Let N be the set of natural numbers and a relation R on N be defined by = \(x,y) in N x N:x^3 - 3x^2y - xy^2 + 3y^3 = 0\. Then the relation R is

(a) : Reflexive : Let (x,x) in R \ => x^3 - 3x^3 - x^3 + 3x^3 = 0 \ Therefore R is reflexive. Symmetric : Let (x,y) in R => x^3 - 3x^2y - xy^2 + 3y^3 = 0 \ => 3x^3 - yx^2 - 3y^2x + y^3 = 0 So, R is not symmetric. Similarly, R is not transitive.

Let N be the set of natural numbers and a relation R on... | Relation (Mathematics)

JEE Mains ( $27^{th\ }$ July $2^{nd\ }$ Shift 2021)

Mathematics

Let

N

be the set of natural numbers and a relation

R

N

be defined by

= \{(x,y) \in N \times N:x^{3} - 3x^{2}y -

xy^{2} + 3y^{3} = 0\}

. Then the relation

R

reflexive but neither symmetric nor transitive

an equivalence relation

reflexive and symmetric, but not transitive

symmetric but neither reflexive nor transitive

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Correct Answer:

Explanation

(a) : Reflexive : Let $(x,x) \in R \\$ $\Rightarrow x^{3} - 3x^{3} - x^{3} + 3x^{3} = 0 \\ \therefore R$ is reflexive. Symmetric : Let $(x,y) \in R \\$ $\Rightarrow x^{3} - 3x^{2}y - xy^{2} + 3y^{3} = 0 \\ \Rightarrow 3x^{3} - yx^{2} - 3y^{2}x + y^{3} = 0 \\$ So, $R$ is not symmetric. $\\$ Similarly, $R$ is not transitive.

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