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# STEP 2016 Solutions watch

1. (Original post by Ewanclementson)
For the f(x), g(x), h(x) would an acceptable answer to the final part be the observation that it can be arranged to have (h(x)+x)+(h(1/1-x)+(1/1-x))=1 and so each of these pairs of terms are 'something' +the function of that 'something' and so it is clear that h(x)=1/2-x would be the solution as the -x part would cancel the 'something' and then the two +1/2 would add to give the one. Note: the 'something' is a function of x
No, that is not showing that it is the only solution.
2. (Original post by Mathemagicien)
STEP III Question 13
Is this a valid method for part i?

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3. (Original post by gasfxekl)
For the last part shouldnt you also say that p1,2m+2=p1,2m+1?
Yes. Well spotted.
4. It was a day filled with mixed emotions. I got 3 full solutions in the first 90 mins then couldn't do much for the last 90 mins.
Score predictions anyone?
Q1 - Full
Q2 - Did the first part, said that the first part also holds for r and formed equations for OP and QR and attempted to solve for intersection points. Could not proceed through that though.
Q4 - Full except for the last part where I did not simplify and let the final answer be 2 (sech y + whatever the ans in 2nd to last part)
Q5 - Did the first part, Claimed that (2m+1)! covers Pm+1, 2m+1 and the remaining primes will cancel with m! for the 2nd part. 3rd part I stated that it holds when 2m+1 is composite and couldn't proceed for prime. 4th part I attempted induction with even n (wasn;t successful though) and stated that if it is true for even m then it is true for m+1.
Q8 - Full.
Q6 - Had 10 minutes left when I began this. Stated the values of R and gamma, and used the formula for arctanh for the conditions for A and B. I then solved for the intersection point, obtaining the result as the instructor said 'STOP WRITING'.
5. (Original post by Devinky)
It was a day filled with mixed emotions. I got 3 full solutions in the first 90 mins then couldn't do much for the last 90 mins.
Score predictions anyone?
Q1 - Full
Q2 - Did the first part, said that the first part also holds for r and formed equations for OP and QR and attempted to solve for intersection points. Could not proceed through that though.
Q4 - Full except for the last part where I did not simplify and let the final answer be 2 (sech y + whatever the ans in 2nd to last part)
Q5 - Did the first part, Claimed that (2m+1)! covers Pm+1, 2m+1 and the remaining primes will cancel with m! for the 2nd part. 3rd part I stated that it holds when 2m+1 is composite and couldn't proceed for prime. 4th part I attempted induction with even n (wasn;t successful though) and stated that if it is true for even m then it is true for m+1.
Q8 - Full.
Q6 - Had 10 minutes left when I began this. Stated the values of R and gamma, and used the formula for arctanh for the conditions for A and B. I then solved for the intersection point, obtaining the result as the instructor said 'STOP WRITING'.
Your in the 70/80s.

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6. (Original post by Mathemagicien)
I feel like there should be a new thread for all these "estimate my score" posts, it'd clutter this place less, and you could compare to other people easier
its called the step prep thread
7. Hi guys I can't find a solution to 7 on here, so here's mine. This question seems surprisingly unpopular. Anyone want to guess how many an answer with half of part (i) missing and all else full (but maybe not 100% rigorous) would be worth?
Attached Images
8. STEP III 2016 Q7.compressed.pdf (731.3 KB, 109 views)
9. Anyone up for estimating a mark scheme collaboratively? Like splitting the marks up for each part and then if people want to go even more detailed where they think method marks would be etc. For the part breakdowns we could just do it on a shared google docs or something.

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10. (Original post by jjsnyder)
Anyone up for estimating a mark scheme collaboratively? Like splitting the marks up for each part and then if people want to go even more detailed where they think method marks would be etc. For the part breakdowns we could just do it on a shared google docs or something.

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I suppose if enough people think it's a good idea, then I don't mind helping out with it after exams. (So after Tuesday)
11. There inherent problem with a "mark scheme" is that so many questions genuinely can be done in numerous different ways - as evidenced here, with quite a number of the solutions being different to my own.

I'm sure we can get out an unofficial suggested mark scheme, but I'd also worry that would attract more questions. I also worry about telling someone they got a 1 (with 72 marks), only for it to transpire that they did worse than this. We might insight a ridiculous number of official mark appeals.
12. (Original post by jjsnyder)
Anyone up for estimating a mark scheme collaboratively? Like splitting the marks up for each part and then if people want to go even more detailed where they think method marks would be etc. For the part breakdowns we could just do it on a shared google docs or something.

