A) \[\left( -\infty ,\frac{1}{3} \right)\]
B) \[\left( \frac{1}{3},5 \right)\]
C) \[(5,\infty )\]
D) \[\left( -\infty ,\frac{1}{3} \right)\cup (5,\infty )\]
Correct Answer: B
Solution :
| [b] \[\left| 2x-3 \right|<\left| x+2 \right|\] |
| \[\Rightarrow -\left| x+2 \right|<2x-3<\left| x+2 \right|\] ? (i) |
| Case I: \[x+2\ge 0.\] Then by (i), |
| \[-(x+2)<2x-3<x+2\] |
| \[\Rightarrow -x-2<2x-3<x+2\] |
| \[\Rightarrow 1<3x\,\,and\,\,x<5\Rightarrow \frac{1}{3}<x<5\] |
| Case II: \[x+2<0.\] Then by (i), |
| \[(x+2)<2x-3<-(x+2)\] |
| \[\Rightarrow -(x+2)>2x-3>(x+2)\] |
| \[\Rightarrow 1>3x\] and \[x>5\Rightarrow \frac{1}{3}\le x\] and \[x>5,\] not possible. |
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