Op amp with potentiometer
Potentiometer-Resistors Equivalent

To better explain how to proceed with an answer, one must understand how a potentiometer equivalent can be given by two resistors. In the circuit below, a potentiometer of total resistance R is redrawn as two resistors R1 and R2.







simulate this circuit &ndash; Schematic created using CircuitLab

This equivalent circuit is valid as long as \$R = R1+R2\$. Let's assume it is a linear potentiometer. This means that, when the wiper (node B):


is turned "all the way to the left" (the wiper is at A), then \$R1 = 0\$ and \$R2 = R\$;
is turned "all the way to the right" (the wiper is at C), then \$R1 = R\$ and \$R2 = 0\$;
is anywhere in between A and C, R1 and R2 increase and decrease linearly with wiper position. When the wiper is at the exact half-point of the potentiometer, then \$R1=R2=R/2\$.


When you connect the wiper to one of the extremes, then it is the same as shorting out one of the resistors in the equivalent. If the wiper is connected to terminal C:







simulate this circuit

In this equivalent, \$R2\$ can be ignored because it is shorted out and the equivalent resistance between nodes B and C is zero. Therefore, the potentiometer equivalent is just a resistor \$R1\$, that can assume any value in the range 0 (the wiper is at A) to \$R\$ (the wiper is at C).

How to Proceed with the Exercise in Question

As can be seen, the wiper is connected to one of the extremities of the potentiometer. Perform the analysis by replacing the potentiometer with a resistor of unknown value \$xR\$. Since, as explained above, \$xR \in [0,R]\$, then it follows that \$x \in [0,1]\$.

Not delving deeper into the circuit itself, as this seems to be homework.