Explanation

(c)
Given, $f(x) = x^{3} + ax^{2} + bx + c,a^{2} \leq 3b$
On differentiating w.r.t. $x,$ we get
$f^{'}(x) = 3x^{2} + 2ax + b$
Put $f^{'}(x) = 0$
$\Rightarrow \ \ 3x^{2} + 2ax + b = 0$
$\Rightarrow \ \ \ x = \frac{- 2a \pm \ \sqrt{4a^{2} - 12b}}{2 \times 3}$
$= \frac{- 2a \pm 2\sqrt{a^{2} - 3b}}{6}$
Since, $a^{2} \leq 3b$,
$\therefore\ \ \ x$ Has an imaginary value.
Hence, no extreme value of $x\$exist.