Question

If the function f(x) = ax³ + bx² + 11x - 6 satisfies conditions of Rolle’s theorem in [1, 3] and f'(2 + 1/√3) = 0, then values of a and b are respectively.

(a) 1, -6 

(b) -2, 1 

(c) -1, -6 

(d) -1, 6

Answer 1

Correct answer is (a) 1, -6

f(x) = ax³ + bx² + 11x – 6 satisfies the conditions of Rolle’s theorem in [1, 3]

∴ f(1) = f(3)

a(1)³ + b(1)² + 11(1) – 6 = a(3)³ + b(3)³ + 11(3) – 6 

a + b + 11 = 27a + 9b + 33 

26a + 8b = -22 

13a + 4b = -11 

Only a = 1, b = -6 satisfy this equation.