STEP II 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
Ask me ANYTHING - Andrew O'Neill - Buzzcocks comedian, amateur occultist, vegan... 22-05-2013
IMPORTANT: You must wait until midnight (morning exams)/4.30AM (afternoon exams) to discuss Edexcel exams and until 1pm/6pm the following day for STEP and IB exams. Please read before posting, including for rules for practical and oral exams. 28-04-2013
Search Thread
Advanced Search
Sign in to Reply
  1. cpdavis's Avatar
    1 21-06-2012  16:10
    • Section Moderator
    • PS Helper
    • Pixar Fanatic Moderator
    • Location: London
    STEP II 2012 discussion thread
    Post all of your STEP II discussions here.

    Question Paper

    Provided Solutions

    Q1 - DFranklin
    Q2 - desijut
    Q3 - TLRMaths
    Q4 - TheMagicMan
    Q5 - mikelbird
    Q6 Version 1 - ben-smith/Q6 Version 2 - Blutooth
    Q7 - Blutooth
    Q8 - SParm
    Q9 - Tomcrease
    Q10 - Farhan.Hanif93
    Q11 - DFranklin
    Q12 - DFranklin
    Q13 - Farhan.Hanif93

    Well done everyone, we have a complete set of solutions
    Last edited by cpdavis; 25-06-2012 at 22:36.
  2. DFranklin's Avatar
    2 21-06-2012  17:04
    • TSR Royalty
    • Location: London
    • Posts: 18,049
    Re: STEP II 2012 discussion thread
    Bumped as this has just come out of moderation...
  3. cpdavis's Avatar
    3 21-06-2012  17:05
    • Section Moderator
    • PS Helper
    • Pixar Fanatic Moderator
    • Location: London
    Re: STEP II 2012 discussion thread
    (Original post by DFranklin)
    Bumped as this has just come out of moderation...
    Thanks
  4. Alexxh's Avatar
    4 21-06-2012  17:06
    • New Member
    • Posts: 6
    Re: STEP II 2012 discussion thread
    completed 2 questions, another 2 nearly completed (last part didn't work out) and 3 other fragmentary answers :/
    is that a 1?
  5. SParm's Avatar
    5 21-06-2012  17:06
    • Exalted Member
    • Posts: 358
    Re: STEP II 2012 discussion thread
    Here's my solution for question 8:

    Part 1:

    Spoiler:
    Show
    \beta , \alpha , q > 0\\ \beta - \alpha > q\\ (\beta - \alpha )^{2} > q^{2}\\ \beta ^{2} + \alpha ^{2} - 2\beta \alpha > q^{2}\\ \beta ^{2} + \alpha ^{2} - q^{2} - 2\beta\alpha > 0\\ \dfrac{\beta ^{2} + \alpha ^{2} - q^{2}}{\alpha\beta} - 2 > 0\\

    As required.


    Part 2:

    Spoiler:
    Show
    u_{n + 1} = \dfrac{u_{n} ^{2} - q^{2}}{u_{n - 1}}\\ u_{n}(u_{n} + u_{n + 2}) = u_{n} ^{2} + u_{n}u_{n + 2}\: (\tau )

    Note from the first equation:

    u_{n} ^{2} = u_{n - 1}u_{n + 1} + q^{2}

    So (\tau) becomes:

    u_{n - 1}u_{n + 1} + (u_{n}u_{n + 2} + q^{2})

    As the sequence is valid for all n greater than or equal to one, we can say:

    u_{n + 2} = \dfrac{u_{n + 1} ^{2} - q^{2}}{u_{n}}

    Re-arrange to give:

    u_{n}u_{n + 2} + q^{2} = u_{n + 1} ^{2}

    Substitute back into (\tau ):

    u_{n - 1}u_{n + 1} + (u_{n}u_{n + 2} + q^{2}) = u_{n - 1}u_{n + 1} + u_{n + 1} ^{2}

    Factorise this expression:

    u_{n - 1}u_{n + 1} + u_{n + 1} ^{2} = u_{n + 1}(u_{n - 1} + u_{n + 1})

    As required


    Part 3:

    Spoiler:
    Show
    For this part; we can simply find the number p for which this equation works. If p is a real number; then clearly the equation will work. So:

    u_{n + 1} - pu_{n} + u_{n - 1} = 0

    When n = 1

    u_{n - 1} = u_{0} = \alpha\\ u_{n} = u_{1} = \beta\\ u_{n + 1} = u_{2}

    From the sequence equation:

    u_{2} = \dfrac{\beta ^{2} - q^{2}}{\alpha}

    Substitute back into the original equation:

    u_{n + 1} - pu_{n} + u_{n - 1} = \dfrac{\beta ^{2} - q^{2}}{\alpha} - p\beta + \alpha = 0

    Divide by \beta:

    \dfrac{\beta ^{2} - q^{2}}{\alpha\beta} - p + \frac{\alpha}{\beta} = 0

    Note: \dfrac{\alpha}{\beta} = \dfrac{\alpha ^{2}}{\alpha\beta}

    Re-arrange our original equation to get:

    \dfrac{\beta ^{2} + \alpha ^{2} - q^{2}}{\alpha\beta} = p

    \alpha,\: \beta \neq 0\\ \alpha,\: \beta,\: q are real, so p is real, and therefore our original equation works when p is the aforementioned expression.

