Find the domain of defination the following functions.
`f(x)=e^(sin^(-1)(x/2))+tan^(-1) [x/2-1]+ln (sqrt(x-[x]))`

Answer

To find the domain of the function \( f(x) = e^{\sin^{-1}(x/2)} + \tan^{-1}(x/2 - 1) + \ln(\sqrt{x - \lfloor x \rfloor}) \), we will analyze each component of the function separately.

### Step 1: Analyze \( e^{\sin^{-1}(x/2)} \)

The function \( e^{\sin^{-1}(x/2)} \) is defined for all real numbers because the exponential function is defined for all real values. However, we need to consider the domain of \( \sin^{-1}(x/2) \).

The inverse sine function \( \sin^{-1}(y) \) is defined for \( y \) in the range \([-1, 1]\). Therefore, we need:

\[ -1 \leq \frac{x}{2} \leq 1 \]

Multiplying the entire inequality by 2 gives:

\[ -2 \leq x \leq 2 \]

### Step 2: Analyze \( \tan^{-1}(x/2 - 1) \)

The function \( \tan^{-1}(y) \) is defined for all real numbers \( y \). Thus, \( \tan^{-1}(x/2 - 1) \) does not impose any additional restrictions on \( x \).

### Step 3: Analyze \( \ln(\sqrt{x - \lfloor x \rfloor}) \)

The term \( \sqrt{x - \lfloor x \rfloor} \) represents the square root of the fractional part of \( x \). The fractional part \( x - \lfloor x \rfloor \) is defined as:

\[ 0 \leq x - \lfloor x \rfloor < 1 \]

For the logarithm \( \ln(y) \) to be defined, we require \( y > 0 \). Therefore, we need:

\[ \sqrt{x - \lfloor x \rfloor} > 0 \implies x - \lfloor x \rfloor > 0 \]

This means \( x \) cannot be an integer because if \( x \) is an integer, \( x - \lfloor x \rfloor = 0 \).

### Step 4: Combine the Conditions

From Step 1, we found that:

\[ -2 \leq x \leq 2 \]

From Step 3, we determined that \( x \) cannot be an integer. The integers within the interval \([-2, 2]\) are \(-2, -1, 0, 1, 2\).

Thus, we exclude these values from the interval:

The domain of \( f(x) \) is:

\[ [-2, 2] \setminus \{-2, -1, 0, 1, 2\} \]

This can be expressed in interval notation as:

\[ (-2, -1) \cup (-1, 0) \cup (0, 1) \cup (1, 2) \]

### Final Answer

The domain of the function \( f(x) \) is:

\[ (-2, -1) \cup (-1, 0) \cup (0, 1) \cup (1, 2) \]