If A = {x : x^2 - 5x + 6 = 0}, B = {2, 4}, C | vedclass

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(A) First,we solve the equation $x^2 - 5x + 6 = 0$ to find the elements of set $A$.$x^2 - 2x - 3x + 6 = 0 \implies x(x - 2) - 3(x - 2) = 0 \implies (x

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$\{(2, 4), (3, 4)\}$

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(A) First,we solve the equation $x^2 - 5x + 6 = 0$ to find the elements of set $A$.$x^2 - 2x - 3x + 6 = 0 \implies x(x - 2) - 3(x - 2) = 0 \implies (x - 2)(x - 3) = 0$.Thus,$A = \{2, 3\}$.Given $B = \{2, 4\}$ and $C = \{4, 5\}$,the intersection $B \cap C$ is the set of common elements.$B \cap C = \{4\}$.Now,we find the Cartesian product $A 	imes (B \cap C) = \{2, 3\} 	imes \{4\}$.$A 	imes (B \cap C) = \{(2, 4), (3, 4)\}$.Therefore,the correct option is $A$.

If $A = \{x : x^2 - 5x + 6 = 0\}, B = \{2, 4\}, C = \{4, 5\}$,then $A \times (B \cap C)$ is

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