Question

If $R = \{ (x,y):x,y \in I\,and\,{x^2} + {y^2} \leqslant 4\} $ is a relation in I, then the number of elements in domain of R is

Answer 1

Hint: First we will put $x = 0$ in the given equation ${x^2} + {y^2} \leqslant 4$ to get the value of y. And in the same way will put $x = \pm 1$ and $x = \pm 2$ in the given equation and find the value of y. As we will see those values will satisfy the equation so that that will be the domain of the relation.

Complete step by step solution: Given, relation is $R = \{ (x,y):x,y \in I\,and\,{x^2} + {y^2} \leqslant 4\} $
${x^2} + {y^2} \leqslant 4$ (1)
$x,y \in I\,$
Put $x = 0$ in equation (1)
$0 + {y^2} \leqslant 4$
${y^2} \leqslant 4$
After solving, we will get
$ \Rightarrow y = 0, \pm 1, \pm 2$
Put $x = \pm 1$ in equation (1)
${( \pm 1)^2} + {y^2} \leqslant 4$
$1 + {y^2} \leqslant 4$
Subtracting 1 from both the sides
${y^2} \leqslant 3$
After solving, we will get
$ \Rightarrow y = 0, \pm 1$
Put $x = \pm 2$ in equation (2)
${( \pm 2)^2} + {y^2} \leqslant 4$
$4 + {y^2} \leqslant 4$
Subtracting 4 from both the sides
${y^2} \leqslant 0$
After solving, we will get
$ \Rightarrow y = 0$
Now if we put $x > 2$ then $y^2 < 0$ which is not possible.
Hence, Domain of R is $\{ 0, \pm 1, \pm 2\} $
That is domain of R is $\{ - 1, - 2,0,1,2\} $

Therefore, the number of elements in the domain of R is $5$

Note: While solving the question Students should use the accurate value of x. They should choose only integers as it is given in the question that $x,y \in I\,$, and should avoid any other types of numbers.