Ans C) 0.5a (assuming X $\epsilon$ (-$\infty$, $\infty$))
P(X) = ke(-a|x|)
$\therefore \int_{-\infty }^{\infty}$ ke(-a|x|) dx = 1
$\Rightarrow$ 2$\int_{0}^{\infty }$ ke(-a|x|) dx = 1 ($\because$ ke(-a|x|) is even function)
$\Rightarrow$ 2k $\int_{0}^{\infty }$ e-ax dx = 1
$\Rightarrow$ 2k [ $\frac{e^{-ax}}{-a}$ ]0$\infty$ = 1
$\Rightarrow$ $\frac{2k}{-a}$ * (0 - 1) = 1
$\Rightarrow$ k = 0.5a