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OCR (not MEI) C2 - Wednesday 20th May 2015 Watch

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    Maximum of 40 marks... woo...
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    Couldnt find N. Made a quadratic that didnt work but then managed to fumble the right answer using trial and error. N was 24 for anyone wondering
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    What did a = for the coefficient q
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    It was a really weird exam, they pulled questions that Hadn't been used in the whole spec


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    Xsanda's solutions:
    1i) common ratio:[1]
    r=-2
    1ii) 11th term: [2]
    ar¹⁰ = 3072
    1iii) sum to 20: [2]
    Sum to 20 = -1048575

    2i) trapezium rule: [4]
    21.4
    2ii) improvement: [1]
    more/narrower strips

    3i) shaded area: [4]
    32.1
    3ii) perimeter: [5]
    22.7

    4i) Binomial expansion: [4]
    64+192ax+240a²x²+…
    4ii) Find a: [5]
    a=2/3

    5) Double antidifferential: [7]
    y=4x^(3/2)-7x-3

    6i) Factorise: [4]
    (x-2)(x+5)(x-3)
    6ii) Integrate: [4]
    256
    6iii) Problem with integral: [2]
    negative section of integration counts negatively

    7i) u₂₀: [2]
    62
    7ii) Σ 10 to 20 show that: [3]
    517
    7iii) Σ N to 2N: [6]
    N=24

    8a) log 2^(n-3) = 18000: [4]
    n=17.1
    8b) simultaneous logs: [5]
    x=32, y=8

    9ia) Next solution of cos: [1]
    6π-α
    9ib) Solution with negative: [2]
    3π-a
    9ii) Draw graph: [2]
    graph of sin, half height
    9iii) Solve graphs: [4]
    3.32, 12.7
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    (Original post by ben5312)
    What did a = for the coefficient q
    2/3


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    In the time it'd been running


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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: 20.7

    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
    Speedy x_x
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    That has just given me so much confidence! Thanks xandra


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    Very last part of the last section I messed up straight away by dviding 2cos1/3x by cos1/3x.. Carried on though to get two solutions correct to what I was doing.... Any marks at all? How marks was it worth?
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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: 20.7

    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
    I've gotta know... How the heck do you remember all of those answers?!
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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: 20.7

    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
    For the sum to 20 one did you but the negative number in brackets? I got the same as you but put it in brackets and got a more sensible andwer if i remember correctly!
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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: 20.7

    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
    For the expansion I put 240(ax)^2, will I lose a mark for this because it said to simplify so I didn't know if I had to expand out the brackets fml
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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: 20.7

    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
    May just be me, but 12.7 isn't in the range of 0 to 2pi
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    (Original post by eddy29)
    May just be me, but 12.7 isn't in the range of 0 to 2pi
    The range is 0 — 6π, because it was sin(1/3 x) so the range had to be tripled.
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    (Original post by Zallin98)
    For the sum to 20 one did you but the negative number in brackets? I got the same as you but put it in brackets and got a more sensible andwer if i remember correctly!
    I checked it manually, which was stupid, as I ran out of time by q9, but I got the same as he/she did
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    (Original post by eddy29)
    May just be me, but 12.7 isn't in the range of 0 to 2pi
    You had to find the points of intersection on the graph, which covered a greater range than 0 to 2\pi.
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    (Original post by fayecaravelli)
    For the expansion I put 240(ax)^2, will I lose a mark for this because it said to simplify so I didn't know if I had to expand out the brackets fml
    I left mine in the same form you left yours in- should get the marks because that is simpler.
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    Anyone reckon 60 will be an a :/
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    I got the perimeter as 22.7
 
 
 
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