Question

If $f$ and $g$ are continuous functions, then $\underset{\ln \lambda }{\overset{\ln \frac{1}{\lambda }}{\int }}\frac{f\left(\frac{x^2}{4}\right)\left[f\right(x)-f(-x\left)\right]}{g\left(\frac{x^2}{4}\right)\left[g\right(x)+g(-x\left)\right]}dx,\lambda >0,$ is

• A. dependent on $\lambda$ only
• B. a non-zero constant
• C. zero
• D. dependent on $f$ and $g$ only

Answer 1

The correct option is C zero

Let us consider
$H\left(x\right)=\frac{f\left(\frac{x^2}{4}\right)\left[f\right(x)-f(-x\left)\right]}{g\left(\frac{x^2}{4}\right)\left[g\right(x)+g(-x\left)\right]}$
$\Rightarrow H(-x)=\frac{f\left(\frac{x^2}{4}\right)\left[f\right(-x)-f(x\left)\right]}{g\left(\frac{x^2}{4}\right)\left[g\right(-x)+g(x\left)\right]}$
$\Rightarrow H(-x)=\frac{-f\left(\frac{x^2}{4}\right)\left[f\right(x)-f(-x\left)\right]}{g\left(\frac{x^2}{4}\right)\left[g\right(x)+g(-x\left)\right]}$
$\Rightarrow H(-x)=-H\left(x\right)$
So, it is an odd function.

$\therefore \underset{\ln \lambda }{\overset{\ln \frac{1}{\lambda }}{\int }}\frac{f\left(\frac{x^2}{4}\right)\left[f\right(x)-f(-x\left)\right]}{g\left(\frac{x^2}{4}\right)\left[g\right(x)+g(-x\left)\right]}dx$
$=\underset{\ln \lambda }{\overset{-\ln \lambda }{\int }}\frac{f\left(\frac{x^2}{4}\right)\left[f\right(x)-f(-x\left)\right]}{g\left(\frac{x^2}{4}\right)\left[g\right(x)+g(-x\left)\right]}dx$
$=0$