A) \[\left( \frac{9}{2},\infty \right)\]
B) \[\left( -\infty ,\frac{3}{2} \right)\]
C) \[\left( -2,-\frac{3}{2} \right)\]
D) \[\left( -1,\frac{3}{2} \right)\]
Correct Answer: A
Solution :
| [a] The inequality is \[\left| x+2 \right|-\left| x-2 \right|<x-\frac{3}{2}.\]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 |
| \[x>\frac{9}{2}\Rightarrow x\in \left( \frac{9}{2},\infty \right)\] |
| \[\therefore \] Solution set is \[x\in \left( \frac{9}{2},\infty \right)\] |
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