Right answer is (c) (1/2)^nu(n) & -(1/2)^nu(-n-1)
Explanation: The system function of given system is H(z)=\(1-\frac{1}{2} z^{-1}\)
The inverse of the system has a system function as H(z)=\(\frac{1}{1-\frac{1}{2} z^{-1}}\)
Thus it has a zero at origin and a pole at z=1/2.So, two possible cases are |z|>1/2 and |z|<1/2
So, h(n)= (1/2)^nu(n) for causal and stable(|z|>1/2)
and h(n)= -(1/2)^nu(-n-1) for anti causal and unstable for |z|<1/2.