Question

The set of all real numbers satisfying $e^{(\frac{1}{x}-1)}<1$

• A. $(0,\infty )$
• B. $(-\infty ,0)\cup (1,\infty )$
• C. $(-\infty ,\infty )$
• D. $(0,1)$

Answer 1

Answer: (B)

$(-\infty ,0)\cup (1,\infty )$

We have, $e^{(\frac{1}{x}-1)}<1$
Take log both sides on base $e$
$(\frac{1}{x}-1){\log }_{e}e<{\log }_{e}1=0[\because \log a^m=m\log a]$
$\Rightarrow \frac{1-x}{x}<0\Rightarrow \frac{x-1}{x}>0$
$\Rightarrow x\in (-\infty ,0)\cup (1,\infty )$