Question

If $e^x+e^{f\left(x\right)}=e,$ then which among the following options are correct

• A. Domain of $f\left(x\right)$ is $(-\infty ,1)$
• B. Range of $f\left(x\right)$ is $(-\infty ,1)$
• C. Domain of $f\left(x\right)$ is $(-\infty ,0]$
• D. Range of $f\left(x\right)$ is $(-\infty ,0]$

Answer 1

Answer: (A), (B)

The correct options are
A Domain of $f\left(x\right)$ is $(-\infty ,1)$
B Range of $f\left(x\right)$ is $(-\infty ,1)$

Given,
$e^x+e^{f\left(x\right)}=e\Rightarrow e^{x-1}+e^{f\left(x\right)-1}=1\Rightarrow e^{f\left(x\right)-1}=1-e^{x-1}$

Taking natural $\log$ on both sides, we get
$\Rightarrow \ln e^{f\left(x\right)-1}=\ln (1-e^{x-1})$
$\Rightarrow f\left(x\right)-1=\ln (1-e^{x-1})$
$\Rightarrow f\left(x\right)=1+\ln (1-e^{x-1})$

Now, for $f\left(x\right)$ to be defined
$\Rightarrow 1-e^{x-1}>0$
$\Rightarrow e^{x-1}<1$
$\Rightarrow e^{x-1}<e^{\ln 1}$
$\Rightarrow x-1<\ln 1$
$\Rightarrow x<1$
Hence, the domain of $f\left(x\right)$ is $(-\infty ,1).$

$\because x\in (-\infty ,1)$
$\Rightarrow x-1\in (-\infty ,0)$
$\Rightarrow e^{x-1}\in (0,1)$
$\Rightarrow \ln (1-e^{x-1})\in (-\infty ,0)$
$\Rightarrow 1+\ln (1-e^{x-1})\in (-\infty ,1)$
Hence, the range of $f\left(x\right)$ is $(-\infty ,1).$