Question

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 $f(x)=x+\underset{0}{\overset{\pi /2}{\int }}\sin (x+y)f(y)dy$. Then $(a+b)$ is equal to

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

Answer 1

Answer: (C)

$f(x)=x+\underset{0}{\overset{\pi /2}{\int }}(\sin x\cos y+\cos x\sin y)f(y)dy$
$f(x)=x+\underset{0}{\overset{\pi /2}{\int }}((\cos yf(y)dy)\sin x+(\sin yf(y)dy)\cos x).......$(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
$\Rightarrow \frac{a}{{\pi }^2-4}=\underset{0}{\overset{\pi /2}{\int }}\cos yf(y)dy.....$(2)
$\Rightarrow \frac{b}{{\pi }^2-4}=\underset{0}{\overset{\pi /2}{\int }}\sin yf(y)dy.....$(3)
Add (2) and (3)
$\frac{a+b}{{\pi }^2-4}=\underset{0}{\overset{\pi /2}{\int }}(\sin y+\cos y)f(y)dy\dots (4)$
$\frac{a+b}{{\pi }^2-4}=\underset{0}{\overset{\pi /2}{\int }}(\sin y+\cos y)f\left(\frac{\pi }{2}-y\right)dy\dots (5)$
Add (4) and (5)
$\frac{2(a+b)}{{\pi }^2-4}=\underset{0}{\overset{\pi /2}{\int }}(\sin y+\cos y)\left(\frac{\pi }{2}+\frac{(a+b)}{{\pi }^2-4}(\sin y+\cos y)\right)dy$
$=\pi +\frac{a+b}{{\pi }^2-4}\left(\frac{\pi }{2}+1\right)$
$(a+b)=-2\pi (\pi +2)$