Let [x] denote the greatest integer less than | vedclass

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(B) The function $f(x) = \sec^{-1}(u)$ is defined for $|u| \geq 1$,which means $u \leq -1$ or $u \geq 1$.Here,$u = 2[x] + 1$.So,$2[x] + 1 \leq -1$ or

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$(-\infty, \infty)$

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(B) The function $f(x) = \sec^{-1}(u)$ is defined for $|u| \geq 1$,which means $u \leq -1$ or $u \geq 1$.Here,$u = 2[x] + 1$.So,$2[x] + 1 \leq -1$ or $2[x] + 1 \geq 1$.Case $1$: $2[x] + 1 \leq -1 \Rightarrow 2[x] \leq -2 \Rightarrow [x] \leq -1$. This implies $x Case $2$: $2[x] + 1 \geq 1 \Rightarrow 2[x] \geq 0 \Rightarrow [x] \geq 0$. This implies $x \geq 0$.Combining both cases,$x \in (-\infty, 0) \cup [0, \infty) = (-\infty, \infty)$.

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

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