The domain of `f(x)=sqrt(2{x}^2-3{x}+1),` where {.} denotes the fractional part in `[-1,1]` is `[-1,1]-(1/(2,1))` `[-1,-1/2]uu[(0,1)/2]uu{1}` `[-(1,1)/2]` (d) `[-1/2,1]`

Options

A. `[-1,1] ~((1)/(2),1)`
B. `[-1,-(1)/(2)] cup [0,(1)/(2)] cup {1}`
C. `[-1,(1)/(2)]`
D. `[-(1)/(2),1]`

Explanation

To find the domain of the function \( f(x) = \sqrt{2\{x^2\} - 3\{x\} + 1} \), where \(\{x\}\) denotes the fractional part of \(x\), we need to ensure that the expression inside the square root is non-negative.

### Step-by-Step Solution:

1. **Understanding the Fractional Part**:
The fractional part \(\{x\}\) is defined as:
\[ \{x\} = x - \lfloor x \rfloor \]
This means \(\{x\}\) is always in the interval \([0, 1)\).

2. **Setting Up the Inequality**:
We need to solve the inequality:
\[ 2\{x^2\} - 3\{x\} + 1 \geq 0 \]

3. **Finding the Expression for \(\{x^2\}\)**:
Since \(x\) can be expressed as \(n + \{x\}\) where \(n\) is an integer, we have:
\[ x^2 = (n + \{x\})^2 = n^2 + 2n\{x\} + \{x\}^2 \]
Thus, the fractional part \(\{x^2\}\) can be expressed as:
\[ \{x^2\} = x^2 - \lfloor x^2 \rfloor \]
However, for our inequality, we will focus on the values of \(\{x\}\) and \(\{x^2\}\) in the interval \([0, 1)\).

4. **Analyzing the Quadratic**:
The inequality can be rewritten as:
\[ 2\{x^2\} - 3\{x\} + 1 \geq 0 \]
This is a quadratic in terms of \(\{x\}\). We can find its roots by using the quadratic formula:
\[ \{x\} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} \]
\[ = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4} \]
This gives us the roots:
\[ \{x\} = 1 \quad \text{and} \quad \{x\} = \frac{1}{2} \]

5. **Finding Intervals**:
The quadratic opens upwards (since the coefficient of \(\{x\}^2\) is positive), so it will be non-negative outside the roots:
- The intervals for \(\{x\}\) are:
- \(\{x\} \leq \frac{1}{2}\)
- \(\{x\} \geq 1\) (not possible since \(\{x\} < 1\))

6. **Considering the Range of \(\{x\}\)**:
Since \(\{x\}\) is always in \([0, 1)\), we focus only on:
\[ 0 \leq \{x\} \leq \frac{1}{2} \]

7. **Finding Corresponding \(x\) Values**:
The fractional part \(\{x\} = 0\) when \(x\) is an integer. The fractional part \(\{x\} = \frac{1}{2}\) occurs at \(x = n + \frac{1}{2}\) where \(n\) is an integer. Thus, the valid \(x\) values are:
\[ x \in [n, n + \frac{1}{2}] \]
for any integer \(n\).

8. **Final Domain**:
Collecting all intervals, we find the domain of \(f(x)\) is:
\[ [-1, -\frac{1}{2}] \cup [0, \frac{1}{2}] \cup [1] \]

### Conclusion:
The domain of the function \( f(x) = \sqrt{2\{x^2\} - 3\{x\} + 1} \) is:
\[ [-1, -\frac{1}{2}] \cup [0, \frac{1}{2}] \cup \{1\} \]