Correct answer: 1
Case n=0: |2f-1| = 2f. If f >= 1/2: 2f-1 = 2f => -1=0 (no solution). If f < 1/2: 1-2f = 2f => f=1/4, so a=1/4. Case n=-1: |2(-1+f)-1| = 3(-1)+2f = -3+2f. For valid RHS: -3+2f >= 0 => f >= 3/2, impossible. Case n=1: |2(1+f)-1| = 3+2f => |2f+1| = 3+2f => 2f+1 = 3+2f => 1=3 (no). No other cases give solutions. Wait, let me also try negative: n=-1 gives RHS = -3+2f which needs to be non-negative, impossible. n=0 gives a=1/4. But what about 2a-1 < 0 for n>=0 case? For n=0, f<1/2: already done, a=1/4. Sum S = {1/4}, 4*sum = 4*(1/4) = 1. But answer options suggest more elements. Let me try n=1 again: |3+2f| but wait, |2(1+f)-1| = |1+2f| = 1+2f (since positive). Equation: 1+2f = 3+2f => 1=3 (no). For n=-1 and a < 0: |2a-1| = |-2+2f-1| = |2f-3| = 3-2f (since f<1, 2f<2<3). RHS = 3(-1)+2f = -3+2f. So 3-2f = -3+2f => 6=4f => f=3/2 > 1, invalid. Try n=0, a could be negative if... no, n=0,f in [0,1) means a in [0,1). Sum = 1/4, 4*1/4 = 1. Answer: 1.