STEP III 2012 Discussion Thread

Maths exam discussion - share revision tips in preparation for GCSE, A Level and other maths exams and discuss how they went afterwards.

Announcements Posted on
Important: please read these guidelines before posting about exams on The Student Room 28-04-2013
Search Thread
Advanced Search
Sign in to Reply
  1. MrDD's Avatar
    81 29-06-2012  00:39
    • Junior Member
    • Posts: 31
    Re: STEP III 2012 Discussion Thread
    (Original post by Alexxh)
    Done question 1 completely, (also got e^something for the second part :P)
    oh and I don't think any information was missing! Although the phrasing of the question was certainly awkward...
    I'm sure something was missing in Q1. The first expression is referred to as a 'result', and therefore needs something preceding it for it to have resulted from.

    I'm also fairly sure that the missing first part was something very similar to

    If z=(y^n) (dy/dx)^2, show that

    The question is then about differential equations in y and x, for which we are using z as a substitution to help solve them and reach the form y=f(x)

    for part (ii), i got the solution y=e^(0.5x^2) using the substitution

    z=(1/y) (dy/dx)
  2. MrDD's Avatar
    82 29-06-2012  00:49
    • Junior Member
    • Posts: 31
    Re: STEP III 2012 Discussion Thread
    (Original post by Alexxh)
    I think this is how I did part i) of question 1)
    Does that seem right?


    Solution to assumed version of Q1
    Attached Files
  3. File Type: pdf 2012-III-1.pdf (202.3 KB, 451 views)
    4. Last edited by MrDD; 30-06-2012 at 00:53.
  4. Euclid's Avatar
    83 29-06-2012  03:28
    • Respected Member
    Re: STEP III 2012 Discussion Thread
    (Original post by MrDD)
    yeuk

    i'll post my solution again since people seem to be missing it

    Solution to assumed version of Q1
    Why do you make the assumption z=y^n(\frac{dy}{dx})^2? It follows from integrating the given expression wrt x;
    (n=1)
    \frac{dz}{dx}=\frac{dy}{dx}[(\frac{dy}{dx})^{2}+2y\frac{d^{2 }y}{dx^{2}}]... integrating by parts.... \int(\frac{dy}{dx})^{3}{dx}={y}( \frac{dy}{dx})^{2}-\int{2y}\frac{dy}{dx}\frac{d^{2} y}{dx^{2}}{dx}...so z=y(\frac{dy}{dx})^2.
  5. MrDD's Avatar
    84 29-06-2012  06:55
    • Junior Member
    • Posts: 31
    Re: STEP III 2012 Discussion Thread
    (Original post by Euclid)
    Why do you make the assumption z=y^n(\frac{dy}{dx})^2? It follows from integrating the given expression wrt x;
    (n=1)
    \frac{dz}{dx}=\frac{dy}{dx}[(\frac{dy}{dx})^{2}+2y\frac{d^{2 }y}{dx^{2}}]... integrating by parts.... \int(\frac{dy}{dx})^{3}{dx}={y}( \frac{dy}{dx})^{2}-\int{2y}\frac{dy}{dx}\frac{d^{2} y}{dx^{2}}{dx}...so z=y(\frac{dy}{dx})^2.
    just that i think that was intended to be a part of the question, but something went wrong at the printing stage

    i've emailed them about this question, so hopefully we'll soon find out what was intended
  6. mikelbird's Avatar
    85 29-06-2012  07:24
    • Respected Member
    • Location: Gloucestershire
    • Posts: 208
    Re: STEP III 2012 Discussion Thread
    Answers to 2 and 5
    Attached Files
  7. File Type: pdf Step2012Paper3Question2.pdf (59.0 KB, 264 views)
  8. File Type: pdf Step2012Paper3Question5.pdf (67.2 KB, 116 views)
    9. Post rating:
    2
  9. TheMagicMan's Avatar
    86 29-06-2012  07:28
    • Overlord in Training
    • Posts: 3,017
    (Original post by Euclid)
    Why do you make the assumption z=y^n(\frac{dy}{dx})^2? It follows from integrating the given expression wrt x;
    (n=1)
    \frac{dz}{dx}=\frac{dy}{dx}[(\frac{dy}{dx})^{2}+2y\frac{d^{2 }y}{dx^{2}}]... integrating by parts.... \int(\frac{dy}{dx})^{3}{dx}={y}( \frac{dy}{dx})^{2}-\int{2y}\frac{dy}{dx}\frac{d^{2} y}{dx^{2}}{dx}...so z=y(\frac{dy}{dx})^2.
    The reason that people think that there was a bit missing is not because the question was too hard but rather because the question was very blunt: they introduced z with no explanation, they didn't ask you to do anything with the first equation etc.


