_1}){\rho _1}({x_3},{\text{ }}{y_3})\;\;\;\)hence transitive A1
\({\rho _2}\)
\(({x_1},{\text{ }}{y_1}){\rho _2}({x_1},{\text{ }}{y_1}) \Rightarrow \sqrt {2x_1^2} \leqslant \sqrt {2y_1^2} \;\;\;\)This is not true in the case of (3,1)
hence not reflexive. R1
\(({x_1},{\text{ }}{y_1}){\rho _2}({x_2},{\text{ }}{y_2}) \Rightarrow \sqrt {x_1^2 + x_2^2} \leqslant \sqrt {y_1^2 + y_2^2} \)
\( \Rightarrow \sqrt {x_2^2 + x_1^2} \leqslant \sqrt {y_2^2\_y_1^2} \Rightarrow ({x_2},{\text{ }}{x_2}){\rho _2}({x_1},{\text{ }}{y_1})\;\;\;\)hence symmetric. A1
it is not transitive. A1
attempt to find a counterexample (M1)
for example \((1,{\text{ }}0){\rho _2}(0,{\text{ 1)}}\) and \((0,{\text{ }}1){\rho _2}(1,{\text{ 0)}}\) A1
however, it is not true that \((1,{\text{ }}0){\rho _2}(1,{\text{ 0)}}\) A1
\({\rho _1}\) is an equivalence relation A1
the equivalence classes for \({\rho _1}\) form a family of curves of the form
\({y^2} - {x^2} = k\) A1
Examiners report
Most candidates attempted this question with many showing correctly that \({\rho _1}\) is an equivalence relation. Most candidates, however, were unable to find a counterexample to show that \({\rho _2}\) is not transitive although many suspected that was the case. Most candidates were unable to describe the equivalence classes.
Most candidates attempted this question with many showing correctly that \({\rho _1}\) is an equivalence relation. Most candidates, however, were unable to find a counterexample to show that \({\rho _2}\) is not transitive although many suspected that was the case. Most candidates were unable to describe the equivalence classes.
