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STEP I 1997 Help

Hello guys, if possible I was wondering someone could offer me a hint on question 5 on STEP I 1997 paper. I firstly do not understand the question, does the modulus symbol imply the absolute value or does it imply the magnitude( the length) of the rod. Thanks a lot.
Reply 1
Original post by TheEpiphany
Hello guys, if possible I was wondering someone could offer me a hint on question 5 on STEP I 1997 paper. I firstly do not understand the question, does the modulus symbol imply the absolute value or does it imply the magnitude( the length) of the rod. Thanks a lot.


I've just looked at my PDF copy of that paper and unless I'm going mad it explicity tells you that |PR| denotes the length of PR etc :smile:
Reply 2
Original post by davros
I've just looked at my PDF copy of that paper and unless I'm going mad it explicity tells you that |PR| denotes the length of PR etc :smile:


It mentions that for the next stage in the question in brackets, so do I assume the same for the first part of the question as well ?
Reply 3
Original post by TheEpiphany
It mentions that for the next stage in the question in brackets, so do I assume the same for the first part of the question as well ?


Yes, I think the comment is meant to apply to the whole question - they just haven't laid it out as clearly as they could!
Reply 4
Original post by davros
Yes, I think the comment is meant to apply to the whole question - they just haven't laid it out as clearly as they could!


Thanks a lot. Stuck on question 7 on the same paper :/
Reply 5
Original post by TheEpiphany
Thanks a lot. Stuck on question 7 on the same paper :/


Is this the integration of a cubic polynomial question?

I did this a while ago and from what I remember it's just a brute-force algebra slog!

I'll be offline soon, but post your working if still stuck and someone will take a look :smile:
Reply 6
Original post by davros
Is this the integration of a cubic polynomial question?

I did this a while ago and from what I remember it's just a brute-force algebra slog!

I'll be offline soon, but post your working if still stuck and someone will take a look :smile:


I just have no idea how to approach it, I am going to try specific cases and see if I spot a pattern or something. This is kinda hard, I seriously gotta get better at this sort of question.
Reply 7
Original post by TheEpiphany
I just have no idea how to approach it, I am going to try specific cases and see if I spot a pattern or something. This is kinda hard, I seriously gotta get better at this sort of question.


Don't you just take a 'typical' cubic polynomial, say px^3 + qx^2 + rx + s, bung it into the integral, see what comes out and solve some simultaneous equations or get some retrictions on the coefficients by comparing with the other expression they give you for the result? If I'm remembering correctly, the integration goes from -1 to +1 (or something similar) so odd functions (like x) just integrate to 0 across the range!
Reply 8
Original post by davros
Don't you just take a 'typical' cubic polynomial, say px^3 + qx^2 + rx + s, bung it into the integral, see what comes out and solve some simultaneous equations or get some retrictions on the coefficients by comparing with the other expression they give you for the result? If I'm remembering correctly, the integration goes from -1 to +1 (or something similar) so odd functions (like x) just integrate to 0 across the range!


That's probably the best idea, though on 'nice' way I've seen involves realising that it must work for y=1, y=x, y=x^2, y=x^3 and working from there, which gets rid of some algebra. I did the first the method and it worked out fine.

This is probably my favourite pure section from step 1 - a lot of nice questions. I hate the coin counting one though.


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Reply 9
Original post by davros
Don't you just take a 'typical' cubic polynomial, say px^3 + qx^2 + rx + s, bung it into the integral, see what comes out and solve some simultaneous equations or get some retrictions on the coefficients by comparing with the other expression they give you for the result? If I'm remembering correctly, the integration goes from -1 to +1 (or something similar) so odd functions (like x) just integrate to 0 across the range!


Completely agree man, especially the question with pi and 22/7, that question was beautiful. Thanks for the hint, I am going to try using that method. I should have realised that a general form of cubic equation should be used. But i thought i would end up with far too many variables and would confuse myself or suthin :redface:. I have started prep far too early i think, i mean i haven't even done my AS exams yet :/

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