Correct Answer - Option 4 :
\(\rm \frac{a}{2}\)
Concept:
Definite Integrals:
\(\rm \int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx\).
If f(x) = f(2a - x), then \(\rm \int_0^{2a}f(x)\ dx=2\int_0^af(x)\ dx\).
A function f(x) is:
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Even, if f(-x) = f(x). And \(\rm \int_{-a}^ {\ \ a}f(x)\ dx=2\int_{0}^af(x)\ dx\).
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Odd, if f(-x) = -f(x). And \(\rm \int_{-a}^ {\ \ a}f(x)\ dx=0\).
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Periodic, if f(np ± x) = f(x), for some number p and n ∈ Z.
Calculation:
We know that \(\rm \int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx\).
Let I = \(\rm \int_0^a \frac{f(a-x)}{f(x)+f(a-x)}\ dx\)
⇒ \(\rm \int_0^a \frac{f[(a+0)-(a-x)]}{f[(a+0)-x]+f[(a+0)-(a-x)]}\ dx\)
⇒ \(\rm \int_0^a \frac{f(x)}{f(a-x)+f(x)}\ dx\)
⇒ 2I = \(\rm \int_0^a \frac{f(a-x)}{f(x)+f(a-x)}\ dx\) + \(\rm \int_0^a \frac{f(x)}{f(a-x)+f(x)}\ dx\)
⇒ 2I = \(\rm \int_0^a 1\ dx\)
⇒ 2I = a.
∴ I = \(\rm\frac a2\).
