Question-121175

Question Number 121175 by Ar Brandon last updated on 05/Nov/20

Answered by MJS_new last updated on 05/Nov/20

x < 1

g (x) < 1

Commented by Ar Brandon last updated on 05/Nov/20

Any calculations or explanations, Sir ?��

Commented by MJS_new last updated on 05/Nov/20

e^{x} + e^{g (x)} = e \Leftrightarrow g (x) = \ln (e - e^{x})

e - e^{x} > 0 \Rightarrow e > e^{x} \Rightarrow x < 1

\lim_{x \to 1^{-}} g (x) = - \infty

\lim_{x \to - \infty} g (x) = 1

\Rightarrow g (x) < 1

Commented by Ar Brandon last updated on 05/Nov/20

It's clear now. Thanks Sir

Commented by MJS_new last updated on 05/Nov/20

{you}^{'} re welcome

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