| Date | November 2008 | Marks available | 10 | Reference code | 08N.3srg.hl.TZ0.3 |
| Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ0 |
| Command term | Find and Show that | Question number | 3 | Adapted from | N/A |
Question
Two functions, F and G , are defined on \(A = \mathbb{R}\backslash \{ 0,{\text{ }}1\} \) by
\[F(x) = \frac{1}{x},{\text{ }}G(x) = 1 - x,{\text{ for all }}x \in A.\]
(a) Show that under the operation of composition of functions each function is its own inverse.
(b) F and G together with four other functions form a closed set under the operation of composition of functions.
Find these four functions.
Markscheme
(a) the following two calculations show the required result
\(F \circ F(x) = \frac{1}{{\frac{1}{x}}} = x\)
\(G \circ G(x) = 1 - (1 - x) = x\) M1A1A1
[3 marks]
(b) part (a) shows that the identity function defined by I(x) = x belongs to S A1
the two compositions of F and G are:
\(F \circ G(x) = \frac{1}{{1 - x}};\) (M1)A1
\(G \circ F(x) = 1 - \frac{1}{x}\left( { = \frac{{x - 1}}{x}} \right)\) (M1)A1
the final element is
\(G \circ F \circ G(x) = 1 - \frac{1}{{1 - x}}\left( { = \frac{x}{{x - 1}}} \right)\) (M1)A1
[7 marks]
Total [10 marks]
Examiners report
This question was generally well done. In part(a), the quickest answer involved showing that squaring the function gave the identity. Some candidates went through the more elaborate method of finding the inverse function in each case.
