Q39 / 75 · JEE Main 2026 (22 January Shift 2) · Mathematics
Let $f(x)=[x]^{2}-[x+3]-3, x \in \mathbf{R}$, where [] is the greatest integer funtion. Then
- A$\int_{0}^{2} f(x) \mathrm{d} x=-6$
- Correct answerB$f(x)<0$ only for $x \in[-1,3)$
- C$f(x)>0$ only for $x \in[4, \infty)$
- D$f(x)=0$ for finitely many values of $x$
Solution
For $x \in [n, n+1)$ where $n$ is an integer, $[x] = n$ and $[x+3] = n+3$.
Thus $f(x) = n^2 - (n+3) - 3 = n^2 - n - 6 = (n-3)(n+2)$.
For $f(x) < 0$, we need $(n-3)(n+2) < 0$, which gives $-2 < n < 3$, i.e., $n \in \{-1, 0, 1, 2\}$.
This corresponds to $x \in [-1, 3)$.
We can verify:
For $n = -1$: $f = 1 - 2 - 3 = -4 < 0$
For $n = 0,1$: $f = -6 < 0$
For $n = 2$: $f = 4 - 5 - 3 = -4 < 0$
For $n = 3$: $f = 9 - 6 - 3 = 0$
For $n \geq 4$, $(n-3)(n+2) > 0$.
Therefore $f(x) < 0$ only for $x \in [-1, 3)$.
Thus $f(x) = n^2 - (n+3) - 3 = n^2 - n - 6 = (n-3)(n+2)$.
For $f(x) < 0$, we need $(n-3)(n+2) < 0$, which gives $-2 < n < 3$, i.e., $n \in \{-1, 0, 1, 2\}$.
This corresponds to $x \in [-1, 3)$.
We can verify:
For $n = -1$: $f = 1 - 2 - 3 = -4 < 0$
For $n = 0,1$: $f = -6 < 0$
For $n = 2$: $f = 4 - 5 - 3 = -4 < 0$
For $n = 3$: $f = 9 - 6 - 3 = 0$
For $n \geq 4$, $(n-3)(n+2) > 0$.
Therefore $f(x) < 0$ only for $x \in [-1, 3)$.
Hence, the correct answer is option: 'B'.