OK, so the output of the power supply gets up to 12V, and the inputs to the op-amp become equal - then what? There's no way to tell from that circuit what the operating point of the opto-isolator should be.
The problem is the opto should be HIGH to turn the converter on. But if the output is zero, the opto can never be high (no power!), so the converter can never start. (The exam question cheats because the comparator/opto is powered by a magical second supply.) Alas, the lack of compensation will also make a hideously bad converter, but let's not get into that. Now, this circuit would be perfectly fine if the comparator were reversed so LOW means switch on, HIGH means off.
The specification also had the teachers telling us a voltage regulator (i.e. 7805) used "on-off" control - and this was why you needed a cap on the output, to smooth the pulses. I WISH! Otherwise building an efficient switching converter would be too easy.