In figure, the variation of potential energy of a particle of mass m = 2 kg is represented w.r.t its x-coordinate. The particle moves under the effect of the conservative force along the x-axis. Which of the following statements is incorrect about the particle?

If the particle is released at the origin, it will try to go in the direction of force. Here dU/dx is positive and hence force is negative; as a result, it will move towards negative x-axis.When the particle is released at x = 2 + Δ, it will reach the least possible potential point of energy (-15 J) where it will have maximum kinetic energy12mvmax 2=25vmax =5 ms−1The particle will now perform oscillatory motion with x = 5 as mean position.In (3), Ei=ui+ki=15+6=21 JAt x=10, Uf=20    Kf=1≠0So the particle crosses x = 10.