A) \[(-5,\,\,-2)\cup (-1,\,\,\infty )\]
B) \[(-5,\,\,-2)\]
C) \[(-1,\,\,\infty )\]
D) None of these
Correct Answer: A
Solution :
We have, \[\frac{|x+3|+x}{x+2}>1\] Now, two cases arise: Case I When\[x+3\ge 0,\,\,i.e.\,\,x\ge -3\] In this case, we have \[|x+3|\,\,=x+3\] Now, \[\frac{|x+3|-2}{x+2}>0\] \[\Rightarrow \] \[\frac{x+3-2}{x+2}>0\] \[\Rightarrow \] \[\frac{x+1}{x+2}>0\] \[\Rightarrow \] \[x\in (-\infty ,\,\,-2)\cup (-1,\,\,\infty )\] But, \[x\ge -3\] \[\therefore \] \[x\in [-3,\,\,-2)\cup (-1,\,\,\infty )\] Case II When\[x+3<0,\,\,i.e.,\,\,x<-3\] In this case, we have \[|x+3|\,\,=-(x+3)\] Now, \[\frac{|x+3|-2}{x+2}>0\] \[\Rightarrow \] \[\frac{-(x+3)-2}{x+2}>0\] \[\Rightarrow \] \[\frac{-x-5}{x+2}>0\] \[\Rightarrow \] \[\frac{x+5}{x+2}<0\] \[\Rightarrow \] \[-5<x<-2\] But\[x<-3\]. \[\therefore \] \[x\in (-5,\,\,-3)\] Hence, the solution set of the given inequation is \[(-5,\,\,-2)\cup (-1,\,\,\infty )\]
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