Question

Let $f\left(x\right)=\frac{x}{1-x}$ and let $\alpha$ be a real number. If $x_0=\alpha ,x_1=f\left(x_0\right),x_2=f\left(x_1\right),....$ and $x_{2011}=-\frac{1}{2012}$ then the value of $\alpha$ is

• A. $\frac{2011}{2012}$
• B. $1$
• C. $2011$
• D. $-1$

Answer 1

The correct option is D $-1$

$f\left(x\right)=\frac{x}{1-x}$ also $x_0=\alpha$

$x_1=f\left(x_0\right)=f\left(\alpha \right)=\frac{\alpha }{1-\alpha }$

$x_2=f\left(x_1\right)=f\left(\frac{\alpha }{1-\alpha }\right)=\frac{\frac{\alpha }{1-\alpha }}{1-\frac{\alpha }{1-\alpha }}=\frac{\alpha }{1-2\alpha }$

$x_3=f\left(x_2\right)=f\left(\frac{\alpha }{1-2\alpha }\right)=\frac{\frac{\alpha }{1-2\alpha }}{1-\frac{\alpha }{1-2\alpha }}=\frac{\alpha }{1-3\alpha }$

Analysis the values $x_1,x_2,x_3$, we can say that $x_{2011}=\frac{\alpha }{1-20011\alpha }=-\frac{1}{2012}$

$\Rightarrow 2012\alpha =-1+2011\alpha$

$\Rightarrow 2012\alpha -2011\alpha =-1$

$\Rightarrow \alpha =-1$