Explanation

(b)
$f(x) = \frac{4^{x}}{4^{x} + 2}$
$\therefore\ \ \ \ \ f(1 - x) + f(x) = \frac{4^{1 - x}}{4^{1 - x} + 2} + \frac{4^{x}}{4^{x} + 2}$
$\ \ \ \ \ \ \ \ = \frac{4}{4 + {2.4}^{x}} + \frac{4^{x}}{4^{x} + 2} = \frac{2}{2 + 4^{x}} + \frac{4^{x}}{4^{x} + 2} = 1$
By putting $x = \frac{1}{97},\ \frac{2}{97},\ \frac{3}{97},\ \ldots.\frac{48}{97}$
And adding, we get
$\ \ \ \ \ \ f\left( \frac{1}{97} \right) + f\left( \frac{2}{97} \right) + \ldots + f\left( \frac{96}{97} \right) = 48$