Question

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

• A. $[-3,-2.5]\cap [-2.5,-2]$
• B. $[-2,0)\cup (0,1)$
• C. $[0,1]$
• D. None of the above

Answer 1

The correct option is B $[-2,0)\cup (0,1)$

The given function $f\left(x\right)=\frac{1}{{\log }_{10}(1-x)}+\sqrt{x+2}$ is defined if

$x+2\ge 0$, i.e., $x\ge -2$ or $-2\le x$ and
${\log }_{10}(1-x)\ne 0$
$\Rightarrow 1-x\ne 1\Rightarrow x\ne 0$

Again, $1-x>0$

Combining all the results for values of $x$, we get
$-2\le x<0$ and $0<x<1$

i.e. $x\in [-2,0)$ or $x\in (0,1)$

Hence, $[-2,0)\cup (0,1)$ is the domain for $f\left(x\right)$.