A) \[0<x\le 1\]
B) \[0\le x\le 1\]
C) \[-\infty <x\le 0\]
D) \[-\infty <x<1\]
Correct Answer: D
Solution :
[d] We have, \[{{2}^{x}}+{{2}^{y}}=2\Rightarrow {{2}^{y}}=2-{{2}^{x}}\] \[\Rightarrow y={{\log }_{2}}(2-{{2}^{x}})\Rightarrow 2-{{2}^{x}}>0\] (\[\therefore \] y is real) \[\Rightarrow {{2}^{x}}<2\Rightarrow {{2}^{x-1}}<1\Rightarrow x-1<0\Rightarrow x<1\] \[\Rightarrow -\infty <x<1\]
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