A) \[(-\infty ,\,\,3]\]
B) \[(-\infty ,\,\,-2)\cup (2,\,\,\infty )\]
C) \[[-2,\,\,2]\]
D) \[[-3,\,\,\infty )\]
Correct Answer: B
Solution :
\[{{x}^{4}}+{{x}^{2}}=1={{({{x}^{2}}+1)}^{2}}-{{x}^{2}}\] \[=({{x}^{2}}+x+1)({{x}^{2}}-x+1)\] \[{{x}^{2}}+x+1={{\left( x+\frac{1}{2} \right)}^{2}}+\frac{3}{4}\ne 0\] \[\forall \,\,x\] Now,\[(a-1){{({{x}^{2}}+x+1)}^{2}}=(a+1)({{x}^{4}}+{{x}^{2}}+1)\] \[(a-1)({{x}^{2}}+x+1)=(a+1)({{x}^{2}}-x+1)\] \[\Rightarrow {{x}^{2}}-ax+1=0\]which has real and distinct roots \[\therefore \] \[\Delta ={{a}^{2}}-4>0\] \[\Rightarrow \] \[{{a}^{2}}>4\] \[\Rightarrow \] \[a\in (-\infty ,\,\,-2)\cup (2,\,\,\infty )\]
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