A) 9 :1
B) 3 :1
C) 2:1
D) 1:1
Correct Answer: C
Solution :
\[\frac{{{\operatorname{I}}_{max}}}{{{\operatorname{I}}_{min}}}=\frac{3}{1}\] \[\frac{{{\left( \sqrt{{{\operatorname{I}}_{1}}}+\sqrt{{{\operatorname{I}}_{1}}} \right)}^{2}}}{{{\left( \sqrt{{{\operatorname{I}}_{1}}}-\sqrt{{{\operatorname{I}}_{1}}} \right)}^{2}}}=\frac{3}{1}\] \[\frac{\sqrt{{{\operatorname{I}}_{1}}}+\sqrt{{{\operatorname{I}}_{2}}}}{\sqrt{{{\operatorname{I}}_{1}}}-\sqrt{{{\operatorname{I}}_{2}}}}=\frac{\sqrt{3}}{1}\] \[\sqrt{3}\sqrt{{{\operatorname{I}}_{1}}}-\sqrt{3}\sqrt{{{\operatorname{I}}_{2}}}=\sqrt{{{\operatorname{I}}_{1}}}+\sqrt{{{\operatorname{I}}_{2}}}\] \[\sqrt{3}\sqrt{{{\operatorname{I}}_{1}}}-\sqrt{{{\operatorname{I}}_{1}}}=\sqrt{{{\operatorname{I}}_{2}}}+\sqrt{3}\sqrt{{{\operatorname{I}}_{2}}}\] \[\left( \sqrt{3}-1 \right)\sqrt{{{\operatorname{I}}_{1}}}=\sqrt{{{\operatorname{I}}_{2}}}\left( \sqrt{3}+1 \right)\] \[\frac{\sqrt{{{\operatorname{I}}_{1}}}}{\sqrt{{{\operatorname{I}}_{2}}}}=\frac{{{A}_{1}}}{{{A}_{2}}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}\]
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