Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f(x)=[x^2-x]+|-x+[x]|$, where $x\in \mathbb{R}$ and $[t]$ denotes the greatest integer less than or equal to $t$. Then, $f$ is

Unmasking the Function

Welcome, future engineer. Today, we are not just solving a problem; we are peeling back the layers of a function that seems designed to confuse.

When you see a combination of the Greatest Integer Function $[x]$ and the Modulus function $|x|$, it is natural to feel a moment of hesitation. But remember, in the world of JEE Advanced, intimidation is just a signal that you are about to learn something profound.

Let us break down $f(x)=[x^2-x]+|-x+[x]|$ step by step.

The Great Simplification

The first step in any complex problem is to simplify the landscape. We have a modulus term: $|-x+[x]|$.

This looks messy, but let us look closer. If we factor out a negative sign, we get $|-(x-[x])|$.

Now, recall the fundamental definition of the fractional part of $x$: $\{x\}=x-[x]$. This is the heartbeat of this function.

Substituting this in, our term becomes $|-\{x\}|$. Since the absolute value of a negative number is the same as the absolute value of its positive counterpart, this is simply $|\{x\}|$.

And here is the beauty: the fractional part $\{x\}$ is, by definition, always in the interval $[0,1)$. It is never negative!

Therefore, the modulus is redundant. We can strip it away entirely. Our function is now elegantly simplified to:

This is the key that unlocks the entire problem.

The Cliff at $x=0$

Now, let us investigate the behavior at $x=0$. First, we find the value of the function at the point:

But continuity requires the limit to match the value. We must check the Right Hand Limit (RHL) as $x\to 0^{+}$.

Imagine $x$ is a tiny positive number, like $0.001$. What happens to $x^2-x$?

Since $x$ is very small, $x^2$ is even smaller, making $x^2-x$ a small negative number (e.g., $0.000001-0.001=-0.000999$).

The greatest integer of a number just below zero is $-1$. Meanwhile, the fractional part $\{x\}$ as $x\to 0^{+}$ is simply $0$.

So, our RHL is $-1+0=-1$. Since the RHL ($-1$) does not equal the function value ($0$), the function is discontinuous at $x=0$.

The Balance at $x=1$

Do not let your guard down yet. We must check $x=1$. First, the value:

Now, for the limits. Let us look at the RHL as $x\to 1^{+}$.

If $x$ is slightly larger than $1$ (e.g., $1.1$), then $x^2$ is significantly larger than $x$. Thus, $x^2-x$ is a small positive number.

The greatest integer of a small positive number is $0$. The fractional part $\{x\}$ as $x\to 1^{+}$ is $0$. So, the RHL is $0+0=0$.

Now, the LHL as $x\to 1^{-}$. If $x$ is slightly less than $1$ (e.g., $0.9$), then $x^2$ is smaller than $x$, making $x^2-x$ a small negative number.

The greatest integer of a small negative number is $-1$. However, the fractional part $\{x\}$ as $x\to 1^{-}$ approaches $1$.

When we add these together: $-1+1=0$. The LHL is $0$, the RHL is $0$, and the function value is $0$. They all converge!

The function is continuous at $x=1$.

We have navigated the traps. We saw how the function breaks at $x=0$ because the floor function drops to $-1$, and we saw how it heals at $x=1$ because the fractional part rises to $1$, perfectly compensating for the floor function's drop.

This is the elegance of calculus—it is not just about calculation; it is about understanding the balance of forces. You have successfully analyzed the continuity of this function. Keep this analytical mindset, and no problem will ever be too daunting for you.