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

Finding Extreme Values via Derivatives

The function $y = \alpha \log x + \beta x^3 - x$ has extreme values where its first derivative $\frac{dy}{dx} = 0$. Differentiate term by term:

$\frac{dy}{dx} = \frac{\alpha}{x} + 3\beta x^2 - 1$

Given extreme points at $x = 1$ and $x = -1$, evaluate $\frac{dy}{dx}$ at both points:

At $x = 1$: $\frac{\alpha}{1} + 3\beta (1)^2 - 1 = 0 \Rightarrow \alpha + 3\beta = 1$

At $x = -1$: $\frac{\alpha}{-1} + 3\beta (-1)^2 - 1 = 0 \Rightarrow -\alpha + 3\beta = 1$

Add the two equations to eliminate $\alpha$:

$(\alpha + 3\beta) + (-\alpha + 3\beta) = 1 + 1 \Rightarrow 6\beta = 2 \Rightarrow \beta = \frac{1}{3}$

Substitute $\beta = \frac{1}{3}$ into $\alpha + 3\beta = 1$:

$\alpha + 3 \cdot \frac{1}{3} = 1 \Rightarrow \alpha + 1 = 1 \Rightarrow \alpha = 0$

The solution $\alpha = 0$, $\beta = \frac{1}{3}$ corresponds to option A.

Final Answer: $\boxed{A}$