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Integral Calculus: JEE Main Mathematics Question with Solution

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+\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

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

-2 \pi(\pi-2)

-2 \pi(\pi+2)

-\pi(\pi-2)

-\pi(\pi+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 $a+b = -2\pi(\pi+2)$ .

Option analysis

Why each option works or fails

A ·

-2 \pi(\pi-2)

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

B ·

-2 \pi(\pi+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 $a+b = -2\pi(\pi+2)$ . No correction needed.

C ·

-\pi(\pi-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 ·

-\pi(\pi+2)

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

Reviewed route

Solution

StepWorking

01given

$f(x) = x + \frac{a}{\pi^2-4}\sin x + \frac{b}{\pi^2-4}\cos x$ and $f(x) = x + \int_0^{\pi/2} \sin(x+y)f(y)\,dy$ .

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 $\frac{a}{\pi^2-4}$ and $\frac{b}{\pi^2-4}$ as definite integrals, then add them and apply King's property $\int_0^a g(y)\,dy = \int_0^a g(a-y)\,dy$ .

03execute

Expanding the integral: $f(x) = x + \sin x \int_0^{\pi/2} f(y)\cos y\,dy + \cos x \int_0^{\pi/2} f(y)\sin y\,dy$ . Comparing with $f(x) = x + \frac{a}{\pi^2-4}\sin x + \frac{b}{\pi^2-4}\cos x$ gives: $\frac{a}{\pi^2-4} = \int_0^{\pi/2} f(y)\cos y\,dy$ and $\frac{b}{\pi^2-4} = \int_0^{\pi/2} f(y)\sin y\,dy$ . Summing both equations: let $K = \frac{a+b}{\pi^2-4} = \int_0^{\pi/2} (\sin y + \cos y)f(y)\,dy$ .

04execute

Using King's property on $K = \int_0^{\pi/2} (\sin y + \cos y)f(y)\,dy$ : Since $\sin(\pi/2 - y) + \cos(\pi/2 - y) = \cos y + \sin y$ , we also have $K = \int_0^{\pi/2} (\sin y + \cos y)f\left(\frac{\pi}{2}-y\right)dy$ . Adding the two expressions: $2K = \int_0^{\pi/2} (\sin y + \cos y)\left[f(y) + f\left(\frac{\pi}{2}-y\right)\right]dy$ . Now compute $f(y) + f\left(\frac{\pi}{2}-y\right)$ : $f(y) + f\left(\frac{\pi}{2}-y\right) = y + \left(\frac{\pi}{2}-y\right) + \frac{a}{\pi^2-4}(\sin y + \cos y) + \frac{b}{\pi^2-4}(\cos y + \sin y) = \frac{\pi}{2} + K(\sin y + \cos y)$ .

05execute

Substitute this sum into the integral: $2K = \int_0^{\pi/2} (\sin y + \cos y)\left[\frac{\pi}{2} + K(\sin y + \cos y)\right]dy$ $= \frac{\pi}{2} \int_0^{\pi/2} (\sin y + \cos y)\,dy + K \int_0^{\pi/2} (\sin y + \cos y)^2\,dy$ . Calculate the two standard integrals: $\int_0^{\pi/2} (\sin y + \cos y)\,dy = [-\cos y + \sin y]_0^{\pi/2} = 1 - (-1) = 2$ . $\int_0^{\pi/2} (\sin y + \cos y)^2\,dy = \int_0^{\pi/2} (1 + \sin 2y)\,dy = [y - \frac{1}{2}\cos 2y]_0^{\pi/2} = \frac{\pi}{2} - \frac{1}{2}(-1 - 1) = \frac{\pi}{2} + 1$ . Thus: $2K = \frac{\pi}{2}(2) + K\left(\frac{\pi}{2} + 1\right) = \pi + K\left(\frac{\pi}{2} + 1\right)$ .

06execute

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

✓verify

Checking dimensions and sign: $K = \frac{a+b}{\pi^2-4}$ . Since $\pi \approx 3.14 > 2$ , $2 - \pi < 0$ , which makes $K < 0$ , giving $a+b < 0$ . The factor $(\pi^2-4)$ splits into $(\pi-2)(\pi+2)$ , cancelling the $(\pi-2)$ in the denominator and leaving $-2\pi(\pi+2)$ . Matches Option (1).

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 $\int_0^{\pi/2} \sin(x+y) f(y) dy$ be rewritten?
$\sin x \int_0^{\pi/2} f(y) \cos y \, dy + \cos x \int_0^{\pi/2} f(y) \sin y \, dy$
Let $A = \int_0^{\pi/2} f(y) \cos y \, dy$ and $B = \int_0^{\pi/2} f(y) \sin y \, dy$ . Comparing $f(x) = x + A \sin x + B \cos x$ with the given form $f(x) = x + \frac{a}{\pi^2-4} \sin x + \frac{b}{\pi^2-4} \cos x$ , how are $a$ and $b$ related to $A$ and $B$ ?
$a = (\pi^2-4)A$ and $b = (\pi^2-4)B$
Substituting $f(y) = y + A \sin y + B \cos y$ into the definitions of $A$ and $B$ , which linear system governs $A$ and $B$ ?
$A - \frac{\pi}{4} B = \frac{\pi}{2} - 1$ and $-\frac{\pi}{4} A + B = 1$
Solving the system $\begin{pmatrix} 1 & -\pi/4 \\ -\pi/4 & 1 \end{pmatrix} \begin{pmatrix} A \\ B \end{pmatrix} = \begin{pmatrix} \frac{\pi-2}{2} \\ 1 \end{pmatrix}$ and finding $a+b = (\pi^2-4)(A+B)$ , what is the final value of $a+b$ ?
$-2\pi(\pi+2)$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

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 $a+b = -2\pi(\pi+2)$ .

