A) \[(-5,-2)\]
B) \[(-1,\infty )\]
C) \[(-5,-2)\cup (-1,\infty )\]
D) None of these
Correct Answer: C
Solution :
| [c] We have \[\frac{\left| x+3 \right|+x}{x+2}>1\] |
| \[\Rightarrow \frac{\left| x+3 \right|+x}{x+2}-1>0\Rightarrow \frac{\left| x+3 \right|-2}{x+2}>0\] |
| Now, two cases arise: |
| Case I: When \[x+3\ge 0,i.e.x\ge -3\] then, |
| \[\frac{\left| x+3 \right|-2}{x+2}>0\Rightarrow \frac{x+3-2}{x+2}>0\] |
| \[\Rightarrow \frac{x+1}{x+2}>0\] |
| \[\Rightarrow \{(x+1)>0\,\,and\,\,x+2>0\}\] |
| or \[\{x+1<0\,\,and\,\,x+2<0\}\] |
| \[\Rightarrow \{x>-1\,\,and\,\,x>-2\}\] or \[\{x<-1\,\,and\,\,x<-2\}\] |
| \[\Rightarrow x>-1\,\,or\,\,x<-2\] |
| \[\Rightarrow x\in (-1,\infty )\,\,or\,\,x\in (-\infty ,-2)\] |
| \[\Rightarrow x\in (-3,-2)\cup (-1,\infty )\][Since \[x\ge -3\]] ? (i) |
| Case II: When \[x+3<0,i.e.x<-3\] |
| \[\frac{\left| x+3 \right|-2}{x+2}>0\Rightarrow \frac{-x-3-2}{x+2}>0\] |
| \[\Rightarrow \frac{-(x+5)}{x+2}>0\Rightarrow \frac{x+5}{x+2}<0\] |
| \[\Rightarrow (x+5<0\,\,and\,\,x+2>0)\,\,\,or\,\,\,(x+5>0)\] |
| 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 |
| \[x\in (-5,-2)\cup (-1,\infty )\] |
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