Step-by-step Derivation

1. Find the derivative of the function. The given function is $f(x) = \frac{\sin(\pi x)}{x^2}$ for $x > 0$. Using the quotient rule for differentiation, $f'(x) = \frac{d}{dx} \left( \frac{\sin(\pi x)}{x^2} \right)$, we get: $f'(x) = \frac{x^2 \cdot \frac{d}{dx}(\sin(\pi x)) - \sin(\pi x) \cdot \frac{d}{dx}(x^2)}{(x^2)^2}$ $f'(x) = \frac{x^2(\pi \cos(\pi x)) - \sin(\pi x)(2x)}{x^4}$ For $x > 0$, we can simplify by factoring out $x$ from the numerator: $f'(x) = \frac{x(\pi x \cos(\pi x) - 2 \sin(\pi x))}{x^4} = \frac{\pi x \cos(\pi x) - 2 \sin(\pi x)}{x^3}$

2. Find the critical points. Critical points occur where $f'(x) = 0$. Setting the numerator to zero: $\pi x \cos(\pi x) - 2 \sin(\pi x) = 0$ Assuming $\cos(\pi x) \neq 0$, we can divide by it: $\pi x = 2 \frac{\sin(\pi x)}{\cos(\pi x)} \implies \tan(\pi x) = \frac{\pi x}{2}$ Let $z = \pi x$. The equation becomes $\tan(z) = \frac{z}{2}$.

3. Locate the roots of $\tan(z) = z/2$. We analyze the solutions by considering the graphs of $y = \tan(z)$ and $y = z/2$ for $z>0$.

   • For $z \in (k\pi - \pi/2, k\pi)$ for $k \ge 1$, $\tan(z) < 0$ while $z/2 > 0$. So, there are no solutions in these intervals.
   • For $z \in (k\pi, k\pi + \pi/2)$ for $k \ge 1$, $\tan(z)$ is positive and increases from $0$ to $+\infty$. The line $y = z/2$ is also positive. At $z=k\pi$, $\tan(k\pi) = 0 < k\pi/2 = z/2$. As $z \to (k\pi + \pi/2)^-$, $\tan(z) \to +\infty$, which is greater than the finite value of $z/2$. By the Intermediate Value Theorem, there must be a root in this interval. Since the derivative of $\tan(z) - z/2$ is $\sec^2(z) - 1/2 = \tan^2(z) + 1/2 > 0$, the function $\tan(z) - z/2$ is strictly increasing, so the root is unique in each interval. Let the sequence of positive roots be $z_1 < z_2 < z_3 < \dots$, where $z_k \in (k\pi, k\pi + \pi/2)$. The critical points of $f(x)$ are $c_k = z_k/\pi$, which satisfy $c_k \in (k, k + 1/2)$.

4. Classify the critical points. The sign of $f'(x)$ is determined by the sign of $\pi x \cos(\pi x) - 2 \sin(\pi x) = \cos(\pi x)(\pi x - 2\tan(\pi x))$. Let $g(x) = \pi x - 2\tan(\pi x)$. The critical points $c_k$ are the roots of $g(x)=0$. The derivative $g'(x) = \pi - 2\pi \sec^2(\pi x) = \pi(1-2\sec^2(\pi x)) < 0$. So $g(x)$ is a decreasing function, changing sign from $+$ to $-$ at each critical point $c_k$.

   • Local Maxima: A local maximum occurs when $f'(x)$ changes from $+$ to $-$. This requires $\cos(\pi x) > 0$. This happens when $\pi x$ is in the first or fourth quadrant, i.e., $x \in (2n-1/2, 2n+1/2)$ for some integer $n$. The critical points $c_k$ are in $(k, k+1/2)$. For $\cos(\pi c_k) > 0$, we need $k$ to be even. Let $k = 2n$ for $n \ge 1$. Then $c_{2n} \in (2n, 2n+1/2)$, where $\cos(\pi x) > 0$. At $c_{2n}$, the sign of $f'(x)$ changes as $g(x)$ does, from $+$ to $-$. So, $x_n = c_{2n}$ are the points of local maximum.

