Question

Let \( \mathbb{R}_+ \) denote the set of all non-negative real numbers. Then the function \( f : \mathbb{R}_+ \to \mathbb{R}_+ \) defined as \( f(x) = x^2 + 1 \) 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:

For one-one and onto functions: \item Use \( f(a) = f(b) \) to verify the one-one property. \item Analyze the range of \( f(x) \) and compare it to the codomain to check if it is onto.

Answer 1

The Correct Option is A

Step 1: Check if \( f(x) \) is one-one.
For \( f(x) \) to be one-one, \( f(a) = f(b) \) must imply \( a = b \). Let: \[ f(x) = x^2 + 1. \] Assume \( f(a) = f(b) \): \[ a^2 + 1 = b^2 + 1. \] Simplify: \[ a^2 = b^2 \quad \Rightarrow \quad a = b \, {(since \( x \in \mathbb{R}_+ \))}. \] Thus, \( f(x) \) is one-one. Step 2: Check if \( f(x) \) is onto.
For \( f(x) \) to be onto, every \( y \in \mathbb{R}_+ \) must have a corresponding \( x \in \mathbb{R}_+ \) such that \( f(x) = y \). Let: \[ f(x) = y \quad \Rightarrow \quad x^2 + 1 = y \quad \Rightarrow \quad x^2 = y - 1. \] For \( x^2 \geq 0 \), \( y - 1 \geq 0 \), or \( y \geq 1 \). Thus, \( f(x) \) is not onto because it cannot produce values in \( [0, 1) \).
Final Answer: \( \boxed{ {(A)}} \)