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  4. Let f(x)=(x-3/x+1), x ≠-1. Then f2010(2014) (where fn(x)= fof ldots of (x)(n times .)) is

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Q. Let $f(x)=\frac{x-3}{x+1}, x \neq-1$. Then $f^{2010}(2014)$ (where $f^n(x)=$ fof $\ldots$ of $(x)(n$ times $\left.)\right)$ is

Relations and Functions

Solution:

$f^2(x)=f o f(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 $f^{3 k }(x) =x$
$f^{2010}(2014) =f^{3.670}(2014)=2014$