A) 0
B) \[\frac{2009}{2010}\]
C) \[\frac{1}{2009}\]
D) \[\frac{1}{2010}\]
Correct Answer: D
Solution :
[d] \[{{x}_{0}}=a,{{x}_{1}}=f(x)=\frac{{{x}_{0}}}{1-{{x}_{0}}}=\frac{a}{1-a};\] \[{{x}_{2}}=f({{x}_{1}})=\frac{{{x}_{1}}}{1-{{x}_{1}}}=\frac{\frac{a}{1-a}}{1-\frac{a}{1-a}}=\frac{a}{1-2a};\] \[\therefore {{x}_{2009}}=\frac{a}{1-2009a}=1\Rightarrow 1-2009a=a\] \[\Rightarrow a=\frac{1}{2010}\]
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