| If \[a+\frac{1}{a}=\sqrt{3},\]then the value of \[{{a}^{6}}-\frac{1}{{{a}^{6}}}+2\]will be |
A) \[3\sqrt{3}\]
B) \[5\]
C) \[1\]
D) \[2\]
Correct Answer: D
Solution :
| \[a+\frac{1}{a}=\sqrt{3}\] (i) |
| On squaring both sides, we get |
| \[\Rightarrow \] \[{{a}^{2}}+\frac{1}{{{a}^{2}}}+2=3\] |
| \[\Rightarrow \] \[{{a}^{2}}+\frac{1}{{{a}^{2}}}=1\] (ii) |
| Now, multiplying Eqs. (i) and (Ii), we get |
| \[\left( a+\frac{1}{a} \right)\left( {{a}^{2}}+\frac{1}{{{a}^{2}}} \right)=\sqrt{3}\] |
| \[\Rightarrow \] \[{{a}^{3}}+\frac{a}{{{a}^{2}}}+\frac{{{a}^{2}}}{a}+\frac{1}{{{a}^{3}}}=\sqrt{3}\] |
| \[\Rightarrow \] \[{{a}^{3}}+\frac{1}{{{a}^{3}}}+\left( \frac{1}{a}+a \right)=\sqrt{3}\] |
| \[\Rightarrow \] \[{{a}^{3}}+\frac{1}{{{a}^{3}}}+\sqrt{3}=\sqrt{3}\] [from Eq. (i)] |
| \[\Rightarrow \] \[{{a}^{3}}+\frac{1}{{{a}^{3}}}=0\]\[\Rightarrow \]\[{{a}^{6}}=-\,\,1\] |
| \[\therefore \] \[{{a}^{6}}=\frac{1}{{{a}^{6}}}+2={{(-\,\,1)}^{6}}-\frac{1}{{{(-\,\,1)}^{6}}}+2\] |
| \[=1-1+2=2\] |
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