Understanding the Concept:
A continuous-time system is classified as memoryless (or static) if its output $y(t)$ at any given time instance $t = t_0$ depends *only* on the value of the input signal $x(t)$ at that exact same instantaneous time $t = t_0$. If the output equation requires values of the input from the past (e.g., $t-1$) or from the future (e.g., $t+2$), the system possesses memory and is classified as a dynamic system.
Step 1: Testing Option (A)
Given system: $y(t) = x(2t) + x(t)$.
Let us evaluate the system behavior at a specific time instance, such as $t = 2$:
\[
y(2) = x(2 \cdot 2) + x(2) = x(4) + x(2)
\]
To calculate the current output at $t = 2$, the system needs to know the value of the input signal at a future time instance, $t = 4$. Because it depends on future inputs, this system requires memory.
Step 2: Testing Option (B)
Given system: $y(t) = x(t) + 2x(t)$.
This expression simplifies directly to:
\[
y(t) = 3x(t)
\]
Let us check this system at any time instance $t = t_0$:
\[
y(t_0) = 3x(t_0)
\]
The output at any moment $t_0$ depends exclusively on the input signal at that exact same instant $t_0$. It does not require any past values or future values of the input. Therefore, this system is completely memoryless.
Step 3: Testing Option (C)
Given system: $y(t) = -x(t) + x(1-t)$.
Let us evaluate this system at the time instance $t = 0$:
\[
y(0) = -x(0) + x(1 - 0) = -x(0) + x(1)
\]
The output at $t = 0$ requires the value of the input at a future time $t = 1$. This means the system is not memoryless.
Step 4: Testing Option (D)
Given system: $y(t) = x(t) + 2x(t+2)$.
Let us evaluate this system at the time instance $t = 0$:
\[
y(0) = x(0) + 2x(0+2) = x(0) + 2x(2)
\]
The output at $t = 0$ requires the future input value at $t = 2$. This system requires memory.