Explanation

(a)
Here, $f(x) = x^{3} + bx^{2} + cx + d$
$\Rightarrow f^{'}(x) = 3x^{2} + 2bx + c$
(As we know, if $ax^{2} + bx + c > 0\$for all $x$
$\Rightarrow a > 0\ and\ D < 0)$
Now, $D = 4b^{2} - 12c = 4\left( b^{2} - c \right) - 8c$
(where$b^{2} - c < 0\ and\ c > 0)$
$\therefore D = ( - ve)or\ D < 0.$
$\Rightarrow f^{'}(x) = 3x^{2} + 2bx + c > 0$for all $x \in ( - \infty,\infty)$
Hence, $f(x)$is strictly increasing function.