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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Sets and Relations

NTA Abhyas 2020 · Mathematics · Sets and Relations

If R = \ (x, y) | x, y ∈ ℤ, x^ 2 + y^ 2 ≤ 4\ is a relation in Z ,...

If R={(x,y)∣x,y∈Z,x2+y2≤4}R = \left\{\left(x, y\right) | x, y \in \mathbb{Z}, x^{2} + y^{2} \le 4\right\} is a relation in ZZ , then domain of RR is

Options

  1. A{0,1,2}\left\{0, 1, 2\right\}
  2. B{0,−1,−2}\left\{0, - 1, - 2\right\}
  3. C{−2,−1,0,1,2}\left\{- 2, - 1, 0, 1, 2\right\}
  4. DNone of these

Correct answer

C. {−2,−1,0,1,2}\left\{- 2, - 1, 0, 1, 2\right\}

Step-by-step solution

∵R={(x,y)x,y∈Z,x2+y2≤4}∵R = \left\{\left(x, y\right)x, y \in \mathbb{Z}, x^{2} + y^{2} \le 4\right\}

R={(−2,0),(−1,0),(0,−1),(−1,1),(1,−1),(0,−1)(0,1),(0,2),(0,−2),(1,0),(0,1),(1,1),(−1,−1),(2,0),(0,0)}R = \left\{\left(- 2, 0\right), \left(- 1, 0\right), ( 0, - 1 ) , \left(- 1, 1\right), ( 1, - 1 ) , \left(0, - 1\right)\left(0, 1\right), \left(0, 2\right), \left(0, - 2\right), \left(1, 0\right), (0,1) , \left(1, 1\right), ( - 1, - 1 ) , \left(2, 0\right), (0,0)\right\}

Hence, Domain of R={−2,−1,0,1,2}R = \left\{- 2, - 1, 0, 1, 2\right\}
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