Question

${\int }_0^{\pi /2}\frac{f\left(x\right)}{f\left(x\right)+f(\frac{\pi }{2}-x)}dx,$where $\int f\left(x\right)\ne -f(\frac{\pi }{2}-x);for$ $0\le x\le \frac{\pi }{2}$ has the value

• A. $f\left(0\right)$
• B. $f\left(\frac{\pi }{2}\right)$
• C. $\frac{\pi }{2}$
• D. none of these

Answer 1

Answer: (D)

none of these

$I={\int }_0^{\pi /2}\frac{f\left(x\right)}{f\left(x\right)+f(\frac{\pi }{2}-x)}dx$ ...(1)
Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$\therefore I={\int }_0^{\pi /2}\frac{f(\frac{\pi }{2}-x)}{f(\frac{\pi }{2}-x)+f\left(x\right)}dx$ ...(2)
From (1) and (2), we get
$2I={\int }_0^{\pi /2}dx=\frac{\pi }{2}\Rightarrow I=\frac{\pi }{4}$