If $f (x)$ is defined on $(0, 1)$ then the domain of definition of $f (e^{x}) + f (ln | x |)$ is

(- e, - 1)

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(- e, - 1) \cup (1, e)

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(- 5, - 1) \cup (1, \infty)

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(- e, e)

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Solution

The correct option is B $(- e, - 1)$
$g = f {(e)}^{x} + f (I n | x |)$
Domain of g $== 0 < e^{x} < 1$
$x \in (- \infty, 0)$ ----------------------- $(i)$
And $,$
$0 (I n | x |) < 1$
$| x | < e$
$x \in (- e, - 1) \cup (1, e)$ ------------ $(i i)$
from $(i)$ and $(i i),$ we have
$x \in (- e, - 1)$
Hence $,$ option $(A)$ is correct $.$

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