Right option is (c) System is Linear, time-invariant and stable
The explanation is: y1 (t) = |v (t)|, y2 (t) = |k v (t)|= |k|y1 (t)
If k is negative, |k| y1 (t) ≠k y1 (t)
Since it is not homogeneous, so non-linear system.
Again, y1 (t) = |v (t)|, y2 (t) = |y (t-t0)| = y1 (t-t0)
∴ System is time invariant.
The response at any time t=t0, depends only on the excitation at that time and not on the excitation at any later time, so causal system.
If x (t) is bounded then y (t) is also bounded, so stable system.