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  • Let <img alt="\mathrm{f(x)=x+\frac{a}{\pi^2-4} \sin x+\frac{b}{\pi^2-4} \cos x, x \in \mathbb{R}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bf%28x%29%3Dx&plus;%5Cfrac%7Ba

Let \mathrm{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 \mathrm{f(x)=x+\int_0^{\pi / 2} \sin (x+ y) f(y) d y}. Then (\mathrm{a}+\mathrm{b}) is equal to

 

Option: 1

-2 \pi(\pi-2)
 


Option: 2

-2 \pi(\pi+2)
 


Option: 3

-\pi(\pi-2)
 


Option: 4

-\pi(\pi+2)


Answers (1)

best_answer

\begin{aligned} & f(x)=x+\int_0^{\frac{\pi}{2}}(\sin x \cos y+\cos x \sin y) f(y) d y \\ \end{aligned}

\begin{aligned} & \left.f(x)=x+\int_0^{\frac{\pi}{2}}(\cos y f(y) d y) \sin x+(\sin y f(y) d y) \cos x\right) \end{aligned}

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

by comparing (1) and (2)

\Rightarrow \frac{\mathrm{a}}{\pi^2-4}=\int_0^{\frac{\pi}{2}} \cos y \mathrm{f}(\mathrm{y}) \mathrm{dy}....................(3)
and \: \frac{b}{\pi^2-4}=\int_0^{\frac{\pi}{2}} \sin y f(y) d y..............(4)
adding (3) and (4)
\begin{aligned} & \frac{a+b}{\pi^2-4}=\int_0^{\frac{\pi}{2}}(\sin y+\cos y) f(y) d y \\ \end{aligned}................(5)

\begin{aligned} & \frac{a+b}{\pi^2-4}=\int_0^{\frac{\pi}{2}}(\sin y+\cos y) f\left(\frac{\pi}{2}-y\right) d y \end{aligned}................(6)

Additing (5) and (6)

\begin{aligned} & \frac{2(a+b)}{\pi^2-4}=\int_0^{\frac{\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) \\ & \Rightarrow a+b=-2 \pi(\pi+2) \end{aligned}

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Nehul

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