A) \[{{E}_{1}}{{r}_{1}}>{{E}_{2}}({{r}_{1}}+R)\]
B) \[{{n}_{1}}\]
C) \[{{n}_{1}}({{n}_{1}}>{{n}_{2}}).\]
D) \[{{\alpha }_{\max }}\]
Correct Answer: C
Solution :
We have,\[{{2}^{x}}+{{2}^{y}}={{2}^{x+y}}\] On differentiating both sides w.r.t. x, we get \[{{2}^{x}}\log 2+{{2}^{y}}\log 2\frac{dy}{dx}\] \[={{2}^{x+y}}\log 2\left( 1+\frac{dy}{dx} \right)\] \[\Rightarrow \]\[({{2}^{y}}-{{2}^{x+y}})\frac{dy}{dx}={{2}^{x+y}}-{{2}^{x}}\] \[\Rightarrow \]\[\frac{dy}{dx}=\frac{{{2}^{x}}({{2}^{y}}-1)}{{{2}^{y}}(1-{{2}^{x}})}={{2}^{x-y}}\left( \frac{{{2}^{y}}-1}{1-{{2}^{x}}} \right)\]
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