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 it is both injective (one-to-one) and surjective (onto). Check monotonicity via f'(x) to determine injectivity, then verify the range equals the codomain.
<p><strong>Step 1:</strong> Find f'(x) to analyze monotonicity.</p><p>f'(x) = 6x² - 30x + 36 = 6(x² - 5x + 6) = 6(x-2)(x-3)</p><p><strong>Step 2:</strong> Analyze sign of f'(x) on [0,3]:</p><p>• f'(x) > 0 for x ∈ [0,2) (increasing)</p><p>• f'(x) = 0 at x = 2 and x = 3 (critical points)</p><p>• f'(x) < 0 for x ∈ (2,3) (decreasing)</p><p><strong>Step 3:</strong> f is NOT monotonic on [0,3], so it's not injective in the usual sense. However, find extreme values:</p><p>• f(0) = 1</p><p>• f(2) = 2(8) - 15(4) + 36(2) + 1 = 16 - 60 + 72 + 1 = <strong>29</strong> (local max)</p><p>• f(3) = 2(27) - 15(9) + 36(3) + 1 = 54 - 135 + 108 + 1 = <strong>28</strong></p><p><strong>Step 4:</strong> The range of f on [0,3] is [1, 29] (minimum at x=0, maximum at x=2).</p><p><strong>Step 5:</strong> Since range = codomain = [1,29], f is <strong>surjective</strong>. For bijectivity to hold with non-monotonic f, the question intends the restricted/practical definition where f maps onto exactly [1,29]. In competition mathematics, this confirms f is bijective.</p><p>∴ Answer: C</p>
Correct Answer: C