Key Concepts and Formulas

1. Trigonometric Identity: $\sin(A+B) = \sin A \cos B + \cos A \sin B$. This identity is crucial for expanding the integrand.
2. Properties of Definite Integrals:
   • $\int_a^b (g(x) + h(x)) dx = \int_a^b g(x) dx + \int_a^b h(x) dx$
   • $\int_a^b c \cdot g(x) dx = c \cdot \int_a^b g(x) dx$ (where $c$ is a constant with respect to the integration variable).
3. Substitution and Coefficient Matching: After manipulating the functional equation to express $f(x)$ in a specific form, we equate the coefficients of corresponding terms (like $\sin x$ and $\cos x$) with the given explicit form of $f(x)$.

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Step-by-Step Solution

Step 1: Expand the Integral and Express $f(x)$ in terms of unknown constants. We are given the functional equation: $f(x) = x + \int\limits_0^{\pi/2} {\sin (x + y)f(y)dy}$ Using the trigonometric identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$, we expand the term $\sin(x+y)$: $f(x) = x + \int\limits_0^{\pi/2} {(\sin x \cos y + \cos x \sin y)f(y)dy}$ Since the integration is with respect to $y$, $\sin x$ and $\cos x$ are treated as constants within the integral. We can therefore split the integral and pull these terms out: $f(x) = x + \sin x \int\limits_0^{\pi/2} {\cos y f(y)dy} + \cos x \int\limits_0^{\pi/2} {\sin y f(y)dy}$ The integrals $\int\limits_0^{\pi/2} {\cos y f(y)dy}$ and $\int\limits_0^{\pi/2} {\sin y f(y)dy}$ are definite integrals over a fixed interval $[0, \pi/2]$. Therefore, their values are constants, independent of $x$. Let's define these constants: Let $C_1 = \int\limits_0^{\pi/2} {\cos y f(y)dy}$ Let $C_2 = \int\limits_0^{\pi/2} {\sin y f(y)dy}$ Substituting these constants back into the equation for $f(x)$, we get: $f(x) = x + C_1 \sin x + C_2 \cos x \quad \quad (*)$ This equation gives us the structure of $f(x)$ in terms of $x$, $\sin x$, $\cos x$, and two unknown constants $C_1$ and $C_2$.

Step 2: Relate the unknown constants $C_1$ and $C_2$ to the given coefficients $a$ and $b$. We are given the explicit form of the function: $f(x) = x + \frac{a}{\pi^2-4} \sin x + \frac{b}{\pi^2-4} \cos x$ By comparing this given form with the derived form in Step 1 ($f(x) = x + C_1 \sin x + C_2 \cos x$), we can equate the coefficients of $\sin x$ and $\cos x$: $C_1 = \frac{a}{\pi^2-4} \quad \quad (Eq. 1)$ $C_2 = \frac{b}{\pi^2-4} \quad \quad (Eq. 2)$ Now, we substitute the definition of $C_1$ and $C_2$ back into these equations: $\int\limits_0^{\pi/2} {\cos y f(y)dy} = \frac{a}{\pi^2-4} \quad \quad (Eq. 3)$ $\int\limits_0^{\pi/2} {\sin y f(y)dy} = \frac{b}{\pi^2-4} \quad \quad (Eq. 4)$

Step 3: Substitute the explicit form of $f(y)$ into the integral equations and solve for $a$ and $b$. We will use the form $f(y) = y + \frac{a}{\pi^2-4} \sin y + \frac{b}{\pi^2-4} \cos y$ in Equations 3 and 4.

Substitute $f(y)$ into Equation 3: $\int\limits_0^{\pi/2} {\cos y \left(y + \frac{a}{\pi^2-4} \sin y + \frac{b}{\pi^2-4} \cos y\right)dy} = \frac{a}{\pi^2-4}$ $\int\limits_0^{\pi/2} {y \cos y dy} + \frac{a}{\pi^2-4} \int\limits_0^{\pi/2} {\sin y \cos y dy} + \frac{b}{\pi^2-4} \int\limits_0^{\pi/2} {\cos^2 y dy} = \frac{a}{\pi^2-4}$ Let's evaluate each integral separately: Integral 1: $\int\limits_0^{\pi/2} {y \cos y dy}$ We use integration by parts: $\int u dv = uv - \int v du$. Let $u=y$ and $dv=\cos y dy$. Then $du=dy$ and $v=\sin y$. $\int\limits_0^{\pi/2} {y \cos y dy} = [y \sin y]_0^{\pi/2} - \int\limits_0^{\pi/2} {\sin y dy}$ $= \left(\frac{\pi}{2} \sin\left(\frac{\pi}{2}\right) - 0 \cdot \sin(0)\right) - [-\cos y]_0^{\pi/2}$ $= \left(\frac{\pi}{2} \cdot 1 - 0\right) - (-\cos\left(\frac{\pi}{2}\right) - (-\cos(0)))$ $= \frac{\pi}{2} - (0 - (-1)) = \frac{\pi}{2} - 1$

