Question

lf $2f\left(x\right)+3f(\frac{1}{x})=\frac{1}{x}-2,x\ne 0$, 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. $-\frac{2}{5}$ log2

Answer 1

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

Given, $2f\left(x\right)+3f(\frac{1}{x})=\frac{1}{x}-2,x\ne 0$ ..........(1)
Replacing $x$ by $\frac{1}{x}$ , we have :
$2f(\frac{1}{x})+3f(x)=x-2$ ..............(2)

From (1) and (2) , on solving for $f\left(x\right)$ we have $f\left(x\right)=\frac{1}{5}(3x-\frac{2}{x}-2)$
Now ${\int }_1^{2}f\left(x\right)dx=\frac{1}{5}{\int }_1^2(3x-\frac{2}{x}-2)dx$

On integrating the above equation we get ${\int }_1^{2}f\left(x\right)dx=\frac{-2}{5}log2+\frac{1}{2}$
Hence, option 'A' is correct.