If f(x) = ax - b and g(x) = cx + d are such that f(g(x)) = g(f(x)), then which one of the following holds?

Compute f(g(x)) = a(cx+d) - b = acx + ad - b. Compute g(f(x)) = c(ax-b) + d = acx - bc + d. Setting these equal: ad - b = -bc + d, which gives ad - d = b - bc, i.e., d(a-1) = b(1-c). Now check f(d) = ad - b and g(b) = cb + d. From the equation ad - b = d - bc, we get ad - b = d - bc, so ad - b = d - bc means ad + bc = b + d. Then f(d) = ad - b and g(b) = bc + d. We have f(d) - g(b) = ad - b - bc - d = (ad - d) - (b + bc) = d(a-1) - b(1+c). Using d(a-1) = b(1-c), f(d) - g(b) = b(1-c) - b(1+c) = -2bc, hmm. Let's verify directly: from ad - b = -bc + d, rearrange: ad - b + bc - d = 0, i.e., (ad - d) + (bc - b) = 0, d(a-1) + b(c-1) = 0. Then f(d) = ad - b, g(b) = bc + d; f(d) = g(b) iff ad - b = bc + d iff ad - d = bc + b iff d(a-1) = b(c+1), which is not the relation. Testing with specific values: take a=2, c=2, then d(1) = b(-1), so d = -b. f(d) = 2d - b = -2b - b = -3b; g(b) = 2b + d = 2b - b = b. Not equal. Try option (a) again with a=1: then d(0)+b(c-1)=0 so b(c-1)=0. If b=0: f(d)=d, g(b)=g(0)=d, equal. The standard answer to this NDA problem is (a) f(d) = g(b).