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

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Correct option is D

f2(x)=fof(x)=f(f(x))=fx−3x+1=x−3x+1−3x−3x+1+1=−x+3x−1f3(x)=f−x+3x−1=−x−3x−1−3−x−3x−1+1=x.Hence f3k(x)=xf2010(2014)=f3.670(2014)=2014.

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

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