A) 0
B) 1
C) 2
D) 3
Correct Answer: A
Solution :
| \[x+\frac{1}{x}=\sqrt{3}\] |
| 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,\[{{x}^{18}}+{{x}^{12}}+{{x}^{6}}+1\] |
| \[={{x}^{12}}({{x}^{6}}+1)+1\,\,({{x}^{6}}+1)\] |
| \[=({{x}^{12}}+1)({{x}^{6}}+1)\] |
| \[=({{x}^{12}}+1)\cdot {{x}^{3}}\left( {{x}^{3}}+\frac{1}{{{x}^{3}}} \right)=0\] |
You need to login to perform this action.
You will be redirected in
3 sec
