1. Tardigrade
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  4. For x ∈ R , x ≠ 0, x ≠ 1, let f0(x) = (1/1-x) and fn+1 (x) | = f0 (fn(x)), n = 0 , 1 , 2 , ... Then the value of f100(3) + f1 ((2/3) ) + f2 ( (3/2) ) is equal to :

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Q. For $x \, \in \, R , x \neq 0, x \neq 1,$ let $f_0(x) = \frac{1}{1-x}$ and $f_{n+1} (x) | = f_0 (f_n(x)), n = 0 , 1 , 2 , ...$ Then the value of $f_{100}(3) + f_1 \left(\frac{2}{3} \right) + f_2 \left( \frac{3}{2} \right)$ is equal to :

JEE MainJEE Main 2016Relations and Functions - Part 2

Solution:

$f_{1} \left(x\right)=f_{0+1}\left(x\right)=f_{0}\left(f_{0}\left(x\right)\right)=\frac{1}{1-\frac{1}{1+-x}}=\frac{x-1}{x}$
$f_{2}\left(x\right)=f_{1+1}\left(x\right)=f_{0}\left(f_{1}\left(x\right)\right)=\frac{1}{1-\frac{x-1}{x}}=x$
$f_{3}\left(x\right)=f_{2+1}\left(x\right)=f_{0}\left(f_{2}\left(x\right)\right)=f_{0}\left(x\right)=\frac{1}{1-x}$
$f_{4}\left(x\right)=f_{3+1}\left(x\right)=f_{0}\left(f_{0}\left(x\right)\right)=\frac{x-1}{x}$
$\therefore f_{0}=f_{3}=f_{6}=.........=\frac{1}{1-x}$
$f_{1}=f_{4}=f_{7}=f_{10}=......=\frac{x-1}{x}$
$f_{2}=f_{5}=f_{8}=.......=x$
$f_{100}\left(3\right)=\frac{3-1}{3}=\frac{2}{3}f_{1}\left(\frac{2}{3}\right)=\frac{\frac{2}{3}-1}{\frac{2}{3}}=-\frac{1}{2}$
$f_{2}\left(\frac{3}{2}\right)=\frac{3}{2}$
$\therefore f_{100}\left(3\right)+f_{1}\left(\frac{2}{3}\right)+f_{2}\left(\frac{3}{2}\right)=\frac{5}{3}$