If the function
\[ f(x) = \begin{cases} \frac{(e^x - 1) \sin kx}{4 \tan x}, & x \neq 0 \\ P, & x = 0 \end{cases} \]
is differentiable at \( x = 0 \), then:
If
\[ A = \{ P(\alpha, \beta) \mid \text{the tangent drawn at P to the curve } y^3 - 3xy + 2 = 0 \text{ is a horizontal line} \} \]
and
\[ B = \{ Q(a, b) \mid \text{the tangent drawn at Q to the curve } y^3 - 3xy + 2 = 0 \text{ is a vertical line} \} \]
then \( n(A) + n(B) = \)