A) \[0\]
B) \[1\]
C) \[2\]
D) \[3\]
Correct Answer: C
Solution :
[c] \[f({{x}^{2}}+1)=\frac{2}{f({{2}^{x}})-1}\] \[\Rightarrow \,\,\,\,\,\underset{x\to 0}{\mathop{\lim }}\,\,\,f({{x}^{2}}+1)=\underset{x\to 0}{\mathop{\lim }}\,\frac{2}{f({{2}^{x}})-1}\] \[\Rightarrow \,\,\,\,\,L=\frac{2}{L-1}\] (where \[L=\underset{x\to 1}{\mathop{\lim }}\,\,\,f(x)\]) \[\Rightarrow \,\,\,\,\,{{L}^{2}}-L-2=0\] \[\Rightarrow \,\,\,\,\,(L-2)(L+1)=0\] \[\therefore \,\,\,\,\,\,\,\,L=2\,\,\,(\,\,\,f(x)>0)\]
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