The domain and range of the relation R given | vedclass

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

Question

(D) Given the relation $R = \{(x, y) : y = x + \frac{6}{x}, x, y \in N, x < 6\}$.We test each value of $x \in \{1, 2, 3, 4, 5\}$ to see if $y$ is a na

Vedclass · Accepted Answer

Domain $= \{1, 2, 3\}$,Range $= \{5, 7\}$.

Vedclass · Answer

(D) Given the relation $R = \{(x, y) : y = x + \frac{6}{x}, x, y \in N, x We test each value of $x \in \{1, 2, 3, 4, 5\}$ to see if $y$ is a natural number $(N)$:For $x = 1, y = 1 + \frac{6}{1} = 7 \in N$.For $x = 2, y = 2 + \frac{6}{2} = 2 + 3 = 5 \in N$.For $x = 3, y = 3 + \frac{6}{3} = 3 + 2 = 5 \in N$.For $x = 4, y = 4 + \frac{6}{4} = 4 + 1.5 = 5.5 
otin N$.For $x = 5, y = 5 + \frac{6}{5} = 5 + 1.2 = 6.2 
otin N$.Thus,the relation $R = \{(1, 7), (2, 5), (3, 5)\}$.The domain is the set of first elements: $\{1, 2, 3\}$.The range is the set of second elements: $\{5, 7\}$.

The domain and range of the relation $R$ given by $R = \{(x, y) : y = x + \frac{6}{x}, x, y \in N \text{ and } x < 6\}$ are

Similar Questions

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:

The domain and range for the function $f(x) = e^{|x| \sin x}$ are:

The range of the function $f(x) = \frac{x^2+x+1}{x^2-x+1}$ is

The least integer in the range of $f(x) = \sqrt{(x + 4)(1 - x)} - \log_2 x$ is

Let $f: A \rightarrow B$ be defined as $f(x) = \frac{1}{2} - \tan \left(\frac{\pi x}{2}\right)$ and $g: B \rightarrow C$ be defined as $g(x) = \sqrt{3 + 4x - 4x^2}$. If $A, B, C$ are subsets of $\mathbb{R}$ and $f$ is an onto function, then the range of the function $f(x)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module