A) 12
B) 16
C) 14
D) 10
Correct Answer: C
Solution :
Ans. \[x=2+\sqrt{3,}\,\,\,y=2-\sqrt{3}\] \[\therefore \] \[x+y=2+\sqrt{3}+2-\sqrt{3}=4\] and \[xy=(2+\sqrt{3})\,(2-\sqrt{3})\,=4-3=1\] now, \[\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}=\frac{{{x}^{2}}+{{y}^{2}}}{{{(xy)}^{2}}}\] \[=\frac{{{(x+y)}^{2}}-2xy}{{{(xy)}^{2}}}\] \[=\frac{{{(4)}^{2}}-2(1)}{{{(1)}^{2}}}=\frac{16-2}{1}=14\]You need to login to perform this action.
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