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

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(D) Given $f(x) = \frac{2x}{x-1}$.To check for injectivity,we find the derivative: $f'(x) = \frac{(x-1)(2) - 2x(1)}{(x-1)^2} = \frac{-2}{(x-1)^2}$.Sin

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Injective but not surjective.

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(D) Given $f(x) = \frac{2x}{x-1}$.To check for injectivity,we find the derivative: $f'(x) = \frac{(x-1)(2) - 2x(1)}{(x-1)^2} = \frac{-2}{(x-1)^2}$.Since $f'(x) To check for surjectivity,let $f(x) = y$. Then $y = \frac{2x}{x-1} \Rightarrow yx - y = 2x \Rightarrow x(y-2) = y \Rightarrow x = \frac{y}{y-2}$.For $f$ to be surjective,for every $y \in R$,there must exist an $x \in A$ such that $f(x) = y$.If $y = 2$,then $x = \frac{2}{0}$,which is undefined. Thus,$2$ has no pre-image in $A$.Also,if $y = 4$,$x = \frac{4}{4-2} = 2$. However,$2$ is a positive integer,so $2 
otin A$.Since there exist elements in the codomain $R$ that do not have a pre-image in $A$,$f$ is not surjective.Therefore,$f$ is injective but not surjective.

Let $A = \{x \in R \mid x \text{ is not a positive integer}\}$. Let a function $f$ be defined as $f: A \rightarrow R$ such that $f(x) = \frac{2x}{x-1}$. Then $f$ is:

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