Question

Solve

$|x-1|+|x-2|\ge 4,xϵR$

Answer 1

Putting x-1=0 and x-2=0, we get x=1 and x=2 as the critical points. These points divide the whole real line into three parts, namely $(-\infty ,1)[1,2)and[2,\infty )$. So, we consider the following three cases.

Case I

When $-\infty <x<1$.

In this case, $x-1<0andx-2<0$

$\therefore |x-1|=-(x-1)=-x+1and|x-2|=-(x-2)=-x+2$

Now, $|x-1|+|x-2|\ge 4$

$\Rightarrow -x+1-x+2\ge 4$

$\Rightarrow -2x+3\ge 4\Rightarrow -2x\ge 4-3\Rightarrow -2x\ge 1\Rightarrow x\le \frac{-1}{2}$

$\Rightarrow xϵ(-\infty ,\frac{-1}{2}]$.

But, $-\infty <x<1$.

$\therefore$ solution set in this case $=(-\infty ,\frac{-1}{2}]\cap (-\infty ,1)=(-\infty ,\frac{-1}{2}]$.

Case II When $1\le x<2$

In this case, $x-1\ge 0andx-2<0$

$\therefore |x-1|=x-1and|x-2|=-(x-2)=-x+2$

Now, $|x-1|+|x-2|\ge 4$

$\Rightarrow x-1-x+2\ge 4\Rightarrow -1\ge 4$, which is absurd.

So, the given inequation has no solution in [1,2).

Case III

When $x\le x<\infty$

In this case, $x-2\ge 0andx-1>0$

$\therefore |x-2|=x-2and|x-1|=x-1$

Now, $|x-1|+|x-2|\ge 4$

$\Rightarrow x-1+x-2\ge 4\Rightarrow 2x-3\ge 4\Rightarrow 2x\ge 7\Rightarrow x\ge \frac{7}{2}$

Also, in this case, we have $x\ge 2$

$\therefore$ solution set in this case $=[\frac{7}{2},\infty )\cap [2,\infty ]=[\frac{7}{2},\infty )$

Hence, from all the above cases, we have

Solution set $=[-\infty ,\frac{-1}{2},)\cup [\frac{7}{2},\infty )$