If the input x(t) and output y(t) of a system are related as y(t) = max (0, x(t)), then the system is

The correct option is: Non-linear and time-invariant.

To determine the nature of the system, we analyze it for both linearity and time-invariance.

1. Linearity Test:
A system is linear if it satisfies both the additivity and homogeneity properties (superposition principle). Let us test the additivity property using counterexamples.
Let an input $x_1\left(t\right)=1$ produce the output:
$y_1\left(t\right)=max(0,1)=1$
Let another input $x_2\left(t\right)=-2$ produce the output:
$y_2\left(t\right)=max(0,-2)=0$
Now, let the combined input be $x_3\left(t\right)=x_1\left(t\right)+x_2\left(t\right)=1+(-2)=-1$. The output for this combined input is:
$y_3\left(t\right)=max(0,-1)=0$
However, the sum of the individual outputs is:
$y_1\left(t\right)+y_2\left(t\right)=1+0=1$
Since $y_3\left(t\right)\ne y_1\left(t\right)+y_2\left(t\right)$, the superposition principle does not hold. Thus, the system is non-linear.

2. Time-Invariance Test:
A system is time-invariant if a time shift in the input signal results in an identical time shift in the output signal.
Let the input signal be delayed by a time shift $t_0$, such that $x_d\left(t\right)=x(t-t_0)$. The response to this delayed input is:
$y_d\left(t\right)=max(0,x_d(t\left)\right)=max(0,x(t-t_0\left)\right)$
Now, if we delay the original output $y\left(t\right)$ by $t_0$, we obtain:
$y(t-t_0)=max(0,x(t-t_0\left)\right)$
Since $y_d\left(t\right)=y(t-t_0)$, the system is time-invariant.

Conclusion:
Combining the two properties, the system is non-linear and time-invariant.