Question

If the function $$f(x)={x}^{3}-6{x}^{2}+ax+b$$ defined on $$[1,3]$$ satisfies the rolle's theorem for $$c=\cfrac{2\sqrt{3}+1}{\sqrt{3}}$$ then

A
a=11,b=6
B
a=11,b=6
C
a=11,bR
D
None of these

Solution

The correct option is C $$a=11,b\in R$$Given that$$f(x)=x^3-6x^2+ax+b$$$$\Rightarrow f'(x)=3x^2-12x+a$$$$\Rightarrow f'(c)=3c^2-12c+a$$According to Rolle's theorem$$\Rightarrow f'(c)=0$$$$\Rightarrow 3c^2-12c+a=0$$Substitute the given value of $$c$$$$\Rightarrow 3(\dfrac{2\sqrt {3}+1}{\sqrt 3})^2-12(\dfrac{2\sqrt 3 +1}{\sqrt 3})+a=0$$$$\Rightarrow 13+4\sqrt 3-24-4\sqrt 3=-a$$$$\Rightarrow a=11$$ and $$b$$ belongs to $$R$$Mathematics

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