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

(C) symmetric but neither reflexive nor transitive

(D) reflexive and symmetric, but not transitive

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Sets Relations and Functions

Let

N

be the set of natural numbers and a relation

R

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 :

Answer: (A)

The defining expression factorises:

x^{3} - 3 x^{2} y - x y^{2} + 3 y^{3} = x^{2} (x - 3 y) - y^{2} (x - 3 y) = (x - 3 y) (x - y) (x + y)

For

x, y \in N

the factor

x + y

is positive, so

(x, y) \in R

exactly when

x = y

x = 3 y

.
Every natural number satisfies

x = x

, so

R

is reflexive. Next,

(3, 1) \in R

because

3 = 3 \cdot 1

, while

(1, 3) \in / R

because

1 \neq = 3

and

1 \neq = 9

, so

R

is not symmetric. Finally

(9, 3) \in R

and

(3, 1) \in R

, but

(9, 1) \in / R

because

9 \neq = 1

and

9 \neq = 3

, so

R

is not transitive. Thus

R

is reflexive but neither symmetric nor transitive.

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Single correct option