• # question_answer If $f\,(x)=\int\limits_{1}^{x}{\frac{x\,f\,(x)+1}{{{x}^{2}}}}\,\,d\,x$ and $f\,(2)=\frac{3}{4},$ then $f'\,(1)$ is A) 1 B) $\frac{1}{2}$ C) $-\frac{1}{2}$ D) 0

 $f\,(x)=\int\limits_{1}^{x}{\frac{x\,f\,(x)+1}{{{x}^{2}}}}\,\,dx$ $f\,(x)=\frac{x\,f\,(x)+1}{{{x}^{2}}}=\frac{1}{x}\,f\,(x)+\frac{1}{{{x}^{2}}}$ i.e.        $f'\,(x)-\frac{1}{x}f\,(x)=\frac{1}{{{x}^{2}}}$
 Which is linear differential equation on solving it, we get $\therefore$      $\frac{1}{x}f\,(x)=\int{\frac{1}{{{x}^{3}}}}\,\,dx+c=-\frac{1}{2{{x}^{2}}}+c$ i.e.        $f\,(x)=-\frac{1}{2x}+cx$ $\frac{3}{4}=f\,(2)=-\frac{1}{4}+2c$   $\Rightarrow$   $c=\frac{1}{2}$ $f'\,(x)=-\frac{1}{2{{x}^{2}}}+c$ $\therefore$      $f'\,(1)=\frac{1}{2}+\frac{1}{2}=1$