Sets, Relations & Functions
Injective and Surjective Functions
Grade 11
Question:
<p>Let \(A = \{x \in R : x\text{ is not a positive integer}\}\). Define a function \(f : A \to R\) as \(f(x) = \dfrac{2x}{x-1}\), then <i>f</i> is:</p>
<p>not injective</p>
<p>neither injective nor surjective</p>
<p>surjective but not injective</p>
<p>injective but not surjective</p>
Step-by-Step Solution
Key Concept: Analyze injectivity and surjectivity by examining the function's behavior on its domain A (all reals except positive integers) and determining if every real value in R is achieved.
<p><strong>Step 1: Understand the domain</strong></p><p>A = ℝ \ {1, 2, 3, 4, ...} (all reals except positive integers)</p><p><strong>Step 2: Check Injectivity</strong></p><p>Let f(x₁) = f(x₂):</p><p>2x₁/(x₁-1) = 2x₂/(x₂-1)</p><p>2x₁(x₂-1) = 2x₂(x₁-1)</p><p>x₁x₂ - x₁ = x₂x₁ - x₂</p><p>x₁ = x₂</p><p>∴ f is <strong>injective</strong></p><p><strong>Step 3: Check Surjectivity</strong></p><p>Let y = 2x/(x-1). Solving for x:</p><p>y(x-1) = 2x</p><p>yx - y = 2x</p><p>x(y-2) = y</p><p>x = y/(y-2), defined for y ≠ 2</p><p><strong>Step 4: Critical observation</strong></p><p>When y = 2, we get x = 2/0 (undefined). So y = 2 has no pre-image in domain A.</p><p>For any y ≠ 2: x = y/(y-2) must be checked if it lies in A.</p><p>Since x = y/(y-2) is never a positive integer for y ≠ 2, all such values ARE in A.</p><p>However, since 2 ∉ Range(f), f is <strong>not surjective onto R</strong>.</p><p>∴ Answer: <strong>D (Injective but not surjective)</strong></p>
Correct Answer: D