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