Explanation

(a)
$\because\ \ f(x) = \frac{1}{4x^{2} + 2x + 1}$
On differentiating w.r.t. $x$, we get
$f^{'}(x) = \frac{- (8x + 2)}{\left( 4x^{2} + 2x + 1 \right)^{2}}$ …(i)
For maxima or minima put $f^{'}(x) = 0$
$\Rightarrow \ \ \ 8x + 2 = 0\ \ \ \Rightarrow \ \ x = - \frac{1}{4}$
Again differentiating w.r.t. $x$ of Eq. (i), we get
$f^{''}(x) = - \frac{\left\lbrack \begin{array}{r}
\left( 4x^{2} + 2x + 1 \right)^{2}(8) - (8x - 2)2 \times \\
\left( 4x^{2} + 2x + 1 \right)(8x + 2)
\end{array} \right\rbrack}{\left( 4x^{2} + 2x + 1 \right)^{4}}$
At $x = - \frac{1}{4},\ f^{''}\left( - \frac{1}{4} \right) = - ve$
$f(x)$ is maximum at $x = - \frac{1}{4}$
$\therefore$ maximum value of $f(x)$
$f\left( - \frac{1}{4} \right)_{\max} = \frac{1}{4 \times \frac{1}{16} - 2 \times \frac{1}{4} + 1}$
$= \frac{1}{\frac{1}{4} - \frac{2}{4} + 1}$
$= \frac{4}{1 - 2 + 4} = \frac{4}{3}$