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    (Original post by DominoMatrix227)
    Was there anyone who did not complete all the questions?
    Me. I think I got 1-8 all right though so I'll survive
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    (Original post by xsanda)
    Xsanda's solutions:
    R=-2
    ar¹⁰ = 3072
    Sum to 20 = -1048575

    trapezium: 21.4
    narrower strips

    triangle area: 32.1
    perimeter: 22.7

    expansion: 64+192ax+240a²x²+…
    a=2/3

    y=4x^(3/2)-7x-3
    (x-2)(x+5)(x-3)
    256
    negative section of integration counts negatively

    u₂₀ = 62
    Σ = 517 (show that)

    N=24
    n=17.1
    x=32, y=8

    6π-α
    3π-a
    graph of sin, half height
    3.32, 12.7
    Xsanda >>> Alfred W.
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    Why did the alpha thing have to be in radians?
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    (Original post by smoothER)
    Why did the alpha thing have to be in radians?
    Bc x was in rads
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    (Original post by frinldy dinosawr)
    UNOFFICIAL MARK SCHEME


    Q1) Fairly basic arithmetic sequences
    i) -2
    ii) 3072
    iii)-1048575

    Q2) Trapezium rule 21.4 units^2

    Q3) i) Area of shaded region 32.1cm^2
    ii) Perimeter of segment 22.7cm

    Q4) i)64+192ax+240a^2x^2
    ii)2/3

    Q5) i)4x^3/2-7x-3


    Q6) i) 62
    ii) Its a show that question: split up the sum of into two parts and verify

    Q7) i) N=24
    ii) n=17.1
    iii) y=8 x=32

    Q8) i)(x-2)(x-3)(x+5) ii) 256 units^2 iv) Reasoning + Graph Q9) i) 6pi - a ii) 3pi-a
    iii) Graph
    iv) 3.32 and 12.7 and -0.44 and +0.44 or something like that, feel free to correct
    All good apart the last one, there's no -0.44 and +0.44, it only intersects the line twice in that entire section.

    Posted from TSR Mobile
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    N was equal to 24, stupidly I had the right workings and added one number wrong so I'll lose 1 mark ffs
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    (Original post by MartoEnchev)
    I got a quadratic and the 2 solutions were n=-44 and n=(a positive number not sure the exact value). n can't be negative though. However I don't think that was the right way to solve it
    obviously not as its rong but will we get any marks
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    (Original post by AlistairMcDonald)
    Bc x was in rads
    How come? I thought it just said 2/3cos1/3x or something. Am I supposed to know from that?

    Do you think any method marks if in degrees?
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    Can someone please write out a full solution to the aritmethic 2N and N 6 marker, as I have now uploaded the paper. Thanks
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    (Original post by frinldy dinosawr)
    UNOFFICIAL MARK SCHEME


    Q1) Fairly basic arithmetic sequences
    i) -2
    ii) 3072
    iii)-1048575

    Q2) Trapezium rule 21.4 units^2

    Q3) i) Area of shaded region 32.1cm^2
    ii) Perimeter of segment 22.7cm

    Q4) i)64+192ax+240a^2x^2
    ii)2/3

    Q5) i)4x^3/2-7x-3


    Q6) i) 62
    ii) Its a show that question: split up the sum of into two parts and verify

    Q7) i) N=24
    ii) n=17.1
    iii) y=8 x=32

    Q8) i)(x-2)(x-3)(x+5) ii) 256 units^2 iv) Reasoning + Graph Q9) i) 6pi - a ii) 3pi-a
    iii) Graph
    iv) 3.32 and 12.7 and -0.44 and +0.44 or something like that, feel free to correct
    Sorry what was question 7ii?

    Edit - sorry, I mean wasnt it a show that question so whats the 17.2 for?
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    For the integration question about area, were you supposed to answer 256 or give the negative answer? (can't remember if it was -206.5 or -207.5). Because the next question asked why this was not the area for -5<x<3 and this would be 256.
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    Posted from TSR Mobile

    Last question Y coordinates 0.894 and - 0.894
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    (Original post by chris2015)
    A=53, B= 46. This was the trickiest paper I have ever came across.harder than any previous years.
    Reckon it's still easier than last years though :/ what was last year's boundary?

    Posted from TSR Mobile
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    (Original post by frinldy dinosawr)
    UNOFFICIAL MARK SCHEME



    iii)-1048575 a=3 r=-2
    a=3 r=-2
    s= 3(1-(-2^10))/1--2 = 1025?
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    (Original post by AlistairMcDonald)
    Just a heads up, not sure that you're allowed to post that here!!
    Yeah I wasnt sure, removed it now anyway. Thank you for warning me
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    (Original post by chris2015)
    A=53, B= 46. This was the trickiest paper I have ever came across.harder than any previous years.
    This would be lovely boundaries


    Posted from TSR Mobile
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    (Original post by frinldy dinosawr)
    UNOFFICIAL MARK SCHEME


    Q1) Fairly basic arithmetic sequences
    i) -2
    ii) 3072
    iii)-1048575

    Q2) Trapezium rule 21.4 units^2

    Q3) i) Area of shaded region 32.1cm^2
    ii) Perimeter of segment 22.7cm

    Q4) i)64+192ax+240a^2x^2
    ii)2/3

    Q5) i)4x^3/2-7x-3


    Q6) i) 62
    ii) Its a show that question: split up the sum of into two parts and verify

    Q7) i) N=24
    ii) n=17.1
    iii) y=8 x=32

    Q8) i)(x-2)(x-3)(x+5) ii) 256 units^2 iv) Reasoning + Graph Q9) i) 6pi - a ii) 3pi-a
    iii) Graph
    iv) 3.32 and 12.7 and -0.44 and +0.44 or something like that, feel free to correct
    can u update it and say how many marks each q was please
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    We summon you, Mr M! Please bless us with your mark scheme with mark-breakdown for each question, oh lord!
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    How many marks would you get for the sum2N - sumN question for obtaining an incorrect quadratic and attempting to complete the square?
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    for the simultaneous log one i got x=0.5 and y=512 would i get any marks as when you substitute it into log2x + log2y=8 you get 8?
 
 
 
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