Date | November 2016 | Marks available | 13 | Reference code | 16N.3.AHL.TZ0.Hsrg_2 |
Level | Additional Higher Level | Paper | Paper 3 | Time zone | Time zone 0 |
Command term | Sketch, Show that, and Find | Question number | Hsrg_2 | Adapted from | N/A |
Question
Let be the set . Let be the set .
A function is defined by .
Let be the set .
A function is defined by .
(i) Sketch the graph of and hence justify whether or not is a bijection.
(ii) Show that is a group under the binary operation of multiplication.
(iii) Give a reason why is not a group under the binary operation of multiplication.
(iv) Find an example to show that is not satisfied for all .
(i) Sketch the graph of and hence justify whether or not is a bijection.
(ii) Show that for all .
(iii) Given that and are both groups, explain whether or not they are isomorphic.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
(i) A1
Notes: Award A1 for general shape, labelled asymptotes, and showing that .
graph shows that it is injective since it is increasing or by the horizontal line test R1
graph shows that it is surjective by the horizontal line test R1
Note: Allow any convincing reasoning.
so is a bijection A1
(ii) closed since non-zero real times non-zero real equals non-zero real A1R1
we know multiplication is associative R1
identity is 1 A1
inverse of is A1
hence it is a group AG
(iii) does not have an identity A2
hence it is not a group AG
(iv) whereas is one counterexample A2
hence statement is not satisfied AG
[13 marks]
award A1 for general shape going through (0, 1) and with domain A1
graph shows that it is injective since it is increasing or by the horizontal line test and graph shows that it is surjective by the horizontal line test R1
Note: Allow any convincing reasoning.
so is a bijection A1
(ii) and M1A1
hence AG
(iii) since is a bijection and the homomorphism rule is obeyed R1R1
the two groups are isomorphic A1
[8 marks]