Right answer is (b) \(\frac{1-a^2}{1-2acosω+a^2}\)
Explanation: First we observe x(n) can be expressed as
x(n)=x1(n)+x2(n)
where x1(n)= a^n, n>0
=0, elsewhere
x2(n)=a^-n, n<0
=0, elsewhere
Now applying Fourier transform for the above two signals, we get
X1(ω)=\(\frac{1}{1-ae^{-jω}}\) and X2(ω)=\(\frac{ae^{jω}}{1-ae^{jω}}\)
Now, X(ω)=X1(ω)+ X2(ω)=\(\frac{1}{1-ae^{-jω}}+\frac{ae^{jω}}{1-ae^{jω}}=\frac{1-a^2}{1-2acosω+a^2}\).