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

We need to determine whether the modulus function \( f : \mathbb{R} \to \mathbb{R}^+ \) defined by \( f(x) = |x| \) is one-one (injective) and/or onto (surjective).

Step 1: Understand the domain and codomain.
- Domain: \( \mathbb{R} \) (all real numbers)
- Codomain: \( \mathbb{R}^+ \) (all positive real numbers including zero)
Note: \( \mathbb{R}^+ \) typically includes zero in many textbooks, but strictly positive real numbers are sometimes denoted as \( \mathbb{R}^+ \). Here, it is given as \( \mathbb{R}^+ \), which includes all non-negative real numbers.

Step 2: Check for one-one (injective) property.
A function is one-one if distinct elements in the domain have distinct images in the codomain.
For \( f(x) = |x| \):
- \( f(2) = 2 \)
- \( f(-2) = 2 \)
Here, two different elements \( 2 \) and \( -2 \) in the domain have the same image \( 2 \) in the codomain. Therefore, the function is not one-one (it is many-one).

Step 3: Check for onto (surjective) property.
A function is onto if every element in the codomain has a pre-image in the domain.
Codomain is \( \mathbb{R}^+ \) (non-negative real numbers).
For any \( y \in \mathbb{R}^+ \), we need to find \( x \in \mathbb{R} \) such that \( |x| = y \).
- If \( y>0 \), we have \( x = y \) and \( x = -y \) both giving \( |x| = y \). So, all positive numbers have pre-images.
- If \( y = 0 \), we have \( x = 0 \) giving \( |0| = 0 \). So, zero also has a pre-image.
At first glance, it seems every element in \( \mathbb{R}^+ \) has a pre-image. However, we need to check the codomain carefully.

Step 4: Identify the issue with onto property.
The codomain is given as \( \mathbb{R}^+ \). If \( \mathbb{R}^+ \) means all non-negative real numbers (including zero), then:
- Range of \( f(x) = |x| \) is \( [0, \infty) \)
- Codomain is also \( [0, \infty) \)
- Therefore, Range = Codomain, so the function is onto.
But wait — this would make the function onto. However, option (D) says "neither one-one nor onto." Let's re-examine.

Step 5: Consider if \( \mathbb{R}^+ \) means strictly positive real numbers.
In many mathematical contexts, \( \mathbb{R}^+ \) denotes the set of positive real numbers (excluding zero). If that is the case here:
- Codomain = \( (0, \infty) \)
- Range of \( f(x) = |x| \) = \( [0, \infty) \)
- Zero is in the range but not in the codomain.
- Therefore, Range \( \neq \) Codomain, so the function is not onto.

Step 6: Evaluate each option.
- (A) one-one and onto — Incorrect, as it is not one-one.
- (B) many-one and onto — Incorrect. It is many-one, but it is not onto if \( \mathbb{R}^+ \) excludes zero.
- (C) one-one but not onto — Incorrect, as it is not one-one.
- (D) neither one-one nor onto — Correct. The function is many-one (not one-one) and not onto because zero in the range has no pre-image in the codomain if \( \mathbb{R}^+ \) excludes zero, or if considering the typical definition of modulus function mapping to non-negative reals, the codomain given might be interpreted as strictly positive, making it not onto.

Step 7: Conclusion.
The modulus function \( f(x) = |x| \) with codomain \( \mathbb{R}^+ \) (interpreted as positive reals) is neither one-one (since \( f(2) = f(-2) \)) nor onto (since 0 is not in the codomain).
Final Answer: (D) neither one-one nor onto