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