Question

If $2f\left(x\right)-3f\left(\frac{1}{x}\right)=x,$then${\int }_1^{2}f\left(x\right)dx=$

• A. $\frac{3}{5}\log 2$
• B. $-\frac{3}{5}(1+\log 2)$
• C. $-\frac{3}{5}\log 2$
• D. None of these

Answer 1

The correct option is B

$-\frac{3}{5}(1+\log 2)$

The explanation for the correct options:

Step1. Given function:

Given that,

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

$x\to \frac{1}{x}$

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

Adding $equation\left(i\right)\times 2+equation\left(ii\right)\times 3$

$$\begin{gathered} 2\left[2f\left(x\right)-3f\left(\frac{1}{x}\right)\right]+3\left[2f\left(\frac{1}{x}\right)-3f\left(x\right)\right]=2x+\frac{3}{x} \\ 4f\left(x\right)-6f\left(\frac{1}{x}\right)+6f\left(\frac{1}{x}\right)-9f\left(x\right)=2x+\frac{3}{x} \\ -5f\left(x\right)=2x+\frac{3}{x} \end{gathered}$$

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

Step2. Find the integration:

So, ${\int }_1^{2}f\left(x\right)dx=$$-\frac{1}{5}{\int }_1^2\left(\frac{2x^2+3}{x}\right)dx$ $\left[\because \int xdx=\frac{x^2}{2}+c\right]\left[\because \int \frac{1}{x}dx=logx+c\right]$

$=-\frac{1}{5}{[x^2+3\ln x]}_1^2$

$=-\frac{3}{5}(1+\ln 2)$

Hence, Option(B) is the correct answer.