| Date | May 2016 | Marks available | 4 | Reference code | 16M.1.hl.TZ0.1 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Find and State | Question number | 1 | Adapted from | N/A |
Question
The set \(P\) contains all prime numbers less than 2500.
The set \(Q\) is the set of all subsets of \(P\).
The set \(S\) contains all positive integers less than 2500.
The function \(f:{\text{ }}S \to Q\) is defined by \(f(s)\) as the set of primes exactly dividing \(s\), for \(s \in S\).
For example \(f(4) = \{ 2\} ,{\text{ }}f(45) = \{ 3,{\text{ }}5\} \).
Explain why only one of the following statements is true
(i) \(17 \subset P\);
(ii) \(\{ 7,{\text{ }}17,{\text{ }}37,{\text{ }}47,{\text{ }}57\} \in Q\);
(iii) \(\phi \subset Q\) and \(\phi \in Q\), where \(\phi \) is the empty set.
(i) State the value of \(f(1)\), giving a reason for your answer.
(ii) Find \(n\left( {f(2310)} \right)\).
Determine whether or not \(f\) is
(i) injective;
(ii) surjective.
Markscheme
(i) 17 is an element not a subset of \(P\) R1
(ii) 57 is not a prime number R1
(iii) any demonstration that this is the true statement A1
because every set contains the empty set as a subset R1
[4 marks]
(i) \(f(1) = \phi \) A1
because 1 has no prime factors R1
(ii) \(f(2310) = f(2 \times 3 \times 5 \times 7 \times 11){\text{ }}\left( { = \{ 2,{\text{ }}3,{\text{ }}5,{\text{ }}7,{\text{ }}11\} } \right)\) A1
\(n\left( {f(2310)} \right) = 5\) A1
[4 marks]
(i) not injective A1
because, for example, \(f(2) = f(4) = \{ 2\} \) R1
(ii) not surjective A1
\({f^{ - 1}}(2,{\text{ }}3,{\text{ }}5,{\text{ }}7,{\text{ }}11,{\text{ }}13)\) does not belong to S because
\(2 \times 3 \times 5 \times 7 \times 11 \times 13 > 2500\) R1
Note: Accept any appropriate example.
[4 marks]
Examiners report
The question caused a number of problems for candidates. In part (a) a number of candidates thought part (i) was correct as they did not realise it was an element and a number thought part (ii) was correct as they did not recognise 57 as a prime number. In both of these two cases, candidates then suggested part (iii) was false giving a variety of incorrect justifications. Part (b) was more successful for most candidates with many wholly correct answers seen. Part (c) again saw many correct answers, but some candidates tried to argue the opposite, incorrect viewpoint or in other cases gave no reason for their decisions, showing a complete misunderstanding of the command term “determine”.
The question caused a number of problems for candidates. In part (a) a number of candidates thought part (i) was correct as they did not realise it was an element and a number thought part (ii) was correct as they did not recognise 57 as a prime number. In both of these two cases, candidates then suggested part (iii) was false giving a variety of incorrect justifications. Part (b) was more successful for most candidates with many wholly correct answers seen. Part (c) again saw many correct answers, but some candidates tried to argue the opposite, incorrect viewpoint or in other cases gave no reason for their decisions, showing a complete misunderstanding of the command term “determine”.
The question caused a number of problems for candidates. In part (a) a number of candidates thought part (i) was correct as they did not realise it was an element and a number thought part (ii) was correct as they did not recognise 57 as a prime number. In both of these two cases, candidates then suggested part (iii) was false giving a variety of incorrect justifications. Part (b) was more successful for most candidates with many wholly correct answers seen. Part (c) again saw many correct answers, but some candidates tried to argue the opposite, incorrect viewpoint or in other cases gave no reason for their decisions, showing a complete misunderstanding of the command term “determine”.