    This was posted from The Student Room's iPhone/iPad App
  10. LucasJ94's Avatar
    87 29-06-2012  07:49
    • New Member
    • Posts: 6
    Re: STEP III 2012 Discussion Thread
    (Original post by LucasJ94)
    Felt awful walking out of the exam, I normally like STEP III papers, but I'm good at calculus and the big trig questions, which there was none of :/.

    I just want to stop thinking about this paper now, so can someone let me know how they think I did so I can have a bit of closure.

    I only did 5 questions :/.

    1 - Used n=1 and n=-2 on each part. And I built a relationship between z and y using the result they gave at the top. But I couldn't solve the equation I got for z in x, for either case. So barely any real progress.

    3 - Drew all graphs, got up to the final show that equation, and couldn't do that, didn't state relevance to initial graphs.

    5 - Completed.

    7 - Up to the final summation, think my result for Z2(t) was wrong (looked a little too elaborate)

    8 - Completed, but unsure of my result for the part about there being 2 values. I just found the second value and stated conditions on n, without any real proof or complete method.

    How bad's the damage?
    I don't like repeating myself, especially on forums. But I had another night of bad sleep due to this, it's eating up my time even now when I'm finished with it.

    Can anyone give me some peace of mind? Whether it's a 3, 2 or 1, I just want to have a guide as to what I can expect so that hopefully I can stop caring until the day of results.

    Thank you.
  11. MrDD's Avatar
    88 29-06-2012  07:54
    • Junior Member
    • Posts: 31
    Re: STEP III 2012 Discussion Thread
    (Original post by mikelbird)
    Answers to 2 and 5
    i don't think the last part of 5 is quite right, if you have SQRT(6) instead of SQRT(2) then you have a 6-rational point, not a 2-rational point.

    Also, I think your second squared bracket should have a + in the middle, otherwise your sinhxcoshx terms don't cancel.

    i think expanding (3coshx + SQRT(2)sinhx)^2 - (3sinhx + SQRT(2)coshx)^2 might work, although in the exam i used sec and tan instead of cosh and sinh
    Post rating:
    1
  12. fruktas's Avatar
    89 29-06-2012  07:56
    • Benevolent Member
    • Location: London
    • Posts: 713
    Re: STEP III 2012 Discussion Thread
    (Original post by Rahul.S)
    induction ofc could work....I decided to do a direct proof instead. They didnt mention a specific method so its fine.

    your AEA should be S standard :lol:

    oh yh how did your bredwin do?
    AEA won't matter if Cambridge accept me I so hope I have done enough! Safmaster did very similar to me
  13. fruktas's Avatar
    90 29-06-2012  07:58
    • Benevolent Member
    • Location: London
    • Posts: 713
    Re: STEP III 2012 Discussion Thread
    (Original post by hassi94)
    I had to come back to say I did exact this Though I feel I dud well enough even if I get 0 on that one
    they better award us somehow, I wasted a lot of time on that question But tbh, I think I have done enough for a 1 without it, so fingers crossed!
  14. mikelbird's Avatar
    91 29-06-2012  08:27
    • Respected Member
    • Location: Gloucestershire
    • Posts: 208
    Re: STEP III 2012 Discussion Thread
    (Original post by MrDD)
    i don't think the last part of 5 is quite right, if you have SQRT(6) instead of SQRT(2) then you have a 6-rational point, not a 2-rational point.

