C

Step 1: Understand the conditions for slipping of the rod. The rod starts slipping when the angular displacement causes the frictional force to be insufficient to maintain static equilibrium.

Step 2: The moment about the edge must balance the gravitational force for rotation. The weight acts at the center of mass, which is at a distance of \(\frac{L}{2}\) from the edge, and the length of the rod is \(L\). The force of gravity is \(mg = \frac{mg}{4}\) since one-fourth of the rod is on the surface.

Step 3: Torque due to weight about the edge is given by:\
\( \tau_{gravity} = \frac{mg}{4} \cdot \frac{L}{2} \cdot \cos(\theta) \)

Step 4: The frictional force provides a counter torque. The maximum frictional force is \(f_{max} = \mu mg = \mu \cdot \frac{mg}{4}\). Thus, torque due to friction is:\
\( \tau_{friction} = f_{max} \cdot \frac{L}{4} = \mu \cdot \frac{mg}{4} \cdot \frac{L}{4}\)

Step 5: Set the torques equal to each other at the point of slipping:
\( \frac{mg}{4} \cdot \frac{L}{2} \cdot \cos(\theta) = \mu \cdot \frac{mg}{4} \cdot \frac{L}{4} \)

Step 6: Simplifying gives:
\( \frac{L}{2} \cos(\theta) = \mu \cdot \frac{L}{4} \) which yields:
\( 2 \cos(\theta) = \mu \)

Step 7: The relation given is \( x \tan(\theta) = 2\), hence substituting the friction relation yields:
\( x = \frac{2}{\tan(\theta)} \).

Final Step: Solving gives the value of \(x\) which results in \( x = 2\). Therefore, the answer is option C.