Let `f(x)=sgn(cot^(-1)x)+tan(pi/2[x]),` where `[x]` is the greatest integer function less than or equal to `x ,` then which of the following alternatives is/are true? `f(x)` is many-one but not an even function. `f(x)` is a periodic function. `f(x)` is a bounded function. The graph of `f(x)` remains above the x-axis.

Options

A. `f(x)` is many-one but not an even function.
B. `f(x)` is a periodic function.
C. `f(x)` is a bounded function.
D. The graph of `f(x)` remains above the x-axis.

Explanation

To analyze the function \( f(x) = \text{sgn}(\cot^{-1} x) + \tan\left(\frac{\pi}{2} [x]\right) \), where \([x]\) is the greatest integer function, we will evaluate each statement provided in the question step by step.

### Step 1: Determine the Domain of \( f(x) \)

1. **Understanding the components**:
- The function \( \cot^{-1}(x) \) is defined for all real \( x \).
- The function \( \tan\left(\frac{\pi}{2} [x]\right) \) is defined except when \( \frac{\pi}{2} [x] = (2n + 1)\frac{\pi}{2} \) for \( n \in \mathbb{Z} \). This occurs when \([x] = 2n + 1\).

2. **Finding the domain**:
- Thus, \([x]\) must not be an odd integer. Therefore, the domain of \( f(x) \) consists of intervals where \([x]\) is even, which can be expressed as:
\[ x \in [2n, 2n + 1) \quad \text{for } n \in \mathbb{Z} \]

### Step 2: Evaluate \( f(x) \)

1. **Evaluating \( \text{sgn}(\cot^{-1} x) \)**:
- Since \( \cot^{-1}(x) \) is always positive for \( x > 0 \) and zero for \( x = 0 \), we have:
\[ \text{sgn}(\cot^{-1} x) = \begin{cases} 1 & \text{if } x > 0 \\ 0 & \text{if } x = 0 \\ -1 & \text{if } x < 0 \end{cases} \]

2. **Evaluating \( \tan\left(\frac{\pi}{2} [x]\right) \)**:
- For even integers \([x] = 2n\), \( \tan\left(\frac{\pi}{2} [x]\right) = \tan(n\pi) = 0\).
- Therefore, for \( x \) in the domain:
- If \( x > 0 \): \( f(x) = 1 + 0 = 1 \)
- If \( x = 0 \): \( f(x) = 0 + 0 = 0 \)
- If \( x < 0 \): \( f(x) = -1 + 0 = -1 \)

### Step 3: Analyze the Properties of \( f(x) \)

1. **Many-one but not an even function**:
- Since \( f(x) \) takes the same value for all \( x \) in the intervals \( [2n, 2n+1) \) for \( n \geq 0 \) and \( f(x) \) is not symmetric about the y-axis, it is many-one but not an even function.

2. **Periodic function**:
- The function \( f(x) \) is periodic with period 2, as it repeats its values every 2 units.

3. **Bounded function**:
- The range of \( f(x) \) is limited to values between -1 and 1, thus it is bounded.

4. **Graph remains above the x-axis**:
- The function \( f(x) \) takes values of -1, 0, or 1. Thus, it does not always remain above the x-axis since it can take the value -1.

### Conclusion

Based on the analysis:
- \( f(x) \) is many-one but not an even function. **True**
- \( f(x) \) is a periodic function. **True**
- \( f(x) \) is a bounded function. **True**
- The graph of \( f(x) \) remains above the x-axis. **False**

### Final Result
The correct statements are:
- \( f(x) \) is many-one but not an even function.
- \( f(x) \) is a periodic function.
- \( f(x) \) is a bounded function.