STEP Prep Thread 2017

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Four things that unis think matter more than league tables 08-12-2016
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    (Original post by Mathemagicien)
    Well, after doing some STEP problems about complex mappings (the mobius transformation?), it mapped circles and lines to either a circle or a line. And I thought to myself, why not consider lines as circles too? They have the same style of definition in the complex plane. That'd save me the effort of calling them different things. So I looked it up, and found some guys on the internet who agreed with me (https://en.wikipedia.org/wiki/Generalised_circle)
    Do you think you could refer me to those STEP problems?

    Btw, I love your profile picture. :-)
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    Inversion.
    Coolest thing to come out of geometry.
    Can solve some very hard problems in matter of lines.


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    (Original post by Injective)
    Do you think you could refer me to those STEP problems?
    Probably not - but I'd guess its STEP 3, and there will likely be some of them in the old set (1987-1990s) of STEP papers.

    Btw, I love your profile picture. :-)
    Are you suggesting that some people do not like Glorious Leader???
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    (Original post by Mathemagicien)
    Probably not - but I'd guess its STEP 3, and there will likely be some of them in the old set (1987-1990s) of STEP papers.



    Are you suggesting that some people do not like Glorious Leader???
    I've noticed it in the old set too, part of the old specification I think. Cheers will have a look.

    And not at all...
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    (Original post by injective)
    i've noticed it in the old set too, part of the old specification i think. Cheers will have a look.

    And not at all...
    1996, iii
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    (Original post by Zacken)
    1996, iii
    Did you know that off head, or did you use the search database?


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    (Original post by Injective)
    Lol, btw I know he is joking.
    I was hoping you were lol. Otherwise
    2017 is Kms.
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    (Original post by Zacken)
    1996, iii
    Dude, did u get into Kings?
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    (Original post by drandy76)
    Did you know that off head, or did you use the search database?


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    search database doesn't work for me idk why

    (Original post by STRANGER2)
    Dude, did u get into Kings?
    Yeah
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    (Original post by Zacken)
    search database doesn't work for me idk why
    My boy Siklos don't rate you no more


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    (Original post by physicsmaths)
    I was hoping you were lol. Otherwise
    2017 is Kms.
    Was I joking? Who knows....


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    http://files.physicsandmathstutor.co...20STEP%201.pdf

    For question 1

    For the last part iii you use that:

    x^3 = 99+-70(sqrt2)

    But I only know that x = 3-2(sqrt2) from ii

    So how am I to assume that it works out for x=3+2(sqrt2)
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    (Original post by girlwonder17)
    http://files.physicsandmathstutor.co...20STEP%201.pdf

    For question 1

    For the last part iii you use that:

    x^3 = 99+-70(sqrt2)

    But I only know that x = 3-2(sqrt2) from ii

    So how am I to assume that it works out for x=3+2(sqrt2)
    It's pretty intuitive that it works for the positive solution as well. It's very easy to quickly verify this in your head; just think about how the signs work out when you cube the binomial.
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    (Original post by girlwonder17)
    http://files.physicsandmathstutor.co...20STEP%201.pdf

    For question 1

    For the last part iii you use that:

    x^3 = 99+-70(sqrt2)

    But I only know that x = 3-2(sqrt2) from ii

    So how am I to assume that it works out for x=3+2(sqrt2)
    Spoiler:
    Show
    You are supposed to solve the equation for x, then you should end up with \ x= \sqrt[3]{99 \pm 70 \sqrt{2}}
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    (Original post by zetamcfc)
    Spoiler:
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    You are supposed to solve the equation for x, then you should end up with \ x= \sqrt[3]{99 \pm 70 \sqrt{2}}
    I don't think this answers their question, since they got that far but were wondering what the justification was for the result of (ii) in the case of the positive solution.
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    (Original post by IrrationalRoot)
    I don't think this answers their question, since they got that far but were wondering what the justification was for the result of (ii) in the case of the positive solution.
    It follows from working out the +ve case.
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    (Original post by zetamcfc)
    It follows from working out the +ve case.
    Well ideally you shouldn't 'work out' the +ve case like you did in (ii); you can just deduce it from the -ve case.
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    (Original post by IrrationalRoot)
    It's pretty intuitive that it works for the positive solution as well. It's very easy to quickly verify this in your head; just think about how the signs work out when you cube the binomial.
    I suppose you are right yea; but I would probably do the whole part b for the positive sign.

    Am I right in thinking that this would work for any similar type of cubic.

    thx
    (Original post by zetamcfc)
    Spoiler:
    Show
    You are supposed to solve the equation for x, then you should end up with \ x= \sqrt[3]{99 \pm 70 \sqrt{2}}
    I get that thnx but I don't get how you are just supposed to figure out that it works out to be positive, like irrational said check the sign I guess, just seems not like i am being complete idk

    To make it clear the 99-70(sqrt 2) makes sense it is the 99+70(sqrt 2 ) that is not implied in previous parts.

    Also unabomber?! He is cute, I agree with his views on technology, but killed people :/
    Spoiler:
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    Mountain men are always free
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    STEP seems exhausting.

    Its not that the content is hard (yet at least), just stretches your understanding.


    Although that paper I 2004 doesn't seem too hard.
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    (Original post by girlwonder17)
    I suppose you are right yea; but I would probably do the whole part b for the positive sign.

    Am I right in thinking that this would work for any similar type of cubic.

    thx


    I get that thnx but I don't get how you are just supposed to figure out that it works out to be positive, like irrational said check the sign I guess, just seems not being complete idk

    To make it clear the 99-70(sqrt 2) makes sense it is the 99+70(sqrt 2 ) that is not implied in previous parts.

    Also unabomber?! He is cute, I agree with his views on technology, but killed people :/
    Spoiler:
    Show
    Mountain men are always free
    Doing part (ii) again is unnecessary; you definitely don't want to waste time making unnecessary justifications in STEP - it should all be as concise as possible.

    The 'looking at signs method' can easily be shown to work with all such cubes of binomials, you can quickly prove this is you like since it's very easy.

    It seems that you think this isn't rigorous enough, but it really does just follow from the fact the magnitude of each term in the binomial expansion will clearly be the same as before, but the signs are switched every other term due to the powers of -1 (and the rational and irrational terms alternate).
 
 
 
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