| Date | November 2011 | Marks available | 3 | Reference code | 11N.3srg.hl.TZ0.5 |
| Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ0 |
| Command term | Show that | Question number | 5 | Adapted from | N/A |
Question
Consider the functions \(f:A \to B\) and \(g:B \to C\).
Show that if both f and g are injective, then \(g \circ f\) is also injective.
Show that if both f and g are surjective, then \(g \circ f\) is also surjective.
Show, using a single counter example, that both of the converses to the results in part (a) and part (b) are false.
Markscheme
let s and t be in A and \(s \ne t\) M1
since f is injective \(f(s) \ne f(t)\) A1
since g is injective \(g \circ f(s) \ne g \circ f(t)\) A1
hence \(g \circ f\) is injective AG
[3 marks]
let z be an element of C
we must find x in A such that \(g \circ f(x) = z\) M1
since g is surjective, there is an element y in B such that \(g(y) = z\) A1
since f is surjective, there is an element x in A such that \(f(x) = y\) A1
thus \(g \circ f(x) = g(y) = z\) R1
hence \(g \circ f\) is surjective AG
[4 marks]
converses: if \(g \circ f\) is injective then g and f are injective
if \(g \circ f\) is surjective then g and f are surjective (A1)
A2
Note: There will be many alternative counter-examples.
[3 marks]
Examiners report
This question was found difficult by a large number of candidates and no fully correct solutions were seen. A number of students made thought-through attempts to show it was surjective, but found more difficulty in showing it was injective. Very few were able to find a single counter example to show that the converses of the earlier results were false. Candidates struggled with the abstract nature of the question.
This question was found difficult by a large number of candidates and no fully correct solutions were seen. A number of students made thought-through attempts to show it was surjective, but found more difficulty in showing it was injective. Very few were able to find a single counter example to show that the converses of the earlier results were false. Candidates struggled with the abstract nature of the question.
This question was found difficult by a large number of candidates and no fully correct solutions were seen. A number of students made thought-through attempts to show it was surjective, but found more difficulty in showing it was injective. Very few were able to find a single counter example to show that the converses of the earlier results were false. Candidates struggled with the abstract nature of the question.
