If f : R - R is a function defined by f(x) = [x] cos ((2x-1)/2) pi, where [x] denotes the greatest

Q: Wait, why do we reject the discontinuity of the greatest integer function here?

That is a fantastic observation! Usually, the greatest integer function $[x]$ is indeed discontinuous at every integer $n$. However, continuity is defined by the limit existing and equaling the function value. In this specific product, the greatest integer function is being multiplied by $\sin(x\pi)$. As $x$ approaches any integer $n$, the $\sin(x\pi)$ term approaches $\sin(n\pi)$, which is $0$. This zero acts as a 'neutralizer,' forcing the limit of the entire product to be zero regardless of the jump in the greatest integer function. It is a beautiful example of how one part of a function can 'mask' the behavior of another.

Q: Can we just use the product rule for limits directly?

You absolutely can, but you must be careful. The product rule states that $\lim_{x \to n} [f(x)g(x)] = \lim_{x \to n} f(x) \cdot \lim_{x \to n} g(x)$ only if both individual limits exist. Here, $\lim_{x \to n} [x]$ does not exist because the left-hand limit is $n-1$ and the right-hand limit is $n$. Therefore, you cannot simply multiply the limits. You must evaluate the limit of the product as a whole, which is exactly what we did by checking the left-hand and right-hand limits separately.

Q: Is it always true that $[x] \cdot \sin(x\pi)$ is continuous?

In this specific case, yes, because the sine function vanishes at every integer. If the function were, say, $[x] \cdot \cos(x\pi)$, the result would be different because $\cos(n\pi)$ is $\pm 1$, not $0$. The continuity here is a special property of the specific trigonometric term provided. Always check the behavior of the product at the points of discontinuity of the step function rather than assuming a general rule.

Question

If $f : R 	o R$ is a function defined by $f(x) = [x] \cos \left( \frac{2x-1}{2} ight) \pi$, where $[x]$ denotes the greatest integer function, then $f$ is

Options:

(A) continuous for every real x.

(B) discontinuous only at x = 0

(C) discontinuous only at non-zero integral values of x.

(D) continuous only at x = 0.

Sigma Percentile · Accepted Answer

### The Illusion of Discontinuity

Imagine you are standing on a path, and suddenly, the ground beneath you seems to break. That is the feeling many students get when they encounter the greatest integer function, $[x]$.

It is notorious for its jumps, its breaks, and its refusal to be smooth. When you see $f(x) = [x] \cos \left( \frac{2x-1}{2} ight) \pi$, your intuition might scream, "Discontinuity!"

But in mathematics, as in life, things are not always what they seem. Let us peel back the layers of this function together.

### The Trigonometric Transformation

Before we panic about the greatest integer function, let us look at the trigonometric term: $\cos \left( \frac{2x-1}{2} \pi ight)$. This looks a bit messy, so let us distribute the $\pi$ and the denominator:

$$ \cos \left( x\pi - \frac{\pi}{2} ight) $$

Now, recall the property that $\cos(-	heta) = \cos(	heta)$. We can flip the terms inside the cosine to get $\cos \left( \frac{\pi}{2} - x\pi ight)$.

Suddenly, the fog clears. We are looking at the complementary angle identity: $\cos \left( \frac{\pi}{2} - 	heta ight) = \sin(	heta)$.

With a stroke of algebraic elegance, our intimidating cosine term collapses into $\sin(x\pi)$. Our function is now:

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

### The Great Neutralizer

Now, we return to the greatest integer function. We know that $[x]$ is discontinuous at every integer $n$.

To check for continuity at any integer $n$, we must verify if the Left Hand Limit (L.H.L.), the Right Hand Limit (R.H.L.), and the function value $f(n)$ are all equal.

Let us look at the L.H.L. as $x 	o n^-$:

$$ \lim_{x 	o n^-} [x] \sin(x\pi) $$

As $x$ approaches $n$ from the left, $[x]$ is $n-1$. However, the sine term $\sin(x\pi)$ approaches $\sin(n\pi)$, which is $0$.

Thus, the L.H.L. is $(n-1) \cdot 0 = 0$.

Now, consider the R.H.L. as $x 	o n^+$:

$$ \lim_{x 	o n^+} [x] \sin(x\pi) $$

As $x$ approaches $n$ from the right, $[x]$ is $n$. Again, the sine term approaches $\sin(n\pi) = 0$.

The R.H.L. is $n \cdot 0 = 0$.

Finally, the function value at $x=n$ is:

$$ f(n) = [n] \sin(n\pi) = n \cdot 0 = 0 $$

### The Conclusion

Look at what happened! The jump in the greatest integer function was completely neutralized by the zero of the sine function.

The L.H.L., the R.H.L., and the function value are all $0$. Therefore, the function is **continuous at every integer**.

Since it is also continuous at every non-integer, we have proven that $f(x)$ is **continuous for every real $x$**.

This problem teaches us a vital lesson: never judge a function by its most intimidating part. Sometimes, the solution lies in how the different parts of the function interact.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $f : R \to R$ is a function defined by $f (x) = [x] cos (\frac{2 x - 1}{2}) π$ , where $[x]$ denotes the greatest integer function, then $f$ is

Select Answer:

Visualized Solution

Analyzing the Function $f (x)$

Simplifying the Trigonometric Term

Applying Trigonometric Identities

The Simplified Function

Setting up the Left Hand Limit (L.H.L.)

Evaluating the L.H.L.

Setting up the Right Hand Limit (R.H.L.)

Evaluating the R.H.L.

Exact Function Value at $x = n$

Conclusion on Continuity

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

The Illusion of Discontinuity

The Trigonometric Transformation

The Great Neutralizer

The Conclusion

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