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    (Original post by davon806)
    I got all exactly the same as you!
    In addition,for the trapezium rule one,I got 7.12.
    For the sum of 12 terms(the last part of the first question),I got 159.96
    For the obtuse angle of the sector,I got 1.87(to 3sf)
    And for the transformation problem,I guess it is stretching the original graph
    surd(8x^3 -1) in the x-direction of scale factor 2

    Anyone else can share their answers to us ?
    I agree with those! But I think I made a silly mistake on the transformation question, and did the transformation the wrong way (s.f. = 1/2).
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    (Original post by Alifff)
    It does the job haha but do you remember the transformation of the circle?
    from (X-5)^2 + (Y-7)^2 = 49 to (X+1)^2 + Y^2 = 49?
    yeah (-6,-7)
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    Last question, values was 20/7 and 4 The integration question was 53/2 and the area between tangent and curve was 14. That's what I got.
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    (Original post by Andrreewwww)
    yeah (-6,-7)
    I put (-6, 7) damn it but i think i still get 2 marks?
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    (Original post by Alifff)
    I put (-6, 7) damn it but i think i still get 2 marks?

    You had to Plus 7 to the (Y-7) to get to (y) so you might be right, I thought the question was (y+7)^2 though, but my memory isn't great
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    (Original post by Alifff)
    I put (-6, 7) damn it but i think i still get 2 marks?
    My answer was, transformation (6,-7)
    Not sure if i got it right :-/
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    (Original post by stuart_aitken)
    This is all I want to know:

    For core 1

    Integration question:
    Area under curve was 524/15
    Shaded area was 961/15

    Final question:
    20/7 <= k <= 4

    Correct or not...?

    Posted from TSR Mobile
    I thought the intergration question on C1 was something /6?
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    (Original post by Andrreewwww)
    You had to Plus 7 to the (Y-7) to get to (y) so you might be right, I thought the question was (y+7)^2 though, but my memory isn't great
    Yh I'm not so sure either!
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    (Original post by alig4594)
    I thought the intergration question on C1 was something /6?
    It was 162/6, if my answer is right ( which equals 27)
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    (Original post by timeteo9)
    The log question was:

    log2(x+7)2 - log2(x+5) - 3 = 0

    →log2(x2+14x+49) = 3 + log2(x+5)
    →log2(x2+14x+49) = log223 + log2(x+5)
    →log2(x2+14x+49) = log2(8x+40)
    →x2+14x+49 = 8x+40
    →x2+6x+9 = 0
    →(x+3)2

    So x = -3

    (However, for some reason I thought 14-8=8 so messed up the last bit... :mad:)
    I did all that, then wrote x = 3, instead of minus 3 How manh marks will I lose?

    Posted from TSR Mobile
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    (Original post by alig4594)
    I thought the intergration question on C1 was something /6?
    it was 63/6 then u subtract from the area of trapezium
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    (Original post by Hart1995)
    I did all that, then wrote x = 3, instead of minus 3 How manh marks will I lose?

    Posted from TSR Mobile
    I did the same too ... how mayn marks will i lose
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    (Original post by alig4594)
    I thought the intergration question on C1 was something /6?
    I got something /6 I think it was 189/6
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    (Original post by Hart1995)
    I did all that, then wrote x = 3, instead of minus 3 How manh marks will I lose?

    Posted from TSR Mobile
    You'll probably lose only one mark, I might lose two (hopefully not 3) for that ghastly subtraction error. It was an exam full of silly mistakes for me, forgot the angle was obtuse, thought it was (2+x-2)3 - (2+x-2)3) instead of (2+x-2)3 - (2-x-2)3) at first; fortunately I spotted that mistake and corrected it, although I did have to ask for more paper
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    (Original post by Hart1995)
    I did all that, then wrote x = 3, instead of minus 3 How manh marks will I lose?

    Posted from TSR Mobile
    I guess you would probably lose 1 mark
    log2(x+7)2 - log2(x+5) - 3 = 0

    →log2(x2+14x+49) = 3 + log2(x+5)
    →log2(x2+14x+49) = log223 + log2(x+5) ------> 1 mark for law of log
    →log2(x2+14x+49) = log2(8x+40) ------> 1 mark for law of log
    →x2+14x+49 = 8x+40 ------> 1 mark for eliminating log
    →x2+6x+9 = 0 ----------> 1 mark for rearranging term
    →(x+3)2 ---------> 1 mark for factorization

    So x = -3 ---------> 1 mark for answer

    That's how I think the marks distribution would be
    :rolleyes:
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    For the people who did C2 and got "stretch x direction scale factor 2" how did you get that? I got SF 8 because the coeff of x^3 goes from 8 to 1 which means you multiplied the coeff by 1/8 which then means you stretched by SF 3? Or am I barking up the wrong tree?
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    (Original post by elixiroflife.)
    for the people who did c2 and got "stretch x direction scale factor 2" how did you get that? I got sf 8 because the coeff of x^3 goes from 8 to 1 which means you multiplied the coeff by 1/8 which then means you stretched by sf 3? Or am i barking up the wrong tree?
    i meant "stretched by sf 8"
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    grade boundaries for core 2?
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    wasnt it scale factor 1/8?
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    (Original post by ElixirOfLife.)
    For the people who did C2 and got "stretch x direction scale factor 2" how did you get that? I got SF 8 because the coeff of x^3 goes from 8 to 1 which means you multiplied the coeff by 1/8 which then means you stretched by SF 3? Or am I barking up the wrong tree?
    Yes, the coefficient of x^3 goes from 8 to 1, but 8x3 = (2x)3, so the coefficient of x goes from 2 to 1, so s.f. = 2
 
 
 
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