Date | May 2017 | Marks available | 3 | Reference code | 17M.3srg.hl.TZ0.1 |
Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ0 |
Command term | Verify and Hence | Question number | 1 | Adapted from | N/A |
Question
The set A contains all positive integers less than 20 that are congruent to 3 modulo 4.
The set B contains all the prime numbers less than 20.
The set C is defined as C={7, 9, 13, 19}.
Write down all the elements of A and all the elements of B.
Determine the symmetric difference, AΔB, of the sets A and B.
Write down all the elements of A∩B, A∩C and B∪C.
Hence by considering A∩(B∪C), verify that in this case the operation ∩ is distributive over the operation ∪.
Markscheme
the elements of A are: 3, 7, 11, 15, 19 A1
the elements of B are 2, 3, 5, 7, 11, 13, 17, 19 A1
Note: Accept A={3, 7, 11, 15, 19} and B={2, 3, 5, 7, 11, 13, 17, 19}
[2 marks]
attempt to determine A∖B∪B∖A or (A∪B)∩(A∩B)′ (M1)
symmetric difference ={2, 5, 13, 15, 17} A1
Note: Allow (M1)A1FT.
[2 marks]
the elements of A∩B are 3, 7, 11 and 19 A1
the elements of A∩C are 7 and19 A1
the elements of B∪C are 2, 3, 5, 7, 9, 11, 13, 17 and 19 A1
Note: Accept A∩B={3, 7, 11, 19}, A∩C={7, 19} and B∪C={2, 3, 5, 7, 9, 11, 13, 17, 19}.
[3 marks]
we need to show that
A∩(B∪C)=(A∩B)∪(A∩C) (M1)
A∩(B∪C)={3, 7, 11, 19} A1
(A∩B)∪(A∩C)={3, 7, 11, 19} A1
hence showing the required result
Note: Allow (M1)A1FTA1FT.
[3 marks]