Definite Integration Question 10
Question 10 - 29 January - Shift 1
Let $f(x)=x+\frac{a}{\pi^{2}-4} \sin x+\frac{b}{\pi^{2}-4} \cos x$, $x \in \mathbb{R}$ be a function which satisfies
$f(x)=x+\int_0^{\pi / 2} \sin (x+y) f(y) d y$. Then $(a+b)$ is equal to
(1) $-\pi(\pi+2)$
(2) $-2 \pi(\pi+2)$
(3) $-2 \pi(\pi-2)$
(4) $-\pi(\pi-2)$
Show Answer
Answer: (2)
Solution:
Formula: Trigonometric Functions of Sum or Difference of Two Angles, Properties of definite integral
$f(x)=x+\int_0^{\pi / 2}(\sin x \cos y+\cos x \sin y) f(y) d y$
$f(x)=x+\int_0^{\pi / 2}((\cos y f(y) d y) \sin x+(\sin y f(y) d y) \cos x) \ldots (1)$
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
$\Rightarrow \frac{a}{\pi^{2}-4}=\int_0^{\pi / 2} \cos y f(y) d y \ldots (2)$
$\Rightarrow \frac{b}{\pi^{2}-4}=\int_0^{\pi / 2} \sin y f(y) d y \ldots (3)$
Add $(2)$ and $(3)$
$\frac{a+b}{\pi^{2}-4}=\int_0^{\pi / 2}(\sin y+\cos y) f(y) d y \ldots (4)$
$\frac{a+b}{\pi^{2}-4}=\int_0^{\pi / 2}(\sin y+\cos y) f(\frac{\pi}{2}-y) d y \ldots (5)$
Add $(4)$ and $(5)$
$ \begin{aligned} \frac{2(a+b)}{\pi^{2}-4} & =\int_0^{\pi / 2}(\sin y+\cos y)(\frac{\pi}{2}+\frac{(a+b)}{\pi^{2}-4}(\sin y+\cos y)) d y \\ & =\pi+\frac{a+b}{\pi^{2}-4}(\frac{\pi}{2}+1) \\ (a+b) & =-2 \pi(\pi+2) \end{aligned} $
