What is the domain of the function f(x) = sec⁻¹(2[x]+1)?

The domain of $\sec^{-1}(y)$ is $y \in (-\infty, -1] \cup [1, \infty)$. This means that the argument of the inverse secant function must be less than or equal to -1, or greater than or equal to 1.

In our function, the argument is $2[x]+1$. So, we must have:

The condition $[x] \le -1$ means that the greatest integer less than or equal to $x$ must be $-1$ or any integer less than $-1$ (e.g., $-2, -3, \dots$). This implies that $x$ must be less than 0. For example, if $x = -0.5$, $[x] = -1$. If $x = -1.5$, $[x] = -2$. If $x = -1$, $[x] = -1$.
So, the solution for this inequality is $x \in (-\infty, 0)$.

The condition $[x] \ge 0$ means that the greatest integer less than or equal to $x$ must be 0 or any integer greater than 0 (e.g., $1, 2, \dots$). This implies that $x$ must be greater than or equal to 0. For example, if $x = 0.5$, $[x] = 0$. If $x = 1.5$, $[x] = 1$. If $x = 0$, $[x] = 0$.
So, the solution for this inequality is $x \in [0, \infty)$.

The domain of $f(x)$ is the union of the solutions from Step 3 and Step 4:

This union covers all real numbers.

Comparing this with the given options:
(1) $(-\infty, -1] \cup [1, \infty)$
(2) $(-\infty, -1] \cup [0, \infty)$
(3) $(-\infty, \infty)$
(4) $(-\infty, \infty) - \{0\}$

The calculated domain matches option (3).

The final answer is $\text{(3)}$.