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For $x\,\, \in \,R\,,x\, \ne \,0,$ 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
$\frac {8}{3}$
$\frac {4}{3}$
$\frac {5}{3}$
$\frac {1}{3}$
Solution
${f_1}\left( x \right) = {f_{0 + 1}}\left( x \right) = {f_0}\left( {{f_0}\left( x \right)} \right) = \frac{1}{{1 – \frac{1}{{1 – x}}}} = \frac{{x – 1}}{x}$
${f_2}\left( x \right) = {f_{1 + 1}}\left( x \right) = {f_0}\left( {{f_1}\left( x \right)} \right) = \frac{1}{{1 – \frac{{x – 1}}{x}}} = x$
${f_3}\left( x \right) = {f_{2 + 1}}\left( x \right) = {f_0}\left( {{f_2}\left( x \right)} \right) = {f_0}\left( x \right) = \frac{1}{{1 – x}}$
${f_4}\left( x \right) = {f_{3 + 1}}\left( x \right) = {f_0}\left( {{f_3}\left( x \right)} \right) = \frac{{x – 1}}{x}$
${f_0} = {f_3} = {f_6} = …….. = \frac{1}{{1 – x}}$
${f_1} = {f_4} = {f_7} = …….. = \frac{{x – 1}}{x}$
${f_2} = {f_5} = {f_8} = …….. = x$
${f_{100}}\left( 3 \right) = \frac{{3 – 1}}{3} = \frac{2}{3}{f_1}\left( {\frac{2}{3}} \right) = \frac{{\frac{2}{3} – 1}}{{\frac{2}{3}}} = – \frac{1}{2}$
${f_2}\left( {\frac{3}{2}} \right) = \frac{3}{2}$
$\therefore {f_{100}}\left( 3 \right) + {f_1}\left( {\frac{2}{3}} \right) + {f_2}\left( {\frac{3}{2}} \right) = \frac{5}{3}$
