| Date | None Specimen | Marks available | 2 | Reference code | SPNone.3srg.hl.TZ0.2 |
| Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ0 |
| Command term | Determine | Question number | 2 | Adapted from | N/A |
Question
The binary operations \( \odot \) and \( * \) are defined on \({\mathbb{R}^ + }\) by
\[a \odot b = \sqrt {ab} {\text{ and }}a * b = {a^2}{b^2}.\]
Determine whether or not
\( \odot \) is commutative;
\( * \) is associative;
\( * \) is distributive over \( \odot \) ;
\( \odot \) has an identity element.
Markscheme
\(a \odot b = \sqrt {ab} = \sqrt {ba} = b \odot a\) A1
since \(a \odot b = b \odot a\) it follows that \( \odot \) is commutative R1
[2 marks]
\(a * (b * c) = a * {b^2}{c^2} = {a^2}{b^4}{c^4}\) M1A1
\((a * b) * c = {a^2}{b^2} * c = {a^4}{b^4}{c^2}\) A1
these are different, therefore \( * \) is not associative R1
Note: Accept numerical counter-example.
[4 marks]
\(a * (b \odot c) = a * \sqrt {bc} = {a^2}bc\) M1A1
\((a * b) \odot (a * c) = {a^2}{b^2} \odot {a^2}{c^2} = {a^2}bc\) A1
these are equal so \( * \) is distributive over \( \odot \) R1
[4 marks]
the identity e would have to satisfy
\(a \odot e = a\) for all a M1
now \(a \odot e = \sqrt {ae} = a \Rightarrow e = a\) A1
therefore there is no identity element A1
[3 marks]
