What happens to the inductive reactance and the current in a purely inductive circuit if the frequency is halved?

The inductive reactance is given by $$X_L = 2\pi f L = \omega L$$. If the frequency is halved, the new reactance is $$X_L' = 2\pi \left(\frac{f}{2}\right) L = \frac{X_L}{2}$$. So the inductive reactance is halved.

In a purely inductive circuit, the current is $$I = \frac{V}{X_L}$$. When the reactance is halved, the current becomes $$I' = \frac{V}{X_L/2} = \frac{2V}{X_L} = 2I$$. So the current is doubled.

Therefore, when the frequency is halved, the inductive reactance is halved and the current is doubled.