Question

The domain of the function $f\left(x\right)=\frac{1}{{\log }_{10}(1-x)}+\sqrt{x+2}$ is

• A. $[-3,-2.5]\cup [-2.5,-2]$
• B. $[-2,0]\cup [0,1]$
• C. $[0,1]$
• D. None of these

Answer 1

The correct option is B

$[-2,0]\cup [0,1]$

$x+2\ge 0$ i.e., $x\ge -2$
$\because {\log }_{10}(1-x)\ne 0\Rightarrow 1-x\ne 1\Rightarrow x\ne 0$
Again $1-x>0\Rightarrow 1>x\Rightarrow x<1$
All these can be combined as $-2\le x<0$ and $0<x<1$.