It is given that for the function f(x) = \(x^3 – 6x^2 + ax + b\) on [1, 3], Rolles’s theorem holds with c = \(2 +{1\over \sqrt{3}}\). Find the values of a and b, if f(1) = f(3) = 0.

Solution :

We are given that f(1) = f(3) = 0.

\(\therefore\) \(1^3 – 6 \times 1 + a + b\) = \(3^3 – 6 \times 3^2 + 3a + b\) = 0

\(\implies\) a + b = 5 and 3a + b = 27

Solving these two equations for a and b, f'(c) is zero or not.

We have,

f(x) = \(x^3 – 6x^2 + ax + b\)

\(\implies\) f(x) = \(x^3 – 6x^2 + 11x – 6\)

\(\implies\) f'(x) = \(3x^2 – 12x + 11\)

\(\therefore\) f'(c) = \(3c^2 – 12c + 11\) = 3\((2 +{1\over \sqrt{3}})^2\) – 12(\(2 +{1\over \sqrt{3}}\)) + 11

= 12 + \(12\over \sqrt{3}\) + 1 – 24 – \(12\over \sqrt{3}\) + 11 = 0

Hence, a = 11 and b = -6.