Question

The set of all real numbers satisfying the inequation $x^2−âˆ\pounds x+2âˆ\pounds +x>0$ is

• A. $[−2,−\sqrt{2})∪(\sqrt{2},∞)$
• B. $(−∞,−2)∪(2,∞)$
• C. $(−∞,−\sqrt{2})∪(\sqrt{2},∞)$
• D. $(−∞,−2)∪(\sqrt{2},∞)$

Answer 1

Answer: (C)

$x^2−âˆ\pounds x+2âˆ\pounds +x>0$
Case $I$ If $x+2â‰\yen 0$
$x^2−(x+2)+x>0\left\{âˆ\mu âˆ\pounds xâˆ\pounds =\{\begin{array}{ll}<br/><br/>x,& xâ‰\yen 0\\ <br/><br/>−x,& x<0<br/><br/>\end{array}\right\}$
$⇒x^2−2>0$
$⇒(x−\sqrt{2})(x+\sqrt{2})=0$

$x∈[−2,−\sqrt{2}]∪[\sqrt{2},∞]$
Case II $x+2<0$
$x^2+(x+2)+x>0$
$⇒x^2+2x+2>0$
$⇒(x+1{)}^2+1>0$
$⇒x<−2$
Hence, the solution set is
$x∈(−∞,−\sqrt{2})∪(\sqrt{2},∞)$