Let f be a non-constant continuous function f | vedclass

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(C) Let $I = \int_{0}^{a} \frac{dx}{1+f(x)} \quad \dots (i)$ Using the property $\int_{0}^{a} g(x) dx = \int_{0}^{a} g(a-x) dx$, we have: $I = \int_{0

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$\frac{a}{2}$

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(C) Let $I = \int_{0}^{a} \frac{dx}{1+f(x)} \quad \dots (i)$ Using the property $\int_{0}^{a} g(x) dx = \int_{0}^{a} g(a-x) dx$, we have: $I = \int_{0}^{a} \frac{dx}{1+f(a-x)}$ Given $f(x) f(a-x) = 1$, we can write $f(a-x) = \frac{1}{f(x)}$. Substituting this into the integral: $I = \int_{0}^{a} \frac{dx}{1+\frac{1}{f(x)}} = \int_{0}^{a} \frac{f(x) dx}{f(x)+1} \quad \dots (ii)$ Adding equations $(i)$ and $(ii)$: $2I = \int_{0}^{a} \frac{dx}{1+f(x)} + \int_{0}^{a} \frac{f(x) dx}{1+f(x)}$ $2I = \int_{0}^{a} \frac{1+f(x)}{1+f(x)} dx = \int_{0}^{a} 1 dx$ $2I = [x]_{0}^{a} = a$ $I = \frac{a}{2}$

Let $f$ be a non-constant continuous function for all $x \geq 0$. Let $f$ satisfy the relation $f(x) f(a-x)=1$ for some $a \in R^{+}$. Then, $I=\int_{0}^{a} \frac{d x}{1+f(x)}$ is equal to

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