Equations And Inequalities Question 18
Question: If $ \frac{| x+3 |+x}{x+2}>1, $ then $ x\in $
Options:
A) $ (-5,-2) $
B) $ (-1,\infty ) $
C) $ (-5,-2)\cup (-1,\infty ) $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] 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)>0andx+2>0} $
or $ {x+1<0andx+2<0} $
$ \Rightarrow {x>-1andx>-2} $ or $ {x<-1andx<-2} $
$ \Rightarrow x>-1orx<-2 $
$ \Rightarrow x\in (-1,\infty )orx\in (-\infty ,-2) $
$ \Rightarrow x\in (-3,-2)\cup (-1,\infty ) $ [Since $ x\ge -3 $ ] ? (i)
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>0) $
and $ x+2<0) $
$ \Rightarrow (x<-5andx>-2) $
Or $ (x>-5andx<-2) $ it is not possible.
$ \Rightarrow x\in (-5,-2) $ (ii) Combining (i) and (ii), the required solution is $ x\in (-5,-2)\cup (-1,\infty ) $
BETA
