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Date	May 2022	Marks available	4	Reference code	22M.1.AHL.TZ1.7
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 1
Command term	Find	Question number	7	Adapted from	N/A

Question

The continuous random variable $X$ has probability density function

$f (x) = \{\begin{cases} \frac{k}{\sqrt{4 - 3 x^{2}}}, & 0 \leq x \leq 1 \\ 0, & otherwise . \end{cases}$

Find the value of $k$ .

[4]

Find $E (X)$ .

[4]

Markscheme

attempt to integrate $\frac{k}{\sqrt{4 - 3 x^{2}}}$ (M1)

$= k ⌊\frac{1}{\sqrt{3}} arcsin (\frac{\sqrt{3}}{2} x)⌋$ A1

Note: Award (M1)A0 for $arcsin (\frac{\sqrt{3}}{2} x)$ .
Condone absence of $k$ up to this stage.

equating their integrand to $1$ M1

$k {[\frac{1}{\sqrt{3}} arcsin (\frac{\sqrt{3}}{2} x)]}_{0}^{1} = 1$

$k = \frac{3 \sqrt{3}}{π}$ A1

[4 marks]

$E (X) = \frac{3 \sqrt{3}}{π} \int_{0}^{1} \frac{x}{\sqrt{4 - 3 x^{2}}} d x$ A1

Note: Condone absence of limits if seen at a later stage.

EITHER

attempt to integrate by inspection (M1)

$= \frac{3 \sqrt{3}}{π} \times - \frac{1}{6} \int - 6 x {(4 - 3 x^{2})}^{- \frac{1}{2}} d x$

$= \frac{3 \sqrt{3}}{π} {[- \frac{1}{3} \sqrt{4 - 3 x^{2}}]}_{0}^{1}$ A1

Note: Condone the use of $k$ up to this stage.

for example, $u = 4 - 3 x^{2} \Rightarrow \frac{d u}{d x} = - 6 x$

Note: Other substitutions may be used. For example $u = - 3 x^{2}$ .

$= - \frac{\sqrt{3}}{2 π} \int_{4}^{1} u^{- \frac{1}{2}} d u$ M1

Note: Condone absence of limits up to this stage.

$= - \frac{\sqrt{3}}{2 π} {[2 \sqrt{u}]}_{4}^{1}$ A1

Note: Condone the use of $k$ up to this stage.

THEN

$= \frac{\sqrt{3}}{π}$ A1

Note: Award A0M1A1A0 for their $k [- \frac{1}{3} \sqrt{4 - 3 x^{2}}]$ or $k [- 2 \sqrt{u}]$ for working with incorrect or no limits.

[4 marks]

Examiners report

Most candidates who attempted part (a) knew that the integrand must be equated to 1 and only a small proportion of these managed to recognize the standard integral involved here. The effect of 3 in 3x² was missed by many resulting in very few completely correct answers for this part. Part (b) proved to be challenging for vast majority of the candidates and was poorly done in general. Stronger candidates who made good progress in part (a) were often successful in part (b) as well. Most candidates used a substitution, however many struggled to make progress using this approach. Often when using a substitution, the limits were unchanged. If the function was re-written in terms of x, this did not result in an error in the final answer.

[N/A]

Syllabus sections

Topic 5 —Calculus » SL 5.10—Indefinite integration, reverse chain, by substitution

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Markscheme

Examiners report

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