Question

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

• A. dependent on $\lambda$
• B. a non zero constant
• C. zero
• D. ${\lambda }^2$

Answer 1

The correct option is C zero

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

$=\underset{{\ln }_{\lambda }}{\overset{-\ln \lambda }{\int }}\frac{f\left(\frac{x^2}{4}\right)\left[f\right(x)-f(-x\left)\right]}{g\left(\frac{x^2}{4}\right)\left[g\right(x)+g(-x\left)\right]}dx=0$
(as function inside the integration is odd)
because $f\left(\frac{x^2}{4}\right)$ is even
$g\left(\frac{x^2}{4}\right)$ is even
$g\left(x\right)+g(-x)$ is even

$f\left(x\right)-f(-x)$ is odd $[\because f(-x)-f(x)=-(f\left(x\right)-f(-x)\left)\right]$

So total function is odd

$\therefore \frac{f\left(\frac{(-x{)}^2}{4}\right)\left[f\right(-x)-f(x\left)\right]}{g\left(\frac{(-x{)}^2}{4}\right)\left[g\right(-x)+g(x\left)\right]}=-\frac{f\left(\frac{x^2}{4}\right)\left[f\right(x)-f(-x\left)\right]}{g\left(\frac{x^2}{4}\right)\left[g\right(x)+g(-x\left)\right]}$