Question

Let $f(x)=\frac{x-3}{x+1},x\ne -1$. Then $f^{2010}(2014)$ (where $f^n(x)=$ fof $\dots$ of $(x)(n$ times $))$ is

• A. 2010
• B. 4020
• C. 4028
• D. 2014

Answer 1

Answer: (D)

$f^2(x)=fof(x)=f(f(x))=f\left(\frac{x-3}{x+1}\right)$
$=\frac{\frac{x-3}{x+1}-3}{\frac{x-3}{x+1}+1}=-\frac{x+3}{x-1}$
$f^3(x)=f\left(-\frac{x+3}{x-1}\right)=\frac{\frac{-x-3}{x-1}-3}{\frac{-x-3}{x-1}+1}=x$
Hence $f^{3k}(x)=x$
$f^{2010}(2014)=f^{3.670}(2014)=2014$