Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $S=\{t\in \mathbb{R}:f(x)=|x-\pi |\cdot (e^{|x|}-1)\sin |x|\text{ is not differentiable at }t\}$. Then the set $S$ is equal to :

A Calculus Journey

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dissect a function that, at first glance, looks like a nightmare of sharp corners and jagged edges.

We are looking at the function:

When you see modulus signs, your instinct might be to panic, expecting a graph full of V-shaped kinks. But in JEE Advanced, we don't panic; we investigate. Let's peel back the layers of this function and discover the hidden smoothness beneath.

Identifying the Suspects

Calculus is a detective game. We are looking for points where the function fails to be differentiable.

A function is non-differentiable where it has a 'kink'—a point where the slope changes abruptly. In our function, the modulus terms $|x|$ and $|x-\pi |$ are the prime suspects.

They are defined differently on either side of their roots, $x=0$ and $x=\pi$. These are the only two points where the 'DNA' of our function changes. Everywhere else, it is a product of smooth, well-behaved functions.

So, our mission is clear: we must investigate the behavior of $f(x)$ specifically at $x=0$ and $x=\pi$.

The Origin ($x=0$)

Let's stand at the origin. As $x$ approaches $0$, the term $|x-\pi |$ is perfectly well-behaved; it just approaches $|0-\pi |=\pi$. It is a constant, a silent observer that doesn't affect the differentiability.

The real action happens with $(e^{|x|}-1)$ and $\sin |x|$. Recall your standard limits: for small $x$, $e^{|x|}-1\approx |x|$ and $\sin |x|\approx |x|$.

If we substitute these into our function, we get:

And here is the magic: $|x{|}^2$ is exactly $x^2$. The sharp corner of the first $|x|$ is perfectly neutralized by the second $|x|$.

The function locally behaves like a smooth parabola $y=\pi x^2$. A parabola is the definition of smoothness! Its derivative at $x=0$ is $0$. The suspect is cleared; the function is differentiable at $x=0$.

The $\pi$ Point ($x=\pi$)

Now, let's move to our second suspect, $x=\pi$. As $x$ approaches $\pi$, the term $(e^{|x|}-1)$ approaches $(e^{\pi }-1)$, which is just a non-zero constant. Again, it is harmless.

The troublemakers are $|x-\pi |$ and $\sin |x|$. Since we are near $x=\pi$, $x$ is positive, so $\sin |x|=\sin x$.

Using the trigonometric identity $\sin x=-\sin (x-\pi )$, we can see that as $x\to \pi$, $\sin x$ behaves like $-(x-\pi )$. Now, look at the product: $|x-\pi |\cdot -(x-\pi )$.

Let $h=x-\pi$. The expression becomes $-h|h|$. Just like $x|x|$, the function $h|h|$ is smooth at the origin with a derivative of $0$.

It creates a beautiful, inverted, smooth curve with a horizontal tangent. The sharp corner of $|x-\pi |$ is once again smoothed out by the linear behavior of the sine term. The second suspect is cleared!

The Empty Set

We have interrogated both suspects. At $x=0$, the function behaves like a parabola. At $x=\pi$, it behaves like a smooth cubic-like curve.

Everywhere else, it is a product of smooth functions. There are no points of non-differentiability.

The set $S$ of points where $f(x)$ is non-differentiable is, quite elegantly, the empty set, $\varphi$.

Isn't that beautiful? What looked like a jagged, broken path was actually a perfectly smooth journey all along. Keep this intuition in your toolkit: multiplication can be a powerful tool for smoothing out the sharpest of corners. Happy problem solving!