Question

If the domain of $f(x)$ is $(0,1)$, then the domain of $y=f(e^{x})+f(\ln|x|)$ is:

• A. $(-1,-\frac{1}{e})$
• B. $(\frac{1}{e},1)$
• C. $(-e,-1)$
• D. $(-e,-1)\cup(1,e)$

Hint:

For sums of functions like $$ f(g(x))+f(h(x)), $$ both functions must exist simultaneously. Always take the intersection of the individual domains.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The domain of a composite function \( f(g(x)) \) is determined by the values of \( x \) for which \( g(x) \) falls within the defined domain of \( f \).
We are given that the domain of \( f(t) \) is \( 0<t<1 \).
For the sum of two functions \( f(g_1(x)) + f(g_2(x)) \) to be defined, \( x \) must simultaneously satisfy the domain constraints for both \( g_1(x) \) and \( g_2(x) \).
This means we need to find the intersection of the two individual domains.
Step 2: Key Formula or Approach:
1. Find domain of \( f(e^x) \): Set \( 0<e^x<1 \).
2. Find domain of \( f(\ln|x|) \): Set \( 0<\ln|x|<1 \).
3. Determine the intersection of the resulting sets.
Step 3: Detailed Explanation:
Let's analyze the first component, \( f(e^x) \):
The input to \( f \) is \( e^x \). According to the given domain of \( f \), we must have:
\[ 0<e^x<1 \]
Since \( e^x \) is always greater than 0 for all real \( x \), the left inequality \( e^x>0 \) is always satisfied.
For the right inequality, \( e^x<1 \), we take the natural logarithm on both sides:
\[ \ln(e^x)<\ln(1) \implies x<0 \]
So, the domain for the first part is \( D_1 = (-\infty, 0) \).
Now, let's analyze the second component, \( f(\ln|x|) \):
The input to \( f \) is \( \ln|x| \). We must have:
\[ 0<\ln|x|<1 \]
To solve this, we take the exponential of all sides (since \( e^x \) is an increasing function):
\[ e^0<e^{\ln|x|}<e^1 \]
\[ 1<|x|<e \]
This absolute value inequality represents two intervals on the number line:
Case 1: \( 1<x<e \)
Case 2: \( -e<x<-1 \)
So, the domain for the second part is \( D_2 = (-e, -1) \cup (1, e) \).
Now, we find the intersection \( D_1 \cap D_2 \):
\[ (-\infty, 0) \cap [(-e, -1) \cup (1, e)] \]
The interval \( (1, e) \) is entirely composed of positive numbers, so it has no overlap with \( (-\infty, 0) \).
The interval \( (-e, -1) \) is entirely composed of negative numbers, so it is a subset of \( (-\infty, 0) \).
Thus, the intersection is \( (-e, -1) \).
Step 4: Final Answer:
The combined domain is \( (-e, -1) \), which is option (C).