Explanation

(c)
Given, $f(x) = x^{3} - 3x$
$\therefore\ \ \ \ f^{'}(x) = 3x^{2} - 3$
For maxima, $f^{'}(x) = 6x$
$\Rightarrow \ \ \ \ 3x^{2} - 3 = 0\ \ \ \Rightarrow \ \ \ \ x = \pm 1$
$\because\ \ \ \ x = 1 \in \lbrack 0,\ 2\rbrack$
At $x = 1,\ \ f^{''}(x) > 0$, minima
$f(0) = 0,\ \ f(1) = - 2$ and $f(2) = 2$
Hence, maximum value is 2