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 is defined as $R = \{ (x, y) | x, y \in Z, x^2 + y^2 \le 4 \}$.To find the domain,we identify all possible values of $x \in Z$ such t

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The relation is defined as $R = \{ (x, y) | x, y \in Z, x^2 + y^2 \le 4 \}$.To find the domain,we identify all possible values of $x \in Z$ such that there exists at least one $y \in Z$ satisfying $x^2 + y^2 \le 4$.If $x = 0$,then $y^2 \le 4 \implies y \in \{-2, -1, 0, 1, 2\}$.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 $x^2 > 4$,so $x^2 + y^2 > 4$ for any $y \in Z$.Thus,the set of all possible values of $x$ is $\{-2, -1, 0, 1, 2\}$.Therefore,the domain of $R$ 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

If $5x - 12y + 10 = 0$ and $12y - 5x + 16 = 0$ are two tangents to a circle,then the radius of the circle is

Which of the following is not a peroxy acid?

The wavelength of a $1 \ keV$ photon is $1.24 \times 10^{-9} \ m$. What is the frequency of a $1 \ MeV$ photon?

The principle of immunization or vaccination is based on the property of ..........

Vedclass Test Series

Exam Paper Generator

Online Exam Module