If R = { (x, y) | x, y in Z, x^2 + y^2 le 4 } | vedclass

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

Question

(C) The relation $R$ is defined as $x^2 + y^2 \le 4$,where $x, y \in Z$.The domain is the set of all values of $x$ for which there exists at least one

Vedclass · Accepted Answer

$\{-2, -1, 0, 1, 2\}$

Vedclass · Answer

(C) The relation $R$ is defined as $x^2 + y^2 \le 4$,where $x, y \in Z$.The domain is the set of all values of $x$ for which there exists at least one $y \in Z$ such that $x^2 + y^2 \le 4$.If $x = 0$,then $0^2 + y^2 \le 4 \implies y^2 \le 4 \implies y \in \{-2, -1, 0, 1, 2\}$.If $x = 1$,then $1^2 + y^2 \le 4 \implies y^2 \le 3 \implies y \in \{-1, 0, 1\}$.If $x = -1$,then $(-1)^2 + y^2 \le 4 \implies y^2 \le 3 \implies y \in \{-1, 0, 1\}$.If $x = 2$,then $2^2 + y^2 \le 4 \implies 4 + y^2 \le 4 \implies y^2 \le 0 \implies y = 0$.If $x = -2$,then $(-2)^2 + y^2 \le 4 \implies 4 + y^2 \le 4 \implies y^2 \le 0 \implies y = 0$.If $|x| > 2$,then $x^2 > 4$,which means no integer $y$ exists such that $x^2 + y^2 \le 4$.Thus,the set of all possible values for the domain is $\{-2, -1, 0, 1, 2\}$.

If $R = \{ (x, y) | x, y \in Z, x^2 + y^2 \le 4 \}$ is a relation in $Z$,then the domain of $R$ is:

Similar Questions

The domain of the function defined by $f(x) = \frac{-5}{4x^2+1} + \sqrt{x^2-4}$ is

The domain of the function $f(x) = \frac{1}{\sqrt{[x]^2 - [x] - 2}}$ is, where $[x]$ denotes the greatest integer function.

The domain of the function $f(x) = \sqrt{x-1} + \sqrt{6-x}$ is

The sum of all the elements in the range of $f(x) = \text{Sgn}(\sin x) + \text{Sgn}(\cos x) + \text{Sgn}(\tan x) + \text{Sgn}(\cot x)$, where $x \neq \frac{n\pi}{2}, n \in \mathbb{Z}$ and $\text{Sgn}(t) = \begin{cases} 1, & \text{if } t > 0 \\ -1, & \text{if } t < 0 \end{cases}$, is

The domain of $f(x) = \sin \log \left( \frac{\sqrt{4-x^2}}{1-x} \right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module