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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<π2. 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 cosαsinα<α<sinαcosα.

[5]
a.

Hence show that limα0αsinα=1.

[2]
b.

Let zn=1,zC,nN,n5. 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 P0,P1,P2,Pn1 placed in order in an anticlockwise direction round the circle, starting on the positive x-axis. Show the positions of P0,P1,P2 and Pn1.

[1]
d.

Show that the length of the line segment P0P1 is 2sinπn.

[4]
e.

Hence, write down the total length of the perimeter of the regular n sided polygon P0P1P2Pn1P0.

[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 OPQ=12cosαsinα       A1

Area sector =1212α       A1

Area triangle OPR=121tanα       A1

So looking at the diagram 12cosαsinα<12α<12sinαcosα       M1

cosαsinα<α<sinαcosα       AG

[5 marks]

a.

Hence cosα<αsinα<1cosα and as α0,cosα1  we have     M1R1

limα0αsinα=1      AG

[2 marks]

b.

(rcisθ)n=1cis0rncisnθ=1cisθ      M1A1M1A1

rn=1r=1        nθ=0+2πk,kZ       A1A1

θ=2πkn,0kn1       A1

z=cis2πkn,0kn1       A1

[8 marks]

c.

     A1

[1 mark]

d.

Bisecting the triangle OP0P1 to form two right angle triangles         M1

    

Length of P0P1=2t where t=sin(2πn2)      M1A1A1

So length is 2sinπn      AG

[4 marks]

e.

Length of perimeter is 2nsinπn       A1

[1 mark]

f.

2nsinπn=2πnπsinπn2π as n      M1A1

[2 marks]

g.

Area of OP0P1=121×1sin2πn   so total area is n2sin2πn.   M1A1A1

[3 marks]

h.

n2sin2πn=πn2πsin2πnπ 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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