FP3 (Not MEI) - Friday June 1 2012, AM

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  1. Killjoy-'s Avatar
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    Re: FP3 (Not MEI) - Friday June 1 2012, AM
    (Original post by ElMoro)
    For the june 09 paper, question (7)(ii)(b)

    How do you justify whether \tan \frac{1}{12} \pi is 2 - \sqrt{3} or 2 + \sqrt{3}?

    Thanks
    \tan 3 \theta = 1 so \theta = \frac{\pi}{12}, \frac{1}{3}(\frac{\pi}{4}+ \pi), \frac{1}{3}(\frac{\pi}{4}+ 2\pi). (Using the periodic property for tan.)
    Note that tan of \frac{1}{3}(\frac{\pi}{4}+ 3\pi)=\frac{1}{3}\frac{\pi}{4}+ \pi is the same as the tan of \frac{\pi}{12}, so we do have three roots here.

    But also \tan \theta cannot be the same for all the values of \theta above. Can you see why?

    So the other two roots give two more values of \tan \theta which satisfy the equation we formed.
    If you solve the cubic you will get \tan \theta for all three  \theta.

    We know the values of theta for the roots, we just need to match them to tan's.
    Now without using a calculator you can decide which values of t correspond to \thetas by looking at how the graph of tan changes.

    I hope my somewhat convoluted explanation makes sense.
  2. Ree69's Avatar
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    Re: FP3 (Not MEI) - Friday June 1 2012, AM
    (Original post by Satsui no hado)
    Thanks now what about between a point and a line? xD
    Let point A have position vector a1 and line B have equation a2 + λb, where λ is a scalar parameter.

    The perpendicular distance is \frac{|(a_2-a_1)\ \times\ b|}{|b|}.
  3. Sgt.Incontro's Avatar
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    (Original post by ElMoro)
    Messed this exam up in january because I focused on other exams and didn't practice enough? Guess what I've been doing this time 'round?
    Are you on a gap year? Good luck for this exam.

    This was posted from The Student Room's Android App on my HTC Sensation Z710e
  4. Anon 17's Avatar
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    Re: FP3 (Not MEI) - Friday June 1 2012, AM
    (Original post by ElMoro)
    Think groups and complex numbers&trig are the hardest topics

    Groups because it's just so out there and some of those trig questions really need you to think out the box imo

    Think vectors and differential equations are fairly standard.

    What do you guys think?
    Generally yeah, although Jan 12 had a nasty vectors question. I think groups are alright though if you don't get a really bad question, and the trig is fine if it's standardish but some of the (even simpler) questions like Jun 11 question 2 are just awkward if you don't know what you're doing.
  5. MrProb's Avatar
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    Re: FP3 (Not MEI) - Friday June 1 2012, AM
    i'm really struggling with jan 12 q4 on vectors. eurgghh i hate vectors so much. i even made a thread about it

    http://www.thestudentroom.co.uk/show...9#post37838339

    could someone please help , i'd really appreciate it!!
  6. Anon 17's Avatar
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    Re: FP3 (Not MEI) - Friday June 1 2012, AM
    Any day before tips or questions?

    My tip is make sure you know all the methods as well as what is stated on the spec - The spec leaves out a few things you'll need to know, mainly solving some of the complex number problems. Remember if it asks for some value of C and S from one equation (e.g. the sum of C + iS or something) then expand any exponential forms into polar form and then you can simply take C to be the real part of the equation, and S to be the imaginary part.
  7. bananadude's Avatar
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    Re: FP3 (Not MEI) - Friday June 1 2012, AM
    For finding the distance between two skew lines do people generally derive it from the vectors or just sub it all in to the formula?
  8. Killjoy-'s Avatar
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    Re: FP3 (Not MEI) - Friday June 1 2012, AM
    (Original post by bananadude)
    For finding the distance between two skew lines do people generally derive it from the vectors or just sub it all in to the formula?
    I sub it into the formula but I find its easier to use if you can see where it comes from using the derivation.
  9. Anon 17's Avatar
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    Re: FP3 (Not MEI) - Friday June 1 2012, AM
    I find using the formula quicker.

    Jan 12, question 3 (the first groups one). I just did the paper and got everything right except the question (excluding an RTFQ moment). I honestly am at a loss to how you'd go about approaching a question like this - How does everyone approach these groups problems?

    http://www.thestudentroom.co.uk/show....php?t=2018041
    Last edited by Anon 17; 31-05-2012 at 16:21.
  10. Killjoy-'s Avatar
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    Re: FP3 (Not MEI) - Friday June 1 2012, AM
    (Original post by Anon 17)
    I find using the formula quicker.

    Jan 12, question 3 (the first groups one). I just did the paper and got everything right except the question (excluding an RTFQ moment). I honestly am at a loss to how you'd go about approaching a question like this - How does everyone approach these groups problems?
    For i) I started with yx and showed that if you combine it with y^{-1} first and then with x^{-1} you end up getting the identity e.

    For ii) You have to use a 'shifting technique' (that's what I call it anyway, lol).
    I'll post the full, unsimplified working but if you prefer to do it yourself please read the hint under the spoiler.
    Spoiler:
    Show
    x^{n}y^{n}=(xy)^{n}
    x^{n}y^{n}y^{-1}=(xy)^{n-1}xyy^{-1}
    x^{n}y^{n-1}=(xy)^{n-1}x
    x^{-1}x^{n}y^{n-1}=x^{-1}(xy)^{n-1}x
    x^{n-1}y^{n-1}=y(xy)^{n-2}x
    x^{n-1}y^{n-1}=(yx)^{n-1}

    Remember that you don't know the group is commutative so you can't use that.

