A) 4: 3
B) 49:1
C) 64 : 27
D) 81: 49
Correct Answer: B
Solution :
\[\frac{{{\operatorname{I}}_{max}}}{{{\operatorname{I}}_{min}}}=\frac{{{\left( {{\sqrt{\operatorname{I}}}_{1}}+{{\sqrt{\operatorname{I}}}_{2}} \right)}^{2}}}{{{\left( {{\sqrt{\operatorname{I}}}_{1}}-{{\sqrt{\operatorname{I}}}_{2}} \right)}^{2}}}=\frac{16}{9}\] \[\frac{{{\sqrt{\operatorname{I}}}_{1}}+{{\sqrt{\operatorname{I}}}_{2}}}{{{\sqrt{\operatorname{I}}}_{1}}-{{\sqrt{\operatorname{I}}}_{2}}}=\frac{4}{3}\] \[3{{\sqrt{\operatorname{I}}}_{1}}+3{{\sqrt{\operatorname{I}}}_{2}}=4{{\sqrt{\operatorname{I}}}_{1}}-4{{\sqrt{\operatorname{I}}}_{2}}\] \[7{{\sqrt{\operatorname{I}}}_{1}}={{\sqrt{\operatorname{I}}}_{1}}\] \[\frac{{{\operatorname{I}}_{1}}}{{{\operatorname{I}}_{2}}}=\frac{49}{1}\]
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