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Date	May 2022	Marks available	5	Reference code	22M.1.SL.TZ1.9
Level	Standard Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 1
Command term	Find	Question number	9	Adapted from	N/A

Question

Consider $f (x) = 4 \cos x (1 - 3 \cos 2 x + 3 \cos^{2} 2 x - \cos^{3} 2 x)$ .

Expand and simplify ${(1 - a)}^{3}$ in ascending powers of $a$ .

[2]

a.i.

By using a suitable substitution for $a$ , show that $1 - 3 \cos 2 x + 3 \cos^{2} 2 x - \cos^{3} 2 x = 8 \sin^{6} x$ .

[4]

a.ii.

Show that $\int_{0}^{m} f (x) d x = \frac{32}{7} \sin^{7} m$ , where $m$ is a positive real constant.

[4]

b.i.

It is given that $\int_{m}^{\frac{π}{2}} f (x) d x = \frac{127}{28}$ , where $0 \leq m \leq \frac{π}{2}$ . Find the value of $m$ .

[5]

b.ii.

Markscheme

EITHER

attempt to use binomial expansion (M1)

$1 + {}_{3}C_{1} \times 1 \times (- a) + {}_{3}C_{2} \times 1 \times {(- a)}^{2} + 1 \times {(- a)}^{3}$

$(1 - a) (1 - a) (1 - a)$

$= (1 - a) (1 - 2 a + a^{2})$ (M1)

THEN

$= 1 - 3 a + 3 a^{2} - a^{3}$ A1

[2 marks]

a.i.

$a = \cos 2 x$ (A1)

So, $1 - 3 \cos 2 x + 3 \cos^{2} 2 x - \cos^{3} 2 x =$

${(1 - \cos 2 x)}^{3}$ A1

attempt to substitute any double angle rule for $\cos 2 x$ into ${(1 - \cos 2 x)}^{3}$ (M1)

$= {(2 \sin^{2} x)}^{3}$ A1

$= 8 \sin^{6} x$ AG

Note: Allow working RHS to LHS.

[4 marks]

a.ii.

recognizing to integrate $\int (4 \cos x \times 8 \sin^{6} x) d x$ (M1)

EITHER

applies integration by inspection (M1)

$32 \int (\cos x \times {(\sin x)}^{6}) d x$

$= \frac{32}{7} \sin^{7} x (+ c)$ A1

${[\frac{32}{7} \sin^{7} x]}_{0}^{m} (= \frac{32}{7} \sin^{7} m - \frac{32}{7} \sin^{7} 0)$ A1

$u = \sin x \Rightarrow \frac{d u}{d x} = \cos x$ (M1)

$\int 32 \cos x (\sin^{6} x) d x = \int 32 u^{6} d u$

$= \frac{32}{7} u^{7} (+ c)$ A1

${[\frac{32}{7} \sin^{7} x]}_{0}^{m}$ OR ${[\frac{32}{7} u^{7}]}_{0}^{\sin m} (= \frac{32}{7} \sin^{7} m - \frac{32}{7} \sin^{7} 0)$ A1

THEN

$= \frac{32}{7} \sin^{7} m$ AG

[4 marks]

b.i.

EITHER

$\int_{m}^{\frac{π}{2}} f (x) d x (= {[\frac{32}{7} \sin^{7} x]}_{m}^{\frac{π}{2}}) = \frac{32}{7} \sin^{7} \frac{π}{2} - \frac{32}{7} \sin^{7} m$ M1

$\frac{32}{7} \sin^{7} \frac{π}{2} - \frac{32}{7} \sin^{7} m = \frac{127}{28}$ OR $\frac{32}{7} (1 - \sin^{7} m) = \frac{127}{28}$ (M1)

$\int_{0}^{\frac{π}{2}} f (x) d x = \int_{0}^{m} f (x) d x + \int_{m}^{\frac{π}{2}} f (x) d x$ M1

$\frac{32}{7} = \frac{32}{7} \sin^{7} m + \frac{127}{28}$ (M1)

THEN

$\sin^{7} m = \frac{1}{128} (= \frac{1}{2^{7}})$ (A1)

$\sin m = \frac{1}{2}$ (A1)

$m = \frac{π}{6}$ 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.

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b.i.

[N/A]

b.ii.