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# STEP II 2012 discussion thread

1. (Original post by hassi94)
Hey - thought I might ask you:

I got the ln(1+x) one fully done and perfect as far as I can see so I'll say 20

20

As well as this; I did the above which you quoted so we're in agreement at about 10.

10 (30 running total)

I also did the first question (binomial) but as far as I remember I did the first bit, I did x^24 in the first bit but I can't remember if my method was correct or not (I assume so as I got the right answer and I'm pretty certain I didn't try just to 'fit in' an answer) - and then did x^25 right (though that was barely worth a mark probably). I got x^66 wrong because I summed all the triangular numbers required, rather than doing 2 times all but the last. I did however use the correct triangle numbers.

Say 13-17 depending on whether marker believes you understand exactly what you were doing (43 - 47 running total)

I also did the integral question where I got everything except for didn't finish the last part. I had pretty much gotten it at f(x+sqrt(x^2+1)) - in fact I had at one point, then got to and wrote 1/t^2 + 1/t^4, thought it was wrong and put a line through it so we'll say that I haven't but I got very close. Everything else was right.

Again say 13 - 17 (running total 56 - 64)

I also did the first part of the graph sketching question and worked out the stationary points. I did the b > a+2 (I think it was something like that) one where there were 3 alternating 'humps' but did not find the stationary points - it was correctly drawn. I didn't get onto the b = a + 2

Probably 5 - 7 for part one; say 5 for second part (running total 66 - 76)

Got through the very first part of 8 (blah - 2 > 0) and a tiny bit of the next part.

Not very much say 2 - 3 (running total 68 - 79)

Where do you think this lands me, roughly? I know this is a lot to ask so if you don't feel like putting the effort into this (perfectly understandable ) then just don't reply I feel like I'm on the wrong side of the 1/2 boundary but thought I'd get someone else's opinion.

Thanks!
See comments after each part - but who knows exactly what the markers think. Somewhere in range 68-79 ?!? At top end of range should be a 1 in most years; at the bottom end of the range possibly might be 50/50 depending on the overall level of the paper. This seems mixed from comments some people loved it, some hated it - too difficult to call.
2. I have a few questions regarding question marking, although I don't need to get a grade 1 I am curious as to what mark I got.

On question one, I have very clumsily found the general term of (1-x^6)^2 to be ((n+1)/n)x^6n, but then I've got the correct general term for the two other expansions, and I've got the correct formulae for the coefficient of x^n (except I've used n+1/n instead of just n+1). This yielded the wrong coefficient of x^24, and then on part (ii) I used a similar method but I didn't get 55. I then didn't attempt the remaining parts.

On Q3 I did all integrals except the last one and I nearly got it in the form f(x+root(x^2+1)), I didn't try to split it into two functions.

For question 4 I did parts (i) and (iii) correctly but I didn't get anywhere with (ii).

On 5 I did (i) correctly, and in (ii) I sketched both graphs correctly but without correct calculation of the stationary points.

On 6 I showed the expression for 16Q^2, then on question 9 I showed the first two inequalities.
3. I've estimated my score to be around 75-85, but it would be great to get some opinions I did:

Q1: Completed bar the x^66 coefficient I miss calculated to be 630 instead of the correct 650

Q2: Up to equating coefficients on ii) where I found my equations to be inconsistent, so I stopped

Q3: Completed

Q4: Completed bar the part in ii) where you have to consider (1/y) to show the required result

Q5: The first part fully but none of ii)

Q9: I got the first and second result fine, but I think with the second I was technically incorrect with my signs in the working
4. (Original post by desijut)
I posted this is in the STEP thread, i thought i'd post it here: Also, can anyone post a solution for Q1 and Q13?

Question 2 solution
th order

(i)
will be of order and is of order

since , will be of the highest order on the LHS, and the highest order of the LHS is clearly 1, so so

let

so

comparing coefficients:
(1)
(2)

(1)
gives 0 so factoring m-1 out gives

(2)

so c is a constant and or

or where c is a constant

(ii)... Got lazy, i solved this in the exam, couldnt be bothered to do it again

let

(same process as before)

and comparing coefficients

many equations later

and
or
and

or
I agree with your results..and just for completeness sake....
Attached Images
5. Step2012Paper2Question2.pdf (69.4 KB, 138 views)
6. (Original post by B Jack)
I've estimated my score to be around 75-85, but it would be great to get some opinions I did:

Q1: Completed bar the x^66 coefficient I miss calculated to be 630 instead of the correct 650

Q2: Up to equating coefficients on ii) where I found my equations to be inconsistent, so I stopped

Q3: Completed

Q4: Completed bar the part in ii) where you have to consider (1/y) to show the required result

Q5: The first part fully but none of ii)

Q9: I got the first and second result fine, but I think with the second I was technically incorrect with my signs in the working
My (somewhat conservative) take on this is mid to high 70s (73 to 78ish from what you've put). Of course it depends whether you've done anything that's not quite correct or anything that is good that you didn't pick up on and will be given credit for.

Sounds like it went pretty well, good luck for III.
7. Question 5 done
Attached Images
8. Step2012Paper2Question5.pdf (86.1 KB, 535 views)
9. Step2012Paper2Question5.pdf (92.2 KB, 336 views)
10. I found a quick way of doing Q1 (without some details). Basically, for the first part, you need to find coefficients in the expansion of (1-x^6)^-2 (1-x^3)^-1, so why not rewrite this as (1+x^3) (1-x^6)^-3 using difference of two squares? This gets only one infinite product, simplifying the algebra.

