Range of `f(x)=[1/(log(x^(2)+e))]+1/sqrt(1+x^(2)),` where `[*]` denotes greatest integer function, is

Options

A. `(0,(e+1)/e) cup {2}`
B. (0,1)
C. `(0,1] cup {2}`
D. `(0,1) cup {2}`

Answer

To find the range of the function \( f(x) = \left[ \frac{1}{\log(x^2 + e)} \right] + \frac{1}{\sqrt{1 + x^2}} \), where \([\cdot]\) denotes the greatest integer function, we will analyze the two components of the function separately.

### Step 1: Analyze \( f_1(x) = \frac{1}{\log(x^2 + e)} \)

1. **Determine the domain of \( \log(x^2 + e) \)**:
- Since \( x^2 + e > e > 0 \) for all \( x \in \mathbb{R} \), \( \log(x^2 + e) \) is defined for all real \( x \).

2. **Find the minimum value of \( \log(x^2 + e) \)**:
- The minimum occurs when \( x^2 \) is minimized, which is at \( x = 0 \):
\[ \log(0^2 + e) = \log(e) = 1. \]
- As \( |x| \to \infty \), \( x^2 + e \to \infty \) and thus \( \log(x^2 + e) \to \infty \).

3. **Determine the range of \( \frac{1}{\log(x^2 + e)} \)**:
- Since \( \log(x^2 + e) \) ranges from \( 1 \) to \( \infty \):
\[ \frac{1}{\log(x^2 + e)} \text{ ranges from } \frac{1}{\infty} \text{ to } \frac{1}{1} \text{, i.e., } (0, 1]. \]

4. **Apply the greatest integer function**:
- Thus, \( f_1(x) = \left[ \frac{1}{\log(x^2 + e)} \right] \) can take values:
- \( 0 \) (for \( \frac{1}{\log(x^2 + e)} \) in \( (0, 1) \))
- \( 1 \) (for \( \frac{1}{\log(x^2 + e)} = 1 \) when \( x = 0 \)).

### Step 2: Analyze \( f_2(x) = \frac{1}{\sqrt{1 + x^2}} \)

1. **Determine the range of \( \frac{1}{\sqrt{1 + x^2}} \)**:
- The minimum value occurs as \( |x| \to \infty \):
\[ \frac{1}{\sqrt{1 + x^2}} \to 0. \]
- The maximum value occurs at \( x = 0 \):
\[ \frac{1}{\sqrt{1 + 0^2}} = 1. \]
- Thus, \( f_2(x) \) ranges from \( 0 \) to \( 1 \).

### Step 3: Combine the results

1. **Calculate \( f(x) = f_1(x) + f_2(x) \)**:
- If \( x = 0 \):
\[ f(0) = f_1(0) + f_2(0) = 1 + 1 = 2. \]
- If \( x \neq 0 \):
- \( f_1(x) \) can be \( 0 \) or \( 1 \).
- \( f_2(x) \) ranges from \( 0 \) to \( 1 \).
- Therefore, \( f(x) \) can take values:
- \( 0 + (0, 1) = (0, 1) \) (when \( f_1(x) = 0 \))
- \( 1 + (0, 1) = (1, 2) \) (when \( f_1(x) = 1 \))

### Final Result

The overall range of \( f(x) \) is:
\[ f(x) \in [0, 1) \cup [2]. \]