Correct Answer - `(-2,-1]uu[1,2)`
Case I When `x ge 0`.
Then, `|x|=x and so (1)/(2-|x|)ge1 rArr (1)/(2-x)-1ge0 rArr (x-1)/(2-x)ge0`.
`therefore (x-1 ge0 and 2-xgt0) or(x-1le0 and 2-xlt0)`
`rArr(xge1 and xlt2) or
(xle1 and x gt2)`
`rArr (1lex lt2) rArr x in[1,2)`
Case II When `xlt0`.
Then, `|x| = -x. So , (1)/(2-|x|)ge 1 rArr (1)/(2+x) -1ge 0 rArr (-1-x)/(2+x) ge 0 rArr (1+x)/(2+x) le 0`
`rArr (1+x le0 and 2+xgt0) or (1+x ge0 and 2+x gt0)`
`rArr (-2ltxle-1) rArr x in(-2,-1]`.
`therefore " solution set " =(-2,-1]uu[1,2)`.
