Step 1: Since $AB \parallel EF$, by Basic Proportionality Theorem (Thales' Theorem):
$$\frac{DA}{AE} = \frac{DB}{BF}$$
Step 2: Substitute the values.
$$\frac{2x}{3x+1} = \frac{x}{2x - \frac{1}{2}}$$
Step 3: Cross-multiply.
$$2x\left(2x - \frac{1}{2}\right) = x(3x+1)$$
$$4x^2 - x = 3x^2 + x \Rightarrow x^2 - 2x = 0 \Rightarrow x(x-2) = 0$$
Step 4: $x = 0$ or $x = 2$. But $x = 0$ makes the sides zero, which is not possible.
Hence $x = 2$ only.

Correct option : (B)