Let $f(x)=x+\frac{a}{\pi^{2}-4} \sin x+\frac{b}{\pi^{2}-4} \cos x$, $x \in \mathbb{R}$ be a function which satisfies $f(x)=x+\int_{0}^{\pi / 2} \sin (x+y) f(y) d y$. Then $(\mathrm{a}+\mathrm{b})$ is equal to

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Definite Integrals

Let

f (x) = x + \frac{a}{π ^{2} - 4} sin x + \frac{b}{π ^{2} - 4} cos x

x \in R

be a function which satisfies

f (x) = x + \int_{0}^{π /2} sin (x + y) f (y) d y

. Then

(a + b)

is equal to

Answer: (B)

Expanding

sin (x + y) = sin x cos y + cos x sin y

, the relation

f (x) = x + \int_{0}^{π /2} sin (x + y) f (y) d y

shows

f (x) = x + I sin x + J cos x

with

I = \int_{0}^{π /2} cos y f (y) d y

and

J = \int_{0}^{π /2} sin y f (y) d y

, matching the given form so that

I = \frac{a}{π ^{2} - 4}, J = \frac{b}{π ^{2} - 4}

. Substituting

f (y) = y + I sin y + J cos y

and evaluating the integrals gives

\frac{I}{2} - \frac{π}{4} J = \frac{π}{2} - 1

and

\frac{J}{2} - \frac{π}{4} I = 1

; adding these,

(I + J) \frac{2 - π}{4} = \frac{π}{2}

, so

I + J = \frac{2 π}{2 - π}

. Then

a + b = (π^{2} - 4) (I + J) = (π - 2) (π + 2) \cdot \frac{2 π}{2 - π} = - 2 π (π + 2)

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Single correct option