A function f from the set of natural numbers to integers defined by f(n)={cases} n-1/2, & {when } n { is odd} \\ -n/2, & {when } n { is even} {cases}. is [AIEEE 2003]

Correct Answer is : one-one and onto both

Correct Answer is : one-one and onto both

Test Index

Functions Part 1

Section: Mathematics

Question : 95 of 100

Marks: +1, -0

A function

f

f(n)=\begin{cases} \frac{n-1}{2}, & \text{when } n \text{ is odd} \\ -\frac{n}{2}, & \text{when } n \text{ is even} \end{cases}.

Solution:

We have

f: N \rightarrow I

x

and

y

are two even natural numbers,

then

f(x)=f(y) \Rightarrow \;\frac{-x}{2}=\;\frac{-y}{2} \Rightarrow x=y

Again if

x

and

y

are two odd natural numbers then

f(x)=f(y) \Rightarrow \;\frac{x-1}{2}=\;\frac{y-1}{2} \Rightarrow x=y

\therefore f

is onto.

Also each negative integer is an image of even natural number and each positive integer is an image of odd natural number.

\therefore f

is onto.

Hence

f

is one one and onto both.

Go to Question:

Prev Question

Next Question