`f(x)=(cosec^-1x)/sqrt(x^2-[x]^2` Find the domain

Answer

To find the domain of the function \( f(x) = \frac{\csc^{-1}(x)}{\sqrt{x^2 - [x]^2}} \), we need to consider the conditions under which both the numerator and the denominator are defined.

### Step 1: Analyze the numerator
The numerator is \( \csc^{-1}(x) \). The domain of the cosecant inverse function is:
\[ x \leq -1 \quad \text{or} \quad x \geq 1 \]
This means that \( x \) must be either less than or equal to -1 or greater than or equal to 1.

### Step 2: Analyze the denominator
The denominator is \( \sqrt{x^2 - [x]^2} \). For this expression to be defined and real, the following conditions must hold:
1. \( x^2 - [x]^2 > 0 \) (since the square root must be positive).
2. The expression inside the square root must not equal zero.

Using the property of the greatest integer function, we can express \( x \) as:
\[ [x] = n \quad \text{where } n \text{ is an integer} \]
Thus, we can rewrite the denominator:
\[ x^2 - [x]^2 = x^2 - n^2 = (x - n)(x + n) \]
For this product to be greater than zero, we need:
\[ (x - n)(x + n) > 0 \]

### Step 3: Determine when \( (x - n)(x + n) > 0 \)
This inequality holds true in the following intervals:
- \( x < -n \)
- \( x > n \)

### Step 4: Combine conditions
Now we combine the conditions from the numerator and the denominator:
1. From the numerator: \( x \leq -1 \) or \( x \geq 1 \)
2. From the denominator: \( x < -n \) or \( x > n \) (where \( n \) is an integer).

### Step 5: Exclude integer values
Since \( [x] \) is the greatest integer function, we must exclude integer values from our domain. This means that we cannot include any integers in our intervals.

### Final Domain
Thus, the domain of \( f(x) \) can be expressed as:
\[ (1, \infty) \setminus \mathbb{Z} \quad \text{and} \quad (-\infty, -1) \setminus \mathbb{Z} \]
This means that \( x \) can take any real value greater than 1 or less than -1, except for integer values.

### Summary of Domain
The final domain of the function \( f(x) \) is:
\[ \text{Domain: } (1, \infty) \setminus \mathbb{Z} \cup (-\infty, -1) \setminus \mathbb{Z} \]