A) 0
B) 1
C) 2
D) \[-\,2\]
Correct Answer: A
Solution :
\[x+\frac{1}{x}=\sqrt{3}\] On squaring both sides, \[{{x}^{2}}+\frac{1}{{{x}^{2}}}+2=3\] \[\Rightarrow \]\[{{x}^{2}}+\frac{1}{{{x}^{2}}}=1\] \[\Rightarrow \]\[{{x}^{4}}-{{x}^{2}}+1=0\] \[\therefore \] \[{{x}^{24}}+{{x}^{18}}={{x}^{18}}({{x}^{6}}+1)\] \[={{x}^{18}}({{({{x}^{2}})}^{3}}+{{1}^{3}})\] \[={{x}^{18}}({{x}^{2}}+1)({{x}^{4}}-{{x}^{2}}+1)=0\]
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