A) 34
B) 40
C) 25
D) 38
Correct Answer: A
Solution :
\[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{{{(\sqrt{2I}+\sqrt{I})}^{2}}}{{{(\sqrt{2I}-\sqrt{I})}^{2}}}={{\left( \frac{\sqrt{2}+1}{\sqrt{2}-1} \right)}^{2}}\] \[={{\left[ \frac{(\sqrt{2}+1)(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)} \right]}^{2}}\] \[={{\left( \frac{2+1+2\sqrt{2}}{2-1} \right)}^{2}}\] \[=\frac{9+8+12\sqrt{2}}{1}=34\]
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