It is given that for the function f(x) = x³ + bx²+ ax + 5 on [1, 3], Rolle’s theorem holds with . Find values of a and b.

Given: f(x) = x³ + bx²+ ax + 5

and for this f(x) Rolle’s theorem holds on [1, 3] with

To find: Value of a and b

Proof:

Firstly, we find f(1) and f(5)

We have, f(x) = x³ + bx²+ ax + 5

Put x = 1

f(1) = (1)³ + b(1)² + a(1) + 5

= 1 + b + a + 5

= a + b + 6

Put x = 3

f(3) = (3)³ + b(3)² + a(3) + 5

= 27 + 9b + 3a + 5

= 3a + 9b + 32

We know the Conditions of Rolle’s theorem:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

c) f(a) = f(b)

If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0

Here, Rolle’s theorem hold for given f(x)

⇒ f(1) = f(3)

⇒ a + b + 6 = 3a + 9b + 32

⇒ 3a – a + 9b – b + 32 – 6 = 0

⇒ 2a + 8b + 26 = 0

⇒ a + 4b + 13 = 0 …(i)

Now, differentiate the given f(x) = x³ + bx²+ ax + 5 with respect to x, we get

f’(x) = 3x² + 2bx + a

⇒ f’(c) = 3c² + 2bc + a

According to Rolle’s theorem f’(c) = 0

…(ii)

[from eq. (i) a + 4b + 13 = 0]

⇒ 12 + 2b = 0

⇒ 2b = -12

⇒ b = -6

Putting the value of b in eq. (i), we get

a + 4(-6) + 13 = 0

⇒ a – 24 + 13 = 0

⇒ a – 11 = 0

⇒ a = 11

Hence, the value of a = 11 and b = -6