For, x∈R,x≠0,x≠1, let f0(x)=11−xand fn+1(x)=f0fn(x),n=0,1,2,….Then the value of f100(3)+f123+f232 is equal to :

For, $x \in R, x \neq 0, x \neq 1$ , 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 $f_{100} (3) + f_{1} (\frac{2}{3}) + f_{2} (\frac{3}{2})$ is equal to :

Solution:

$f_{0} (x) = \frac{1}{1 - x} f_{1} (x) = f_{0} (\frac{1}{1 - x}) = \frac{x - 1}{x} f_{2} (x) = x; f_{100} (3) + f_{1} (\frac{2}{3}) + f_{2} (\frac{3}{2}) = \frac{2}{3} + \frac{\frac{2}{3} - 1}{\frac{2}{3}} + \frac{3}{2} = \frac{5}{3}$

Related content

SRMJEEE Online Test Series – Practice Papers

KIITEE Sample Papers 2024 – Practice Paper PDF Download

Linking Verbs

NEET Exam Centres in Chhattisgarh 2024

Letter To Your Friend About Your Hobby

NEET Exam Centres in Haryana 2024

NEET Exam Centres in West Bengal 2024

Finite Verbs

CBSE Class 8 Language Worksheets – PDF Download

NEET Exam Centres in Uttarakhand 2024