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Q.
Let $f(x)=\frac{x-1}{x+1}, x \in R-\{0,-1,1)$. If $f^{n+1}(x)=f\left(f^{n}(x)\right)$ for all $n \in N$, then $f^{6}(6)+f^{7}(7)$ is equal to:
JEE MainJEE Main 2022Relations and Functions - Part 2
Solution:
$f ( x )=\frac{ x -1}{ x +1}$
$\Rightarrow f ^{2}( x )= f ( f ( x ))=\frac{\frac{ x -1}{ x +1}-1}{\frac{ x -1}{ x +1}+1}=-\frac{1}{ x }$
$f ^{3}( x )= f \left( f ^{2}( x )\right)= f \left(-\frac{1}{ x }\right)=\frac{ x +1}{1- x }$
$\Rightarrow f ^{4}( x )= f \left(\frac{ x +1}{1- x }\right)=-\frac{1}{ x }$
$\Rightarrow f ^{6}( x )=-\frac{1}{ x } \Rightarrow f ^{6}(6)=-\frac{1}{8}$
$f ^{7}( x )=\left(-\frac{1}{ x }\right)=\frac{ x +1}{1- x }$
$\Rightarrow f ^{7}(7)=\frac{8}{-6}=-\frac{4}{3}$
$\therefore-\frac{1}{6}+-\frac{4}{3}=-\frac{3}{2}$