Question

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 :

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

Answer 1

The Correct Option is B

The correct answer is B:\(\frac{5}{3}\)
\(f_0(x)=\frac{1}{1-x}\) for \(x∈R\),\(x≠0,x≠1\)
\(f_{n+1}(x)=f_0(f_n(x)),n=0,1,2......\)
 Then, \(f_{1} \left(x\right)=f_{0+1}\left(x\right)=f_{0}\left(f_{0}\left(x\right)\right)=\frac{1}{1-\frac{1}{1+-x}}=\frac{x-1}{x}\)
\(f_{2}\left(x\right)=f_{1+1}\left(x\right)=f_{0}\left(f_{1}\left(x\right)\right)=\frac{1}{1-\frac{x-1}{x}}=x\)
\(f_{3}\left(x\right)=f_{2+1}\left(x\right)=f_{0}\left(f_{2}\left(x\right)\right)=f_{0}\left(x\right)=\frac{1}{1-x}\)
\(f_{4}\left(x\right)=f_{3+1}\left(x\right)=f_{0}\left(f_{0}\left(x\right)\right)=\frac{x-1}{x}\)
\(\therefore f_{0}=f_{3}=f_{6}=.........=\frac{1}{1-x}\)
\(f_{1}=f_{4}=f_{7}=f_{10}=......=\frac{x-1}{x}\)
\(f_{2}=f_{5}=f_{8}=.......=x\)
\(f_{100}\left(3\right)=\frac{3-1}{3}=\frac{2}{3}f_{1}\left(\frac{2}{3}\right)=\frac{\frac{2}{3}-1}{\frac{2}{3}}=-\frac{1}{2}\)
\(f_{2}\left(\frac{3}{2}\right)=\frac{3}{2}\)
\(\therefore f_{100}\left(3\right)+f_{1}\left(\frac{2}{3}\right)+f_{2}\left(\frac{3}{2}\right)=\frac{5}{3}\)