MHT CET202611 April 2026Evening ShiftMathematicsApplication of DerivativesActual
If the function f(x) = ax^3 + bx^2 + 11x - 6 , defined on [1, 3] , satisfies all the conditions of Rolle's theorem for c = 2 + 1 3 , then
Options
- Aa = -1, b = 1 2
- Ba = 1, b = -6
- Ca = -1, b = 6
- Da = -2, b = 1
Correct answer
B. a = 1, b = -6
Step-by-step solution
Since f(x) satisfies Rolle's theorem on [1, 3] , we must have f(1) = f(3) . f(1) = a(1)^3 + b(1)^2 + 11(1) - 6 = a + b + 5 f(3) = a(3)^3 + b(3)^2 + 11(3) - 6 = 27a + 9b + 27 Equating f(1) and f(3) : a + b + 5 = 27a + 9b + 27 26a + 8b + 22 = 0 13a + 4b + 11 = 0 By Rolle's theorem, f'(c) = 0 for c = 2 + 1 3 . f'(x) = 3ax^2 + 2bx + 11 f' (2 + 1 3 ) = 3a (2 + 1 3 )^2 + 2b (2 + 1 3 ) + 11 = 0 3a (4 + 1 3 + 4 3 ) + 4b + 2b 3 + 11 = 0 13a + 4 3 a + 4b + 2b 3 + 11 = 0 (13a + 4b + 11) + 1 3 (12a + 2b) = 0 Substituting 13a +