Question: [![enter image description here](https://i.stack.imgur.com/CUnSG.jpg)](https://i.stack.imgur.com/CUnSG.jpg)

According to my analysis, none of the options are correct. But the answer mentioned in my workbook is option B. 
[![enter image description here](https://i.stack.imgur.com/evnkb.png)](https://i.stack.imgur.com/evnkb.png)

In positive half cycle, I'm getting vo = vi.

[![enter image description here](https://i.stack.imgur.com/VTIjL.png)](https://i.stack.imgur.com/VTIjL.png)

And in negative half, vo = vi/2. This doesn't match any options.

Answer: 
1. Yes, the Zener breaks down because of the 10V offset. The sine wave is to allow you to measure the impedance of the Zener model at the test current.
2. From the datasheet you can usually find information that will help. Zx (dynamic impedance), for example, but it has to be specified at or near the operating current. This particular type of Zener from Rohm is rather poorly specified- perhaps the model parameters were extracted from actual measurements, or perhaps they're not all that realistic.
Actually your voltage regulator works. 2,8Vpp input variation causes only 6mVpp variation in output. Not bad for a circuit this simple.

This regulator principle is not good, if you want to use the regulated voltage to supply some load which takes plenty of current and that current is not constant, but varies widely.

This principle of course works also in high current systems, but it wastes energy all the time. The zenerdiode sucks just as much as is needed to drop the voltage to about 6,2V when part of the resistor's current is taken by the possible load and the zenerdiode sucks the rest.