As $A=\{x \in R: x$ is not a positive integer $\}$ $f: A \rightarrow R$ given by $f(x)=\frac{2 x}{x-1}$ $$ f\left(x_1\right)=f\left(x_2\right) \Leftrightarrow x_1=x_2 $$ So, $f$ is one-one. As $\mathrm{f}(\mathrm{x}) \neq 2$ for any $\mathrm{x} \in \mathrm{A} \Rightarrow \mathrm{f}$ is not onto. $\therefore \mathrm{f}$ is injective but not surjective.