Let the relation R be defined on the set of n | vedclass

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

Question

(A) The relation is defined as $3 a + 2 b = 27$,where $a, b \in N$ (natural numbers). Since $b = \frac{27 - 3 a}{2}$,for $b$ to be a natural number,$(

Vedclass · Accepted Answer

$\{(1, 12), (3, 9), (5, 6), (7, 3)\}$

Vedclass · Answer

(A) The relation is defined as $3 a + 2 b = 27$,where $a, b \in N$ (natural numbers). Since $b = \frac{27 - 3 a}{2}$,for $b$ to be a natural number,$(27 - 3 a)$ must be even and positive. If $a = 1$,$b = \frac{27 - 3}{2} = 12$. If $a = 3$,$b = \frac{27 - 9}{2} = 9$. If $a = 5$,$b = \frac{27 - 15}{2} = 6$. If $a = 7$,$b = \frac{27 - 21}{2} = 3$. If $a = 9$,$b = \frac{27 - 27}{2} = 0$,which is not in $N$. Thus,the relation $R = \{(1, 12), (3, 9), (5, 6), (7, 3)\}$. Therefore,option $A$ is correct.

Let the relation $R$ be defined on the set of natural numbers $N$ by $a R b$ if $3 a + 2 b = 27$. Then $R$ is:

Similar Questions

Which of the following structures contains its own $DNA$?

If $x=\log _{0.1} 0.001$ and $y=\log _9 81$,then $\sqrt{x-2 \sqrt{y}}$ is equal to

For the stationary wave $y = 4 \sin \left( \frac{\pi x}{15} \right) \cos (96 \pi t)$,the distance between a node and the next antinode is

Ex situ conservation is carried out through:

The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and $l^2+m^2-n^2=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module