If `f(x) = {{:(|1-4x^(2)|",",0 le x lt 1),([x^(2)-2x]",",1 le x lt 2):}`, where [] denotes the greatest integer function, then

A. f(x) is continuous for all `x in [0, 2)`
B. f(x) is differentiable for all `x in [0, 2) - {1}`
C. f(X) is differentiable for all `x in [0, 2)-{(1)/(2),1}`
D. None of these

Options

A. f(x) is continuous for all `x in [0, 2)`
B. f(x) is differentiable for all `x in [0, 2) - {1}`
C. f(X) is differentiable for all `x in [0, 2)-{(1)/(2),1}`
D. None of these

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined in two parts:

1. For \( 0 \leq x < 1 \): \( f(x) = |1 - 4x^2| \)
2. For \( 1 \leq x < 2 \): \( f(x) = [x^2 - 2x] \) where \( [ ] \) denotes the greatest integer function.

### Step 1: Check Continuity at \( x = 1 \)

To check if \( f(x) \) is continuous at \( x = 1 \), we need to evaluate the left-hand limit, right-hand limit, and the function value at that point.

- **Left-hand limit as \( x \to 1^- \)**:
\[ f(1^-) = |1 - 4(1)^2| = |1 - 4| = | -3 | = 3 \]

- **Right-hand limit as \( x \to 1^+ \)**:
\[ f(1^+) = [1^2 - 2(1)] = [1 - 2] = [-1] = -1 \]

- **Function value at \( x = 1 \)**:
\[ f(1) = [1^2 - 2(1)] = [1 - 2] = [-1] = -1 \]

Since \( f(1^-) = 3 \) and \( f(1^+) = -1 \), we see that:
\[ f(1^-) \neq f(1^+) \]
Thus, \( f(x) \) is not continuous at \( x = 1 \).

### Step 2: Check Differentiability in the intervals

#### For \( 0 \leq x < 1 \):
The function \( f(x) = |1 - 4x^2| \) is differentiable except where the expression inside the absolute value is zero.

- Set \( 1 - 4x^2 = 0 \):
\[ 4x^2 = 1 \implies x^2 = \frac{1}{4} \implies x = \frac{1}{2} \]

Thus, \( f(x) \) is not differentiable at \( x = \frac{1}{2} \).

#### For \( 1 \leq x < 2 \):
The function \( f(x) = [x^2 - 2x] \) is a piecewise constant function, and it is differentiable everywhere except at points where the greatest integer function changes value.

- Calculate \( x^2 - 2x \) at the endpoints:
- At \( x = 1 \): \( f(1) = -1 \)
- At \( x = 2 \): \( f(2) = 0 \)

The function \( x^2 - 2x \) is continuous and differentiable in the interval \( [1, 2) \) but not at \( x = 1 \) because of the transition from the first piece to the second piece.

### Conclusion:
- \( f(x) \) is not continuous at \( x = 1 \).
- \( f(x) \) is not differentiable at \( x = \frac{1}{2} \) and also at \( x = 1 \).

### Final Answer:
- A. False: \( f(x) \) is not continuous for all \( x \in [0, 2) \).
- B. False: \( f(x) \) is not differentiable at \( x = 1 \).
- C. True: \( f(x) \) is not differentiable at \( x = \frac{1}{2} \) and \( x = 1 \).
- D. False: One of the options is correct.

Thus, the correct option is **C**.