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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Sets and Relations

NTA Abhyas 2020 · Mathematics · Sets and Relations

The relation& R ~given by& \ (x, y) : x^ 2 - 3xy + 2y^ 2 = 0, x, y...

The relation& RR~given by& \left\{\left(x, y\right): x^{2} - 3xy + 2y^{2} = 0, \forall x, y \in R\right\} is

Options

  1. Areflexive but not symmetric
  2. Bsymmetric but not transitive
  3. Csymmetric and transitive
  4. Dan equivalence relation

Correct answer

A. reflexive but not symmetric

Step-by-step solution

∵x2−3xy+2y2=0⇒x2−xy−2xy+2y2=0∵ x^{2} - 3xy + 2y^{2} = 0 ⇒ x^{2} - xy - 2xy + 2y^{2} = 0
⇒x(x−y)−2y(x−y)=0⇒ x \left(x - y\right) - 2y \left(x - y\right) = 0
⇒(x−2y)(x−y)=0⇒ \left(x - 2y\right) \left(x - y\right) = 0
⇒x=yorx=2y⇒ x = y or x = 2y
Now, ∵inR∵ in R all ordered pairs \left(x, x\right)arepresent<br/>are present<br />∴Itisreflexive<br/>Now,(4,2)∈RIt is reflexive<br />Now, \left(4, 2\right)\in \mathbb{R} as 4=2(2)<br/>but(2,4)∉R4 = 2\left(2\right)<br />but \left(2,4\right)\notin R as 2≠2(4)<br/>2 \ne 2\left(4\right)<br />∴It is not symmetric<br />Also \left(4,2\right) & \left(2,1\right)\in \mathbb{R} but \left(4,1\right)\notin R
∴∴ It is not transitive
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