A) \[\left( -\infty ,-2 \right)\cup \left( 2,\infty \right)\]
B) \[\left( -\infty ,-\sqrt{2} \right)\cup \left( \sqrt{2},\infty \right)\]
C) \[\left( -\infty ,-1 \right)\cup \left( 1,\infty \right)\]
D) \[\left( \sqrt{2},\infty \right)\]
Correct Answer: B
Solution :
For \[\operatorname{x}\ge -\,2,\,\,{{x}^{2}}-x-2+x>0\] \[\Rightarrow \,\,\,{{x}^{2}}>2\,\,\Rightarrow \,\,x\in (-\infty ,-\sqrt{2})\cup \,\,(\sqrt{2},\,\,\infty )\] \[\Rightarrow \,\,\,\operatorname{x}\in \left[ -\,2,-\sqrt{2} \right)\cup \left( \sqrt{2},\,\,\infty \right) For\,\,x<-\,2\] \[{{\operatorname{x}}^{2}}+x+2+x>0 \,or\,\,{{x}^{2}}+2x+2>0\] which is true for all x. Hence \[\operatorname{x}\,\in (-\,\infty ,\,-\sqrt{2})\,\cup (\sqrt{2},\,\infty )\]
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