A) \[\alpha >2\]
B) \[\alpha <-2\]
C) \[|\alpha |\,\,>2\]
D) \[\alpha =2\]
Correct Answer: C
Solution :
We have,\[f(x)=\frac{1-x}{1+x}\] \[\Rightarrow \] \[f(f(x))=f\left( \frac{1-x}{1+x} \right)\] \[=\frac{1-\frac{1-x}{1+x}}{1+\frac{1-x}{1+x}}=x\] Again, \[f(x)=\frac{1-x}{1+x}\] \[\Rightarrow \] \[f(1/x)=\frac{1-\frac{1}{x}}{1+\frac{1}{x}}\] \[=\frac{x-1}{x+1}\] \[\Rightarrow \] \[f(f(1/x))=f\left( \frac{x-1}{x+1} \right)\] \[=\frac{1-\frac{x-1}{x+1}}{1+\frac{x-1}{x+1}}\] \[=1/x\] \[\therefore \] \[\alpha =f(f(x))+f(f(1/x))\] \[\Rightarrow \] \[\alpha =x+\frac{1}{x}\] \[\Rightarrow \] \[|\alpha |=\left| x+\frac{1}{x} \right|\ge 2\] \[\Rightarrow \] \[|\alpha |\,\,\ge 2\]
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