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Question: If the function \(f(x) = x^{3} - 6x^{2} + ax + b\) satisfies Rolle's theorem in the interval [1, 3]a...

If the function f(x)=x3−6x2+ax+bf(x) = x^{3} - 6x^{2} + ax + b satisfies Rolle's theorem in the interval [1, 3]and f′(23+13)=0f^{'}\left( \frac{2\sqrt{3} + 1}{\sqrt{3}} \right) = 0 then

A

a=11a = 11

B

a=−6a = - 6

C

a=6a = 6

D

a=1a = 1

Answer

a=11a = 11

Explanation

Solution

f(x)=x3−6x2+ax+bf(x) = x^{3} - 6x^{2} + ax + b ⇒ f′(x)=3x2−12x+af^{'}(x) = 3x^{2} - 12x + a

⇒ f′(c)=0f^{'}(c) = 0 ⇒ f′(2+13)=0f^{'}\left( 2 + \frac{1}{\sqrt{3}} \right) = 0

⇒ 3(2+13)2−12(2+13)+a=03\left( 2 + \frac{1}{\sqrt{3}} \right)^{2} - 12\left( 2 + \frac{1}{\sqrt{3}} \right) + a = 0

⇒ 3(4+13+43)−12(2+13)+a=03\left( 4 + \frac{1}{3} + \frac{4}{\sqrt{3}} \right) - 12\left( 2 + \frac{1}{\sqrt{3}} \right) + a = 0

⇒ 12+1+43−24−43+a=012 + 1 + 4\sqrt{3} - 24 - 4\sqrt{3} + a = 0 ⇒ a=11a = 11.