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STEP Prep Thread 2016 (Mark. II)

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Reply 1660
Original post by sweeneyrod
I did a similar thing in Q1, and I'm inclined to be a bit more generous, maybe 10-14 marks. I feel like the first three parts of Q6 will be worth more than 12 marks, though I'm not sure exactly what. And I think you'd lose at most 1 mark in Q3 for dubious justification (not that that will make much of a difference).


Ah, cheers. Hoping Q1 will be a bit generous, but I can't see the parts that I've done being worth very much. If only I'd solved that cubic properly, arghhhhh.

(at this point, 1 mark is going to make a lot of a difference :tongue: )

Thanks! :smile:
Original post by Zacken
x


For question 3, you don't need to investigate the possibility of repeated roots. f_n(x) and f'_n(x) can't both be 0 unless x = 0, and there is obviously no root there.
Original post by Remyxomatosis
Can an S in STEP 1 make up for a high 2 in STEP 2? That is, supposing that STEP 3 goes to plan...


Yes, I think you would be likely to get in with S, 2, 1. Around 75% of people with 2, 1 in II and III get in (according to the stats people posted earlier) and most of those people won't have an S in STEP I, so your chances are even higher.
Original post by Zacken
Ah, cheers. Hoping Q1 will be a bit generous, but I can't see the parts that I've done being worth very much. If only I'd solved that cubic properly, arghhhhh.

(at this point, 1 mark is going to make a lot of a difference :tongue: )

Thanks! :smile:


I've got an awful feeling that I didn't even state the cubic properly, I don't think I squared y. My thinking is that going from x and y to 4y^2 = 3x - 1 is just a bit of algebra, and even if you didn't have correct x and y you shouldn't lose too much for that.
Reply 1664
Original post by StrangeBanana
For question 3, you don't need to investigate the possibility of repeated roots. f_n(x) and f'_n(x) can't both be 0 unless x = 0, and there is obviously no root there.


Yeah, I reckon I needed to mention that.

Basically my justification went as such, as x tends to +- infinity, f_n(x) tends to +infinity since the x^n term dominates, so since there is at most one repeated root, f_n(x) can't possible have a root. But then I didn't mention that f_n(x) and its derivative can't possibly be 0 which might have been needed?
Original post by Zacken
Yeah, I reckon I needed to mention that.

Basically my justification went as such, as x tends to +- infinity, f_n(x) tends to +infinity since the x^n term dominates, so since there is at most one repeated root, f_n(x) can't possible have a root. But then I didn't mention that f_n(x) and its derivative can't possibly be 0 which might have been needed?


): damnit I'd wish I'd written it down; perhaps its implied by my sketch o:

I just said complex roots come in pairs, because the coefficients of the polynomial are real. so if there's at most 1 real root then for even n there's 0 and for odd n there's 1.
(edited 7 years ago)
Could anyone tell me roughly what I got? I was thinking maybe low 60's?

Q1 Full
Q3 showed up to f'(a)f'(b) < 0
Q5 did parts i, ii couldn't do iii
Q6 couldn't do last part but verified it for a few simple cases
Q7 made a few mistakes probably 14 in total
Q8 did sketch, showed first two approximations, made arithmetic slip showing 3rd, found the error in approximation but didn't get started on the last bit

Thanks, bit annoyed I didn't get enough time to finish Q3, ahh well, exam conditions
Reply 1667
Original post by StrangeBanana
): damnit I'd wish I'd written it down; perhaps its implied by my sketch o:

I just said complex roots come in pairs, because the coefficients of the polynomial are real. so if there's at most 1 real root then for even n there's 0 and for odd n there's 1.


That's elegant, does it nicely - reckon you don't need any additional justification for that. :biggrin:
Ok so just to make sure this is definitely a 1 or more:

Spoiler

Reply 1669
Original post by mpaterson
Could anyone tell me roughly what I got? I was thinking maybe low 60's?

