Let `f(x) = {{:(1-x ,"If" 0 le x le 1),(0, "if" 1 lt x le 2),((2-x)^(2),"if"2 lt x le 3):}` and function `F(x) = int_(0)^(x)f(t) dt`. If number of points of discountinuity in [0,3] and non-differentiablity in (0,2) and `F(x)` are `alpha` and `beta` respectively , then `(alpha - beta)` is equal to .

Answer

To solve the problem, we need to analyze the function \( f(x) \) and the function \( F(x) \) defined as follows:

1. **Function Definition**:
\[ f(x) = \begin{cases} 1 - x & \text{if } 0 \leq x \leq 1 \\ 0 & \text{if } 1 < x \leq 2 \\ (2 - x)^2 & \text{if } 2 < x \leq 3 \end{cases} \]

2. **Function \( F(x) \)**:
\[ F(x) = \int_{0}^{x} f(t) \, dt \]

### Step 1: Determine Points of Discontinuity of \( f(x) \)

To find the points of discontinuity, we need to check the behavior of \( f(x) \) at the boundaries of the intervals defined in the piecewise function.

- **At \( x = 1 \)**:
- Left-hand limit: \( \lim_{x \to 1^-} f(x) = 1 - 1 = 0 \)
- Right-hand limit: \( \lim_{x \to 1^+} f(x) = 0 \)
- Since both limits are equal, \( f(x) \) is continuous at \( x = 1 \).

- **At \( x = 2 \)**:
- Left-hand limit: \( \lim_{x \to 2^-} f(x) = 0 \)
- Right-hand limit: \( \lim_{x \to 2^+} f(x) = (2 - 2)^2 = 0 \)
- Since both limits are equal, \( f(x) \) is continuous at \( x = 2 \).

Thus, there are **no points of discontinuity** in the interval \([0, 3]\).

### Step 2: Determine Points of Non-Differentiability of \( F(x) \)

Next, we check where \( F(x) \) is non-differentiable in the interval \( (0, 2) \).

Using the Fundamental Theorem of Calculus:
\[ F'(x) = f(x) \]

We will check the differentiability at the points \( x = 1 \) and \( x = 2 \):

- **At \( x = 1 \)**:
- Left-hand derivative: \( \lim_{x \to 1^-} F'(x) = \lim_{x \to 1^-} (1 - x) = 0 \)
- Right-hand derivative: \( \lim_{x \to 1^+} F'(x) = \lim_{x \to 1^+} 0 = 0 \)
- Since both derivatives are equal, \( F(x) \) is differentiable at \( x = 1 \).

- **At \( x = 2 \)**:
- Left-hand derivative: \( \lim_{x \to 2^-} F'(x) = 0 \)
- Right-hand derivative: \( \lim_{x \to 2^+} F'(x) = \lim_{x \to 2^+} (2 - x)^2 = 0 \)
- Since both derivatives are equal, \( F(x) \) is differentiable at \( x = 2 \).

Thus, there are **no points of non-differentiability** in the interval \( (0, 2) \).

### Conclusion

- The number of points of discontinuity \( \alpha = 0 \).
- The number of points of non-differentiability \( \beta = 0 \).

Therefore, the final result is:
\[ \alpha - \beta = 0 - 0 = 0 \]