Explanation

(a)
$f(a) = \frac{e^{a}}{1 + e^{a}\ }\$ …(i)
and $f( - a) = \frac{e^{- a}}{1 + e^{- a}} = \frac{1}{1 + e^{a}}$ …(ii)
On adding Eqs.(i) and (ii), we get
$f(a) + f( - a) = 1$
$\Rightarrow \ f(a) = 1 - f( - a)$
Let $f( - a) = t,\$then$\ f(a) = 1 - t$
Now, $I_{1} = \int_{t}^{1 - t}{xg\ \left\lbrack x(1 - x) \right\rbrack dx\ \ \ldots(i)}$
$\Rightarrow I_{1} = \int_{t}^{1 - t}{(1 - x)g\left\lbrack x(1 - x) \right\rbrack dx}\ \ \ \ \ldots(ii)$
On adding Eqs. (i) and (ii). We get
$2I_{1} = \int_{t}^{1 - t}{g\left\lbrack x(1 - x) \right\rbrack dx = I_{2}}$
$\Rightarrow \frac{I_{2}}{I_{1}} = 2$