The correct answer is:
(D) $y(x)$ changes sign on $(a, b)$.
Explanation:
Given that $y(a) = y(b) = 0$, and $y(x)$ is continuous, this implies that there exists at least one point $c \in (a, b)$ such that $y(c) = 0$. This is because $y(x)$ is continuous on $[a, b]$ and satisfies $y(a) = y(b) = 0$, so it must cross the $x$-axis at least once in the interval $(a, b)$.
Now, we need to determine whether $y(x)$ changes sign on $(a, b)$ or not. For this, we can use the Intermediate Value Theorem. Since $y(x)$ is continuous and $y(c) = 0$ for some $c \in (a, b)$, there must be some point $d \in (a, c)$ and some point $e \in (c, b)$ such that $y(d) < 0$ and $y(e) > 0$, respectively. This is because $y(x)$ must pass through all values between negative and positive as it crosses the $x$-axis.
Hence, option (D) is correct: $y(x)$ changes sign on $(a, b)$.
This answer is AI generated but verified by expert.