Question

Let the function $f\left(x\right)=\sqrt{x^2-|x+2|+x}$ be a real valued function, then the domain of $f\left(x\right)$ is

• A. $(-\infty ,-\sqrt{2}]\cup [\sqrt{2},\infty )$
• B. $R$
• C. $[-2,\infty )$
• D. $R-[-\sqrt{2},\sqrt{2}]$

Answer 1

The correct option is A $(-\infty ,-\sqrt{2}]\cup [\sqrt{2},\infty )$

$f\left(x\right)=\sqrt{x^2-|x+2|+x}$
For the square root to exist,
$x^2-|x+2|+x\ge 0$

Case $I:x<-2$
$x^2+(x+2)+x\ge 0\Rightarrow x^2+2x+2\ge 0$
$\Rightarrow (x+1{)}^2+1\ge 0\Rightarrow x\in R\therefore x\in (-\infty ,-2)\cdots (1)$

Case $II:x\ge -2$
$x^2-(x+2)+x\ge 0\Rightarrow x^2-2\ge 0$
$\Rightarrow x\in (-\infty ,-\sqrt{2}]\cup [\sqrt{2},\infty )\therefore x\in [-2,-\sqrt{2}]\cup [\sqrt{2},\infty )\cdots \left(2\right)$

Hence, from equation $\left(1\right)$ and $\left(2\right)$,
$x\in (-\infty ,-\sqrt{2}]\cup [\sqrt{2},\infty )$