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: Understanding the Concept:
This question refers to the Boundedness Theorem (a consequence of the Extreme Value Theorem) from real analysis, which describes a fundamental property of continuous functions defined on closed and bounded intervals.
Step 2: Detailed Explanation:
The Boundedness Theorem states that if a function \(f\) is continuous on a closed and bounded interval \([a, b]\), then \(f\) is bounded on that interval.
A function is "bounded" if it is both bounded above and bounded below.
Bounded above means there exists a real number M such that \(f(x) \le M\) for all \(x\) in \([a, b]\).
Bounded below means there exists a real number m such that \(f(x) \ge m\) for all \(x\) in \([a, b]\).
Let's analyze the given statements:
(A) Constant: This is not necessarily true. For example, \(f(x) = x\) is continuous on \([0, 1]\) but is not constant.
(B) Bounded above: This is true, according to the Boundedness Theorem.
(C) Bounded below: This is also true, according to the Boundedness Theorem.
(D) Unbounded: This is false and directly contradicts the theorem.
Step 3: Final Answer:
The theorem guarantees that the function will be both bounded above and bounded below. Therefore, statements (B) and (C) are correct.