QJEEM23Mab61d: JEE Main 2023 Mathematics - Class 12 JEEM23Mab61d Solution: Let be a function which satisfies . then is equal to

Question

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QJEEM23Mab61d: JEE Main 2023 Mathematics - Class 12 JEEM23Mab61d Solution: Let be a function which satisfies . then is equal to

Let $f(x) = x + {a \over {{\pi ^2} - 4}}\sin x + {b \over {{\pi ^2} - 4}}\cos x,x \in R$ be a function which

satisfies $f(x) = x + \int\limits_0^{\pi /2} {\sin (x + y)f(y)dy}$ . then $(a+b)$ is equal to

Multiple Choice Options

A $- 2\pi (\pi + 2)$
B $- \pi (\pi - 2)$
C $- \pi (\pi + 2)$
D $- 2\pi (\pi - 2)$

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Accepted Answer

f(x)=x+\int\limits_{0}^{\pi / 2}(\sin x \cos y+\cos x \sin y) f(y) d y

f(x)=x+\int\limits_{0}^{\pi / 2}((\cos y f(y) d y) \sin x+(\sin y f(y) d y) \cos x)\quad....(1)

On comparing with

f(x)=x+\frac{a}{\pi^{2}-4} \sin x+\frac{b}{\pi^{2}-4} \cos x, x \in \mathbb{R}

then

\Rightarrow \frac{a}{\pi^{2}-4}=\int_{0}^{\pi / 2} \cos y f(y) d y \quad....(2)

\Rightarrow \frac{b}{\pi^{2}-4}=\int_{0}^{\pi / 2} \sin y f(y) d y \quad....(3)

Add (2) and (3)

\frac{a+b}{\pi^{2}-4}=\int_{0}^{\pi / 2}(\sin y+\cos y) f(y) d y \quad....(4)

\frac{a+b}{\pi^{2}-4}=\int_{0}^{\pi / 2}(\sin y+\cos y) f\left(\frac{\pi}{2}-y\right) d y\quad....(5)

Add (4) and (5)

\frac{2(a+b)}{\pi^{2}-4}=\int_{0}^{\pi / 2}(\sin y+\cos y)\left(\frac{\pi}{2}+\frac{(a+b)}{\pi^{2}-4}(\sin y+\cos y)\right) d y

=\pi+\frac{a+b}{\pi^{2}-4}\left(\frac{\pi}{2}+1\right)

(a+b)=-2 \pi(\pi+2)

QJEEM23Mab61d: JEE Main 2023 Mathematics - Class 12 JEEM23Mab61d Solution: Let be a function which satisfies . then is equal to

Multiple Choice Options

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