JEE MAIN - Mathematics (2025 - 28th January Evening Shift - No. 10)
Let [x] denote the greatest integer less than or equal to x. Then the domain of $ f(x) = \sec^{-1}(2[x] + 1) $ is:
$(-\infty, \infty)$
$(-\infty, \infty)- \{0\}$
$(-\infty, -1] \cup [0, \infty)$
$(-\infty, -1] \cup [1, \infty)$
Explanation
$$\begin{aligned}
& 2[\mathrm{x}]+1 \leq-1 \text { or } 2[\mathrm{x}]+1 \geq 1 \\
& \Rightarrow[\mathrm{x}] \leq-1 \cup[\mathrm{x}] \geq 0 \\
& \Rightarrow \mathrm{x} \in(-\infty, 0) \cup \mathrm{x} \in[0, \infty) \\
& \Rightarrow \mathrm{x} \in(-\infty, \infty)
\end{aligned}$$
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