Sets relations and functions jee mains questions | JEE Main Mathematics Sets Relations And Functions

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Solution

Q.
Consider the relations $R_{1}$ and $R_{2}$ defined as $a R_{1} b \Leftrightarrow a^{2} + b^{2} = 1$ for all $a, b \in R$ and $(a, b) R_{2} (c, d) \Leftrightarrow a + d = b + c$ for all $(a, b), (c, d) \in N \times N .$ Then [2024]

1 $R_{1}$ and $R_{2}$ both are equivalence relations

2 Only $R_{1}$ is an equivalence relation

3 Only $R_{2}$ is an equivalence relation

4 Neither $R_{1}$ nor $R_{2}$ is an equivalence relation

Ans.
(3)

   $Consider, (x, x) \in R_{1}$

   $Then, x^{2} + x^{2} \neq 1 for any x \in R$

   $∴$    $R_{1} is not reflexive$

   $Hence, R_{1} is not an equivalence relation .$

   $For R_{2}$

   $(i) L e t (x, y) R (x, y) \Rightarrow x + y = y + x, which is true for all x, y \in N .$

   $∴ R_{2} is reflexive .$

   $(i i) Let (x, y) R (z, p) \Rightarrow x + p = y + z \Rightarrow p + x = z + y$

$\Rightarrow z + y = p + x \Rightarrow (z, p) R (x, y)$

   $∴ R_{2} i s s y m m e t r i c .$

   $(i i i) Let (a, b) R (c, d) \Rightarrow a + d = b + c$ ...(i)

$(c, d) R (e, f) \Rightarrow c + f = d + e$ ...(ii)

   $From (i) and (ii), we have a + d + c + f = b + c + d + e$

   $a + f = b + e \Rightarrow (a, b) R (e, f)$

   $∴ R_{2} is transitive .$

   $S o, R_{2} is an equivalence relation .$

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