Question

If a continuous function $f$ on $[0,a]$ satisfy $f\left(x\right)f(a-x)=1,$ for $a>0$

then ${\int }_0^a\frac{dx}{1+f\left(x\right)}$ is equal to?

• A. $0$
• B. $a$
• C. $a/2$
• D. none of these

Answer 1

The correct option is C $a/2$

Let $I={\int }_0^a\frac{dx}{1+f\left(x\right)}\Rightarrow \left(1\right)$
Now using ${\int }_a^{b}g\left(x\right)dx={\int }_a^{b}g(a+b-x)dx$
$I={\int }_0^a\frac{dx}{1+f(a-x)}={\int }_0^a\frac{dx}{1+\frac{1}{f\left(x\right)}}[\because f(x\left)f\right(a-x)=1(\text{given}\left)\right]$
$\Rightarrow I={\int }_0^a\frac{f\left(x\right)dx}{1+f\left(x\right)}\Rightarrow \left(2\right)$
Now adding (1) and (2) we get,
$2I={\int }_0^{a}dx=[x{]}_0^a=a$
$\Rightarrow I=\frac{a}{2}$