Question

Every continuous real valued function on [a, b] is
(A). Constant.
(B). Bounded above.
(C). Bounded below.
(D). Unbounded.
Choose the correct answer from the options given below:

• A. (A) only.
• B. (B) and (C) only.
• C. (D) only.
• D. (A), (B) and (C) only.

Hint:

The keywords are "continuous" and "closed interval [a, b]". When you see these together, think of a piece of string held between two points. You can't draw it without lifting your pen (continuous), and it has defined endpoints. Such a string will always have a highest point and a lowest point; it can't go to infinity.

Answer 1

The Correct Option is B

Step 1: Core Idea:
This question concerns the Boundedness Theorem, derived from the Extreme Value Theorem in real analysis. It defines a crucial characteristic of continuous functions on closed and bounded intervals.
Step 2: Explanation:
The Boundedness Theorem asserts that if a function \(f\) is continuous on a closed, bounded interval \([a, b]\), then \(f\) is bounded on that interval.
A function is "bounded" if it has both an upper and a lower bound.
Bounded above means a real number M exists such that \(f(x) \le M\) for all \(x\) in \([a, b]\).
Bounded below means a real number m exists such that \(f(x) \ge m\) for all \(x\) in \([a, b]\).
Analyzing the statements:
(A) Constant: Incorrect. For instance, \(f(x) = x\) is continuous on \([0, 1]\) but isn't constant.
(B) Bounded above: Correct, based on the Boundedness Theorem.
(C) Bounded below: Correct, according to the Boundedness Theorem.
(D) Unbounded: Incorrect and contradicts the theorem.
Step 3: Conclusion:
The theorem guarantees the function is bounded both above and below. Therefore, statements (B) and (C) are correct.