Question

The modulus function \( f : \mathbb{R} \to \mathbb{R}^+ \) given by \( f(x) = |x| \) is

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

Hint:

For the modulus function \( f(x) = |x| \):
- It is many-one because \( f(a) = f(-a) \) for \( a \neq 0 \)
- It is onto only if the codomain is \( [0, \infty) \). If the codomain is \( (0, \infty) \) (strictly positive), then 0 has no pre-image, making it not onto.

Answer 1

The Correct Option is D

To determine whether the function \( f(x) = |x| \) is one-one or onto, we need to analyze its properties:

1. One-One Function (Injective):
   • A function \( f \) is one-one if, for every \( a, b \in \mathbb{R} \), \( f(a) = f(b) \) implies \( a = b \).
   • Consider \( f(x) = |x| \). Suppose \( f(a) = f(b) \), which means \( |a| = |b| \).
   • This implies \( a = b \) or \( a = -b \). If \( a = -b \), then \( a \neq b \), hence \( f(x) \) is not one-one.
2. Onto Function (Surjective):
   • A function \( f : \mathbb{R} \to \mathbb{R}^+ \) is onto if, for every \( y \in \mathbb{R}^+ \), there is an \( x \in \mathbb{R} \) such that \( f(x) = y \).
   • Given \( f(x) = |x| \), the range of \( f(x) \) is the set of non-negative real numbers \([0, \infty)\).
   • Because \( f(x) \) does not produce negative values, \( f \) does not cover the entire set \(\mathbb{R}^+\), thus it is not onto.

Since the function \( f(x) = |x| \) is neither one-one nor onto, the correct option is neither one-one nor onto.