Let f ( x ) = x + a 2 - 4 sin x + b 2 - 4 cos x , x be a function which satisfies f ( x ) = x + 0 / 2 sin ( x + y ) f ( y ) ? d y . Then ( a + b ) is equal to [2023]

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Solution

Q.
$Let f (x) = x + \frac{a}{π^{2} - 4} \sin x + \frac{b}{π^{2} - 4} \cos x, x \in ℝ$ be a function which satisfies $f (x) = x + \int_{0}^{π / 2} \sin (x + y) f (y) d y$ . Then $(a + b)$ is equal to [2023]

1 $- π (π + 2)$

2 $- π (π - 2)$

3 $- 2 π (π + 2)$

4 $- 2 π (π - 2)$

Ans.
(3)

$f (x) = x + \int_{0}^{π / 2} (\sin x \cos y + \cos x \sin y) f (y) d y$

$= x + \int_{0}^{π / 2} ((\cos y f (y) d y) \sin x + (\sin y f (y) d y) \cos x)$ ...(i)

On comparing with $f (x) = x + \frac{a}{π^{2} - 4} \sin x + \frac{b}{π^{2} - 4} \cos x$

$\Rightarrow \frac{a}{π^{2} - 4} = \int_{0}^{π / 2} \cos y f (y) d y$ ...(ii)

$\Rightarrow \frac{b}{π^{2} - 4} = \int_{0}^{π / 2} \sin y f (y) d y$ ...(iii)

Adding (ii) and (iii), we get

$\frac{a + b}{π^{2} - 4} = \int_{0}^{π / 2} (\sin y + \cos y) f (y) d y$ ...(iv)

$\frac{a + b}{π^{2} - 4} = \int_{0}^{π / 2} (\sin y + \cos y) f (\frac{π}{2} - y) d y$ ...(v)

Adding (iv) and (v), we get

$\frac{2 (a + b)}{π^{2} - 4} = \int_{0}^{π / 2} (\sin y + \cos y) [\frac{π}{2} + \frac{(a + b)}{π^{2} - 4} (\sin y + \cos y)] d y$

$= π + \frac{a + b}{π^{2} - 4} (\frac{π}{2} + 1)$ $\Rightarrow (a + b) = - 2 π (π + 2)$

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