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(D) To check if the function is invertible,we need to show it is both one-to-one (injective) and onto (surjective).$1$. One-to-one: Let $f(x_1) = f(x_

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invertible and $f^{-1}(y) = \frac{2y - 1}{y - 1}$

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(D) To check if the function is invertible,we need to show it is both one-to-one (injective) and onto (surjective).$1$. One-to-one: Let $f(x_1) = f(x_2)$.$\frac{x_1 - 1}{x_1 - 2} = \frac{x_2 - 1}{x_2 - 2}$$(x_1 - 1)(x_2 - 2) = (x_2 - 1)(x_1 - 2)$$x_1x_2 - 2x_1 - x_2 + 2 = x_1x_2 - 2x_2 - x_1 + 2$$-2x_1 - x_2 = -2x_2 - x_1$$x_2 = x_1$. Thus,$f$ is one-to-one.$2$. Onto: Let $y = \frac{x - 1}{x - 2}$.$y(x - 2) = x - 1$$yx - 2y = x - 1$$yx - x = 2y - 1$$x(y - 1) = 2y - 1$$x = \frac{2y - 1}{y - 1}$.Since for every $y \in R - \{1\}$,there exists an $x \in R - \{2\}$,the function is onto.Since $f$ is both one-to-one and onto,it is invertible,and $f^{-1}(y) = \frac{2y - 1}{y - 1}$.

Let $f : A \to B$ be a function defined as $f(x) = \frac{x - 1}{x - 2}$,where $A = R - \{2\}$ and $B = R - \{1\}$. Then $f$ is

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