Question

$\underset{0}{\overset{t^2}{\int }}xF(x)dx=\frac{2}{5}{t}^5$, then $F\left(\frac{4}{25}\right)=$

• A. $\frac{1}{5}$
• B. $\frac{1}{25}$
• C. $\frac{2}{5}$
• D. $\frac{4}{25}$

Answer 1

The correct option is C $\frac{2}{5}$

$\begin{array}{c}\underset{0}{\overset{t^2}{\int }}xF(x)dx=\frac{2}{5}{t}^5\\ \frac{d}{dt}\left(\underset{0}{\overset{t^2}{\int }}xF(x)dx\right)=\frac{2}{5}\frac{d}{dt}\left(t^5\right)\\ \Rightarrow t^{2}F\left(t^2\right).\left(2t\right)=\frac{2}{5}.5t^4=2t^4\\ \Rightarrow F\left(t^2\right)=t\\ when,t=\frac{2}{5}\\ F\left(t^2\right)=F\left(\frac{4}{25}\right)\\ F\left(\frac{4}{25}\right)=\frac{2}{5}\end{array}$