Functions Question 444
Question: Let $ f(x)=\frac{x}{1-x} $ and ?a? be a real number. If $ x_0=a,x_1=f(x_0),x_2=f(x_1),x_3=f(x_2)… $ If $ x_{2009}=1, $ then the value of a is
Options:
A) 0
B) $ \frac{2009}{2010} $
C) $ \frac{1}{2009} $
D) $ \frac{1}{2010} $
Show Answer
Answer:
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} $
BETA
