Let A = {x in R : x text{ is not a positive i | vedclass

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(A) Given $f(x) = \frac{2x}{x - 1}$ for $x \in A$,where $A = R \setminus \{1, 2, 3, \dots\}$.To check for injectivity (one-one): Let $f(x_1) = f(x_2)$

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injective but not surjective

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(A) Given $f(x) = \frac{2x}{x - 1}$ for $x \in A$,where $A = R \setminus \{1, 2, 3, \dots\}$.To check for injectivity (one-one): Let $f(x_1) = f(x_2)$.$\frac{2x_1}{x_1 - 1} = \frac{2x_2}{x_2 - 1}$$x_1(x_2 - 1) = x_2(x_1 - 1)$$x_1x_2 - x_1 = x_2x_1 - x_2$$-x_1 = -x_2 \Rightarrow x_1 = x_2$.Thus,$f$ is injective (one-one).To check for surjectivity (onto): Let $y = \frac{2x}{x - 1}$.$y(x - 1) = 2x$$yx - y = 2x$$x(y - 2) = y$$x = \frac{y}{y - 2}$.For $f$ to be surjective,for every $y \in R$,there must exist an $x \in A$ such that $f(x) = y$. If $y = 2$,$x$ is undefined. Also,if $x$ is a positive integer,$x 
otin A$. For example,if $y = 4$,$x = \frac{4}{4 - 2} = 2$. Since $2 
otin A$,there is no $x \in A$ such that $f(x) = 4$. Therefore,$f$ is not surjective.

Let $A = \{x \in R : x \text{ is not a positive integer}\}$. Define a function $f : A \to R$ as $f(x) = \frac{2x}{x - 1}$. Then $f$ is

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