    Also, I think your second squared bracket should have a + in the middle, otherwise your sinhxcoshx terms don't cancel.

    i think expanding (3coshx + SQRT(2)sinhx)^2 - (3sinhx + SQRT(2)coshx)^2 might work, although in the exam i used sec and tan instead of cosh and sinh
    ooop!! you are quite right!!
  15. desijut's Avatar
    92 29-06-2012  08:30
    • Peer Of The TSR Realm
    • Posts: 1,578
    Re: STEP III 2012 Discussion Thread
    (Original post by MrDD)
    yeuk

    i'll post my solution again since people seem to be missing it

    Solution to assumed version of Q1
    Haha, got e^(^x^)^2 by accident, 19 marks?
    Last edited by desijut; 29-06-2012 at 08:31.
  16. desijut's Avatar
    93 29-06-2012  08:33
    • Peer Of The TSR Realm
    • Posts: 1,578
    Re: STEP III 2012 Discussion Thread
    (Original post by Rahul.S)
    I dont remember alot but it required a similar approach to the previous parts.
    Yeah, i know that, but the locus was a lot more ****ed
  17. Alexxh's Avatar
    94 29-06-2012  08:39
    • New Member
    • Posts: 6
    Re: STEP III 2012 Discussion Thread
    (Original post by MrDD)
    yeuk

    i'll post my solution again since people seem to be missing it

    Solution to assumed version of Q1
    I didn't miss your solution at all. I just thought as long as there isn't anything wrong with mine, there's no reason not to post it. Also, I find my method more straight forward..
  18. mikelbird's Avatar
    95 29-06-2012  08:44
    • Respected Member
    • Location: Gloucestershire
    • Posts: 208
    Re: STEP III 2012 Discussion Thread
    Ignore (silly mistake).
    Attached Files
  19. File Type: pdf Step2012Paper3Question5.pdf (66.8 KB, 40 views)
    20. Last edited by mikelbird; 29-06-2012 at 08:54. Reason: mistake
  20. mikelbird's Avatar
    96 29-06-2012  08:54
    • Respected Member
    • Location: Gloucestershire
    • Posts: 208
    Re: STEP III 2012 Discussion Thread
    This time i will get it right!!!! (All I can say is dont try and do mathematics if you have insomnia or have been drinking (or both)!!!
    Attached Files
  21. File Type: pdf Step2012Paper3Question5.pdf (67.4 KB, 127 views)
    22. Last edited by mikelbird; 29-06-2012 at 14:05. Reason: correction
  22. cpdavis's Avatar
    97 29-06-2012  09:28
    • Section Moderator
    • PS Helper
    • Pixar Fanatic Moderator
    • Location: London
    (Original post by mikelbird)
    This time i will get it right!!!!
    I knew I was doing something wrong when working on this last night. The reasons why doing maths at 2am is a bad idea after an evening of jitsu.

    On my way home, will get some solutions out and latex if they are not up yet


    This was posted from The Student Room's iPhone App
  23. maksokio's Avatar
    98 29-06-2012  10:06
    • New Member
    • Posts: 11
    Re: STEP III 2012 Discussion Thread
    (Original post by maksokio)
    how many marks do you guys think I will get for doing the whole of question 7 except part (iii)?
    could someone help with with that?
  24. DarkMatter_22's Avatar
    99 29-06-2012  10:28
    • Junior Member
    • Location: Bulgaria
    • Posts: 44
    Re: STEP III 2012 Discussion Thread
    (Original post by maksokio)
    how many marks do you guys think I will get for doing the whole of question 7 except part (iii)?
    Maybe about 12. It was the hardest part of the question so I think it will worth 7-9 marks. But nobody knows...
  25. B Jack's Avatar
    100 29-06-2012  10:50
    • Full Member
    • Location: Banbury
    • Posts: 90
    Re: STEP III 2012 Discussion Thread
    Here's a solution to question 8.

