For minimum and maximum no of tuples resulted due to the natural join and in which we have a common attribute P which is foreign key in relation R2 referencing to R1's primary key P.So since the common attribute is a foreign key here , we have to take care of integrity constraints of a foreign key as well.So let us consider the 2 extremes :
Case 1 : All tuples of R2 have same value of P
In that case , all tuples of R2 will match with exactly one tuple of R1 as in natural join we join the tuples based on the value of common attribute if they are found to be same.So in this case all the 2000 tuples of R2 will combine with exactly one of the tuples of R1.
Hence no of tuples that will result in R1 natural join R2 = 2000
Case 2 : P value is uniformly distributed over R2 tuples :
By that we mean every 2 tuple of R2 is matched with one tuple of R1 , so all the 1000 values of P in R1 is uniformly distributed over R2.
Therefore due to one value of P in R1 , we have 2 matchings with R2 , natural join results 2 tuples .
So for entire natural join of R1 and R2 = 2 * 1000
= 2000 tuples
Hence in any case we are going to have 2000 tuples in the natural join only since a tuple of R2 is going to match with at least one tuple of R1 as the common attribute P of R2 referencees to R1 and also P is non null in R2 so it is guaranteed that matching of each tuple in R2 is going to happen.
Hence , maximum = minimum no of tuples = 2000 tuples
Therefore , X + 2Y = 2000 + 4000
= 6000