If `f(x) = log_(e^2x) ((2lnx+2)/(-x))` and `g(x) = {x}` then range of `g(x)` for existance of `f(g(x))` is

Options

A. `(0, 2//e)`
B. `(0, 1//e) - {1//e^(2)}`
C. `(0, 3//e)`
D. none of these

Explanation

To solve the problem, we need to analyze the functions \( f(x) \) and \( g(x) \) and determine the range of \( g(x) \) for which \( f(g(x)) \) exists.

### Step-by-Step Solution

1. **Identify the functions**:
- Given \( f(x) = \log_{e^{2x}} \left( \frac{2\ln x + 2}{-x} \right) \)
- Given \( g(x) = \{ x \} \) (the fractional part of \( x \))

2. **Determine the domain of \( f(x) \)**:
- The base of the logarithm \( e^{2x} \) must be positive and not equal to 1.
- \( e^{2x} > 0 \) for all \( x \) (always true).
- \( e^{2x} \neq 1 \) implies \( 2x \neq 0 \) or \( x \neq 0 \).
- Therefore, \( x \) must be greater than 0: \( x > 0 \).

3. **Determine when the argument of the logarithm is positive**:
- We need \( \frac{2\ln x + 2}{-x} > 0 \).
- This simplifies to \( 2\ln x + 2 < 0 \) (since \(-x < 0\) for \(x > 0\)).
- Thus, \( \ln x < -1 \).
- Exponentiating both sides gives \( x < \frac{1}{e} \).

4. **Combine the conditions**:
- From the above, we have two conditions:
- \( x > 0 \)
- \( x < \frac{1}{e} \)
- Therefore, the domain of \( f(x) \) is \( 0 < x < \frac{1}{e} \).

5. **Determine the range of \( g(x) \)**:
- The function \( g(x) = \{ x \} \) (fractional part of \( x \)) takes values in the interval \( [0, 1) \).

6. **Find the intersection of the range of \( g(x) \) and the domain of \( f(x) \)**:
- The range of \( g(x) \) is \( [0, 1) \).
- The domain of \( f(x) \) is \( (0, \frac{1}{e}) \).
- The intersection of these intervals is \( (0, \frac{1}{e}) \).

7. **Conclusion**:
- The range of \( g(x) \) for which \( f(g(x)) \) exists is \( (0, \frac{1}{e}) \).

### Final Answer
The range of \( g(x) \) for the existence of \( f(g(x)) \) is \( (0, \frac{1}{e}) \).

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