InterviewSolution
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InterviewSolution
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Solve |x-1|+|x-2|ge4,x in R. |
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Answer» Solution : Putting `x -1=0 and x-2=0`, we get `x =1 and x =2` as the critical points, These points DIVIDE the whole realline 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)`. |
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