Let f(x) be a function satisfying f(x)+f(π-x)=π², ∀ x ∈ ℝ. Then ∫0^π f(x) sin x d x is equal to :

Solution

Maths sir · 3 steps

Use the King property, x → π - x, then add the two forms so that f disappears into π².

1. King property. Call the integral I. Replacing x by π - x does not change a definite integral from 0 to π. Also, sin(π - x) = sin x, so I = ∫₀^π f(π - x)sin x dx.
2. Add both forms. Add this to the original I. The bracket becomes f(x) + f(π - x), which the question tells us equals π².

   2I = ∫₀^π [f(x) + f(π - x)]sin x dx = π² ∫₀^π sin x dx = π² × 2

3. So. ∫₀^π sin x dx = [-cos x]₀^π = 1 + 1 = 2. That gives 2I = 2π², so I = π², which is option (a).

The idea to keep. When you are told f(x) + f(a - x) is a constant, apply x → a - x on ∫₀ᵃ and add. The unknown f is replaced by that constant.

Answer from the NTA final answer key. Solution by PrepWiser.