Equations-And-Inequalities Question 59
Question: Solution of $ | 2x-3 |<| x+2 | $ is
Options:
A) $ ( -\infty ,\frac{1}{3} ) $
B) $ ( \frac{1}{3},5 ) $
C) $ (5,\infty ) $
D) $ ( -\infty ,\frac{1}{3} )\cup (5,\infty ) $
Correct Answer: B $ \Rightarrow -| x+2 |<2x-3<| x+2 | $ ? (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<3xandx<5\Rightarrow \frac{1}{3}<x<5 $ $ \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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Answer:
Solution:
Case II: $ x+2<0. $ Then by (i),
$ (x+2)<2x-3<-(x+2) $
BETA
