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Date	May 2019	Marks available	3	Reference code	19M.1.AHL.TZ2.H_10
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 2
Command term	Find	Question number	H_10	Adapted from	N/A

Question

The random variable $X$ has probability density function $f$ given by

$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {k\left( {\pi - {\text{arcsin}}\,x} \right)}&{0 \leqslant x \leqslant 1} \\ 0&{{\text{otherwise}}} \end{array}} \right.,\,\,{\text{where }}k{\text{ is a positive constant}}{\text{.}}$

Given that $y = \left( {\frac{{{x^2}}}{2}} \right){\text{arcsin}}\,x - \left( {\frac{1}{4}} \right){\text{arcsin}}\,x + \left( {\frac{x}{4}} \right)\sqrt {1 - {x^2}}$ , show that

State the mode of $X$ .

[1]

Find $\int {{\text{arcsin}}\,x\,{\text{d}}x}$ .

[3]

b.i.

Hence show that $k = \frac{2}{{2 + \pi }}$ .

[3]

b.ii.

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = x\,{\text{arcsin}}\,x$ .

[4]

c.i.

${\text{E}}\left( X \right) = \frac{{3\pi }}{{4\left( {\pi + 2} \right)}}$ .

[5]

c.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

mode is 0 A1

[1 mark]

attempt at integration by parts (M1)

$\frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{1}{{\sqrt {1 - {x^2}} }}{\text{,}}\,\,{\text{d}}v = {\text{d}}x$

$= x\,{\text{arcsin}}\,x - \int {\frac{{x{\text{d}}x}}{{\sqrt {1 - {x^2}} }}}$ A1

$= x\,{\text{arcsin}}\,x + \sqrt {1 - {x^2}} \left( { + c} \right)$ A1

[3 marks]

b.i.

$\int\limits_0^1 {\left( {\pi - {\text{arcsin}}\,x} \right)} \,{\text{d}}x = \left[ {\pi x - x\,{\text{arcsin}}\,x - \sqrt {1 - {x^2}} } \right]_0^1$ A1

$= \left( {\pi - \frac{\pi }{2} - 0} \right) - \left( {0 - 0 - 1} \right) = \frac{\pi }{2} + 1$

$= \frac{{\pi + 2}}{2}$ A1

$\int\limits_0^1 k \left( {\pi - {\text{arcsin}}\,x} \right){\text{d}}x = 1$ (M1)

Note: This line can be seen (or implied) anywhere.

Note: Do not allow FT A marks from bi to bii.

$k\left( {\frac{{\pi + 2}}{2}} \right) = 1$

$\Rightarrow k = \frac{2}{{2 + \pi }}$ AG

[3 marks]

b.ii.

attempt to use product rule to differentiate M1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = x\,{\text{arcsin}}\,x + \frac{{{x^2}}}{{2\sqrt {1 - {x^2}} }} - \frac{1}{{4\sqrt {1 - {x^2}} }} - \frac{{{x^2}}}{{4\sqrt {1 - {x^2}} }} + \frac{{\sqrt {1 - {x^2}} }}{4}$ A2

Note: Award A2 for all terms correct, A1 for 4 correct terms.

$= x\,{\text{arcsin}}\,x + \frac{{2{x^2}}}{{4\sqrt {1 - {x^2}} }} - \frac{1}{{4\sqrt {1 - {x^2}} }} - \frac{{{x^2}}}{{4\sqrt {1 - {x^2}} }} + \frac{{1 - {x^2}}}{{4\sqrt {1 - {x^2}} }}$ A1

Note: Award A1 for equivalent combination of correct terms over a common denominator.

$= x\,{\text{arcsin}}\,x$ AG

[4 marks]

c.i.

${\text{E}}\left( X \right) = k\int\limits_0^1 {x\left( {\pi - {\text{arcsin}}\,x} \right)} \,{\text{d}}x$ M1

$= k\int\limits_0^1 {\left( {\pi x - x\,{\text{arcsin}}\,x} \right)} \,{\text{d}}x$

$= k\left[ {\frac{{\pi {x^2}}}{2} - \frac{{{x^2}}}{2}{\text{arcsin}}\,x + \frac{1}{4}{\text{arcsin}}\,x - \frac{x}{4}\sqrt {1 - {x^2}} } \right]_0^1$ A1A1