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Date November 2016 Marks available 6 Reference code 16N.3ca.hl.TZ0.3
Level HL only Paper Paper 3 Calculus Time zone TZ0
Command term Determine Question number 3 Adapted from N/A

Question

Consider the infinite spiral of right angle triangles as shown in the following diagram.

N16/5/MATHL/HP3/ENG/TZ0/SE/03b

The nthnth triangle in the spiral has central angle θnθn, hypotenuse of length anan and opposite side of length 1, as shown in the diagram. The first right angle triangle is isosceles with the two equal sides being of length 1.

Consider the series n=1θnn=1θn.

Using l’Hôpital’s rule, find limx(arcsin(1(x+1))1x)limxarcsin(1(x+1))1x.

[6]
a.

(i) Find a1a1 and a2a2 and hence write down an expression for anan.

(ii) Show that θn=arcsin1(n+1)θn=arcsin1(n+1).

[3]
b.

Using a suitable test, determine whether this series converges or diverges.

[6]
c.

Markscheme

limx(arcsin(1(x+1))1x)limxarcsin(1(x+1))1x is of the form 00

and so will equal the limit of 12(x+1)321(1x+1)12x32     M1M1A1A1

 

Note: M1 for attempting differentiation of the top and bottom, M1A1 for derivative of top (only award M1 if chain rule is used), A1 for derivative of bottom.

 

=limx(x(x+1))32xx+1=limx(xx+1)    M1

 

Note: Accept any intermediate tidying up of correct derivative for the method mark.

 

=1    A1

[6 marks]

a.

(i)     a1=2, a2=3     A1

an=n+1    A1

(ii)     sinθn=1an=1n+1     A1

 

Note: Allow θn=arcsin(1an) if an=n+1 in b(i).

 

so θn=arcsin1(n+1)     AG

[3 marks]

b.

for n=1arcsin1(n+1) apply the limit comparison test (since both series of positive terms)     M1

with n=11n     A1

from (a) limnarcsin1(n+1)1n=1, so the two series either both converge or both diverge     M1R1

n=112 diverges (as is a p-series with p=12)     A1

hence n=1θn diverges     A1

[6 marks]

c.

Examiners report

[N/A]
a.
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b.
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c.

Syllabus sections

Topic 9 - Option: Calculus » 9.2 » Convergence of infinite series.
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