Is it mathematically valid to just pick in a JEE question?

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

JEE MainMathematics

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Integral Calculus: JEE Main Mathematics Question with Solution

Let

\mathrm{f}(\mathrm{x})

be a function satisfying

\mathrm{f}(\mathrm{x})+\mathrm{f}(\pi-\mathrm{x})=

\pi^2, \forall \mathrm{x} \in \mathrm{R}

. Then

\int_0^\pi \mathrm{f}(\mathrm{x}) \sin \mathrm{x} \mathrm{dx}

is equal to

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\frac{\pi^2}{4}

\frac{\pi^2}{2}

2 \pi^2

\pi^2

Step-by-step solutionView

Correct answer

Applying King's property or choosing the constant function $f(x) = \frac{\pi^2}{2}$ gives $\int_0^\pi f(x) \sin x \, dx = \pi^2$ .

Option analysis

Why each option works or fails

A ·

\frac{\pi^2}{4}

Dividing by 2 twice—once from setting $f(x) = \frac{\pi^2}{2}$ and mistakenly again when integrating $\sin x$ , or forgetting that $\int_0^\pi \sin x \, dx = 2$ . Evaluate the trigonometric integral carefully: $\int_0^\pi \sin x \, dx = [-\cos x]_0^\pi = 1 - (-1) = 2$ , not $\frac{1}{2}$ .

B ·

\frac{\pi^2}{2}

Taking $\int_0^\pi \sin x \, dx = 1$ instead of $2$ when computing $I = \frac{\pi^2}{2} \int_0^\pi \sin x \, dx$ . Remember that the area under one half-period of the sine wave from $0$ to $\pi$ is $2$ , which cancels the $\frac{1}{2}$ factor from symmetry.

C ·

2 \pi^2

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

D ·

\pi^2

None. This is the correct value. Correctly applying King's property yields $2I = \pi^2 \int_0^\pi \sin x \, dx = 2\pi^2$ , so $I = \pi^2$ .

Reviewed route

Solution

StepWorking

01given

$f(x) + f(\pi - x) = \pi^2, \forall x \in \mathbb{R}$ and $I = \int_0^\pi f(x) \sin x \, dx$ .

02approach

Apply King's property $\int_a^b g(x) \, dx = \int_a^b g(a+b-x) \, dx$ to use the functional equation $f(x) + f(\pi - x) = \pi^2$ .

03execute

Using King's property with $a=0, b=\pi$ : $I = \int_0^\pi f(\pi - x) \sin(\pi - x) \, dx = \int_0^\pi f(\pi - x) \sin x \, dx$ Adding the two expressions for $I$ : $2I = \int_0^\pi [f(x) + f(\pi - x)] \sin x \, dx = \int_0^\pi \pi^2 \sin x \, dx$ $2I = \pi^2 [-\cos x]_0^\pi = \pi^2 [ -(-1) - (-1) ] = 2\pi^2$ $I = \pi^2$

04verify

Substitute a constant function satisfying the relation: let $f(x) = \frac{\pi^2}{2}$ . Then $f(x) + f(\pi-x) = \pi^2$ . The integral is $\int_0^\pi \frac{\pi^2}{2} \sin x \, dx = \frac{\pi^2}{2} \cdot 2 = \pi^2$ , which matches option (3).

05given

$f(x) + f(\pi - x) = \pi^2$

06approach

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

07execute

$I = \int_0^\pi \frac{\pi^2}{2} \sin x \, dx = \frac{\pi^2}{2} [-\cos x]_0^\pi = \frac{\pi^2}{2} \times 2 = \pi^2$

✓verify

Since the question specifies the integral has a unique value independent of $f$ , this particular choice gives the unique correct value $\pi^2$ .

✓ 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: 7 September 2026

Quick checks

Students also ask

Why does

\sin(\pi - x)

stay

\sin x

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

Is it mathematically valid to just pick

f(x) = \frac{\pi^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 $f(x) = \frac{\pi^2}{2}$ gives $\int_0^\pi f(x) \sin x \, dx = \pi^2$ .

