A) \[\frac{25}{16}\]
B) \[\frac{16}{26}\]
C) \[\frac{1}{9}\]
D) \[\frac{9}{1}\]
Correct Answer: A
Solution :
\[\frac{{{I}_{\max }}}{{{I}_{\min }}}={{\left( \frac{\frac{{{a}_{1}}}{{{a}_{2}}}+1}{\frac{{{a}_{1}}}{{{a}_{2}}}-1} \right)}^{2}}={{\left( \frac{\frac{1}{9}+1}{\frac{1}{9}-1} \right)}^{2}}={{\left( \frac{5}{4} \right)}^{2}}=\frac{25}{16}.\]
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