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Mathematics Question on Operations on Sets

The set of all real numbers x for which x2−∣x+2∣+x>0x^2-|x+2|+x>0 is

A

(−8,−2)∪(2,8)(-8,-2)\cup(2,8)

B

(−8,−2)∪(2,8)(-8,-\sqrt2)\cup(\sqrt2,8)

C

(−8,−1)∪(1,8)(-8,-1)\cup(1,8)

D

(2,8)(\sqrt2,8)

Answer

(−8,−2)∪(2,8)(-8,-\sqrt2)\cup(\sqrt2,8)

Explanation

Solution

Given, x2−∣x+2∣+x>0           ...(i)x^2-|x+2|+x>0\, \, \, \, \, \, \, \, \, \, \, ...(i)
Case I When \hspace20mmx+2\ge0
\therefore\hspace20mmx^2-x-2+x>0 \Rightarrow x^2-2>0
\Rightarrow\hspace25mmx\sqrt2
\Rightarrow\hspace25mmx\in(-2,-\sqrt2)\cup(\sqrt2,8)\, \, \, \, ...(ii)
Case II When x+2<0x+2<0
∴       x2+x+2+x>0\therefore\, \, \, \, \, \, \, x^2+x+2+x>0
⇒         x2+2x+2>0\Rightarrow\, \, \, \, \, \, \, \, \, x^2+2x+2>0
⇒         (x+1)2+1>0\Rightarrow\, \, \, \, \, \, \, \, \, (x+1)^2+1>0
which is true for all x.
∴       x≤−2  or x∈(−8,−2)         ...(iii)\therefore\, \, \, \, \, \, \, x\le-2 \, \, or\, x\in(-8,-2) \, \, \, \, \, \, \, \, \, ...(iii)
From Eqs. (ii) and (iii), we get
\hspace25mmx\in(-8,-\sqrt2)\cup(\sqrt2,8)