Likewise, for the next part, rewrite (1-x^6)^-2 (1-x^3)^-1 (1-x)^-1 as (1-x^6)^-4 P(x), where P(x) is (1+x+x^2+2x^3+2x^4+2x^5+x^6+x^7+ x^8).

Expanding the binomial is easy enough, and then it's a simple matter to extract the required coefficients at your leisure. Note that to work out coefft(x^66) I only needed to add (13 choose 3) and (14 choose 3), which is easy enough if you keep them split into prime factors.
11. (Original post by mikelbird)
Question 5 done
If you don't use numbers then the stationary points are:

b>a+2. -- (a+b)/2 i.e. mid-point as to be expected by symmetry; (a+b) +/- root((b-a)^2+4) all over 2. Symmetric about mid-point.

b = a+2 -- (a+1) +/- root (2). Symmetric about mid-point as expected.

12. (Original post by mikelbird)
I agree with your results..and just for completeness sake....
cheers, i was hoping someone would do this
13. Guys, I was wondering if you could give me some tips about taking step? Like how to start preparing for it? I will hopefully be taking it next year.
14. (Original post by Mr Dependable xD)
Guys, I was wondering if you could give me some tips about taking step? Like how to start preparing for it? I will hopefully be taking it next year.
You might start by going to these two threads.

http://www.thestudentroom.co.uk/show....php?t=1310974

http://www.thestudentroom.co.uk/show....php?t=1886802
15. (Original post by GreenLantern1)
I would say it was slightly easier personally. Though a lot of people seem to have struggled and I owuld imagine these are the ones that are restricting themselves to the pure questions!
I think there are a couple of people who have said it was slightly on the easy side but most seem to have found it quite hard
16. (Original post by Flibberdyjib)
I found a quick way of doing Q1 (without some details). Basically, for the first part, you need to find coefficients in the expansion of (1-x^6)^-2 (1-x^3)^-1, so why not rewrite this as (1+x^3) (1-x^6)^-3 using difference of two squares? This gets only one infinite product, simplifying the algebra.

Likewise, for the next part, rewrite (1-x^6)^-2 (1-x^3)^-1 (1-x)^-1 as (1-x^6)^-4 P(x), where P(x) is (1+x+x^2+2x^3+2x^4+2x^5+x^6+x^7+ x^8).

Expanding the binomial is easy enough, and then it's a simple matter to extract the required coefficients at your leisure. Note that to work out coefft(x^66) I only needed to add (13 choose 3) and (14 choose 3), which is easy enough if you keep them split into prime factors.
This is a nice solution but not something I would ever have come up with in the exam..If I can bang something out I usually just go with the 'brute force' method . Question 7 too has an elegant approach which I would never have looked for in an exam
17. (Original post by TheMagicMan)
I think there are a couple of people who have said it was slightly on the easy side but most seem to have found it quite hard
Yeh. I imagine the examiners are try to push people out of their comfor zone of just doin gPure questions. As personally the applied were very easy. Q12 really needed not much more than GCSE and the Mechanics questions included a load of show that's so you knew you had reached the right answer.
18. I'd consider myself an applied guy but, regardless, I thought this was a hard paper.
19. (Original post by ben-smith)
I'd consider myself an applied guy but, regardless, I thought this was a hard paper.
Agreed. I'm an applied guy too - Q9, Q11 were OK but Q10 took some slogging through. Q6 and Q8 did come out OK but from a first glance I didn't think it was exactly obvious how to solve them - with most questions I can 'see' the route to the end on just reading through.

The other pure questions seemed just to be grinding through - not easy to get out in half an hour without a mistake in the algebra somewhere.
20. (Original post by msmith2512)
Agreed. I'm an applied guy too - Q9, Q11 were OK but Q10 took some slogging through. Q6 and Q8 did come out OK but from a first glance I didn't think it was exactly obvious how to solve them - with most questions I can 'see' the route to the end on just reading through.

The other pure questions seemed just to be grinding through - not easy to get out in half an hour without a mistake in the algebra somewhere.
3 and 4 really weren't a grind at all...less than a page of A4 to solve each
21. (Original post by msmith2512)
Agreed. I'm an applied guy too - Q9, Q11 were OK but Q10 took some slogging through. Q6 and Q8 did come out OK but from a first glance I didn't think it was exactly obvious how to solve them - with most questions I can 'see' the route to the end on just reading through.

The other pure questions seemed just to be grinding through - not easy to get out in half an hour without a mistake in the algebra somewhere.
Whenever I see an angle of friction question in an exam where there is choice in the questions, I run a mile because 9 times out of 10 it is an absolute algebraic slog (relative to almost every other question on the paper). That was the case for Q10. Obviously, you could do it with resolving, taking moments etc etc but that won't be much easier...

Q9, 11 and 13 were friendly targets IMO; I thought the applied was pretty generous, to be fair.

(Original post by TheMagicMan)
3 and 4 really weren't a grind at all...less than a page of A4 to solve each
Agreed.
22. (Original post by TheMagicMan)
3 and 4 really weren't a grind at all...less than a page of A4 to solve each
I'd like to see q3 done in a page of A4 - must be small writing. The solution on this thread takes up 3 pages and plenty of scope for small errors to carry forward to make solutions impossible - with a need to go back, review and correct.
23. (Original post by TheMagicMan)
This is a nice solution but not something I would ever have come up with in the exam..If I can bang something out I usually just go with the 'brute force' method . Question 7 too has an elegant approach which I would never have looked for in an exam
Oh, what was the elegant approach to q7?

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