# Consider the following relations:$\dpi{100} R=\left \{ (x,y) \left | x,y\; are\; real\; numbers\; and\; x=wy\, for\, some\, rational\, number\, w \right \};$ are integers such that$\dpi{100} n,q\neq 0\: and \: qm = pn \left \ \right \}$then Option 1) $R$ is an equivalence relation but $S$  is not an equivalence relation Option 2) neither $R$ nor $S$ is an equivalence relation Option 3) $S$is an equivalence relation but  $R$ is not an equivalence relation Option 4) $R$ and$S$ both are equivalence relations

As we learnt in

Equivalence relation -

Any relation which is reflexive, symmetric and transitive is called an equivalence relation

-

$S=\left \{\left(\frac{m}{n}, \frac{p}{q} \right ) \right \}$

given that qm = pn , $\frac{m}{n}=\frac{p}{q}$

Is is always true for (a, a) reflexive.

$\therefore\ \; f^{-1}(x)=\pm \sqrt{x+1}-1$

$\therefore\ \; (x+1)^{2} -1=\sqrt{x+1}-1$

$(x+1)^{4}=(x+1)$

$(x+1)=0\ \Rightarrow\ \; (x+1)^{3}=1, \;x+1=1, \; x=0$

x = - 1

$\therefore$    {0,  -1}

Correct option is 4.

Option 1)

$R$ is an equivalence relation but $S$  is not an equivalence relation

This is an incorrect option.

Option 2)

neither $R$ nor $S$ is an equivalence relation

This is an incorrect option.

Option 3)

$S$is an equivalence relation but  $R$ is not an equivalence relation

This is an incorrect option.

Option 4)

$R$ and$S$ both are equivalence relations

This is the correct option.

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