Question

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

• 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

The correct option is B

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

Explanation of correct answer :

Finding the domain of the function:

Given, $f\left(x\right)=\sqrt{1+{\log }_e\left(1-x\right)}$

As we know, for ${\log }_{a}b$, $b$ must always be positive, i.e.$b>0$.

Then, $$\begin{gathered} \left(1-x\right)>0 \\ \Rightarrow x<1 \\ \Rightarrow x\in (-\infty ,1) \end{gathered}$$

As we know, The term under a root must always be positive, i.e.

$$\begin{gathered} \Rightarrow 1+{\log }_e\left(1-x\right)>0 \\ \Rightarrow {\log }_e\left(1-x\right)>-1 \\ \Rightarrow 1-x>e^{-1}\left(\text{if},{\log }_{a}b=c,\text{then}b=a^c\right) \\ \Rightarrow x<1-e^{-1} \\ \Rightarrow x\in \left(-\infty ,\frac{e-1}{e}\right) \end{gathered}$$

We have got two values of $x$, to get the final domain of the function, we will do intersection of both domains.

Thus, the domain of the function is $-\infty <x\le \frac{e-1}{e}$

Hence, the correct answer is Option(B).