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: 
Any voltage above 6.2V will cause the Zener to break down.
Peak current will simply be (11.4 - 6.2)/100 = 52 mA
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.