Correct Answer - Option 2 : f(-x) + f(x)
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:
-
Even, if f(-x) = f(x). And \(\rm \int_{-a}^ {\ \ a}f(x)\ dx=2\int_{0}^af(x)\ dx\).
-
Odd, if f(-x) = -f(x). And \(\rm \int_{-a}^ {\ \ a}f(x)\ dx=0\).
-
Periodic, if f(np ± x) = f(x), for some number p and n ∈ Z.
Calculation:
We know that for an even function,
f(-x) = f(x) and \(\rm \int_{-a}^ {\ \ a}f(x)\ dx=2\int_{0}^af(x)\ dx\).
For the given condition,
\(\rm \int_0^a \left[f(x)+f(-x)\right]dx=\int_{-a}^{\ \ a} g(x)\ dx\) to be true,
f(x) and g(x) both must be even functions,
i.e. f(x) = f(-x) and g(x) = f(x) + f(-x).
