The number of values of x such that `x, [x] and {x}` are in arithmetic progression is equal to (where `[.]` denotes the greatest integer function and `{.}` denotes the fractional part function)

Options

A. 0
B. 1
C. 2
D. 4

Answer

To solve the problem, we need to find the number of values of \( x \) such that \( x \), \( [x] \), and \( \{x\} \) are in arithmetic progression. Here, \( [x] \) denotes the greatest integer function (floor function) and \( \{x\} \) denotes the fractional part function.

### Step-by-Step Solution:

1. **Understanding the Functions**:
- Let \( x = n + f \), where \( n = [x] \) (the greatest integer part of \( x \)) and \( f = \{x\} \) (the fractional part of \( x \)). Here, \( n \) is an integer and \( f \) satisfies \( 0 \leq f < 1 \).

2. **Setting Up the Arithmetic Progression**:
- For \( x \), \( [x] \), and \( \{x\} \) to be in arithmetic progression, we must have:
\[ 2[x] = x + \{x\} \]
- Substituting \( x = n + f \) into the equation gives:
\[ 2n = (n + f) + f \]
- Simplifying this, we get:
\[ 2n = n + 2f \implies n = 2f \]

3. **Finding the Range of \( f \)**:
- Since \( f \) is the fractional part, we know \( 0 \leq f < 1 \).
- From \( n = 2f \), we can express \( f \) in terms of \( n \):
\[ f = \frac{n}{2} \]
- Therefore, we have:
\[ 0 \leq \frac{n}{2} < 1 \implies 0 \leq n < 2 \]

4. **Determining Possible Values of \( n \)**:
- The integer \( n \) can take values in the range \( [0, 2) \). Thus, the possible integer values for \( n \) are:
- \( n = 0 \)
- \( n = 1 \)

5. **Calculating Corresponding Values of \( x \)**:
- For \( n = 0 \):
\[ f = \frac{0}{2} = 0 \implies x = 0 + 0 = 0 \]
- For \( n = 1 \):
\[ f = \frac{1}{2} \implies x = 1 + \frac{1}{2} = \frac{3}{2} \]

6. **Conclusion**:
- The values of \( x \) that satisfy the condition are \( 0 \) and \( \frac{3}{2} \).
- Therefore, the total number of values of \( x \) is **2**.

### Final Answer:
The number of values of \( x \) such that \( x, [x], \) and \( \{x\} \) are in arithmetic progression is **2**.