    (I’ve assumed I can do this because of the wording of the question which asks you to prove that statement for some number “p”; and doesn’t put any limitations on what “p” can be; just asks for it in terms of \alpha,\: \beta,\: q, so I thought that you could go the other way and show that such a “p” exists and therefore the statement is true for that “p” which is basically the same as what they’re asking.)


    Part 4:

    Spoiler:
    Show
    \beta > \alpha + q\\ \beta - \alpha > q

    So we can immediately see from the first part of the question:

    \dfrac{\beta ^{2} + \alpha ^{2} - q^{2}}{\alpha\beta} - 2 > 0

    Note that:

    \dfrac{\beta ^{2} + \alpha ^{2} - q^{2}}{\alpha\beta} = p

    Re-arrange previous inequality and we get:

    p > 2, so p = 2 + \mu where \mu > 0

    So:

    u_{n + 1} - pu_{n} + u_{n - 1} = u_{n + 1} - (2 + \mu)u_{n} + u_{n - 1} = 0

    Re-arrange this expression:

    u_{n + 1} - u_{n} = (1 + \mu)u_{n} - u_{n - 1}

    We can see from this that:

    u_{n + 1} - u_{n} > u_{n} - u_{n - 1}

    And if we consider the difference of the first two terms, u_{0} and u_{1}, we get:

    u_{1} - u_{0} = \beta - \alpha > q > 0

    The first difference is greater than 0 and the subsequent differences are greater than the previous differences so the differences between two adjacent terms is always greater than 0, and so the sequence is strictly increasing, as required.

    When \beta = \alpha + q, p = 2, so:

    u_{n + 1} - u_{n} = u_{n} - u_{n - 1}, and we can see the sequence increases constantly.
    Last edited by SParm; 21-06-2012 at 18:16. Reason: Typo
    Post rating:
    1
    1
  6. Farhan.Hanif93's Avatar
    6 21-06-2012  17:07
    • Section Moderator
    • PS Helper
    • TSR Idol
    • Location: Cambridge
    Re: STEP II 2012 discussion thread
    (Original post by 8inchestall)
    question 4/5, i didnt know what it meant by expand in powers of 1/y, i changed the ln|(2+y)/(2-y)| to ln|(1+(2y/2-y))| but had no idea how to go on
    Responding to you here, I believe you were supposed to consider \ln \left(1 + \dfrac{1}{2y}\right) - \ln \left(1 - \dfrac{1}{2y}\right) from the start.
  7. ben-smith's Avatar
    7 21-06-2012  17:08
    • Overlord in Training
    • Location: Hilbert Space
    • Posts: 2,367
    Re: STEP II 2012 discussion thread
    Did anyone do 7? Found it kind of strange
  8. TheMagicMan's Avatar
    8 21-06-2012  17:08
    • Overlord in Training
    • Posts: 2,940
    Re: STEP II 2012 discussion thread
    (Original post by SParm)
    Here's my solution for question 8:

    Part 1:

    Spoiler:
    Show
    \beta , \alpha , q > 0\\ \beta - \alpha > q\\ (\beta - \alpha )^{2} > q^{2}\\ \beta ^{2} + \alpha ^{2} - 2\beta \alpha > q^{2}\\ \beta ^{2} + \alpha ^{2} - q^{2} - 2\beta\alpha > 0\\ \dfrac{\beta ^{2} + \alpha ^{2} - q^{2}}{\alpha\beta} - 2 > 0\\

    As required.


