`f(x)=sin^(-1)[e^(x)]+sin^(-1)[e^(-x)]` where [.] greatest integer function then

Options

A. (a) domain of `f(x) " is " (-log_(e)2,log_(e)2)`
B. (b) range of `f(x) ={pi}`
C. (c) Range of `f(x) " is " {(pi)/(2),pi}`
D. (d) `f(x) =cos^(-1)x` has only one solution

Explanation

To solve the problem \( f(x) = \sin^{-1}[\lfloor e^x \rfloor] + \sin^{-1}[\lfloor e^{-x} \rfloor] \), we will follow these steps:

### Step 1: Understand the Greatest Integer Function
The greatest integer function, denoted as \( \lfloor x \rfloor \), gives the largest integer less than or equal to \( x \). Thus, we need to evaluate \( \lfloor e^x \rfloor \) and \( \lfloor e^{-x} \rfloor \).

### Step 2: Determine the Range of \( e^x \) and \( e^{-x} \)
- As \( x \) varies over all real numbers, \( e^x \) ranges from \( 0 \) to \( \infty \).
- Consequently, \( e^{-x} \) ranges from \( \infty \) to \( 0 \).

### Step 3: Analyze \( \lfloor e^x \rfloor \) and \( \lfloor e^{-x} \rfloor \)
- For \( x < 0 \), \( e^x < 1 \) so \( \lfloor e^x \rfloor = 0 \).
- For \( x = 0 \), \( e^0 = 1 \) so \( \lfloor e^0 \rfloor = 1 \).
- For \( x > 0 \), \( e^x \) will be greater than \( 1 \) and will take integer values starting from \( 1 \).

- For \( e^{-x} \):
- For \( x < 0 \), \( e^{-x} > 1 \) so \( \lfloor e^{-x} \rfloor \) will take values starting from \( 1 \).
- For \( x = 0 \), \( e^{-0} = 1 \) so \( \lfloor e^{-0} \rfloor = 1 \).
- For \( x > 0 \), \( e^{-x} < 1 \) so \( \lfloor e^{-x} \rfloor = 0 \).

### Step 4: Evaluate \( f(x) \) for Different Intervals
1. **For \( x < 0 \):**
- \( \lfloor e^x \rfloor = 0 \)
- \( \lfloor e^{-x} \rfloor \geq 1 \)
- Thus, \( f(x) = \sin^{-1}(0) + \sin^{-1}(\lfloor e^{-x} \rfloor) = 0 + \sin^{-1}(1) = \frac{\pi}{2} \).

2. **For \( x = 0 \):**
- \( \lfloor e^0 \rfloor = 1 \)
- \( \lfloor e^{-0} \rfloor = 1 \)
- Thus, \( f(0) = \sin^{-1}(1) + \sin^{-1}(1) = \frac{\pi}{2} + \frac{\pi}{2} = \pi \).

3. **For \( x > 0 \):**
- \( \lfloor e^x \rfloor \geq 1 \)
- \( \lfloor e^{-x} \rfloor = 0 \)
- Thus, \( f(x) = \sin^{-1}(\lfloor e^x \rfloor) + \sin^{-1}(0) = \sin^{-1}(\lfloor e^x \rfloor) + 0 = \sin^{-1}(\lfloor e^x \rfloor) \).

### Step 5: Determine the Domain and Range
- **Domain:** The function \( f(x) \) is defined for all \( x \in \mathbb{R} \).
- **Range:**
- For \( x < 0 \), \( f(x) = \frac{\pi}{2} \).
- For \( x = 0 \), \( f(0) = \pi \).
- For \( x > 0 \), as \( e^x \) increases, \( \lfloor e^x \rfloor \) takes values starting from \( 1 \) and goes to \( \infty \), thus \( f(x) \) can take values from \( \frac{\pi}{2} \) to \( \frac{\pi}{2} + \frac{\pi}{2} \).

### Conclusion
- **Domain:** \( \mathbb{R} \)
- **Range:** \( \left[ \frac{\pi}{2}, \pi \right] \)