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

To determine the continuity of the function \( f(x) = [x - 1]\cos\!\left( \frac{2x - 1}{2}\pi \right) \), we will analyze both components of the function: the integer part \([x - 1]\) and the trigonometric function \(\cos\!\left( \frac{2x - 1}{2}\pi \right)\).

1. The function \([x - 1]\) represents the greatest integer less than or equal to \(x - 1\). This is known to be discontinuous at integral values of \(x\) because the value jumps at these points.
2. The function \(\cos\!\left( \frac{2x - 1}{2}\pi \right)\) is continuous for all real numbers because cosine is a continuous function over its entire domain.
3. For \(f(x)\) to be continuous, the product of these two functions must also be continuous. A discontinuity would occur if there is a combination of their behaviors that leads to a jump in values.
4. However, at each integer point \(x=n\), \([x-1]\) becomes a constant value (since it jumps to the closest lower integer), and the cosine part remains continuous with specific values depending on \(n\). Thus, the product \([x - 1]\cos\!\left( \frac{2x - 1}{2}\pi \right)\) is continuous across all real numbers as long as the value of the product does not lead to any abrupt jump.
5. In essence, because the cosine function does not interfere with the greatest integer function at these points (the cosine value is bounded between -1 and 1), any discontinuity from \([x - 1]\) is smoothed out.

After evaluating this reasoning, we can conclude that:

• At every real \(x\), the product \([x - 1]\cos\!\left( \frac{2x - 1}{2}\pi \right)\) is continuous, because either the \([x - 1]\) maintains the integer value without causing the function to jump due to bounded cosine values and vice-versa propensities where cosine fluctuations are uniform across \(x\).

Therefore, the function f(x) is continuous for every real \(x\).