JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Definite Integral with a Functional Reflection Symmetry

Let f(x) be a function satisfying f(x)+f(π-x)= π^2, ∀ x ∈ R. Then integral from 0 to π of f(x) sin x dx is equal to

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A(π^2)/4B(π^2)/2C2 π^2Dπ^2

Step-by-step solutionView

Correct answer

Applying King's property or choosing the constant function $f (x) = \frac{π^{2}}{2}$ gives $\int_{0}^{π} f (x) \sin x d x = π^{2}$ .

Option analysis

Why each option works or fails

A · (π^2)/4

Dividing by 2 twice—once from setting $f (x) = \frac{π^{2}}{2}$ and mistakenly again when integrating $\sin x$ , or forgetting that $\int_{0}^{π} \sin x d x = 2$ . Evaluate the trigonometric integral carefully: $\int_{0}^{π} \sin x d x = [- \cos x]_{0}^{π} = 1 - (- 1) = 2$ , not $\frac{1}{2}$ .

B · (π^2)/2

Taking $\int_{0}^{π} \sin x d x = 1$ instead of $2$ when computing $I = \frac{π^{2}}{2} \int_{0}^{π} \sin x d x$ . Remember that the area under one half-period of the sine wave from $0$ to $π$ is $2$ , which cancels the $\frac{1}{2}$ factor from symmetry.

C · 2 π^2

Multiplying the constant sum $π^{2}$ by the integral of $\sin x$ without dividing by $2$ when combining $I + I = 2 I$ . When adding $I$ and its reflected form, the left side is $2 I$ ; you must divide the combined integral by $2$ to solve for $I$ .

D · π^2

This is the correct value. Correctly applying King's property yields $2 I = π^{2} \int_{0}^{π} \sin x d x = 2 π^{2}$ , so $I = π^{2}$ .

Reviewed route

Solution

StepWorking

01Given

$f (x) + f (π - x) = π^{2}, \forall x \in R$ and $I = \int_{0}^{π} f (x) \sin x d x$ .

02Approach

Apply King's property $\int_{a}^{b} g (x) d x = \int_{a}^{b} g (a + b - x) d x$ to use the functional equation $f (x) + f (π - x) = π^{2}$ .

03Execute

Using King's property with $a = 0, b = π$ : $I = \int_{0}^{π} f (π - x) \sin (π - x) d x = \int_{0}^{π} f (π - x) \sin x d x$ Adding the two expressions for $I$ : $2 I = \int_{0}^{π} [f (x) + f (π - x)] \sin x d x = \int_{0}^{π} π^{2} \sin x d x$ $2 I = π^{2} [- \cos x]_{0}^{π} = π^{2} [- (- 1) - (- 1)] = 2 π^{2}$ $I = π^{2}$

✓Verify

Substitute a constant function satisfying the relation: let $f (x) = \frac{π^{2}}{2}$ . Then $f (x) + f (π - x) = π^{2}$ . The integral is $\int_{0}^{π} \frac{π^{2}}{2} \sin x d x = \frac{π^{2}}{2} \cdot 2 = π^{2}$ , which matches option D.

✓ 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: 7 September 2026

Quick checks

Students also ask

Why does

\sin (π - x)

stay

\sin x

By the standard trigonometric identity in the second quadrant, $\sin (π - x) = \sin x$ .

Is it mathematically valid to just pick

f (x) = \frac{π^{2}}{2}

in a JEE question?

Yes, because if the problem is well-posed and the options are independent of $f (x)$ , any valid function satisfying the constraint must give the correct answer.

Answer

Applying King's property or choosing the constant function $f (x) = \frac{π^{2}}{2}$ gives $\int_{0}^{π} f (x) \sin x d x = π^{2}$ .

Why each option works or fails

A: (π^2)/4 - Dividing by 2 twice—once from setting $f (x) = \frac{π^{2}}{2}$ and mistakenly again when integrating $\sin x$ , or forgetting that $\int_{0}^{π} \sin x d x = 2$ . Evaluate the trigonometric integral carefully: $\int_{0}^{π} \sin x d x = [- \cos x]_{0}^{π} = 1 - (- 1) = 2$ , not $\frac{1}{2}$ .
B: (π^2)/2 - Taking $\int_{0}^{π} \sin x d x = 1$ instead of $2$ when computing $I = \frac{π^{2}}{2} \int_{0}^{π} \sin x d x$ . Remember that the area under one half-period of the sine wave from $0$ to $π$ is $2$ , which cancels the $\frac{1}{2}$ factor from symmetry.
C: 2 π^2 - Multiplying the constant sum $π^{2}$ by the integral of $\sin x$ without dividing by $2$ when combining $I + I = 2 I$ . When adding $I$ and its reflected form, the left side is $2 I$ ; you must divide the combined integral by $2$ to solve for $I$ .
D · correct: π^2 - This is the correct value. Correctly applying King's property yields $2 I = π^{2} \int_{0}^{π} \sin x d x = 2 π^{2}$ , so $I = π^{2}$ .

Step-by-step solution

Given: $f (x) + f (π - x) = π^{2}, \forall x \in R$ and $I = \int_{0}^{π} f (x) \sin x d x$ .
Approach: Apply King's property $\int_{a}^{b} g (x) d x = \int_{a}^{b} g (a + b - x) d x$ to use the functional equation $f (x) + f (π - x) = π^{2}$ .
Execute: Using King's property with $a = 0, b = π$ : $I = \int_{0}^{π} f (π - x) \sin (π - x) d x = \int_{0}^{π} f (π - x) \sin x d x$ Adding the two expressions for $I$ : $2 I = \int_{0}^{π} [f (x) + f (π - x)] \sin x d x = \int_{0}^{π} π^{2} \sin x d x$ $2 I = π^{2} [- \cos x]_{0}^{π} = π^{2} [- (- 1) - (- 1)] = 2 π^{2}$ $I = π^{2}$
Verify: Substitute a constant function satisfying the relation: let $f (x) = \frac{π^{2}}{2}$ . Then $f (x) + f (π - x) = π^{2}$ . The integral is $\int_{0}^{π} \frac{π^{2}}{2} \sin x d x = \frac{π^{2}}{2} \cdot 2 = π^{2}$ , which matches option D.

Shortcut:

When to use it: When the relation must hold for any such function f(x), picking the simplest constant function f(x) = c directly yields the answer.

Given: $f (x) + f (π - x) = π^{2}$

Approach: Choose the constant function $f (x) = \frac{π^{2}}{2}$ , which satisfies the given condition identically for all $x$ .

Execute: $I = \int_{0}^{π} \frac{π^{2}}{2} \sin x d x = \frac{π^{2}}{2} [- \cos x]_{0}^{π} = \frac{π^{2}}{2} \times 2 = π^{2}$

Verify: Since the question specifies the integral has a unique value independent of $f$ , this particular choice gives the unique correct value $π^{2}$ .

Definite Integral with a Functional Reflection Symmetry

What feels right?

Applying King's property or choosing the constant function $f (x) = \frac{π^{2}}{2}$ gives $\int_{0}^{π} f (x) \sin x d x = π^{2}$ .

Why each option works or fails

Solution

01Given

02Approach

03Execute

✓Verify

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01Given

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Applying King's property or choosing the constant function $f (x) = \frac{π^{2}}{2}$ gives $\int_{0}^{π} f (x) \sin x d x = π^{2}$ .