Question

If f(x) is a differentiable function and ${\int }_0^{t^2}xf\left(x\right)\mathrm{dx}=\frac{2}{5}{t}^5$, then $f\left(\frac{4}{25}\right)$ equals

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

Answer 1

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

We have,
${\int }_0^{t^2}xf\left(x\right)\mathrm{dx}=\frac{2}{5}{t}^5\mathrm{Differentiating}\mathrm{both}\mathrm{sides}w.r.t.x,\mathrm{we}\mathrm{get}\Rightarrow t^{2}f\left(t^2\right)\times 2t-0=\frac{2}{5}\times 5t^4\Rightarrow f\left(t^2\right)=t\mathrm{Putting}t=\frac{2}{5},\mathrm{we}\mathrm{get}\Rightarrow f\left(\frac{4}{25}\right)=\frac{2}{5}$