If R - x - y - x - y - Z - x 2- y 2- 4 is a relation in Z- then domain of R isA- -2-1-0-1-2B- 0-1-2C- 0-1-2D- -2-1-1-2

The correct option is A { 2, 1,0,1,2}As R={(x,y)| x,y ∈ℤ , x^2+y^2 ≤ 4 } nbsp; If y=0⇒ x=0,±1,±2 If y=±1⇒ x=0,±1 If y=±2⇒ x=0 ∴domain of R={ 2, 1,0,1,2} ...

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Standard XII

Mathematics

Relation Definition

If R = x , y ...

Question

If $R = {(x, y) | x, y \in Z, x^{2} + y^{2} \leq 4}$ is a relation in $Z$ , then domain of $R$ is

{0, 1, 2}

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{0, - 1, - 2}

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

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{- 2, - 1, 1, 2}

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Solution

The correct option is C ${- 2, - 1, 0, 1, 2}$
As $R = {(x, y) | x, y \in Z, x^{2} + y^{2} \leq 4}$

If $y = 0 \Rightarrow x = 0, \pm 1, \pm 2$
If $y = \pm 1 \Rightarrow x = 0, \pm 1$
If $y = \pm 2 \Rightarrow x = 0$
$∴ domain of R = {- 2, - 1, 0, 1, 2}$

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Relation Definition

Standard XII Mathematics

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If R={(x,y)|x,y∈Z,x2+y2≤4} is a relation in Z, then domain of R is

The correct option is C {−2,−1,0,1,2}As R={(x,y)|x,y∈Z,x2+y2≤4} If y=0⇒x=0,±1,±2 If y=±1⇒x=0,±1 If y=±2⇒x=0 ∴domain of R={−2,−1,0,1,2}

If $R = {(x, y) | x, y \in Z, x^{2} + y^{2} \leq 4}$ is a relation in $Z$ , then domain of $R$ is

The correct option is C ${- 2, - 1, 0, 1, 2}$
As $R = {(x, y) | x, y \in Z, x^{2} + y^{2} \leq 4}$

If $y = 0 \Rightarrow x = 0, \pm 1, \pm 2$
If $y = \pm 1 \Rightarrow x = 0, \pm 1$
If $y = \pm 2 \Rightarrow x = 0$
$∴ domain of R = {- 2, - 1, 0, 1, 2}$