The domain of `f(x) is (0,1)` .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 find the domain of the function \( f(e^x) + f(\ln|x|) \), given that the domain of \( f(x) \) is \( (0, 1) \), we need to analyze the conditions under which both \( e^x \) and \( \ln|x| \) fall within the interval \( (0, 1) \).

### Step 1: Analyze \( f(e^x) \)

The function \( f(e^x) \) is defined when \( e^x \) is in the interval \( (0, 1) \).

- The expression \( e^x < 1 \) implies:
\[ x < 0 \]
- Since \( e^x \) is always positive, the condition \( e^x > 0 \) is always satisfied for all real \( x \).

Thus, the domain for \( f(e^x) \) is:
\[ (-\infty, 0) \]

### Step 2: Analyze \( f(\ln|x|) \)

Next, we consider \( f(\ln|x|) \), which is defined when \( \ln|x| \) is in the interval \( (0, 1) \).

- The condition \( 0 < \ln|x| < 1 \) can be split into two inequalities:
1. \( \ln|x| > 0 \) implies \( |x| > 1 \) (or \( x < -1 \) or \( x > 1 \)).
2. \( \ln|x| < 1 \) implies \( |x| < e \) (or \( -e < x < e \)).

Combining these inequalities, we find:
- For \( x < -1 \): \( |x| > 1 \) and \( |x| < e \) gives \( -e < x < -1 \).
- For \( x > 1 \): \( |x| > 1 \) and \( |x| < e \) gives \( 1 < x < e \).

Thus, the domain for \( f(\ln|x|) \) is:
\[ (-e, -1) \cup (1, e) \]

### Step 3: Combine the Domains

Now, we need to find the intersection of the domains from steps 1 and 2:
1. Domain of \( f(e^x) \): \( (-\infty, 0) \)
2. Domain of \( f(\ln|x|) \): \( (-e, -1) \cup (1, e) \)

The intersection with \( (-\infty, 0) \) gives:
- From \( (-e, -1) \), we have \( (-e, -1) \).
- The interval \( (1, e) \) does not intersect with \( (-\infty, 0) \).

Thus, the combined domain is:
\[ (-e, -1) \]

### Conclusion

The domain of \( f(e^x) + f(\ln|x|) \) is \( (-e, -1) \).

### Final Answer

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