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    Did it say give the period in radians??
    I know the rest of the Q was in radians but for part i I just wrote 360/a
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    (Original post by Heyitsmerhi)
    I did -1 to 0 and 0 to 3 as otherwise it wouldn't give the exact area would it? Not 100% sure but I got something like 179/5
    it should give u the same answer if you split it, but really you are supposed to do it from -1 to 3.
    you only have to split it when the part of the area is under the *X* axis, which in this case it wasnt
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    Does anyone remember roughly what the questions were for these answers?5bi) -6/a+2/a^2+4 (4m) 8iv) x=6.73 (3m)
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    (Original post by morrissies)
    Did it say give the period in radians??
    I know the rest of the Q was in radians but for part i I just wrote 360/a
    Check the mark scheme last year for 9i and ii.
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    (Original post by AlfieH)
    My teacher has confirmed that the answer to 9iii is Pi/3a and 4Pi/3a.
    But 4Pi/3a gives a negative solution when substituted into the original equation, so it cannot be a right answer.

    sin(a*4Pi/3a)=sqrt3*cos(a*4Pi/3a)=-sqrt3/2
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    (Original post by abrahammurciano)
    But 4Pi/3a gives a negative solution when substituted into the original equation, so it cannot be a right answer.

    sin(a*4Pi/3a)=sqrt3*cos(a*4Pi/3a)=-sqrt3/2
    Maybe it meant the first two positive solutions in respect to the x axis.
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    (Original post by student.1997)
    Does anyone remember roughly what the questions were for these answers?5bi) -6/a+2/a^2+4 (4m) 8iv) x=6.73 (3m)
    5bi was something like integrate 6x^-2+4x^-3 between A and 1 (or -1).

    8iv was 180=3^(n-2); solve for n.
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    Anyone got an unofficial mark scheme ??
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    For the last question how many marks do you think I would get,
    I squared everything to get sin^2ax=3cos^2ax
    Then used the identies to get 1-cos^2ax=3cos^2ax
    Then rearranged to get 4cos^2ax=1 ,then cos^2ax=1/4
    then square rooted to get cosax=+ or - 1/2
    Then somehow I ended up with pi/3a and 2pi/3a as my final answer.
    Thanks very much
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    (Original post by Sabreak10)
    Anyone got an unofficial mark scheme ??
    1i) 10cm

    1ii) 5.04cm

    2i) 3/10π

    2ii) 20.4cm

    3i) 27+27kx+9k^2x^2+k^3x^3

    3ii) +/-√3

    4i) log base 3 x^2/x+4

    4ii) x=12

    5a) x^4/2-x^3+2x^2-6x

    5bi) -6/a+2/a^2+4

    5bii) 4

    6i) k=91

    6ii) Sum of terms = 978

    6iii) N=38

    7i) R=0, quotient = x^2-4x+3

    7ii) (x+1), (x-1), (x-3)

    7iii) Prove

    7iv) 512/15

    8i) Translation 2 units in positive x direction

    8ii) Stretch scale factor 1/9 in y direction

    8iii) Sketch, crosses y at (0,1/9)

    8iv) x=6.73

    8v) Area =9.60

    9i) 2pi/a

    9ii) a=5/3 and k=√3/2

    Someone else came up with this one
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    (Original post by micycle)
    5bi was something like integrate 6x^-2+4x^-3 between A and 1 (or -1).

    8iv was 180=3^(n-2); solve for n.
    Thank you!!
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    On this mark scheme they haven't included the +c on the end of 5a, don't u need it for full marks
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    What were the questions 8)iv and 8)v
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    (Original post by Sabreak10)
    Anyone got an unofficial mark scheme ??
    https://docs.google.com/document/d/1...dEsoArVk0/edit
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    I got 1/9 and 0 for question 3)ii but the mark scheme says sqrt3
    Maybe I misread the question... What did everyone else get?

    Edit: Never mind, turns out I didn't read the question properly -_- I had so much time to check as well! Oh well
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    Wow awesome, know I got between 65 and 69/72- definitely an A . Hopefully that will cancel out my mediocre performance in C1.
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    For the one where k is +/-root3, people say they equated k to 27 - where's the 27 come from and how is it the constant
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    (Original post by Bobrey)
    For the one where k is +/-root3, people say they equated k to 27 - where's the 27 come from and how is it the constant
    I think it's a constant as it stays the same whatever the k value is and is unaffected by it and it's got no x's with it, I may be wrong though
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    (Original post by morrissies)
    not word for word but it was:
    sum of first 20 terms of AP > sum to infinity of GP

    (for the AP a=5 d=1.5) (for the GP a=1.2 (?) r=0.9)
    Wasn't A for the g.p. 120
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    (Original post by Bobby21231)
    I think it's a constant as it stays the same whatever the k value is and is unaffected by it and it's got no x's with it, I may be wrong though
    You are correct.
 
 
 
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