QJEEM23Ma1a74: JEE Main 2023 Mathematics - Class 11 JEEM23Ma1a74 Solution: Let f ( x ) be a function satisfying f ( x ) + f ( π − x ) = π…

Question

Verified Solution • 2025–26 Session moderate

Let $f (x)$ be a function satisfying $f (x) + f (π - x) = π^{2}, \forall x \in R$ . Then $\int_{0}^{π} f (x) \sin x d x$ is equal to :

Accepted Answer

Let

I=\int\limits_0^\pi f(x) \sin x d x

..........(i)

\begin{aligned} = \int_{0}^{π} f (π - x) \sin (π - x) d x \\ = \int_{0}^{π} f (π - x) \sin x d x . . . . . . . . (i i) \end{aligned}

On adding Equations (i) and (ii), we get

\begin{aligned} 2 I = \int_{0}^{π} [f (x) + f (π - x)] \sin x d x \\ \Rightarrow 2 I = \int_{0}^{π} π^{2} \sin x d x = π^{2} \int_{0}^{π} \sin x d x \\ = π^{2} (- \cos x)_{0}^{π} = - π^{2} (- 1 - 1) = 2 π^{2} \\ \Rightarrow I = π^{2} \end{aligned}

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