Question

A function \( f : \mathbb{R}_+ \to \mathbb{R} \) (where \( \mathbb{R}_+ \) is the set of all non-negative real numbers) defined by \( f(x) = 4x + 3 \) is:

• A. one-one but not onto
• B. onto but not one-one
• C. both one-one and onto
• D. neither one-one nor onto

Hint:

To check if a function is one-one, test \( f(x_1) = f(x_2) \implies x_1 = x_2 \). For onto, ensure the range matches the codomain.

Answer 1

The Correct Option is A

The given function is \( f(x) = 4x + 3 \), where \( x \in \mathbb{R}_+ \).
Step 1: Check for one-one
A function is one-one if \( f(x_1) = f(x_2) \implies x_1 = x_2 \). For \( f(x) = 4x + 3 \): \[ f(x_1) = 4x_1 + 3, \quad f(x_2) = 4x_2 + 3. \] Equating, \[ 4x_1 + 3 = 4x_2 + 3 \implies 4x_1 = 4x_2 \implies x_1 = x_2. \] Thus, \( f(x) \) is one-one.
Step 2: Check for onto
A function is onto if for every \( y \in \mathbb{R} \), there exists \( x \in \mathbb{R}_+ \) such that \( f(x) = y \). Rearranging \( f(x) = 4x + 3 \), \[ x = \frac{y - 3}{4}. \] For \( x \in \mathbb{R}_+ \), \( y - 3 \geq 0 \), i.e., \( y \geq 3 \). Thus, \( f(x) \) maps \( \mathbb{R}_+ \) to \( [3, \infty) \), and hence is not onto as it does not cover all of \( \mathbb{R} \). Final Answer: (A) one-one but not onto.