Question

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

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

Answer 1

The correct option is D $\frac{5}{3}$

$f_1\left(x\right)=f_0\left(f_0\right(x\left)\right)=\frac{1}{1-f_0\left(x\right)}=\frac{1}{1-\frac{1}{1-x}}$

$=1-\frac{1}{x}$
$f_2\left(x\right)=f_0\left(f_1\right(x\left)\right)=\frac{1}{1-f_1\left(x\right)}=\frac{1}{1-(1-\frac{1}{x})}=x$
$f_3\left(x\right)=f_0\left(f_2\right(x\left)\right)=f_0\left(x\right)$
Thus $f_{3k}\left(x\right)=f_0\left(x\right),f_{3k+1}\left(x\right)=f_1\left(x\right)$

so $f_{100}\left(x\right)=f_1\left(x\right)$
$f_{3k+2}\left(x\right)=f_2\left(x\right)$ for all $x\in R\sim \{0,1\},k\ge 0$
So, $f_{100}\left(3\right)+f_1\left(\frac{2}{3}\right)+f_2\left(\frac{3}{2}\right)$
$=f_1\left(3\right)+f_1\left(\frac{2}{3}\right)+f_2\left(\frac{3}{2}\right)$
$=1-\frac{1}{3}+1-\frac{3}{2}+\frac{3}{2}=\frac{5}{3}$.