Key Concepts and Formulas

• Domain of $\operatorname{cosec}^{-1}x$: The inverse cosecant function, $\operatorname{cosec}^{-1}x$, is defined for all real numbers $x$ such that $|x| \ge 1$, which means $x \in (-\infty, -1] \cup [1, \infty)$.
• Domain of $\sqrt{A}$: For the square root function $\sqrt{A}$ to yield a real number, the radicand $A$ must be non-negative, i.e., $A \ge 0$.
• Domain of $\frac{N}{D}$: For a fraction $\frac{N}{D}$ to be defined, the denominator $D$ must be non-zero, i.e., $D \ne 0$.
• Fractional Part $\{x\}$: The expression $x - [x]$ represents the fractional part of $x$, denoted as $\{x\}$. It is defined as $0 \le \{x\} < 1$ for all real numbers $x$. The fractional part $\{x\}$ is equal to 0 if and only if $x$ is an integer.

Step-by-Step Solution

The given function is $f(x) = {{\cos e{c^{ - 1}}x} \over {\sqrt {x - [x]} }}$. To find the domain of this real-valued function, we need to ensure that all its components are well-defined and that no division by zero occurs.

Step 1: Domain of the Numerator ($\operatorname{cosec}^{-1}x$) The numerator of the function is $\operatorname{cosec}^{-1}x$. According to the definition of the inverse cosecant function, it is defined for all real numbers $x$ such that $|x| \ge 1$. This condition translates to: $x \le -1$ or $x \ge 1$. In interval notation, this is $x \in (-\infty, -1] \cup [1, \infty)$.

Step 2: Domain of the Denominator ($\sqrt{x - [x]}$) - Square Root Condition The denominator contains a square root term, $\sqrt{x - [x]}$. For this square root to be defined in the real numbers, the expression under the square root must be non-negative. So, we must have: $x - [x] \ge 0$. The expression $x - [x]$ is the fractional part of $x$, denoted by $\{x\}$. By definition, the fractional part of any real number is always between 0 (inclusive) and 1 (exclusive): $0 \le \{x\} < 1$. Therefore, the condition $x - [x] \ge 0$ is always satisfied for all real numbers $x$. This condition does not impose any additional restrictions on the domain.

Step 3: Domain of the Denominator ($\sqrt{x - [x]}$) - Non-Zero Condition The term $\sqrt{x - [x]}$ is in the denominator of the function. For the function to be defined, the denominator cannot be zero. So, we must have: $\sqrt{x - [x]} \ne 0$. Squaring both sides (which is permissible since $\sqrt{x - [x]}$ is already established to be non-negative), we get: $x - [x] \ne 0$. As established in Step 2, $x - [x] = \{x\}$. Thus, the condition becomes: $\{x\} \ne 0$. The fractional part of a number is zero if and only if the number is an integer. For example, $\{3\} = 3 - [3] = 3 - 3 = 0$, and $\{-5\} = -5 - [-5] = -5 - (-5) = 0$. Therefore, the condition $\{x\} \ne 0$ means that $x$ cannot be an integer. $x \notin \mathbb{Z}$.

Step 4: Combining All Conditions To find the domain of the function $f(x)$, we must satisfy all the conditions derived in the previous steps simultaneously.

1. From the numerator: $x \in (-\infty, -1] \cup [1, \infty)$.
2. From the denominator (non-zero condition): $x \notin \mathbb{Z}$ (x is not an integer).

We need to find the intersection of these two sets. The set $(-\infty, -1] \cup [1, \infty)$ includes all real numbers less than or equal to -1 and all real numbers greater than or equal to 1. This set includes the integers -1 and 1. Since we must exclude all integers ($x \notin \mathbb{Z}$), we need to remove the integers -1 and 1 from the set $(-\infty, -1] \cup [1, \infty)$. Removing $-1$ from $(-\infty, -1]$ results in $(-\infty, -1)$. Removing $1$ from $[1, \infty)$ results in $(1, \infty)$. Combining these, the domain of the function is: $x \in (-\infty, -1) \cup (1, \infty)$.

This set can be described as all real numbers except those in the closed interval $[-1, 1]$.

Common Mistakes & Tips

• Strict Inequality for Denominators: When a square root is in the denominator, the expression inside the square root must be strictly positive ($>0$), not just non-negative ($\ge 0$), because the square root itself cannot be zero.
• Understanding $[x]$ and $\{x\}$: Be clear about the properties of the greatest integer function and the fractional part. $\{x\} = 0$ is a direct indicator that $x$ is an integer.
• Combining Intervals: Carefully combine conditions by taking the intersection of the sets of allowed values. Exclude any integers from the final domain if they are not allowed.

Summary

The domain of the function $f(x) = {{\cos e{c^{ - 1}}x} \over {\sqrt {x - [x]} }}$ is determined by two primary conditions: the domain of the inverse cosecant function and the requirement that the denominator is non-zero. The inverse cosecant function requires $|x| \ge 1$, meaning $x \in (-\infty, -1] \cup [1, \infty)$. The denominator $\sqrt{x - [x]}$ requires $x - [x] > 0$ (since it's in the denominator), which means $\{x\} > 0$, so $x$ cannot be an integer. Combining these, we exclude all integers from $(-\infty, -1] \cup [1, \infty)$, resulting in the domain $(-\infty, -1) \cup (1, \infty)$. This is equivalent to all real numbers except the interval $[-1, 1]$.

The final answer is $\boxed{\text{(D)}}$.