Let f(x) =|x-1|+|x-2|
Hence solution is `x in (-oo, -1//2] cup [7//2,oo)`.
Alternative Method :
Consider `|x-1|+|x-2| lt 4`
Now , `|(x-1)+(x-2)|le |x-1|+|x-2|`
`therefor | (x-1)+(x-2)| gt 4`
`rArr |2x-3| lt 4`
`rArr -4 lt 2x -3 lt 4 `
`rArr -1 lt 2x lt 7 `
`-1//2 lt x lt 7//2`
Therefore , for `|x-1| + | x-2 | ge 4` we have `x in ( -oo,-1//2 ] cup [ 7//2 , oo)`
