If R = {(x, y) : x, y $\in$ Z, x2 + 3y2 $\le$ 8} is a relation on the set of integers Z, then the domain of R–1 is :
Solution
Given R = {(x, y) : x, y
$\in$ Z, x<sup>2</sup> + 3y<sup>2</sup>
$\le$ 8}
<br><br>So R = {(0,1), (0,–1), (1,0), (–1,0), (1,1), (1,-1)
<br>(-1,1), (-1,-1), (2,0), (-2,0), (-2,0), (2,1), (2,-1), (-2,1), (-2,-1)}
<br><br>$\Rightarrow$ R : { -2, -1, 0, 1, 2} $\to$ {-1, 0, 1}
<br><br>$\therefore$ R<sup>-1</sup> : {-1, 0, 1} $\to$ { -2, -1, 0, 1, 2}
<br><br>$\therefore$ Domain of R<sup>–1</sup> = {-1, 0, 1}
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
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