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7 b) iii) I could see straight away how to do it, but I just didn't have time to write it out

8. a) iii) ? I wrote cot as 1/tan, and since x/2^n -> 0 as n->inf, then tan(x/2^n) -> x/2^n which then just simplifies to what they give

I was hoping to as well, but the chances of that are looking very, very slim now :( I'm so annoyed that I missed seeing one of the questions until the very end, I had plenty of time earlier that I could have done it!

I didn't bother with testing the points. I just said something like "the first stationary point corresponds to the maximum length of the rope", but I don't think I will get full marks for this question

sorry, I meant the minimum height, which corresponds to the maximum displacement

was an easy paper, but: "I would have lost a few marks in dodgy/not thorough enough explanations, numerical errors, and a question which i didn't notice until the last minute and would be lucky to get 1/2 for..

I got all the paper out, but would have lost a few marks in dodgy/not thorough enough explanations, numerical errors, and a question which i didn't notice until the last minute and would be lucky to get 1/2 for.. 110/120 maybe?

yeah differentiate it, to get the stationary points pretty sure i made a numerical error (or multiple - to many numbers!) in there somewhere but I got about 119 as the minimum points

but writing up the proof sucks...

I enjoy these ones a lot. Burzum has been my most played band lately. tru dat

put a space between the less than sign and stuff either side a2 + b2 > = 2ab so a3 + ab2 > = 2a2b and ba2 + b3 > = 2ab2 adding together gives a3 + ab2 + ba2 + b3 > = 2a2b + 2ab2 a3 + b3 >= a2b + ab2 (3/8)(a3 + b3)+(1/8)(a3 + b3) >= 3/8 (a2b + ab2) + 1/8 (a3 + b3) (a3 + b3)/2 >= 1/8 (a+b)3 =...

it's not fixed though ? how is this any different to having four people arranged in a circle? instead, it's four colours arranged in a circle

ok.. suppose you had a cube with a blue top, and a red bottom. you then have four other colours to colour the other 4 sides with. in how many ways can this be done, so that no two cubes are alike?

have you done questions with people sitting in a circle before? consider the remaining 4 colours as being in a circle; so, by rotating the cube, even though the colours are now in different positions, it is still the same arrangement do a similar thing with the remaining 4 colours if you look...

differentiate the function, f(x) = sin x - x + (1/6)x^3 f'(x) = cosx - 1 + x^2/2 this is equal to zero when cosx = 0, so this is the minimum or maximum value f(x) in this range and f''(x) = -sinx + x > 0 for x >0 so this is the minimum value .'. f(x) > f(0) for x > 0 sin x...

maybe not, I didn't really give too much thought to the question. can you something wrong with it?

morbid angel would have been cool :(

if you look at, say, the red square, then there are 5 colours that could go on the opposite face of the cube. you now have 4 remaining faces, which form the equivalent of a circle, so there are 3! = 6 ways to paint these, so 5*6=30 ways

some stay dry but others feel the pain :(

wouldn't you at least want to try and find out what these states are (and thereby increase the probability of being right)? give that these states exist within our immediate perception, and they present themselves to us as good or bad, I think we can have a pretty good idea about what they are