The function `f(x)=sgn(x-1). Cot^(-1)[x-1]` is `([x] "denotes greatest integer function of x")`

Options

A. Discontinuous at `x=1`
B. Continuous and differentiable at `x=1`
C. Not defined at `x=1`
D. Continuous but not differentiable at `x=1`

Explanation

To solve the problem, we need to analyze the function \( f(x) = \text{sgn}(x-1) \cdot \cot^{-1}(x-1) \) and determine its continuity, particularly at the point \( x = 1 \).

### Step-by-Step Solution:

1. **Understanding the Signum Function**:
The signum function \( \text{sgn}(x-1) \) is defined as:
- \( 1 \) if \( x > 1 \)
- \( 0 \) if \( x = 1 \)
- \( -1 \) if \( x < 1 \)

2. **Evaluating the Function at \( x = 1 \)**:
We first find \( f(1) \):
\[ f(1) = \text{sgn}(1-1) \cdot \cot^{-1}(1-1) = \text{sgn}(0) \cdot \cot^{-1}(0) = 0 \cdot \frac{\pi}{2} = 0 \]

3. **Finding the Right-Hand Limit as \( x \to 1^+ \)**:
We calculate the limit from the right:
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \text{sgn}(x-1) \cdot \cot^{-1}(x-1) \]
For \( x > 1 \), \( \text{sgn}(x-1) = 1 \):
\[ \lim_{x \to 1^+} f(x) = 1 \cdot \cot^{-1}(0) = 1 \cdot \frac{\pi}{2} = \frac{\pi}{2} \]

4. **Finding the Left-Hand Limit as \( x \to 1^- \)**:
Now we calculate the limit from the left:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \text{sgn}(x-1) \cdot \cot^{-1}(x-1) \]
For \( x < 1 \), \( \text{sgn}(x-1) = -1 \):
\[ \lim_{x \to 1^-} f(x) = -1 \cdot \cot^{-1}(0) = -1 \cdot \frac{\pi}{2} = -\frac{\pi}{2} \]

5. **Conclusion on Continuity**:
We compare the function value and the limits:
- \( f(1) = 0 \)
- \( \lim_{x \to 1^+} f(x) = \frac{\pi}{2} \)
- \( \lim_{x \to 1^-} f(x) = -\frac{\pi}{2} \)

Since \( f(1) \), \( \lim_{x \to 1^+} f(x) \), and \( \lim_{x \to 1^-} f(x) \) are not equal, the function \( f(x) \) is discontinuous at \( x = 1 \).

### Final Answer:
The function \( f(x) \) is discontinuous at \( x = 1 \).