Animated Solution for Mathematics - Limits, Continuity and Differentiability: The function $f(x)=[x]\cos \left(\frac{2x-1}{2}\right)\pi$, $[.]$ denotes the greatest integer function, is discontinuous at

Welcome, fellow traveler of the JEE landscape. Today, we are going to dismantle a problem that often trips up even the brightest minds. We are looking at the function $f(x)=[x]\cos \left(\frac{2x-1}{2}\right)\pi$.

At first glance, it looks like a nightmare of discontinuities. We see the Greatest Integer Function, $[x]$, which is notorious for its jumps at every integer $n$. Your intuition might scream that it must be discontinuous at all integers, but in mathematics, things are rarely as they seem.

Let us start by dissecting our subject. We have a product of two distinct mathematical entities: the step-like, rigid $[x]$ and the smooth, oscillating cosine wave.

The Greatest Integer Function is defined as the largest integer less than or equal to $x$. It is perfectly flat between integers, but at every integer $n$, it leaps from $n-1$ to $n$. This is our primary suspect for discontinuity.

However, we must look at the second factor: $\cos \left(\frac{2x-1}{2}\pi \right)$. Let us simplify this using the identity $\cos (\theta -\frac{\pi }{2})=\sin (\theta )$. By rewriting the argument as $(x-\frac{1}{2})\pi =x\pi -\frac{\pi }{2}$, we transform our function into the much more elegant form:

Now, imagine you are standing at an integer $n$. We know that $[x]$ is about to jump. But look at the sine term: $\sin (n\pi )$.

For any integer $n$, $\sin (n\pi )$ is exactly $0$. This is the "secret weapon" of our function. To test for continuity at $x=n$, we must check if the limit as $x$ approaches $n$ equals the function value $f(n)$.

First, the function value:

Now, let us approach from the left ($x\to n^{-}$). As $x$ approaches $n$ from the left, $[x]$ is $n-1$. So, the limit is:

Finally, approach from the right ($x\to n^{+}$). As $x$ approaches $n$ from the right, $[x]$ is $n$. The limit becomes:

Look at what has happened! The Left Hand Limit is $0$, the Right Hand Limit is $0$, and the function value is $0$. The jump in the Greatest Integer Function has been completely neutralized by the zero of the sine function.

It is as if the sine function reaches out and "pins" the graph to the x-axis at every integer point, effectively healing the discontinuity.

Since the function is continuous at all non-integer points (where both components are continuous) and we have just proven it is continuous at all integer points, we reach a surprising conclusion: the function is continuous everywhere. The set of points of discontinuity is empty.

This problem teaches us a vital lesson for the JEE Advanced: never judge a function by its parts. Always look at how the components interact. Sometimes, the most "discontinuous" looking functions are the most harmonious when they work together.