Question

For $x \in R$ , $x \ne 0$ , $x \ne 1$ , let ${f_0}\left( x \right) = \dfrac{1}{{1 - x}}$ and ${f_{n + 1}}\left( x \right) = {f_0}\left( {{f_n}\left( x \right)} \right)$ , n = 0, 1, 2, ….Then the value of ${f_{100}}\left( 3 \right) + {f_1}\left( {\dfrac{2}{3}} \right) + {f_2}\left( {\dfrac{3}{2}} \right)$ is equal to
A. $\dfrac{8}{3}$
B. $\dfrac{5}{3}$
C. $\dfrac{1}{3}$
D. $\dfrac{4}{3}$

Answer 1

Hint: First, we will find ${f_1}\left( x \right)$ by using ${f_{n + 1}}\left( x \right) = {f_0}\left( {{f_n}\left( x \right)} \right)$ and substitute $x = \dfrac{2}{3}$ in ${f_1}\left( x \right)$ to get ${f_1}\left( {\dfrac{2}{3}} \right)$ .Similarly, we will find the value of ${f_2}\left( {\dfrac{3}{2}} \right)$ and ${f_{100}}\left( 3 \right)$ and after summing all three, we can get the correct answer.

Complete step-by-step answer:
We have been given that
 ${f_0}\left( x \right) = \dfrac{1}{{1 - x}}$
Also, ${f_{n + 1}}\left( x \right) = {f_0}\left( {{f_n}\left( x \right)} \right)$ .
So, ${f_1}\left( x \right) = {f_0}\left( {{f_0}\left( x \right)} \right)$
Hence, using the given formula, we can write ${f_1}\left( x \right)$ as
     \[
   = \dfrac{1}{{1 - {f_0}\left( x \right)}} \\
   = \dfrac{1}{{1 - \left( {\dfrac{1}{{1 - x}}} \right)}} \\
   = \dfrac{1}{{\left( {\dfrac{{1 - x - 1}}{{1 - x}}} \right)}} \\
 \]
  \[
   = \dfrac{{1 - x}}{{ - x}} \\
   = \dfrac{{x - 1}}{x} \\
 \]
 $\therefore {f_1}\left( {\dfrac{2}{3}} \right) = \dfrac{{\dfrac{2}{3} - 1}}{{\dfrac{2}{3}}}$
  $
   = \dfrac{{\dfrac{{2 - 3}}{3}}}{{\dfrac{2}{3}}} \\
   = \dfrac{{ - 1}}{2} \\
 $
Now, ${f_2}\left( x \right) = {f_0}\left( {{f_1}\left( x \right)} \right)$
         $
   = \dfrac{1}{{1 - {f_1}\left( x \right)}} \\
   = \dfrac{1}{{1 - \left( {\dfrac{{x - 1}}{x}} \right)}} \\
   = \dfrac{1}{{\left( {\dfrac{{x - x + 1}}{x}} \right)}} \\
   = \dfrac{x}{1} \\
   = x \\
 $
 \[\therefore {f_2}\left( {\dfrac{3}{2}} \right) = \dfrac{3}{2}\]
Now, ${f_3}\left( x \right) = {f_0}\left( {{f_2}\left( x \right)} \right)$
         $
   = \dfrac{1}{{1 - {f_2}\left( x \right)}} \\
   = \dfrac{1}{{1 - x}} \\
 $
Similarly, ${f_{100}}\left( x \right) = \dfrac{{x - 1}}{x}$
 $\therefore {f_{100}}\left( 3 \right) = \dfrac{{3 - 1}}{3}$
\[\] $ = \dfrac{2}{3}$
Now, adding ${f_{100}}\left( 3 \right)$ , ${f_1}\left( {\dfrac{2}{3}} \right)$ and ${f_2}\left( {\dfrac{3}{2}} \right)$ .
 \[
   = {f_{100}}\left( 3 \right) + {f_1}\left( {\dfrac{2}{3}} \right) + {f_2}\left( {\dfrac{3}{2}} \right) \\
   = \dfrac{2}{3} - \dfrac{1}{2} + \dfrac{3}{2} \\
   = \dfrac{{4- 3 + 9}}{6} \\
   = \dfrac{{10}}{6} \\
   = \dfrac{5}{3} \\
 \]
Thus, Option (B) $\dfrac{5}{3}$ is the correct answer.

Note: The ${f_{n + 1}}\left( x \right) = {f_0}\left( {{f_n}\left( x \right)} \right)$ can be done as $fog\left( x \right) = f\left( {g\left( x \right)} \right)$ , in which we need to put value of \[g\left( x \right)\] first and then we can find the value of \[f\left( {g\left( x \right)} \right)\] i.e. $fog(x)$
For example, let $f\left( x \right) = 1 + {x^2}$ and $g\left( x \right) = {x^3}$
$\therefore fog\left( x \right) = f\left( {g\left( x \right)} \right)$
$
   = 1 + {\left( {{x^3}} \right)^2} \\
   = 1 + {x^6} \\
 $