A) \[-4<x<0\]
B) \[0<x<1\]
C) \[-100<x<100\]
D) \[-\infty <x<\infty \]
Correct Answer: D
Solution :
\[{{x}^{12}}-{{x}^{9}}+{{x}^{4}}-x+1>0,\]three cases arix Case I: When \[x\le 0\] \[{{x}^{12}}>0,-{{x}^{9}}>0,{{x}^{4}}>0,-x>0\] \[\Rightarrow \] \[{{x}^{12}}-{{x}^{9}}+{{x}^{4}}-x+1>0,\forall x\le 0\] ...(i) Case II: When\[0<x\le 1\] \[{{x}^{9}}<{{x}^{4}},x<1\]\[\Rightarrow \]\[-{{x}^{9}}+{{x}^{4}}>0\]and\[1-x>0\] \[\therefore \]\[{{x}^{12}}-{{x}^{9}}+{{x}^{4}}-x+1>0,\forall 0<x\le 1\] ...(ii) Case III: When \[x>1\] \[{{x}^{12}}>{{x}^{9}},\text{ }{{x}^{4}}>x\] \[\Rightarrow \] \[{{x}^{12}}-{{x}^{9}}\text{+}{{x}^{4}}-x+1>0,\forall x>1\] ...(iii) \[\therefore \] From Eqs. (i), (ii) and (iii) the above equation hold for\[x\in R\]. Hence, option is the correct answer.
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