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    (Original post by HecTic_H)
    The question in the unofficial mark scheme as you've written differentiation? Does anyone have any idea?

    Also, the k question was (I think) Q8, and the coordinate geometry was perhaps 4?
    There was also a disguised quadratic with x = -8 and 27 or something like that.
    The marks allocated in the order you have written your questions for the unofficlal mark scheme were I think:

    1. 1) 4 + 4root3 [3]
    2) can't really describe the graph, if you know what y=1/x looks like then reflect in either axis. [2]
    Translation of 2 in x was -1/(x-2) [1]
    The translation they wanted you to state was a stretch of 1/3 in either axis. [2]
    3) 5^8 [1]
    5^(-1/4) [2]
    5^(9/2) [2]
    4) Disguised quadratic
    x = -8 and 27 [5?]5) Line AB = ? [2]
    Line was 6x-8y-29=0 [7]
    5) differentiation? Can't remember the answer to this, was reasonably straight forward though if you expanded brackets.?
    6) simultaneous equation - x=1, y=2 or x=14/3, y=-13/3 [5]
    7) differentiate x^(-1/3)? Should get out -1/48 when you put x in. [4]
    8) Draw the graph [4]
    X < -1, X > 3/2 for y=0 [2]
    k < - 25/8 [3]
    9) a=13 [5]
    Second derivative was positive therefore min point [2]?
    X=1/3 [2]?
    10) centre (5,-2) radius 5 [3]
    Prove tangent [5]
    Area of triangle = 40 [4]


    I tried to amend it, but I could be completely wrong lol
    Ah thank you! Completely forgot about that quadratic one, the answers I got were -8 and 27 though so I'll go with these as being right Can't remember the differentiation question for the life of me, it was something like f(x)=(x^2+7)(8-x) and you had to find f'(x). I'll edit my post now though, thanks again
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    (Original post by K-King121)
    Correct working as in the correct coordinates and correct working for area of triangle?
    Both; you should have this anyway if you got 40, unless you guessed 40 and wrote nothing else!
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    Hi, can anyone break down what they think the 4 marks would be for for question 7? Gradient of x^-1/3 when x=-8.
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    Somehow got 10root5 for last question but had the coordinates right... How many marks will I drop? :confused:
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    (Original post by chloe-jessica)
    In the original post, click 'show spoiler'. The questions aren't in the right order as I can't remember the order (I have said people can correct me but nobody has) but the answers are there.
    Sorry but I can't find it ???
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    (Original post by Charlieukharris)
    Sorry but I can't find it ???
    1. 1) 4 + 4root3 [3]
    2) can't really describe the graph, if you know what y=1/x looks like then reflect in either axis. [2]
    Translation of 2 in x was -1/(x-2) [1]
    The translation they wanted you to state was a stretch of 1/3 in either axis. [2]
    3) 5^8 [1]
    5^(-1/4) [2]
    5^(9/2) [2]
    4) Disguised quadraticx = -8 and 27 [5?]
    5) Line AB = ? [2]
    Line was 6x-8y-29=0 [7]
    5) differentiation? Can't remember the answer to this, was reasonably straight forward though if you expanded brackets.?
    6) simultaneous equation - x=1, y=2 or x=14/3, y=-13/3 [5]
    7) differentiate x^(-1/3)? Should get out -1/48 when you put x in. [4]
    8) Draw the graph [4]
    X < -1, X > 3/2 for y=0 [2]
    k < - 25/8 [3]
    9) a=13 [5]
    Second derivative was positive therefore min point [2]?
    X=1/3 [2]?
    10) centre (5,-2) radius 5 [3]
    Prove tangent [5]
    Area of triangle = 40 [4]
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    (Original post by Rr_rolo)
    I wrote 40 but no units squared would i lose marks
    No, I highly doubt it
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    (Original post by jamespaine)
    Hi, can anyone break down what they think the 4 marks would be for for question 7? Gradient of x^-1/3 when x=-8.
    Yep.