Integral 2: $\int\limits_0^{\pi/2} {\sin y \cos y dy}$ We can use the identity $\sin(2y) = 2 \sin y \cos y$, so $\sin y \cos y = \frac{1}{2} \sin(2y)$. $\int\limits_0^{\pi/2} {\frac{1}{2} \sin(2y) dy} = \frac{1}{2} \left[-\frac{1}{2} \cos(2y)\right]_0^{\pi/2}$ $= -\frac{1}{4} [\cos(\pi) - \cos(0)] = -\frac{1}{4} [-1 - 1] = -\frac{1}{4} (-2) = \frac{1}{2}$

Integral 3: $\int\limits_0^{\pi/2} {\cos^2 y dy}$ We use the identity $\cos^2 y = \frac{1 + \cos(2y)}{2}$. $\int\limits_0^{\pi/2} {\frac{1 + \cos(2y)}{2} dy} = \frac{1}{2} \left[y + \frac{1}{2} \sin(2y)\right]_0^{\pi/2}$ $= \frac{1}{2} \left[\left(\frac{\pi}{2} + \frac{1}{2} \sin(\pi)\right) - \left(0 + \frac{1}{2} \sin(0)\right)\right]$ $= \frac{1}{2} \left[\frac{\pi}{2} + 0 - 0\right] = \frac{\pi}{4}$

Now, substitute these integral values back into the equation derived from Equation 3: $\left(\frac{\pi}{2} - 1\right) + \frac{a}{\pi^2-4} \left(\frac{1}{2}\right) + \frac{b}{\pi^2-4} \left(\frac{\pi}{4}\right) = \frac{a}{\pi^2-4}$ Multiply by $(\pi^2-4)$: $(\pi^2-4)\left(\frac{\pi}{2} - 1\right) + \frac{a}{2} + \frac{b\pi}{4} = a$ $(\pi^2-4)\frac{\pi-2}{2} + \frac{a}{2} + \frac{b\pi}{4} = a$ $(\pi+2)(\pi-2)\frac{\pi-2}{2} + \frac{a}{2} + \frac{b\pi}{4} = a$ $\frac{(\pi+2)(\pi-2)^2}{2} + \frac{a}{2} + \frac{b\pi}{4} = a$ Multiply by 4 to clear denominators: $2(\pi+2)(\pi-2)^2 + 2a + b\pi = 4a$ $2(\pi+2)(\pi-2)^2 + b\pi = 2a \quad \quad (Eq. 5)$

Now, substitute $f(y)$ into Equation 4: $\int\limits_0^{\pi/2} {\sin y \left(y + \frac{a}{\pi^2-4} \sin y + \frac{b}{\pi^2-4} \cos y\right)dy} = \frac{b}{\pi^2-4}$ $\int\limits_0^{\pi/2} {y \sin y dy} + \frac{a}{\pi^2-4} \int\limits_0^{\pi/2} {\sin^2 y dy} + \frac{b}{\pi^2-4} \int\limits_0^{\pi/2} {\sin y \cos y dy} = \frac{b}{\pi^2-4}$ We already know $\int\limits_0^{\pi/2} {\sin y \cos y dy} = \frac{1}{2}$. We need to evaluate the other two integrals:

Integral 4: $\int\limits_0^{\pi/2} {y \sin y dy}$ Use integration by parts: $u=y$, $dv=\sin y dy$. Then $du=dy$, $v=-\cos y$. $\int\limits_0^{\pi/2} {y \sin y dy} = [-y \cos y]_0^{\pi/2} - \int\limits_0^{\pi/2} {(-\cos y) dy}$ $= \left(-\frac{\pi}{2} \cos\left(\frac{\pi}{2}\right) - (-0 \cdot \cos(0))\right) + \int\limits_0^{\pi/2} {\cos y dy}$ $= \left(-\frac{\pi}{2} \cdot 0 - 0\right) + [\sin y]_0^{\pi/2}$ $= 0 + (\sin\left(\frac{\pi}{2}\right) - \sin(0)) = 1 - 0 = 1$

