Question

Let $f$ be a non-constant continous function for all $x\ge 0$. Let $f$ satisfy the relation $f\left(x\right)f(a-x)=1$ for some $a\in R^{+}$. Then $I={\int }_0^a\frac{dx}{1+f\left(x\right)}$ is equal to

• A. $a$
• B. $\frac{a}{4}$
• C. $\frac{a}{2}$
• D. $f\left(a\right)$

Answer 1

The correct option is C $\frac{a}{2}$

$I={\int }_0^a\frac{dx}{1+f\left(x\right)}$
$\Rightarrow {\int }_0^a\frac{dx}{1+f(a-x)}={\int }_0^a\frac{f\left(x\right)dx}{1+f\left(x\right)}\Rightarrow 2I={\int }_0^{a}dx\Rightarrow I=\frac{a}{2}$