JEE MainMathematics

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Solving a Fredholm Integral Equation with a Separable Kernel

Q: Why do we add the integral to its reflected version using King's rule instead of integrating f(y) directly?

Integrating f(y) directly requires integrating y*cos(y) and y*sin(y) by parts separately for a and b. King's rule exploits the symmetry of (sin y + cos y). It directly couples f(y) + f(pi/2 - y). This eliminates the variable y since y + pi/2 - y = pi/2. It turns a tedious by-parts integration into a simple one-line step.

Let f(x)=x+a/(π^2-4) sin x+b/(π^2-4) cos x, x ∈ ℝ be a function which satisfies f(x)=x+integral from 0 to π / 2 of sin (x+y) f(y) d y. Then (a+b) is equal to

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A-2 π(π-2)B-2 π(π+2)C-π(π-2)D-π(π+2)

Step-by-step solutionView

Correct answer

Expanding $\sin (x + y) = \sin x \cos y + \cos x \sin y$ leads to a linear system for the unknown integral coefficients, yielding $a + b = - 2 π (π + 2)$ .

Option analysis

Why each option works or fails

A · -2 π(π-2)

Believing that the denominator $π^{2} - 4$ factorizes with an inverted sign or introducing a sign error when solving the $2 \times 2$ linear system for $A$ and $B$ , leading to $(π - 2)$ instead of $(π + 2)$ . Carefully track negative signs when evaluating definite integrals involving $\cos y$ and $\sin y$ by parts and check the matrix inversion step.

B · -2 π(π+2)

This is the correct option: expressing the integral as $A \sin x + B \cos x$ , setting up the system for $A$ and $B$ , and matching coefficients to find $a$ and $b$ gives $a + b = - 2 π (π + 2)$ . No correction needed.

C · -π(π-2)

Dropping a factor of 2 when summing $a$ and $b$ alongside an internal sign error during the solution of the linear system. Ensure coefficients from integration by parts are not inadvertently halved or omitted when combining $a$ and $b$ .

D · -π(π+2)

Missing the factor of 2 that arises when adding $a = - (π^{2} + 2 π)$ and $b = - (π^{2} + 2 π)$ or equivalent expressions. Compute $a$ and $b$ fully before adding them: $a = - 2 π - π^{2}$ and $b = - 2 π - π^{2}$ , so their sum is $- 2 π (π + 2)$ rather than $- π (π + 2)$ .

Reviewed route

Solution

StepWorking

01Given

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

02Approach

Expand $\sin (x + y) = \sin x \cos y + \cos x \sin y$ inside the integral to separate variables in terms of $\sin x$ and $\cos x$ . Compare coefficients of $\sin x$ and $\cos x$ with the given form to express $\frac{a}{π^{2} - 4}$ and $\frac{b}{π^{2} - 4}$ as definite integrals, then add them and apply King's property $\int_{0}^{a} g (y) d y = \int_{0}^{a} g (a - y) d y$ .

03Execute

Expanding the integral: $f (x) = x + \sin x \int_{0}^{π / 2} f (y) \cos y d y + \cos x \int_{0}^{π / 2} f (y) \sin y d y$ . Comparing with $f (x) = x + \frac{a}{π^{2} - 4} \sin x + \frac{b}{π^{2} - 4} \cos x$ gives: $\frac{a}{π^{2} - 4} = \int_{0}^{π / 2} f (y) \cos y d y$ and $\frac{b}{π^{2} - 4} = \int_{0}^{π / 2} f (y) \sin y d y$ . Summing both equations: let $K = \frac{a + b}{π^{2} - 4} = \int_{0}^{π / 2} (\sin y + \cos y) f (y) d y$ .

04Execute

Using King's property on $K = \int_{0}^{π / 2} (\sin y + \cos y) f (y) d y$ : Since $\sin (π / 2 - y) + \cos (π / 2 - y) = \cos y + \sin y$ , we also have $K = \int_{0}^{π / 2} (\sin y + \cos y) f (\frac{π}{2} - y) d y$ . Adding the two expressions: $2 K = \int_{0}^{π / 2} (\sin y + \cos y) [f (y) + f (\frac{π}{2} - y)] d y$ . Now compute $f (y) + f (\frac{π}{2} - y)$ : $f (y) + f (\frac{π}{2} - y) = y + (\frac{π}{2} - y) + \frac{a}{π^{2} - 4} (\sin y + \cos y) + \frac{b}{π^{2} - 4} (\cos y + \sin y) = \frac{π}{2} + K (\sin y + \cos y)$ .

