Question

The function $f(x)=[x]\cdot \cos\left(\frac{2x-1}{2}\pi\right)$ where $[\cdot]$ denotes the greatest integer function, is discontinuous at

• A. all irrational numbers x.
• B. no x.
• C. all integers.
• D. all rational numbers.

Hint:

Calculus Tip: When testing the product of a discontinuous function (like $[x]$) and a continuous function (like $\cos$), if the continuous function evaluates to 0 exactly at the points where the other jumps, the product function becomes perfectly continuous!

Answer 1

The Correct Option is B

Concept: Calculus - Continuity and Limits (Greatest Integer Function).

Step 1: Identify potential points of discontinuity. The function is $f(x) = [x] \cdot \cos\left(\frac{2x-1}{2}\pi\right)$. The cosine function, $\cos\left(\frac{2x-1}{2}\pi\right)$, is a trigonometric function and is continuous for all real numbers $x \in \mathbb{R}$. The greatest integer function, $[x]$, is continuous for all non-integer values but has jump discontinuities at every integer $x \in \mathbb{Z}$. Therefore, we only need to check the continuity of $f(x)$ at integer points.

Step 2: Set up the limit check at an arbitrary integer $x=n$. Let $n$ be any integer ($n \in \mathbb{Z}$). To prove continuity at $x=n$, we must show that the Left Hand Limit (LHL), Right Hand Limit (RHL), and the function's value at $n$ are all exactly equal.

Step 3: Evaluate the Left Hand Limit (LHL) as $x \to n^-$. As $x$ approaches $n$ from the left (slightly less than $n$), the greatest integer $[x]$ evaluates to $n-1$. The cosine term evaluates to $\lim_{x\to n^-} \cos\left(\frac{2x-1}{2}\pi\right) = \cos\left(\frac{2n-1}{2}\pi\right)$.
We can simplify the angle: $\frac{2n-1}{2}\pi = n\pi - \frac{\pi}{2}$.
Using trigonometric properties, $\cos(n\pi - \frac{\pi}{2}) = 0$ for any integer $n$. Thus, LHL $= (n-1) \cdot 0 = 0$.

Step 4: Evaluate the Right Hand Limit (RHL) and function value at $x=n$. As $x$ approaches $n$ from the right (slightly greater than $n$), the greatest integer $[x]$ evaluates to $n$. The cosine term again evaluates to $\cos\left(n\pi - \frac{\pi}{2}\right) = 0$. Thus, RHL $= n \cdot 0 = 0$. For the exact function value at $x=n$, $f(n) = [n] \cdot \cos\left(\frac{2n-1}{2}\pi\right) = n \cdot 0 = 0$.

Step 5: Compare limits and conclude. Since LHL = RHL = $f(n) = 0$ at every integer $n$, the function $f(x)$ is continuous at all integer points. Because we established in Step 1 that it is naturally continuous at all non-integer points, the function is continuous for all real numbers $x$. Therefore, there are no points of discontinuity. $$ \therefore \text{The function is discontinuous at no } x. $$