Which of the following relations are functions? Give reasons. If it is a function,determine its domain and range.
$\{(1,3), (1,5), (2,5)\}$
$\{(1,3), (1,5), (2,5)\}$
(N/A) The given relation is $R = \{(1,3), (1,5), (2,5)\}$.
In a relation,for it to be a function,every element of the domain must have a unique image in the codomain.
Here,the element $1$ in the domain is associated with two different images,$3$ and $5$.
Since the same first element $1$ corresponds to two different images,this relation is not a function.
In a relation,for it to be a function,every element of the domain must have a unique image in the codomain.
Here,the element $1$ in the domain is associated with two different images,$3$ and $5$.
Since the same first element $1$ corresponds to two different images,this relation is not a function.
Explore More
Similar Questions
Given $A = \{1, 2, 3, 4, 5\}$ and $B = \{1, 4, 5\}$. If $R$ is a relation from $A$ to $B$ such that $(x, y) \in R$ with $x > y$,then the range of $R$ is:
$A$ relation on the set $A = \{x : |x| < 3, x \in Z\}$,where $Z$ is the set of integers,is defined by $R = \{(x, y) : y = |x|, x \neq -1\}$. Then the number of elements in the power set of $R$ is
DifficultJEE MAIN 2014
View SolutionIf $A = \{1, 2, 3\}$ and $B = \{1, 4, 6, 9\}$,and $R$ is a relation from $A$ to $B$ defined by "$x$ is greater than $y$",then the range of $R$ is:
Easy
View SolutionLet $A = \{1, 2, 3\}$. The total number of distinct relations that can be defined over $A$ is
Easy
View SolutionGiven two finite sets $A$ and $B$ such that $n(A) = 2$ and $n(B) = 3$. Then the total number of relations from $A$ to $B$ is:
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo