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

Given: $a, b \in \mathbb{R}$ with $a < b$

Step 1: Analyzing each option:

(A) There exists a continuous function $f: [a,b] \to (a,b)$ such that $f$ is one-one

We can construct such a function by scaling and shifting. Consider:

$$f(x) = a + \frac{b-a}{2} + \frac{b-a}{\pi}\arctan\left(x - \frac{a+b}{2}\right)$$

This function is continuous on $[a,b]$, strictly increasing (hence one-one), and maps into $(a,b)$ since $\arctan$ has range $(-\frac{\pi}{2}, \frac{\pi}{2})$.

TRUE

(B) There exists a continuous function $f: [a,b] \to (a,b)$ such that $f$ is onto

By the Extreme Value Theorem, any continuous function $f: [a,b] \to \mathbb{R}$ attains its minimum and maximum values on the closed interval $[a,b]$.

If $f$ were onto $(a,b)$, then $\inf f([a,b]) = a$ and $\sup f([a,b]) = b$. However, by the EVT, $f$ must actually attain these extreme values at some points in $[a,b]$, meaning $a, b \in f([a,b])$. This contradicts the requirement that the range is the open interval $(a,b)$.

FALSE

(C) There exists a continuous function $f: (a,b) \to [a,b]$ such that $f$ is one-one

Consider a function that monotonically increases from values near $a$ to values near $b$:

$$f(x) = a + (b-a)\left(\frac{1}{2} + \frac{1}{\pi}\arctan\left(\frac{x - \frac{a+b}{2}}{(b-a)/4}\right)\right)$$

This function is continuous on $(a,b)$, strictly increasing (hence one-one), and its range is contained in $[a,b]$ since the expression approaches $a$ as $x \to a^+$ and approaches $b$ as $x \to b^-$.

TRUE

(D) There exists a continuous function $f: (a,b) \to [a,b]$ such that $f$ is onto

We construct a continuous onto function using a piecewise approach. Define:

$$f(x) = \begin{cases} a + 2(x-a) & \text{if } a < x \leq \frac{a+b}{2} \ b - 2(b-x) & \text{if } \frac{a+b}{2} < x < b \end{cases}$$

with appropriate scaling factors. This maps $(a,b)$ onto $[a,b]$ continuously by covering the entire target interval through the two pieces meeting in the middle.

TRUE

Answer: (A), (C), and (D) are true