Explanation

(d) Potential energy $V = \frac{x^{4}}{4} - \frac{x^{2}}{2}$
For maximum kinetic energy, potential energy of a particle should be minimum
For minimum value of $V,\frac{dV}{dx} = 0$ and $\frac{d^{2}V}{dx^{2}} > 0$
Force $F = - \left( \frac{dV}{dx} \right) = \frac{4x^{3}}{4} - \frac{2x}{2} = 0 \Rightarrow x^{3} - x = 0$
$\Rightarrow x\left( x^{2} - 1 \right) = 0$
$i.e.$ at $x = 0,\ x = + 1$ and $x = - 1$ for on the particle will be zero
Now $\frac{d^{2}V}{dx^{2}} = 3x^{2} - 1$
For $x = + 1$ and $x = - 1\frac{d^{2}V}{dx^{2}} > 1$
It means the potential energy of the particle will be minimum at $x = 1$ and $x = - 1$
Now substituting these values in expression of potential energy
Energy $V_{\min} = \left\lbrack \frac{(1)^{4}}{4} - \frac{(1)^{2}}{2} \right\rbrack J = \left\lbrack \frac{1}{4} - \frac{1}{2} \right\rbrack J = - \frac{1}{4}J$
$(Kinetic\ energy)_{\max} = Total\ energy - (potential\ energy)_{\min} = 2 - \left( - \frac{1}{4} \right)$
$\frac{1}{2}\ mv_{\max}^{2} = \frac{9}{4} \Rightarrow v_{\max}^{2} = \frac{9}{2} \Rightarrow v_{\max} = \frac{3}{\sqrt{2}}\ m/sec$