For $$x \in R - \left\{ {0,1} \right\},$$    let $$\,{f_1}\left( x \right) = \frac{1}{x},\,{f_2}\left( x \right) = 1 - x$$       and $${f_3}\left( x \right) = \frac{1}{{1 - x}}$$    be three given functions. If a function, $$J\left( x \right)$$  satisfies $$\left( {{f_2}oJo{f_1}} \right)\left( x \right) = {\text{ }}{f_3}\left( x \right)$$     then $$J\left( x \right)$$  is equal to

Solution :
The given relation is
$$\eqalign{ & \left( {{f_2}oJo{f_1}} \right)\left( x \right) = {f_3}\left( x \right) = \frac{1}{{1 - x}} \cr & \Rightarrow \left( {{f_2}oJ} \right)\left( {{f_1}\left( x \right)} \right) = \frac{1}{{1 - x}} \cr & \Rightarrow \left( {{f_2}oJ} \right) = \left( {\frac{1}{x}} \right) = \frac{1}{{1 - x}}\,\left[ {\because {f_1}\left( x \right) = \frac{1}{x}} \right] \cr & \Rightarrow {f_2}\left( {J\left( {\frac{1}{x}} \right)} \right) = \frac{1}{{1 - x}} \cr & \Rightarrow \left( {{f_2}J\left( x \right)} \right) = \frac{1}{{1 - \frac{1}{x}}} = \frac{x}{{x - 1}}\left[ {\frac{1}{x}{\text{ is replaced by }}x} \right] \cr & \Rightarrow 1 - J\left( x \right) = \frac{x}{{x - 1}}\left[ {\lambda {f_2}\left( x \right) = 1 - x} \right] \cr & \therefore J\left( x \right) = 1 - \frac{x}{{x - 1}} = \frac{1}{{1 - x}} = {f_3}\left( x \right) \cr} $$