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`(-oo,1),[1,2) and [2,oo)`. So, we consider the following three cases.
Case I When `-ooltxlt1`.
In this case, `x-1lt0 and x-2lt0`.
`therefore |x-1|=-(x-1)=-x+1 and|x-2|=-(x-2)=-x+2`.
Now, `|x-1|+|x-2|ge4`
`rArr-x+1-x+2ge4`
`rArr -2x+3ge4rArr-2xge4-3rArr=-2xge1rArrxle(-1)/(2)`
`rArr x in(-oo,(-1)/(2)]`.
But, `-ooltxlt1.`
`therefore` solution set in this case `=(-oo,(-1)/(2)]cap(-oo,1)=(-oo,(-1)/(2)]`.
Case II When `1le x lt 2`.
In this, case, `x-1ge0 and x-2lt0`.
`therefore |x-1|=x-1and|x-2|=-(x-2)=-x+2`.
Now, `|x-1|+|x-2|ge4`
`rArr x-1-x+2ge4rArr-1ge4`, which is absured.
So, the given inequation has no solution in `[1,2)`.
Case III When `2 le x lt oo`.
In this case, `x-2ge0and x-1gt0`.
`therefore |x-2|=x-2 and |x-1|=x-1.`
Now, `|x-1|+|x-2|ge4`
`rArr x-1+x-2ge4rArr2x-3ge4 rArr2xge7 rArrxge(7)/(2)`.
Also, in this case, we have `xge 2`.
`therefore` solution set in this case `=[(7)/(2),oo)nn[2,oo),=[(7)/(2),oo)`.
Hence, from all the above cases, we have
Solution set `=(-oo,(-1)/(2)]uu[(7)/(2),oo)`.
