Explanation

(a)
$\because y = a\left( 1 - \cos x \right)$
On differentiating w.r.t. $x$, we get
$y^{'} = a\sin x$ …(i)
Put $y^{'} = 0$ for maxima or minima
$\Rightarrow \ \ \sin x = 0$
$\Rightarrow \ \ \ \ x = 0,\ \pi$
Again differentiating w.r.t. $x$ of Eq. (i) we get
$y^{''} = a\cos x\ \ \Rightarrow \ \ y^{''}(0) = 0$ and $y^{''}(\pi) = - a$
Hence, $y$ is maximum when $x = \pi$