The correct option is B injective but not surjective
Given f(x)=2xx−1
f(x)=2+2x−1
Let x1,x2∈A such that f(x1)=f(x2)
⇒1x1−1=1x2−1⇒x1=x2
Hence, f is an injective function.
Let y=f(x)
⇒y=2xx−1
⇒x=yy−2
If y=3, then x=33−2=3>0
But it is given, x is not a positive integer.
So, function is not a surjective function.