A) 193
B) 194
C) 195
D) 196
Correct Answer: B
Solution :
Given \[x=7+4\sqrt{3}\] \[a\times b=b\times a\] \[(a+b)+c=a+(b+c)\] And \[x+\frac{1}{x}=7+4\sqrt{3}+7-4\sqrt{3}=14\] Now, \[{{x}^{2}}+\frac{1}{{{x}^{2}}}={{\left( x+\frac{1}{x} \right)}^{2}}-2={{(14)}^{2}}-2=194\]
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