InterviewSolution
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InterviewSolution
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`(|x-2|-1)/(|x-2|-2)le0` |
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Answer» We have either `x-2ge0orx-2lt0`. Case I When `x-2 ge0`. In this case, `x ge 2 and |x-2|=x-2` So, the given inequation becomes `(x-2-1)/(x-2-2)le0rArr(x-3)/(x-4)le0`. `therefore (x-3ge0andx-4lt0)or(x-3le0andx-4gt0)` `rArr (xge3 andxlt4)or(xlt3 andx gt4)` `rArr 3lexlt4" "[thereforexlt3 andx gt4 " isnot possible"]` `rArr x in[3,4)" "["this include " xge2]`. Case II When `x-2lt0` In this case, `xlt 2 and |x-2|=-(x-2)=-x+2`. So, the given inequation becomes `(-x+2-1)/(-x+2-2)le0rArr(-x+1)/(-x)le0` `rArr(x-1)/(x)le0` [mulitplying num. and denom.by -1] `therefore(x-1le0andxgt0)or(x-ge0 and xlt0)` `rArr(xle1 andxgt0)or(xge1andx lt0)` `rArr 0ltxle1" "[thereforex ge1andx lt0 " is not possible"]` `rArrx in(0,1]" "["this include " xlt2]`. Hence, from the above two cases, we get Solution set `=(0,1]uu[3,4)`. |
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