JEE Main 2026 Jan 23 Shift 2, Mathematics Q23: Differentiation Under the Integral Sign

JEE Main2026Jan 23, Shift 2Mathematics

The number of elements in the set $S = {x : x \in [0, 100] and \int_{0}^{x} t^{2} sin (x - t) d t = x^{2}}$ is ________.

Your answer

Solution

Expand $sin (x - t) = sin x cos t - cos x sin t$ . The integral becomes

$sin x \int_{0}^{x} t^{2} cos t d t - cos x \int_{0}^{x} t^{2} sin t d t$ .

Using integration by parts (or directly differentiating the convolution form), one finds

$\int_{0}^{x} t^{2} sin (x - t) d t = x^{2} - 2 + 2 cos x$ .

(Quick check via differentiation: differentiating both sides w.r.t. $x$ twice consistently yields $sin x$ inside the integrand convolution.)

Setting this equal to $x^{2}$ : $- 2 + 2 cos x = 0 \Rightarrow cos x = 1 \Rightarrow x = 2 k π$ , $k = 0, 1, 2, \dots$ .

Require $x \in [0, 100]$ : $2 k π \leq 100 \Rightarrow k \leq \frac{50}{π} \approx 15.91$ .

So $k = 0, 1, 2, \dots, 15$ , giving 16 values.

Concept behind this question

Concept-wise practice with instant solutions on Fundamenthol.