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Date November 2016 Marks available 8 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 Explain Question number Hsrg_2 Adapted from N/A

Question

Let A be the set { x | x R ,   x 0 } . Let B be the set { x | x ] 1 ,   + 1 [ ,   x 0 } .

A function f : A B is defined by f ( x ) = 2 π arctan ( x ) .

Let D be the set { x | x R ,   x > 0 } .

A function g : R D is defined by g ( x ) = e x .

(i)     Sketch the graph of y = f ( x ) and hence justify whether or not f is a bijection.

(ii)     Show that A is a group under the binary operation of multiplication.

(iii)     Give a reason why B is not a group under the binary operation of multiplication.

(iv)     Find an example to show that f ( a × b ) = f ( a ) × f ( b ) is not satisfied for all a ,   b A .

[13]
a.

(i)     Sketch the graph of y = g ( x ) and hence justify whether or not g is a bijection.

(ii)     Show that g ( a + b ) = g ( a ) × g ( b ) for all a ,   b R .

(iii)     Given that { R ,   + } and { D ,   × } are both groups, explain whether or not they are isomorphic.

[8]
b.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(i)     N16/5/MATHL/HP3/ENG/TZ0/SG/M/02.a.i     A1

 

Notes: Award A1 for general shape, labelled asymptotes, and showing that x 0 .

 

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 f 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 x is 1 x ( x 0 )      A1

hence it is a group     AG

(iii)     B does not have an identity     A2

hence it is not a group     AG

(iv)     f ( 1 × 1 ) = f ( 1 ) = 1 2  whereas f ( 1 ) × f ( 1 ) = 1 2 × 1 2 = 1 4  is one counterexample     A2

hence statement is not satisfied     AG

[13 marks]

a.

N16/5/MATHL/HP3/ENG/TZ0/SG/M/02.b

award A1 for general shape going through (0, 1) and with domain R      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 g   is a bijection     A1

(ii)     g ( a + b ) = e a + b  and g ( a ) × g ( b ) = e a × e b = e a + b      M1A1

hence g ( a + b ) = g ( a ) × g ( b )      AG

(iii)     since g is a bijection and the homomorphism rule is obeyed     R1R1

the two groups are isomorphic     A1

[8 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 2—Functions » SL 2.5—Composite functions, identity, finding inverse
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