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CBSE Class 12 Math 2022 Term I Solved Paper

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Question : 15 of 50
Marks: +1, -0
If R={(x,y);x,y∈Z,x2+y2≤4}R = \{(x, y) ; x, y \in Z, x^2+y^2 \leq 4\} is a relation in set ZZ, then domain of RR is
Solution:     👈: Video Solution
Explanation: Given,
R={(x,y):x,y∈Z,x2+y2≤4}R = \{(x, y): x, y \in Z, x^2+y^2 \leq 4\}
 Since, y=f(x)\text{ Since, } \begin{aligned} & y = f(x) \end{aligned}
y=4−x2y = \sqrt{4-x^2}
 For, x=0,±1,±2,⇒y=±2,±3,0\text{ For, } x=0, \pm 1, \pm 2, \Rightarrow y= \pm 2, \pm \sqrt{3}, 0
Since,
Hence, the required domain will be 0±1,±20 \pm 1, \pm 2
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