Velocity is maximum when $KE$ of particle is maximum. ie potential energy of particle is minimum
$\large\frac{dV}{dx}$$=0=>x^3-x=0=>x=\pm 1$
minimum $PE$ is when $x=1$
min $PE=\large\frac{1}{4}-\frac{1}{2}=\frac{-1}{4}$$J$
Given $KE_{\large \max}$$+PE_{\large min}$$=2\;J$
$KE_{\large max }=2 +\large\frac{1}{4}=\frac{9}{4}$
$\large\frac{9}{4}=\frac{1}{2} $$mv^2_{\large max} \qquad (m=1 kg)$
$ v^2 _{\large max}=\large\frac{9}{2}$
$v_{\large max}=\large\frac{3}{\sqrt 2}$$\;m/s$
Hence a is the correct answer.