UPSC NDA I 2021 Mathematics Solved Paper

Concept:The property of definite integrals states: $\int_{-a}^{a} g(x) \, dx = \int_{0}^{a} \left[ g(x) + g(-x) \right] dx$.Explanation:We are given $\int_{0}^{a} \left[ f(x) + f(-x) \right] dx = \int_{-a}^{a} g(x) \, dx$.Using the integral property on the right side: $\int_{-a}^{a} g(x) \, dx = \int_{0}^{a} \left[ g(x) + g(-x) \right] dx$.Thus, $\int_{0}^{a} \left[ f(x) + f(-x) \right] dx = \int_{0}^{a} \left[ g(x) + g(-x) \right] dx$.For this equality to hold for the arbitrary function $f$, we must have $g(x) + g(-x) = f(x) + f(-x)$.One simple solution is $g(x) = f(x)$. Substituting $g(x)=f(x)$ gives $\int_{-a}^{a} f(x) \, dx$.Now $\int_{-a}^{a} f(x) \, dx = \int_{-a}^{0} f(x) \, dx + \int_{0}^{a} f(x) \, dx$.In the first integral, substitute $x = -t$, then $\int_{-a}^{0} f(x) \, dx = \int_{a}^{0} f(-t) (-dt) = \int_{0}^{a} f(-t) \, dt = \int_{0}^{a} f(-x) \, dx$.So $\int_{-a}^{a} f(x) \, dx = \int_{0}^{a} f(-x) \, dx + \int_{0}^{a} f(x) \, dx = \int_{0}^{a} \left[ f(x) + f(-x) \right] dx$.Hence $g(x) = f(x)$ satisfies the given equation.The other options do not satisfy: Option B gives twice the required value; Option C gives the negative.Answer:A. f(x)