Question

Let $f$ be a non-constant continuous function for all $x\ge 0$. Let $f$ satisfy the relation $f\left(x\right)=f(a-x)=1$ for some $aϵR^{+}$. 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

Answer: (C)

$\frac{a}{2}$

Given : $f\left(x\right)=f(a-x)=1$

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

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

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

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

$⟹2I={\int }_0^{a}dx$

$⟹2I=|x{|}_0^a$

$⟹I=\frac{a}{2}$