For the mapping f: R-{1} rightarrow R-{2}, gi | vedclass

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(D) To check for 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_1x_2 - x_2$$-x_1 =

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$f$ is both one-one and onto

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(D) To check for 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_1x_2 - x_2$$-x_1 = -x_2 \Rightarrow x_1 = x_2$.Thus, $f$ is one-one.To check for onto: Let $y = \frac{2x}{x-1}$.$y(x-1) = 2x \Rightarrow yx - y = 2x \Rightarrow x(y-2) = y \Rightarrow x = \frac{y}{y-2}$.Since $y \in R-\{2\}$, $x$ is always defined and $x 
eq 1$. Thus, for every $y$ in the codomain, there exists an $x$ in the domain.Therefore, $f$ is onto.Hence, $f$ is both one-one and onto.

For the mapping $f: R-\{1\} \rightarrow R-\{2\}$, given by $f(x)=\frac{2x}{x-1}$, which of the following is correct?

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