Q. If $f:R \rightarrow R$ and $g:R \rightarrow R$ are one to one, real valued functions, then the value of the integral $\int_{- \pi}^{\pi}{\left\lbrack f(x) + f( - x) \right\rbrack\left\lbrack g(x) - g( - x) \right\rbrack dx}$ is
A) 0
B) $\pi$
C) 1
D) None of these
Explanation
(a)
Let $\phi(x) = \left\lbrack f(x) + f( - x) \right\rbrack\left\lbrack g(x) - g( - x) \right\rbrack,$
then $\phi( - x) = \left\lbrack f( - x) + f(x) \right\rbrack\lbrack g( - x) - g(x)\rbrack$
$\phi( - x) = - \phi(x)$
$\Rightarrow \$ It is an odd function.
$\therefore\ \int_{- \pi}^{\pi}{\phi(}x)dx = 0$
$\Rightarrow \ \int_{- \pi}^{\pi}{\left\lbrack f(x) + f( - x) \right\rbrack\left\lbrack g(x) - g( - x) \right\rbrack dx = 0}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.