Correct Answer - C
Case I:
` 0 lt |x|-1 lt 1 " or " 1 lt |x| lt 2`
Then `x^(2)+4x+4 le 1`
or `x^(2)+4x+3 le 0`
or `-3 le x le -1`
So, `x in (-2,-1) " (1)" `
Case II:
` |x|-1 gt 1 " or " |x| gt 2`
Then `x^(2)+4x+4 ge 1`
or `x^(2)+4x+3 ge 0`
or `x ge -1 " or " x le -3`
So, ` x in (-oo, -3] cup (2,oo)`
So, ` x in (-oo, -3] cup (2,oo) " (2) " `
From (1) and (2), ` x in (-oo, -3] cup (-2, -1) cup (2,oo)`.