For every real number x neq -1,let f(x) = fra | vedclass

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

Question

(A) Given $f(x) = \frac{x}{x+1}$.Calculating the first few terms:$f_1(x) = \frac{x}{x+1}$$f_2(x) = f(f(x)) = \frac{\frac{x}{x+1}}{\frac{x}{x+1} + 1} =

Vedclass · Accepted Answer

$\frac{2^n}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1)}$

Vedclass · Answer

(A) Given $f(x) = \frac{x}{x+1}$.Calculating the first few terms:$f_1(x) = \frac{x}{x+1}$$f_2(x) = f(f(x)) = \frac{\frac{x}{x+1}}{\frac{x}{x+1} + 1} = \frac{x}{x + x + 1} = \frac{x}{2x+1}$$f_3(x) = f(f_2(x)) = \frac{\frac{x}{2x+1}}{\frac{x}{2x+1} + 1} = \frac{x}{x + 2x + 1} = \frac{x}{3x+1}$By induction,$f_n(x) = \frac{x}{nx+1}$.Evaluating at $x = -2$:$f_n(-2) = \frac{-2}{n(-2)+1} = \frac{-2}{-2n+1} = \frac{2}{2n-1}$.The product is $P = f_1(-2) \cdot f_2(-2) \cdot \ldots \cdot f_n(-2) = \prod_{k=1}^{n} \frac{2}{2k-1} = \frac{2^n}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n-1)}$.

For every real number $x \neq -1$,let $f(x) = \frac{x}{x+1}$. Define $f_1(x) = f(x)$ and for $n \geq 2$,$f_n(x) = f(f_{n-1}(x))$. Then the product $f_1(-2) \cdot f_2(-2) \cdot \ldots \cdot f_n(-2)$ is equal to:

Similar Questions

If $f(x) = x^3 - x$ and $g(x) = \sin^2 x$,then $f\left(g\left(\frac{\pi}{6}\right)\right) = $

If $f(x)=\sqrt{x}$ $(x \geq 0)$ and $g(x)=1+x^2$,then $(f \circ g)^{\prime}(1)=$

If $g(x)=x^{2}+x-1$ and $(g \circ f)(x)=4 x^{2}-10 x+5,$ then $f\left(\frac{5}{4}\right)$ is equal to

Let $f: R \rightarrow R$ be a function defined by $f(x) = \left(2\left(1 - \frac{x^{25}}{2}\right)\left(2 + x^{25}\right)\right)^{\frac{1}{50}}$. If the function $g(x) = f(f(f(x))) + f(f(x))$,then the greatest integer less than or equal to $g(1)$ is

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are two functions defined by $f(x)=2x-3$ and $g(x)=x^{3}+5$,then $(fog)^{-1}(x) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module