If a relation R on the set of natural numbers | vedclass

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

Question

(A) The relation $R$ is defined on the set of natural numbers $N = \{1, 2, 3, \dots\}$ by the equation $x + 2y = 8$.We can rewrite this as $x = 8 - 2y

Vedclass · Accepted Answer

$\{2, 4, 6\}$

Vedclass · Answer

(A) The relation $R$ is defined on the set of natural numbers $N = \{1, 2, 3, \dots\}$ by the equation $x + 2y = 8$.We can rewrite this as $x = 8 - 2y$.Since $x \in N$,we must have $x > 0$,which implies $8 - 2y > 0$,so $2y Since $y \in N$,the possible values for $y$ are $1, 2, 3$.Now,we find the corresponding values of $x$ for each $y$:If $y = 1$,$x = 8 - 2(1) = 6$.If $y = 2$,$x = 8 - 2(2) = 4$.If $y = 3$,$x = 8 - 2(3) = 2$.Thus,the relation $R$ is the set of ordered pairs: $R = \{(6, 1), (4, 2), (2, 3)\}$.The domain of $R$ is the set of all first elements of the ordered pairs in $R$.Therefore,the domain is $\{2, 4, 6\}$.

If a relation $R$ on the set of natural numbers $N$ is defined by $x + 2y = 8$,then the domain of $R$ is:

Similar Questions

The range of the polynomial $P(x) = 4x^3 - 3x$ as $x$ varies over the interval $\left(-\frac{1}{2}, \frac{1}{2}\right)$ is

For the function $f(x) = \left[ \frac{1}{[x]} \right]$,where $[x]$ denotes the greatest integer less than or equal to $x$,which of the following statements is true?

If $A$ is the domain and $B$ is the range of the function $f(x) = \begin{cases} 3x-1, & x > 1 \\ x^2+1, & x \leq 1 \end{cases}$ then $A-B=$

The domain and range of a real valued function $f(x) = \cos x - 3$ are respectively.

The range of the function $f(x) = \sqrt{9 - x^2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module