The set of values of x for which the inequality `|x-1|+|x+1|lt 4` always holds true, is

Options

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

Answer

A

Explanation

On the LHS of the given inequation there are two terms `|x-1| " and " |x+1|`. On equating x -1 and x-1 to zero, we get: x=-1 and 1 as critical points. These points divide the real line into three regions viz, `(-oo, -1),[-1,1)" and "[1, oo)`. So, we consider the following cases:
CASE I when `-oo lt x lt -1`
In this case, we have
`|x-1| =-(x-1) " and"|x+1| =-(x-1)`
`therefore |x-1|+|x+1| lt 4`
`rArr -(x-1)-(x+1) lt 4 rArr -2x lt 4 rArr x gt -2`
But, `-oo lt x lt -1`
`therefore x in (-2, -1)`
CASE II When `-1 le x lt 1`
In this case, we have
`|x-1|=-(x-1) " and" |x+1|=x+1`
`therefore |x-1| + |x+1| lt 4`
`rArr-(x-1) + x+1 lt 4`
`rArr 2 lt 4`, which is true for all `x in [-1, 1]).`
`therefore x in [-1, 1)`
CASE III When `1 le x lt oo`
In this case, we have
`|x-1| = x-1 " and "|x+1| =x-1`
`therefore|x-1|+|x+1|lt 4`
`rArr x-1+x+1 lt 4 rArr x lt 2`
`rArr x in [1, 2) " " [because 1 le x lt oo]`
Hence, `x in (-2, -1)cup[-1, 1)cup[1, 2)"or", x in (-2, 2).`