The function f(x) = x^3 - 6x^2 + ax + b satis | vedclass

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(A) For Rolle's theorem to hold on $[1, 3]$,the function must satisfy $f(1) = f(3)$.$f(1) = 1^3 - 6(1)^2 + a(1) + b = 1 - 6 + a + b = a + b - 5$.$f(3)

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$11, -6$

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(A) For Rolle's theorem to hold on $[1, 3]$,the function must satisfy $f(1) = f(3)$.$f(1) = 1^3 - 6(1)^2 + a(1) + b = 1 - 6 + a + b = a + b - 5$.$f(3) = 3^3 - 6(3)^2 + a(3) + b = 27 - 54 + 3a + b = 3a + b - 27$.Equating $f(1) = f(3)$:$a + b - 5 = 3a + b - 27$$2a = 22 \implies a = 11$.Also,Rolle's theorem requires $f'(c) = 0$ for some $c \in (1, 3)$.$f'(x) = 3x^2 - 12x + a = 3x^2 - 12x + 11$.Setting $f'(c) = 0$:$3c^2 - 12c + 11 = 0$.The roots are $c = \frac{12 \pm \sqrt{144 - 132}}{6} = \frac{12 \pm \sqrt{12}}{6} = 2 \pm \frac{\sqrt{3}}{3}$.Since $2 - \frac{\sqrt{3}}{3} \approx 2 - 0.577 = 1.423$,which lies in $(1, 3)$,the condition is satisfied for $a = 11$.However,Rolle's theorem does not uniquely determine $b$. Given the options,we check if $b$ is constrained by other implicit conditions or if the question implies a specific context. Re-evaluating the options,if $a=11$,only option $A$ matches $a=11$.

The function $f(x) = x^3 - 6x^2 + ax + b$ satisfies the conditions of Rolle's theorem in $[1, 3]$. Then the values of $a$ and $b$ are respectively

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