The number of solutions of the equation: $x^2 + (x+1)\sin\frac{\pi x}{6} = \frac{3+x}{2}$; $-2 \leq x \leq 0$
Step-by-Step Solution
Key Concept: Factorization combined with graphical analysis of transcendental equations eliminates extraneous solutions outside the domain.
The equation $\left(x^2 + \frac{3+x}{2}\right) + (x+1)\sin\frac{\pi x}{6} = 0$ simplifies to $(x+1)\left[(2x-3) + 2\sin\frac{\pi x}{6}\right] = 0$. This gives $x = -1$ or $2\sin\frac{\pi x}{6} = 3-2x$. Graphing both sides, the intersection point $x = 1, y = \frac{1}{2}$ lies outside the given domain $-2 \leq x \leq 0$. Therefore, $x = -1$ is the only solution.
Correct Answer: 2