If y = a log x + b x^2 + x has its extreme va | vedclass

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(C) Given the function $y = a \log x + b x^2 + x$. First,find the derivative with respect to $x$: $\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$. Since the e

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$-\frac{17}{4}$

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(C) Given the function $y = a \log x + b x^2 + x$. First,find the derivative with respect to $x$: $\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$. Since the extreme values occur at $x = -1$ and $x = 2$,the derivative must be zero at these points. For $x = -1$: $\frac{a}{-1} + 2b(-1) + 1 = 0 \Rightarrow -a - 2b + 1 = 0 \Rightarrow a = 1 - 2b$. For $x = 2$: $\frac{a}{2} + 2b(2) + 1 = 0 \Rightarrow \frac{a}{2} + 4b + 1 = 0 \Rightarrow a + 8b + 2 = 0$. Substitute $a = 1 - 2b$ into the second equation: $(1 - 2b) + 8b + 2 = 0 \Rightarrow 6b + 3 = 0 \Rightarrow 6b = -3 \Rightarrow b = -\frac{1}{2}$. Now,find $a$: $a = 1 - 2(-\frac{1}{2}) = 1 + 1 = 2$. Finally,calculate $\left(\frac{a}{b} + \frac{b}{a}\right) = \left(\frac{2}{-1/2} + \frac{-1/2}{2}\right) = (-4 - \frac{1}{4}) = -\frac{17}{4}$.

If $y = a \log x + b x^2 + x$ has its extreme values at $x = -1$ and $x = 2$,then the value of $\left(\frac{a}{b} + \frac{b}{a}\right)$ is

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