Explanation

(b)
Given, $f(x) = 2x^{3} - 3x^{2} - 12x + 4$
On differentiating w.r.t. $x$, we get
$f(x) = 6x^{2} - 6x - 12$
For maxima or minima put $f^{'}(x) = 0$
$\Rightarrow \ \ x^{2} - x - 2 = 0\ \ \Rightarrow \ \ x = 2,\ - 1$
Again differentiating, we get $f^{''}(x) = 12x - 6$
$\Rightarrow \ \ f^{''}(2) = + ve,\ \ f^{''}( - 1) = - ve$
$\therefore$ Given function has one maximum and one minimum