The solution set of the inequation `(|x+3|+x)/(x+2) gt 1`, is

Options

A. `(-5, -2) cup (-1, oo)`
B. `(-5, -2)`
C. `(-1, oo)`
D. none of these

Answer

A

Explanation

We have
`(|x+3|+x)/(x+2) gt 1`
`rArr (|x+3|+x)/(x+2)-1 gt 0`
`rArr (|x+3|+x-x-2)/(x+2) gt 0 rArr (|x+3|-2)/(x+2) gt 0`
Now two cases arise.
CASE I When `x + 3 ge 0 i.e. x ge -3`
In this case, we have `|x+3| = x+3`
`therefore (|x+3|-2)/(x+2) gt 0`
`rArr (x+3-2)/(x+2) gt 0`
`rArr (x+1)/(x+2) gt 0 rArr x in (-oo, -2) cup(-1, oo)`
But `x ge -3.`
` therefore x in [-3, -2) cup (-1, oo)`
CASE II When `x + 3 lt 0 i.e. x lt -3`
In this case, we have `|x + 3| = -(x+3)`
`therefore(|x+3|-2)/(x+2) gt 0`
`rArr (-(x+3)-2)/(x+2) gt 0`
`rArr (-x-5)/(x+2) gt 0 rArr (x+5)/(x+2) lt 0 rArr -5 lt x lt -2`
But ` x lt -3. "So," x in (-5, -3)`.
Hence, the solution set of the given inequation is
`(-5, -2)cup (-1, oo)`