Which of the following relations are functions? Give reasons. If it is a function,determine its domain and range.
$\{(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)\}$
$\{(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)\}$
(N/A) The given relation is $R = \{(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)\}$.
$A$ relation is a function if every element of the domain has a unique image in the codomain.
In this relation,each first element $(2, 4, 6, 8, 10, 12, 14)$ is associated with exactly one unique second element ($1, 2, 3, 4, 5, 6, 7$ respectively).
Therefore,the given relation is a function.
The domain is the set of all first elements: $\text{Domain} = \{2, 4, 6, 8, 10, 12, 14\}$.
The range is the set of all second elements: $\text{Range} = \{1, 2, 3, 4, 5, 6, 7\}$.
$A$ relation is a function if every element of the domain has a unique image in the codomain.
In this relation,each first element $(2, 4, 6, 8, 10, 12, 14)$ is associated with exactly one unique second element ($1, 2, 3, 4, 5, 6, 7$ respectively).
Therefore,the given relation is a function.
The domain is the set of all first elements: $\text{Domain} = \{2, 4, 6, 8, 10, 12, 14\}$.
The range is the set of all second elements: $\text{Range} = \{1, 2, 3, 4, 5, 6, 7\}$.
Explore More
Similar Questions
Which of the following represents a function?
Examine the following relation and state whether it is a function or not,giving reasons: $R = \{(2, 2), (2, 4), (3, 3), (4, 4)\}$
Easy
View SolutionIf $Q$ denotes the set of all rational numbers and $f\left(\frac{p}{q}\right)=\sqrt{p^2-q^2}$ for any $\frac{p}{q} \in Q$, then observe the following statements.
$I$. $f\left(\frac{p}{q}\right)$ is real for each $\frac{p}{q} \in Q$.
$II$. $f\left(\frac{p}{q}\right)$ is a complex number for each $\frac{p}{q} \in Q$.
Which of the following is correct?
$I$. $f\left(\frac{p}{q}\right)$ is real for each $\frac{p}{q} \in Q$.
$II$. $f\left(\frac{p}{q}\right)$ is a complex number for each $\frac{p}{q} \in Q$.
Which of the following is correct?
Let $f: X \rightarrow Y$ be a function and $A_y = f^{-1}(\{y\})$ for $y \in Y$. Then $A_i \cap A_j = \phi$ $(i \neq j)$ for all $i, j \in Y$ and $\bigcup_{y \in Y} A_y = X$, if
If $f$ is a relation from the set of positive real numbers $R_{+}$ to the set of positive real numbers $R_{+}$ defined by $f(x) = 3x^2 - 2$, then $f$ is
Vedclass 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