If A is the set of real values of x such that e1x-1<1, then A=

e1x−1 0∴ x 1∴ x∈−∞, 0∪1, ∞

If A is the set of real values of x such that e1x-1<1, then A=

e1x−1 0∴ x 1∴ x∈−∞, 0∪1, ∞

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$I f A i s t h e s e t o f r e a l v a l u e s o f x s u c h t h a t e^{\frac{1}{x} - 1} < 1, t h e n A =$

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

$(- \infty, 0)$

$(1, \infty)$

$(0, 1)$

answer is A.

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Detailed Solution

$\begin{array}{l} e^{\frac{1}{x} - 1} < 1 \\ \Rightarrow \frac{1}{x} - 1 < \log_{e}^{1} \\ \Rightarrow \frac{1}{x} - 1 < 0 \\ \Rightarrow \frac{1 - x}{x} < 0 \\ \Rightarrow x (1 - x) < 0 \\ \Rightarrow x (x - 1) > 0 \\ ∴ x < 0 o r x > 1 \\ ∴ x \in (- \infty, 0) \cup (1, \infty) \end{array}$