Explanation

(c)
$\frac{dy}{dx} = a^{2} - 3x^{2} = 0\ \ \Leftrightarrow \ \ x = \pm \frac{a}{\sqrt{3}}$
Since, $\frac{d^{2y}}{dx^{2}} = - 6x$, so $y$ is minimum for $x = - \frac{a}{\sqrt{3}}$
Since, $x^{2} + x + 2 > 0$ for all$x$, so for $\frac{x^{2} + x + 2}{x^{2} + 5x + 6} \leq 0$, we must have $x^{2} + 5x + 6 < 0$. If $x = - \frac{a}{\sqrt{3}}$, we have,
$\frac{a^{2}}{3} - \frac{5a}{\sqrt{3}} + 6 < 0$
$\Rightarrow \ \ a^{2} - 5\sqrt{3}a + 18 < 0$
$\Rightarrow \ \ \left( a - 2\sqrt{3} \right)\left( a - 3\sqrt{3} \right) < 0$
$\Rightarrow \ \ \ a \in (2\sqrt{3},\ 3\sqrt{3})$.