The function f(x) = frac{sec^{-1}x}{sqrt{x - | vedclass

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(B) The function $f(x)$ is defined if the following conditions are met:$1$. The domain of $\sec^{-1}x$ is $(-\infty, -1] \cup [1, \infty)$,which can b

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$R - ((-1, 1) \cup \{n \mid n \in Z\})$

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(B) The function $f(x)$ is defined if the following conditions are met:$1$. The domain of $\sec^{-1}x$ is $(-\infty, -1] \cup [1, \infty)$,which can be written as $R - (-1, 1)$.$2$. The denominator $\sqrt{x - [x]}$ must be defined and non-zero. The expression inside the square root must be strictly positive: $x - [x] > 0$.$3$. Since $x - [x] = \{x\}$ (the fractional part of $x$),the condition $x - [x] > 0$ implies that $x$ cannot be an integer. Thus,$x \in R - Z$.$4$. Combining these two conditions,the function is defined for $x \in (R - (-1, 1)) \cap (R - Z)$.$5$. This intersection is equivalent to $R - ((-1, 1) \cup \{n \mid n \in Z\})$.

Function	Range
$A. f(x) = \|x\|$	$I. [0, \infty)$
$B. f(x) = x^2$	$II. \mathbb{R}$
$C. f(x) = x^3$	$III. [0, \infty)$
$D. f(x) = \text{sgn}(x)$	$IV. \{-1, 0, 1\}$

The function $f(x) = \frac{\sec^{-1}x}{\sqrt{x - [x]}}$,where $[.]$ denotes the greatest integer less than or equal to $x$,is defined for all $x$ belonging to:

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