seanieg89
Well-Known Member
- Joined
- Aug 8, 2006
- Messages
- 2,671
- Gender
- Male
- HSC
- 2007
Ah right, yep. Good thing Carrotsticks asked it then!The one where you had to do arrangements of the word holomorphic, where you had to have no vowels together.
Ah right, yep. Good thing Carrotsticks asked it then!The one where you had to do arrangements of the word holomorphic, where you had to have no vowels together.
well the angle between two lines formula is an absolute value, so you have to justify it's removal by saying that the denominator was positive. So maybe lose a mark.Can someone that knows how they mark explain this. For 16b) i), it gave that |x|, |y| <1. So obviously you're meant to deal with this restriction when you take the tan inverse etc. and get +kpi to show that k=0 and hence the result follows so I did that. My question is though, its only one mark- if I didn't deal with the restriction, would I get 0/1 or 1/1?
Fuck.Can someone that knows how they mark explain this. For 16b) i), it gave that |x|, |y| <1. So obviously you're meant to deal with this restriction when you take the tan inverse etc. and get +kpi to show that k=0 and hence the result follows so I did that. My question is though, its only one mark- if I didn't deal with the restriction, would I get 0/1 or 1/1?
I would think 0, but you will only know when BoS release their marking centre notes. These are the sort of pitfalls in Q15/16 I have been talking about, they are a lot sneakier than simply having a complicated integration by parts that you either get or you don't.Can someone that knows how they mark explain this. For 16b) i), it gave that |x|, |y| <1. So obviously you're meant to deal with this restriction when you take the tan inverse etc. and get +kpi to show that k=0 and hence the result follows so I did that. My question is though, its only one mark- if I didn't deal with the restriction, would I get 0/1 or 1/1?
If P(k) < P(k-1) how can P(k) be the maximum of P? Sign issue. Also I am not sure how pedantic they are going to be about the 'closest integer' issue.Anyone get 16)c)iii)?
For 16)c)iv):
P(k) is greatest when k is the largest integer that satisfies P(k)<P(k-1) i.e. the largest k for which k^2-k-n>0. If k^2-k-n>0, then the postulate for part (iii) is correct is true and thus, root(n) > k - (1/2). Now we want the largest k for which k < root(n) + (1/2). Thus, k is the closest integer to root(n).
Great, thanks. I actually didn't get iii) or iv) in the test - the ideas were all there but I couldn't put them together coherently enough. Everything always seems to crystallise after the test is over, when your mind isn't racing.For ciii) The thing you are given is equivalent to n > k^2 - k after squaring and simplifying. But if one integer is bigger than another, it must be bigger by at least one! So we can add 1/4 to the RHS whilst preserving the inequality.
This gives n > k(k-1) +1/4 = (k-1/2)^2. Now take square roots of both sides, observing that they are both positive by conditions on n and k.
Yep, this is always the case. Luckily not everything in life is so "against the clock".Great, thanks. I actually didn't get iii) or iv) in the test - the ideas were all there but I couldn't put them together coherently enough. Everything always seems to crystallise after the test is over, when your mind isn't racing.
This restriction is there to ensure that the combined inverse tangent still lies between -pi/2 and pi/2. It is possible that the sum of the inverse tangents could end up above pi/2 for example, so this restriction takes care of that problem.Can someone that knows how they mark explain this. For 16b) i), it gave that |x|, |y| <1. So obviously you're meant to deal with this restriction when you take the tan inverse etc. and get +kpi to show that k=0 and hence the result follows so I did that. My question is though, its only one mark- if I didn't deal with the restriction, would I get 0/1 or 1/1?
Yeah exactly- I mentioned this. My question is though- if you didn't, do you still get the mark?This restriction is there to ensure that the combined inverse tangent still lies between -pi/2 and pi/2
Doubt you would lose a mark. It's just a technicality that you are addressing to ensure rigour, not part of the proof itself.Yeah exactly- I mentioned this. My question is though- if you didn't, do you still get the mark?
Probably 2/3If I got very close to proving the k + 1 step for the induction proof, just missing out on fixing up the algebra, do you think I'd get 1 or 2 out of 3?
by which I mean, I'd used assumption, had (k + 1)/(k + 2) on bottom but the polynomials they were multiplied by were not the same...
Also for ciii) I had squared both sides and made a decent attempt at getting sqrt(n) on one side, but just didn't realise you could add the quarter to it and have the inequality still hold, I'd say 1/2? Surely that's where most people got to, what does everyone else think?