If f: Z rightarrow Z is defined by f(x)=begin | vedclass

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

Question

(A) Given,$f: Z ightarrow Z$ defined as $f(x)=\begin{cases} \frac{x}{2}, & 	ext{if } x 	ext{ is even} \ 0, & 	ext{if } x 	ext{ is odd} \end{cas

Vedclass · Accepted Answer

onto but not one-to-one

Vedclass · Answer

(A) Given,$f: Z ightarrow Z$ defined as $f(x)=\begin{cases} \frac{x}{2}, & 	ext{if } x 	ext{ is even} \ 0, & 	ext{if } x 	ext{ is odd} \end{cases}$.For one-to-one check: Consider $x_1 = 1$ and $x_2 = 3$. Both are odd,so $f(1) = 0$ and $f(3) = 0$. Since $f(1) = f(3)$ but $1 
eq 3$,the function is not one-to-one.For onto check: The range of $f$ consists of all integers. For any integer $y \in Z$,we can find an $x \in Z$ such that $f(x) = y$. If $y = 0$,$f(1) = 0$. If $y 
eq 0$,let $x = 2y$,which is even,then $f(2y) = \frac{2y}{2} = y$. Since the range equals the codomain $(Z)$,the function is onto.Therefore,$f$ is onto but not one-to-one.

If $f: Z \rightarrow Z$ is defined by $f(x)=\begin{cases} \frac{x}{2}, & \text{if } x \text{ is even} \\ 0, & \text{if } x \text{ is odd} \end{cases}$,then $f$ is

Similar Questions

$f: N \rightarrow N, f(x) = x^3$ is . . . . . . .

Let $f : X \rightarrow Y$ be a function such that $f(x) = \sqrt{x - 2} + \sqrt{4 - x} .$ Then the set of $X$ and $Y$ for which $f(x)$ is both injective as well as surjective,is-

The function $f(x) = \frac{x^2+2x-15}{x^2-4x+9}$,$x \in R$ is

The function $f: N \rightarrow Z$ defined by $f(n) = \begin{cases} \frac{n}{2} & , n \text{ is even} \\ -\left(\frac{n-1}{2}\right) & , n \text{ is odd} \end{cases}$ is . . . . . . .

If $f:[0, \infty) \to [0, \infty)$ and $f(x) = \frac{x}{1+x}$,then $f$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module