Let `x in R,` then `[x/3]+[(x+1)/3]+[(x+2)/3],` where [*] denotes the greatest integer function is equal to ...

Options

A. [x]
B. [x]+1
C. [x]-1
D. None of these

Explanation

To solve the problem, we need to evaluate the expression \([x/3] + [(x+1)/3] + [(x+2)/3]\) for different ranges of \(x\) where \([*]\) denotes the greatest integer function (also known as the floor function).

### Step-by-Step Solution:

1. **Understanding the Greatest Integer Function**:
The greatest integer function \([y]\) gives the largest integer less than or equal to \(y\). For example, \([2.5] = 2\) and \([3] = 3\).

2. **Analyzing the Expression**:
We need to evaluate the expression for different ranges of \(x\):
\[ f(x) = \left[\frac{x}{3}\right] + \left[\frac{x+1}{3}\right] + \left[\frac{x+2}{3}\right] \]

3. **Case 1: \(0 \leq x < 1\)**:
- \(\left[\frac{x}{3}\right] = 0\) (since \(0 \leq \frac{x}{3} < \frac{1}{3}\))
- \(\left[\frac{x+1}{3}\right] = 0\) (since \(0 \leq \frac{x+1}{3} < \frac{2}{3}\))
- \(\left[\frac{x+2}{3}\right] = 0\) (since \(0 \leq \frac{x+2}{3} < 1\))
- Therefore, \(f(x) = 0 + 0 + 0 = 0\).

4. **Case 2: \(1 \leq x < 2\)**:
- \(\left[\frac{x}{3}\right] = 0\) (since \(1/3 \leq \frac{x}{3} < 2/3\))
- \(\left[\frac{x+1}{3}\right] = 0\) (since \(2/3 \leq \frac{x+1}{3} < 1\))
- \(\left[\frac{x+2}{3}\right] = 1\) (since \(1 \leq \frac{x+2}{3} < 1.666...\))
- Therefore, \(f(x) = 0 + 0 + 1 = 1\).

5. **Case 3: \(2 \leq x < 3\)**:
- \(\left[\frac{x}{3}\right] = 0\) (since \(2/3 \leq \frac{x}{3} < 1\))
- \(\left[\frac{x+1}{3}\right] = 1\) (since \(1 \leq \frac{x+1}{3} < 1.333...\))
- \(\left[\frac{x+2}{3}\right] = 1\) (since \(1.333... \leq \frac{x+2}{3} < 2\))
- Therefore, \(f(x) = 0 + 1 + 1 = 2\).

6. **Case 4: \(3 \leq x < 4\)**:
- \(\left[\frac{x}{3}\right] = 1\) (since \(1 \leq \frac{x}{3} < 1.333...\))
- \(\left[\frac{x+1}{3}\right] = 1\) (since \(1.333... \leq \frac{x+1}{3} < 1.666...\))
- \(\left[\frac{x+2}{3}\right] = 2\) (since \(1.666... \leq \frac{x+2}{3} < 2\))
- Therefore, \(f(x) = 1 + 1 + 2 = 4\).

7. **General Pattern**:
From the cases above, we can summarize:
- For \(0 \leq x < 1\), \(f(x) = 0\)
- For \(1 \leq x < 2\), \(f(x) = 1\)
- For \(2 \leq x < 3\), \(f(x) = 2\)
- For \(3 \leq x < 4\), \(f(x) = 3\)
- This pattern continues such that \(f(x) = [x]\) for \(x \in \mathbb{R}\).

### Final Answer:
Thus, we conclude that:
\[ f(x) = [x] \]