A) reflexive but neither symmetric nor transitive
B) symmetric and transitive.
C) reflexive and symmetric.
D) reflexive and transitive.
Correct Answer: D
Solution :
\[R=\{(x,y):x,y\in N\,and\,{{x}^{2}}-4xy+3{{y}^{2}}=0\}\] Now, \[{{x}^{2}}-4xy+3{{y}^{2}}=0\] \[\Rightarrow\] \[(x-y)(x-3y)=0\] \[\therefore\] \[x=y\,or\,x=3y\] \[\therefore\] \[R=\{(1,1),\}(3,1),(2,2),(6,2),(3,3),(9,3),....\}\] Since (1,1), (2,2), (3,3),...... are present in the relation, therefore R is reflexive. Since (3,1) is an element of R but (1,3) is not the element of R, therefore R is not symmetric Here(3,1) \[\in\] R and (1,1) \[\in\] \[R\Rightarrow\] (3,1) \[R\Rightarrow\] (6,2) \[R\Rightarrow\] and (2,2) \[R\Rightarrow\] (6,2) \[R\Rightarrow\] For all such (a, b) \[R\Rightarrow\] and (b, c) \[R\Rightarrow\] \[\Rightarrow\] (a,c) \[\in R\] Hence R is transitive.
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