    y = x-1/3

    dy/dx = -1/3 x-4/3 [2]
    dy/dx|x=-8 = -1/3 (3√-8)4 [1]
    = -1/3 times 1/16
    = -1/48 [1]
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    (Original post by TCO)
    625 root 5 I think

    Posted from TSR Mobile
    it was
    (5xroot5) ^3

    5 x root5 can be seen as 5 ^3/2

    therefore using indicises laws, multiply the two and you get:

    5^9/2 or 5^4.5
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    (Original post by mountaindewwwww)
    That's wrong on so many levels, go back over your gcse work.
    i think you need to go over gcses it is definitely a third
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    probably 2 What did you guys get for the value of k?
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    Do you think I'd lose a mark for putting 4root3 + 4 rather than 4+4root3?

    Nothing big but i'm aiming for full raw marks
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    (Original post by ChoccyPhilly)
    Do you think I'd lose a mark for putting 4root3 + 4 rather than 4+4root3?

    Nothing big but i'm aiming for full raw marks
    Not at all haha
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    (Original post by chloe-jessica)
    1. 1) 4 + 4root3 [3]
    2) can't really describe the graph, if you know what y=1/x looks like then reflect in either axis. [2]
    Translation of 2 in x was -1/(x-2) [1]
    The translation they wanted you to state was a stretch of 1/3 in either axis. [2]
    3) 5^8 [1]
    5^(-1/4) [2]
    5^(9/2) [2]
    4) Disguised quadraticx = -8 and 27 [5?]
    5) Line AB = ? [2]
    Line was 6x-8y-29=0 [7]
    5) differentiation? Can't remember the answer to this, was reasonably straight forward though if you expanded brackets.?
    6) simultaneous equation - x=1, y=2 or x=14/3, y=-13/3 [5]
    7) differentiate x^(-1/3)? Should get out -1/48 when you put x in. [4]
    8) Draw the graph [4]
    X < -1, X > 3/2 for y=0 [2]
    k < - 25/8 [3]
    9) a=13 [5]
    Second derivative was positive therefore min point [2]?
    X=1/3 [2]?
    10) centre (5,-2) radius 5 [3]
    Prove tangent [5]
    Area of triangle = 40 [4]
    I think the differentiation one was f'(x)=-3x^2+10x-3 and f(x)=-x^3+5x^2-3x+15? Which is f(x)=(5-x)(x^2+3)
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    (Original post by Renzhi10122)
    I think the differentiation one was f'(x)=-3x^2+10x-3 and f(x)=-x^3+5x^2-3x+15? Which is f(x)=(5-x)(x^2+3)
    Yeah that's right; how do you even remember that...
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    (Original post by Renzhi10122)
    I think the differentiation one was f'(x)=-3x^2+10x-3 and f(x)=-x^3+5x^2-3x+15? Which is f(x)=(5-x)(x^2+3)
    That sounds right to me thank you!
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    (Original post by Madhutty)
    Yeah that's right; how do you even remember that...
    Haha, I just worked backwards
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    (Original post by Madhutty)
    Yep.

    y = x-1/3

    dy/dx = -1/3 x-4/3 [2]
    dy/dx|x=-8 = -1/3 (3√-8)4 [1]
    = -1/3 times 1/16
    = -1/48 [1]
    Would I get any marks if I misread it as y= x^1/3? so did

    dy/dx = 1/3x^-2/3

    1/3(-8)^-2/3 = 1/12
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    (Original post by jamespaine)
    Would I get any marks if I misread it as y= x^1/3? so did

    dy/dx = 1/3x^-2/3

    1/3(-8)^-2/3 = 1/12
    2/4 for correct differentiation, in theory 3/4 for correct substitution and arithmetic, but you'd definitely lose the final [A1] mark.
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    (Original post by jamespaine)
    Would I get any marks if I misread it as y= x^1/3? so did

    dy/dx = 1/3x^-2/3

    1/3(-8)^-2/3 = 1/12
    Possibly a mark for attempting differentiation but that's all i can think of
 
 
 
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