Let rho be a relation defined on the set of n | vedclass

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

Question

(A) Given the relation $ho = \{(x, y) \in N 	imes N: 2x + y = 41\}$.Since $x, y \in N$, we have $x \geq 1$ and $y \geq 1$.From $2x + y = 41$, we ge

Vedclass · Accepted Answer

$A \subset \{x \in N: 1 \leq x \leq 20\}$ and $B \subset \{y \in N: 1 \leq y \leq 39\}$

Vedclass · Answer

(A) Given the relation $ho = \{(x, y) \in N 	imes N: 2x + y = 41\}$.Since $x, y \in N$, we have $x \geq 1$ and $y \geq 1$.From $2x + y = 41$, we get $y = 41 - 2x$.Since $y \geq 1$, we have $41 - 2x \geq 1$, which implies $2x \leq 40$, so $x \leq 20$.Thus, $x \in \{1, 2, 3, \dots, 20\}$.For each $x$, $y = 41 - 2x$. Substituting values of $x$:If $x = 1, y = 39$.If $x = 20, y = 1$.So, the domain $A = \{1, 2, 3, \dots, 20\}$ and the range $B = \{1, 3, 5, \dots, 39\}$.Both $A$ and $B$ are subsets of the sets given in option $A$.

Let $\rho$ be a relation defined on the set of natural numbers $N$, as $\rho = \{(x, y) \in N \times N: 2x + y = 41\}$. Then the domain $A$ and range $B$ are:

Similar Questions

The function $f: R \rightarrow R$ is defined by $f(x) = \cos^2 x + \sin^4 x$ for $x \in R$. Then $f(R)$ is equal to:

Let $D = \{x \in R : f(x) = \sqrt{\frac{x-|x|}{x-[x]}} \text{ is defined} \}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4+x^2}$. Then $D \cap C =$

The range of $f(x) = \frac{x^2 + 34x - 71}{x^2 + 2x - 7}$ is

Find the domain of the function $f(x) = \frac{1}{\sqrt{(x + 1)(e^x - 1)(x - 4)(x + 5)(x - 6)}}$.

The range of the function $y=3 \sin \left(\sqrt{\frac{\pi^{2}}{16}-x^{2}}\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module