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Date November 2017 Marks available 2 Reference code 17N.2.AHL.TZ0.H_11
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term Express Question number H_11 Adapted from N/A

Question

Consider the function f ( x ) = 2 sin 2 x + 7 sin 2 x + tan x 9 ,   0 x < π 2 .

Let u = tan x .

Determine an expression for f ( x ) in terms of x .

[2]
a.i.

Sketch a graph of y = f ( x ) for 0 x < π 2 .

[4]
a.ii.

Find the x -coordinate(s) of the point(s) of inflexion of the graph of y = f ( x ) , labelling these clearly on the graph of y = f ( x ) .

[2]
a.iii.

Express sin x in terms of μ .

[2]
b.i.

Express sin 2 x in terms of u .

[3]
b.ii.

Hence show that f ( x ) = 0 can be expressed as u 3 7 u 2 + 15 u 9 = 0 .

[2]
b.iii.

Solve the equation f ( x ) = 0 , giving your answers in the form arctan k where k Z .

[3]
c.

Markscheme

f ( x ) = 4 sin x cos x + 14 cos 2 x + sec 2 x (or equivalent)     (M1)A1

[2 marks]

a.i.

N17/5/MATHL/HP2/ENG/TZ0/11.a.ii/M     A1A1A1A1

 

Note:     Award A1 for correct behaviour at x = 0 , A1 for correct domain and correct behaviour for x π 2 , A1 for two clear intersections with x -axis and minimum point, A1 for clear maximum point.

 

[4 marks]

a.ii.

x = 0.0736     A1

x = 1.13     A1

[2 marks]

a.iii.

attempt to write sin x in terms of u only     (M1)

sin x = u 1 + u 2     A1

[2 marks]

b.i.

cos x = 1 1 + u 2     (A1)

attempt to use sin 2 x = 2 sin x cos x   ( = 2 u 1 + u 2 1 1 + u 2 )     (M1)

sin 2 x = 2 u 1 + u 2     A1

[3 marks]

b.ii.

2 sin 2 x + 7 sin 2 x + tan x 9 = 0

2 u 2 1 + u 2 + 14 u 1 + u 2 + u 9   ( = 0 )     M1

2 u 2 + 14 u + u ( 1 + u 2 ) 9 ( 1 + u 2 ) 1 + u 2 = 0 (or equivalent)     A1

u 3 7 u 2 + 15 u 9 = 0     AG

[2 marks]

b.iii.

u = 1 or u = 3     (M1)

x = arctan ( 1 )     A1

x = arctan ( 3 )     A1

 

Note:     Only accept answers given the required form.

 

[3 marks]

c.

Examiners report

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

Syllabus sections

Topic 5—Calculus » SL 5.1—Introduction of differential calculus
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