• # question_answer Let$f$be a differential function such that $f'(x)=7-\frac{3}{4}\frac{f(x)}{x},(x>0)$and$f(1)\ne 4.$Then$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,xf\left( \frac{1}{x} \right):$ A) exists and equals$\frac{4}{7}.$ B) exists and equals 4. C) does not exist. D) exists and equals 0.

 $f'(x)=7-\frac{3}{4}.\frac{f(x)}{x},x>0$ $\therefore$      $f'(x)+\frac{3}{4x}f(x)=7$              (Linear) $f(x).{{e}^{\int{\frac{3}{4x}dx}}}=\int{7.{{e}^{\int{\frac{3}{4x}dx}}}}+c$ $f(x).{{x}^{3/4}}=\int{7.{{x}^{3/4}}}+c$
 $=7\frac{{{x}^{7/4}}}{\frac{7}{4}}+c$ $\therefore$      $f(x)=4x+c{{x}^{-3/4}}$ $\therefore$      $f\left( \frac{1}{x} \right)=\frac{4}{x}+c{{x}^{3/4}}$ $\therefore$      $\underset{x\to {{0}^{+}}}{\mathop{Lt}}\,xf\left( \frac{1}{x} \right)=\underset{x\to {{0}^{+}}}{\mathop{Lt}}\,4+c{{x}^{7/4}}=4.$