Let R = {(1, 3), (2, 2), (3, 2)} and S = {(2, | vedclass

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(C) The composition of relations $RoS$ is defined as the set of all pairs $(x, z)$ such that there exists $y \in A$ where $(x, y) \in S$ and $(y, z) \

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$\{(2, 3), (3, 2), (2, 2)\}$

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(C) The composition of relations $RoS$ is defined as the set of all pairs $(x, z)$ such that there exists $y \in A$ where $(x, y) \in S$ and $(y, z) \in R$.Given $S = \{(2, 1), (3, 2), (2, 3)\}$ and $R = \{(1, 3), (2, 2), (3, 2)\}$.We check each element $(x, y) \in S$:$1$. For $(2, 1) \in S$,we look for $(1, z) \in R$. We find $(1, 3) \in R$,so $(2, 3) \in RoS$.$2$. For $(3, 2) \in S$,we look for $(2, z) \in R$. We find $(2, 2) \in R$,so $(3, 2) \in RoS$.$3$. For $(2, 3) \in S$,we look for $(3, z) \in R$. We find $(3, 2) \in R$,so $(2, 2) \in RoS$.Thus,$RoS = \{(2, 3), (3, 2), (2, 2)\}$.

List-$I$	List-$II$
$A$. Range of $\sec ^{-1}\left[1+\cos ^2 x\right]$, where $[.]$ denotes the greatest integer function	$I$. Odd function
$B$. Domain of $f(x)$ where $f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2}$	$II$. $\left\{0, \frac{1}{2}\right\}$
$C$. $f(x+y)=f(x)+f(y) ; f(1)=5$	$III$. $\left\{\sec ^{-1} 5, \sec ^{-1} 4\right\}$
$D$. $\sin ^{-1} x-\cos ^{-1} x+\sin ^{-1}(1-x)=0 \Rightarrow x \in$	$IV$. $R$
	$V$. $\left\{\sec ^{-1} 1, \sec ^{-1} 2\right\}$

Let $R = \{(1, 3), (2, 2), (3, 2)\}$ and $S = \{(2, 1), (3, 2), (2, 3)\}$ be two relations on set $A = \{1, 2, 3\}$. Then $RoS =$

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