A) \[\frac{4}{3}\]
B) \[\frac{1}{3}\]
C) \[\frac{5}{3}\]
D) \[\frac{8}{3}\]
Correct Answer: C
Solution :
\[{{f}_{0}}(x)=\frac{1}{1-x}\] \[{{f}_{1}}(x)={{f}_{0}}({{f}_{0}}(x))=\frac{1}{1-{{f}_{0}}(x)};{{f}_{0}}(x)\ne 1\] \[=\frac{1}{1-\frac{1}{1-x}}x\ne 0\] \[=\frac{1-x}{-x}\] \[{{f}_{2}}(x)={{f}_{0}}({{f}_{1}}(x))=\frac{1}{1-{{f}_{1}}(x)};{{f}_{1}}(x)\ne 1\] \[=\frac{1}{1+\frac{1-x}{x}}=x\] similarly \[{{f}_{3}}(x)={{f}_{0}}(x)\] \[{{f}_{4}}(x)={{f}_{1}}(x)\]??? \[{{f}_{100}}(3)+{{f}_{1}}\left( \frac{2}{3} \right)+{{f}_{2}}\left( \frac{3}{2} \right)={{f}_{1}}(3)+{{f}_{1}}\left( \frac{2}{3} \right)+\frac{3}{2}\] \[=1-\frac{1}{3}+1-\frac{3}{2}+\frac{3}{2}=\frac{5}{3}\]
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