Question

The set of all real numbers $x$ for which $x^2−âˆ\pounds x+2âˆ\pounds +x>0$ is

• A. $(−∞,−2)∪(2,∞)$
• B. $(−∞,−\sqrt{2})∪(\sqrt{2},∞)$
• C. $(−∞,−1)∪(1,∞)$
• D. $(\sqrt{2},∞)$

Answer 1

Answer: (B)

Given, $x^2−âˆ\pounds x+2âˆ\pounds +x>\mathrm{0...}(i)$
Case I When $x+2â‰\yen 0$
$∴x^2−x−2+x>0⇒x^2−2>0$
$⇒x<−\sqrt{2}$ or $x>\sqrt{2}$
$⇒x∈(−2,−\sqrt{2})∪(\sqrt{2},∞)...(ii)$
Case II When $x+2<0$
$∴x^2+x+2+x>0$
$⇒x^2+2x+2>0$
$⇒(x+1{)}^2+1>0$
which is true for all $x$.
$∴x≤−2$ or $x∈(−∞,−2)...(iii)$
From Eqs. (ii) and (iii), we get
$x∈(−∞−\sqrt{2})∪(\sqrt{2},∞)$