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Sets and Relations question

2021 · 27 Jul · Shift 2 · Q35
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  5. /2021 · 27 Jul · Shift 2 · Q35

Sets and Relations question

2021 · 27 Jul · Shift 2 · Q35

JEE MainMathematicsSets and RelationsMCQ+4 / −1
Let N be the set of natural numbers and a relation R on N be defined by R={(x,y)∈N×N:x3−3x2y−xy2+3y3=0}R = \{ (x,y) \in N \times N:{x^3} - 3{x^2}y - x{y^2} + 3{y^3} = 0\}R={(x,y)∈N×N:x3−3x2y−xy2+3y3=0}. Then the relation R is :
  1. A
    symmetric but neither reflexive nor transitive
  2. B
    reflexive but neither symmetric nor transitive
  3. C
    reflexive and symmetric, but not transitive
  4. D
    an equivalence relation
View written solutionFree

Correct answer: B

  1. Given relation

We have

R={(x,y)∈N×N:x3−3x2y−xy2+3y3=0}.R=\{(x,y)\in N\times N: x^3-3x^2y-xy^2+3y^3=0\}.R={(x,y)∈N×N:x3−3x2y−xy2+3y3=0}.

We must determine whether RRR is reflexive, symmetric, and transitive.

  1. Factor the given expression

Let

f(x,y)=x3−3x2y−xy2+3y3.f(x,y)=x^3-3x^2y-xy^2+3y^3.f(x,y)=x3−3x2y−xy2+3y3.

Group terms:

(x3−3x2y)+(−xy2+3y3)=x2(x−3y)−y2(x−3y).(x^3-3x^2y)+(-xy^2+3y^3)=x^2(x-3y)-y^2(x-3y).(x3−3x2y)+(−xy2+3y3)=x2(x−3y)−y2(x−3y).

So,

f(x,y)=(x2−y2)(x−3y).f(x,y)=(x^2-y^2)(x-3y).f(x,y)=(x2−y2)(x−3y).

Now use

x2−y2=(x−y)(x+y).x^2-y^2=(x-y)(x+y).x2−y2=(x−y)(x+y).

Hence

x3−3x2y−xy2+3y3=(x−y)(x+y)(x−3y).x^3-3x^2y-xy^2+3y^3=(x-y)(x+y)(x-3y).x3−3x2y−xy2+3y3=(x−y)(x+y)(x−3y).

Therefore,

(x,y)∈R  ⟺  (x−y)(x+y)(x−3y)=0.(x,y) \in R \iff (x-y)(x+y)(x-3y)=0.(x,y)∈R⟺(x−y)(x+y)(x−3y)=0.

Since x,y∈Nx,y\in Nx,y∈N, we have x+y≠0x+y\neq 0x+y=0. So the condition reduces to

x−y=0orx−3y=0.x-y=0 \quad \text{or} \quad x-3y=0.x−y=0orx−3y=0.

Thus,

(x,y)∈R  ⟺  x=yorx=3y.(x,y)\in R \iff x=y \quad \text{or} \quad x=3y.(x,y)∈R⟺x=yorx=3y.
  1. Check reflexivity

A relation is reflexive if (x,x)∈R(x,x)\in R(x,x)∈R for all x∈Nx\in Nx∈N.

Put y=xy=xy=x. Then clearly

x=y,x=y,x=y,

so (x,x)∈R(x,x)\in R(x,x)∈R for every x∈Nx\in Nx∈N.

Hence, RRR is reflexive.

  1. Check symmetry

A relation is symmetric if

(x,y)∈R  ⟹  (y,x)∈R.(x,y)\in R \implies (y,x)\in R.(x,y)∈R⟹(y,x)∈R.

Take an example: let y=1y=1y=1, x=3x=3x=3. Then

x=3y,x=3y,x=3y,

so (3,1)∈R(3,1)\in R(3,1)∈R.

Now check (1,3)(1,3)(1,3):

  • 1≠31\neq 31=3
  • 1≠3⋅3=91\neq 3\cdot 3=91=3⋅3=9

So (1,3)∉R(1,3)\notin R(1,3)∈/R.

Hence, RRR is not symmetric.

  1. Check transitivity

A relation is transitive if

(x,y)∈R and (y,z)∈R  ⟹  (x,z)∈R.(x,y)\in R \text{ and } (y,z)\in R \implies (x,z)\in R.(x,y)∈R and (y,z)∈R⟹(x,z)∈R.

We know:

(x,y)∈R  ⟺  x=y or x=3y.(x,y)\in R \iff x=y \text{ or } x=3y.(x,y)∈R⟺x=y or x=3y.

Also,

(y,z)∈R  ⟺  y=z or y=3z.(y,z)\in R \iff y=z \text{ or } y=3z.(y,z)∈R⟺y=z or y=3z.

Now check all cases:

Case 1: x=yx=yx=y and y=zy=zy=z

Then x=zx=zx=z, so (x,z)∈R(x,z)\in R(x,z)∈R.

Case 2: x=yx=yx=y and y=3zy=3zy=3z

Then x=3zx=3zx=3z, so (x,z)∈R(x,z)\in R(x,z)∈R.

Case 3: x=3yx=3yx=3y and y=zy=zy=z

Then x=3zx=3zx=3z, so (x,z)∈R(x,z)\in R(x,z)∈R.

Case 4: x=3yx=3yx=3y and y=3zy=3zy=3z

Then

x=3(3z)=9z.x=3(3z)=9z.x=3(3z)=9z.

For (x,z)(x,z)(x,z) to be in RRR, we need either

x=zorx=3z.x=z \quad \text{or} \quad x=3z.x=zorx=3z.

But here x=9zx=9zx=9z, which is neither zzz nor 3z3z3z for natural zzz. So (x,z)∉R(x,z)\notin R(x,z)∈/R.

A concrete counterexample:

(9,3)∈Rsince 9=3⋅3,(9,3)\in R \quad \text{since } 9=3\cdot 3,(9,3)∈Rsince 9=3⋅3, (3,1)∈Rsince 3=3⋅1,(3,1)\in R \quad \text{since } 3=3\cdot 1,(3,1)∈Rsince 3=3⋅1,

but

(9,1)∉R,(9,1)\notin R,(9,1)∈/R,

since 9≠19\neq 19=1 and 9≠3⋅19\neq 3\cdot 19=3⋅1.

Hence, RRR is not transitive.

  1. Conclusion

So the relation is:

  • reflexive: Yes
  • symmetric: No
  • transitive: No

Therefore, the correct option is

B: reflexive but neither symmetric nor transitive\boxed{\text{B: reflexive but neither symmetric nor transitive}}B: reflexive but neither symmetric nor transitive​
  1. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So the stored answer is correct.

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