Why each option works or fails

A: $-2 \pi(\pi-2)$ - Believing that the denominator $\pi^2-4$ factorizes with an inverted sign or introducing a sign error when solving the $2 \times 2$ linear system for $A$ and $B$ , leading to $(\pi-2)$ instead of $(\pi+2)$ . Carefully track negative signs when evaluating definite integrals involving $\cos y$ and $\sin y$ by parts and check the matrix inversion step.
B · correct: $-2 \pi(\pi+2)$ - 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 $a+b = -2\pi(\pi+2)$ . No correction needed.
C: $-\pi(\pi-2)$ - 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: $-\pi(\pi+2)$ - Missing the factor of 2 that arises when adding $a = -(\pi^2 + 2\pi)$ and $b = -(\pi^2 + 2\pi)$ or equivalent expressions. Compute $a$ and $b$ fully before adding them: $a = -2\pi - \pi^2$ and $b = -2\pi - \pi^2$ , so their sum is $-2\pi(\pi+2)$ rather than $-\pi(\pi+2)$ .

Step-by-step solution

given: $f(x) = x + \frac{a}{\pi^2-4}\sin x + \frac{b}{\pi^2-4}\cos x$ and $f(x) = x + \int_0^{\pi/2} \sin(x+y)f(y)\,dy$ .
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 $\frac{a}{\pi^2-4}$ and $\frac{b}{\pi^2-4}$ as definite integrals, then add them and apply King's property $\int_0^a g(y)\,dy = \int_0^a g(a-y)\,dy$ .
execute: Expanding the integral: $f(x) = x + \sin x \int_0^{\pi/2} f(y)\cos y\,dy + \cos x \int_0^{\pi/2} f(y)\sin y\,dy$ . Comparing with $f(x) = x + \frac{a}{\pi^2-4}\sin x + \frac{b}{\pi^2-4}\cos x$ gives: $\frac{a}{\pi^2-4} = \int_0^{\pi/2} f(y)\cos y\,dy$ and $\frac{b}{\pi^2-4} = \int_0^{\pi/2} f(y)\sin y\,dy$ . Summing both equations: let $K = \frac{a+b}{\pi^2-4} = \int_0^{\pi/2} (\sin y + \cos y)f(y)\,dy$ .
execute: Using King's property on $K = \int_0^{\pi/2} (\sin y + \cos y)f(y)\,dy$ : Since $\sin(\pi/2 - y) + \cos(\pi/2 - y) = \cos y + \sin y$ , we also have $K = \int_0^{\pi/2} (\sin y + \cos y)f\left(\frac{\pi}{2}-y\right)dy$ . Adding the two expressions: $2K = \int_0^{\pi/2} (\sin y + \cos y)\left[f(y) + f\left(\frac{\pi}{2}-y\right)\right]dy$ . Now compute $f(y) + f\left(\frac{\pi}{2}-y\right)$ : $f(y) + f\left(\frac{\pi}{2}-y\right) = y + \left(\frac{\pi}{2}-y\right) + \frac{a}{\pi^2-4}(\sin y + \cos y) + \frac{b}{\pi^2-4}(\cos y + \sin y) = \frac{\pi}{2} + K(\sin y + \cos y)$ .
execute: Substitute this sum into the integral: $2K = \int_0^{\pi/2} (\sin y + \cos y)\left[\frac{\pi}{2} + K(\sin y + \cos y)\right]dy$ $= \frac{\pi}{2} \int_0^{\pi/2} (\sin y + \cos y)\,dy + K \int_0^{\pi/2} (\sin y + \cos y)^2\,dy$ . Calculate the two standard integrals: $\int_0^{\pi/2} (\sin y + \cos y)\,dy = [-\cos y + \sin y]_0^{\pi/2} = 1 - (-1) = 2$ . $\int_0^{\pi/2} (\sin y + \cos y)^2\,dy = \int_0^{\pi/2} (1 + \sin 2y)\,dy = [y - \frac{1}{2}\cos 2y]_0^{\pi/2} = \frac{\pi}{2} - \frac{1}{2}(-1 - 1) = \frac{\pi}{2} + 1$ . Thus: $2K = \frac{\pi}{2}(2) + K\left(\frac{\pi}{2} + 1\right) = \pi + K\left(\frac{\pi}{2} + 1\right)$ .
execute: Rearranging the equation for $K$ : $K\left(2 - \left(\frac{\pi}{2} + 1\right)\right) = \pi \implies K\left(1 - \frac{\pi}{2}\right) = \pi \implies K\left(\frac{2-\pi}{2}\right) = \pi \implies K = \frac{2\pi}{2-\pi} = -\frac{2\pi}{\pi-2}$ . Since $K = \frac{a+b}{\pi^2-4}$ , we get: $a+b = K(\pi^2-4) = -\frac{2\pi}{\pi-2}(\pi-2)(\pi+2) = -2\pi(\pi+2)$ .
verify: Checking dimensions and sign: $K = \frac{a+b}{\pi^2-4}$ . Since $\pi \approx 3.14 > 2$ , $2 - \pi < 0$ , which makes $K < 0$ , giving $a+b < 0$ . The factor $(\pi^2-4)$ splits into $(\pi-2)(\pi+2)$ , cancelling the $(\pi-2)$ in the denominator and leaving $-2\pi(\pi+2)$ . Matches Option (1).

Integral Calculus: JEE Main Mathematics Question with Solution

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 $a+b = -2\pi(\pi+2)$ .

Why each option works or fails

Solution

01given

02approach

03execute

04execute

05execute

06execute

✓verify

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Integral Calculus: JEE Main Mathematics Question with Solution

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Students also ask

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