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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Mathematics
Chapter
Sets and Relations

BITSAT 2015 · Mathematics · Sets and Relations

If =\ (x, y) x^ 2 +y^ 2 =1 ; x, y ∈ ℝ\ . Then, is

If ρ={(x,y)∣x2+y2=1;x,y∈R}\rho=\left\{(x, y) \mid x^{2}+y^{2}=1 ; x, y \in \mathbb{R}\right\}. Then, ρ\rho is

Options

  1. Areflexive
  2. Bsymmetric
  3. Ctransitive
  4. Danti-symmetric

Correct answer

B. symmetric

Step-by-step solution

Obviously, the relation is not reflexive and transitive, but it is symmetric, because x2+x2=2x2≠1x^{2}+x^{2}=2 x^{2} \neq 1 and x2+y2=1,y2+z2=1⇒x2+z2=1 But x2+y2=1⇒y2+x2=1\begin{array}{l}\text{ and } x^{2}+y^{2}=1, y^{2}+z^{2}=1 \\ \Rightarrow x^{2}+z^{2}=1 \\ \text{ But } x^{2}+y^{2}=1 \Rightarrow y^{2}+x^{2}=1 \end{array}
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