A) 0
B) 1
C) 2
D) 3
Correct Answer: A
Solution :
| [a] \[x+\frac{1}{x}=\sqrt{3}\] On cubing both sides, we get \[{{x}^{3}}+\frac{1}{{{x}^{3}}}+3\left( x+\frac{1}{x} \right)={{(\sqrt{3})}^{3}}\] \[\Rightarrow \] \[{{x}^{3}}+\frac{1}{{{x}^{3}}}+3\sqrt{3}=3\sqrt{3}\] \[\Rightarrow \] \[{{x}^{3}}+\frac{1}{{{x}^{3}}}=0\] Now, \[{{x}^{18}}+{{x}^{12}}+{{x}^{6}}+1\] \[={{x}^{12}}({{x}^{6}}+1)+1\,({{x}^{6}}+1)\] \[=({{x}^{12}}+1)({{x}^{6}}+1)\] \[=({{x}^{2}}+1).{{x}^{3}}\left( {{x}^{3}}+\frac{1}{{{x}^{3}}} \right)=0\] |
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