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

Q: Wait, why does the Greatest Integer Function $[x-1]$ not cause a discontinuity here?

Great question! Usually, the GIF creates a jump discontinuity at every integer. However, continuity is a property of the *entire* function. Here, the GIF is multiplied by $\sin(x\pi)$. At any integer $n$, the GIF takes a specific value, but the $\sin(x\pi)$ term approaches zero. Since the limit of the product is the product of the limits (if they exist), the 'jump' of the GIF is multiplied by zero, effectively 'healing' the discontinuity.

Q: Is $\sin(x\pi)$ always zero at integers?

Yes, absolutely. For any integer $n$, $\sin(n\pi)$ is always zero. This is a fundamental property of the sine function. Because our function is $f(x) = [x-1] \sin(x\pi)$, at any integer $x=n$, the value becomes $[n-1] \cdot 0$, which is always zero. This is the key to why the function remains continuous.

Q: Do I need to check non-integer points for continuity?

You don't need to perform a formal limit check for non-integers. For any non-integer $x$, the function $[x-1]$ is locally constant (it doesn't change value in a small neighborhood around $x$), and $\sin(x\pi)$ is a continuous function. The product of a constant and a continuous function is continuous. Therefore, the only 'suspicious' points are the integers, which we have already verified.

Question

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

Options:

(A) discontinuous only at $x = 1$

(B) discontinuous at all integral values of $x$ except at $x = 1$

(C) continuous only at $x = 1$

(D) continuous for every real $x$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorer of calculus! Today, we are going to dissect a function that, at first glance, might seem like a chaotic mess of jumps and oscillations. We are looking at the function:

$$f(x) = [x - 1] \cos\left(\frac{2x-1}{2} \piight)$$

When you see the Greatest Integer Function (GIF), your 'continuity radar' should indeed start beeping. The GIF is notorious for its step-like jumps at every integer. However, mathematics is often about finding hidden order in apparent complexity.

### Simplifying the Trigonometric Landscape

The angle inside the cosine function, $\frac{2x-1}{2} \pi$, can be simplified. We rewrite it as $(x - \frac{1}{2})\pi$, which expands to $x\pi - \frac{\pi}{2}$.

Now our function looks like:

$$f(x) = [x - 1] \cos\left(x\pi - \frac{\pi}{2}ight)$$

Since cosine is an even function, $\cos(-	heta) = \cos(	heta)$, we can flip the inside to get $\cos(\frac{\pi}{2} - x\pi)$. Using the complementary angle identity $\cos(\frac{\pi}{2} - 	heta) = \sin(	heta)$, our function transforms into the much friendlier form:

$$f(x) = [x - 1] \sin(x\pi)$$

### The Magic of the Zero Product

Now, consider this product of two functions: the step function $[x - 1]$ and the smooth, oscillating sine wave $\sin(x\pi)$. We know the sine wave is continuous everywhere.

The only potential trouble spots are the integers, where the GIF jumps. Let us test what happens at any integer $n$:

$$f(n) = [n - 1] \sin(n\pi)$$

Since $\sin(n\pi) = 0$ for any integer $n$, the entire expression collapses to zero. The sine function acts as a 'neutralizer,' effectively pinning the function to the x-axis at every integer.

### The Rigorous Proof

To be absolutely certain, let us check the limits at an arbitrary integer $n$.

For the Left Hand Limit (LHL) as $x 	o n^-$:

$$\lim_{x 	o n^-} f(x) = \lim_{x 	o n^-} [x-1] \sin(x\pi) = (n-2) \cdot 0 = 0$$

For the Right Hand Limit (RHL) as $x 	o n^+$:

$$\lim_{x 	o n^+} f(x) = \lim_{x 	o n^+} [x-1] \sin(x\pi) = (n-1) \cdot 0 = 0$$

Since the LHL, the RHL, and the function value $f(n)$ all equal zero, the function is perfectly continuous at every integer. We have successfully navigated the traps and found that **$f(x)$ is continuous for every real $x$**.

Keep this intuition: sometimes, a function that looks like it should break is actually held together by the elegant geometry of trigonometry.

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

Select Answer:

Visualized Solution

Analyzing the Function $f (x)$

Simplifying the Angle

Applying Trigonometric Identity

The Sine Transformation

Identifying Critical Points

Evaluating Function at Integers

Setting up Left Hand Limit (LHL)

Evaluating the LHL

Setting up Right Hand Limit (RHL)

Evaluating the RHL

Final Conclusion

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

Analyzing the Setup

Simplifying the Trigonometric Landscape

The Magic of the Zero Product

The Rigorous Proof

Similar Questions

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

The function $f (x) = [x] cos (\frac{2 x - 1}{2}) π$ , $[.]$ denotes the greatest integer function, is discontinuous at

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 ?

The function $f (x) = [x]^{2} - [x^{2}]$ (where $[y]$ is the greatest integer less than or equal to $y$ ), is discontinuous at

Let $f (x) = [x] sin (\frac{π}{[ x + 1 ]})$ , where $[∙]$ denotes the greatest integer function. The domain of $f$ is... and the points of discontinuity of $f$ in the domain are.....

Let $f (x) = lim_{θ \to 0} (\frac{c o s π x - x ^{(2/ θ)} s i n ( x - 1 )}{1 + x ^{(2/ θ)} ( x - 1 )}), x \in R$ . Consider the following two statements : (I) $f (x)$ is discontinuous at $x = 1$ . (II) $f (x)$ is continuous at $x = - 1$ . Then,

Let $[t]$ denote the greatest integer $\leq t$ and $lim_{x \to 0} x [\frac{4}{x}] = A$ . Then the function, $f (x) = [x^{2}] sin (π x)$ is discontinuous, when $x$ is equal to :

Let $f (x) = [x^{2} - x] + ∣ - x + [x] ∣$ , where $x \in R$ and $[t]$ denotes the greatest integer less than or equal to $t$ . Then, $f$ is

Let $[t]$ denotes the greatest integer $\leq t$ and $lim_{x \to 0} x [\frac{4}{x}] = A$ . Then the function, $f (x) = [x^{2}] sin π x$ is discontinuous, when $x$ is equal to

Let $α \in R$ be such that the function $f (x) = {\frac{c o s ^{- 1} ( 1 - { x } ^{2} ) s i n ^{- 1} ( 1 - { x })}{{ x } - { x } ^{3}}, α, x \neq = 0 x = 0$ is continuous at $x = 0$ , where ${x} = x - [x]$ , $[x]$ is the greatest integer less than or equal to $x$ . Then :