A) 5I and I
B) 5I and 3I
C) 9I and I
D) 9I and 3I
Correct Answer: C
Solution :
\[{{I}_{\max }}={{(\sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}})}^{2}}={{(\sqrt{I}+\sqrt{4I})}^{2}}=9I\] \[{{I}_{\min }}={{(\sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}})}^{2}}={{(\sqrt{I}-\sqrt{4I})}^{2}}=I\]
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