The range of $a$ for which the points $(a, 2 + a)$ and $(\frac{3}{2}, a, a^2)$ lie on opposite sides of the line $2x + 3y = 6$ can lie in intervals:
Step-by-Step Solution
Key Concept: Intersection points are found by solving the system of equations simultaneously, then parameter constraints come from geometric conditions on the intersecting curves.
For points $A$ and $B$ on the curves $2x + 3\left(\frac{x}{y}\right)^2 = 6$ and $y = x + 2$, $y = -4x^2$, substitute to get $2x^2 + 3x - 9 = 0$, which factors as $(2x-3)(x+3) = 0$. This gives $A = (-3, -4)$ and $B = \left(\frac{3}{2}, -1\right)$. For parameter $\alpha$: $\frac{3}{2}\alpha = -3$ gives $\alpha = -2$, and $\frac{3}{2}\alpha = \frac{3}{2}$ gives $\alpha = 1$. Therefore $\alpha \in (-\infty, -2) \cup (0, 1)$.
Correct Answer: 1,3