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 its linearity and time-invariance properties separately based on the given input-output relationship:

$$y\left(t\right)=max(0,x(t\left)\right)$$

1. Test for Linearity:
A system is linear if and only if it satisfies both the additivity and scaling (homogeneity) properties. Let's check the scaling property. If we scale the input by a constant factor c, the new input is:
$x_1\left(t\right)=c·x\left(t\right)$
The corresponding output for this scaled input is:
$y_1\left(t\right)=max(0,c·x(t\left)\right)$
For the system to be linear, the output must also scale by the same factor:
$c·y\left(t\right)=c·max(0,x(t\left)\right)$
Let us test this with a counterexample by setting c = -1 and x(t) = 1:
The output for the scaled input is:
$y_1\left(t\right)=max(0,-1)=0$
However, the scaled output is:
$-1·y\left(t\right)=-1·max(0,1)=-1$
Since the two results are not equal ($0\ne -1$), the homogeneity property does not hold. Thus, the system is non-linear.

2. Test for Time-Invariance:
A system is time-invariant if a time shift in the input results in the exact same time shift in the output. Let the input be delayed by a time shift t₀:
$x_2\left(t\right)=x(t-t_0)$
The response of the system to this delayed input is:
$y_2\left(t\right)=max(0,x(t-t_0\left)\right)$
Now, delaying the original output y(t) by t₀ yields:
$y(t-t_0)=max(0,x(t-t_0\left)\right)$
Because the output due to the shifted input is identical to the shifted output ($y_2\left(t\right)=y(t-t_0)$), the system is time-invariant.

Conclusion:
Since the system violates the scaling condition of linearity but satisfies the criteria for time shift behavior, the system is classified as non-linear and time-invariant.