# 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.

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