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* Honours Maths paper 1 * AFTERMATH
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Can somebody explain this bell curve concept to me?
Not the most helpful, but the best illustration of it on Google Images (that I could see). I'm not Jennie...
It means to keep things even keel, they will alter the marking schemes to make sure the majority of people will get C-B marks, and less get As, etc.. Supposedly it's to keep the college system from crippling, as one teacher put it.0 -
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Dunno about that now. The C part 2 of Q7 maybe (and that was grand - I found it far easier than the C part 1, you just had to understand your differenciation a bit, which an A1 student should be able to do).partyatmygaff wrote: »The SEC pulled a fast one on us. The project maths paper is stupidly easy in comparison to ours. Not only that but it looks as if that Q7 and Q8 was originally written for Project Maths
The Q8 C part, I dunno, is that on PM?
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It's not the question itself... It's the way they asked the question. If we had been asked if the curve was simply increasing or decreasing most people would have had no problem with that question. It was the fact they asked people to choose from three graphs while giving reasons that threw people off.Dunno about that now. The C part 2 of Q7 maybe (and that was grand - I found it far easier than the C part 1, you just had to understand your differenciation a bit, which an A1 student should be able to do).
The Q8 C part, I dunno, is that on PM?0 -
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Annd.. why did you have to do that? D=
I let U equal the top. And the bottom was half Du. so I got U over one.
Which was just U
Then I had U^2 over two. Ended up with 24 at the end.
I don't think that it works if du is on the bottom line. I have a friend who divided it by long division too, it works out the same as the way I said. Here's mine (bear in mind I could be wrong too):
1)
2)
3) Divide both by (x+1)
4) u = (x+1), therefore
therefore
du = dx
5) Sub in u for (x+1) and du for dx
And integrate and change limits from there.
I messed up because I left out the u by itself:(0 -
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}{x+1}&space;dx)
^2+1}{x+1}&space;dx)
&space;+&space;\frac{1}{x+1})dx)
&space;+&space;\frac{1}{u})du)
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