Question

The modulus function f: R \( → \) 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:

To quickly determine if a function is one-one, use the Horizontal Line Test on its graph. The graph of \( y = |x| \) is a 'V' shape with its vertex at the origin. Any horizontal line \( y = c \) (where \( c>0 \)) will intersect the graph at two points. This confirms the function is many-one.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to determine if the function \( f(x) = |x| \) with domain \( \mathbb{R} \) (all real numbers) and codomain \( \mathbb{R}^+ \) (non-negative real numbers, i.e., \( [0, \infty) \)) is one-one (injective) and/or onto (surjective).
- A function is one-one (injective) if every distinct element in the domain maps to a distinct element in the codomain. That is, if \( f(x_1) = f(x_2) \), then \( x_1 = x_2 \).
- A function is onto (surjective) if its range is equal to its codomain. That is, for every element \( y \) in the codomain, there exists at least one element \( x \) in the domain such that \( f(x) = y \).
Step 2: Detailed Explanation:
Checking for one-one property:
Let's take two distinct elements from the domain \( \mathbb{R} \), for example, \( x_1 = -2 \) and \( x_2 = 2 \).
Now, let's find their images under \( f \):
\[ f(x_1) = f(-2) = |-2| = 2 \] \[ f(x_2) = f(2) = |2| = 2 \] Here, we have \( f(x_1) = f(x_2) \) but \( x_1 \neq x_2 \).
Since two different inputs (-2 and 2) have the same output (2), the function is not one-one. It is a many-one function.
Checking for onto property:
The codomain is given as \( \mathbb{R}^+ \), which represents the set of non-negative real numbers \( [0, \infty) \).
The range of the function \( f(x) = |x| \) is the set of all possible output values. Since the absolute value of any real number is always non-negative, the range of \( f(x) \) is also \( [0, \infty) \).
Since Range = Codomain (\( [0, \infty) = [0, \infty) \)), the function is onto.
Step 3: Final Answer:
The function is many-one and onto.