Correct answer is (b) False
To explain I would say: a^n u(n) ↔ \(\frac{1}{1-az^{-1}} = \frac{z}{z-a}\); ROC:|z|>|a|
-a^n u(-n-1) ↔ \(\frac{1}{1-az^{-1}} = \frac{z}{z-a}\); ROC:|z|<|a|
Hence, x(n) = a^n u(n) and x(n) = -a^n u(-n-1) have the same X(Z) and differ only in ROC.