$\frac{I_{max}}{I_{min}}=\frac{(\sqrt {I_1}+\sqrt {I_2})^2}{(\sqrt {I_1}-\sqrt {I_2})^2}=\bigg(\frac{\sqrt{I_1 / I_2}+1}{\sqrt{I_1 / I_2}-1}\bigg)^2=9$ (Given)
Solving this, we have $\frac{I_1}{I_2}=4 \, \, but \, \, \, I \propto A^2$
$\therefore \hspace25mm \frac{A_1}{A_2}=2$
$\therefore$ Correct options are (b) and (d).