Let `f(x)=x/(1+x^2) and g(x)=(e^(-x))/(1+[x])` (where [.] denote greatest integer function), then

Options

A. `f(x)` is monotonically decreasing in `[1,3.8]`
B. `f(x)` is monotonically increasing i `[1,3.8]`
C. the greatest value of `f(x)` is `1/3xx14.44`
D. the least value of `f(x)` is `2`

Explanation

To solve the problem, we need to analyze the functions \( f(x) \) and \( g(x) \) and determine their properties, including domain and range.

### Step 1: Identify the functions
We have:
- \( f(x) = \frac{x}{1 + x^2} \)
- \( g(x) = \frac{e^{-x}}{1 + [x]} \) (where \([x]\) is the greatest integer function)

### Step 2: Determine the domain of \( f(x) \)
The function \( f(x) \) is defined for all real numbers \( x \) since the denominator \( 1 + x^2 \) is never zero. Therefore, the domain of \( f(x) \) is:
\[ \text{Domain of } f(x) = \mathbb{R} \]

### Step 3: Determine the range of \( f(x) \)
To find the range of \( f(x) \), we analyze its behavior:
- As \( x \to \infty \), \( f(x) \to 0 \)
- As \( x \to -\infty \), \( f(x) \to 0 \)
- At \( x = 0 \), \( f(0) = 0 \)

To find critical points, we differentiate \( f(x) \):
\[ f'(x) = \frac{(1 + x^2)(1) - x(2x)}{(1 + x^2)^2} = \frac{1 + x^2 - 2x^2}{(1 + x^2)^2} = \frac{1 - x^2}{(1 + x^2)^2} \]
Setting \( f'(x) = 0 \) gives \( 1 - x^2 = 0 \) or \( x = \pm 1 \).

Evaluating \( f(x) \) at these points:
- \( f(1) = \frac{1}{2} \)
- \( f(-1) = -\frac{1}{2} \)

Thus, the maximum value is \( \frac{1}{2} \) and the minimum value is \( -\frac{1}{2} \). Therefore, the range of \( f(x) \) is:
\[ \text{Range of } f(x) = \left[-\frac{1}{2}, \frac{1}{2}\right] \]

### Step 4: Determine the domain of \( g(x) \)
The function \( g(x) \) is defined for \( 1 + [x] \neq 0 \). This means:
\[ [x] \neq -1 \implies [x] \geq 0 \text{ or } [x] \leq -2 \]
Thus, \( g(x) \) is defined for all \( x \) except when \( x \) is in the interval \([-1, 0)\).

### Step 5: Determine the range of \( g(x) \)
As \( x \to \infty \), \( g(x) \to 0 \). As \( x \to -\infty \), \( g(x) \to 0 \) as well. The function \( g(x) \) is positive for all \( x \) where it is defined. Therefore, we need to evaluate \( g(x) \) at points where \( [x] \) takes integer values.

For \( [x] = n \) (where \( n \) is an integer):
\[ g(x) = \frac{e^{-x}}{1 + n} \]
This shows that \( g(x) \) can take values in the range \( (0, \infty) \) depending on the value of \( n \).

### Conclusion
The domain of \( f(x) \) is \( \mathbb{R} \) and the range is \( \left[-\frac{1}{2}, \frac{1}{2}\right] \). The domain of \( g(x) \) is \( \mathbb{R} \setminus [-1, 0) \) and the range is \( (0, \infty) \).

### Final Answer
- Domain of \( f(x) \): \( \mathbb{R} \)
- Range of \( f(x) \): \( \left[-\frac{1}{2}, \frac{1}{2}\right] \)
- Domain of \( g(x) \): \( \mathbb{R} \setminus [-1, 0) \)
- Range of \( g(x) \): \( (0, \infty) \)