Question

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

• A. dependent on $\lambda$
• B. a non-zero constant
• C. zero
• D. None of these

Answer 1

Answer: (A), (C)

The correct options are
A dependent on $\lambda$
C zero

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

$={\int }_{\log \lambda }^{-\log \lambda }\frac{f\left(x^2/4\right)\left[f\right(x)-f(-x\left)\right]}{g\left(x^2/4\right)\left[g\right(x)+g(-x\left)\right]}=0$

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

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

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

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

$\therefore h(-x)=-h\left(x\right)$

Thus, the function is odd.

$\therefore I=0$