Question

If $f\left(\frac{1}{x}\right)+x^{2}f\left(x\right)=0$ for $x>0,$

and $I={\int }_{1/x}^{x}f\left(z\right)dz,\frac{1}{2}\le x\le 2$

then $I$ is?

• A. $f\left(2\right)-f\left(1/2\right)$
• B. $f\left(1/2\right)-f\left(2\right)$
• C. $0$
• D. None of these

Answer 1

Answer: (C)

$0$

Substitute $z=\frac{1}{t}$
$I={\int }_x^{\frac{1}{t}}f\left(\frac{1}{t}\right).\left(\frac{-1}{t^2}\right)dt={\int }_x^{\frac{1}{x}}(-t^2.f\left(t\right))\left(\frac{-1}{t^2}\right)$
$={\int }_x^{\frac{1}{x}}f\left(t\right)dt=-{\int }_{\frac{1}{x}}^{x}f\left(t\right)dt=-I\therefore 2I=0$