If `f(x)={|1-4x^2|,0lt=x<1 and [x^2-2x],1lt=x<2` where [.] denotes the greatest integer function, then

Options

A. differentiable for all x
B. continnous at x=1
C. non- differntiable at x=1
D. none of these

Answer

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

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

### Step 1: Analyze the first part of the function \( f(x) = |1 - 4x^2| \)

For \( 0 < x < 1 \):
- We need to find when \( 1 - 4x^2 = 0 \).
- Solving \( 1 - 4x^2 = 0 \) gives:
\[ 4x^2 = 1 \implies x^2 = \frac{1}{4} \implies x = \frac{1}{2} \quad (\text{since } x > 0) \]
- Thus, \( f(x) \) changes at \( x = \frac{1}{2} \).

**Evaluate \( f(x) \) in the intervals:**
- For \( 0 < x < \frac{1}{2} \):
\[ 1 - 4x^2 > 0 \implies f(x) = 1 - 4x^2 \]
- For \( \frac{1}{2} < x < 1 \):
\[ 1 - 4x^2 < 0 \implies f(x) = -(1 - 4x^2) = 4x^2 - 1 \]

### Step 2: Check continuity at \( x = 1 \)

For \( x = 1 \):
- Left-hand limit as \( x \to 1^- \):
\[ f(1^-) = 4(1)^2 - 1 = 4 - 1 = 3 \]
- Right-hand limit as \( x \to 1^+ \):
\[ f(1^+) = [1^2 - 2(1)] = [1 - 2] = [-1] = -1 \]
- Since \( \lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x) \), \( f(x) \) is not continuous at \( x = 1 \).

### Step 3: Analyze the second part of the function \( f(x) = [x^2 - 2x] \)

For \( 1 < x < 2 \):
- Calculate \( x^2 - 2x \):
\[ x^2 - 2x = x(x - 2) \]
- At \( x = 1 \):
\[ f(1) = [1^2 - 2(1)] = [1 - 2] = [-1] \]
- At \( x = 2 \):
\[ f(2) = [2^2 - 2(2)] = [4 - 4] = [0] = 0 \]

### Step 4: Check differentiability at \( x = 1 \)

- Since \( f(x) \) is not continuous at \( x = 1 \), it cannot be differentiable there.

### Conclusion

- \( f(x) \) is not continuous at \( x = 1 \).
- \( f(x) \) is differentiable at \( x = \frac{1}{2} \) but not at \( x = 1 \).

### Final Answer
The correct option is that \( f(x) \) is **not continuous at \( x = 1 \)** and **not differentiable at \( x = 1 \)**.