Is potential energy always inversely proportional to separation? This is a question from Class 11 Physics NCERT. 
The answer is that only graph (v) is correct and the reason given here is that potential energy varies inversely with separation. But does that always have to be the case? 
Also, I had learnt that during elastic collision, all the kinetic energy momentarily converts to elastic potential energy due to deformation, which means that potential energy should be maximum. But here the graph shows that the potential energy is minimum during collision. 
I'm certain I am missing a key concept here. Can anyone help me understand where I'm going wrong? Thank you.
 A: 
But here the graph shows that the potential energy is minimum during collision.

But it doesn't. Keep in mind that $r$ becomes smaller as the balls get closer together and then becomes larger as they move apart. The horizontal axis is not time. So the potential energy actually increases to a maximum at the minimum separation, and then it goes back down as the balls separate.
As for the other points, it makes sense that you want $V$ to increase as $r$ decreases. This is because
$$F=-\frac{\text dV}{\text dr}$$
The inverse relationship between $V$ and $r$ means that the force is pushing the balls in contact towards being farther apart. 
None of the other answer choices give you the correct force behavior (check for yourself).
To answer your title question, this does not always have to be the case. But for your system we don't want any attraction between the balls, so we do want this behavior here. It doesn't necessarily have to be inversely proportional, but it does need to monotonically decrease as $r$ increases (i.e. an inverse relationship).
