Question

The set of all real x for which $x^2-|x+2|+x\ge 0$ is:

• A. $(-\infty ,-2)\bigcup (2,\infty )$
• B. $(-\infty ,-\sqrt{2}]\bigcup [\sqrt{2},\infty )$
• C. $(-\infty ,-1)\bigcup (1,\infty )$
• D. $(\sqrt{2},\infty )$

Answer 1

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

$x<-2\Rightarrow x^2+x+2+x\ge 0$
or $x^2+2x+2\ge 0$
or $(x+1{)}^2+1\ge 0$ which is true for all x
$\therefore x<-2$
$x\ge -2\Rightarrow x^2-x-2+x\ge 0$
or $x^2-2\ge 0$
or $x\le -\sqrt{2}$ or $x\le \sqrt{2}$
$\therefore xϵ(-\infty ,-\sqrt{2}]\bigcup [\sqrt{2},+\infty )$
$\therefore$ Solution is:
$xϵ(-\infty ,2)\bigcup (-\infty ,-\sqrt{2}]\bigcup [\sqrt{2},\infty )$
$i.e.,(-\infty ,-\sqrt{2}]\bigcup [\sqrt{2},+\infty )$