Explanation

(a)
Let $f(x) = x\sqrt{1 - x^{2}}$
$\Rightarrow \ \ \ \ f^{'}(x) = \frac{1 - 2x^{2}}{\sqrt{1 - x^{2}}}$
$\Rightarrow \$ For maxima or minima, put $\frac{dy}{dx} = 0$
$x = \pm \frac{1}{\sqrt{2}}$
But $x > 0$, therefore we have $x = \frac{1}{\sqrt{2\ }}$
Now, $f^{''}(x) = \frac{\sqrt{1 - x^{2}}( - 4x) - \left( 1 - 2x^{2} \right)\frac{- x}{\sqrt{1 - x^{2}}}}{(1 - x^{2})}$
$= \frac{2x^{3} - 3x}{\left( 1 - x^{2} \right)^{3/2}}$
$\Rightarrow \ \ \ f''\left( \frac{1}{\sqrt{2}} \right)$=$-$ve, maximum.