Question

Let \(f(x)=x+\frac{a}{\pi^2-4} sinx+\frac{b}{\pi^2-4} cos x\), \(x∈R\) be a function which satisfies \(f(x)=x+∫_0^{\frac{\pi}{2}} sin(x+y) f(y)dy\). Then (a+b) equal to

• A. -π(π+2)
• B. -π(π-2)
• C. -2π(π+2)
• D. -2π(π-2)

Answer 1

The Correct Option is C

We are given that: \[ f(x) = x + \int_0^{\frac{\pi}{2}} \sin(x + y) f(y) \, dy. \] Step 1: Expanding the equation. 
First, rewrite the given function \( f(x) \) as: \[ f(x) = x + \int_0^{\frac{\pi}{2}} (\cos y \sin(x) + \sin y \cos(x)) f(y) \, dy. \] This can be rewritten as: \[ f(x) = x + \left( \int_0^{\frac{\pi}{2}} \cos y f(y) \, dy \right) \sin(x) + \left( \int_0^{\frac{\pi}{2}} \sin y f(y) \, dy \right) \cos(x). \] Comparing this with the original form of \( f(x) \), which is: \[ f(x) = x + \frac{a}{\frac{\pi^2}{4}} \sin x + \frac{b}{\frac{\pi^2}{4}} \cos x, \] we obtain the following relationships: \[ \frac{a}{\frac{\pi^2}{4}} = \int_0^{\frac{\pi}{2}} \cos y f(y) \, dy, \quad \frac{b}{\frac{\pi^2}{4}} = \int_0^{\frac{\pi}{2}} \sin y f(y) \, dy. \] This leads to: \[ a = \frac{\pi^2}{4} \int_0^{\frac{\pi}{2}} \cos y f(y) \, dy, \quad b = \frac{\pi^2}{4} \int_0^{\frac{\pi}{2}} \sin y f(y) \, dy. \quad \cdots (1) \] Step 2: Solving for the value of \( a + b \). Add equations (1) for \( a \) and \( b \): \[ a + b = \frac{\pi^2}{4} \left( \int_0^{\frac{\pi}{2}} \cos y f(y) \, dy + \int_0^{\frac{\pi}{2}} \sin y f(y) \, dy \right). \] This simplifies to: \[ a + b = \frac{\pi^2}{4} \int_0^{\frac{\pi}{2}} (\cos y + \sin y) f(y) \, dy. \] Step 3: Substituting the values and simplifying. Using the identity for \( f(y) \), we integrate to get the final value: \[ a + b = -2\pi (\pi + 2). \] Thus, the correct value of \( (a + b) \) is \( -2\pi (\pi + 2) \), and the correct answer is option (3).