Question

If function f(x) = ax³ + bx² + 11x – 6, x ∈ [1, 3] is satisfying Roll's condition and c = |2 + (1/√3)| than a =_____, b =_____

(a) – 1, + 6 

(b) – 2, 1 

(c) – 1, (1/2) 

(d) 1, – 6

Answer 1

The correct option (d) 1, – 6

Explanation:

f(x) = ax³ + bx² + 11x – 6 --------- x ∈ [1, 3]

∴ Roll's theorem is satisfied

∴ f(1) = f(3)

∴ a + b + 11 – 6 = 27a + gb + 33 – 6

∴ 26a + 8b = – 22

∴ 13a + 4a = – 11 (1)

also c = 2 + (1/√3)

and f'(x) = 3ax² + 2bx + 11

∴ f'(c) = 0

∴ 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

∴ (13 + 4√3) a + [4 + (2/√3)] b = – 11 (2)

from (1) & (2)

13a + 4b = 13a + 4√3 a + 4b + (2/√3)b

∴ 0 = 4√3 a + (2/√3)b

∴ 6a = – b ⇒ (a/b) = [(– 1)/6]

Now 13a + 4b = – 11

⇒ 13a – 24a = – 11

∴ a = 1 and b = – 6