Dogtanian

I think you have the picture back-to-front. From the graph/diagram given with the question the areplane is flying towards the origin' date=' not away from it. So P(0) will give you the height above the ground when the plane has landed. (This idea seams rather counter-intuitive to me, like I think you have done, it would be better to assume the plane styarts at x=0 and heads towards x=L , but it so happens to be going the other way round ) Try this agian now you've re done the first part again...hopefully the answer should fall out of the sky to you...lets hope unlike the plane...

You are given a general cubic polynomial and have can find it's derivative. You can find the values of c and d by sustituing x=0 into both P and P'. by looking at the "shape" of the curve at x=L you can see what P'(L) will be. You also know what P(L) equals. Use these facts to find 2 simultaneous equations and solve them to find a and b in terms of L and h. I hope this helps.

Yes, what we see as a shadow is a projection of the 3-d area into which the light cannot get on to a surface. I would go far to say the shdow is 2-d, even if it is bent over a 3-d object in 3-d space, here I would just take it would still be on a 2-d surface, only the surface wouldn't be "flat".

In my semigroups course I had a question the other week where I had to draw the multiplication table for the semigroup of just 2 elements...I guess at all levels there can be some need to use the tables of the simplest groups etc.

The probability that one hole has 2 pigeons is 1. Surely it's the probablity of a particular hole has 2 that is 1/7, which is nothing directly related to the Pigeon Hole Principle. If by definition you mean a more formal statement of the Principle then it would be hard to find anything better than what has already been said. It basically goes as follows: If we put N+1 pigeons in N pigeon-holes, then some pigeon-hole contains at least two pigeons. The statment doesn't say anthing about needing to put at least one pigeon in each hole, but just that if you have more pigeons than holes and randomly place all the pigeons in a hole, then there will be at least one hole with at least 2 pigeons.