Equations And Inequalities Question 50
Question: Solve for $ x,,\frac{| x+3 |+x}{x+2}>1 $
Options:
A) $ x\in (-5,-2)\cup (-1,\infty ) $
B) $ x\in (5,2)\cup (-1,\infty ) $
C) $ x\in (5,2) $
D) $ x\in (-1,\infty ) $
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] We have $ \frac{| x+3 |+x}{x+2}>1 $
$ \Rightarrow \frac{| x+3+x |}{x+2}-1>0 $
$ \Rightarrow \frac{| x+3 |-2}{x+2}>0 $ Now two cases arise:
Case I: When $ x+3\ge 0, $ i.e., $ x\ge -3. $
Then $ \frac{| x+3 |-2}{x+2}>0\Rightarrow \frac{x+3-2}{x+2}>0 $
$ \Rightarrow \frac{x+1}{x+2}>0\Rightarrow {(x+1)>0,and,x+2>0} $
or $ {x+1<0andx+2<0} $
$ \Rightarrow {x>-1andx>-2} $ or $ {x<-1andx<-2} $
$ \Rightarrow x>-1 $ or $ x<-2\Rightarrow x\in (-1,\infty ) $
or $ x\in (-\infty ,-2) $
$ \Rightarrow x\in (-3,-2)\cup (-1,\infty ) $ [Since $ x\ge -3 $ ] ? (1)
Case II: When $ x+3<0, $ i.e., $ x<-3 $
$ \frac{| x+3 |-2}{x+2}>0\Rightarrow \frac{-x-3-2}{x+2}>0 $
$ \Rightarrow \frac{-(x+5)}{x+2}>0\Rightarrow \frac{x+5}{x+2}<0 $
$ \Rightarrow (x+5<0andx+2>0) $
or $ (x+5>0andx+2<0) $
$ \Rightarrow (x<-,5,and,x>-,2) $
or $ (x>-,5,and,x<-,2) $ It is not possible.
$ \Rightarrow x\in (-,5,-,2) $
Combining (1) and (2), the required solution is $ x\in (-5,-2)\cup (-1,\infty ) $
BETA
