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Let $$[x]$$ denote the greatest integer less than or equal to $$x$$. Then the domain of $$f(x)=sec^{-1}(2[x]+1)$$ is:
The argument of $$\sec^{-1}(u)$$ must satisfy $$|u| \ge 1$$.
So, $$|2[x]+1| \ge 1$$
Case 1: $$2[x]+1 \ge 1 \implies 2[x] \ge 0 \implies [x] \ge 0 \implies x \in [0, \infty)$$.
Case 2: $$2[x]+1 \le -1 \implies 2[x] \le -2 \implies [x] \le -1 \implies x \in (-\infty, 0)$$.
Combining these: $$(-\infty, 0) \cup [0, \infty) = (-\infty, \infty)$$
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