Question

If $f\left(\frac{1}{x}\right)+x^{2}f\left(x\right)=0$, x> 0 and $I={\int }_{1/x}^{x}f\left(z\right)dz$, $\frac{1}{2}\le x\le 2$, then I is

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

Answer 1

Answer: (C)

0

$f\left(\frac{1}{x}\right)+x^{2}f\left(x\right)=0,x>0$ & $I={\int }_{\frac{1}{x}}^{x}f\left(z\right)dz;x\in \left[\frac{1}{2}\right]$ ----(1)
$f\left(\frac{1}{x}\right)=-x^{2}f\left(x\right)$
$x=\frac{1}{x}$
$f\left(x\right)=\frac{-1}{x^2}+\left(\frac{1}{x}\right)$
$z=\frac{1}{p}$
$dz=\frac{-dp}{p^2}$
$I={\int }_{\frac{1}{x}}^x-f\left(\frac{1}{p}\right)\frac{dp}{p^2}$
$I={\int }_{\frac{1}{x}}^x-f\left(p\right)dp$
$={\int }_{\frac{1}{x}}^{x}f\left(z\right)dz$ --------(2)
adding 1 & 2
we get
$2I={\int }_{\frac{1}{x}}^{x}f\left(z\right)dz+{\int }_{\frac{1}{x}}^x-f\left(z\right)dz$
$2I=0$
$I=0$