Question

It is given that Rolle's theorem holds good for the function :

f(x) = x³ + ax² + bx, x ∈ [1,2] at the point x = 4/3.

Find the values of a and b.

Answer 1

Given,

f(x) = x³ + ax² + bx, x ∈ [1,2]

f (x) being a polynomial, is continuous on [1,2]

f'(x) = 3x² + 2ax + b exists on x ∈ [1,2]

∴ f(x) is derivable on

f(1) = a + b + 1

f(2) = 8 + 4a + 2b

Since Rolle's theorem holds for f(x) on [1,2]

so, f(1) = f(2)

a + b + 1 = 8 + 4a + 2b

⇒ 3a + b + 7 = 0

3a + b = -7

Also, Rolle's theorem holds at the point x = 4/3

⇒ f'(c) = 0

\(f' (\frac{4}{3})=3\times\frac{16}{9}+2a\times\frac{4}{3}+b=0\)

⇒ 8a + 3b = -16

Multiply equation (1) by 3 and subtracting equation (2), we get

Put a = -5 in (1), we get

3a + b = -7

3 x (-5) + b = -7

-15 + b = -7

b = -7 + 15

b = 8

∴ a = -5, b = 8