1. Condition for the inverse secant to exist

For real values, $\sec^{-1}(y)$ is defined when $|y| \ge 1.$ So for $f(x)=\sec^{-1}(2[x]+1),$ we need $|2[x]+1|\ge 1.$

2. Let $n=[x]$

Since $[x]$ is the greatest integer function, $n=[x]$ is always an integer. Thus the argument of $\sec^{-1}$ becomes $2n+1,$ which is always an odd integer.

So possible values are $\ldots,-5,-3,-1,1,3,5,\ldots$

3. Check whether these values satisfy the domain condition

For every odd integer $2n+1$, we have $|2n+1|\ge 1.$ Indeed, $2n+1$ can never be $0$.

Hence $\sec^{-1}(2[x]+1)$ is defined for every real $x$.

4. Therefore the domain

Since no real $x$ is excluded, $\text{Domain}=(-\infty,\infty).$

5. Option check

• A: $(-\infty,\infty)$ ✅
• B: excludes $0$, unnecessary ❌
• C: excludes $(-1,0)$, unnecessary ❌
• D: excludes $(-1,1)$, unnecessary ❌

Therefore, the correct option is $\boxed{A}.$