Step 1: Understanding the Concept:
- A function is Injective (one-to-one) if every distinct input \(x\) has a distinct output \(f(x)\). Formally, \(f(x_1) = f(x_2) \implies x_1 = x_2\).
- A function is Surjective (onto) if the Range of the function equals its Codomain. In this case, the Codomain is \(\mathbb{R}\). For surjectivity, every real number \(y\) must have at least one \(x\) such that \(f(x) = y\).
Step 2: Key Formula or Approach:
Analyze the mapping and range of the absolute value function \(f(x) = |x|\).
Step 3: Detailed Explanation:
1. Testing Injectivity:
Consider two distinct elements in the domain \(\mathbb{R}\): \(x_1 = 1\) and \(x_2 = -1\).
\(f(1) = |1| = 1\)
\(f(-1) = |-1| = 1\)
Since \(f(1) = f(-1)\) but \(1 \neq -1\), the function maps multiple inputs to the same output. Therefore, it is not injective.
2. Testing Surjectivity:
The codomain is \(\mathbb{R}\) (all real numbers from \(-\infty\) to \(+\infty\)).
By definition, the absolute value of any real number is non-negative: \(|x| \geq 0\).
Therefore, the Range of \(f(x)\) is \([0, \infty)\).
Negative real numbers in the codomain (e.g., \(-5\)) have no pre-image in the domain because \(|x| = -5\) has no real solution.
Since Range \([0, \infty) \neq\) Codomain \(\mathbb{R}\), the function is not surjective.
Conclusion: The function is neither injective nor surjective.
Step 4: Final Answer:
The absolute value function fails both the horizontal line test (not injective) and the range-matching test (not surjective). Correct option is (D).