A) 12
B) 14
C) 16
D) 10
Correct Answer: B
Solution :
\[a=2+\sqrt{3}\] \[\frac{1}{a}=\frac{1}{2+\sqrt{3}}=\frac{1}{(2+\sqrt{3})}\times \frac{2-\sqrt{3}}{(2-\sqrt{3})}\] \[=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}\] \[\therefore \]\[{{a}^{2}}+\frac{1}{{{a}^{2}}}={{\left( a+\frac{1}{a} \right)}^{2}}-2\] \[={{(2+\sqrt{3}+2-\sqrt{3})}^{2}}-2\] \[=16-2=14\]You need to login to perform this action.
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