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

If $f(x) = \frac{x^2 - 1}{x^2 + 1}$ for every real number $x$,then the minimum value of $f$ is:

If the domain of the function $f(x) = \log_e \left( \frac{2x+3}{4x^2+x-3} \right) + \cos^{-1} \left( \frac{2x-1}{x+2} \right)$ is $(\alpha, \beta]$,then the value of $5\beta - 4\alpha$ is equal to

If $[x]^2-5[x]+6=0$,where $[x]$ denotes the greatest integer function,then

The range of the function $f(x) = [x] - x$ is

The domain of the function $f(x) = \sin^{-1}[2x^2 - 3] + \log_2(\log_{1/2}(x^2 - 5x + 5))$,where $[t]$ is the greatest integer function,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module