Question

Let \( a, b \in \mathbb{R} \) and \( a < b. \) Which of the following statement(s) is/are true?

• A. There exists a continuous function \( f : [a, b] \to (a, b) \) such that \( f \) is one-one
• B. There exists a continuous function \( f : [a, b] \to (a, b) \) such that \( f \) is onto
• C. There exists a continuous function \( f : (a, b) \to [a, b] \) such that \( f \) is one-one
• D. There exists a continuous function \( f : (a, b) \to [a, b] \) such that \( f \) is onto

Hint:

Endpoints matter: A continuous function from a closed interval cannot be onto an open one, but an open-to-closed mapping can be surjective.

Answer 1

The Correct Option is A, C, D

Step 1: Examine option (A).
Define \( f(x) = \frac{a + b}{2} + \frac{b - a}{2}\sin\left(\frac{\pi(x - a)}{b - a}\right). \) This function maps \([a, b]\) into \((a, b)\) and is one-one because \(\sin\) is monotonic in \([0, \pi]\). Thus, (A) is TRUE.

Step 2: Check (B).
No continuous function from a closed interval \([a, b]\) to an open interval \((a, b)\) can be onto, since endpoints \(a\) and \(b\) in the domain have no images equal to the open interval's endpoints. Hence, (B) is FALSE.

Step 3: Check (C).
A continuous one-one function from an open interval \((a, b)\) to a closed interval \([a, b]\) would have to take boundary values, which is impossible. Hence, (C) is FALSE.

Step 4: Check (D).
Define \( f(x) = a + (b - a)x^2 \) with domain \( (0, 1) \). This function maps \((a, b)\) onto \([a, b]\) (for a suitable linear transformation). Hence, (D) is TRUE.

Final Answer: (A) and (D).