Question

If \( f(x) = [x - 1]\cos\!\left( \frac{2x - 1}{2}\pi \right) \), then \( f \) is :

• A. discontinuous only at x=1
• B. discontinuous at all integral values except x=1
• C. continuous only at x=1
• D. continuous for every real x

Hint:

A function $f(x) = g(x)h(x)$ is continuous at $a$ if $g(x)$ has a jump discontinuity but $h(a) = 0$ and $h(x)$ is continuous.

Answer 1

The Correct Option is D

Step 1: Potential points of discontinuity are integers $n$.
Step 2: At $x=n$, the term $\cos(\frac{2n-1}{2})\pi = \cos(n\pi - \pi/2) = 0$.
Step 3: Since the factor multiplying the step function is 0 at every point where the step function jumps, the limit is always 0.
Step 4: Thus $f(x)$ is continuous everywhere.