Answer :

**Rolle’s theorem:** If f: [a, b] → R is continuous on [a, b] and differentiable on (a, b) such that f(a) = f(b), then there exists some c in (a, b) such that f’(c) = 0

Now, Given Rolle’s theorem holds for

f(x) = x^{3} – 6x^{2} + ax + b in [1, 3] and for

Now, f (1) = f(3)

⇒ (1)^{3} – 6(1)^{2} + a (1) + b = (3)^{3} – 6(3)^{2} + a (3) + b

⇒ 1 – 6 + a (1) + b = 27 – 54 + 3a + b

⇒-5 + a = 27 + 3a

⇒ 2a = 22

⇒ a = 11

As there is no equation for b, so b can take any values.

Also, f’(c) = 0

As, f(x) = x^{3} – 6x^{2} + ax + b

⇒ f’(x) = 3x^{2} – 12x + a

= 0

So, for any value of ‘b’ and a = 11, the given equation will hold rolle’s theorem in [1, 3]

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