Date | May 2017 | Marks available | 2 | Reference code | 17M.3srg.hl.TZ0.1 |
Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ0 |
Command term | Determine | 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 \(C = \{ 7,{\text{ }}9,{\text{ }}13,{\text{ }}19\} \).
Write down all the elements of \(A\) and all the elements of \(B\).
Determine the symmetric difference, \(A\Delta B\), of the sets \(A\) and \(B\).
Write down all the elements of \(A \cap B,{\text{ }}A \cap C\) and \(B \cup C\).
Hence by considering \(A \cap (B \cup C)\), verify that in this case the operation \( \cap \) is distributive over the operation \( \cup \).
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 \(A = \{ 3,{\text{ }}7,{\text{ }}11,{\text{ }}15,{\text{ }}19\} \) and \(B = \{ 2,{\text{ }}3,{\text{ }}5,{\text{ }}7,{\text{ }}11,{\text{ }}13,{\text{ }}17,{\text{ }}19\} \)
[2 marks]
attempt to determine \(A\backslash B \cup B\backslash A\) or \((A \cup B) \cap (A \cap B)'\) (M1)
symmetric difference \( = \{ 2,{\text{ }}5,{\text{ }}13,{\text{ }}15,{\text{ }}17\} \) A1
Note: Allow (M1)A1FT.
[2 marks]
the elements of \(A \cap B\) are 3, 7, 11 and 19 A1
the elements of \(A \cap C\) are 7 and19 A1
the elements of \(B \cup C\) are 2, 3, 5, 7, 9, 11, 13, 17 and 19 A1
Note: Accept \(A \cap B = \{ 3,{\text{ }}7,{\text{ }}11,{\text{ }}19\} ,{\text{ }}A \cap C = \{ 7,{\text{ }}19\} \) and \(B \cup C = \{ 2,{\text{ }}3,{\text{ }}5,{\text{ }}7,{\text{ }}9,{\text{ }}11,{\text{ }}13,{\text{ }}17,{\text{ }}19\} \).
[3 marks]
we need to show that
\(A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\) (M1)
\(A \cap (B \cup C) = \{ 3,{\text{ }}7,{\text{ }}11,{\text{ }}19\} \) A1
\((A \cap B) \cup (A \cap C) = \{ 3,{\text{ }}7,{\text{ }}11,{\text{ }}19\} \) A1
hence showing the required result
Note: Allow (M1)A1FTA1FT.
[3 marks]