To find the acceleration of the particle as a function of $x,$ we start with the given velocity function:$\\ v(x) = \beta x^{-2n} \\$The acceleration $a(x)$ is the derivative of velocity with respect to time:$\\ a = \frac{dv}{dt} \\$Using the chain rule, we can express this as:$\\ a = \frac{dv}{dx} \cdot \frac{dx}{dt} \\$Since $\frac{dx}{dt} = v,$ we have:$\\ a = v \cdot \frac{dv}{dx} \\$First, we find $\frac{dv}{dx}: \\ v = \beta x^{-2n} \\ \frac{dv}{dx} = \frac{d}{dx} (\beta x^{-2n}) = \beta \cdot (-2n) \cdot x^{-2n-1} \\ \frac{dv}{dx} = -2n\beta x^{-2n-1} \\$Now, substitute $v$ and $\frac{dv}{dx}$ into the expression for acceleration:$\\ a = v \cdot \frac{dv}{dx} \\ a = (\beta x^{-2n}) \cdot (-2n\beta x^{-2n-1}) \\$Simplify the expression:$\\ a = -2n\beta^2 x^{-2n} \cdot x^{-2n-1} \\ a = -2n\beta^2 x^{-4n-1} \\$Thus, the acceleration of the particle as a function of $x$ is:$\\ a(x) = -2n\beta^2 x^{-4n-1} \\$This corresponds to Option 1.