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Date May Example questions Marks available 2 Reference code EXM.3.AHL.TZ0.2
Level Additional Higher Level Paper Paper 3 Time zone Time zone 0
Command term Find Question number 2 Adapted from N/A

Question

This question will explore connections between complex numbers and regular polygons.

The diagram below shows a sector of a circle of radius 1, with the angle subtended at the centre O being α , 0 < α < π 2 . A perpendicular is drawn from point P  to intersect the x -axis at Q . The tangent to the circle at P  intersects the x -axis at  R .

By considering the area of two triangles and the area of the sector show that cos α sin α < α < sin α cos α .

[5]
a.

Hence show that lim α 0 α sin α = 1 .

[2]
b.

Let  z n = 1 , z C , n N , n 5 . Working in modulus/argument form find the n solutions to this equation.

[8]
c.

Represent these n solutions on an Argand diagram. Let their positions be denoted by  P 0 , P 1 , P 2 , P n 1  placed in order in an anticlockwise direction round the circle, starting on the positive x -axis. Show the positions of  P 0 , P 1 , P 2 and  P n 1 .

[1]
d.

Show that the length of the line segment  P 0 P 1 is 2 sin π n .

[4]
e.

Hence, write down the total length of the perimeter of the regular n  sided polygon  P 0 P 1 P 2 P n 1 P 0 .

[1]
f.

Using part (b) find the limit of this perimeter as n .

[2]
g.

Find the total area of this n sided polygon.

[3]
h.

Using part (b) find the limit of this area as n .

[2]
i.

Markscheme

Area triangle  O P Q = 1 2 cos α sin α        A1

Area sector  = 1 2 1 2 α        A1

Area triangle  O P R = 1 2 1 tan α        A1

So looking at the diagram  1 2 cos α sin α < 1 2 α < 1 2 sin α cos α        M1

cos α sin α < α < sin α cos α        AG

[5 marks]

a.

Hence  cos α < α sin α < 1 cos α and as  α 0 , cos α 1   we have     M1R1

lim α 0 α sin α = 1      AG

[2 marks]

b.

( r c i s θ ) n = 1 c i s 0 r n c i s n θ = 1 c i s θ       M1A1M1A1

r n = 1 r = 1          n θ = 0 + 2 π k , k Z        A1A1

θ = 2 π k n , 0 k n 1        A1

z = c i s 2 π k n , 0 k n 1        A1

[8 marks]

c.

     A1

[1 mark]

d.

Bisecting the triangle  O P 0 P 1 to form two right angle triangles         M1

    

Length of  P 0 P 1 = 2 t where t = sin ( 2 π n 2 )       M1A1A1

So length is  2 sin π n       AG

[4 marks]

e.

Length of perimeter is 2 n sin π n        A1

[1 mark]

f.

2 n sin π n = 2 π n π sin π n 2 π as  n       M1A1

[2 marks]

g.

Area of  O P 0 P 1 = 1 2 1 × 1 sin 2 π n    so total area is  n 2 sin 2 π n .   M1A1A1

[3 marks]

h.

n 2 sin 2 π n = π n 2 π sin 2 π n π as  n       M1A1

[2 marks]

i.

Examiners report

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Syllabus sections

Topic 1—Number and algebra » AHL 1.13—Polar and Euler form
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Topic 1—Number and algebra

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