Local Extrema and Sequence of Critical Points of a Damped Trigonometric Function

To determine the correct options, let us analyze the local extrema of the function $f(x) = \frac{\sin \pi x}{x^2}, \quad x > 0.$

Step 1: Finding the critical points

Differentiating $f(x)$ with respect to $x$: $f'(x) = \frac{\pi x^2 \cos \pi x - 2x \sin \pi x}{x^4} = \frac{\pi x \cos \pi x - 2 \sin \pi x}{x^3} = \frac{\cos \pi x}{x^3} (\pi x - 2 \tan \pi x).$

Setting $f'(x) = 0$ gives: $\tan \pi x = \frac{\pi x}{2}.$

Let $u = \pi x > 0$. The condition for critical points becomes: $\tan u = \frac{u}{2}.$

Step 2: Classifying local maxima and local minima

Consider the intervals where $\tan u > 0$, namely $u \in \left(k\pi, k\pi + \frac{\pi}{2}\right)$ for integers $k \ge 1$:

1. When $k$ is odd ($k = 2n-1$ for $n \ge 1$):

   • $u \in \left((2n-1)\pi, (2n-1)\pi + \frac{\pi}{2}\right)$, which means $\cos u < 0$.
   • For $u$ slightly less than the root, $\tan u < \frac{u}{2} \implies u - 2\tan u > 0$, so $f'(x) < 0$.
   • For $u$ slightly greater than the root, $\tan u > \frac{u}{2} \implies u - 2\tan u < 0$, so $f'(x) > 0$.
   • Hence, $f'(x)$ changes sign from negative to positive, which corresponds to a local minimum.
   • Let the points of local minimum be denoted by $y_n$. Then: $\pi y_n = (2n-1)\pi + \phi_n, \quad \text{where } \phi_n \in \left(0, \frac{\pi}{2}\right) \text{ and } \tan \phi_n = \frac{(2n-1)\pi + \phi_n}{2}.$ Thus, $y_n \in \left(2n-1, 2n - \frac{1}{2}\right)$.

2. When $k$ is even ($k = 2n$ for $n \ge 1$):

   • $u \in \left(2n\pi, 2n\pi + \frac{\pi}{2}\right)$, which means $\cos u > 0$.
   • For $u$ slightly less than the root, $\tan u < \frac{u}{2} \implies u - 2\tan u > 0$, so $f'(x) > 0$.
   • For $u$ slightly greater than the root, $\tan u > \frac{u}{2} \implies u - 2\tan u < 0$, so $f'(x) < 0$.
   • Hence, $f'(x)$ changes sign from positive to negative, which corresponds to a local maximum.
   • Let the points of local maximum be denoted by $x_n$. Then: $\pi x_n = 2n\pi + \theta_n, \quad \text{where } \theta_n \in \left(0, \frac{\pi}{2}\right) \text{ and } \tan \theta_n = \frac{2n\pi + \theta_n}{2}.$ Thus, $x_n \in \left(2n, 2n + \frac{1}{2}\right)$.

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Step 3: Evaluating the Options

• Option (A): For $n = 1$: $y_1 \in \left(1, \frac{3}{2}\right) \quad \text{and} \quad x_1 \in \left(2, \frac{5}{2}\right) \implies y_1 < x_1.$ Therefore, $x_1 < y_1$ is incorrect.

• Option (B): We have $\tan \theta_n = n\pi + \frac{\theta_n}{2}$. As $n$ increases, $n\pi + \frac{\theta_n}{2}$ strictly increases, which implies $\theta_n$ is a strictly increasing sequence in $\left(0, \frac{\pi}{2}\right)$. Thus, $\theta_{n+1} > \theta_n$. Now, $x_{n+1} - x_n = \frac{(2n+2)\pi + \theta_{n+1} - (2n\pi + \theta_n)}{\pi} = 2 + \frac{\theta_{n+1} - \theta_n}{\pi}.$ Since $\theta_{n+1} - \theta_n > 0$, we have: $x_{n+1} - x_n > 2 \quad \text{for every } n.$ Therefore, Option (B) is correct.

• Option (C): As established, $\pi x_n = 2n\pi + \theta_n$ with $\theta_n \in \left(0, \frac{\pi}{2}\right)$, which gives: $x_n = 2n + \frac{\theta_n}{\pi} \in \left(2n, 2n + \frac{1}{2}\right) \quad \text{for every } n.$ Therefore, Option (C) is correct.

• Option (D): Comparing the equations for $\theta_n$ and $\phi_n$: $\tan \theta_n = \frac{2n\pi + \theta_n}{2} > \frac{(2n-1)\pi + \phi_n}{2} = \tan \phi_n \implies \theta_n > \phi_n.$ Now, $x_n - y_n = \frac{(2n\pi + \theta_n) - ((2n-1)\pi + \phi_n)}{\pi} = 1 + \frac{\theta_n - \phi_n}{\pi}.$ Since $\theta_n > \phi_n$, it follows that: $|x_n - y_n| = x_n - y_n > 1 \quad \text{for every } n.$ Therefore, Option (D) is correct.

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Conclusion

The correct options are (B), (C), and (D).