    I started by expanding (xy)^{n} so I could notice xyxyxyxy... then manipulate that to get ...yxyxyx....

    For iii) the question is asking you if the two statements are logically equivalent.
    All the steps you took in ii) are reversible, you could work backwards using a similar method to show you end up where you started off, or you may state the steps are reversible. (MS allows this, but to be fair it was only worth 2 marks!)
  11. Anon 17's Avatar
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    Re: FP3 (Not MEI) - Friday June 1 2012, AM
    (Original post by Killjoy-)
    Spoiler:
    Show

    x^{-1}x^{n}y^{n-1}=x^{-1}(xy)^{n-1}x
    x^{n-1}y^{n-1}=y(xy)^{n-2}x
    x^{n-1}y^{n-1}=(yx)^{n-1}
    I'm unsure how you managed to get y all the way to the right in the second last step, and how you got the last step to work.

    Seen as it isn't in the spec, or the book... Can anyone post some rules as to what we can and can't do with groups algebra? I think it's the only bit of FP3 that I'm still confused with.

    EDIT - Nevermind, san_M explained it in my topic on the question. I can't believe I didn't realise you can do (xy)n = (xyxyxyxy....xy) for groups. Still, the question on groups algebra rules remains. Thanks though, that was a great help anyway =)
    Last edited by Anon 17; 31-05-2012 at 16:54.
  12. SecondHand's Avatar
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    Re: FP3 (Not MEI) - Friday June 1 2012, AM
    If the question states that what you are given is truely a group then you can only assume a few things.

    1) Closure - Elements combined under the operation return an element in the set
    2) Associative - (ab)c = a(bc)
    3) Inverse - Each element has an inverse such that A*A^-1 = A^-1*A = e
    4) Identity - The group has an identity

    aaaaa=a^5 (etc)

    You can assume nothing else.

    If the group is abelian then it is also commutative. If the group is cyclic then it is isomorphic to addition modulo n. All prime groups must be cyclic as groups can only have subgroups with orders of factors of the order of the group.

    I love groups so please fire all questions in this direction.
  13. Killjoy-'s Avatar
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    Re: FP3 (Not MEI) - Friday June 1 2012, AM
    (Original post by Anon 17)
    I'm unsure how you managed to get y all the way to the right in the second last step, and how you got the last step to work.
    You mean here?
    x^{n-1}y^{n-1}=y(xy)^{n-2}x
    I took xy out so that it was in front of (xy)^{n-2} on the RHS and then used the pre-multiplied inverse of x to get rid of x, ending up with only y in front of (xy)^{n-2} so that I could shift it along to get (yx)^{n-1} in the last step (and get rid of the x on the end in doing so).
  14. SecondHand's Avatar
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    Re: FP3 (Not MEI) - Friday June 1 2012, AM
    Also this formula you talk about for distance between two skew lines? Is it..

    \mathbf{n}=\mathbf{d_1} \times \mathbf{d_2}
    \frac{\mathbf{n}.(\mathbf{a_1}-\mathbf{a_2})}{|n|}
  15. Anon 17's Avatar
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    Re: FP3 (Not MEI) - Friday June 1 2012, AM
    Yeah, that's the one. I'm not 100% sure about this but I always apply the modulus to the top as well, to avoid a negative distance.

    SecondHand: a is a fixed element of G, x is any element of G. axa = x-1 is given.

    Prove that a = a-1.

    I'm thinking this isn't actually done by algebra, as all my attempts come out as xax = a-1.
  16. SecondHand's Avatar
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    Re: FP3 (Not MEI) - Friday June 1 2012, AM
    Do you know anything else about G? Where is the question from?

    About vector questions I always just draw the diagram and then say I need to work out (b-a)cosx, change the cosx to the dot product and so on.
  17. Anon 17's Avatar
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    Re: FP3 (Not MEI) - Friday June 1 2012, AM
    It's from the FP2/FP3 OCR book, question 5 of the practice exam 1 for FP3.

    Quoting them:

    G is a multiplicative group with identity element e, and a is a fixed element of G for which axa = x-1 for all elements x that belong to G. Prove that:

    (i) a = a-1
    (ii) ax = (ax)-1 for all x in G
    (iii) x = x-1 for all x in G
    (iv) xy = yx for all x, y in G
    I love how the one part of the one topic I'm stuck with has no information for it... Anywhere... I can't find any help on how I'm meant to approach or do these ridiculous questions. Even the other groups questions are relatively easy, but these types of questions I just can't get at all...
    Last edited by Anon 17; 31-05-2012 at 17:34.
  18. SecondHand's Avatar
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    Re: FP3 (Not MEI) - Friday June 1 2012, AM
    I can't see the first one, apologies.
  19. Anon 17's Avatar
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    Re: FP3 (Not MEI) - Friday June 1 2012, AM
    Fair one, I'm finding it tough.

    I managed to get the second, any idea about the others?

    The second one:

    axa = x-1
    ax = x-1a-1
    ax = (ax)-1

    As:

    x-1a-1 = (ax)-1
    axx-1a-1 = (ax)(ax)-1
    aa-1 = (ax)(ax)-1
    e = e
  20. ElMoro's Avatar
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    Re: FP3 (Not MEI) - Friday June 1 2012, AM
    For the June 11 paper, question 5 part ii: why must k = -1?

    EDIT: Nevermind, I was being an idiot :facepalm2:
    Last edited by ElMoro; 31-05-2012 at 17:51.
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