A) \[(-\infty ,\infty )\]
B) \[[0,\infty )\]
C) \[(-\infty ,0]\]
D) \[R\backslash [0,1]\]
Correct Answer: A
Solution :
| [a] Given \[f(x)=\sqrt{{{x}^{14}}-{{x}^{11}}+{{x}^{6}}-{{x}^{3}}+{{x}^{2}}+1}\] |
| For \[f(x)\] to be defined, |
| \[{{x}^{14}}-{{x}^{11}}+{{x}^{6}}-{{x}^{3}}+{{x}^{2}}+1\ge 0\] |
| Case 1: \[x\ge 1\] |
| \[{{x}^{14}}-{{x}^{11}}+{{x}^{6}}+{{x}^{3}}+{{x}^{2}}+1\] |
| \[=({{x}^{14}}-{{x}^{11}})+({{x}^{6}}-{{x}^{3}})+({{x}^{2}}+1)>0\] |
| Case 2: \[0\le x\le 1\] |
| \[{{x}^{14}}-{{x}^{11}}+{{x}^{6}}-{{x}^{3}}+{{x}^{2}}+1\] |
| \[={{x}^{14}}\{({{x}^{11}}-{{x}^{11}})+({{x}^{3}}-{{x}^{2}})+1>0\] |
| \[\{\because \,\,\,{{x}^{11}}-{{x}^{6}}\le 0,{{x}^{3}}-{{x}^{2}}\le 0\}\] |
| Case 3: \[x<0\] |
| \[{{x}^{14}}-{{x}^{11}}+{{x}^{6}}-{{x}^{3}}+{{x}^{2}}+1>0\] |
| \[(\because \,\,\,\,{{x}^{11}}<0,{{x}^{3}}<0,{{x}^{14}},{{x}^{6}},{{x}^{2}}>0)\] |
| Thus for all area, |
| \[x,\,\,{{x}^{14}}-{{x}^{11}}+{{x}^{6}}-{{x}^{3}}+{{x}^{2}}+1\ge 0\] |
| Hence the domain of \[f(x)=R=(-\infty ,\infty )\] |
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