Question

Let \([t]\) represent the greatest integer not more than \(t\). Then the number of discontinuous points of \(f(x) = \left[\frac{1}{x}\right]\) in \((0, \infty)\) is

• A. \(0\)
• B. \(1\)
• C. \(2\)
• D. \(\infty\)

Hint:

The greatest integer function \([x]\) is discontinuous at integer values of \(x\). For \(\left[\frac{1}{x}\right]\), discontinuities occur at \(x = \frac{1}{n}\), making them infinite in number on \((0, \infty)\).

Answer 1

The Correct Option is B

We are given \(f(x) = \left[\frac{1}{x}\right]\), the greatest integer function. The function \(\left[\frac{1}{x}\right]\) is discontinuous at values of \(x\) where \(\frac{1}{x}\) is an integer, say \(n\). That occurs at \(x = \frac{1}{n}\), for all \(n \in \mathbb{N}\). So there are infinitely many discontinuities in \((0, \infty)\), but we are interested in points where the function is **actually discontinuous**. However, due to the greatest integer function behavior, these discontinuities happen at each such value. Upon re-checking the actual image, the number of discontinuities in \((0, \infty)\) is **infinite**, but the marked answer is (2) — which is **incorrect**. The **correct answer** should be: % Correct Answer % Correct Answer Correct Answer:} (4) \(\infty\)