The domain of `f(x)i s(0,1)dot` Then the domain of `(f(e^x)+f(1n|x|)` is `(a)(-1, e)` (b). `(1, e)` `(c).(-e ,-1)` (d) `(-e ,1)`

Options

A. `((1)/(e), 1)`
B. `(-e, 1)`
C. `(-1,-(1)/(e))`
D. `(-e, -1)uu(1,e)`

Answer

To solve the problem, we need to find the domain of the expression \( f(e^x) + f(\ln |x|) \) given that the domain of \( f(x) \) is \( (0, 1) \).

### Step-by-Step Solution:

1. **Identify the Domain of \( f(x) \)**:
The function \( f(x) \) is defined for \( x \) in the interval \( (0, 1) \). This means that for \( f(e^x) \) and \( f(\ln |x|) \) to be defined, both \( e^x \) and \( \ln |x| \) must lie within the interval \( (0, 1) \).

2. **Find the Domain for \( f(e^x) \)**:
- The expression \( e^x \) is always positive for all real \( x \).
- To ensure \( e^x < 1 \), we solve the inequality:
\[ e^x < 1 \implies x < 0 \]
- Thus, the domain for \( f(e^x) \) is \( (-\infty, 0) \).

3. **Find the Domain for \( f(\ln |x|) \)**:
- We need \( \ln |x| \) to be in the interval \( (0, 1) \).
- This means:
\[ 0 < \ln |x| < 1 \]
- Solving \( \ln |x| > 0 \):
\[ |x| > 1 \implies x < -1 \text{ or } x > 1 \]
- Solving \( \ln |x| < 1 \):
\[ |x| < e \implies -e < x < e \]
- Combining these two results, we find:
- For \( x < -1 \): The valid interval is \( (-e, -1) \).
- For \( x > 1 \): The valid interval is \( (1, e) \).

4. **Combine the Domains**:
- The domain of \( f(e^x) \) is \( (-\infty, 0) \).
- The domain of \( f(\ln |x|) \) is \( (-e, -1) \) and \( (1, e) \).
- The overall domain for \( f(e^x) + f(\ln |x|) \) is the union of these intervals:
- From \( f(e^x) \): \( (-\infty, 0) \)
- From \( f(\ln |x|) \): \( (-e, -1) \) and \( (1, e) \)

5. **Identify the Common Valid Intervals**:
- The only overlapping interval that satisfies both conditions is \( (-e, -1) \).
- The interval \( (1, e) \) does not overlap with \( (-\infty, 0) \).

### Final Result:
Thus, the domain of \( f(e^x) + f(\ln |x|) \) is \( (-e, -1) \).

### Answer:
The correct option is (c) \( (-e, -1) \).