The potential energy of a particle is given by formula $U=100-5x+100x^2$, where U and x are in SI units. If mass of the particle is $0.1 \mathrm{kg}$, then magnitude of its acceleration 

$F=-\frac{dU}{dx}=5-200x$

At origin, $x=0$

$$\begin{gathered} F=5 \mathrm{N} \\ a=\frac{F}{m}=\frac{5}{0.1}=50 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Mean position is at F=0

or at, $x=\frac{5}{200}=0.025 \mathrm{m}$

$a=\frac{F}{m}=\frac{5-200x}{0.1}=(50-2000x)$ At $0.05 \mathrm{m}$ from the origin,

$x=+0.05 \mathrm{m}$ or $x=-0.05 \mathrm{m}$

Substituting in Eq. (i), we have

$$\begin{gathered} \left|a\right|=150 \mathrm{m}/{\mathrm{s}}^2 \\ or \\ \left|a\right|=50 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

At $0.05 \mathrm{m}$ from the mean position means,

$$\begin{gathered} x=0.075 \\ or \\ x=-0.025 \mathrm{m} \end{gathered}$$

Substituting in Eq. (i), we have

$\left|a\right|=100 \mathrm{m}/{\mathrm{s}}^2$