Explanation

(b)
Let the given expression by $y$.
$\therefore y = \ \frac{x + 2}{2x^{2} + 3x + 6}$
$\Rightarrow 2x^{2}y + (3y - 1)x + (6y - 2) = 0$
If $y \neq 0$, then $\mathrm{\Delta}\ \geq 0$ for real $x$.
$ie,\ b^{2} - 4ac \geq 0$
$\therefore(3y - 1)^{2} - 8y(6y - 2) \geq 0$
$\Rightarrow \ - 39y^{2} + 10y + 1 \geq 0$
$\Rightarrow (13y + 1)(3y - 1) \leq 0$
$\Rightarrow \ - \frac{1}{13} \leq y \leq \frac{1}{3}$
If $y = 0$, then $x = - 2$ which is real and this value of $y$ is included in the above range.