Relations & Functions
Injective and Surjective Functions
Grade 12
Question:
<p>Given \( f(x) = \left|1 - \dfrac{1}{x}\right| \). Which of the following statements is correct about <i>f(x)</i>?</p>
<p><i>f(x)</i> is injective and surjective</p>
<p><i>f(x)</i> is not injective but it is surjective (with co-domain \([0,\infty)\))</p>
<p><i>f(x)</i> is injective but not surjective</p>
<p><i>f(x)</i> is neither injective nor surjective</p>
Step-by-Step Solution
Key Concept: Analyze the absolute value function by breaking it into cases based on the sign of (1 - 1/x), then examine domain restrictions and behavior in each region.
<p><strong>Step 1: Identify domain and critical points</strong></p><p>Domain: x ≠ 0 (since 1/x is undefined at x = 0)</p><p>Critical point: 1 - 1/x = 0 ⟹ x = 1</p><p><strong>Step 2: Analyze the function by regions</strong></p><p>For x > 1: 1/x < 1, so (1 - 1/x) > 0 ⟹ f(x) = 1 - 1/x</p><p>For 0 < x < 1: 1/x > 1, so (1 - 1/x) < 0 ⟹ f(x) = 1/x - 1</p><p>For x < 0: 1/x < 0, so (1 - 1/x) > 1 > 0 ⟹ f(x) = 1 - 1/x</p><p><strong>Step 3: Determine properties</strong></p><p>• On (0,1): f(x) = 1/x - 1 is decreasing from +∞ to 0</p><p>• On (1,∞): f(x) = 1 - 1/x is increasing from 0 to 1</p><p>• On (-∞,0): f(x) = 1 - 1/x is increasing from 1 to +∞</p><p>• Range: [0, ∞), with f(x) = 0 only at x = 1</p><p>• f is neither even nor odd; f is discontinuous at x = 0</p><p>∴ Answer: B</p>
Correct Answer: B