A) 3 : 1
B) 4 : 1
C) 2 : 1
D) 9 : 1
Correct Answer: B
Solution :
\[\frac{{{I}_{\max }}}{{{I}_{\min }}}={{\left( \frac{{{a}_{1}}+{{a}_{2}}}{{{a}_{1}}-{{a}_{2}}} \right)}^{2}}={{\left( \frac{3a+a}{3a-a} \right)}^{2}}=\frac{4}{1}\]
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