If $$(a, a^2)$$ falls inside the angle made by the lines $$y = \dfrac{x}{2}, x > 0$$ and $$y = 3x, x > 0$$, then $$a$$ belongs to

The point is $$P(a,\,a^{2})$$.

The two boundary lines are
  $$y=\tfrac{x}{2} \quad\text{and}\quad y=3x,$$
with the condition $$x\gt 0$$ (first-quadrant rays).

For any point whose abscissa is $$x=a\gt 0$$, the ordinates of the two boundary lines are
  Lower ray: $$y_L=\tfrac{a}{2},$$
  Upper ray: $$y_U=3a.$$

For the point $$P(a,a^{2})$$ to lie strictly inside the angle, its ordinate must satisfy
$$y_L \lt a^{2} \lt y_U.$$

That gives the double inequality
$$\frac{a}{2} \lt a^{2} \lt 3a.$$

Because $$a\gt 0$$, divide the two parts separately by $$a$$ (which keeps the inequality signs unchanged):

Left part: $$\frac{a^{2}}{a} \gt \frac{a}{2a}\; \Longrightarrow\; a \gt \frac{1}{2}.$$

Right part: $$\frac{a^{2}}{a} \lt \frac{3a}{a}\; \Longrightarrow\; a \lt 3.$$

Combining, we obtain
$$\frac{1}{2} \lt a \lt 3.$$

Hence $$a$$ lies in the open interval $$\left(\dfrac{1}{2},\,3\right).$$

Option C which is: $$\left(\dfrac{1}{2}, 3\right)$$