<p>Let <em>f</em>(<em>x</em>) = 0 be an equation of degree six, having integer coefficients and whose one root is \(2\cos\dfrac{\pi}{18}\). Then, the sum of all the roots of <em>f</em>'(<em>x</em>) = 0, is</p>
Step-by-Step Solution
Key Concept: If 2cos(π/18) is a root of a degree-6 polynomial with integer coefficients, then all its conjugate roots (obtained via minimal polynomial and algebraic relations) must also be roots. The sum of roots of f'(x)=0 relates to the roots of f(x)=0 through Rolle's theorem and derivative properties.
<p><strong>Step 1:</strong> Find the minimal polynomial of 2cos(π/18).</p><p>Let θ = π/18, so 3θ = π/6. Then cos(3θ) = cos(π/6) = √3/2.</p><p>Using cos(3θ) = 4cos³θ - 3cosθ: 4cos³(π/18) - 3cos(π/18) = √3/2</p><p>Let x = 2cos(π/18), then cos(π/18) = x/2:</p><p>4(x/2)³ - 3(x/2) = √3/2 → x³/2 - 3x/2 = √3/2 → x³ - 3x = √3</p><p>Squaring: (x³ - 3x)² = 3 → x⁶ - 6x⁴ + 9x² - 3 = 0</p><p><strong>Step 2:</strong> Identify the complete polynomial f(x).</p><p>The minimal polynomial of 2cos(π/18) is x⁶ - 6x⁴ + 9x² - 3 = 0 (degree 6 with integer coefficients).</p><p>So f(x) = x⁶ - 6x⁴ + 9x² - 3</p><p><strong>Step 3:</strong> Find f'(x) and sum of its roots.</p><p>f'(x) = 6x⁵ - 24x³ + 18x = 6x(x⁴ - 4x² + 3)</p><p>f'(x) = 6x(x² - 1)(x² - 3)</p><p><strong>Step 4:</strong> Calculate sum of all roots of f'(x) = 0.</p><p>The roots of f'(x) = 0 are: 0, ±1, ±√3</p><p>Sum = 0 + 1 + (-1) + √3 + (-√3) = <strong>0</strong></p><p>∴ Answer: <strong>0</strong></p>
Correct Answer: 0