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    (Original post by some-student)
    I'm applying next year but I'm doing MAT papers for fun really, but I feel an idiot now because I will have literally none to revise for next year when I apply. However I have told myself to not look at the 2016 paper until I start year 13. We've only started C2 now but I am looking forward to the harder modules.
    I'm sure you'll be okay - if you can do MAT at this point, I doubt you'll struggle next year. I certainly would keep the 2016 paper for practice, but last minute cramming is less effective than people think. If you want stuff to save for next year, UKMT SMC stuff is a bit more geometry based but still quite interesting, STEP papers are much harder but good fun and so are AEA papers.

    What are you applying for? I think I'll go for CS or Maths & CS. Good luck to you and everyone else!
    Maths and CS.
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    (Original post by ttva)
    how long should i spend on the MCs? I can solve about 3~3.5 questions entirely out of the 4 free response ones. Even though I am not exactly sure how much time I take as I haven't do the free response by it self.
    Arpund 40-45 minutes it was for me but i ended up staying much longer in the real exam!
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    (Original post by qeyoo)
    Just got 57 on 2015 as my mock, not very impressed... Mostly it was that I messed up the multiple choice I think, normally I get 32-36, and I got 24 this time.

    Don't really know what to do now other than collapse into a ball of disappointment.

    Posted from TSR Mobile
    if it makes you feel better im still scraping the high 40's no clue where or how im going to get up just seem to be hitting a brick wall something which has never happened in maths
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    (Original post by 11234)
    I started with polynomial long division to get 1 + 2a/a-x. Then it should look like the graph of 1/x but moved about cos of transformations. I think the asymotote is then at x=a and y=1
    How did you get the 2a/a-x ? i got 2x/a-x am i doing something stupidly wrong?
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    Guys, I am legitimately scared. I know what I need to know for the exam but the moment I look at the past papers everything looks much harder than it is. Do you have any cure for that?
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    (Original post by denkata)
    Guys, I am legitimately scared. I know what I need to know for the exam but the moment I look at the past papers everything looks much harder than it is. Do you have any cure for that?
    Keep doing them.
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    (Original post by rtc16)
    How did you get the 2a/a-x ? i got 2x/a-x am i doing something stupidly wrong?
    No, you're right.
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    (Original post by rtc16)
    How did you get the 2a/a-x ? i got 2x/a-x am i doing something stupidly wrong?
    a+x/a-x = (-(a-x) + 2a)/a-x = -1 +2a/a-x
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    Hey Zacken just found out we're classmates - which college are you at?
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    (Original post by GrungeGirl)
    How do you draw the graph for Q4(i) in 2013 paper? Attachment 589766

    Thanks
    When you get any rational function (of the general form f(x)=(a'x+b)/(cx+d)) you should consider only its axis intercepts and its asymptotes.

    The vertical asymptote is x=-d/c, the horizontal asymptote is y=a'/c (as x approaches infinity b and d become insignificant) and the intercepts are (-b/a', 0) and (0, b/d) respectively. Obviously look out for cases such as d=0 or a'=0 where there is no intercept.

    Then you only need to consider which "quadrants" (defined by the two asymptotes) the curve lies in. It will lie in diagonally opposite quadrants (1st and 3rd or 2nd and 4th). This could be done by substituting values. All rational functions have the same general shape (see y=1/x).

    In this case, a'=1, b=a, c=-1, d=a. So the vertical asymptote is x=a, the horizontal asymptote is y=-1 (not required) and the axis intercepts are (0,1) and (-a, 0). The question only requires us to draw the graph in the "second quadrant", which the curve indeed lies in. This means that it also lies in "quadrant 4" (where it doesn't cross any axis).
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    (Original post by Insecure)
    Hey Zacken just found out we're classmates - which college are you at?
    King's. You?

    (Original post by 11234)
    a+x/a-x = (-(a-x) + 2a)/a-x = -1 +2a/a-x
    You wrote -1 + in your last post, hence his confusion.
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    Trinity bruh
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    (Original post by Zacken)
    King's. You?



    You wrote -1 + in your last post, hence his confusion.
    sorry i got lazy....plus cant do latex too much effort
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    (Original post by Insecure)
    Trinity bruh
    hi im applying trinity this year. Any tips? Do you happen to be on any IMO or olympiad team...cos they nearly scared me into not applying
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    (Original post by 11234)
    hi im applying trinity this year. Any tips? Do you happen to be on any IMO or olympiad team...cos they nearly scared me into not applying
    Hey, there's this huge misconception about IMO at Trinity. It's true that there are many IMO contestants here and that most IMO contestants apply to Trinity, but a bit less than half of us here (including me) have no experience in maths olympiads at all. I would, however, advise you not to apply to Trinity.

    Tip would be just to work hard. Feel free to ask more questions.
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    (Original post by Zacken)
    No, that's really not what I'm talking about, this is what algebraic geometry is: https://www.dpmms.cam.ac.uk/study/II...5-2016/HW1.pdf



    Do you mean how I got u\int_{-1}^1 3x^2 \, \mathrm{d}x = 2u? If so: what's the value of \int_{-1}^1 3x^2 \, \mathrm{d}x.
    It's more how you could just multiply them together like that.

    Also, would the integral of f(-x) not be -u?
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    (Original post by RuairiMorrissey)
    It's more how you could just multiply them together like that.
    Read above, I've explained this three times now. The integral from -1 to 1 of f(x) dx or f(t) dt is just a random number. It's some constant u. It's not a function, it's not a variable. It's just a way of writing a number. It represents a constant. You can pull constants outside integrals.

    Also, would the integral of f(-x) not be -u?
    No. Draw a sketch. Reflect it in the y-axis. If you still don't believe me, use the substitution v = -x and see what the integral simplifies down to.
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    (Original post by Insecure)
    ...
    Have I met you?
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    (Original post by Zacken)
    Read above, I've explained this three times now. The integral from -1 to 1 of f(x) dx or f(t) dt is just a random number. It's some constant u. It's not a function, it's not a variable. It's just a way of writing a number. It represents a constant. You can pull constants outside integrals.



    No. Draw a sketch. Reflect it in the y-axis. If you still don't believe me, use the substitution v = -x and see what the integral simplifies down to.
    Ah sorry, there's been about 3 new pages since i last checked so I must have missed your posts.

    Thanks for the explanation.
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    2014 Q1 Part G?
 
 
 
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