Let R be the set of real numbers and f: R rig | vedclass

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

Question

(D) Given $f(x) = \frac{\{x\}}{1+[x]^2}$.$I.$ For any $x \in R$,let $n = [x]$,then $x = n + \{x\}$ where $0 \le \{x\} < 1$. Thus $f(x) = \frac{\{x\}}{

Vedclass · Accepted Answer

None of $I, II$ and $III$

Vedclass · Answer

(D) Given $f(x) = \frac{\{x\}}{1+[x]^2}$.$I.$ For any $x \in R$,let $n = [x]$,then $x = n + \{x\}$ where $0 \le \{x\} $II.$ At $x = n$ (an integer),$\lim_{x 	o n^-} f(x) = \lim_{x 	o n^-} \frac{x-[x]}{1+[x]^2} = \frac{n-(n-1)}{1+(n-1)^2} = \frac{1}{1+(n-1)^2}$,while $f(n) = \frac{n-n}{1+n^2} = 0$. Since the limit does not equal the function value at integers,$f$ is discontinuous at all integers. So,statement $II$ is false.$III.$ Note that $f(0) = \frac{0}{1+0^2} = 0$ and $f(1) = \frac{1-1}{1+1^2} = 0$. Since $f(0) = f(1)$ but $0 
eq 1$,$f$ is not one-one. So,statement $III$ is false.Therefore,none of the statements are true.

Similar Questions

If $f : R \rightarrow R$ such that $f(x) = 5x - 3\cos x - 4\sin x$,then the function $f(x)$ is

Function $f: R \rightarrow R$ defined by $f(x) = x^3$ is . . . . . . .

$A$ mapping from $\mathbb{N}$ to $\mathbb{N}$ is defined as follows: $f: \mathbb{N} \rightarrow \mathbb{N}$ where $f(n) = (n+5)^2$ for all $n \in \mathbb{N}$ (where $\mathbb{N}$ is the set of natural numbers). Then:

The function $f: R \rightarrow R$ defined by $f(x)=\frac{x}{\sqrt{1+x^2}}$ is

If $f(x) = \begin{cases} x, & \text{when } x \text{ is rational} \\ 0, & \text{when } x \text{ is irrational} \end{cases}$ and $g(x) = \begin{cases} 0, & \text{when } x \text{ is rational} \\ x, & \text{when } x \text{ is irrational} \end{cases}$,then $(f - g)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module