Zeros of Derivative of Product of Linear and Sine Functions

To determine where the derivative $f'(x)$ vanishes, we start by differentiating the function: $f(x) = x \sin(\pi x), \quad x > 0$

Using the product rule, the first derivative is: $f'(x) = \sin(\pi x) + \pi x \cos(\pi x)$

Setting $f'(x) = 0$: $\sin(\pi x) + \pi x \cos(\pi x) = 0$

First, consider the case where $\cos(\pi x) = 0$. This occurs at: $\pi x = n\pi + \frac{\pi}{2} \implies x = n + \frac{1}{2}, \quad n \in \mathbb{N}$ Evaluating $f'(x)$ at these points gives: $f'\left(n + \frac{1}{2}\right) = \sin\left(n\pi + \frac{\pi}{2}\right) + 0 = (-1)^n \neq 0$ Hence, $f'(x) \neq 0$ at $x = n + \frac{1}{2}$, and we can safely divide by $\cos(\pi x)$ for any zero of $f'(x)$.

Dividing by $\cos(\pi x)$, the condition $f'(x) = 0$ is equivalent to: $\tan(\pi x) = -\pi x$

Let $u = \pi x$. Since $x > 0$, we have $u > 0$. The equation becomes: $\tan u = -u$

Now, consider any natural number $n \in \mathbb{N}$. The interval $x \in (n, n + 1)$ corresponds to $u \in (n\pi, (n+1)\pi)$. We analyze the solutions in two sub-intervals:

1. For $x \in \left(n, n + \frac{1}{2}\right)$ (i.e., $u \in \left(n\pi, n\pi + \frac{\pi}{2}\right)$): In this interval, $u$ lies in the first quadrant relative to $n\pi$, so: $\tan u > 0$ However, since $u > 0$, we have: $-u < 0$ Since a strictly positive quantity cannot equal a strictly negative quantity, there are no solutions to $\tan u = -u$ in the interval $\left(n, n + \frac{1}{2}\right)$.

2. For $x \in \left(n + \frac{1}{2}, n + 1\right)$ (i.e., $u \in \left(n\pi + \frac{\pi}{2}, (n+1)\pi\right)$): Define the function: $g(u) = \tan u + u$ The derivative of $g(u)$ is: $g'(u) = \sec^2 u + 1$ Since $\sec^2 u \ge 1$, we have $g'(u) \ge 2 > 0$ for all $u \in \left(n\pi + \frac{\pi}{2}, (n+1)\pi\right)$. Thus, $g(u)$ is strictly increasing on this interval.

   Evaluating the limits at the endpoints: $\lim_{u \to \left(n\pi + \frac{\pi}{2}\right)^+} g(u) = -\infty + \left(n\pi + \frac{\pi}{2}\right) = -\infty$ $\lim_{u \to (n+1)\pi^-} g(u) = 0 + (n+1)\pi = (n+1)\pi > 0$

   By the Intermediate Value Theorem and strict monotonicity, $g(u) = 0$ has exactly one unique root in the interval $\left(n\pi + \frac{\pi}{2}, (n+1)\pi\right)$.

Consequently:

• In the interval $\left(n + \frac{1}{2}, n + 1\right)$, $f'(x)$ vanishes at a unique point.
• In the entire interval $(n, n + 1)$, since there are no roots in $\left(n, n + \frac{1}{2}\right]$ and exactly one root in $\left(n + \frac{1}{2}, n + 1\right)$, $f'(x)$ vanishes at a unique point.

Therefore, statements (B) and (C) are correct.