Correct Answer - D
If the particle is released at the origin, it will try to go in the direction of force. Hence `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+Detla`, it will reach at point of least possible potential energy `(-15J)` where it will have maximum kinetic energy.
`1/2mv_(max)^2=25`
`v_(max)=5ms^-1`
The particle will now perform oscillatory motion will `x=5` as mean position.
In (c), `E_i=u_i+k_i=15+6=21J`
At `x=10, U_f=20`
`K_f=1!=0`
So the particle crosses `x=10`.