Explanation

(a)
Here, $\int_{0}^{t^{2}}{\{ x}\ f(x)\} dx = \frac{2}{5}t^{5}$
(Using Newton Leibnitz formula): differentiating both sides, we get
$t^{2}\left\{ f(t^{2}) \right\}.\left\{ \frac{d}{dt}(t^{2}) \right\} - 0.f(0)\left\{ \frac{d}{dt}(0) \right\} = 2t^{4}$
$\Rightarrow \ t^{2}\ f\left( t^{2} \right).2t = 2t^{4}$
$\Rightarrow \ f\left( t^{2} \right) = t$
$\therefore\ f\left( \frac{4}{25} \right) = \pm \frac{2}{5}$ $\left\lbrack putting\ t = \pm \frac{2}{5} \right\rbrack$
$\Rightarrow \ f\left( \frac{4}{25} \right) = \frac{2}{5}$ [neglecting negative]