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    (Original post by TCO)
    Thanks I think I messed up my arithmetic got N to be 125/3 arghh

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    i got exactly the same!! what have we done wrong??
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    (Original post by Jmedi)
    If you calculate it you get 256 and this is incorrect as it should be higher than this;
    some area gets subtracted as a small part of the area is below the x axis
    I got 256 as did my Teacher and most of my class I'm sure it's right?????
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    (Original post by DominoMatrix227)
    What did you get for the 2N thing? I got -44 and 125/3 which I think are both wrong.

    Didn't you do any of 9?
    i got that and everyone seems convinced it is not the case, I'm not sure why!
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    (Original post by Dom1997)
    I got 256 as did my Teacher and most of my class I'm sure it's right?????


    256 is the correct answer for the integral however this is not the true area of the x axis to the graph line as it goes below the x axis. (Hence this was the following part of the question.
    (The true area was around ~258 )
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    (Original post by clarafitzgerald)
    i got exactly the same!! what have we done wrong??
    Calculation of N ; C2 2015 q.pdf
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    (Original post by clarafitzgerald)
    i got that and everyone seems convinced it is not the case, I'm not sure why!
    What's the 125/3 term of a sequence?

    That's why it's wrong.
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    (Original post by Orangeflower360)
    TBH i got the 3.3 wth didnt i add the 180 degrees my dayss idiot right here :' thank you for taking the time!
    no problem
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    (Original post by Parallex)
    What's the 125/3 term of a sequence?

    That's why it's wrong.
    No need to be so pretentious :)
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    really don't understand why that method is used
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    If anyone is looking for an unofficial markscheme, I'll post it again:

    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 DominoMatrix227)
    I have already asked this, I have no idea! How did you fin d the rest of the paper?
    **** hard
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    did any one else get the trapezium rule question as 24.8???
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    (Original post by xsanda)
    If anyone is looking for an unofficial markscheme, I'll post it again:

    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
    How sure are you that these are right? I pretty much got the EXACT answers here! Did a teacher work them out?
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    (Original post by JonoTaylor)
    How sure are you that these are right? I pretty much got the EXACT answers here! Did a teacher work them out?
    I've checked these with several other people who did the test, and so I'm fairly sure they're right.
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    (Original post by DominoMatrix227)
    What did you get for the 2N thing? I got -44 and 125/3 which I think are both wrong.

    Didn't you do any of 9?
    I got N as 24.random numbers and the answer was exactly 24, so hopefully I got a lot of method marks for it.

    I drew the graph.
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    (Original post by Jezzer08)
    did any one else get the trapezium rule question as 24.8???
    I did!

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    OKAy. It was a decent paper. Not too hard to get an A and not too easy to get a D/ high C. I reckon I got around 53/54 out of 72 which is okay. One problem that I had is that through core 3 my teacher made a big deal about not dividing cos(x), sin(x) and tan(x) as to not lose any solutions so when I got to that question, the internalized fuss that he made kicked in so I subtfracted stuff and factorised stuff and ultimately mesed up those 4 marks. Can someone explain some cases where you can didvide those trigonometric functions so that I wont make the same. Istake in core 3

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    (Original post by xsanda)
    If anyone is looking for an unofficial markscheme, I'll post it again:

    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
    for the binomial question could you simplify it by dividing all the values by 8??
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    Is mr m posting mark scheme for c2?


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    I re did the last question and went over it with a teacher, getting 3.3 and 6.1 rad.. I didn't solve it using tan, but did it by squaring both sides - did anyone else do this/get this as an answer?



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