Question

If $2f\left(x\right)+f(-x)=\frac{1}{x}\sin (x-\frac{1}{x})$, then the value of ${\int }_{\frac{1}{e}}^{e}f\left(x\right)dx$, is

• A. $1$
• B. $0$
• C. $e$
• D. $-1$

Answer 1

The correct option is B $0$

Since, $2f\left(x\right)+f(-x)=\frac{1}{x}\sin (x-\frac{1}{x})$ ...(i)
$\therefore 2f(-x)+f\left(x\right)=\frac{1}{x}\sin (x-\frac{1}{x})$ (replace $x$ by $-x$) ...(ii)
$⟹f\left(x\right)=f(-x)$ [subtracting equations (i) and (ii)]
$\therefore 3f\left(x\right)=\frac{1}{x}\sin (x-\frac{1}{x})$
Hence,
$I={\int }_{\frac{1}{e}}^{e}f\left(x\right)dx=\frac{1}{3}{\int }_{\frac{1}{e}}^e\frac{1}{x}\sin (x-\frac{1}{x})dx$
Now, put $x=\frac{1}{t}\Rightarrow dx=-\frac{1}{t^2}dt$
$\therefore I=\frac{1}{3}{\int }_e^{\frac{1}{e}}t\sin (\frac{1}{t}-t).(-\frac{1}{t^2})dt$
$=\frac{1}{3}{\int }_e^{\frac{1}{e}}\frac{1}{t}\sin (t-\frac{1}{t})dt$
$=-\frac{1}{3}{\int }_{\frac{1}{e}}^e\frac{1}{t}\sin (t-\frac{1}{t})dt$
$\Rightarrow I=-I\Rightarrow 2I=0\Rightarrow I=0$
$\therefore {\int }_{\frac{1}{e}}^{e}f\left(x\right)dx=0$.