JEE Mains · Maths · STD 12 - 1. relation and function
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
- A \(\frac {8}{3}\)
- B \(\frac {4}{3}\)
- C \(\frac {5}{3}\)
- D \(\frac {1}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac {5}{3}\)
Step-by-step Solution
Detailed explanation
\({f_1}\left( x \right) = {f_{0 + 1}}\left( x \right) = {f_0}\left( {{f_0}\left( x \right)} \right) = \frac{1}{{1 - \frac{1}{{1 - x}}}} = \frac{{x - 1}}{x}\)…
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