The domain of the function `f (x) = sin^(-1)((1+x^3)/(2x^(3/2)))+sqrt(sin(sinx))+log_(3{x}+1)(x^2+1)`, where {.} represents fractional part function, is:

Options

A. `xepsilon{1}`
B. `xepsilonR-{1,-1}`
C. `xgt3,x!=1`
D. `xepsilon phi`

Explanation

To find the domain of the function

\[ f(x) = \sin^{-1}\left(\frac{1+x^3}{2x^{3/2}}\right) + \sqrt{\sin(\sin x)} + \log_{3\{x\}+1}(x^2 + 1) \]

we need to analyze each component of the function separately.

### Step 1: Analyze the Inverse Sine Function

The first part of the function is

\[ \sin^{-1}\left(\frac{1+x^3}{2x^{3/2}}\right) \]

For the inverse sine function to be defined, the argument must be in the range \([-1, 1]\). Therefore, we need:

\[ -1 \leq \frac{1+x^3}{2x^{3/2}} \leq 1 \]

**Sub-step 1a: Solve the left inequality**

\[ \frac{1+x^3}{2x^{3/2}} \geq -1 \]

This simplifies to:

\[ 1 + x^3 \geq -2x^{3/2} \]

or

\[ x^3 + 2x^{3/2} + 1 \geq 0 \]

Since \(x^3 + 2x^{3/2} + 1\) is a sum of non-negative terms for \(x \geq 0\), this inequality holds for all \(x \geq 0\).

**Sub-step 1b: Solve the right inequality**

\[ \frac{1+x^3}{2x^{3/2}} \leq 1 \]

This simplifies to:

\[ 1 + x^3 \leq 2x^{3/2} \]

or

\[ x^3 - 2x^{3/2} + 1 \leq 0 \]

Let \(y = x^{3/2}\), then \(x = y^{2/3}\). The inequality becomes:

\[ y^{2} - 2y + 1 \leq 0 \]

This factors to:

\[ (y-1)^2 \leq 0 \]

This implies \(y = 1\) or \(x = 1\). Thus, the only solution for this part is \(x = 1\).

### Step 2: Analyze the Square Root Function

The second part of the function is

\[ \sqrt{\sin(\sin x)} \]

For the square root to be defined, we require:

\[ \sin(\sin x) \geq 0 \]

Since \(\sin x\) is always between \(-1\) and \(1\), \(\sin(\sin x)\) is non-negative when \(\sin x \in [0, \pi]\). Therefore, \(x\) must be in the interval:

\[ [0, \pi] \]

### Step 3: Analyze the Logarithmic Function

The third part of the function is

\[ \log_{3\{x\}+1}(x^2 + 1) \]

For the logarithm to be defined, we need:

1. \(3\{x\} + 1 > 0\)
2. \(x^2 + 1 > 0\)

The second condition is always satisfied for all \(x\). The first condition requires that \(3\{x\} + 1 > 0\), which implies:

\[ \{x\} \neq 0 \]

This means \(x\) cannot be an integer.

### Step 4: Combine All Conditions

From the analysis, we have:

1. \(x = 1\) from the inverse sine function.
2. \(x \in [0, \pi]\) from the square root function.
3. \(x \in \mathbb{R} \setminus \mathbb{Z}\) from the logarithmic function.

The only value that satisfies all these conditions is \(x = 1\), which is not an integer.

### Conclusion

Thus, the domain of the function \(f(x)\) is:

\[ \boxed{\{1\}} \]