A) \[14,8\sqrt{3}\]
B) \[-14,-8\sqrt{3}\]
C) \[14,-8\sqrt{3}\]
D) \[-14,\,8\sqrt{3}\]
Correct Answer: C
Solution :
We have, \[x=2-\sqrt{3}\] \[\therefore \]\[\frac{1}{x}=\frac{1}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3}\] \[\therefore \]\[\left( x-\frac{1}{x} \right)=2-\sqrt{3}+2+\sqrt{3}=4\] Also, \[\left( x-\frac{1}{x} \right)=2-\sqrt{3}-2-\sqrt{3}=-2\sqrt{3}\] Now, \[{{\left( x+\frac{1}{x} \right)}^{2}}={{x}^{2}}+\frac{1}{{{x}^{2}}}+2\] \[\Rightarrow \]\[{{(4)}^{2}}={{x}^{2}}+\frac{1}{{{x}^{2}}}+2\Rightarrow {{x}^{2}}+\frac{1}{{{x}^{2}}}=14\] and \[{{x}^{2}}-\frac{1}{{{x}^{2}}}=\left( x+\frac{1}{x} \right)\left( x-\frac{1}{x} \right)\] \[=4\times (-2\sqrt{3})=-8\sqrt{3}\]
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