Take the breakdown voltage of the zener diode used in the given circuit as 6V. For the input voltage shown in the figure below, the time variation of the output voltage is: (Graphs drawn are schematic and not to the scale)

In a back-to-back Zener diode clipping circuit, one diode is forward-biased ($$V_F \approx 0\text{ V}$$) while the other undergoes breakdown ($$V_Z = 6\text{ V}$$) during each half-cycle, capping the output magnitude at $$V_Z$$.

For the positive half-cycle ($$V_{\text{in}} > 0$$):

$$\text{Top diode: forward biased } (0\text{ V}),\ \text{Bottom diode: breakdown } (6\text{ V}) \implies V_o = +6\text{ V max}$$

For the negative half-cycle ($$V_{\text{in}} < 0$$):

$$\text{Top diode: breakdown } (-6\text{ V}),\ \text{Bottom diode: forward biased } (0\text{ V}) \implies V_o = -6\text{ V max}$$

Output clipping behavior:

$$|V_{\text{in}}| \le 6\text{ V} \implies V_o = V_{\text{in}}$$

$$|V_{\text{in}}| > 6\text{ V} \implies |V_o| = 6\text{ V}$$

Answer: Option (C): Symmetrical clipping occurs at $$+6\text{ V}$$ and $$-6\text{ V}$$.