Functions – Greatest Integer Function – JEE Main 22 Jan 2026 Shift 2

Solution:

Simplify the function using the property $[x+n] = [x]+n$ for integer $n$:
$$ f(x) = [x]^2 – ([x]+3) – 3 $$
$$ f(x) = [x]^2 – [x] – 6 $$
Let $[x] = t$. Then $f(x) = t^2 – t – 6 = (t-3)(t+2)$.

Analyze the options:

**Check Option (2):** $f(x) < 0$ $$ (t-3)(t+2) < 0 \Rightarrow -2 < t < 3 $$ $$ -2 < [x] < 3 $$ Since $[x]$ is an integer, possible values are $-1, 0, 1, 2$. 1. $[x] = -1 \Rightarrow -1 \le x < 0$ 2. $[x] = 0 \Rightarrow 0 \le x < 1$ 3. $[x] = 1 \Rightarrow 1 \le x < 2$ 4. $[x] = 2 \Rightarrow 2 \le x < 3$ Combining these intervals: $x \in [-1, 3)$. Thus, $f(x) < 0$ only for $x \in [-1, 3)$. This option is correct.
**Check Option (1):** $f(x) > 0$
$$ (t-3)(t+2) > 0 \Rightarrow t > 3 \text{ or } t < -2 $$ $[x] > 3 \Rightarrow [x] \ge 4 \Rightarrow x \ge 4$.
$[x] < -2 \Rightarrow [x] \le -3 \Rightarrow x < -2$. $x \in (-\infty, -2) \cup [4, \infty)$. Option (1) says "only for $[4, \infty)$", so it is incorrect.
**Check Option (3):** $\int_0^2 f(x) dx$
$$ \int_0^2 ([x]^2 – [x] – 6) dx = \int_0^1 (-6) dx + \int_1^2 (1 – 1 – 6) dx $$
$$ = [-6x]_0^1 + [-6x]_1^2 = -6 + (-12 – (-6)) = -6 – 6 = -12 $$
Option (3) is incorrect.

**Check Option (4):** $f(x) = 0$
$$ (t-3)(t+2) = 0 \Rightarrow [x] = 3 \text{ or } [x] = -2 $$
$x \in [3, 4) \cup [-2, -1)$.
There are infinitely many solutions. Option (4) is incorrect.

Ans. (2)