Let A = {x : x in R, x text{ is not a positiv | vedclass

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(A) Given the function $f(x) = \frac{2x}{x-1}$ where $A = \{x \in R : x 
eq 1, 2, 3, \dots\}$.To check for injectivity,we find the derivative $f'(x)$

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injective but not surjective.

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(A) Given the function $f(x) = \frac{2x}{x-1}$ where $A = \{x \in R : x 
eq 1, 2, 3, \dots\}$.To check for injectivity,we find the derivative $f'(x)$:$f'(x) = \frac{(x-1)(2) - 2x(1)}{(x-1)^2} = \frac{2x - 2 - 2x}{(x-1)^2} = \frac{-2}{(x-1)^2}$.Since $f'(x) To check for surjectivity,let $y = \frac{2x}{x-1}$.$y(x-1) = 2x \implies yx - y = 2x \implies x(y-2) = y \implies x = \frac{y}{y-2}$.For $f$ to be surjective,for every $y \in R$,there must exist an $x \in A$. If $y = 2$,$x$ is undefined. Furthermore,we must ensure $x$ is not a positive integer. If $y = 0$,$x = 0$,which is allowed. However,if we choose $y$ such that $x$ becomes a positive integer (e.g.,if $x=2$,$y = \frac{2(2)}{2-1} = 4$),then $y=4$ has a pre-image $x=2$,but $2 
otin A$. Thus,$f$ is not surjective.

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

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