If x=-1 and x=2 are extreme points of f(x)=al | vedclass

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(C) Given $f(x) = \alpha \log |x| + \beta x^2 + x$.The derivative is $f'(x) = \frac{\alpha}{x} + 2\beta x + 1$.Since $x = -1$ and $x = 2$ are extreme

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$\alpha=2, \beta=-\frac{1}{2}$

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(C) Given $f(x) = \alpha \log |x| + \beta x^2 + x$.The derivative is $f'(x) = \frac{\alpha}{x} + 2\beta x + 1$.Since $x = -1$ and $x = 2$ are extreme points, $f'(-1) = 0$ and $f'(2) = 0$.For $x = -1$: $f'(-1) = \frac{\alpha}{-1} + 2\beta(-1) + 1 = 0 \Rightarrow -\alpha - 2\beta + 1 = 0 \Rightarrow \alpha + 2\beta = 1$ ...$(i)$For $x = 2$: $f'(2) = \frac{\alpha}{2} + 2\beta(2) + 1 = 0 \Rightarrow \frac{\alpha}{2} + 4\beta + 1 = 0 \Rightarrow \alpha + 8\beta = -2$ ...(ii)Subtracting $(i)$ from (ii): $(\alpha + 8\beta) - (\alpha + 2\beta) = -2 - 1 \Rightarrow 6\beta = -3 \Rightarrow \beta = -\frac{1}{2}$.Substituting $\beta = -\frac{1}{2}$ into $(i)$: $\alpha + 2(-\frac{1}{2}) = 1 \Rightarrow \alpha - 1 = 1 \Rightarrow \alpha = 2$.

If $x=-1$ and $x=2$ are extreme points of $f(x)=\alpha \log |x|+\beta x^2+x, (x \neq 0)$, then

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