The probability density function of a random variable X is defined as:

$$f(x) = \left\{ \begin{array}{r}ax\cos x,\\0,\end{array} \right.\begin{array}{*{20}{l}}{0 \le x \le {\textstyle{\pi  \over 2}},{\rm{where }}\,a \in \mathbb{R}}\\{{\rm{elsewhere}}}\end{array}$$

(a)     Show that $a = \frac{2}{{\pi  - 2}}$.

(b)     Find ${\text{P}}\left( {X < \frac{\pi }{4}} \right)$.

(c)     Find:

          (i)     the mode of X;

          (ii)     the median of X.

(d)     Find ${\text{P}}\left( {X < \frac{\pi }{8}|X < \frac{\pi }{4}} \right)$.

(a)     $a\int_0^{\frac{\pi }{2}} {x\cos x{\text{d}}x = 1}$     (M1)

integrating by parts:

$u = x$     $v' = \cos x$     M1

$u' = 1$     $v = \sin x$

$\int {x\cos x{\text{d}}x = x\sin x + \cos x}$     A1

$\left[ {x\sin x + \cos x} \right]_0^{\frac{\pi }{2}} = \frac{\pi }{2} - 1$     A1

$a = \frac{1}{{\frac{\pi }{2} - 1}}$     A1

$= \frac{2}{{\pi  - 2}}$     AG

[5 marks]

(b)     ${\text{P}}\left( {X < \frac{\pi }{4}} \right) = \frac{2}{{\pi  - 2}}\int_0^{\frac{\pi }{4}} {x\cos x{\text{d}}x = 0.460}$     (M1)A1

Note:     Accept $\frac{2}{{\pi  - 2}}{\text{ }}\left( { = \frac{{\pi \sqrt 2 }}{8} + \frac{{\sqrt 2 }}{2} - 1} \right)$ or equivalent

[2 marks]

(c)     (i)     ${\text{mode}} = 0.860$     A1

          (x-value of a maximum on the graph over the given domain)

          (ii)     $\frac{2}{{\pi  - 2}}\int_0^m {x\cos x{\text{d}}x = 0.5}$     (M1)

          $\int_0^m {x\cos x{\text{d}}x = \frac{{\pi  - 2}}{4}}$

          $m\sin m + \cos m - 1 = \frac{{\pi  - 2}}{4}$     (M1)

          ${\text{median}} = 0.826$     A1

Note:     Do not accept answers containing additional solutions.

[4 marks]

(d)     ${\text{P}}\left( {X < \frac{\pi }{8}|X < \frac{\pi }{4}} \right) = \frac{{{\text{P}}\left( {X < \frac{\pi }{8}} \right)}}{{{\text{P}}\left( {X < \frac{\pi }{4}} \right)}}$     M1

$= \frac{{0.129912}}{{0.459826}}$

$= 0.283$     A1

[2 marks]

Total [13 marks]