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

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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