Sets, Relations & Functions
Types of functions
Grade 11
Question:
<p>The function \(f:[0,3] \to [1,29]\), defined by \(f(x) = 2x^3 - 15x^2 + 36x + 1\), is:</p>
<p>(a) one-one and onto</p>
<p>(b) one-one but not onto</p>
<p>(c) onto but not one-one</p>
<p>(d) neither one-one nor onto</p>
Step-by-Step Solution
Key Concept: A function is bijective if and only if it is both injective (one-to-one) and surjective (onto). Check injectivity via monotonicity using derivatives, and verify the range matches the codomain by finding extrema.
<p><strong>Step 1:</strong> Find the derivative to check monotonicity.</p><p>f'(x) = 6x² - 30x + 36 = 6(x² - 5x + 6) = 6(x - 2)(x - 3)</p><p><strong>Step 2:</strong> Analyze the sign of f'(x) on [0,3].</p><p>For x ∈ [0,2): f'(x) > 0 (function increasing)</p><p>For x ∈ (2,3): f'(x) < 0 (function decreasing)</p><p>f'(2) = 0 and f'(3) = 0, so x = 2 is a local maximum and x = 3 is a critical point.</p><p><strong>Step 3:</strong> Evaluate f at critical points and endpoints.</p><p>f(0) = 1</p><p>f(2) = 2(8) - 15(4) + 36(2) + 1 = 16 - 60 + 72 + 1 = 29</p><p>f(3) = 2(27) - 15(9) + 36(3) + 1 = 54 - 135 + 108 + 1 = 28</p><p><strong>Step 4:</strong> Determine the range.</p><p>Maximum value = 29 at x = 2; Minimum value = 1 at x = 0</p><p>Range = [1, 29], which equals the codomain.</p><p><strong>Step 5:</strong> Verify injectivity and surjectivity.</p><p>Although f is not strictly monotonic, it is injective on [0,3] because the decreasing part [2,3] maps [29,28] and the increasing part [0,2] maps [1,29], with only f(0) = 1 common at boundaries. The function achieves all values in [1,29].</p><p>∴ f is bijective. Answer: C</p>
Correct Answer: C