1. Given function and domain

We have $A=\{x\in \mathbb R: x \text{ is not a positive integer}\}$ and $f:A\to \mathbb R,\qquad f(x)=\frac{2x}{x-1}.$

Since $1$ is a positive integer, $1\notin A$, so the formula is well-defined on all of $A$.

We must decide whether $f$ is injective/surjective.

2. Check injectivity

Assume $f(x_1)=f(x_2)$ for some $x_1,x_2\in A$.

Then $\frac{2x_1}{x_1-1}=\frac{2x_2}{x_2-1}.$

Cross-multiplying, $2x_1(x_2-1)=2x_2(x_1-1).$ Cancel $2$: $x_1x_2-x_1=x_1x_2-x_2.$ Hence, $-x_1=-x_2\implies x_1=x_2.$

Therefore, $f$ is injective.

3. Check surjectivity onto $\mathbb R$

Let $y\in \mathbb R$. We try to solve $y=\frac{2x}{x-1}$ for $x$.

Rearranging, $y(x-1)=2x$ $yx-y=2x$ $x(y-2)=y.$

If $y\neq 2$, then $x=\frac{y}{y-2}.$ So for every $y\neq 2$, there is a candidate preimage.

Now check $y=2$: $\frac{2x}{x-1}=2 \implies \frac{x}{x-1}=1 \implies x=x-1,$ which is impossible. So $y=2$ is not in the range.

Hence $f$ is not surjective onto $\mathbb R$.

4. Make sure domain restriction does not affect injectivity conclusion

The domain excludes positive integers. That does not create any duplicate outputs, so injectivity remains true.

Also, surjectivity already fails because $2$ is never attained, regardless of domain.

Thus, the function is injective but not surjective.

5. Option check

• A: not injective — false
• B: neither injective nor surjective — false
• C: surjective but not injective — false
• D: injective but not surjective — true

So the correct option is: $\boxed{D}$