Solve `1 le |x-2| le 3`

Options

A. `[-1, 5]`
B. `[3, 5]`
C. `[-1, 1]`
D. `[-1,1]cup[3, 5]`

Answer

D

Explanation

We know that
`a le |x-c|leb hArr x in [-b +c, -a+c] cup [a +c, b+c]`
`therefore 1 le |x-2|le 3 hArr x in [-3+2, -1+2]cup[1+2, 3+2]`
`hArr x in [-1, 1] cup [3, 5]`