Which of the following relations are functions? Give reasons. If it is a function,determine its domain and range.
$\{(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)\}$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(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: $\text{Range} = \{1, 2, 3, 4, 5, 6, 7\}$.

Explore More

Similar Questions

$A$ function $f$ is defined by $f(x) = 2x - 5$. Find the values of $f(0)$,$f(7)$,and $f(-3)$.

$A$ function $f$ is defined by $f(x) = 2x - 5$. Find the value of $f(7)$.

The relation $f$ is defined by $f(x) = \begin{cases} x^2, & 0 \le x \le 3 \\ 3x, & 3 \le x \le 10 \end{cases}$. The relation $g$ is defined by $g(x) = \begin{cases} x^2, & 0 \le x \le 2 \\ 3x, & 2 \le x \le 10 \end{cases}$. Show that $f$ is a function and $g$ is not a function.

Let $f$ be the subset of $Z \times Z$ defined by $f = \{(ab, a+b) : a, b \in Z\}$. Is $f$ a function from $Z$ to $Z$? Justify your answer.

Let $A = \{a, b, c, d\}$ and $B = \{1, 2, 3\}$. The relations $R_1, R_2, R_3, R_4$ are defined as follows:
$R_1 = \{(a, 1), (b, 2), (c, 1), (d, 2)\}$
$R_2 = \{(a, 1), (b, 1), (c, 1), (d, 1)\}$
$R_3 = \{(a, 2), (b, 3), (c, 2), (d, 2)\}$
$R_4 = \{(a, 1), (b, 2), (a, 2), (d, 3)\}$
Which of the following is true?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo