The set of all real numbers `x` for which `x^2-|x+2|+x >0` is `(-oo,-2)` b. `(-oo,-sqrt(2))uu(sqrt(2),oo)` c. `(-oo,-1)uu(1,oo)` d.`(sqrt(2),oo)`

Options

A. `(-oo, -2)cup (2, oo)`
B. `(-oo, -sqrt(2))cup (sqrt(2), oo)`
C. `(-oo, -1) cup (1, oo)`
D. `( sqrt(2), oo)`

Answer

B

Explanation

CASE I When `x+2 ge 0`
In this case, we have
`|x+2| = x +2`
`therefore x^(2) - |x+2| +x gt 0`
`rArr x^(2)- (x+2) +x gt 0`
`rArr x^(2) -2 gt 0`
`rArr x lt -sqrt(2) "or" x gt sqrt(2) rArr x in (-oo, -sqrt(2) cup (sqrt(2), oo)`
But `x +2 ge 0 i.e., x ge -2`
`therefore x in [-2, -sqrt(2)) cup (sqrt(2), oo)`
CASE II When `x +2 lt 0`
In this case, we have
`|x + 2| = -(x+2)`
`therefore x^(2) -|x + 2|+x gt 0`
`rArr x^(2) +x +2 +x gt 0 rArr x^(2) + 2x +2 gt 0`
This is true for all real values of x as the discriminant of `x^(2) + 2x +2` is less than zero.
So, `x +2 lt 0 i.e. x in (-oo, -2)` is the solution set.
Hence, the solution set is
`[-2, -sqrt(2)) cup (sqrt(2), oo) cup (-oo, -2) "or", (-oo, -sqrt(2)) cup (sqrt(2), oo)`.