Question:
If $\int_{0}^{x} f(t) d t=x^{2}+\int_{x}^{1} t^{2} f(t) d t$, then $f^{\prime}(1 / 2)$ is :
Correct Option: , 2
Solution:
$\int_{0}^{x} f(t) d t=x^{2}+\int_{x}^{1} t^{2} f(t) d t$
$f^{\prime}\left(\frac{1}{2}\right)=?$
Differentiate w.r.t. ' $x$ '
$f(x)=2 x+0-x^{2} f(x)$
$f(x)=\frac{2 x}{1+x^{2}} \Rightarrow f^{\prime}(x)=\frac{\left(1+x^{2}\right) 2-2 x(2 x)}{\left(1+x^{2}\right)^{2}}$
$f^{\prime}(x)=\frac{2 x^{2}-4 x^{2}+2}{\left(1+x^{2}\right)^{2}}$
$f^{\prime}\left(\frac{1}{2}\right)=\frac{2-2\left(\frac{1}{4}\right)}{\left(1+\frac{1}{4}\right)^{2}}=\frac{\left(\frac{3}{2}\right)}{\frac{25}{16}}=\frac{48}{50}=\frac{24}{25}$
Option (2)