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Question

lf $$\mathrm{f}(\mathrm{x})= a \log \mathrm{x}+\mathrm{b}\mathrm{x}^{2}+\mathrm{x}$$ has extreme values at  $$\mathrm{x}=-1,\ \mathrm{x}=2$$ then $$\mathrm{a}=\ldots..,\mathrm{b}=\ldots\ldots$$


A
2,12
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B
12,2
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C
12,2
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D
2,12
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Solution

The correct option is A $$2,\displaystyle \frac{-1}{2}$$
$$f'(x)=\dfrac {a}{x}+2bx+1$$
$$f'(x)$$ is 0 at $$x=-1$$ & $$x=2$$
(1) $$0=-a-2b+1$$
$$a+2b=1$$
(2) $$0=\dfrac {a}{2}+4b+1$$
$$a+8b+2=0$$
So $$a=2, b=\dfrac {-1}{2}$$

Mathematics

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