`f(x)={(|2x-3|[x],0lexle2),(x^(2)/2,2ltxle3):}`. If f(x) is defined in (0,3) then ([.] denotes greatest integer function)

A. Point of discontiuity at `x=1, 1/2,2`
B. Point of discontinuity at x=1 and 2
C. Point of non-differentiability at `x=1, 1/2,2`
D. Point of non-differentiability at `x=1, 3/2, 2`

Options

A. Point of discontiuity at `x=1, 1/2,2`
B. Point of discontinuity at x=1 and 2
C. Point of non-differentiability at `x=1, 1/2,2`
D. Point of non-differentiability at `x=1, 3/2, 2`

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined in the intervals \( (0, 3) \). The function is given as:

\[ f(x) = \begin{cases} |2x - 3| [x] & \text{for } 0 \leq x < 2 \\ \frac{x^2}{2} & \text{for } 2 < x \leq 3 \end{cases} \]

Here, \([x]\) denotes the greatest integer function.

### Step 1: Analyze the first piece of the function \( f(x) \) for \( 0 \leq x < 2 \)

1. **For \( 0 \leq x < 1 \)**:
- The greatest integer function \([x] = 0\).
- Therefore, \( f(x) = |2x - 3| \cdot 0 = 0 \).

2. **For \( 1 \leq x < 2 \)**:
- The greatest integer function \([x] = 1\).
- Therefore, \( f(x) = |2x - 3| \cdot 1 = |2x - 3| \).

- Now, we need to evaluate \( |2x - 3| \):
- At \( x = 1 \): \( |2(1) - 3| = |2 - 3| = 1 \).
- At \( x = 2 \): \( |2(2) - 3| = |4 - 3| = 1 \).

- The expression \( |2x - 3| \) changes at \( x = 1.5 \):
- For \( 1 \leq x < 1.5 \): \( |2x - 3| = 3 - 2x \).
- For \( 1.5 < x < 2 \): \( |2x - 3| = 2x - 3 \).

### Step 2: Analyze the second piece of the function \( f(x) \) for \( 2 < x \leq 3 \)

- Here, \( f(x) = \frac{x^2}{2} \).
- At \( x = 2 \): \( f(2) = \frac{2^2}{2} = 2 \).
- At \( x = 3 \): \( f(3) = \frac{3^2}{2} = 4.5 \).

### Step 3: Identify points of discontinuity and non-differentiability

1. **At \( x = 1 \)**:
- Left limit: \( f(1^-) = 0 \).
- Right limit: \( f(1^+) = 1 \).
- Since \( f(1^-) \neq f(1^+) \), there is a discontinuity at \( x = 1 \).

2. **At \( x = 2 \)**:
- Left limit: \( f(2^-) = 1 \) (from the first piece).
- Right limit: \( f(2^+) = 2 \) (from the second piece).
- Since \( f(2^-) \neq f(2^+) \), there is a discontinuity at \( x = 2 \).

3. **At \( x = 1.5 \)**:
- The function changes from \( 3 - 2x \) to \( 2x - 3 \).
- The left-hand derivative is \( -2 \) and the right-hand derivative is \( 2 \).
- Since the left and right derivatives are not equal, \( f(x) \) is non-differentiable at \( x = 1.5 \).

### Conclusion

The points of discontinuity are at \( x = 1 \) and \( x = 2 \), and the point of non-differentiability is at \( x = 1, 1.5, 2 \).

Thus, the answer is:
- **Points of discontinuity**: \( x = 1, 2 \) (Option B).
- **Points of non-differentiability**: \( x = 1, 1.5, 2 \) (Option D).