Question

If $e^x+e^{f\left(x\right)}=e$ then for $f\left(x\right)$

• A. domain is $(-\infty ,1)$
• B. range is $(-\infty ,1)$
• C. domain is $(-\infty ,0]$
• D. range is $(-\infty ,1]$

Answer 1

Answer: (A), (B)

domain is $(-\infty ,1)$
range is $(-\infty ,1)$

Given,

$e^{x-1}+e^{f\left(x\right)-1}=1$
$e^{f\left(x\right)-1}=1-e^{x-1}$
Taking natural $\log$ on both sides, we get
$f\left(x\right)-1=\ln (1-e^{x-1})$
$f\left(x\right)=1+\ln (1-e^{x-1})$
$1-e^{x-1}>0$
$e^{x-1}<1$
$x-1<\log \left(1\right)$
$x-1<0$
$x<1$
Hence domain is $(-\infty ,1)$.
Now, we can see that as $x\to -\infty$ $f\left(x\right)\to 1$
and as $x\to 1$ $f\left(x\right)\to -\infty$
Hence the range of $f\left(x\right)$ is $(-\infty ,1)$.