InterviewSolution
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InterviewSolution
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Solve: `|x+1|+|x|gt3,x in R` |
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Answer» Correct Answer - `(-oo,-2)uu(1,oo)` Putting x +1 = 0 and x = 0, we get x = -1 and x = 0 as the critical points. These points divide the whole real line three parts, namely`(-oom-1),[-1,0)` and `[0,oo)`. Case I When`-oo lt x lt -1`. In this case,`x+1lt 0 and xlt0`. `therefore |x+1|=-(x+1) =-x-1 and |x|=-x`. `So,|x+1|+|x|gt3 rArr -x-1-xgt3 rArr -2xgt4rArr x lt-2`. `therefore " solution set in the case " =(-oo,-2)" "[therefore -oolt x lt-1]` Case II When `=1lexlt0`. In the case, `x+1ge0andxlt0`. `therefore |x+1|=x+1 and |x|=-x`. `So, |x+1|+|x|gt3 rArr x+1-xgt3 rArr 1 gt3`, which is abured. Case III When `0 le x lt oo`. In this case, `x+1 gt 0 and x ge 0`. `therefore |x+1| = x+1 and |x| = x`. So, `|x+1| +|x|gt 3 rArr x+1+xgt 3 rArr 2x gt2 rArr x gt 1`. ` therefore` solution set in the case `=(1,oo)`. Hence, solution set `=(-oo,-2)uu(1,00)`. |
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