Question

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

• A. $[-2,0)\cup (0,1)$
• B. $[-2,0)\cup (0,1]$
• C. $(-2,0)\cup (0,1]$
• D. $(-2,0)\cup [0,1]$

Answer 1

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

Given,

$f\left(x\right)=\frac{1}{{\log }_{10}(1-x)}+\sqrt{x+2}$

$f\left(x\right)$ will be defined if, $1-x>0,\ne 1$ and $x+2\ge 0$

Or $x<1,\ne 0$ and $x\ge -2$

Combining these conditions we get, $x\in [-2,0)\cup (0,1)$