A) \[\pi /6\]
B) \[\pi /4\]
C) \[\pi /3\]
D) \[\pi /2\]
Correct Answer: B
Solution :
\[\tan \,(\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\] Þ \[\tan (\alpha +\beta )=\frac{\frac{1}{1+\frac{1}{{{2}^{x}}}}+\frac{1}{1+{{2}^{x+1}}}}{1-\frac{1}{1+1/{{2}^{x}}}\frac{1}{1+{{2}^{x+1}}}}\] Þ \[\tan (\alpha +\beta )=\frac{{{2}^{x}}+{{2.2}^{x+x}}+{{2}^{x}}+1}{1+{{2}^{x}}+{{2.2}^{x}}+{{2.2}^{x+x}}-{{2}^{x}}}\] Þ \[\tan (\alpha +\beta )=1\]\[\Rightarrow \alpha +\beta =\frac{\pi }{4}\].
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