JEE Main 2026 — 22 January, Evening Shift — Question 18

Given $f(x) = [x]^2 - [x+3] - 3$. Since $[x+3] = [x] + 3$, we have $f(x) = [x]^2 - ([x] + 3) - 3 = [x]^2 - [x] - 6$. Factor: $f(x) = ([x] + 2)([x] - 3)$. $f(x) > 0$ when $[x] < -2$ or $[x] > 3$. $[x] < -2$ gives $x \in (-\infty, -2)$. $[x] > 3$ gives $x \in [4, \infty)$. $f(x) < 0$ when $-2 < [x] < 3$, i.e., $[x] \in \{-1,0,1,2\}$. This corresponds to $x \in [-1,3)$. $\int_0^2 f(x) dx = \int_0^1 (0^2 - 0 - 6) dx + \int_1^2 (1^2 - 1 - 6) dx = \int_0^1 (-6) dx + \int_1^2 (-6) dx = -6 - 6 = -12$. $f(x) = 0$ when $[x] = -2$ or $[x] = 3$, which gives infinitely many $x$ (e.g., $x \in [-2,-1)$ and $x \in [3,4)$). Thus only option B is correct.