   • Local Minima: A local minimum occurs when $f'(x)$ changes from $-$ to $+$. This requires $\cos(\pi x) < 0$. This happens when $x \in (2n+1/2, 2n+3/2)$. For $\cos(\pi c_k) < 0$, we need $k$ to be odd. Let $k = 2n-1$ for $n \ge 1$. Then $c_{2n-1} \in (2n-1, 2n-1+1/2)$, where $\cos(\pi x) < 0$. At $c_{2n-1}$, the sign of $f'(x)$ changes opposite to $g(x)$, i.e., from $-$ to $+$. So, $y_n = c_{2n-1}$ are the points of local minimum.

5. Evaluate the options.

   • C: $x_1 < y_1$ $x_1 = c_2$ and $y_1 = c_1$. Since the sequence of critical points is increasing, $c_1 < c_2$, which implies $y_1 < x_1$. Therefore, option C is incorrect.

   • D: $x_n \in (2n, 2n + 1/2)$ for every n We found that the points of local maximum are $x_n = c_{2n}$. We also established that $c_k \in (k, k+1/2)$. For $k=2n$, we have $c_{2n} \in (2n, 2n+1/2)$, so $x_n \in (2n, 2n+1/2)$. Therefore, option D is correct.

   • A: $|x_n - y_n| > 1$ for every n We have $x_n = c_{2n}$ and $y_n = c_{2n-1}$. Let $z_{2n} = \pi x_n$ and $z_{2n-1} = \pi y_n$. These are roots of $\tan(z) = z/2$. Let $h(z) = \tan(z) - z/2$. Then $h(z_{2n})=0$ and $h(z_{2n-1})=0$. We know $h(z)$ is strictly increasing on $(k\pi, k\pi+\pi/2)$. Both $z_{2n}$ and $z_{2n-1}+\pi$ are in the interval $(2n\pi, 2n\pi+\pi/2)$. $h(z_{2n-1}+\pi) = \tan(z_{2n-1}+\pi) - \frac{z_{2n-1}+\pi}{2} = \tan(z_{2n-1}) - \frac{z_{2n-1}}{2} - \frac{\pi}{2}$. Since $h(z_{2n-1}) = 0$, we have $\tan(z_{2n-1}) = z_{2n-1}/2$. So, $h(z_{2n-1}+\pi) = -\pi/2 < 0$. Since $h(z_{2n-1}+\pi) < h(z_{2n})$ and $h$ is increasing, we have $z_{2n-1}+\pi < z_{2n}$. $\pi y_n + \pi < \pi x_n \implies y_n + 1 < x_n \implies x_n - y_n > 1$. Thus, $|x_n - y_n| > 1$. Option A is correct.

   • B: $x_{n+1} - x_n > 2$ for every n We have $x_{n+1} = c_{2n+2}$ and $x_n = c_{2n}$. Let $z_{2n+2}=\pi x_{n+1}$ and $z_{2n}=\pi x_n$. Both $z_{2n+2}$ and $z_{2n}+2\pi$ are in the interval $((2n+2)\pi, (2n+2)\pi+\pi/2)$. Consider $h(z_{2n}+2\pi) = \tan(z_{2n}+2\pi) - \frac{z_{2n}+2\pi}{2} = \tan(z_{2n}) - \frac{z_{2n}}{2} - \pi = 0 - \pi = -\pi$. We have $h(z_{2n+2})=0$. Since $h(z_{2n}+2\pi) < h(z_{2n+2})$ and $h$ is increasing, we have $z_{2n}+2\pi < z_{2n+2}$. $\pi x_n + 2\pi < \pi x_{n+1} \implies x_n + 2 < x_{n+1} \implies x_{n+1} - x_n > 2$. Option B is correct.

Conclusion

Based on the analysis, options A, B, and D are correct, while option C is incorrect.