• # question_answer If $\int\limits_{0}^{x}{f(t)}dt={{x}^{2}}+\int\limits_{x}^{1}{{{t}^{2}}f(t)}dt,$ then $f'\left( \frac{1}{2} \right)$ is: A) $\frac{24}{25}$ B) $\frac{18}{25}$ C) $\frac{4}{5}$ D) $\frac{6}{25}$

 Differentiability we get $f(x)=2x-{{x}^{2}}f(x)$ $f(x)=\frac{2x}{1+{{x}^{2}}}\Rightarrow f''(x)=2\frac{(1-{{x}^{2}})}{{{(1+{{x}^{2}})}^{2}}}$ $f'\left( \frac{1}{2} \right)=\frac{24}{25}$