Integral 5: $\int\limits_0^{\pi/2} {\sin^2 y dy}$ We use the identity $\sin^2 y = \frac{1 - \cos(2y)}{2}$. $\int\limits_0^{\pi/2} {\frac{1 - \cos(2y)}{2} dy} = \frac{1}{2} \left[y - \frac{1}{2} \sin(2y)\right]_0^{\pi/2}$ $= \frac{1}{2} \left[\left(\frac{\pi}{2} - \frac{1}{2} \sin(\pi)\right) - \left(0 - \frac{1}{2} \sin(0)\right)\right]$ $= \frac{1}{2} \left[\frac{\pi}{2} - 0 - 0\right] = \frac{\pi}{4}$

Now, substitute these integral values back into the equation derived from Equation 4: $1 + \frac{a}{\pi^2-4} \left(\frac{\pi}{4}\right) + \frac{b}{\pi^2-4} \left(\frac{1}{2}\right) = \frac{b}{\pi^2-4}$ Multiply by $(\pi^2-4)$: $(\pi^2-4) + \frac{a\pi}{4} + \frac{b}{2} = b$ $(\pi^2-4) + \frac{a\pi}{4} + \frac{b}{2} = b$ Multiply by 4 to clear denominators: $4(\pi^2-4) + a\pi + 2b = 4b$ $4(\pi^2-4) + a\pi = 2b \quad \quad (Eq. 6)$

Step 4: Solve the system of linear equations for $a$ and $b$. We have two equations: (Eq. 5): $2(\pi+2)(\pi-2)^2 + b\pi = 2a$ (Eq. 6): $4(\pi^2-4) + a\pi = 2b$

From Eq. 6, we can express $2b$ as $4(\pi^2-4) + a\pi$. Substitute this into Eq. 5: $2(\pi+2)(\pi-2)^2 + \frac{1}{\pi} [4(\pi^2-4) + a\pi] = 2a$ $2(\pi+2)(\pi-2)^2 + \frac{4(\pi^2-4)}{\pi} + a = 2a$ $2(\pi+2)(\pi-2)^2 + \frac{4(\pi^2-4)}{\pi} = a$ $a = 2(\pi+2)(\pi-2)^2 + \frac{4(\pi-2)(\pi+2)}{\pi}$ $a = 2(\pi+2)(\pi-2)^2 + \frac{4(\pi^2-4)}{\pi}$

Let's re-evaluate Eq. 5 and Eq. 6 for simplicity. Eq. 5: $2(\pi+2)(\pi-2)^2 + b\pi = 2a$ Eq. 6: $4(\pi+2)(\pi-2) + a\pi = 2b$

From Eq. 6, $2b = 4(\pi^2-4) + a\pi$. Substitute this into Eq. 5: $2(\pi+2)(\pi-2)^2 + \frac{1}{2}(4(\pi^2-4) + a\pi) = 2a$ $2(\pi+2)(\pi-2)^2 + 2(\pi^2-4) + \frac{a\pi}{2} = 2a$ $2(\pi+2)(\pi-2)^2 + 2(\pi-2)(\pi+2) = 2a - \frac{a\pi}{2}$ $2(\pi+2)(\pi-2) [(\pi-2) + 1] = a \left(2 - \frac{\pi}{2}\right)$ $2(\pi+2)(\pi-1) (\pi-1) = a \left(\frac{4-\pi}{2}\right)$ $2(\pi+2)(\pi-1)^2 = a \left(\frac{4-\pi}{2}\right)$ This seems complicated. Let's try solving for $a$ and $b$ in a different way.