    Part 2:

    Spoiler:
    Show
    u_{n + 1} = \dfrac{u_{n} ^{2} - q^{2}}{u_{n - 1}}\\ u_{n}(u_{n} + u_{n + 2}) = u_{n} ^{2} + u_{n}u_{n + 2}\: (\tau )

    Note from the first equation:

    u_{n} ^{2} = u_{n - 1}u_{n + 1} + q^{2}

    So (\tau) becomes:

    u_{n - 1}u_{n + 1} + (u_{n}u_{n + 2} + q^{2})

    As the sequence is valid for all n greater than or equal to one, we can say:

    u_{n + 2} = \dfrac{u_{n + 1} ^{2} - q^{2}}{u_{n}}

    Re-arrange to give:

    u_{n}u_{n + 2} + q^{2} = u_{n + 1} ^{2}

    Substitute back into (\tau ):

    u_{n - 1}u_{n + 1} + (u_{n}u_{n + 2} + q^{2}) = u_{n - 1}u_{n + 1} + u_{n + 1} ^{2}

    Factorise this expression:

    u_{n - 1}u_{n + 1} + u_{n + 1} ^{2} = u_{n + 1}(u_{n - 1} + u_{n - 1})

    As required


    Part 3:

    Spoiler:
    Show
    For this part; we can simply find the number p for which this equation works. If p is a real number; then clearly the equation will work. So:

    u_{n + 1} - pu_{n} + u_{n - 1} = 0

    When n = 1

    u_{n - 1} = u_{0} = \alpha\\ u_{n} = u_{1} = \beta\\ u_{n + 1} = u_{2}

    From the sequence equation:

    u_{2} = \dfrac{\beta ^{2} - q^{2}}{\alpha}

    Substitute back into the original equation:

    u_{n + 1} - pu_{n} + u_{n - 1} = \dfrac{\beta ^{2} - q^{2}}{\alpha} - p\beta + \alpha = 0

    Divide by \beta:

    \dfrac{\beta ^{2} - q^{2}}{\alpha\beta} - p + \frac{\alpha}{\beta} = 0

    Note: \dfrac{\alpha}{\beta} = \dfrac{\alpha ^{2}}{\alpha\beta}

    Re-arrange our original equation to get:

    \dfrac{\beta ^{2} + \alpha ^{2} - q^{2}}{\alpha\beta} = p

    \alpha,\: \beta \neq 0\\ \alpha,\: \beta,\: q are real, so p is real, and therefore our original equation works when p is the aforementioned expression.

    (I’ve assumed I can do this because of the wording of the question which asks you to prove that statement for some number “p”; and doesn’t put any limitations on what “p” can be; just asks for it in terms of \alpha,\: \beta,\: q, so I thought that you could go the other way and show that such a “p” exists and therefore the statement is true for that “p” which is basically the same as what they’re asking.)


    Part 4:

    Spoiler:
    Show
    \beta > \alpha + q\\ \beta - \alpha > q

    So we can immediately see from the first part of the question:

    \dfrac{\beta ^{2} + \alpha ^{2} - q^{2}}{\alpha\beta} - 2 > 0

    Note that:

    \dfrac{\beta ^{2} + \alpha ^{2} - q^{2}}{\alpha\beta} = p

    Re-arrange previous inequality and we get:

    p > 2, so p = 2 + \mu where \mu > 0

    So:

    u_{n + 1} - pu_{n} + u_{n - 1} = u_{n + 1} - (2 + \mu)u_{n} + u_{n - 1} = 0

    Re-arrange this expression:

    u_{n + 1} - u_{n} = (1 + \mu)u_{n} - u_{n - 1}

    We can see from this that:

    u_{n + 1} - u_{n} > u_{n} - u_{n - 1}

    And if we consider the difference of the first two terms, u_{0} and u_{1}, we get:

    u_{1} - u_{0} = \beta - \alpha > q > 0

    The first difference is greater than 0 and the subsequent differences are greater than the previous differences so the differences between two adjacent terms is always greater than 0, and so the sequence is strictly increasing, as required.

    When \beta = \alpha + q, p = 2, so:

