If y = alpha log x + beta x^3 - x has extreme | vedclass

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(A) Given the function $y = \alpha \log x + \beta x^3 - x$. For the function to have extreme values at $x = -1$ and $x = 1$,the derivative $\frac{dy}{

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$0$ and $\frac{1}{3}$

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(A) Given the function $y = \alpha \log x + \beta x^3 - x$. For the function to have extreme values at $x = -1$ and $x = 1$,the derivative $\frac{dy}{dx}$ must be zero at these points. First,find the derivative: $\frac{dy}{dx} = \frac{\alpha}{x} + 3\beta x^2 - 1$. At $x = 1$: $\frac{dy}{dx} = \alpha + 3\beta - 1 = 0 \implies \alpha + 3\beta = 1$. At $x = -1$: $\frac{dy}{dx} = \frac{\alpha}{-1} + 3\beta(-1)^2 - 1 = 0 \implies -\alpha + 3\beta = 1$. Adding the two equations: $(\alpha + 3\beta) + (-\alpha + 3\beta) = 1 + 1 \implies 6\beta = 2 \implies \beta = \frac{1}{3}$. Substituting $\beta = \frac{1}{3}$ into $\alpha + 3\beta = 1$: $\alpha + 3(\frac{1}{3}) = 1 \implies \alpha + 1 = 1 \implies \alpha = 0$. Thus,$\alpha = 0$ and $\beta = \frac{1}{3}$.

If $y = \alpha \log x + \beta x^3 - x$ has extreme values at $x = -1$ and $x = 1$,then $\alpha$ and $\beta$ are respectively

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