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: Determine if \( f(x) \) is one-to-one. A function is one-to-one if \( f(a) = f(b) \) implies \( a = b \). Given \( f(x) = x^2 + 1 \). If \( f(a) = f(b) \), then \( a^2 + 1 = b^2 + 1 \). This simplifies to \( a^2 = b^2 \), which further implies \( a = b \) (since \( x \in \mathbb{R}_+ \)). Therefore, \( f(x) \) is one-to-one.
Step 2: Determine if \( f(x) \) is onto. A function is onto if for every \( y \in \mathbb{R}_+ \), there exists an \( x \in \mathbb{R}_+ \) such that \( f(x) = y \). Setting \( f(x) = y \), we get \( x^2 + 1 = y \), which means \( x^2 = y - 1 \). For \( x^2 \geq 0 \), it must be that \( y - 1 \geq 0 \), so \( y \geq 1 \). Consequently, \( f(x) \) is not onto as it cannot produce values in the interval \( [0, 1) \).
Final Answer: \( \boxed{ {(A)}} \)