Discuss the continuity and differentiability for `f(x) = [sin x]` when `x in [0, 2pi]`, where `[*]` denotes the greatest integer function x.

Answer

To discuss the continuity and differentiability of the function \( f(x) = [\sin x] \) where \( [\cdot] \) denotes the greatest integer function, we will analyze the function over the interval \( x \in [0, 2\pi] \).

### Step-by-Step Solution:

1. **Identify the Function Behavior**:
The function \( f(x) = [\sin x] \) takes the greatest integer value less than or equal to \( \sin x \). We know that \( \sin x \) oscillates between -1 and 1 for \( x \in [0, 2\pi] \).

2. **Evaluate \( f(x) \) at Key Points**:
- At \( x = 0 \): \( \sin(0) = 0 \) → \( f(0) = [0] = 0 \)
- At \( x = \frac{\pi}{2} \): \( \sin\left(\frac{\pi}{2}\right) = 1 \) → \( f\left(\frac{\pi}{2}\right) = [1] = 1 \)
- At \( x = \pi \): \( \sin(\pi) = 0 \) → \( f(\pi) = [0] = 0 \)
- At \( x = \frac{3\pi}{2} \): \( \sin\left(\frac{3\pi}{2}\right) = -1 \) → \( f\left(\frac{3\pi}{2}\right) = [-1] = -1 \)
- At \( x = 2\pi \): \( \sin(2\pi) = 0 \) → \( f(2\pi) = [0] = 0 \)

3. **Determine Intervals**:
- For \( x \in [0, \frac{\pi}{2}) \): \( \sin x \) varies from 0 to just below 1, thus \( f(x) = 0 \).
- For \( x \in (\frac{\pi}{2}, \pi) \): \( \sin x \) decreases from 1 to 0, thus \( f(x) = 0 \).
- For \( x \in (\pi, \frac{3\pi}{2}) \): \( \sin x \) varies from 0 to -1, thus \( f(x) = -1 \).
- For \( x \in (\frac{3\pi}{2}, 2\pi) \): \( \sin x \) increases from -1 to 0, thus \( f(x) = -1 \).

4. **Construct the Piecewise Function**:
We can summarize the function as:
\[ f(x) = \begin{cases} 0 & \text{for } x \in [0, \pi) \\ 1 & \text{for } x = \frac{\pi}{2} \\ -1 & \text{for } x \in (\pi, 2\pi) \\ 0 & \text{for } x = 2\pi \end{cases} \]

5. **Check for Continuity**:
- At \( x = 0 \): \( f(0) = 0 \) (continuous).
- At \( x = \frac{\pi}{2} \): \( f\left(\frac{\pi}{2}\right) = 1 \) but \( \lim_{x \to \frac{\pi}{2}^-} f(x) = 0 \) and \( \lim_{x \to \frac{\pi}{2}^+} f(x) = 0 \) (discontinuous).
- At \( x = \pi \): \( f(\pi) = 0 \) but \( \lim_{x \to \pi^-} f(x) = 0 \) and \( \lim_{x \to \pi^+} f(x) = -1 \) (discontinuous).
- At \( x = \frac{3\pi}{2} \): \( f\left(\frac{3\pi}{2}\right) = -1 \) but \( \lim_{x \to \frac{3\pi}{2}^-} f(x) = 0 \) and \( \lim_{x \to \frac{3\pi}{2}^+} f(x) = -1 \) (discontinuous).
- At \( x = 2\pi \): \( f(2\pi) = 0 \) but \( \lim_{x \to 2\pi^-} f(x) = -1 \) (discontinuous).

6. **Conclusion on Continuity**:
The function \( f(x) \) is discontinuous at \( x = \frac{\pi}{2}, \pi, \frac{3\pi}{2}, \) and \( 2\pi \).

7. **Check for Differentiability**:
Since \( f(x) \) is discontinuous at multiple points, it cannot be differentiable at those points. Therefore, \( f(x) \) is non-differentiable at \( x = \frac{\pi}{2}, \pi, \frac{3\pi}{2}, \) and \( 2\pi \).

### Final Answer:
The function \( f(x) = [\sin x] \) is discontinuous at \( x = \frac{\pi}{2}, \pi, \frac{3\pi}{2}, \) and \( 2\pi \), and it is non-differentiable at these points.