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We are given $A=\{1,2,3\}$ and $B=\{4,5,6\}$. The relation R from A to B is defined as $R=\{(x,y):x+y=6, x\in A, y\in B\}$. We need to find all pairs $(x,y)$ such that $x \in A$, $y \in B$, and $x+y=6$. If $x=1$, then $y=6-1=5$. So, $(1,5) \in R$. If $x=2$, then $y=6-2=4$. So, $(2,4) \in R$. If $x=3$, then $y=6-3=3$. But $3 \notin B$, so $(3,3) \notin R$. Therefore, $R = \{(1,5), (2,4)\}$.
A relation R from A to B is a function if every element in A has a unique image in B. In other words, for every $x \in A$, there should be exactly one $y \in B$ such that $(x,y) \in R$. Here, $R = \{(1,5), (2,4)\}$. The domain of R is $\{1,2\}$. Since $3 \in A$ but there is no $y \in B$ such that $(3,y) \in R$, R is not a function from A to B.
The domain of R is the set of all first elements in the ordered pairs of R. Domain of $R = \{1, 2\}$.
The range of R is the set of all second elements in the ordered pairs of R. Range of $R = \{4, 5\}$.
Final Answer: R = {(1,5), (2,4)}. R is not a function. Domain of R = {1, 2}. Range of R = {4, 5}.