If f: Z rightarrow Z, f(x) = begin{cases} fra | vedclass

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

Question

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

Vedclass · Accepted Answer

onto but not one-to-one

Vedclass · Answer

(A) Given, $f: Z ightarrow Z$, $f(x) = \begin{cases} \frac{x}{2}, & 	ext{if } x 	ext{ is even} \ 0, & 	ext{if } x 	ext{ is odd} \end{cases}$.For $f$ to be one-to-one, $f(x_1) = f(x_2)$ must imply $x_1 = x_2$.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 $1 
eq 3$, $f$ is not one-to-one.For $f$ to be onto, the range must equal the codomain $Z$.If $x$ is even, let $x = 2k$ for some $k \in Z$. Then $f(2k) = \frac{2k}{2} = k$.Since $k$ can be any integer in $Z$, the range of $f$ is $Z$.Thus, $f$ is onto.Therefore, $f$ is onto but not one-to-one.

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

Similar Questions

Show that the modulus function $f : R \rightarrow R$ given by $f(x) = |x|$ is neither one-one nor onto,where $|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x < 0$.

The mapping $f: R \to R$ defined as $f(x) = \cos x, x \in R$ is:

Let $f: R \to R$ be a function defined by $f(x) = \frac{x - m}{x - n}$,where $m \ne n$. Then

If a set $A$ has $n$ elements,then the number of functions defined from $A$ to $A$ that are not one-one is

If a real-valued function $f$ is defined by $f(x) = \frac{ax + \sqrt{a^2 - x^2}}{bx}$,then $f$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module