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Let [x] denote the greatest integer less than or equal to x. Then domain of $f (x) = \sec^{- 1} (2 [x] + 1) is :$

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$(- \infty, \infty] - {0}$

$(- \infty, - 1] \cup [1, \infty)$

$(- \infty, - 1] \cup [0, \infty)$

$(- \infty, - \infty)$

answer is B.

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Detailed Solution

$\begin{array}{l} 2 [x] + 1 \leq - 1 or 2 [x] + 1 \geq 1 \\ \Rightarrow [x] \leq - 1 \cup [x] \geq 0 \\ \Rightarrow x \in (- \infty, 0) \cup x \in [0, \infty) \\ \Rightarrow x \in (- \infty, \infty) \end{array}$

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