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(Original post by Zacken)
I suppose if enough people think it's a good idea, then I don't mind helping out with it after exams. (So after Tuesday)
Its a brilliant idea!!!!!!!!!!!!!! Would love to see it.
13. (Original post by Llewellyn)
There inherent problem with a "mark scheme" is that so many questions genuinely can be done in numerous different ways - as evidenced here, with quite a number of the solutions being different to my own.

I'm sure we can get out an unofficial suggested mark scheme, but I'd also worry that would attract more questions. I also worry about telling someone they got a 1 (with 72 marks), only for it to transpire that they did worse than this. We might insight a ridiculous number of official mark appeals.
That is a good point.

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14. Well, actually, I think it could maybe be done.

https://docs.google.com/document/d/1...it?usp=sharing

Very tricky to know the exact mark allocation... I'm tempted to reduce marks by line for each part and add an overall "get the gist of the question" or "bonus mark(s) at the end for getting a full solution"

I'll add in the rest of the pure section later. Anyone else free to edit/ improve.
15. (Original post by Llewellyn)
Well, actually, I think it could maybe be done.

https://docs.google.com/document/d/1...it?usp=sharing

Very tricky to know the exact mark allocation... I'm tempted to reduce marks by line for each part and add an overall "get the gist of the question" or "bonus mark(s) at the end for getting a full solution"

I'll add in the rest of the pure section later. Anyone else free to edit/ improve.
ur ****in amazing !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! thnx
16. hey guys, got q10 - should i post it - has it already been done yet?
17. (Original post by Mathemagicien)
Don't think anyone has yet... go for it
cool
may not be the most clear
Attached Images

18. (Original post by Farhan.Hanif93)
STEP II Q11 - See attached.
I think your analysis of the last part is incorrect - see attached.
Attached Images
19. Step2016Paper2Question11.pdf (75.5 KB, 74 views)
20. (Original post by mikelbird)
I think your analysis of the last part is incorrect - see attached.
Would you like to point out the mistake in my reasoning? Apart from perhaps elaborating that the intersection of the two lines must take place in the upper right quadrant, I see no issue with my solution as it's only interested in a necessary condition and indeed references the system of part (i) (and gives notably more of an "explanation" than a proof, as requested by the question).

It's clear that the particles in part (ii) intersect at the same time as those in part (i). The cartesian equations of the trajectories from part (i) are correct. And a collision occurs iff the two lines intersect in the upper right quadrant. It is therefore necessary that the gradient of the "initially lower" line (i.e. the line of particle ) must exceed that of the "initially higher" line (i.e. the line of particle ) in order to catch up, hence leading to my conclusion.

I'm certain that it is correct, as is your alternative.
21. (Original post by Farhan.Hanif93)
Would you like to point out the mistake in my reasoning? Apart from perhaps elaborating that the intersection of the two lines must take place in the upper right quadrant, I see no issue with my solution as it's only interested in a necessary condition and indeed references the system of part (i) (and gives notably more of an "explanation" than a proof, as requested by the question).

It's clear that the particles in part (ii) intersect at the same time as those in part (i). The cartesian equations of the trajectories from part (i) are correct. And a collision occurs iff the two lines intersect in the upper right quadrant. It is therefore necessary that the gradient of the "initially lower" line (i.e. the line of particle ) must exceed that of the "initially higher" line (i.e. the line of particle ) in order to catch up, hence leading to my conclusion.

I'm certain that it is correct, as is your alternative.
Perhaps I am not understanding what you mean by cartesian trajectories...the actual trajectories are certainly are not straight lines and so I feel further explanation is needed. I suppose you are making the point about the fact that gravity seems to disappear in the second part when establishing the relationships with regard to theta but then you are not really dealing with a cartesian graph anymore (or the rescaling of the height coordinate needs some explanation).
22. (Original post by mikelbird)
Perhaps I am not understanding what you mean by cartesian trajectories...the actual trajectories are certainly are not straight lines and so I feel further explanation is needed. I suppose you are making the point about the fact that gravity seems to disappear in the second part when establishing the relationships with regard to theta but then you are not really dealing with a cartesian graph anymore (or the rescaling of the height coordinate needs some explanation).
Isn't the point that the "gravitational part" of their trajectories (so to speak) is the same, so you can just forget about it? It's like having the same term present on two sides of an equation, allowing you to cancel.

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