JEE Main 2026 — 23 January, Evening Shift — Question 23

Let $I = \int_0^x t^2 \sin(x-t) \, dt$. Use the property $\int_0^x f(t) \, dt = \int_0^x f(x-t) \, dt$. Then $I = \int_0^x (x-t)^2 \sin t \, dt$. Expand: $I = \int_0^x (x^2 - 2xt + t^2) \sin t \, dt$. Integrate: $I = x^2 \int_0^x \sin t \, dt - 2x \int_0^x t \sin t \, dt + \int_0^x t^2 \sin t \, dt$. Compute: $\int_0^x \sin t \, dt = 1 - \cos x$, $\int_0^x t \sin t \, dt = \sin x - x \cos x$,

$\int_0^x t^2 \sin t \, dt = 2x \sin x - (x^2 - 2) \cos x - 2$. Substitute and simplify: $I = x^2(1-\cos x) - 2x(\sin x - x \cos x) + [2x \sin x - (x^2-2)\cos x - 2]$. Simplify: $I = x^2 - x^2 \cos x - 2x \sin x + 2x^2 \cos x + 2x \sin x - x^2 \cos x + 2 \cos x - 2$. Cancel terms: $I = x^2 + 2 \cos x - 2$. Set $I = x^2$: $x^2 + 2 \cos x - 2 = x^2 \Rightarrow 2 \cos x - 2 = 0 \Rightarrow \cos x = 1$. Solutions: $x = 2n\pi, n \in \mathbb{Z}$. In [0,100], $0 \le 2n\pi \le 100 \Rightarrow 0 \le n \le \frac{100}{2\pi} \approx 15.9$. Thus $n = 0,1,2,\dots,15$ gives 16 solutions. Hence the number of elements in $S$ is 16.