Question
If the domain of $f(x)$ is $(0, 1)$, then the domain of $y = f(e^x) + f(\ln|x|)$ is
This question was previously asked in
WBJEE 2026 Physics and Chemistry Question Paper (24-May-2026)
The correct answer is
$\left(\frac{1}{e}, 1\right)$
To find the domain of the function \( y = f(e^x) + f(\ln|x|) \), we need to ensure both \( f(e^x) \) and \( f(\ln|x|) \) are defined. Given that the domain of \( f(x) \) is \((0, 1)\), we need the following conditions:
- The expression \( e^x \) should be within the domain of \( f(x) \). Thus, \( 0 < e^x < 1 \), which implies:
- \( e^x < 1 \) leads to \( x < 0 \).
- \( e^x > 0 \) holds for all real \( x \), so no restriction here.
- The expression \( \ln|x| \) should be within the domain of \( f(x) \). Thus, \( 0 < \ln|x| < 1 \), which implies:
- \( \ln|x| > 0 \) gives \( |x| > 1 \), meaning \( x > 1 \) or \( x < -1 \).
- \( \ln|x| < 1 \) gives \( |x| < e \), meaning \(-e < x < e\).
Combining these two conditions for \( x \), we get:
- From both \( e^x < 1 \) (which gives \( x < 0 \)) and \( |x| > 1 \) (which requires \( x > 1 \) or \( x < -1 \)), the range for \( x \) is:
- The overlap from the conditions is \( x \in (\frac{1}{e}, 1) \).
Therefore, the domain of the expression \( y = f(e^x) + f(\ln|x|) \) is \(\left(\frac{1}{e}, 1\right)\).
Conclusion: The correct answer is \(\left(\frac{1}{e}, 1\right)\).
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