`if f(x) ={{:(2x-[x]+ xsin (x-[x]),,x ne 0) ,( 0,, x=0):}` where [.] denotes the greatest integer function then

Options

A. f(X) is differentiable at x=0
B. f(x) is differentiable at x=2
C. f(x) is continuous but not differentiable at x=0
D. none of these

Answer

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

\[ f(x) = \begin{cases} 2x - [x] + x \sin(x - [x]) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

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

### Step 1: Check continuity at \( x = 0 \)

To check continuity at \( x = 0 \), we need to evaluate the left-hand limit, right-hand limit, and the function value at \( x = 0 \).

1. **Calculate \( f(0) \)**:
\[ f(0) = 0 \]

2. **Calculate \( f(0^+) \)** (Right-hand limit):
- For \( x \) approaching \( 0 \) from the right (e.g., \( x = 0.1 \)):
\[ f(0^+) = 2(0) - [0] + 0 \cdot \sin(0 - [0]) = 0 - 0 + 0 = 0 \]

3. **Calculate \( f(0^-) \)** (Left-hand limit):
- For \( x \) approaching \( 0 \) from the left (e.g., \( x = -0.1 \)):
\[ f(0^-) = 2(0) - [-0.1] + 0 \cdot \sin(0 - [-0.1]) = 0 - (-1) + 0 = 1 \]

Now we compare the limits:
- \( f(0^+) = 0 \)
- \( f(0^-) = 1 \)
- \( f(0) = 0 \)

Since \( f(0^+) \neq f(0^-) \), the function is discontinuous at \( x = 0 \).

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

Next, we check continuity at \( x = 2 \).

1. **Calculate \( f(2) \)**:
\[ f(2) = 2(2) - [2] + 2 \sin(2 - [2]) = 4 - 2 + 2 \sin(0) = 4 - 2 + 0 = 2 \]

2. **Calculate \( f(2^+) \)** (Right-hand limit):
- For \( x \) approaching \( 2 \) from the right (e.g., \( x = 2.1 \)):
\[ f(2^+) = 2(2) - [2.1] + 2 \sin(2 - [2.1]) = 4 - 2 + 2 \sin(0) = 4 - 2 + 0 = 2 \]

3. **Calculate \( f(2^-) \)** (Left-hand limit):
- For \( x \) approaching \( 2 \) from the left (e.g., \( x = 1.9 \)):
\[ f(2^-) = 2(2) - [1.9] + 2 \sin(2 - [1.9]) = 4 - 1 + 2 \sin(0.1) = 4 - 1 + 2 \cdot 0.09983 \approx 3.19966 \]

Now we compare the limits:
- \( f(2^+) = 2 \)
- \( f(2^-) \approx 3.19966 \)
- \( f(2) = 2 \)

Since \( f(2^+) \neq f(2^-) \), the function is discontinuous at \( x = 2 \).

### Conclusion

Since the function is discontinuous at both \( x = 0 \) and \( x = 2 \), it is also non-differentiable at these points.

### Final Answer

The function \( f(x) \) is discontinuous at both \( x = 0 \) and \( x = 2 \).