Question

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

(a)  1, – 6

(b)  – 2, 1

(c)  – 1, 1/2

(d)  – 1, 6

Answer 1

Correct option  (a) 1, – 6 

Explanation:

By Rolle's theorem f(1) = f(3) gives

a + b + 11 – 6 = 27a + 9b + 33 – 6 

⇒ 26a + 8b + 22 = 0

⇒ 13a + 4b + 11 = 0 

Again f' (x) = 3ax² + 2bx + 11

∴ (1) or (3) are possible

Simplify it, we get b + 6a = 0 .....(2)

Clearly (a) satisfies it only and (c) does not.

Hence a = 1, b = – 6