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Q.
Let $f(x)=x+\frac{a}{\pi^2-4} \sin x+\frac{b}{\pi^2-4} \cos x, x \in R$ be a function which satisfies $f(x)=x+\int\limits_0^{\pi / 2} \sin (x+y) f(y) d y$. Then $(a+b)$ is equal to
$f(x)=x+\int\limits_0^{\pi / 2}(\sin x \cos y+\cos x \sin y) f(y) d y$
$f(x)=x+\int\limits_0^{\pi / 2}((\cos y f(y) d y) \sin x+(\sin y f(y) d y) \cos x) .......$(1)
On comparing with
$f(x)=x+\frac{a}{\pi^2-4} \sin x+\frac{b}{\pi^2-4} \cos x, x \in R$ then
$\Rightarrow \frac{a}{\pi^2-4}=\int\limits_0^{\pi / 2} \cos y f(y) d y .....$(2)
$\Rightarrow \frac{b}{\pi^2-4}=\int\limits_0^{\pi / 2} \sin y f(y) d y .....$(3)
Add (2) and (3)
$ \frac{a+b}{\pi^2-4}=\int\limits_0^{\pi / 2}(\sin y+\cos y) f(y) d y \ldots(4) $
$\frac{a+b}{\pi^2-4}=\int\limits_0^{\pi / 2}(\sin y+\cos y) f\left(\frac{\pi}{2}-y\right) d y \ldots(5) $
Add (4) and (5)
$ \frac{2(a+b)}{\pi^2-4} =\int\limits_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) $
$(a+b) =-2 \pi(\pi+2)$