Sets, Relations & Functions Question - None

Question

Let a function $f : (0, \infty) \to (0, \infty)$ be defined by $f(x) = \left|1 - \dfrac{1}{x}\right|$. Then f is:

Accepted Answer

Step 1: Analyze the function by cases

For x ∈ (0, ∞): f(x) = |1 - 1/x|

When 0 < x < 1: (1 - 1/x) < 0, so f(x) = 1/x - 1

When x ≥ 1: (1 - 1/x) ≥ 0, so f(x) = 1 - 1/x

Step 2: Check Injectivity

For x₁, x₂ ∈ (0,1): f(x₁) = 1/x₁ - 1 and f(x₂) = 1/x₂ - 1. If f(x₁) = f(x₂), then 1/x₁ = 1/x₂, so x₁ = x₂ ✓

For x₁, x₂ ≥ 1: f(x₁) = 1 - 1/x₁ and f(x₂) = 1 - 1/x₂. If f(x₁) = f(x₂), then x₁ = x₂ ✓

For x₁ ∈ (0,1) and x₂ ≥ 1: f(x₁) = 1/x₁ - 1 > 0 and f(x₂) = 1 - 1/x₂ ∈ [0,1), so they cannot be equal ✓

Step 3: Check Surjectivity

For x ∈ (0,1): f(x) = 1/x - 1 ∈ (0, ∞) as x ranges over (0,1)

For x ∈ [1, ∞): f(x) = 1 - 1/x ∈ [0,1)

Range = (0, ∞) ∪ [0,1) = [0, ∞), which is NOT equal to codomain (0, ∞)

Since f is not surjective (0 is not in the range), f is neither injective nor surjective when considering the given codomain.

∴ Answer: A (Neither injective nor surjective)

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