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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 \({f_{n + 1}}...

For x∈R,x≠0,x≠1,x \in R,x \ne 0,x \ne 1,let f0(x)=11−x{f_0}\left( x \right) = \dfrac{1}{{1 - x}} and fn+1(x)=f0(fn(x)),n=0,1,2,....{f_{n + 1}}\left( x \right) = {f_0}\left( {{f_n}\left( x \right)} \right),n = 0,1,2,....Then find the value off100(3)+f1(23)+f2(32){f_{100}}\left( 3 \right) + {f_1}\left( {\dfrac{2}{3}} \right) + {f_2}\left( {\dfrac{3}{2}} \right) .
(A) 13\dfrac{1}{3}
(B) 43\dfrac{4}{3}
(C) 83\dfrac{8}{3}
(D) 53\dfrac{5}{3}

Explanation

Solution

Use the given definition of the function to define f1(x){f_1}\left( x \right) and f2(x){f_2}\left( x \right). Now find the value off3(x){f_3}\left( x \right) and notice that there is a pattern according to the value of nn in the definition of the function. The values repeat themselves for different values of nn.

Complete step-by-step answer:
According to the given data in question the function f0(x){f_0}\left( x \right) is defined as f0(x)=11−x{f_0}\left( x \right) = \dfrac{1}{{1 - x}} and fn+1(x)=f0(fn(x)),n=0,1,2,....{f_{n + 1}}\left( x \right) = {f_0}\left( {{f_n}\left( x \right)} \right),n = 0,1,2,....for allx∈R,x≠0,x≠1x \in R,x \ne 0,x \ne 1, i.e. .. can have any real value except 00 and 11.
With that information, we need to calculate 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).
Since fn+1(x)=f0(fn(x)){f_{n + 1}}\left( x \right) = {f_0}\left( {{f_n}\left( x \right)} \right) therefore for n=0n = 0, we get f0+1(x)=f0(f0(x))⇒f1(x)=f0(f0(x)){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 f0(x){f_0}\left( x \right) in the above relation:
⇒f1(x)=f0(f0(x))=11−f0(x)=11−11−x\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:
⇒f1(x)=11−11−x=11−x−11−x=1−x−x=x−1x\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=1n = 1, by the given definition of fn+1(x){f_{n + 1}}\left( x \right) we get: f2(x)=f0(f1(x)){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 f2(x){f_2}\left( x \right) as:
⇒f2(x)=f0(f1(x))=f0(f0(f0(x)))=f0(x−1x)=11−(x−1)x=xx−x+1=x0+1=x\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 f1(x)=x−1x{f_1}\left( x \right) = \dfrac{{x - 1}}{x} and f2(x)=x{f_2}\left( x \right) = x
Similarly, for n=2n = 2 we get f3(x)=f0(f2(x))=f0(x)=11−x{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:f0(x)=11−x{f_0}\left( x \right) = \dfrac{1}{{1 - x}},f1(x)=x−1x{f_1}\left( x \right) = \dfrac{{x - 1}}{x},f2(x)=x{f_2}\left( x \right) = x,f3(x){f_3}\left( x \right) =11−x = \dfrac{1}{{1 - x}},…….
Here we can notice the pattern in the results, therefore, according to the pattern:
Similarly,f4(x)=x−1x{f_4}\left( x \right) = \dfrac{{x - 1}}{x}, then f5(x)=x{f_5}\left( x \right) = x, then again f6(x)=11−x{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.......n = 0,3,6,9,12,15....... fn(x)=11−x{f_n}\left( x \right) = \dfrac{1}{{1 - x}}
For values n=1,4,7,10,13,16.......n = 1,4,7,10,13,16....... fn(x)=x−1x{f_n}\left( x \right) = \dfrac{{x - 1}}{x}
For values n=2,5,8,11,14,17........n = 2,5,8,11,14,17........ fn(x)=x{f_n}\left( x \right) = x
Therefore, if the value of n=100=(3×33)+1n = 100 = \left( {3 \times 33} \right) + 1 ⇒f100(x)=x−1x \Rightarrow {f_{100}}\left( x \right) = \dfrac{{x - 1}}{x}
Now we can use the values of the function f100(x)=x−1x{f_{100}}\left( x \right) = \dfrac{{x - 1}}{x},f1(x)=x−1x{f_1}\left( x \right) = \dfrac{{x - 1}}{x}and f2(x)=x{f_2}\left( x \right) = x to evaluate the value of the expression 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)
⇒f100(3)+f1(23)+f2(32)=(3−13)+(23−123)+(32)=(23)+(−12)+(32)=23+1=53\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 53\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 xx beyond its domain. Check for a pattern with different values of nn by repeating the pattern at least once.