Q. The domain of the definition of the function $f(x) = \sqrt{1 + \log_{e}{(1 - x)}}$ is
A) $- \infty < x \leq 0$
B) $- \infty < x \leq \frac{e - 1}{e}$
C) $- \infty < x \leq 1$
D) $x \geq 1 - e$
Explanation
(b)
Given, $f(x) = \sqrt{1 + \log_{e}(1 - x)}$
For domain, $(1 - x) > 0$ and $\log_{e}{(1 - x) \geq - 1}$
$\Rightarrow \ \ \ \ x < 1$ and $1 - x \geq e^{- 1}$
$\Rightarrow \ \ \ \ \ \ \ x < 1$ and $x \leq 1 - \frac{1}{e}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ - \infty < x \leq \frac{e - 1}{e}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.