Question

( x^{2}-|x+2|+x>0 )
( |f quad x geq-2 Rightarrow| x+2 mid=x+2 )
( x^{2}-mid x-2+x>0 )
( x^{2}>2 )
( Rightarrow x in(-infty,-sqrt{2}) cup(sqrt{2}, infty) )
( Rightarrow x in[-2,-sqrt{2}) cup(sqrt{2}, infty)-(1) )
If ( x leq-2 Rightarrow|x+2|=-x-2 )
( Rightarrow x^{2}+x+2+x>0 )
( x^{2}+2 x+2>0 )
( (x+1)^{2}+1>0 )
Which is atways true ( Rightarrow n in(-infty,-2] )
( 2 Rightarrow n in(-infty,-sqrt{2}) cup(sqrt{2}, infty) )

# The set of all real numbers x for which x2 - | x + 2 + x > 0, is [IITSC.-2002] (A) (- 00, - 2) U (2,00) (B) (0, -2) U (V2,00) (C) (-0, - 1) U (1, 0) (D) (12,00)

Solution