If A = {P(alpha, beta) mid text{the tangent d | vedclass

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(B) Given curve: $y^3 - 3xy + 2 = 0$. Differentiating with respect to $x$: $3y^2 \frac{dy}{dx} - 3y - 3x \frac{dy}{dx} = 0$. $\frac{dy}{dx} (3y^2 - 3x

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(B) Given curve: $y^3 - 3xy + 2 = 0$. Differentiating with respect to $x$: $3y^2 \frac{dy}{dx} - 3y - 3x \frac{dy}{dx} = 0$. $\frac{dy}{dx} (3y^2 - 3x) = 3y \implies \frac{dy}{dx} = \frac{y}{y^2 - x}$. For a horizontal tangent, $\frac{dy}{dx} = 0 \implies y = 0$. Substituting $y = 0$ into the curve equation: $0^3 - 3x(0) + 2 = 0 \implies 2 = 0$, which is impossible. Thus, $n(A) = 0$. For a vertical tangent, $\frac{dy}{dx} = \infty \implies y^2 - x = 0 \implies x = y^2$. Substituting $x = y^2$ into the curve equation: $y^3 - 3(y^2)y + 2 = 0 \implies y^3 - 3y^3 + 2 = 0 \implies -2y^3 = -2 \implies y^3 = 1 \implies y = 1$. If $y = 1$, then $x = 1^2 = 1$. The point is $(1, 1)$. Thus, $n(B) = 1$. Therefore, $n(A) + n(B) = 0 + 1 = 1$.

If $A = \{P(\alpha, \beta) \mid \text{the tangent drawn at } P \text{ to the curve } y^3 - 3xy + 2 = 0 \text{ is a horizontal line}\}$ and $B = \{Q(a, b) \mid \text{the tangent drawn at } Q \text{ to the curve } y^3 - 3xy + 2 = 0 \text{ is a vertical line}\}$, then $n(A) + n(B) = $

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