Sets, Relations & Functions
Polynomial Functions
Grade 11
Question:
<p>Let <i>f</i> be a polynomial function such that <i>f</i>(3<i>x</i>) = <i>f'</i>(<i>x</i>) · <i>f''</i>(<i>x</i>), for all <i>x</i> ∈ ℝ. Then</p>
<p>\(f(2) - f'(2) + f''(2) = 10\)</p>
<p>\(f''(2) - f(2) = 4\)</p>
<p>\(f''(2) - f'(2) = 0\)</p>
<p>\(f(2) - f'(2) = 28\)</p>
Step-by-Step Solution
Key Concept: Use the constraint f(3x) = f'(x)·f''(x) to determine the degree of f by comparing degrees on both sides, then apply it to find specific coefficients.
<p><strong>Step 1: Determine degree of f</strong></p><p>Let deg(f) = n. Then deg(f(3x)) = n and deg(f'(x)·f''(x)) = (n-1)+(n-2) = 2n-3.</p><p>For equality: n = 2n - 3 ⟹ n = 3</p><p><strong>Step 2: Set up cubic polynomial</strong></p><p>Let f(x) = ax³ + bx² + cx + d (where a ≠ 0)</p><p>Then f'(x) = 3ax² + 2bx + c and f''(x) = 6ax + 2b</p><p><strong>Step 3: Expand f(3x)</strong></p><p>f(3x) = a(27x³) + b(9x²) + c(3x) + d = 27ax³ + 9bx² + 3cx + d</p><p><strong>Step 4: Expand f'(x)·f''(x)</strong></p><p>f'(x)·f''(x) = (3ax² + 2bx + c)(6ax + 2b)</p><p>= 18a²x³ + 6abx² + 12abx² + 4b²x + 6acx + 2bc</p><p>= 18a²x³ + (12ab + 6ab)x² + (12ab + 6ac)x + (4b² + 2bc)</p><p><strong>Step 5: Compare coefficients</strong></p><p>x³: 27a = 18a² ⟹ 27a = 18a² ⟹ 27 = 18a ⟹ a = 3/2</p><p>x²: 9b = 18ab ⟹ 9b = 18(3/2)b ⟹ 9b = 27b (only true if b = 0)</p><p>x¹: 3c = 12ab + 6ac = 12(3/2)(0) + 6(3/2)c = 9c (only true if c = 0)</p><p>x⁰: d = 2bc = 0</p><p><strong>Step 6: Conclusion</strong></p><p>f(x) = (3/2)x³ + constant, satisfying the functional equation.</p><p>∴ Answer: C</p>
Correct Answer: C