    u_{n + 1} - u_{n} = u_{n} - u_{n - 1}, and we can see the sequence increases constantly.
    aka it's an arithmetic progression<----probably not necessary
  9. TRLMaths's Avatar
    9 21-06-2012  17:09
    • Respected Member
    • Posts: 156
    Re: STEP II 2012 discussion thread
    Question 3.
    Attached Thumbnails
    Click image for larger version.  Name: img018.jpg  Views: 376  Size: 490.5 KB  ID: 159044   Click image for larger version.  Name: img019.jpg  Views: 273  Size: 402.8 KB  ID: 159045  
    Post rating:
    1
  10. TheMagicMan's Avatar
    10 21-06-2012  17:09
    • Overlord in Training
    • Posts: 2,940
    Re: STEP II 2012 discussion thread
    (Original post by ben-smith)
    Did anyone do 7? Found it kind of strange
    I spent the last 15 minutes or so on it. Didn't manage to do the last part and can't even remember anything about it other than that it was vectors so I hated it
  11. ben-smith's Avatar
    11 21-06-2012  17:10
    • Overlord in Training
    • Location: Hilbert Space
    • Posts: 2,367
    Re: STEP II 2012 discussion thread
    (Original post by TheMagicMan)
    I spent the last 15 minutes or so on it. Didn't manage to do the last part and can't even remember anything about it other than that it was vectors so I hated it
    I have solution to the last part but it seems way too simple.
  12. TRLMaths's Avatar
    12 21-06-2012  17:10
    • Respected Member
    • Posts: 156
    Re: STEP II 2012 discussion thread
    2012 step ii.
    Attached Thumbnails
    Click image for larger version.  Name: 181400_4213186738826_117386272_n.jpg  Views: 689  Size: 75.0 KB  ID: 159047   Click image for larger version.  Name: 282343_4213185138786_814581670_n.jpg  Views: 589  Size: 35.7 KB  ID: 159048   Click image for larger version.  Name: 285613_4213186218813_638248665_n.jpg  Views: 512  Size: 54.1 KB  ID: 159049   Click image for larger version.  Name: 484356_4213185618798_1269035706_n.jpg  Views: 523  Size: 58.8 KB  ID: 159050   Click image for larger version.  Name: 534204_4213186498820_1380926092_n.jpg  Views: 432  Size: 28.9 KB  ID: 159051   Click image for larger version.  Name: 599440_4213185378792_435877861_n.jpg  Views: 503  Size: 34.4 KB  ID: 159052  

    Click image for larger version.  Name: 599769_4213185938806_1863161601_n.jpg  Views: 473  Size: 27.0 KB  ID: 159053   Click image for larger version.  Name: 600181_4213184938781_1300265847_n.jpg  Views: 307  Size: 62.8 KB  ID: 159054  
    Post rating:
    3
  13. Farhan.Hanif93's Avatar
    13 21-06-2012  17:12
    • Section Moderator
    • PS Helper
    • TSR Idol
    • Location: Cambridge
    Re: STEP II 2012 discussion thread
    (Original post by ben-smith)
    Did anyone do 7? Found it kind of strange
    I saw vectors, and decided it wasn't worth my time. How did you get on?
  14. Peter8837's Avatar
    14 21-06-2012  17:12
    • Full Member
    • Posts: 132
    • Warning points: 5
    Re: STEP II 2012 discussion thread
    What was the coeff of x^24 in Q1 part (i)?
  15. SParm's Avatar
    15 21-06-2012  17:12
    • Exalted Member
    • Posts: 358
    Re: STEP II 2012 discussion thread
    (Original post by TheMagicMan)
    aka it's an arithmetic progression<----probably not necessary
    THAT'S THE ONE !!!! Forgot the name in the heat of the exam.
  16. TheMagicMan's Avatar
    16 21-06-2012  17:12
    • Overlord in Training
    • Posts: 2,940
    Re: STEP II 2012 discussion thread
    (Original post by Farhan.Hanif93)
    Responding to you here, I believe you were supposed to consider \ln \left(1 + \dfrac{1}{2y}\right) - \ln \left(1 - \dfrac{1}{2y}\right) from the start.
    I wasn't really sure what they wanted in the last part to be honest

    I got (1+1/n)^n &lt; e&lt; (1+1/n)^{n+1/2}

    so 1 &lt; e(1+1/n)^{-n}&lt; (1+1/n)^{1/2}

    The RHS has a limit of one so want we want follows by the squeeze theorem...the real question is is there a simpler way (tbh I'm never quite sure how STEp wants you to approach limits)
  17. 8inchestall's Avatar
    17 21-06-2012  17:12
    • Adored and Respected Member
    • Posts: 584
    Re: STEP II 2012 discussion thread
    (Original post by ben-smith)
    I have solution to the last part but it seems way too simple.
    q7 after the first part showing OY=bla, how do you get the hence? and also x1.x2 meant dot product or nay ?
  18. TheMagicMan's Avatar
    18 21-06-2012  17:12
    • Overlord in Training
    • Posts: 2,940
    Re: STEP II 2012 discussion thread
    (Original post by Peter8837)
    What was the coeff of x^24 in Q1 part (i)?
    Wasn't that a show that question? 55 wasn't it?
  19. Peter8837's Avatar
    19 21-06-2012  17:13
    • Full Member
    • Posts: 132
    • Warning points: 5
    Re: STEP II 2012 discussion thread
    (Original post by TheMagicMan)
    Wasn't that a show that question? 55 wasn't it?
    The first part of Q1...
  20. DarkMatter_22's Avatar
    20 21-06-2012  17:14
    • Junior Member
    • Location: Bulgaria
    • Posts: 44
    Re: STEP II 2012 discussion thread
    (Original post by Peter8837)
    The first part of Q1...
    I think it was 15.
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.