The set of real values of x satisfying `||x-1|-1|le 1`, is

Options

A. `[-1, 3]`
B. `[0, 2]`
C. `[-1, 1]`
D. none of these

Answer

A

Explanation

Since x-1 changes its sign as x passes through 1.
So, following cases arise.
CASE I When `x lt 1`
In this case, we have `|x-1|= -(x-1)`
`therefore ||x-1|-1| le1`
`rArr |-(x-1)-1| le 1`
`rArr |-x| le1`
`rArr |x| le 1`
`rArr -1 le x le 1 " "[because |x| le a hArr -a le x le a]`
`rArr -1 le x lt 1 " " [because x lt 1]`
`therefore x in [-1, 1])`
CASE II When `x ge 1`
In this case, we have |x-1| = x -1
`therefore ||x-1|-1| le 1`
`rArr |x-1-1| le 1`
`rArr |x-2| le 1 rArr 2-1 le x le 2 + 1 rArr x in [1, 3]`
Hence, `x in [-1, 3]`