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

Clearly , point $\left(a,a^2\right)$ lies on the opposite sides of 
the two lines .

$\begin{array}{l}\therefore \left(\frac{a}{2}-a^2\right)\left(3a-a^2\right)<0\\ \Rightarrow a^2(1-2a)(3-a)<0\\ \Rightarrow (2a-1)(a-3)<0\Rightarrow \frac{1}{2}<a<3\Rightarrow a\in (1/2,3)\end{array}$