I just want to know that my logic is sound for the above question. So, since the diode is connected to a resistor in parallel, for the voltage values where V is less than the breakdown voltage, the resistor is the only contributor to the current I and thus the graph for the first part follows Ohm's law.
Once the V reached the breakdown voltage, however, the PN diode also starts contributing to the current, but since the resistor is still in play, there isn't the steep rise in current, characteristic of the diode, but it is enough to indicate a non-ohmic presence. Thus, the answer would be A.
Is that correct?