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If $\mathrm{D}$ is the set of all real $\mathrm{x}$ such that $1-e^{(1 / x)-1}$ is positive, then $\mathrm{D}$ is equal to
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Verified Answer
The correct answer is:
$(-\infty, 0) \cup(1, \infty)$
$$
\begin{array}{l}
1-\mathrm{e}^{\frac{1}{x}-1}>0 \\
\Rightarrow \mathrm{e}^{\frac{1}{x}-1} < 1 \Rightarrow \frac{1}{x}-1 < \log 1 \\
\Rightarrow \frac{1}{x}-1 < 0 \Rightarrow \frac{1}{x} < 1 \\
\Rightarrow x \in(-\infty, 0) \cup(1, \infty)
\end{array}
$$
\begin{array}{l}
1-\mathrm{e}^{\frac{1}{x}-1}>0 \\
\Rightarrow \mathrm{e}^{\frac{1}{x}-1} < 1 \Rightarrow \frac{1}{x}-1 < \log 1 \\
\Rightarrow \frac{1}{x}-1 < 0 \Rightarrow \frac{1}{x} < 1 \\
\Rightarrow x \in(-\infty, 0) \cup(1, \infty)
\end{array}
$$
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