| Date | May 2018 | Marks available | 2 | Reference code | 18M.2.hl.TZ0.6 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Find | Question number | 6 | Adapted from | N/A |
Question
The set of all integer s from 0 to 99 inclusive is denoted by S. The binary operations \( * \) and \( \circ \) are defined on S by
\(a * b = \left[ {a + b + 20} \right]\)(mod 100)
\(a \circ b = \left[ {a + b - 20} \right]\)(mod 100).
The equivalence relation R is defined by \(aRb \Leftrightarrow \left( {{\text{sin}}\frac{{\pi a}}{5} = {\text{sin}}\frac{{\pi b}}{5}} \right)\).
Find the identity element of S with respect to \( * \).
Show that every element of S has an inverse with respect to \( * \).
State which elements of S are self-inverse with respect to \( * \).
Prove that the operation \( \circ \) is not distributive over \( * \).
Determine the equivalence classes into which R partitions S, giving the first four elements of each class.
Find two elements in the same equivalence class which are inverses of each other with respect to \( * \).
Markscheme
\(a + e + 20 = a\)(mod 100) (M1)
\(e = - 20\)(mod 100) (A1)
\(e = 80\) A1
[3 marks]
\(a + {a^{ - 1}} + 20 = 80\)(mod 100) (M1)
inverse of \(a\) is \(60 - a\) (mod 100) A1
[2 marks]
30 and 80 A1A1
[2 marks]
\(a \circ \left( {b * c} \right) = a \circ \left( {b + c + 20} \right)\)(mod 100)
\( = a + \left( {b + c + 20} \right) - 20\)(mod 100) (M1)
\( = a + b + c\)(mod 100) A1
\(\left( {a \circ b} \right) * \left( {a \circ c} \right) = \left( {a + b - 20} \right) * \left( {a + c - 20} \right)\)(mod 100) M1
\( = a + b - 20 + a + c - 20 + 20\)(mod 100)
\( = 2a + b + c - 20\)(mod 100) A1
hence we have shown that \(a \circ \left( {b * c} \right) \ne \left( {a \circ b} \right) * \left( {a \circ c} \right)\) R1
hence the operation \( \circ \) is not distributive over \( * \) AG
Note: Accept a counterexample.
[5 marks]
{0,5,10,15...} A1
{1,4,11,14...} A1
{2,3,12,13...} A1
{6,9,16,19...} A1
{7,8,17,18...} A1
[5 marks]
for example 10 and 50, 20 and 40, 0 and 60… A2
[2 marks]
