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Date May 2022 Marks available 7 Reference code 22M.1.AHL.TZ2.11
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 2
Command term State Question number 11 Adapted from N/A

Question

A function f is defined by fx=1x2-2x-3, where x, x-1, x3.

A function g is defined by gx=1x2-2x-3, where x, x>3.

The inverse of g is g-1.

A function h is defined by hx=arctanx2, where x.

Sketch the curve y=f(x), clearly indicating any asymptotes with their equations. State the coordinates of any local maximum or minimum points and any points of intersection with the coordinate axes.

[6]
a.

Show that g-1x=1+4x2+xx.

[6]
b.i.

State the domain of g-1.

[1]
b.ii.

Given that hga=π4, find the value of a.

Give your answer in the form p+q2r, where p, q, r+.

[7]
c.

Markscheme

 

y-intercept 0,-13         A1


Note:
Accept an indication of -13 on the y-axis.


vertical asymptotes x=-1 and x=3          A1

horizontal asymptote y=0          A1

uses a valid method to find the x-coordinate of the local maximum point          (M1)


Note:
For example, uses the axis of symmetry or attempts to solve f'x=0.


local maximum point 1,-14          A1


Note:
Award (M1)A0 for a local maximum point at x=1 and coordinates not given.


three correct branches with correct asymptotic behaviour and the key features in approximately correct relative positions to each other          A1

 

[6 marks]

a.

x=1y2-2y-3           M1


Note: Award M1 for interchanging x and y (this can be done at a later stage).

 

EITHER

attempts to complete the square           M1

y2-2y-3=y-12-4          A1

x=1y-12-4

y-12-4=1xy-12=4+1x          A1

y-1=±4+1x =±4x+1x

 

OR

attempts to solve xy2-2xy-3x-1=0 for y         M1

y=--2x±-2x2+4x3x+12x         A1


Note:
Award A1 even if - (in ±) is missing


=2x±16x2+4x2x         A1

 

THEN

=1±4x2+xx         A1

y>3 and hence y=1-4x2+xx is rejected                R1 

 

Note: Award R1 for concluding that the expression for y must have the ‘+’ sign.
The R1 may be awarded earlier for using the condition x>3.

 

y=1+4x2+xx

g-1x=1+4x2+xx         AG

 

[6 marks]

b.i.

domain of g-1 is x>0         A1

 

[1 mark]

b.ii.

attempts to find hga          (M1)

hga=arctanga2   hga=arctan12a2-2a-3          (A1)

arctanga2=π4   arctan12a2-2a-3=π4

attempts to solve for ga         M1

ga=2  1a2-2a-3=2

 

EITHER

a=g-12         A1

attempts to find their g-12         M1

a=1+422+22         A1

 

Note: Award all available marks to this stage if x is used instead of a.


OR

2a2-4a-7=0         A1

attempts to solve their quadratic equation         M1

a=--4±-42+4274  =4±724         A1


Note: Award all available marks to this stage if x is used instead of a.


THEN

a=1+322  (as a>3)         A1

p=1, q=3, r=2

 

Note: Award A1 for a=1+1218  p=1, q=1, r=18

 

[7 marks]

c.

Examiners report

Part (a) was generally well done. It was pleasing to see how often candidates presented complete sketches here. Several decided to sketch using the reciprocal function. Occasionally, candidates omitted the upper branches or forgot to calculate the y-coordinate of the maximum.

Part (b): The majority of candidates knew how to start finding the inverse, and those who attempted completing the square or using the quadratic formula to solve for y made good progress (both methods equally seen). Otherwise, they got lost in the algebra. Very few explicitly justified the rejection of the negative root.

Part (c) was well done in general, with some algebraic errors seen in occasions.

a.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.

Syllabus sections

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