Question

If for non-zero $x$, $3f\left(x\right)+4f\left(\frac{1}{x}\right)=\frac{1}{x}-10$, then ${\int }_2^{3}f\left(x\right)dx$ is equal to

• A. $\frac{4}{7}ln\frac{2}{3}$
  
• B. $\frac{3}{7}ln\frac{3}{2}$
  
• C. $\frac{3}{7}ln\frac{2}{3}$
  
• D. None of these
  

Answer 1

The correct option is C

$\frac{3}{7}ln\frac{2}{3}$

$3f\left(x\right)+4f\left(\frac{1}{x}\right)=\frac{1}{x}-10...........\left(1\right)$
replace $x$ by $\frac{1}{x}$
$3f\left(\frac{1}{x}\right)+4f\left(x\right)=x-10...........\left(2\right)$
Solving (1) and (2)
$9f\left(x\right)-16f\left(x\right)=\frac{3}{x}-30-4x+40-7f\left(x\right)=10+\frac{3}{x}-4xf\left(x\right)=-\frac{10}{7}-\frac{3}{7x}+\frac{4x}{7}{\int }_2^{3}f\left(x\right)dx={\int }_2^3(-\frac{10}{7}-\frac{3}{7x}+\frac{4x}{7})dx={[\frac{-10x}{7}-\frac{3}{7}ln\left|x\right|+\frac{4x^2}{14}]}_2^3=\frac{3}{7}ln\frac{2}{3}$