NVAMediumJEE 2026Properties of Definite Integrals

JEE Mathematics 2026 Question with Solution

The number of elements in the set S={x:x[0,100] and 0xt2sin(xt)dt=x2}S = \{ x : x \in [0, 100] \text{ and } \int_0^x t^2 \sin(x-t) \, dt = x^2 \} is _____

Answer

Correct answer:16

Step-by-step solution

Standard Method

Given:

I=0xt2sin(xt)dtI = \int_0^x t^2 \sin(x-t) \, dt

and

I=x2,I = x^2,

with x[0,100]x \in [0,100].

Find: The number of elements in the set SS.

Use the substitution u=xtu = x-t. Then

I=0x(xu)2sinuduI = \int_0^x (x-u)^2 \sin u \, du

so

I=0x(x22xu+u2)sinudu.I = \int_0^x (x^2 - 2xu + u^2) \sin u \, du.

Therefore,

I=x20xsinudu2x0xusinudu+0xu2sinudu.I = x^2 \int_0^x \sin u \, du - 2x \int_0^x u \sin u \, du + \int_0^x u^2 \sin u \, du.

Now evaluate the parts shown in the solution:

0xsinudu=1cosx\int_0^x \sin u \, du = 1 - \cos x 0xusinudu=sinxxcosx\int_0^x u \sin u \, du = \sin x - x\cos x 0xu2sinudu=x2cosx+2xsinx+2cosx2.\int_0^x u^2 \sin u \, du = -x^2\cos x + 2x\sin x + 2\cos x - 2.

Substituting these,

I=x2(1cosx)2x(sinxxcosx)+(x2cosx+2xsinx+2cosx2).I = x^2(1-\cos x) - 2x(\sin x - x\cos x) + (-x^2\cos x + 2x\sin x + 2\cos x - 2).

Simplifying,

I=x2+2cosx2.I = x^2 + 2\cos x - 2.

Set I=x2I = x^2:

x2+2cosx2=x2x^2 + 2\cos x - 2 = x^2 2cosx=22\cos x = 2 cosx=1.\cos x = 1.

Hence,

x=2nπx = 2n\pi

for integers nn. Since x[0,100]x \in [0,100],

02nπ1000 \le 2n\pi \le 100 0n50π15.9.0 \le n \le \frac{50}{\pi} \approx 15.9.

Thus,

n{0,1,2,,15},n \in \{0,1,2,\dots,15\},

which gives 1616 values.

Therefore, the number of elements in the set is 1616.

Common mistakes

  • Taking the answer key value 22 without checking the solution working is incorrect. The solution derivation gives cosx=1\cos x = 1 and hence multiple values of xx in [0,100][0,100], so the count must be computed from the interval.

  • Missing the value x=0x = 0 is a common mistake. Since 0=20π0 = 2 \cdot 0 \cdot \pi and the interval is closed, 00 must be included.

  • Using n<50πn < \frac{50}{\pi} but then stopping at n=14n=14 is incorrect. Because 50π15.9\frac{50}{\pi} \approx 15.9, all integers from 00 through 1515 are allowed.

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