Explanation

(c)
$\because\ \ f(x) = \frac{x}{4 + x + x^{2}}$
On differentiating w.r.t. $x$, we get
$f^{'}(x) = \frac{4 + x + x^{2} - x(1 + 2x)}{\left( 4 + x + x^{2} \right)^{2}}$
For maximum, put $f^{'}(x) = 0\ \ \Rightarrow \ \ \frac{4 - x^{2}}{\left( 4 + x + x^{2} \right)^{2}} = 0$
$\Rightarrow \ \ x = 2,\ - 2$
Both the values of $x$ are not in the interval $\lbrack - 1,\ 1\rbrack$
$\therefore\ \ \ f( - 1) = \frac{- 1}{4 - 1 + 1} = \frac{- 1}{4}$
$f(1) = \frac{1}{4 + 1 + 1} = \frac{1}{6}$ (maximum)