If R is a relation defined on the set Z of integers by the rule x- y - R - x2-y2-9- then write domain of R-

We have, (x, y) ϵ x^2 + y^2 = 9 ⇒ y^2 9 x^2 ⇒ y √(9 x^2) Putting x = 3, 0, 3, we get y = 0, ± 3, 0 respectively. For all other values of x, we get y ϵ̸z. ...

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Standard XII

Mathematics

Domain

If R is a rel...

Question

If R is a relation defined on the set Z of integers by the rule $(x, y) ϵ R \Leftrightarrow x^{2} + y^{2} - 9,$ then write domain of R.

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Solution

We have,

$(x, y) ϵ x^{2} + y^{2} = 9$

$\Rightarrow y^{2} - 9 - x^{2}$

$\Rightarrow y - \sqrt{9 - x^{2}}$

Putting x = -3, 0, 3, we get $y = 0, \pm 3, 0$ respectively.

For all other values of x, we get $y / ϵ z .$

$∴$ Domain (R) = {-3, 0, 3}

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If R is a relation defined on the set Z of integers by the rule (x,y)ϵR⇔x2+y2−9, then write domain of R.

We have, (x,y)ϵx2+y2=9 ⇒y2−9−x2 ⇒y−√9−x2 Putting x = -3, 0, 3, we get y=0,±3,0 respectively. For all other values of x, we get y/ϵz. ∴ Domain (R) = {-3, 0, 3}

If R is a relation defined on the set Z of integers by the rule $(x, y) ϵ R \Leftrightarrow x^{2} + y^{2} - 9,$ then write domain of R.

We have,

$(x, y) ϵ x^{2} + y^{2} = 9$

$\Rightarrow y^{2} - 9 - x^{2}$

$\Rightarrow y - \sqrt{9 - x^{2}}$

Putting x = -3, 0, 3, we get $y = 0, \pm 3, 0$ respectively.

For all other values of x, we get $y / ϵ z .$

$∴$ Domain (R) = {-3, 0, 3}