Question

If Rolle's theorem holds for the function f(x) = x³ + bx² + ax + 5, x ∈ [1, 3] with c = `2 + 1/sqrt(3)`, find the values of a and b.

Answer 1

f(x) = x³ + bx² + ax + 5, x ∈ [1, 3]

∴ f'(x) = 3x² + 2bx + a

∴ f'(c) = 3c² + 2bc + a    .....(1)

It is given that Rolle's theorem holds for f(x)

∴ f(3) = f(1)

∴ 27 + 9b + 3a + 5 = 1 + b + a + 5

∴ 2a + 8b = –26

∴ a + 4b = –13   .....(2)

Also f'(c) = 0

∴ From (1): 3c² + 2bc + a = 0

Put c = `2 + 1/sqrt(3)`

∴  `3(2 + 1/sqrt(3))^2 + 2b(2 + 1/sqrt(3)) + a` = 0

∴ `3(4 + 4/sqrt(3) + 1/3) + 4b + (2b)/sqrt(3) + a` = 0

∴ `12 + 12/sqrt(3) + 1 + 4b + (2b)/sqrt(3) + a` = 0

∴ a + 4b = `-13 - 12/sqrt(3) - (2b)/sqrt(3)`   .....(3)

From (2) and (3)

`-13 - 12/sqrt(3) - (2b)/sqrt(3)` = –13

∴ `(2b)/sqrt(3) + 12/sqrt(3)` = 0

∴ 2b + 12 = 0

∴ b = –6

From (2): a + (–24) = –13

∴ a = 11

∴ a = 11, b = –6