If `f: R to R ` is defined by `f(x)=x-[x]-(1)/(2)` for all ` x in R `, where [x] denotes the greatest integer function, then `{x in R: f(x)=(1)/(2)}` is equal to

Options

A. Z
B. N
C. `phi`
D. R

Answer

To solve the problem, we need to analyze the function \( f(x) = x - [x] - \frac{1}{2} \), where \( [x] \) denotes the greatest integer function (also known as the floor function). We want to find the set of \( x \in \mathbb{R} \) such that \( f(x) = \frac{1}{2} \).

### Step-by-Step Solution:

1. **Understanding the Function**:
The function can be rewritten using the definition of the greatest integer function:
\[ f(x) = x - [x] - \frac{1}{2} \]
Here, \( [x] \) is the greatest integer less than or equal to \( x \).

2. **Expressing \( x \)**:
We know that any real number \( x \) can be expressed as:
\[ x = [x] + \{x\} \]
where \( \{x\} = x - [x] \) is the fractional part of \( x \). Thus, we can rewrite \( f(x) \) as:
\[ f(x) = \{x\} - \frac{1}{2} \]

3. **Setting Up the Equation**:
We want to find \( x \) such that:
\[ f(x) = \frac{1}{2} \]
Substituting our expression for \( f(x) \):
\[ \{x\} - \frac{1}{2} = \frac{1}{2} \]

4. **Solving for the Fractional Part**:
Rearranging the equation gives:
\[ \{x\} = \frac{1}{2} + \frac{1}{2} = 1 \]

5. **Analyzing the Fractional Part**:
The fractional part \( \{x\} \) is defined such that \( 0 \leq \{x\} < 1 \). This means that \( \{x\} = 1 \) is not possible for any real number \( x \).

6. **Conclusion**:
Since there are no real numbers \( x \) for which \( \{x\} = 1 \), we conclude that:
\[ \{ x \in \mathbb{R} : f(x) = \frac{1}{2} \} = \emptyset \]

### Final Answer:
The set \( \{ x \in \mathbb{R} : f(x) = \frac{1}{2} \} \) is equal to \( \emptyset \) (the empty set).