Question

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

Given ${\int }_0^{t^2}xf\left(x\right)=\frac{2}{5}{t}^5$.....(i)
Since, the differentiation of definite integral:
$\frac{d}{dt}\left[{\int }_a^{b}h\right(x\left)\right]=\frac{db}{dt}h\left(b\right)-\frac{da}{dt}h\left(a\right)$.....(ii)
Substitute $a=0,b=t^2$ we get
$\frac{d}{dt}\left[{\int }_0^{t^2}h\right(x\left)\right]=\frac{d{t}^2}{dt}h\left(t^2\right)-\frac{d\left(0\right)}{dt}h\left(0\right)$
Now, take $h\left(x\right)=xf\left(x\right)$, we get
$\frac{d}{dt}\left[{\int }_0^{t^2}xf\right(x\left)\right]=2t\left(t^2\right)f\left(t^2\right)-0$
$\frac{d}{dt}\left[{\int }_0^{t^2}xf\right(x\left)\right]=2t^{3}f\left(t^2\right)$
Substitute equation(i) in above equation, we get
$\frac{d}{dt}\left[\frac{2}{5}{t}^5\right]=2t^{3}f\left(t^2\right)$
$\frac{2}{5}\left(5t^4\right)=2t^{3}f\left(t^2\right)$
$2t^4=2t^{3}f\left(t^2\right)$
$f\left(t^2\right)=\frac{t^4}{t^3}$
Thus, $f\left(t^2\right)=t$
Substitute $t=\frac{2}{5}$
$\Rightarrow f\left(\right[\frac{2}{5}{]}^2)=\frac{2}{5}$
$\therefore f\left(\frac{4}{25}\right)=\frac{2}{5}$