Question

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

• A. {0, 1, 2}
• B. {0, -1, -2}
• C. {-2, -1, 0, 1, 2}
• D. None of these

Answer 1

The correct option is C

{-2, -1, 0, 1, 2}

R = {$(x,y):x,yϵZ,x^2+y^2\le 4$}

We know that,

$(-2{)}^2+0^2\le 4\Rightarrow (2{)}^2+0^2\le 4\Rightarrow (-1{)}^2+0^2\le 4\Rightarrow (1{)}^2+0^2\le 4\Rightarrow (-1{)}^2+4\le 0^2+0^2\le 4$

$\Rightarrow (1{)}^2+(1{)}^2\le 4\Rightarrow (-1{)}^2+(-1{)}^2\le 4$

Hence, domain (R) = {-2, -1, 0, 1, 2}