$(a)\;9:2 \\ (b)\;121:49 \\ (c)\;85:77 \\ (d)16:7 $

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Given,

$\large\frac{I_1}{I_2}= \large\frac{{a_1}^2}{{a_2}^2}$

$\large\frac{I_1}{I_2}=\frac{81}{4}$

$\qquad= \large\frac{9^2}{2^2}$

$\therefore \large\frac{a_1}{a_2}=\frac{9}{2}$

$\large\frac{I_{max}}{I_{min}}=\frac{(a_1+a_2)^2}{(a_1-a_2)^2}$

$\qquad= \large\frac{(a+2)^2}{(a-2)}$

$\qquad= \large\frac{121}{49}$

Hence b is the correct answer.

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