NDA Mathematics Question (2026) — Solution

Step-by-step solution

I = _0^4 f(x) g(x) \,dx From the given equation, g(x) = f(x) + f(4-x) Using the property _0^a h(x) \,dx = _0^a h(a-x) \,dx: I = _0^4 f(4-x) g(4-x) \,dx Since g(4-x) = f(4-x) + f(x) = g(x): I = _0^4 f(4-x) g(x) \,dx Adding the two equations for I: 2I = _0^4 f(x) + f(4-x) g(x) \,dx 2I = _0^4 g(x) g(x) \,dx = _0^4 1 \,dx 2I = 4 I = 2 Answer: 2