For x in R, x neq 0, let f_0(x) = frac{1}{1 - | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) $f_1(x) = f_{0+1}(x) = f_0(f_0(x)) = \frac{1}{1 - \frac{1}{1 - x}} = \frac{x - 1}{x}$$f_2(x) = f_{1+1}(x) = f_0(f_1(x)) = \frac{1}{1 - \frac{x - 1

Vedclass · Accepted Answer

$\frac{5}{3}$

Vedclass · Answer

(C) $f_1(x) = f_{0+1}(x) = f_0(f_0(x)) = \frac{1}{1 - \frac{1}{1 - x}} = \frac{x - 1}{x}$$f_2(x) = f_{1+1}(x) = f_0(f_1(x)) = \frac{1}{1 - \frac{x - 1}{x}} = x$$f_3(x) = f_{2+1}(x) = f_0(f_2(x)) = f_0(x) = \frac{1}{1 - x}$Since $f_3(x) = f_0(x)$,the function repeats with a period of $3$.$f_{100}(3) = f_{3 	imes 33 + 1}(3) = f_1(3) = \frac{3 - 1}{3} = \frac{2}{3}$$f_1\left( \frac{2}{3} ight) = \frac{\frac{2}{3} - 1}{\frac{2}{3}} = \frac{-\frac{1}{3}}{\frac{2}{3}} = -\frac{1}{2}$$f_2\left( \frac{3}{2} ight) = \frac{3}{2}$Therefore,$f_{100}(3) + f_1\left( \frac{2}{3} ight) + f_2\left( \frac{3}{2} ight) = \frac{2}{3} - \frac{1}{2} + \frac{3}{2} = \frac{2}{3} + 1 = \frac{5}{3}$

For $x \in R, x \neq 0,$ 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

Similar Questions

If $f(x) = (p - x^n)^{1/n}$,$p > 0$ and $n$ is a positive integer,then $f[f(x)]$ is equal to

If $f(x) = \frac{2x+3}{3x-2}$,$x \neq \frac{2}{3}$,then $(f \circ f)(x)$ is:

If $f$ is the greatest integer function and $g$ is the modulus function,then $(gof)\left( -\frac{5}{3} \right) - (fog)\left( -\frac{5}{3} \right) = $

If $f(x)=e^{|x|}$ and $g(x)=\log x$,then $(g \circ f)(x) =$

If $f: A \rightarrow B$ and $g: B \rightarrow C$ are functions such that $g \circ f: A \rightarrow C$ is onto,then a necessary condition is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module