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    (Original post by Goulden_J)
    Your teachers are *******s then


    It was all in the syllabus
    If you wanted to be finnicky it technically is above the syllabus since the curve was 2cosh2x:smug::smug:

    But nah, it was tricky but not much variation in difficulty to the 2015 paper

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    (Original post by Kira Yagami)
    The lowest it's been is 51,54,55,56. It was 56 last year.

    I think it should be 53/54 for an A , what do you think?
    I've only looked at 90% boundaries before, but assuming 18/18 in coursework I'd guess 59 or 60 for 90 UMS in the exam.
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    I think minimum I got around 63 looking at my answers to the questions a bit more. Along with 18/18 on my coursework would that be a solid A?
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    Your teachers aren't properly qualified if they couldn't do any of it. I thought it was a tricky paper but definitely not too difficult. Expecting 53 for an A and 67 for full UMS.
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    I know I messed up and got the wrong answers for all of the last question And I had to get an A* - I guess that can't happen anymore
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    (Original post by OscarH98)
    Your teachers aren't properly qualified if they couldn't do any of it. I thought it was a tricky paper but definitely not too difficult. Expecting 53 for an A and 67 for full UMS.
    u mean 63
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    For the very last question I put that because g(X) is even that means the function is symmetrical about the y axis. Would I pick up any marks here?
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    (Original post by bethygirl)
    I've seen posts on this thread saying the answer to the first part of the last question was k-1 however literally every body at my college got k+1?
    its k-1, i plotted the function in my calculator with k as 7 and the area was 6. hence k-1
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    (Original post by Studious_Student)
    I think minimum I got around 63 looking at my answers to the questions a bit more. Along with 18/18 on my coursework would that be a solid A?
    Should be around 92-95 UMS in my opinion.
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    (Original post by mathsphys3)
    For the very last question I put that because g(X) is even that means the function is symmetrical about the y axis. Would I pick up any marks here?
    Yes. About y-axis or Oy is fine.
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    (Original post by mathsphys3)
    For the very last question I put that because g(X) is even that means the function is symmetrical about the y axis. Would I pick up any marks here?
    Maybe 1 as that was part of the reasoning for the line of symmetry for f(x).
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    (Original post by jakeshaw97)
    its k-1, i plotted the function in my calculator with k as 7 and the area was 6. hence k-1
    damn, you couldn't leave it as 2k-2?
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    (Original post by lollllll)
    damn, you couldn't leave it as 2k-2?
    2k-2 is not the same as k-1.
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    (Original post by lollllll)
    damn, you couldn't leave it as 2k-2?
    Well, you can't "simplify" 2k - 2 to k - 1. k - 1 was the correct answer. Always check your results on your calculator's integral button, even if it's in terms of a variable.
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    (Original post by lollllll)
    damn, you couldn't leave it as 2k-2?
    no, dont think that would work as 2*7-2=12 not 6. 2k-2 doesnt simplify to k-1, its just double it
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    This exam was definitely harder than last years - there is no doubting it.
    There was lots of parts which would throw people back.
    I expect grade boundaries to be 55-58.
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    I did question 5 with 4root(h^3 - 1) -4 do you reckon I'll get follow through marks cos obvs I got the wrong answer (:
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    (Original post by Leechayy)
    If you wanted to be finnicky it technically is above the syllabus since the curve was 2cosh2x:smug::smug:
    and if you want to be finnickier it was actually 2sqrt(k)cosh2x ^^
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    (Original post by StrangeBanana)
    and if you want to be finnickier it was actually 2sqrt(k)cosh2x ^^
    That's the translated one isn't it?:smug::p:p:holmes:

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    for question 9.ii i showed that the minimum point has x co ordinate of 1/4ln4 and then i subbed that back into f(x) to find out the y co ordinate but i got an incorrect y co ordinate. how many marks in the question will i lose because of this
 
 
 
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