If $e^{x} + e^{f (x)} = e$ , then range of the function of f(x) is

(\infty, 1]

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(- \infty, 1)

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(1, \infty)

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[1, \infty)

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Solution

The correct option is D $(- \infty, 1)$
Given,

$e^{x} + e^{f (x)} = e$
$e^{x} + e^{y} = e$
$\Rightarrow e^{x} = e - e^{y}$
$\Rightarrow x = log (e - e^{y})$
For log to be defined ,
$e - e^{y} > 0$
$\Rightarrow e^{y} < e$
$\Rightarrow y < 1$
$\Rightarrow y \in (- \infty, 1)$

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