Equations And Inequalities Question 2
Question: The solution set of the inequality $ | x+2 |-| x-1 |<x-\frac{3}{2} $ is
Options:
A) $ ( \frac{9}{2},\infty ) $
B) $ ( -\infty ,\frac{3}{2} ) $
C) $ ( -2,-\frac{3}{2} ) $
D) $ ( -1,\frac{3}{2} ) $
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] The inequality is $ | x+2 |-| x-2 |<x-\frac{3}{2}. $ Dividing the problem into three intervals: (i) if $ x<-2, $ then $ -(x+2)+(x-1)<x-\frac{3}{2} $
$ \Rightarrow x>-\frac{3}{2} $
But $ -\frac{3}{2}>-2, $ hence no common values
$ \Rightarrow x\in \phi $
(ii) If $ -2\le x<1, $ then $ (x+2)+(x-1)<x-\frac{3}{2} $
$ \Rightarrow x<-\frac{5}{2} $
But $ -\frac{5}{2}<-2, $ hence no common values
$ \Rightarrow x\in \phi $
(iii) If $ x\ge 1, $ then $ (x+2)-(x-1)<x-\frac{3}{2} $
$ \Rightarrow x>\frac{9}{2} $
$ \because \frac{9}{2}>1. $
$ \Rightarrow $ common solution is $ x>\frac{9}{2}\Rightarrow x\in ( \frac{9}{2},\infty ) $
$ \therefore $ Solution set is $ x\in ( \frac{9}{2},\infty ) $
BETA
