If `f(x) = [x] [sinx]` in `(-1,1)` then f(x) is

Options

A. continuous on (-1,0)
B. differentiable on (-1,1)
C. differentiable at x=0
D. none of these

Explanation

To determine the properties of the function \( f(x) = [x][\sin x] \) in the interval \((-1, 1)\), where \([x]\) denotes the greatest integer function (floor function), we will analyze the function step by step.

### Step 1: Define the function in the given intervals
We need to analyze the function in two cases based on the intervals defined by the greatest integer function.

1. **Case 1:** \( -1 < x < 0 \)
2. **Case 2:** \( 0 \leq x < 1 \)

### Step 2: Evaluate \( f(x) \) in Case 1
In the interval \( -1 < x < 0 \):
- The greatest integer function \([x] = -1\) because \(x\) is negative and less than 0.
- The sine function \(\sin x\) is also negative in this interval, and since \(-1 < \sin x < 0\), we have \([\sin x] = -1\).

Thus, we can calculate:
\[ f(x) = [x][\sin x] = (-1)(-1) = 1 \]
for \( -1 < x < 0 \).

### Step 3: Evaluate \( f(x) \) in Case 2
In the interval \( 0 \leq x < 1 \):
- The greatest integer function \([x] = 0\) because \(x\) is non-negative and less than 1.
- The sine function \(\sin x\) is positive in this interval, and since \(0 < \sin x < 1\), we have \([\sin x] = 0\).

Thus, we can calculate:
\[ f(x) = [x][\sin x] = (0)(0) = 0 \]
for \( 0 \leq x < 1 \).

### Step 4: Summary of the function
From the above evaluations, we summarize:
- For \( -1 < x < 0 \), \( f(x) = 1 \)
- For \( 0 \leq x < 1 \), \( f(x) = 0 \)

### Step 5: Check continuity at \( x = 0 \)
To check continuity at \( x = 0 \):
- The left-hand limit as \( x \) approaches 0 from the left (\( x \to 0^- \)) is:
\[ \lim_{x \to 0^-} f(x) = 1 \]
- The right-hand limit as \( x \) approaches 0 from the right (\( x \to 0^+ \)) is:
\[ \lim_{x \to 0^+} f(x) = 0 \]
Since the left-hand limit and right-hand limit are not equal:
\[ \lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x) \]
Thus, \( f(x) \) is discontinuous at \( x = 0 \).

### Step 6: Check differentiability
Since \( f(x) \) is discontinuous at \( x = 0 \), it cannot be differentiable at that point. However, \( f(x) \) is constant in the intervals \( (-1, 0) \) and \( (0, 1) \), hence it is differentiable in those intervals.

### Conclusion
- \( f(x) \) is continuous on \( (-1, 0) \) and \( (0, 1) \) but not at \( x = 0 \).
- \( f(x) \) is differentiable on \( (-1, 0) \) and \( (0, 1) \) but not at \( x = 0 \).

### Final Answer
- Option A: Continuous on \((-1, 0)\) - **True**
- Option B: Differentiable on \((-1, 1)\) - **False**
- Option C: Differentiable at \( x = 0 \) - **False**
- Option D: None of these - **False**