Let $$f\left( x \right) = \frac{x}{{1 - x}}$$   and $$'a'$$  be a real number. If $${x_0} = a,\,{x_1} = f\left( {{x_0}} \right),\,{x_2} = f\left( {{x_1}} \right),\,{x_3} = f\left( {{x_2}} \right).....$$           If $${x_{2009}} = 1,$$   then the value of $$a$$ is :

Solution :
$$\eqalign{ & {x_0} = a,\,{x_1} = f\left( x \right) = \frac{{{x_0}}}{{1 - {x_0}}} = \frac{a}{{1 - a}}; \cr & {x_2} = f\left( {{x_1}} \right) = \frac{{{x_1}}}{{1 - {x_1}}} = \frac{{\frac{a}{{1 - a}}}}{{1 - \frac{a}{{1 - a}}}} = \frac{a}{{1 - 2a}} \cr & \therefore\, {x_{2009}} = \frac{a}{{1 - 2009\,a}} = 1 \cr & \Rightarrow 1 - 2009\,a = a \cr & \Rightarrow a = \frac{1}{{2010}} \cr} $$