Write the solution set of the inequation `|x-1|geq|x-3|dot`

Options

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

Answer

B

Explanation

Here, x = 1 and x=3 are two critical points which divide the real line into these three parts, namely, `x lt 1, 1 le x lt 3 " and " x ge 3`. So, we discuss the following cases:
CASE I When `x lt 1:`
In this case, we have
`|x-1| = -(x-1) " and " |x-3| = -(x-3)`
`therefore |x-1| ge |x-3|`
`rArr -(x-1) ge -(x-3) rArr 1 ge 3`, which is absurd.
So, the inequation has no solution for `x lt 1`
CASE II When `1 le x lt 3`
In this case, we have
`|x-1| = x-1 " and " |x-3| = -(x-3)`
`therefore |x-1| ge |x-3|`
`x-1 ge - (x-3) rArr 2x-4 ge 0 rArr x ge 2`
But, ` 1 le x lt 3`. Therefore,
`1 le x lt 3 " and " x ge 2 rArr x in [2, 3)`
CASE III When ` x ge 3`
In this case, we have
`|x-1| = x-1 " and " |x-3|= x-3`
`therefore |x-1| ge |x-3|`
`rArr x-1 ge x-3 rArr -1 ge -3,` which is correct.
So, the given inequation has all solutions satisfying `x ge3`.
Hence, the solution set of the given inequation is `[2, oo)`.