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# STEP Prep Thread 2016 (Mark. II)

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1. (Original post by raff97)
Thanks, I did try going back a few pages and couldnt find it
No worries! Edited my post to include a link to the solutions thread if you want to chip in.

(have you finished IA? How was it? )
2. (Original post by Zacken)
Correct me if I'm wrong but Q9's second part was very small, no? Something about changing a positive to a negative?

Maybe 16 marks is fair.
Yes it was, but I think the forces have to be resolved all over again, since the frictional forces change directions, but the idea is the same as part i.
3. (Original post by Zacken)
There wasn't anything clever but from what I've seen of markschemes, they have to reward work and there was a fair bit of work done in the first part, I can see lots of (M1 M1 A1) for each of n=1, 2, 3 and then M1A1 for n=4, totalling 11 marks.
Cool well I hope your right, that would make my half attempt at that question seem slightly more worth it
4. Q3, II , 2006
The last part, how can the result of 2N-1<N+sqroot(N^2-1)<2N be deduced?
5. (Original post by Zacken)
Very much agree. Could be a challenger for 2010 or 2013 as 'easiest paper'. Pity I didn't do as well as I should've!
How well did you do? PM if you want
6. (Original post by shamika)
How well did you do? PM if you want
Spoiler:
Show
This is what I think:

Q1 - not sure if I showed it didn't hold for n=4 convincingly and left my answer for (ii)(b) as ((7^7 +)^3 + 7's)((7^7 + 1)^3 - 7's) so probably dropped 3 or so marks here. 17/20

Q2 - full. 19/20Q3 - didn't realise sin(-4) was negative, so probs cut a mark here. 18/20

Q4 - did some weird stuff? I differentiate first bit correctly, managed to get up to v/sqrt(v^2 + 1) = kx + c where v = f ' (x) by using the first part and then re-arranged for sqrt(v^2 + 1), cubed it to get (v^2 +1)^(3/2) and then plugged it back into the original equation to get f '' (x) / v^3 = k / (kx+c)^3 and then integrated both sides.

Not sure how much to give myself for that? 10 marks?

Q10 - full except that for showing e < 1/3, I made a teeny slip, so cut a mark. 19/20

Q11 - did the first bit about getting the equation, found the maximum value via differentiation, plugged it back into the formula but along the way miscopied from one line to the next and hence didn't get a correct quadratic. (solved the incorrect quadratic as well) but then moved on and got the distance at the last part to be h/cos 2alpha but couldn't plug in my cos 2alpha since I'd got the incorrect version.

Probably 11 marks.

That totals 94... a middling 1?
7. (Original post by Zacken)
No worries! Edited my post to include a link to the solutions thread if you want to chip in.

(have you finished IA? How was it? )
Seems all the pure questions have been done, was never a fan of applied so cant help there. sorry!

(Yeah I have. Was alright I guess, will find out how I did on Saturday... arguable easier than STEP II and III if you put the work in throughout the year. certainly less interesting questions compared to STEP!)
8. So tomorrow do I focus on 2-3 questions the whole time or shall I just poke at questions till I get stuck then move on so I get lots of partial solutions?
9. (Original post by raff97)
Yeah I have. Was alright I guess, will find out how I did on Saturday... arguable easier than STEP II and III if you put the work in throughout the year. certainly less interesting questions compared to STEP!
Wow, You must be nervous. Hope all goes well!
10. (Original post by Geraer100)
Q3, II , 2006
The last part, how can the result of 2N-1<N+sqroot(N^2-1)<2N be deduced?
(N-1)^2 < N^2-1 < N^2, so for N>1...
11. (Original post by Zacken)
Wow, You must be nervous. Hope all goes well!
Cheers! Good luck with STEP II tomorrow and III later on. Looks like you did well in I so you should be fine.
12. Can anyone think of a function such that ? I can't, not sure if I'm missing something obvious or what.
EDIT: I just realised there was some IMO question about iterating polynomials in this way, interesting...
EDIT 2: Except the trivial solutions.
13. (Original post by Farhan.Hanif93)
(N-1)^2 < N^2-1 < N^2, so for N>1...
Thanks!
Sorry missed out the 1/2, It should be 2N-1/2<N+sqroot(N^2-1)<2N
14. (Original post by IrrationalRoot)
Can anyone think of a function such that i.e. a function that is cyclic with period 3? I can't, not sure if I'm missing something obvious or what.
EDIT: I just realised there was some IMO question about iterating polynomials in this way, interesting...
f(x) = x
(possibly not what you want)
15. (Original post by Geraer100)
Thanks!
Sorry missed out the 1/2, It should be 2N-1/2<N+sqroot(N^2-1)<2N
The LHS of that statement is untrue - N=1 shows you that.
16. (Original post by IrrationalRoot)
Can anyone think of a function such that i.e. a function that is cyclic with period 3? I can't, not sure if I'm missing something obvious or what.
EDIT: I just realised there was some IMO question about iterating polynomials in this way, interesting...
Surely if f(x) = 1/x then f(f(f(x))) = 1/x ? Or do you mean something else?
17. (Original post by ComputerMaths97)
So tomorrow do I focus on 2-3 questions the whole time or shall I just poke at questions till I get stuck then move on so I get lots of partial solutions?
The recommended advice is always to focus on getting complete solutions rather than doing lots of separate parts of questions
18. (Original post by smartalan73)
The recommended advice is always to focus on getting complete solutions rather than doing lots of separate parts of questions
Even if you end up with less marks?
19. (Original post by ComputerMaths97)
Surely if f(x) = 1/x then f(f(f(x))) = 1/x ? Or do you mean something else?
Sorry, 'cyclic with period 3' is not the same thing as what I meant lol.
I meant .
So I guess that means cyclic with period 4.
20. (Original post by ComputerMaths97)
Surely if f(x) = 1/x then f(f(f(x))) = 1/x ? Or do you mean something else?
that's period 2

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