Let R be a relation defined on N as a R b is 2 a+3 b is a multiple of 5, a, b ∈ N . Then R is

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1.Relation and Function

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Let $R$ be a relation defined on $N$ as a $R$ b is $2 a+3 b$ is a multiple of $5, a, b \in N$. Then $R$ is

not reflexive

transitive but not symmetric

symmetric but not transitive

an equivalence relation

(JEE MAIN-2023)

Solution

$a R a \Rightarrow 5 a$ is multiple it 5

So reflexive

$a R b \Rightarrow 2 a +3 b =5 \alpha$,

Now b R a

$2 b+3 a=2 b+\left(\frac{5 \alpha-3 b}{2}\right) \cdot 3$

$=\frac{15}{2} \alpha-\frac{5}{2} b=\frac{5}{2}(3 \alpha-b)$

$=\frac{5}{2}(2 a+2 b-2 \alpha)$

$=5(a+b-\alpha)$

Hence symmetric

$\text { a R b } \quad \Rightarrow 2 a+3 b=5 \alpha \text {. }$

$\text { b R c } \quad \Rightarrow 2 b+3 c=5 \beta$

$\text { Now } \quad 2 a+5 b+3 c=5(\alpha+\beta)$

$\Rightarrow 2 a +5 b +3 c =5(\alpha+\beta)$

$\Rightarrow 2 a +3 c =5(\alpha+\beta- b )$

$\Rightarrow a R c$

Hence relation is equivalence relation.

Standard 12

Mathematics

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Let $R$ be a relation defined on $N$ as a $R$ b is $2 a+3 b$ is a multiple of $5, a, b \in N$. Then $R$ is

Solution

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