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