No of different possible marks we can get if 'a' marks will be added for the correct answer and 'b' marks will be deducted if the answer happened to be wrong. An unattempted question is awarded 0 marks. The number of questions will be n.
For example: For n=2 and a=1 and b=1 is -2,-1,0,1,2. I have read this link. Possible marks in a MCQ test
Let $x$ be the total number of correct answers. You've mentioned everything else in the question. So, marks obtained = $$f(x)=ax-(n-x)b$$ Check if it's an injection.
$$f(x)=f(y)$$$$\implies x=y$$or$$a+b=0$$
Case $1$: $x=y \implies injection \implies n+1$ possible number of marks, since $x$ can vary from $0-n$, both inclusive.
The other case, $a+b=0$ is illogical since it implies either the marks gained or marks deducted are negative. But, even if we consider it, $$f(x)=ax+(n-x)a=an,$$ a constant. So, in this case, you'll always get the same marks!