    Spoiler:
    Show
    F_0=0, F_1=1, F_2=1, F_3=2, F_4=3, F_5=5, F_6=8

    i) F_0F_3 - F_1F_2 = 0-1 = -1 = 5-6 = F_2F_5 - F_3F_4 \mathrm{as\ required}

    ii) F_1F_4 - F_2F_3 = F_3F_6 - F_4F_5 = 1

    \mathrm{We\ want\ to\ show\ by\ induction\ that} F_nF_{n+3} - F_{n+1}F_{n+2} = (-1)^{n+1} \forall n \in \mathbb{Z}\ , n \geq 0

    \mathrm{The\ result\ is\ true\ for\ n=0,\ 1,\ 2,\ 3}

    \mathrm{Now\ assume\ the \ result\ is\ true\ for\ n=k}

    \mathrm{So} F_kF_{k+3} - F_{k+1}F_{k+2}\ =(-1)^{k+1}

    \Leftrightarrow (F_{k+2} - F_{k+1})(F_{k+4} - F_{k+2}) - (F_{k+3} - F_{k+2})(F_{k+3} - F_{k+1}) =(-1)^{k+1}

    \Leftrightarrow F_{k+2}F_{k+4} - F_{k+1}F_{k+4} - F_{k+2}^2 - F_{k+3}^2 +F_{k+3}F_{k+1} + F_{k+2}F_{k+3} =(-1)^{k+1}

    \Leftrightarrow F_{k+2}(F_{k+4} - F_{k+2}) - F_{k+3}(F_{k+3} - F_{k+1}) - F_{k+1}F_{k+4} + F_{k+2}F_{k+3} =(-1)^{k+1}

    \Leftrightarrow F_{k+2}F_{k+3} - F_{k+3}F_{k+2} - F_{k+1}F_{k+4} + F_{k+2}F_{k+3} =(-1)^{k+1}

    \Leftrightarrow - F_{k+1}F_{k+4} + F_{k+2}F_{k+3} =(-1)^{k+1}

    \Leftrightarrow F_{k+1}F_{k+4} - F_{k+2}F_{k+3} =(-1)^{k+2}

    \mathrm{which\ is\ the\ result\ we\ require\ with\ n=k+1,\ therefore\ by\ induction\ we\ have}

    F_nF_{n+3} - F_{n+1}F_{n+2} = (-1)^{n+1} \forall n \in \mathbb{Z}\ , n \geq 0

    iii) \mathrm{Note\ for\ n \geq 2,} \arctan \frac{1}{F_n} \leq \frac{\pi}{4}

    \mathrm{so\ we\ can\ show\ the\ result\ by\ simplifying } \tan( \arctan \frac{1}{F_{2r}}) = \tan ({\arctan \frac{1}{F_{2r+1}} + \arctan \frac{1}{F_{2r+2}}})

    \mathrm{and\ using\ the\ previously\ proven\ result}

    \displaystyle\sum_{i=1}^{\infty} \arctan \frac{1}{F_{2r+1}}

    = \displaystyle\sum_{i=1}^{\infty} (\arctan \frac{1}{F_{2r}} - \arctan \frac{1}{F_{2r+2}})

    = \displaystyle\sum_{i=1}^{\infty} \arctan \frac{1}{F_{2r}} - \displaystyle\sum_{i=2}^{\infty} \arctan \frac{1}{F_{2r}}

    = \arctan \frac{1}{F_2} = \frac{\pi}{4}
    Last edited by B Jack; 29-06-2012 at 10:54. Reason: slight mistake :P
    Post rating:
    1
Sign in to Reply
Search Thread
Advanced Search
Share this discussion:  
Useful resources

Articles:

Study Help rules and posting guidelinesStudy Help Member of the Month nominations

Quick Link:

Unanswered Maths Exams Threads

Groups associated with this forum:

View associated groups
Article updates
Moderators

We have a brilliant team of more than 60 volunteers looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is moderated by:

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22

Registered Office: International House, Queens Road, Brighton, BN1 3XE

Shortcuts

Get Started

TSR Group

Using TSR

Info

Connect with TSR

Reputation gems:
The Reputation gems seen here indicate how well reputed the user is, red gem indicate negative reputation and green indicates a good rep.
Post rating score:
These scores show if a post has been positively or negatively rated by our members.