05Execute

Substitute this sum into the integral: $2 K = \int_{0}^{π / 2} (\sin y + \cos y) [\frac{π}{2} + K (\sin y + \cos y)] d y$ $= \frac{π}{2} \int_{0}^{π / 2} (\sin y + \cos y) d y + K \int_{0}^{π / 2} (\sin y + \cos y)^{2} d y$ . Calculate the two standard integrals: $\int_{0}^{π / 2} (\sin y + \cos y) d y = [- \cos y + \sin y]_{0}^{π / 2} = 1 - (- 1) = 2$ . $\int_{0}^{π / 2} (\sin y + \cos y)^{2} d y = \int_{0}^{π / 2} (1 + \sin 2 y) d y = [y - \frac{1}{2} \cos 2 y]_{0}^{π / 2} = \frac{π}{2} - \frac{1}{2} (- 1 - 1) = \frac{π}{2} + 1$ . Thus: $2 K = \frac{π}{2} (2) + K (\frac{π}{2} + 1) = π + K (\frac{π}{2} + 1)$ .

06Execute

Rearranging the equation for $K$ : $K (2 - (\frac{π}{2} + 1)) = π ⟹ K (1 - \frac{π}{2}) = π ⟹ K (\frac{2 - π}{2}) = π ⟹ K = \frac{2 π}{2 - π} = - \frac{2 π}{π - 2}$ . Since $K = \frac{a + b}{π^{2} - 4}$ , we get: $a + b = K (π^{2} - 4) = - \frac{2 π}{π - 2} (π - 2) (π + 2) = - 2 π (π + 2)$ .

✓Verify

Checking dimensions and sign: $K = \frac{a + b}{π^{2} - 4}$ . Since $π \approx 3.14 > 2$ , $2 - π < 0$ , which makes $K < 0$ , giving $a + b < 0$ . The factor $(π^{2} - 4)$ splits into $(π - 2) (π + 2)$ , cancelling the $(π - 2)$ in the denominator and leaving $- 2 π (π + 2)$ . Matches Option B.

Hints that build this answer step by step

Using the angle addition identity $\sin (x + y) = \sin x \cos y + \cos x \sin y$ , how can the integral $\int_{0}^{π / 2} \sin (x + y) f (y) d y$ be rewritten?
$\sin x \int_{0}^{π / 2} f (y) \cos y d y + \cos x \int_{0}^{π / 2} f (y) \sin y d y$
Let $A = \int_{0}^{π / 2} f (y) \cos y d y$ and $B = \int_{0}^{π / 2} f (y) \sin y d y$ . Comparing $f (x) = x + A \sin x + B \cos x$ with the given form $f (x) = x + \frac{a}{π^{2} - 4} \sin x + \frac{b}{π^{2} - 4} \cos x$ , how are $a$ and $b$ related to $A$ and $B$ ?
$a = (π^{2} - 4) A$ and $b = (π^{2} - 4) B$
Substituting $f (y) = y + A \sin y + B \cos y$ into the definitions of $A$ and $B$ , which linear system governs $A$ and $B$ ?
$A - \frac{π}{4} B = \frac{π}{2} - 1$ and $- \frac{π}{4} A + B = 1$
Solving the system $(\begin{array}{cc} 1 & - π / 4 \\ - π / 4 & 1 \end{array}) (\begin{array}{c} A \\ B \end{array}) = (\begin{array}{c} \frac{π - 2}{2} \\ 1 \end{array})$ and finding $a + b = (π^{2} - 4) (A + B)$ , what is the final value of $a + b$ ?
$- 2 π (π + 2)$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 8 September 2026

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Students also ask

Why do we add the integral to its reflected version using King's rule instead of integrating f(y) directly?

Integrating f(y) directly requires integrating y*cos(y) and y*sin(y) by parts separately for a and b. King's rule exploits the symmetry of (sin y + cos y). It directly couples f(y) + f(pi/2 - y). This eliminates the variable y since y + pi/2 - y = pi/2. It turns a tedious by-parts integration into a simple one-line step.

Answer

Expanding $\sin (x + y) = \sin x \cos y + \cos x \sin y$ leads to a linear system for the unknown integral coefficients, yielding $a + b = - 2 π (π + 2)$ .

Why each option works or fails

A: -2 π(π-2) - Believing that the denominator $π^{2} - 4$ factorizes with an inverted sign or introducing a sign error when solving the $2 \times 2$ linear system for $A$ and $B$ , leading to $(π - 2)$ instead of $(π + 2)$ . Carefully track negative signs when evaluating definite integrals involving $\cos y$ and $\sin y$ by parts and check the matrix inversion step.
B · correct: -2 π(π+2) - This is the correct option: expressing the integral as $A \sin x + B \cos x$ , setting up the system for $A$ and $B$ , and matching coefficients to find $a$ and $b$ gives $a + b = - 2 π (π + 2)$ . No correction needed.
C: -π(π-2) - Dropping a factor of 2 when summing $a$ and $b$ alongside an internal sign error during the solution of the linear system. Ensure coefficients from integration by parts are not inadvertently halved or omitted when combining $a$ and $b$ .
D: -π(π+2) - Missing the factor of 2 that arises when adding $a = - (π^{2} + 2 π)$ and $b = - (π^{2} + 2 π)$ or equivalent expressions. Compute $a$ and $b$ fully before adding them: $a = - 2 π - π^{2}$ and $b = - 2 π - π^{2}$ , so their sum is $- 2 π (π + 2)$ rather than $- π (π + 2)$ .

