Question

If $f\left(x\right)$ is defined on $(0,1)$, then the domain of $g\left(x\right)=f\left(e^x\right)+f\left({\log }_e|x|\right)$ is

• A. $(-1,e)$
• B. $(1,e)$
• C. $(-e,-1)$
• D. $(-e,1)$

Answer 1

The correct option is C $(-e,-1)$

Since $f\left(x\right)$ is defined on $(0,1)$
Now , for $f\left(e^x\right)\text{and}f\left({\log }_e|x|\right)$ to be defined,
$0<e^x<1$ and $0<{\log }_e|x|<1$

$\Rightarrow x\in (-\infty ,0)$ and $1<|x|<e$

$\Rightarrow x\in (-\infty ,0)$ and $x\in (-e,-1)\cup (1,e)$

Hence, taking intersection of both the sets we have, $x\in (-e,-1)$