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Question

Rolle's theorem hold for the function $$f(x)=x^3+bx^2+cx, 1 \leq x\leq 2$$ at the point $$4/3$$, the values of b and c are


A
b=8,c=5
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B
b=5,c=8
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C
b=5,c=8
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D
b=5,c=8
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Solution

The correct option is B $$b=-5, c=8$$
$$f(x)=x^3+bx^2+cx, 1 \leq x\leq 2$$
$$f'(x)=3{x}^{2}+2bx+c$$
Since, Rolle's theorem holds for $$c=\dfrac{4}{3}$$ 
$$\Rightarrow f'\left(\dfrac{3}{4}\right)=0$$
$$\Rightarrow 8b+3c=-16$$                     .....(i)
Also, $$f(1)=f(2)$$
$$\Rightarrow 3b+c=-7$$                        ......(ii)
$$\Rightarrow b=-5, c=8$$

Mathematics

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