Question

Let $f(x) = |x - 2|$, where $x$ is a real number. Which one of the following is true?

• A. \( f \text{ is periodic} \)
• B. \( f(x+y) = f(x) + f(y) \)
• C. \( f \text{ is an odd function} \)
• D. \( f \text{ is not a 1-1 function} \)
• E. \( f \text{ is an even function} \)

Hint:

Any function with an absolute value or an even power (like $x^2$) will generally fail to be one-to-one because the negative and positive counterparts often map to the same result.

Answer 1

The Correct Option is D

Concept: A function is one-to-one (1-1) if every unique input yields a unique output. Graphically, this is tested using the Horizontal Line Test. An absolute value function $f(x) = |x - c|$ forms a V-shape, meaning horizontal lines can intersect the graph at two different points.

Step 1: Test for 1-1 property.
Pick a value for the output, say $f(x) = 1$. \[ |x - 2| = 1 \] This leads to two possible values for $x$: 1. $x - 2 = 1 \Rightarrow x = 3$ 2. $x - 2 = -1 \Rightarrow x = 1$ Since $f(3) = 1$ and $f(1) = 1$, different inputs give the same output. Therefore, $f$ is not a 1-1 function.

Step 2: Evaluate other options.
- (A) Periodic: Absolute value functions are not periodic; they don't repeat cycles.
- (C) & (E) Odd/Even: $f(-x) = |-x - 2| = |x + 2|$. This is neither equal to $f(x)$ nor $-f(x)$.
- (B) Additive: $|(1+1) - 2| = 0$, but $|1-2| + |1-2| = 1 + 1 = 2$. Not additive.