For distinct positive integers x and y, where x < y, the function FP(x

Wow . Such a good ques . Kudos to this ques !! Took me >3 min to deduce the logic .
Ans is E IMO .

FP(a, b) +FP(c, d) = FP(e, f) , a, b, c, d, e and f are distinct positive integers ..

Prime + Prime = Prime

we know that addition of 2 prime can only yield prime number if one of them is 2 .

so either a or c must be 1 .

We need to find value of ac .

(1) FP(g, h) = a, where g and h are distinct positive integers

a must be a prime number .
So now c must be 1 . But a can be any prime 3 or 2 .
FP(2, 4) +FP(1, 3) = FP(4, 6) ===> 3 + 2 = 5

or
FP(3, 6) +FP(1, 9) = FP(5,8 ) ===> 5 + 2 = 7

ac can be 2 or 3 .

So statement 1 is not sufficient .

(2) c is less than the minimum possible value of the function FP(x,y).

minimum possible value of the function FP(x,y) is 2 .
Now
C < 2 and C > 0
so C = 1 .
But a can be anything .
So statement 2 is not sufficient .

Now Combining both
we have c=1 and a can be still 2 or 3 .

So statement 1 and 2 , combing is not sufficient .
Ans E