Question:

If the function \(f(x) = x^3 + b x^2 + c x - 6\) satisfies all conditions of Rolle's theorem in \([1,3]\) and \[ f'\left(\frac{2\sqrt{3} + 1}{\sqrt{3}}\right) = 0, \] then find \(bc\).

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Use conditions of Rolle's theorem and derivative roots to find coefficients.
Updated On: Jun 6, 2025
  • 18
  • -66
  • 38
  • -46
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The Correct Option is B

Solution and Explanation

Given \(f(1) = f(3)\) for Rolle's theorem. Calculate: \[ f(1) = 1 + b + c - 6 = b + c - 5, \] \[ f(3) = 27 + 9b + 3c - 6 = 9b + 3c + 21. \] Set equal: \[ b + c - 5 = 9b + 3c + 21 \implies 8b + 2c = -26 \implies 4b + c = -13. \] Derivative: \[ f'(x) = 3x^2 + 2b x + c. \] Set \[ f'\left(\frac{2\sqrt{3}+1}{\sqrt{3}}\right) = 0, \] substitute and solve system of equations to find \(b\) and \(c\). Calculate \(bc = -66\).
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