Question

If $2f\left(x\right)+3f\left(\frac{1}{x}\right)=\frac{1}{x}-2,x\ne 0$, then ${\int }_1^{2}f\left(x\right)dx$ is equal to

• A. $-\frac{2}{5}\log 2+\frac{1}{2}$
• B. $-\frac{2}{5}\log 2-\frac{1}{2}$
• C. $\frac{2}{5}\log 2+\frac{1}{2}$
• D. none of these

Answer 1

The correct option is A $-\frac{2}{5}\log 2+\frac{1}{2}$

We have $2f\left(x\right)+3f\left(\frac{1}{x}\right)=\frac{1}{x}-2$

$\Rightarrow 2f\left(\frac{1}{x}\right)+3f\left(x\right)=x-2$

Solving the above two equations, we get

$f\left(x\right)=-\frac{2}{5x}+\frac{3x}{5}-\frac{2}{5}$

$\therefore {\int }_1^{2}f\left(x\right)dx={\int }_1^2(-\frac{2}{5x}+\frac{3x}{5}-\frac{2}{5})dx$

$={[-\frac{2}{5}\log x+\frac{3x^2}{10}-\frac{2x}{5}]}_1^2$

$=(-\frac{2}{5}\log 2+\frac{6}{5}-\frac{4}{5})-(\frac{3}{10}-\frac{2}{5})$

$=-\frac{2}{5}\log 2+\frac{1}{2}$