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Date May 2022 Marks available 4 Reference code 22M.1.SL.TZ1.9
Level Standard Level Paper Paper 1 (without calculator) Time zone Time zone 1
Command term Show that Question number 9 Adapted from N/A

Question

Consider fx=4cosx1-3cos2x+3cos22x-cos32x.

Expand and simplify (1-a)3 in ascending powers of a.

[2]
a.i.

By using a suitable substitution for a, show that 1-3cos2x+3cos22x-cos32x=8sin6x.

[4]
a.ii.

Show that 0mfxdx=327sin7m, where m is a positive real constant.

[4]
b.i.

It is given that mπ2fxdx=12728, where 0mπ2. Find the value of m.

[5]
b.ii.

Markscheme

EITHER

attempt to use binomial expansion           (M1)

1+C13×1×-a+C23×1×-a2+1×-a3


OR

1-a1-a1-a

=1-a1-2a+a2           (M1)


THEN

=1-3a+3a2-a3          A1

 

[2 marks]

a.i.

a=cos2x                   (A1)

So, 1-3cos2x+3cos22x-cos32x=

1-cos2x3             A1

attempt to substitute any double angle rule for cos2x into 1-cos2x3                   (M1)

=2sin2x3             A1

=8sin6x             AG


Note: Allow working RHS to LHS.

 

[4 marks]

a.ii.

recognizing to integrate 4cosx×8sin6xdx                   (M1)


EITHER

applies integration by inspection                   (M1)

32cosx×sinx6dx

=327sin7x+c             A1

327sin7x0m   =327sin7m-327sin70             A1


OR

u=sinxdudx=cosx                   (M1)

32cosxsin6xdx=32u6du

=327u7+c             A1

327sin7x0m   OR   327u70sinm   =327sin7m-327sin70             A1


THEN

=327sin7m             AG

 

[4 marks]

b.i.

EITHER

mπ2fxdx=327sin7xmπ2=327sin7π2-327sin7m                   M1

327sin7π2-327sin7m=12728  OR  3271-sin7m=12728                   (M1)


OR

0π2fxdx=0mfxdx+mπ2fxdx                   M1

327=327sin7m+12728                   (M1)


THEN

sin7m=1128  =127                   (A1)

sinm=12                   (A1)

m=π6             A1

 

[5 marks]

b.ii.

Examiners report

Many candidates successfully expanded the binomial, with the most common error being to omit the negative sign with a. The connection between (a)(i) and (ii) was often noted but not fully utilised with candidates embarking on unnecessary complex algebraic expansions of expressions involving double angle rules. Candidates often struggled to apply inspection or substitution when integrating. As a 'show that' question, b(i) provided a useful result to be utilised in (ii). So even without successfully completing (i) candidates could apply it in part (ii). Not many managed to do so.

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

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.5—Unit circle definitions of sin, cos, tan. Exact trig ratios, ambiguous case of sine rule
Show 28 related questions
Topic 5 —Calculus » SL 5.10—Indefinite integration, reverse chain, by substitution
Topic 5 —Calculus » SL 5.11—Definite integrals, areas under curve onto x-axis and areas between curves
Topic 3— Geometry and trigonometry
Topic 5 —Calculus

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