InterviewSolution
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InterviewSolution
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Solve: `(|x+2|-x)/(x)lt2,x in R` |
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Answer» Correct Answer - `(-oo,-2)uu(1,oo)` `(|x+2|-x)/(x)-2lt0 rArr(|x+2|-x-2x)/(x)lt0rArr (|x+2|-3x)/(x)lt0` Case I When ` x +2ge 0` `Then, `x ge -2 and |x+2|=x+2`. `therefore (|x+2|-3x)/(x)lt0 rArr (x+2-3x)/(x)lt0rArr(2-2x)/(x)lt0` `rArr (2)/(x)-2lt0 rArr (2)/(x)lt2 rArr 2x lgt2 rArr x gt1`. `therefore (xge-2 and xgt1)rArr x gt1rArrx in(1,oo)`. Case II When ` x+2 lt0`. Then ` x lt -2 and |x+2| = -(x+2)` `therefore (|x+2|-3x)/(x) lt0 rArr (-x-2-3x)/(x)lt0 rArr(-4x-2)/(x)lt0` `rArr (4x+2)/(x)gt0 rArr 4+(2)/(x)gt0rArr (2)/(x)gt-4`. `rArr -4xlt2 rArr x lt(-1)/(2)`. `therefore x lt-2 and x lt (-1)/(2) rArr x lt-2 rArr x in(-oo,-2)`. Hence, solution set `=(-oo,-2)uu(1,oo)`. |
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