JEE MAIN - Mathematics (2021 - 27th July Evening Shift - No. 13)
Let N be the set of natural numbers and a relation R on N be defined by $$R = \{ (x,y) \in N \times N:{x^3} - 3{x^2}y - x{y^2} + 3{y^3} = 0\} $$. Then the relation R is :
symmetric but neither reflexive nor transitive
reflexive but neither symmetric nor transitive
reflexive and symmetric, but not transitive
an equivalence relation
Explanation
$${x^3} - 3{x^2}y - x{y^2} + 3{y^3} = 0$$
$$ \Rightarrow x({x^2} - {y^2}) - 3y({x^2} - {y^2}) = 0$$
$$ \Rightarrow (x - 3y)(x - y)(x + y) = 0$$
Now, x = y $$\forall$$(x, y) $$\in$$N $$\times$$ N so reflexive but not symmetric & transitive.
See, (3, 1) satisfies but (1, 3) does not. Also (3, 1) & (1, $$-$$1) satisfies but (3, $$-$$1) does not.
$$ \Rightarrow x({x^2} - {y^2}) - 3y({x^2} - {y^2}) = 0$$
$$ \Rightarrow (x - 3y)(x - y)(x + y) = 0$$
Now, x = y $$\forall$$(x, y) $$\in$$N $$\times$$ N so reflexive but not symmetric & transitive.
See, (3, 1) satisfies but (1, 3) does not. Also (3, 1) & (1, $$-$$1) satisfies but (3, $$-$$1) does not.
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