Question

Let $f:R\to R$ and $g:R\to R$ be a continous functions. Then the value of the integral
$\underset{-\pi /2}{\overset{\pi /2}{\int }}\left[f\right(x)+f(-x\left)\right]\left[g\right(x)-g(-x\left)\right]dx$ is

• A. $\pi$
• B. $1$
• C. $-1$
• D. $0$

Answer 1

The correct option is D $0$

Let, $h\left(x\right)=\left(f\right(x)+f(-x\left)\right)\left(g\right(x)-g(-x\left)\right)$
$\Rightarrow h(-x)=\left(f\right(-x)+f(x\left)\right)\left(g\right(-x)-g(x\left)\right)$
$\Rightarrow h(-x)=-h\left(x\right)$
$\therefore h\left(x\right)$ is odd function
We know that,
$\underset{-a}{\overset{a}{\int }}f\left(x\right)dx=\{\begin{array}{c}2\underset{0}{\overset{a}{\int }}f\left(x\right)dx,\text{if}f\left(x\right)\text{is even}\\ 0,\text{if}f\left(x\right)\text{is odd}\end{array}$
$\therefore \underset{-\pi /2}{\overset{\pi /2}{\int }}\left(f\right(x)+f(-x\left)\right)\left(g\right(x)-g(-x\left)\right)dx=0$