Question

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

• A. $(-e,-\frac{1}{e})$
• B. $(0,\infty )$
• C. $(e,\infty )$
• D. $R-\left\{0\right\}$

Answer 1

The correct option is A $(-e,-\frac{1}{e})$

Since $f\left(x\right)$ is defined on $(-1,1)$
$\therefore -1<e^x<1,-1<{\log }_e|x|<1$
$x\in (-\infty ,0)\frac{1}{e}<|x|<e$
$\therefore x\in (-e,-\frac{1}{e})\cup (\frac{1}{e},e)$
Hence $x\in (-e,-\frac{1}{e})$