A) \[(-\infty ,\,\,-2)\,\cup (2,\,\infty )\]
B) \[(-\infty ,\,\,-\sqrt{2})\,\cup (\sqrt{2},\,\infty )\]
C) \[(-\infty ,\,\,-1)\,\cup (1,\,\infty )\]
D) \[(\sqrt{2},\,\infty )\]
Correct Answer: B
Solution :
Case I: When \[x+2\ge 0\] i.e. \[x\ge -2,\] Then given inequality becomes \[{{x}^{2}}-(x+2)+x>0\] Þ \[{{x}^{2}}-2>0\Rightarrow \,\,|x|\,>\sqrt{2}\] Þ \[x<-\sqrt{2}\] or \[x>\sqrt{2}\] As \[x\ge -2,\]therefore, in this case the part of the solution set is \[[-2,-\sqrt{2})\cup (\sqrt{2},\infty ).\] Case II: When \[x+2\le 0\] i.e. \[x\le -2,\] Then given inequality becomes \[{{x}^{2}}+(x+2)+x>0\] \[\Rightarrow {{x}^{2}}+2x+2>0\]\[\Rightarrow {{(x+1)}^{2}}+1>0,\] which is true for all real x Hence, the part of the solution set in this case is \[(-\infty ,-2]\]. Combining the two cases, the solution set is \[(-\infty ,-2)\cup ([-2,-\sqrt{2}]\cup (\sqrt{2},\infty )\] \[=(-\infty ,-\sqrt{2})\cup (\sqrt{2},\infty ).\]
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