A) \[x\in (-5,-2)\cup (-1,\infty )\]
B) \[x\in (5,2)\cup (-1,\infty )\]
C) \[x\in (5,2)\]
D) \[x\in (-1,\infty )\]
Correct Answer: A
Solution :
| [a] We have \[\frac{\left| x+3 \right|+x}{x+2}>1\] |
| \[\Rightarrow \frac{\left| x+3+x \right|}{x+2}-1>0\] |
| \[\Rightarrow \frac{\left| x+3 \right|-2}{x+2}>0\] 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\[x\ge -3\]] ? (1) |
| 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 \[(x+5>0\,\,and\,\,x+2<0)\] |
| \[\Rightarrow \,\,\,\,(x<-\,5\,\,\,and\,\,\,x>-\,2)\] |
| or \[(x>-\,5\,\,\,and\,\,\,x<-\,2)\] It is not possible. |
| \[\Rightarrow \,\,\,\,x\in (-\,5,\,\,-\,2)\] |
| Combining (1) and (2), the required solution is \[x\in (-5,-2)\cup (-1,\infty )\] |
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