Question

The set of all real numbers x for which $x^2-[x+2]+x>0,$ is

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

Answer 1

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

$Forx\ge -2,x^2-x-2+x>0\Rightarrow x^2>2\Rightarrow xϵ[-2,-\sqrt{2})\cup (\sqrt{2},\infty )Forx<-2x^2+x+2+x>0or{x}^2+2x+2>0$
Which is true for all x. Hence $xϵ(-\infty ,-\sqrt{2})\cup (\sqrt{2},\infty )$