Question

The domain of definition of the function $f\left(x\right)=\sqrt{1+{\log }_e(1-x)}$ is

• A. $-\infty <x\le 0$
• B. $-\infty <x\le \frac{e-1}{e}$
• C. $-\infty <x\le 1$
• D. $x\ge 1-e$

Answer 1

Answer: (B)

$-\infty <x\le \frac{e-1}{e}$

Given, $f\left(x\right)=\sqrt{1+{\log }_e(1-x)}$
For domain, $(1-x)>0$ and ${\log }_e(1-x)\ge -1$
$\Rightarrow x<1$ and $1-x\ge e^{-1}$
$\Rightarrow x<1$ and $x\le 1-\frac{1}{e}$
$\Rightarrow -\infty <x\le \frac{e-1}{e}$