A) \[{{x}_{1}}={{x}_{3}}\]and \[{{x}_{2}}={{x}_{4}}\]
B) \[{{x}_{2}}={{x}_{4}}\]but \[{{x}_{1}}\ne {{x}_{3}}\]
C) \[{{x}_{1}}{{x}_{2}}=1,{{x}_{3}}{{x}_{4}}\ne 1\]
D) \[{{x}_{3}}{{x}_{4}}=1,{{x}_{1}}{{x}_{2}}\ne 1\]
Correct Answer: A
Solution :
Using \[AM\ge GM\] \[{{x}_{1}}+\frac{1}{{{x}_{2}}}\ge 2\sqrt{\frac{{{x}_{1}}}{{{x}_{2}}}},{{x}_{2}}+\frac{1}{{{x}_{3}}}\ge 2\sqrt{\frac{{{x}_{2}}}{{{x}_{3}}}}\] \[{{x}_{3}}+\frac{1}{{{x}_{4}}}\ge 2\sqrt{\frac{{{x}_{3}}}{{{x}_{4}}}},{{x}_{4}}+\frac{1}{{{x}_{1}}}\ge 2\sqrt{\frac{{{x}_{4}}}{{{x}_{1}}}}\] \[\Rightarrow \]\[\left( {{x}_{1}}+\frac{1}{{{x}_{2}}} \right)\left( {{x}_{2}}+\frac{1}{{{x}_{3}}} \right)\left( {{x}_{3}}+\frac{1}{{{x}_{4}}} \right)\left( {{x}_{4}}+\frac{1}{{{x}_{1}}} \right)\ge {{2}^{4}}\] But\[\left( {{x}_{1}}+\frac{1}{{{x}_{2}}} \right)\left( {{x}_{2}}+\frac{1}{{{x}_{3}}} \right)\left( {{x}_{3}}+\frac{1}{{{x}_{4}}} \right)\left( {{x}_{4}}+\frac{1}{{{x}_{1}}} \right)=16\] \[\therefore \]\[{{x}_{1}}=\frac{1}{{{x}_{2}}},{{x}_{2}}=\frac{1}{{{x}_{3}}},{{x}_{3}}=\frac{1}{{{x}_{4}}},{{x}_{4}}=\frac{1}{{{x}_{1}}}\] \[\Rightarrow \]\[{{x}_{1}}=2,{{x}_{2}}=\frac{1}{2},{{x}_{3}}=2,{{x}_{4}}=\frac{1}{2}\] \[\Rightarrow \]\[{{x}_{1}}={{x}_{3}}\]and\[{{x}_{2}}={{x}_{4}}\]
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