Question:

Step 1: To determine when $f(x) < 0$, we first need to understand the nature of the function $f(x) = [x]^2 - [x+3] - 3$, where $[\cdot]$ denotes the greatest integer function. This function involves the greatest integer function, which means its value changes at integer points. Step 2: Let's break down $f(x)$ for different intervals of $x$ by considering the greatest integer function $[x]$ and $[x+3]$. For $x \in [-1, 0)$, $[x] = -1$ and $[x+3] = 2$, so $f(x) = (-1)^2 - 2 - 3 = 1 - 2 - 3 = -4$. For $x \in [0, 1)$, $[x] = 0$ and $[x+3] = 3$, so $f(x) = 0^2 - 3 - 3 = -6$. For $x \in [1, 2)$, $[x] = 1$ and $[x+3] = 4$, so $f(x) = 1^2 - 4 - 3 = 1 - 4 - 3 = -6$. For $x \in [2, 3)$, $[x] = 2$ and $[x+3] = 5$, so $f(x) = 2^2 - 5 - 3 = 4 - 5 - 3 = -4$. Step 3: We observe that for $x \in [-1, 3)$, $f(x)$ is negative. Now, let's check the behavior of $f(x)$ for $x \geq 3$. For $x \in [3, 4)$, $[x] = 3$ and $[x+3] = 6$, so $f(x) = 3^2 - 6 - 3 = 9 - 6 - 3 = 0$. For $x \geq 4$, $[x] \geq 4$ and $[x+3] \geq 7$, which means $f(x) = [x]^2 - [x+3] - 3 \geq 4^2 - 7 - 3 = 16 - 10 = 6$. Step 4: From the analysis, it's clear that $f(x) < 0$ only for $x \in [-1, 3)$. This matches the statement given in the options. Therefore, the correct answer is the option that states $f(x) < 0$ only for $x \in [-1,3)$. The final answer is: $\boxed{2}$