Q. 14

# It is given that

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) = x3 – 6x2 + 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) = x3 – 6x2 + ax + b

f’(x) = 3x2 – 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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