If the function f(x) = x^3 - 6x^2 + ax + b sa | vedclass

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(D) Given the function $f(x) = x^3 - 6x^2 + ax + b$. First,we find the derivative $f'(x) = 3x^2 - 12x + a$. According to Rolle's theorem,there exists

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$11$

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(D) Given the function $f(x) = x^3 - 6x^2 + ax + b$. First,we find the derivative $f'(x) = 3x^2 - 12x + a$. According to Rolle's theorem,there exists a point $c \in (1, 3)$ such that $f'(c) = 0$. We are given that $f'\left( \frac{2\sqrt{3} + 1}{\sqrt{3}} \right) = 0$,which simplifies to $f'\left( 2 + \frac{1}{\sqrt{3}} \right) = 0$. Substituting $x = 2 + \frac{1}{\sqrt{3}}$ into $f'(x) = 0$: $3\left( 2 + \frac{1}{\sqrt{3}} \right)^2 - 12\left( 2 + \frac{1}{\sqrt{3}} \right) + a = 0$. Expanding the terms: $3\left( 4 + \frac{1}{3} + \frac{4}{\sqrt{3}} \right) - 24 - \frac{12}{\sqrt{3}} + a = 0$. $12 + 1 + 4\sqrt{3} - 24 - 4\sqrt{3} + a = 0$. $13 - 24 + a = 0$. $-11 + a = 0$,which gives $a = 11$.

If the function $f(x) = x^3 - 6x^2 + ax + b$ satisfies Rolle's theorem in the interval $[1, 3]$ and $f'\left( \frac{2\sqrt{3} + 1}{\sqrt{3}} \right) = 0$,then $a = $ ..............

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