A) \[\left( -\,\infty ,\,\,\infty \right)\]
B) \[\left[ 0,\,\,\infty \right)\]
C) \[\left( -\infty ,\,\,0 \right]\]
D) \[\operatorname{R}\,/\left[ 0,\,\,1 \right]\]
Correct Answer: A
Solution :
Given \[\operatorname{f}(x)=\sqrt{{{x}^{14}}-{{x}^{11}}+{{x}^{6}}-{{x}^{3}}+{{x}^{2}}+1}\] 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: \[0 \le x \le 1\] \[{{x}^{14}}-{{x}^{11}}+{{x}^{6}}-{{x}^{3}}+{{x}^{2}}+1\] \[=\,\,\,{{x}^{14}}-\left\{ ({{x}^{11}}+{{x}^{6}})-({{x}^{3}}+{{x}^{2}}) \right\}+1>0\] \[\left\{ \because \,\,\,{{x}^{11}}-{{x}^{6}}\le 0,\,\,{{x}^{3}}-{{x}^{2}}\le 0 \right\}\] Case 3 : \[\operatorname{x}< 0\] \[{{x}^{14}}-{{x}^{11}}+{{x}^{6}}-{{x}^{3}}+{{x}^{2}}+1>0\] \[\left( \because \,\,\,{{x}^{11}}<0,\,\,{{x}^{3}}>0,\,\,{{x}^{14}},\,\,\,{{x}^{6}},\,\,{{x}^{2}}>0 \right)\] Thus for all x, \[{{x}^{14}}-{{x}^{11}}+{{x}^{6}}-\,\,{{x}^{3}}+\,\,{{x}^{2}}+1\,\,\ge \,\,0\] Hence the domain of \[\operatorname{f}(x)= R = \left( -\,\infty , \infty \right)\]
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