Relations & Functions
Injective and Surjective Functions
Grade 12
Question:
<p>Given \( f(x) = \dfrac{2x}{x-1} \). Which of the following statements is correct?</p>
<p><i>f</i> is both injective and surjective</p>
<p><i>f</i> is injective but not surjective</p>
<p><i>f</i> is surjective but not injective</p>
<p><i>f</i> is neither injective nor surjective</p>
Step-by-Step Solution
Key Concept: To analyze function properties, find the domain (where denominator ≠ 0), check if f is injective by solving f(a) = f(b), and determine the range by solving y = f(x) for x in terms of y.
<p><strong>Step 1: Find the Domain</strong></p><p>For f(x) = 2x/(x-1) to be defined, x - 1 ≠ 0, so <strong>Domain = ℝ \ {1}</strong></p><p><strong>Step 2: Check Injectivity</strong></p><p>If f(a) = f(b), then 2a/(a-1) = 2b/(b-1)</p><p>Cross-multiplying: 2a(b-1) = 2b(a-1)</p><p>2ab - 2a = 2ab - 2b ⟹ a = b</p><p>∴ <strong>f is injective (one-to-one)</strong></p><p><strong>Step 3: Find the Range</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)</p><p>This is defined for all y except <strong>y = 2</strong></p><p>∴ <strong>Range = ℝ \ {2}</strong></p><p><strong>Step 4: Conclusion</strong></p><p>f is injective but NOT surjective. f is NOT bijective since 2 ∉ Range.</p><p>∴ Answer: B</p>
Correct Answer: B