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Question: For \(x \in R\) , \(x \ne 0\) , \(x \ne 1\) , let \({f_0}\left( x \right) = \dfrac{1}{{1 - x}}\) and...

For x∈Rx \in R , x≠0x \ne 0 , x≠1x \ne 1 , let f0(x)=11−x{f_0}\left( x \right) = \dfrac{1}{{1 - x}} and fn+1(x)=f0(fn(x)){f_{n + 1}}\left( x \right) = {f_0}\left( {{f_n}\left( x \right)} \right) , n = 0, 1, 2, ….Then the value of f100(3)+f1(23)+f2(32){f_{100}}\left( 3 \right) + {f_1}\left( {\dfrac{2}{3}} \right) + {f_2}\left( {\dfrac{3}{2}} \right) is equal to
A. 83\dfrac{8}{3}
B. 53\dfrac{5}{3}
C. 13\dfrac{1}{3}
D. 43\dfrac{4}{3}

Explanation

Solution

First, we will find f1(x){f_1}\left( x \right) by using fn+1(x)=f0(fn(x)){f_{n + 1}}\left( x \right) = {f_0}\left( {{f_n}\left( x \right)} \right) and substitute x=23x = \dfrac{2}{3} in f1(x){f_1}\left( x \right) to get f1(23){f_1}\left( {\dfrac{2}{3}} \right) .Similarly, we will find the value of f2(32){f_2}\left( {\dfrac{3}{2}} \right) and f100(3){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
f0(x)=11−x{f_0}\left( x \right) = \dfrac{1}{{1 - x}}
Also, fn+1(x)=f0(fn(x)){f_{n + 1}}\left( x \right) = {f_0}\left( {{f_n}\left( x \right)} \right) .
So, f1(x)=f0(f0(x)){f_1}\left( x \right) = {f_0}\left( {{f_0}\left( x \right)} \right)
Hence, using the given formula, we can write f1(x){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} \\\ $