Let [x] be the greatest integer less than or equals to x. Then, at which of the following point(s) t

Q: Wait, why do we use $[x] = n-1$ for the Left Hand Limit? Isn't it just $n$?

Great question! Remember, the Left Hand Limit approaches $n$ from values slightly less than $n$. If $x$ is $2.99$, $[x]$ is $2$ (which is $3-1$). So, for any $x$ in the interval $(n-1, n)$, $[x]$ is always $n-1$. This is the 'jump' that defines the greatest integer function!

Q: Why does the sine term vanish? Is it always zero?

Yes! Because $[x]$ is always an integer, let's call it $k$. The term becomes $\sin(k\pi)$. Since $k$ is an integer, $\sin(k\pi)$ is always $0$ for any integer $k$. This is a beautiful simplification that removes the complexity of the sine function entirely.

Q: Why is the function continuous at $x=0$ but not at $x=1$ or $x=2$?

It comes down to the equation $-n = n$. This forces $2n = 0$, which only happens when $n=0$. At $x=0$, the left and right branches of the function meet perfectly at the origin. At any other integer, the 'jump' in the greatest integer function creates a gap that the function cannot bridge.

Question

Let $[x]$ be the greatest integer less than or equals to $x$. Then, at which of the following point(s) the function $f(x) = x \cos(\pi(x + [x]))$ is discontinuous ?

Options:

(A) $x = -1$

(B) $x = 0$

(C) $x = 2$

(D) $x = 1$

Sigma Percentile · Accepted Answer

### The Dance of the Staircase: Unraveling Discontinuity

Welcome, fellow explorer of the mathematical universe. Today, we are not just solving a problem; we are dissecting the anatomy of a function.

We are looking at $f(x) = x \cos(\pi(x + [x]))$. At first glance, it looks like a standard trigonometric expression, but that $[x]$—the Greatest Integer Function—is a wolf in sheep's clothing.

It is the architect of chaos, the creator of jumps, and the reason we are here today.

### Phase 1: The Algebraic Deconstruction

Before we dive into the calculus, we must simplify our terrain. The expression $f(x) = x \cos(\pi(x + [x]))$ is intimidating because of the nested argument.

Let us use the distributive property to expand the argument inside the cosine: $\pi(x + [x]) = \pi x + \pi [x]$. Now, our function looks like this:

$$f(x) = x \cos(\pi x + \pi [x])$$

This is where we pull a classic tool from our trigonometric toolkit: the addition formula, $\cos(A+B) = \cos A \cos B - \sin A \sin B$. Let $A = \pi x$ and $B = \pi [x]$.

Our function transforms into:

$$f(x) = x [\cos(\pi x)\cos(\pi [x]) - \sin(\pi x)\sin(\pi [x])]$$

Here is the moment of elegance. We know that $[x]$ is always an integer. Let's call this integer $k$.

What happens to $\sin(\pi k)$? It is always zero! And what about $\cos(\pi k)$? It oscillates between $1$ and $-1$, which we can write succinctly as $(-1)^k$.

Thus, our function collapses into a much cleaner form:

$$f(x) = x \cos(\pi x) (-1)^{[x]}$$

### Phase 2: The Anatomy of the Jump

Now that we have a simplified function, we must ask: where can this function break? The term $x$ is continuous, and the term $\cos(\pi x)$ is continuous.

The only source of discontinuity is the Greatest Integer Function, $[x]$. We know that $[x]$ jumps at every integer.

Therefore, we must test the continuity at an arbitrary integer $n$. To be continuous at $x=n$, the Left Hand Limit (LHL) must equal the Right Hand Limit (RHL), and both must equal the function value $f(n)$.

### Phase 3: The Limit Analysis

First, the Left Hand Limit. As $x 	o n^-$, the value of $x$ is slightly less than $n$. Consequently, the greatest integer $[x]$ is $n-1$.

Substituting this into our simplified function:

$$	ext{LHL} = \lim_{x 	o n^-} x \cos(\pi x) (-1)^{n-1}$$

As $x$ approaches $n$, $\cos(\pi x)$ becomes $\cos(n\pi)$, which is $(-1)^n$. So, our LHL becomes:

$$	ext{LHL} = n \cdot (-1)^n \cdot (-1)^{n-1} = n \cdot (-1)^{2n-1}$$

Since $2n-1$ is always an odd integer, $(-1)^{2n-1} = -1$. Thus, **$	ext{LHL} = -n$**.

Now, the Right Hand Limit. As $x 	o n^+$, the value of $x$ is slightly greater than $n$. Here, $[x] = n$.

Substituting this:

$$	ext{RHL} = \lim_{x 	o n^+} x \cos(\pi x) (-1)^n$$

Again, $\cos(n\pi) = (-1)^n$. Our RHL becomes:

$$	ext{RHL} = n \cdot (-1)^n \cdot (-1)^n = n \cdot (-1)^{2n}$$

Since $2n$ is always an even integer, $(-1)^{2n} = 1$. Thus, **$	ext{RHL} = n$**.

### The Conclusion: The Zero Exception

For the function to be continuous, we require $	ext{LHL} = 	ext{RHL}$. This implies $-n = n$, or $2n = 0$, which means $n = 0$.

At $x=0$, the function is continuous. But for any other integer $n 
eq 0$, the LHL is $-n$ and the RHL is $n$.

They are not equal! The function jumps. Therefore, the function is **discontinuous at all integers except zero**.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $[x]$ be the greatest integer less than or equals to $x$ . Then, at which of the following point(s) the function $f (x) = x cos (π (x + [x]))$ is discontinuous ?

Select Answer:

Visualized Solution

Analyze the Function

Expand the Argument

Apply Cosine Addition Formula

Evaluate Integer Multiples of $π$

Simplified Function

Setup Continuity Check

Left Hand Limit (LHL) Setup

Evaluate LHL

Right Hand Limit (RHL) Setup

Evaluate RHL

Condition for Continuity

Visualizing the Graph

Checking the Options

The Sigma Insight: Continuity at a Point and in an Interval

The Dance of the Staircase

Phase 1

Phase 2

Phase 3

The Conclusion

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