Given that f: S rightarrow R is said to have | vedclass

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

Question

(B) To find the fixed point,we set $f(c) = c$.$1 + \sqrt{c} = c$$\sqrt{c} = c - 1$Squaring both sides (where $c \geq 1$):$c = (c - 1)^2$$c = c^2 - 2c

Vedclass · Accepted Answer

$f$ has a unique fixed point in $[1, \infty)$

Vedclass · Answer

(B) To find the fixed point,we set $f(c) = c$.$1 + \sqrt{c} = c$$\sqrt{c} = c - 1$Squaring both sides (where $c \geq 1$):$c = (c - 1)^2$$c = c^2 - 2c + 1$$c^2 - 3c + 1 = 0$Using the quadratic formula $c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$c = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}$Since $c \geq 1$,we check the values:$\frac{3 + \sqrt{5}}{2} \approx \frac{3 + 2.236}{2} = 2.618 \geq 1$ (Valid)$\frac{3 - \sqrt{5}}{2} \approx \frac{3 - 2.236}{2} = 0.382 Thus,there is only one fixed point in the domain $[1, \infty)$.

Given that $f: S \rightarrow R$ is said to have a fixed point at $c \in S$ if $f(c)=c$. Let $f:[1, \infty) \rightarrow R$ be defined by $f(x)=1+\sqrt{x}$. Then:

Similar Questions

Let $g(x) = 1 + x - [x]$ and $f(x) = \begin{cases} -1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0 \end{cases}$. Then for all $x$,$f(g(x))$ is equal to

If $f: \{1, 2, 3, 4\} \to \{1, 2, 3, 4\}$ is a function such that $|f(\alpha) - \alpha| \leqslant 1$ for all $\alpha \in \{1, 2, 3, 4\}$,then the total number of such functions is:

Let a function $f: N \rightarrow N$ be defined by
$f(n) = \begin{cases} 2n, & n = 2, 4, 6, 8, \dots \\ n-1, & n = 3, 7, 11, 15, \dots \\ \frac{n+1}{2}, & n = 1, 5, 9, 13, \dots \end{cases}$
Then,$f$ is

Let $X$ be a set with exactly $5$ elements and $Y$ be a set with exactly $7$ elements. If $\alpha$ is the number of one-one functions from $X$ to $Y$ and $\beta$ is the number of onto functions from $Y$ to $X$,then the value of $\frac{1}{5!}(\beta-\alpha)$ is.

The function $f: (-\infty, \infty) \rightarrow (-\infty, \infty)$ defined by $f(x) = \frac{2^x - 2^{-x}}{2^x + 2^{-x}}$ is :

Vedclass Test Series

Exam Paper Generator

Online Exam Module