A) \[-\infty <x\le 0\]
B) \[-\infty <x\le \frac{e-1}{e}\]
C) \[-\infty <x\le 1\]
D) \[x\ge 1-e\]
Correct Answer: B
Solution :
\[\operatorname{f}(x)=\,\,\sqrt{1+lo{{g}_{e}}(1-x)}\] value of f(x) is real when \[1\,\,+\,\,{{\log }_{e}}\,(1-x)\ge \,\,0\,\,and\,\,1-x>0\] \[\Rightarrow \,\,\,lo{{g}_{e}}(1-x)\ge -1\,\,and\,\,x<1\] \[=\,\,\,{{\log }_{e}} \left( 1- x \right)\ge \,\,lo{{g}_{e}} {{e}^{-}}^{1}\,and x < 1\] \[\Rightarrow \,\,\,1-x\ge \frac{1}{e}\,and\,\,x<1\Rightarrow \,\,\,x\le \frac{e-1}{e}\,\,and\,\,x<1\]
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