Let f(x) = x+a^2-4 x+b^2-4 x, xR, be a function satisfying f(x) = x+_0^/2(x+y)f(y),dy.

Question

Let $f(x) = x+\dfrac{a}{\pi^2-4}\sin x+\dfrac{b}{\pi^2-4}\cos x$, $x\in\mathbb{R}$, be a function satisfying $f(x) = x+\displaystyle\int_0^{\pi/2}\sin(x+y)f(y)\,dy$. Then $(a+b)$ is equal to:

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)

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

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

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

Solutions · 1