From Eq. 6: $a\pi = 2b - 4(\pi^2-4)$ From Eq. 5: $2a = 2(\pi+2)(\pi-2)^2 + b\pi$

Multiply Eq. 6 by 2: $2a\pi = 4b - 8(\pi^2-4)$. Substitute $2a$ from Eq. 5: $\pi [2(\pi+2)(\pi-2)^2 + b\pi] = 4b - 8(\pi^2-4)$ $2\pi(\pi+2)(\pi-2)^2 + b\pi^2 = 4b - 8(\pi^2-4)$ $2\pi(\pi+2)(\pi-2)^2 + 8(\pi^2-4) = 4b - b\pi^2$ $2\pi(\pi+2)(\pi-2)^2 + 8(\pi-2)(\pi+2) = b(4 - \pi^2)$ $2(\pi+2)(\pi-2) [\pi(\pi-2) + 4] = b(4 - \pi^2)$ $2(\pi^2-4) [\pi^2 - 2\pi + 4] = b(4 - \pi^2)$ $2(\pi^2-4) [\pi^2 - 2\pi + 4] = -b(\pi^2 - 4)$ Since $\pi^2-4 \neq 0$, we can divide by it: $2(\pi^2 - 2\pi + 4) = -b$ $b = -2(\pi^2 - 2\pi + 4)$

Now let's solve for $a$. From Eq. 6: $2b = 4(\pi^2-4) + a\pi$. $a\pi = 2b - 4(\pi^2-4)$ $a\pi = 2(-2(\pi^2 - 2\pi + 4)) - 4(\pi^2-4)$ $a\pi = -4(\pi^2 - 2\pi + 4) - 4(\pi^2-4)$ $a\pi = -4\pi^2 + 8\pi - 16 - 4\pi^2 + 16$ $a\pi = -8\pi^2 + 8\pi$ $a = -8\pi + 8 = 8(1-\pi)$

Let's check if these values of $a$ and $b$ are consistent. If $b = -2(\pi^2 - 2\pi + 4)$, then $2b = -4(\pi^2 - 2\pi + 4)$. Eq. 6: $4(\pi^2-4) + a\pi = 2b$ $4(\pi^2-4) + 8(1-\pi)\pi = -4(\pi^2 - 2\pi + 4)$ $4\pi^2 - 16 + 8\pi - 8\pi^2 = -4\pi^2 + 8\pi - 16$ $-4\pi^2 + 8\pi - 16 = -4\pi^2 + 8\pi - 16$. This is consistent.

Let's check Eq. 5: $2(\pi+2)(\pi-2)^2 + b\pi = 2a$ $2(\pi+2)(\pi^2-4\pi+4) + (-2(\pi^2 - 2\pi + 4))\pi = 2(8(1-\pi))$ $2(\pi^3 - 4\pi^2 + 4\pi + 2\pi^2 - 8\pi + 8) - 2\pi^3 + 4\pi^2 - 8\pi = 16(1-\pi)$ $2(\pi^3 - 2\pi^2 - 4\pi + 8) - 2\pi^3 + 4\pi^2 - 8\pi = 16 - 16\pi$ $2\pi^3 - 4\pi^2 - 8\pi + 16 - 2\pi^3 + 4\pi^2 - 8\pi = 16 - 16\pi$ $-16\pi + 16 = 16 - 16\pi$. This is also consistent.

So, we have $a = 8(1-\pi)$ and $b = -2(\pi^2 - 2\pi + 4)$. We need to find $a+b$. $a+b = 8(1-\pi) - 2(\pi^2 - 2\pi + 4)$ $a+b = 8 - 8\pi - 2\pi^2 + 4\pi - 8$ $a+b = -2\pi^2 - 4\pi$ $a+b = -2\pi(\pi + 2)$

This matches option (A).

Common Mistakes & Tips

• Algebraic Errors: Solving the system of equations for $a$ and $b$ can be prone to algebraic mistakes. Double-check all calculations, especially when dealing with fractions and powers.
• Integration Errors: Ensure accurate integration of trigonometric and polynomial terms. Using integration by parts correctly is crucial.
• Coefficient Matching: Be careful when equating coefficients. Ensure that the terms on both sides of the equation have the same variable dependencies.

Summary The problem involves solving a functional equation where $f(x)$ is defined both explicitly and implicitly through an integral. By expanding the integral using trigonometric identities, we express $f(x)$ in a form that allows us to equate coefficients of $\sin x$ and $\cos x$ with the given explicit form of $f(x)$. This leads to a system of two linear equations in terms of the unknown constants $a$ and $b$. By carefully evaluating the definite integrals and solving the system of equations, we find the values of $a$ and $b$ and subsequently compute their sum $a+b$.

The final answer is $\boxed{-2\pi (\pi + 2)}$.