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To check if the function is injective, we need to verify if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in A$. $$f(x_1) = \frac{x_1 - 2}{x_1 - 3}$$ $$f(x_2) = \frac{x_2 - 2}{x_2 - 3}$$ If $f(x_1) = f(x_2)$, then $$\frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3}$$ Cross-multiplying, we get: $$(x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3)$$ $$x_1x_2 - 3x_1 - 2x_2 + 6 = x_1x_2 - 3x_2 - 2x_1 + 6$$ $$-3x_1 - 2x_2 = -3x_2 - 2x_1$$ $$x_2 = x_1$$ Thus, $f(x_1) = f(x_2)$ implies $x_1 = x_2$, so the function is injective.
To check if the function is surjective, we need to verify if for every $y \in B$, there exists an $x \in A$ such that $f(x) = y$. Let $y = \frac{x - 2}{x - 3}$. We need to solve for $x$ in terms of $y$. $$y(x - 3) = x - 2$$ $$yx - 3y = x - 2$$ $$yx - x = 3y - 2$$ $$x(y - 1) = 3y - 2$$ $$x = \frac{3y - 2}{y - 1}$$ Since $y \in B = R - \{1\}$, $y \neq 1$, so the denominator is never zero. Thus, for every $y \in B$, we can find an $x = \frac{3y - 2}{y - 1}$. Now we need to check if $x \in A = R - \{3\}$. If $x = 3$, then $$3 = \frac{3y - 2}{y - 1}$$ $$3(y - 1) = 3y - 2$$ $$3y - 3 = 3y - 2$$ $$-3 = -2$$ This is a contradiction, so $x \neq 3$. Therefore, $x \in A$. Since for every $y \in B$, there exists an $x \in A$ such that $f(x) = y$, the function is surjective.
Since the function is both injective and surjective, it is bijective.
Final Answer: The function is bijective.