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Question

If $$y=f(x)$$ is continuous on $$[0,6]$$. differentiable on $$(0,6)$$ $$f(0)=-2$$ and $$f(6)=16$$, then at some point between $$x=0$$ and $$x=6$$, $$f'(x)$$ must be equal to


A
18
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B
3
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C
3
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D
14
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E
18
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Solution

The correct option is B $$3$$
$$f'(x)=\cfrac { f(b)-f(a) }{ b-a } =\cfrac { 16-(-2) }{ 6-0 } =3$$
(Using Lagrange's mean value theorem)

Mathematics

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