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

Option: 1

12


Option: 2

11


Option: 3

13


Option: 4

15


Answers (1)

best_answer

 Since f(x) satisfies conditions of Rolle's theorem.

\mathrm{\begin{aligned} & \therefore f(1)=f(3) \\ & \Rightarrow 1-6+a+b=27-54+3 a+b \Rightarrow a=11 . \end{aligned} }
Now, \mathrm{f^{\prime}(x)=3 x^2-12 x+a=3 x^2-12 x+11 }.
Clearly, \mathrm{ f^{\prime}\left(\frac{2 \sqrt{3}+1}{\sqrt{3}}\right)=0. Hence, a=11. }

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shivangi.shekhar

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