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# 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 find the value of${f_{100}}\left( 3 \right) + {f_1}\left( {\dfrac{2}{3}} \right) + {f_2}\left( {\dfrac{3}{2}} \right)$ .(A) $\dfrac{1}{3}$ (B) $\dfrac{4}{3}$ (C) $\dfrac{8}{3}$ (D) $\dfrac{5}{3}$

Hint:Use the given definition of the function to define ${f_1}\left( x \right)$ and ${f_2}\left( x \right)$. Now find the value of${f_3}\left( x \right)$ and notice that there is a pattern according to the value of $n$ in the definition of the function. The values repeat themselves for different values of $n$.

According to the given data in question the function ${f_0}\left( x \right)$ is defined as ${f_0}\left( x \right) = \dfrac{1}{{1 - x}}$ and ${f_{n + 1}}\left( x \right) = {f_0}\left( {{f_n}\left( x \right)} \right),n = 0,1,2,....$for all$x \in R,x \ne 0,x \ne 1$, i.e. .. can have any real value except $0$ and $1$.
With that information, we need to calculate the value of ${f_{100}}\left( 3 \right) + {f_1}\left( {\dfrac{2}{3}} \right) + {f_2}\left( {\dfrac{3}{2}} \right)$.
Since ${f_{n + 1}}\left( x \right) = {f_0}\left( {{f_n}\left( x \right)} \right)$ therefore for $n = 0$, we get ${f_{0 + 1}}\left( x \right) = {f_0}\left( {{f_0}\left( x \right)} \right) \Rightarrow {f_1}\left( x \right) = {f_0}\left( {{f_0}\left( x \right)} \right)$
Now we can use the given definition of ${f_0}\left( x \right)$ in the above relation:
$\Rightarrow {f_1}\left( x \right) = {f_0}\left( {{f_0}\left( x \right)} \right) = \dfrac{1}{{1 - {f_0}\left( x \right)}} = \dfrac{1}{{1 - \dfrac{1}{{1 - x}}}}$
It can be further simplified by evaluating the fraction as:
$\Rightarrow {f_1}\left( x \right) = \dfrac{1}{{1 - \dfrac{1}{{1 - x}}}} = \dfrac{1}{{\dfrac{{1 - x - 1}}{{1 - x}}}} = \dfrac{{1 - x}}{{ - x}} = \dfrac{{x - 1}}{x}$
Again, for $n = 1$, by the given definition of ${f_{n + 1}}\left( x \right)$ we get: ${f_2}\left( x \right) = {f_0}\left( {{f_1}\left( x \right)} \right)$ and as per the previous calculations, we can write the value of ${f_2}\left( x \right)$ as:
$\Rightarrow {f_2}\left( x \right) = {f_0}\left( {{f_1}\left( x \right)} \right) = {f_0}\left( {{f_0}\left( {{f_0}\left( x \right)} \right)} \right) = {f_0}\left( {\dfrac{{x - 1}}{x}} \right) = \dfrac{1}{{1 - \dfrac{{\left( {x - 1} \right)}}{x}}} = \dfrac{x}{{x - x + 1}} = \dfrac{x}{{0 + 1}} = x$
Therefore, we get ${f_1}\left( x \right) = \dfrac{{x - 1}}{x}$ and ${f_2}\left( x \right) = x$
Similarly, for $n = 2$ we get ${f_3}\left( x \right) = {f_0}\left( {{f_2}\left( x \right)} \right) = {f_0}\left( x \right) = \dfrac{1}{{1 - x}}$
So, now let’s analyse the results we got:${f_0}\left( x \right) = \dfrac{1}{{1 - x}}$,${f_1}\left( x \right) = \dfrac{{x - 1}}{x}$,${f_2}\left( x \right) = x$,${f_3}\left( x \right)$= \dfrac{1}{{1 - x}},……. Here we can notice the pattern in the results, therefore, according to the pattern: Similarly,{f_4}\left( x \right) = \dfrac{{x - 1}}{x}, then {f_5}\left( x \right) = x, then again {f_6}\left( x \right) = \dfrac{1}{{1 - x}} With the above calculations, we can conclude that: For values n = 0,3,6,9,12,15....... {f_n}\left( x \right) = \dfrac{1}{{1 - x}} For values n = 1,4,7,10,13,16....... {f_n}\left( x \right) = \dfrac{{x - 1}}{x} For values n = 2,5,8,11,14,17........ {f_n}\left( x \right) = x Therefore, if the value of n = 100 = \left( {3 \times 33} \right) + 1$ \Rightarrow {f_{100}}\left( x \right) = \dfrac{{x - 1}}{x}$
Now we can use the values of the function ${f_{100}}\left( x \right) = \dfrac{{x - 1}}{x}$,${f_1}\left( x \right) = \dfrac{{x - 1}}{x}$and ${f_2}\left( x \right) = x$ to evaluate the value of the expression ${f_{100}}\left( 3 \right) + {f_1}\left( {\dfrac{2}{3}} \right) + {f_2}\left( {\dfrac{3}{2}} \right)$
$\Rightarrow {f_{100}}\left( 3 \right) + {f_1}\left( {\dfrac{2}{3}} \right) + {f_2}\left( {\dfrac{3}{2}} \right) = \left( {\dfrac{{3 - 1}}{3}} \right) + \left( {\dfrac{{\dfrac{2}{3} - 1}}{{\dfrac{2}{3}}}} \right) + \left( {\dfrac{3}{2}} \right) = \left( {\dfrac{2}{3}} \right) + \left( {\dfrac{{ - 1}}{2}} \right) + \left( {\dfrac{3}{2}} \right) = \dfrac{2}{3} + 1 = \dfrac{5}{3}$
Hence, the value for the given expression is $\dfrac{5}{3}$.

So, the correct answer is “Option D”.

Note:Use the brackets carefully to evaluate the function definition. Do not put the value of $x$ beyond its domain. Check for a pattern with different values of $n$ by repeating the pattern at least once.