Question

For $x\in R,x\ne 0,x\ne 1,$ 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 :

• A. $\frac{8}{3}$
• B. $\frac{5}{3}$
• C. $\frac{4}{3}$
• D. $\frac{1}{3}$

Answer 1

Answer: (B)

$f_1\left(x\right)=f_{0+1}\left(x\right)=f_0\left(f_0\left(x\right)\right)=\frac{1}{1-\frac{1}{1+-x}}=\frac{x-1}{x}$
$f_2\left(x\right)=f_{1+1}\left(x\right)=f_0\left(f_1\left(x\right)\right)=\frac{1}{1-\frac{x-1}{x}}=x$
$f_3\left(x\right)=f_{2+1}\left(x\right)=f_0\left(f_2\left(x\right)\right)=f_0\left(x\right)=\frac{1}{1-x}$
$f_4\left(x\right)=f_{3+1}\left(x\right)=f_0\left(f_0\left(x\right)\right)=\frac{x-1}{x}$
$\therefore f_0=f_3=f_6=.........=\frac{1}{1-x}$
$f_1=f_4=f_7=f_{10}=......=\frac{x-1}{x}$
$f_2=f_5=f_8=.......=x$
$f_{100}\left(3\right)=\frac{3-1}{3}=\frac{2}{3}{f}_1\left(\frac{2}{3}\right)=\frac{\frac{2}{3}-1}{\frac{2}{3}}=-\frac{1}{2}$
$f_2\left(\frac{3}{2}\right)=\frac{3}{2}$
$\therefore f_{100}\left(3\right)+f_1\left(\frac{2}{3}\right)+f_2\left(\frac{3}{2}\right)=\frac{5}{3}$