Question
If $$f\left( x \right)$$ is differentiable and $$\int\limits_0^{{t^2}} {xf\left( x \right)dx = \frac{2}{5}{t^5},} $$ then $$f\left( {\frac{4}{{25}}} \right)$$ equals-
A.
$$\frac{2}{5}$$
B.
$$ - \frac{5}{2}$$
C.
$$1$$
D.
$$\frac{5}{2}$$
Answer :
$$\frac{2}{5}$$
Solution :
$$\int_0^{{t^2}} {xf\left( x \right)dx = \frac{2}{5}{t^5}\,\,\,\,\,\,\left( {{\text{Here, }}t > 0} \right)} $$
Differentiating both sides w.r.t. $$t$$ [Using Leibnitz theorem]
$$\eqalign{
& \Rightarrow {t^2}f\left( {{t^2}} \right) \times 2t - 0 = \frac{2}{5} \times 5{t^4} \cr
& \Rightarrow f\left( {{t^2}} \right) = t \cr
& {\text{Put }}t = \frac{2}{5}\,\,\,\,\,\,\,\,\,\, \Rightarrow f\left( {\frac{4}{{25}}} \right) = \frac{2}{5} \cr} $$