Why each option works or fails

A: $\frac{\pi^2}{4}$ - Dividing by 2 twice—once from setting $f(x) = \frac{\pi^2}{2}$ and mistakenly again when integrating $\sin x$ , or forgetting that $\int_0^\pi \sin x \, dx = 2$ . Evaluate the trigonometric integral carefully: $\int_0^\pi \sin x \, dx = [-\cos x]_0^\pi = 1 - (-1) = 2$ , not $\frac{1}{2}$ .
B: $\frac{\pi^2}{2}$ - Taking $\int_0^\pi \sin x \, dx = 1$ instead of $2$ when computing $I = \frac{\pi^2}{2} \int_0^\pi \sin x \, dx$ . Remember that the area under one half-period of the sine wave from $0$ to $\pi$ is $2$ , which cancels the $\frac{1}{2}$ factor from symmetry.
C: $2 \pi^2$ - Multiplying the constant sum $\pi^2$ by the integral of $\sin x$ without dividing by $2$ when combining $I + I = 2I$ . When adding $I$ and its reflected form, the left side is $2I$ ; you must divide the combined integral by $2$ to solve for $I$ .
D · correct: $\frac{1}{2}$ - None. This is the correct value. Correctly applying King's property yields $2I = \pi^2 \int_0^\pi \sin x \, dx = 2\pi^2$ , so $I = \pi^2$ .

Step-by-step solution

given: $f(x) + f(\pi - x) = \pi^2, \forall x \in \mathbb{R}$ and $I = \int_0^\pi f(x) \sin x \, dx$ .
approach: Apply King's property $\int_a^b g(x) \, dx = \int_a^b g(a+b-x) \, dx$ to use the functional equation $f(x) + f(\pi - x) = \pi^2$ .
execute: Using King's property with $a=0, b=\pi$ : $I = \int_0^\pi f(\pi - x) \sin(\pi - x) \, dx = \int_0^\pi f(\pi - x) \sin x \, dx$ Adding the two expressions for $I$ : $2I = \int_0^\pi [f(x) + f(\pi - x)] \sin x \, dx = \int_0^\pi \pi^2 \sin x \, dx$ $2I = \pi^2 [-\cos x]_0^\pi = \pi^2 [ -(-1) - (-1) ] = 2\pi^2$ $I = \pi^2$
verify: Substitute a constant function satisfying the relation: let $f(x) = \frac{\pi^2}{2}$ . Then $f(x) + f(\pi-x) = \pi^2$ . The integral is $\int_0^\pi \frac{\pi^2}{2} \sin x \, dx = \frac{\pi^2}{2} \cdot 2 = \pi^2$ , which matches option (3).
given: $f(x) + f(\pi - x) = \pi^2$
approach: Choose the constant function $f(x) = \frac{\pi^2}{2}$ , which satisfies the given condition identically for all $x$ .
execute: $I = \int_0^\pi \frac{\pi^2}{2} \sin x \, dx = \frac{\pi^2}{2} [-\cos x]_0^\pi = \frac{\pi^2}{2} \times 2 = \pi^2$
verify: Since the question specifies the integral has a unique value independent of $f$ , this particular choice gives the unique correct value $\pi^2$ .

Integral Calculus: JEE Main Mathematics Question with Solution

Applying King's property or choosing the constant function f(x)=π22f(x) = \frac{\pi^2}{2}f(x)=2π2​ gives ∫0πf(x)sin⁡x dx=π2\int_0^\pi f(x) \sin x \, dx = \pi^2∫0π​f(x)sinxdx=π2.

01given

02approach

03execute

04verify

05given

06approach

07execute

✓verify

Students also ask

Applying King's property or choosing the constant function $f(x) = \frac{\pi^2}{2}$ gives $\int_0^\pi f(x) \sin x \, dx = \pi^2$ .