Let f(x) be a continuous functions such that f(a-x)+f(x)=0 for all x∈[0,a]. Then ∫0adx1+ef(x) is equal to

I=∫0adx1+ef(x) =∫0adx1+ef(a-x) =∫0adx1+e-f(x) I=∫0aef(x) dxef(x)+1 2I=∫0a1+ef(x)ef(x)+1dx =a I=a2

Let f(x) be a continuous functions such that f(a-x)+f(x)=0 for all x∈[0,a]. Then ∫0adx1+ef(x) is equal to

I=∫0adx1+ef(x) =∫0adx1+ef(a-x) =∫0adx1+e-f(x) I=∫0aef(x) dxef(x)+1 2I=∫0a1+ef(x)ef(x)+1dx =a I=a2

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Let f(x) be a continuous functions such that f(a-x)+f(x)=0 for all $x \in [0, a]$ . Then $\int_{0}^{a} \frac{d x}{1 + e^{f (x)}}$ is equal to

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$\frac{a}{2}$

f(a)

$\frac{1}{2} f (a)$

answer is B.

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Detailed Solution

$I = \int_{0}^{a} \frac{dx}{1 + e^{f (x)}} = \int_{0}^{a} \frac{dx}{1 + e^{f (a - x)}} = \int_{0}^{a} \frac{dx}{1 + e^{- f (x)}} I = \int_{0}^{a} \frac{e^{f (x)} dx}{e^{f (x)} + 1} 2 I = \int_{0}^{a} \frac{1 + e^{f (x)}}{e^{f (x)} + 1} dx = a I = \frac{a}{2}$