Q1 Full
Q3 showed up to f'(a)f'(b) < 0
Q5 did parts i, ii couldn't do iii
Q6 couldn't do last part but verified it for a few simple cases
Q7 made a few mistakes probably 14 in total
Q8 did sketch, showed first two approximations, made arithmetic slip showing 3rd, found the error in approximation but didn't get started on the last bit

Thanks, bit annoyed I didn't get enough time to finish Q3, ahh well, exam conditions


20 + 14 + 15(?) + 13 + 14 + 10 = 85+
Original post by Zacken
Ah, a 1 is really good though. I had the same problem with not getting complete solutions, that last part of the DE question was pure death. Only got fulls to Q7 and 3 and then measly pathetic partials to 4 other questions.


I had a really pathetic partial to Q2 which is annoying. Probably like 6/7 marks idk, it was the whole first part but it seemed like there was a lot more left in that Q.

Later I'll post a brief overview of the Qs I attempted and hopefully someone can tell me approx what mark I got. But yeah my performance was pretty pathetic, I say 1 but idk how the marks work out exactly.

(I've got C4 now lol.)
(edited 7 years ago)
Reply 1671
Original post by EnglishMuon
Ok so just to make sure this is definitely a 1 or more:

Spoiler



19 + 17 + 11 + 19 + 18 + 14 = 98
Original post by Zacken
That's elegant, does it nicely - reckon you don't need any additional justification for that. :biggrin:


I smiled when I saw it ^^

But I think I still needed to talk about there being no repeated roots; I did say f'(x) > 0 or f'(x) < 0 for roots at one point, hopefully that's enough
In question 8 how many marks would finding the approximations (2,5/3,33/20) be worth? I managed to get it to the diophantine equation stage but due to exam mind blanking i couldnt find m and n for the required values of E. Also I dont understand how but I got 1.42 for the last bit instead of 1.08...
Is it wrong to say that E=integral from 1/2 to infinity - sum from 1 to infinity (1/4r^2)?
(edited 7 years ago)
Original post by computerkid
Could anyone possibly tell me how close is this above/below a 1?

Spoiler



@Zacken sorry to bother you marking all these people's, but could you look at what I might have gotten for this?
@Zacken , if my reasoning isn't very right or at least insufficient in a mechanics show that question (I'm talking about Q10), but I got the result they wanted with proper equations, how many marks would I lose?
Reply 1676
Original post by StrangeBanana
I smiled when I saw it ^^

But I think I still needed to talk about there being no repeated roots; I did say f'(x) > 0 or f'(x) < 0 for roots at one point, hopefully that's enough


How'd you do the second bit about a < 0 is a is a real root? I said that for x > 0, f_(n-1) was obviously > 0, so derivative was > 0, so f_n(x) was increasing and f(0) = 1, so f_n(x) > 1, so the only real root could happen for x < 0.
Reply 1677
Original post by computerkid
@Zacken sorry to bother you marking all these people's, but could you look at what I might have gotten for this?


19 + 4 + 3 + 13 + 20 + 20 = 79

Original post by gagafacea1
@Zacken , if my reasoning isn't very right or at least insufficient in a mechanics show that question (I'm talking about Q10), but I got the result they wanted with proper equations, how many marks would I lose?


Not very much, maybe 4?
Original post by Zacken
How'd you do the second bit about a < 0 is a is a real root? I said that for x > 0, f_(n-1) was obviously > 0, so derivative was > 0, so f_n(x) was increasing and f(0) = 1, so f_n(x) > 1, so the only real root could happen for x < 0.


I said that fn(a)= sum of even powers of a with +ve coefficients + sum of odd powers of a with +ve coefficients =0
so since even powers > 0 the sum of odd powers <0, then factorised the sum of odd powers as a*(sum of even powers) < 0 so a<0
Original post by Zacken


Not very much, maybe 4?


Thanks!

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