Let [t] d enote the greatest integer `lt t`. The number of points where the function `f(x) = [x]abs(x^2-1) + sin (pi/([x] + 3)) - [x + 1], x in (-2, 2)` is not continuous is _________.

Answer

To determine the number of points where the function \( f(x) = [x] |x^2 - 1| + \sin\left(\frac{\pi}{[x] + 3}\right) - [x + 1] \) is not continuous for \( x \in (-2, 2) \), we will analyze the function step by step.

### Step 1: Analyze the Greatest Integer Function
The greatest integer function \( [x] \) takes the largest integer less than or equal to \( x \). For the interval \( (-2, 2) \), the values of \( [x] \) can be:
- For \( x \in (-2, -1) \), \( [x] = -2 \)
- For \( x \in [-1, 0) \), \( [x] = -1 \)
- For \( x \in [0, 1) \), \( [x] = 0 \)
- For \( x \in [1, 2) \), \( [x] = 1 \)

### Step 2: Analyze the Modulus Function
The function \( |x^2 - 1| \) changes its behavior at \( x = -1 \) and \( x = 1 \):
- For \( x < -1 \) or \( x > 1 \), \( |x^2 - 1| = x^2 - 1 \)
- For \( -1 < x < 1 \), \( |x^2 - 1| = -(x^2 - 1) = 1 - x^2 \)

### Step 3: Break Down the Function into Intervals
We will analyze the function in the intervals determined by the values of \( [x] \):
1. **Interval \( (-2, -1) \)**:
- \( [x] = -2 \)
- \( f(x) = -2(x^2 - 1) + \sin\left(\frac{\pi}{-2 + 3}\right) - (-2 + 1) \)
- Simplifies to \( f(x) = -2x^2 + 2 + \sin\left(\frac{\pi}{1}\right) + 1 \)

2. **Interval \( [-1, 0) \)**:
- \( [x] = -1 \)
- \( f(x) = -1(1 - x^2) + \sin\left(\frac{\pi}{-1 + 3}\right) - 0 \)
- Simplifies to \( f(x) = -1 + x^2 + \sin\left(\frac{\pi}{2}\right) \)

3. **Interval \( [0, 1) \)**:
- \( [x] = 0 \)
- \( f(x) = 0(1 - x^2) + \sin\left(\frac{\pi}{0 + 3}\right) - 1 \)
- Simplifies to \( f(x) = \sin\left(\frac{\pi}{3}\right) - 1 \)

4. **Interval \( [1, 2) \)**:
- \( [x] = 1 \)
- \( f(x) = 1(x^2 - 1) + \sin\left(\frac{\pi}{1 + 3}\right) - 2 \)
- Simplifies to \( f(x) = x^2 - 1 + \sin\left(\frac{\pi}{4}\right) - 2 \)

### Step 4: Check Continuity at Critical Points
We need to check continuity at the points where \( [x] \) changes, which are \( x = -1, 0, 1 \).

1. **At \( x = -1 \)**:
- Left-hand limit: \( \lim_{x \to -1^-} f(x) = -2(-1)^2 + 2 + 0 = 1 \)
- Right-hand limit: \( \lim_{x \to -1^+} f(x) = -1 + 1 + 1 = 1 \)
- \( f(-1) = 1 \)
- **Continuous at \( x = -1 \)**.

2. **At \( x = 0 \)**:
- Left-hand limit: \( \lim_{x \to 0^-} f(x) = -1 + 0 + 1 = 0 \)
- Right-hand limit: \( \lim_{x \to 0^+} f(x) = \sin\left(\frac{\pi}{3}\right) - 1 = \frac{\sqrt{3}}{2} - 1 \)
- **Not continuous at \( x = 0 \)**.

3. **At \( x = 1 \)**:
- Left-hand limit: \( \lim_{x \to 1^-} f(x) = \frac{\sqrt{3}}{2} - 1 \)
- Right-hand limit: \( \lim_{x \to 1^+} f(x) = 1 - 1 + \frac{1}{\sqrt{2}} - 2 = -1 + \frac{1}{\sqrt{2}} \)
- **Not continuous at \( x = 1 \)**.

### Conclusion
The function \( f(x) \) is not continuous at \( x = 0 \) and \( x = 1 \). Therefore, the number of points where the function is not continuous is **2**.