The potential energy of 1 kg particle free to move along the X-axis is given y $\mathrm{V}\left(\mathrm{x}\right)=\left(\frac{{\mathrm{x}}^4}{4}-\frac{{\mathrm{x}}^2}{2}\right)\mathrm{J}$ The total mechanical energy of the particle is 2 J. Then the maximum speed (in m/s) is
 

$$\begin{gathered} \text{ For minimum value of }\mathrm{V},\mathrm{dv}/\mathrm{dx}=0 \\ \therefore \frac{4{\mathrm{x}}^3}{4}-\frac{2\mathrm{x}}{2}=0\text{ or }\mathrm{x}=0,\mathrm{x}=\pm 1 \\ \frac{{\mathrm{d}}^2\mathrm{v}}{{\mathrm{dx}}^2}=-1\text{ at }\mathrm{x}=0 \\ \text{ and }\frac{{\mathrm{d}}^2\mathrm{v}}{{\mathrm{dv}}^2}=2\text{ at }\mathrm{x}=\pm 1 \\ \mathrm{Hence} V \mathrm{is} \mathrm{minimum} \mathrm{at} \mathrm{x} =\pm \mathrm{l} \mathrm{with} \mathrm{value} \\ {\mathrm{V}}_{min}(\mathrm{x}=\pm 1)=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}\mathrm{J} \\ {\mathrm{K}}_{max}+{\mathrm{V}}_{min}=\text{ Total mechanical energy } \\ {\mathrm{K}}_{max}=-{\mathrm{V}}_{min}+2 \\ {\mathrm{K}}_{max}=+\frac{1}{4}+2=\frac{9}{4} \\ \frac{1}{2}{\mathrm{mv}}^2=\frac{9}{4} \\ \frac{1}{2}\times 1\times {\mathrm{v}}^2=\frac{9}{4}\text{ or }\mathrm{v}=\frac{3}{\sqrt{2}}\mathrm{m}/\mathrm{s} \end{gathered}$$