A function f from the set of natural numbers | vedclass

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

Question

(C) Given the function $f: \mathbb{N} 	o \mathbb{Z}$ defined as:$f(n) = \begin{cases} \frac{n-1}{2}, & 	ext{if } n 	ext{ is odd} \ -\frac{n}{2}, &

Vedclass · Accepted Answer

One-one and onto both

Vedclass · Answer

(C) Given the function $f: \mathbb{N} 	o \mathbb{Z}$ defined as:$f(n) = \begin{cases} \frac{n-1}{2}, & 	ext{if } n 	ext{ is odd} \ -\frac{n}{2}, & 	ext{if } n 	ext{ is even} \end{cases}$Let us calculate the images of the first few natural numbers:$f(1) = \frac{1-1}{2} = 0$$f(2) = -\frac{2}{2} = -1$$f(3) = \frac{3-1}{2} = 1$$f(4) = -\frac{4}{2} = -2$$f(5) = \frac{5-1}{2} = 2$$f(6) = -\frac{6}{2} = -3$$1$. One-one check: For every distinct $n_1, n_2 \in \mathbb{N}$,we get distinct images in $\mathbb{Z}$. Since each input maps to a unique output,the function is one-one.$2$. Onto check: The range of the function is $\{0, -1, 1, -2, 2, -3, 3, \dots\}$,which is the set of all integers $\mathbb{Z}$. Since the range equals the codomain,the function is onto.Therefore,the function is both one-one and onto (bijective).

Column-$I$	Column-$II$
$A$. $f$ is one-one and onto,if	$1$. $A = R^{+}, B = R$
$B$. $f$ is one-one but not onto,if	$2$. $A = B = R$
$C$. $f$ is onto but not one-one,if	$3$. $A = R, B = R^{+}$
$D$. $f$ is neither one-one nor onto,if	$4$. $A = B = R^{+}$

$A$ function $f$ from the set of natural numbers $\mathbb{N}$ to the set of integers $\mathbb{Z}$ is defined by $f(n) = \begin{cases} \frac{n-1}{2}, & \text{if } n \text{ is odd} \\ -\frac{n}{2}, & \text{if } n \text{ is even} \end{cases}$. The function $f$ is:

Similar Questions

Consider a function $f: [0, \frac{\pi}{2}] \rightarrow \mathbb{R}$ given by $f(x) = \sin x$ and $g: [0, \frac{\pi}{2}] \rightarrow \mathbb{R}$ given by $g(x) = \cos x$. Show that $f$ and $g$ are one-one,but $f + g$ is not one-one.

The function $f: R-\{1\} \rightarrow R-\{4\}$ defined by $f(x) = \frac{4x-3}{x-1}$ for $x \in R-\{1\}$ is

$f(x) = x + \sqrt{x^2}$ is a function from $R \to R$,then $f(x)$ is

$f: N \rightarrow N$,is defined by $f(x)=x^6$ then, . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module