Question

Write the solution set of the inequation $|x-1|\ge |x-3|$

Answer 1

$|x-1|\ge |x-3|\Rightarrow |x-1|\ge |x-3|\ge 0\text{By equating the expressions within the modulus to zero,we get x=1,3}\text{These points divide real line in three parts viz.}(-\infty ,1),[1,3)and[3,\infty )Case1:When-\infty <x<1|x-1|=-(x-1)and|x-3|=-(x-3)\therefore |x-1|-|x-3|\ge 0\Rightarrow -(x-1)+(x-3)\ge 0\Rightarrow -2\ge 0whichisnottrue.\text{So,the given inequation has no solution for}xϵ(-\infty ,1)Case2:When1<x<3|x-1|=(x-1)and|x-3|=-(x-3)\therefore |x-1|-|x-3|\ge 0\Rightarrow (x-1)+(x-3)\ge 0\Rightarrow 2x-4\ge 0\Rightarrow x\ge 2But,1\le x<2\text{Therefore,in this case the solution set of the given inequaton is [2,3)}Case3:When3\le x<\infty |x-1|=(x-1)and|x-3|=(x-3)\therefore |x-1|-|x-3|\ge 0\Rightarrow (x-1)-(x-3)\ge 0\Rightarrow 2\ge 0\text{The solution set of the given inequation is}[3,\infty )\text{Combining 1 and 3,we obtain that the solution set of the given inequation is}(2,3)\cup [3,\infty )=[2,\infty )$