The domain of `f(x) is (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)`
B. `(1,e)`
C. `(-e,-1)`
D. `(-e, 1)`

Explanation

To find the domain of \( f(e^x) + f(\ln |x|) \) given that the domain of \( f(x) \) is \( (0, 1) \), we need to determine the values of \( x \) for which both \( e^x \) and \( \ln |x| \) lie within the interval \( (0, 1) \).

### Step 1: Analyze \( e^x \)
The function \( e^x \) is always positive for all real \( x \). We need to find the values of \( x \) such that:
\[ 0 < e^x < 1 \]
This inequality can be solved as follows:
- The inequality \( e^x < 1 \) implies \( x < 0 \) (since \( e^0 = 1 \)).
- The inequality \( e^x > 0 \) is always true for any real \( x \).

Thus, for \( e^x \) to be in the interval \( (0, 1) \), \( x \) must be in the interval:
\[ (-\infty, 0) \]

### Step 2: Analyze \( \ln |x| \)
Next, we analyze \( \ln |x| \). We need to find the values of \( x \) such that:
\[ 0 < \ln |x| < 1 \]
This can be broken down into two parts:
1. \( \ln |x| > 0 \) implies \( |x| > 1 \) (since \( \ln(1) = 0 \)).
2. \( \ln |x| < 1 \) implies \( |x| < e \) (since \( \ln(e) = 1 \)).

Combining these inequalities, we have:
\[ 1 < |x| < e \]
This means:
- For \( x > 0 \): \( 1 < x < e \) (i.e., \( x \in (1, e) \))
- For \( x < 0 \): \( -e < x < -1 \) (i.e., \( x \in (-e, -1) \))

### Step 3: Combine the results
Now we need to find the intersection of the intervals obtained from \( e^x \) and \( \ln |x| \):
- From \( e^x \): \( x \in (-\infty, 0) \)
- From \( \ln |x| \): \( x \in (1, e) \) and \( x \in (-e, -1) \)

The only valid interval that lies within both conditions is:
\[ (-e, -1) \]

### Conclusion
Thus, the domain of \( f(e^x) + f(\ln |x|) \) is:
\[ \boxed{(-e, -1)} \]