[amp_mcq option1=”$${1 \over 3} < \left| z \right| < 3$$" option2="$${2 \over 3} < \left| z \right| < 3$$" option3="$${3 \over 2} < \left| z \right| < 3$$" option4="$${1 \over 3} < \left| z \right| < {2 \over 3}$$" correct="option3"]
The correct answer is $\boxed{{1 \over 3} < \left| z \right| < {2 \over 3}}$.
The region of convergence of a linear combination of two sequences is the intersection of the regions of convergence of the two sequences. In this case, the region of convergence of $x_1[n] + x_2[n]$ is $${1 \over 3} < \left| z \right| < {2 \over 3},$$ so the region of convergence of $x_1[n] – x_2[n]$ must be a subset of this region. Since the region of convergence of a sequence is always a disk centered at the origin, the only possible region of convergence for $x_1[n] – x_2[n]$ is $${1 \over 3} < \left| z \right| < {2 \over 3}.$$
Option A is incorrect because it includes the point $z=3$, which is not in the region of convergence of $x_1[n] – x_2[n]$. Option B is incorrect because it includes the point $z=2$, which is not in the region of convergence of $x_1[n] – x_2[n]$. Option C is incorrect because it includes the point $z=3/2$, which is not in the region of convergence of $x_1[n] – x_2[n]$.