Animated Solution for Mathematics - Differentiation: Let $f(x)=\frac{\sin \pi x}{x^2},x>0$. Let $x_1<x_2<x_3<\cdots <x_n<\dots$ be all the points of local maximum of $f$ and $y_1<y_2<y_3<\cdots <y_n<\dots$ be all the points of local minimum of $f$. Then which of the following options is/are correct ?

Imagine you are standing on a vast, flat plain, watching a pendulum swing. This is a magical pendulum whose amplitude is being crushed by the relentless force of $x^2$. We are analyzing the function:

At first glance, it looks like a simple trigonometric ratio. However, as $x$ grows, the $x^2$ in the denominator forces the oscillations to shrink toward the $x$-axis. This is the beauty of calculus—it allows us to find the 'peaks' and 'valleys' of this decaying motion.

To find the local maxima ($x_n$) and minima ($y_n$), we must find where the slope of our function vanishes. We apply the quotient rule to $f(x)=\frac{\sin \pi x}{x^2}$:

After canceling the common factor of $x$, we arrive at the elegant condition for our critical points:

This is a transcendental equation. It does not yield to simple algebra; instead, it invites us to visualize the intersection of the periodic, explosive branches of $y=\tan \pi x$ and the steady, linear growth of $y=\frac{\pi x}{2}$.

Picture the graph. The tangent function has vertical asymptotes at $x=\frac{1}{2},\frac{3}{2},\frac{5}{2},\dots$. The line $y=\frac{\pi x}{2}$ starts at the origin and climbs steadily. Every time this line crosses a branch of the tangent function, we have found a critical point.

For the first branch, the intersection occurs between $x=1$ and $x=1.5$. Because the slope of the function changes from negative to positive here, this point is a local minimum, $y_1$.

As we move to the next branch, the intersection occurs between $x=2$ and $x=2.5$. Here, the slope changes from positive to negative, marking a local maximum, $x_1$.

As we generalize this, we see a pattern emerge. The $n$-th maximum $x_n$ is trapped in the interval $(2n,2n+\frac{1}{2})$.

This occurs because the tangent function is increasing on this interval, and the line $y=\frac{\pi x}{2}$ is forced to intersect it before it hits the vertical asymptote at $x=2n+\frac{1}{2}$. This confirms one of our core options: $x_n\in (2n,2n+\frac{1}{2})$.

Distances and Asymptotes

Now, let's tackle the distance between these points. Let ${ϵ}_1$ be the distance from a minimum to its asymptote, and ${ϵ}_2$ be the distance from a maximum to its asymptote.

As $x$ increases, the line $y=\frac{\pi x}{2}$ gets higher, meaning it intersects the tangent branch closer to the asymptote. Thus, ${ϵ}_2<{ϵ}_1$.

When we calculate the distance $|x_n-y_n|$, we find it equals $1+({ϵ}_1-{ϵ}_2)$. Since ${ϵ}_1>{ϵ}_2$, the distance is strictly greater than 1.

Similarly, the distance between consecutive maxima $x_{n+1}-x_n$ is $2+({ϵ}_2-{ϵ}_3)$, which is strictly greater than 2. We have navigated the transcendental landscape and proved the properties of these points. You have successfully tamed the damped sine wave!