Step-by-step solution

Given: $f (x) = x + \frac{a}{π^{2} - 4} \sin x + \frac{b}{π^{2} - 4} \cos x$ and $f (x) = x + \int_{0}^{π / 2} \sin (x + y) f (y) d y$ .
Approach: Expand $\sin (x + y) = \sin x \cos y + \cos x \sin y$ inside the integral to separate variables in terms of $\sin x$ and $\cos x$ . Compare coefficients of $\sin x$ and $\cos x$ with the given form to express $\frac{a}{π^{2} - 4}$ and $\frac{b}{π^{2} - 4}$ as definite integrals, then add them and apply King's property $\int_{0}^{a} g (y) d y = \int_{0}^{a} g (a - y) d y$ .
Execute: Expanding the integral: $f (x) = x + \sin x \int_{0}^{π / 2} f (y) \cos y d y + \cos x \int_{0}^{π / 2} f (y) \sin y d y$ . Comparing with $f (x) = x + \frac{a}{π^{2} - 4} \sin x + \frac{b}{π^{2} - 4} \cos x$ gives: $\frac{a}{π^{2} - 4} = \int_{0}^{π / 2} f (y) \cos y d y$ and $\frac{b}{π^{2} - 4} = \int_{0}^{π / 2} f (y) \sin y d y$ . Summing both equations: let $K = \frac{a + b}{π^{2} - 4} = \int_{0}^{π / 2} (\sin y + \cos y) f (y) d y$ .
Execute: Using King's property on $K = \int_{0}^{π / 2} (\sin y + \cos y) f (y) d y$ : Since $\sin (π / 2 - y) + \cos (π / 2 - y) = \cos y + \sin y$ , we also have $K = \int_{0}^{π / 2} (\sin y + \cos y) f (\frac{π}{2} - y) d y$ . Adding the two expressions: $2 K = \int_{0}^{π / 2} (\sin y + \cos y) [f (y) + f (\frac{π}{2} - y)] d y$ . Now compute $f (y) + f (\frac{π}{2} - y)$ : $f (y) + f (\frac{π}{2} - y) = y + (\frac{π}{2} - y) + \frac{a}{π^{2} - 4} (\sin y + \cos y) + \frac{b}{π^{2} - 4} (\cos y + \sin y) = \frac{π}{2} + K (\sin y + \cos y)$ .
Execute: Substitute this sum into the integral: $2 K = \int_{0}^{π / 2} (\sin y + \cos y) [\frac{π}{2} + K (\sin y + \cos y)] d y$ $= \frac{π}{2} \int_{0}^{π / 2} (\sin y + \cos y) d y + K \int_{0}^{π / 2} (\sin y + \cos y)^{2} d y$ . Calculate the two standard integrals: $\int_{0}^{π / 2} (\sin y + \cos y) d y = [- \cos y + \sin y]_{0}^{π / 2} = 1 - (- 1) = 2$ . $\int_{0}^{π / 2} (\sin y + \cos y)^{2} d y = \int_{0}^{π / 2} (1 + \sin 2 y) d y = [y - \frac{1}{2} \cos 2 y]_{0}^{π / 2} = \frac{π}{2} - \frac{1}{2} (- 1 - 1) = \frac{π}{2} + 1$ . Thus: $2 K = \frac{π}{2} (2) + K (\frac{π}{2} + 1) = π + K (\frac{π}{2} + 1)$ .
Execute: Rearranging the equation for $K$ : $K (2 - (\frac{π}{2} + 1)) = π ⟹ K (1 - \frac{π}{2}) = π ⟹ K (\frac{2 - π}{2}) = π ⟹ K = \frac{2 π}{2 - π} = - \frac{2 π}{π - 2}$ . Since $K = \frac{a + b}{π^{2} - 4}$ , we get: $a + b = K (π^{2} - 4) = - \frac{2 π}{π - 2} (π - 2) (π + 2) = - 2 π (π + 2)$ .
Verify: Checking dimensions and sign: $K = \frac{a + b}{π^{2} - 4}$ . Since $π \approx 3.14 > 2$ , $2 - π < 0$ , which makes $K < 0$ , giving $a + b < 0$ . The factor $(π^{2} - 4)$ splits into $(π - 2) (π + 2)$ , cancelling the $(π - 2)$ in the denominator and leaving $- 2 π (π + 2)$ . Matches Option B.

Solving a Fredholm Integral Equation with a Separable Kernel

What feels right?

Expanding $\sin (x + y) = \sin x \cos y + \cos x \sin y$ leads to a linear system for the unknown integral coefficients, yielding $a + b = - 2 π (π + 2)$ .

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

05Execute

06Execute

✓Verify

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