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Q16 (1 Viewer)

lolcakes52

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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?
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.
 

Nooblet94

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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.
 

Sup3rDry

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Hey guys the question asked to find the limit of something i can't remember the one that dealt with inverse tan after the induction one. I got the answer as inverse tan (1), but i just put it like that and forgot to evaluate it as pi on 4. Do you think I got the mark? :(
 

seanieg89

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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.
 

SoresuMakashi

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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, sqrt(n) > k - (1/2). Now we want the largest k for which k < sqrt(n) + (1/2). Thus, k is the closest integer to sqrt(n).

Edit: Fixed incorrect sign, thanks to seanieg89 below.
 
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seanieg89

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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).
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.

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.
 

SoresuMakashi

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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.
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.

Also, Pure Mathematics PhD @ ANU.

So. Jealous.
 
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seanieg89

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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.
Yep, this is always the case. Luckily not everything in life is so "against the clock".
 

Trebla

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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?
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.
 
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deswa1

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This restriction is there to ensure that the combined inverse tangent still lies between -pi/2 and pi/2
Yeah exactly- I mentioned this. My question is though- if you didn't, do you still get the mark?
 

Trebla

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Yeah exactly- I mentioned this. My question is though- if you didn't, do you still get the mark?
Doubt you would lose a mark. It's just a technicality that you are addressing to ensure rigour, not part of the proof itself.
 

NickGero

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If 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?
 
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DAFUQ

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for q16ciii just do P(K+1)>=P(K) and solve. u'll come up with the required equation
 

Sindivyn

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If 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?
Probably 2/3
Generally the marking criteria is 1 mark for proving true for the base case, then 2 marks for the algebraic part. That being said, you have to make significant progress in the algebraic part.
 

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