Let fx=x+1x−1, fofx=f2x; fofofx=f3x, fofofofx=f4x.... then f2008x=

fx=x+1x−1fofx=ffx=fx+1x−1=x+1x−1+1x+1x−1−1⇒ x+1+x−1x+1−x+1=2x2=xfofofx=ffofx=fxfnx is x if n is even , f x if n is odd ∴ f2008(x) is x

Let fx=x+1x−1, fofx=f2x; fofofx=f3x, fofofofx=f4x.... then f2008x=

fx=x+1x−1fofx=ffx=fx+1x−1=x+1x−1+1x+1x−1−1⇒ x+1+x−1x+1−x+1=2x2=xfofofx=ffofx=fxfnx is x if n is even , f x if n is odd ∴ f2008(x) is x

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$L e t f (x) = \frac{x + 1}{x - 1}, f o f (x) = f^{2} (x); f o f o f (x) = f^{3} (x), f o f o f o f (x) = f^{4} (x) .... t h e n f^{2008} (x) =$

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$x$

$\frac{1}{x}$

$f (x)$

answer is A.

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Detailed Solution

$\begin{array}{l} f (x) = \frac{x + 1}{x - 1} \\ f o f (x) = f [f (x)] = f (\frac{x + 1}{x - 1}) = \frac{\frac{x + 1}{x - 1} + 1}{\frac{x + 1}{x - 1} - 1} \Rightarrow \frac{x + 1 + x - 1}{x + 1 - x + 1} = \frac{2 x}{2} = x \\ f o f o f (x) = f [f o f (x)] = f (x) \\ f^{n} (x) i s x i f n i s e v e n, f^{} (x) i f n i s o d d \\ ∴ f^{2008} (x) i s x \end{array}$