Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $[x]$ denote the greatest integer less than or equal to $x$. If $f(x)=[x\sin \pi x]$, then $f(x)$ is

Deconstructing the Greatest Integer Function

Welcome, future engineer. Today, we are going to peel back the layers of a function that often intimidates students: $f(x)=[x\sin \pi x]$.

When you see the greatest integer brackets, your instinct might be to panic or reach for complex derivatives. But I want you to take a deep breath. In JEE Advanced, the greatest integer function is rarely about calculus; it is almost always about understanding the range of the inner expression.

The Inner Beast

We define our inner function as $g(x)=x\sin \pi x$. This is the heart of our problem.

The behavior of $f(x)$ is entirely dictated by the values $g(x)$ takes. If we can show that $g(x)$ stays trapped within a specific interval, the greatest integer function will collapse into a simple constant.

Imagine $g(x)$ as a particle moving on a track. We need to know where it lives.

The Sign Analysis

Let us look at the interval $x\in (-1,1)$. We must be careful here, as many students assume that because sine is negative in the negative domain, the whole function must be negative.

For $x\in (0,1)$, $x$ is positive and $\sin \pi x$ is positive (since $\pi x\in (0,\pi )$). Thus, $g(x)>0$.

Now, consider $x\in (-1,0)$. Here, $x$ is negative. The angle $\pi x$ is in $(-\pi ,0)$, which means $\sin \pi x$ is also negative.

A negative multiplied by a negative is a positive! So, $g(x)>0$ here as well. We have discovered that for all $x\in (-1,1)$ except at $x=0$, our function $g(x)$ is strictly positive. At $x=0$, $g(0)=0$. Thus, $g(x)\ge 0$ for all $x\in (-1,1)$.

The Trap of the Interval

Now, how large does $g(x)$ get? We know $|x|<1$ and $|\sin \pi x|\le 1$.

Therefore, the product $|x\sin \pi x|$ must be strictly less than $1$. The maximum value occurs near $x=\pm 0.5$, where:

Since $0.5<1$, the function never reaches the integer $1$. We have successfully bounded our function: $0\le g(x)<1$ for all $x\in (-1,1)$.

This is the 'Aha!' moment. If the input to the greatest integer function is always between $0$ and $1$, then the output must be $0$.

The Flattening

Because $0\le g(x)<1$, we can conclude that for the entire interval $x\in (-1,1)$, $f(x)=[g(x)]=0$.

The complex trigonometric expression has vanished, replaced by the elegant simplicity of a constant function. A constant function is the most well-behaved creature in calculus.

It is continuous everywhere, and its derivative is zero everywhere. This immediately confirms that $f(x)$ is continuous at $x=0$, continuous in $(-1,0)$, and differentiable in $(-1,1)$.

The Cliff at $x=1$

Finally, we must check the boundary. What happens at $x=1$?

We know $f(1)=[1\cdot \sin \pi ]=[0]=0$. But what happens if we step just a tiny bit to the right, to $x=1+h$?

The term $\sin (\pi +\pi h)$ becomes $-\sin (\pi h)$, which is negative. Thus, $g(1+h)$ becomes slightly negative. The greatest integer of a slightly negative number is $-1$.

We have a jump! The limit from the left is $0$, but the limit from the right is $-1$. The function is discontinuous at $x=1$, and therefore, it cannot be differentiable there.