A) 4
B) \[-6\]
C) \[\sqrt{3}\]
D) 2
Correct Answer: D
Solution :
| [d] Given,\[a+\frac{1}{a}=\sqrt{3}\] \[{{a}^{2}}+\frac{1}{{{a}^{2}}}+2=3\] \[\Rightarrow \] \[{{a}^{2}}+\frac{1}{{{a}^{2}}}=3-2\] \[\Rightarrow \] \[{{a}^{2}}+\frac{1}{{{a}^{2}}}=1\] Again squaring, we get \[{{a}^{4}}+\frac{1}{{{a}^{4}}}+2=1\] \[\Rightarrow \] \[{{a}^{4}}+\frac{1}{{{a}^{4}}}=1-2\] \[\Rightarrow \] \[{{a}^{4}}+\frac{1}{{{a}^{4}}}-2+2=-1\] \[\Rightarrow \] \[{{a}^{4}}+\frac{1}{{{a}^{4}}}-2=-1-2\] \[\Rightarrow \] \[{{\left( {{a}^{2}}-\frac{1}{{{a}^{2}}} \right)}^{2}}={{(\sqrt{-3})}^{2}}\] \[\Rightarrow \] \[{{a}^{2}}-\frac{1}{{{a}^{2}}}=\sqrt{-3}\] Now, taking cubes on both sides, we get \[{{({{a}^{2}})}^{3}}-{{\left( \frac{1}{{{a}^{2}}} \right)}^{3}}-3{{a}^{2}}\times \frac{1}{{{a}^{2}}}\left( {{a}^{2}}\frac{1}{{{a}^{2}}} \right)=-3\sqrt{-3}\] \[\Rightarrow \] \[{{a}^{6}}-\frac{1}{{{a}^{6}}}-3\sqrt{-3}=-3\sqrt{-3}\] \[\Rightarrow \] \[{{a}^{6}}-\frac{1}{{{a}^{6}}=0}\] Adding 2 on both sides, we get \[{{a}^{6}}-\frac{1}{{{a}^{6}}}+2=2\] |
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