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AQA Core 1 June 2010 Question paper & Mark Scheme Watch

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    (Original post by ibysaiyan)
    oh you thank you very much! +rep from me
    i thought u said you had c4 june 2010 ,Pim not doing c1
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    (Original post by Maria1234)
    i thought u said you had c4 june 2010 ,Pim not doing c1
    I have c3 if you want.I need c4 as too (
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    1 bii to find equation AD we know AD is perpendicular to AB hence we use the gradient AB to be the reference gradient. i.e say if gradient AB is -1/2,then gradient of AD will be 2.
    For part c once you got the equation. Use substitution to get the co-ordinate.
    Question 5: r = square root 61 or r^2 = 61.
    I had difficulty with part 6 (i) too for the you need to consider all the sides:
    so you get:3x(y) +4x(y)+5x(y)+3x (4x) = 144
    sort them out and you should end up with the given equation.

    One way of solving part 7 ii is by finding out the determinant.If determinant < 0 then we have no real roots.
    1bii) where do i get the coordinates of AD from cos i am only given D .. do i use that?
    i got 3x-2y+5=0 but when i use that for the next question i cannot get the coordinates using simultaneous equation.

    Also can you explain how to do 7ai please as i got an answer of 2(x-10)^2 -47 but thats not correct i dont think..

    Thank you v much for explaining the others
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    Okay, so I just did this paper and I can explain those two questions.

    For part 1bii) you know that BAD is a right angle. Therefore the gradient of AD is 3/2 as m1*m2=-1 and m1 is -2/3. The line AD passes through both the points A and D so you can calculate the line equation with either of them. As we are given D it is easier to use that point.
    You should get y-7=3/2(x-3),
    which simplifies to give 2x-3y+15=0

    It should then be easy to use for the next question as you can substitute x=5y-6 into this line equation...

    For 7ai) the 2 in front of the brackets means that it's factorised. Therefore you divide what you would normally put in by 2 so by completing the square normally without 2x^2 we would put (x-10)^2 as you've done but becasue there is a coefficient of 2 we would put 2(x-5)^2. You got -47 as 10^2 =100. But as the correct answer is 2(x-5)^2 we would do 5^2 instead - but we also have to multiply that by the 2 in front of the brackets. THis gives 2(x-5)^2-50+53 so if simplifies to give 2(x-5)^2+3.

    Hope that helped, just ask if you are stuck at all!!
    Hey thank you for your help!

    For 1bii) I understand that gradient of AD is 3/2 as it is perpenidcular to AB, Then I used the D coordinates and substitued and got y-7=3/2(x-3) ... but now im confused as to how it simplifies to be 2x-3y+15=0 ... I simplified it and got 3x-2y+5=0 :| Also how do i substitute x=5y-6 to get the coordinates of B? I thought i had to factorise or use simultaneous equations?

    Thankyou for explaining 7ai) .. I understand that now
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    1c - You have to do simultaneous equations -

    (BC) 5y - x = 6
    (AB) 3y + 2x = 14

    From that you can work out x and y.
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    1c - You have to do simultaneous equations -

    (BC) 5y - x = 6
    (AB) 3y + 2x = 14

    From that you can work out x and y.
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    1c - You have to do simultaneous equations -

    (BC) 5y - x = 6
    (AB) 3y + 2x = 14

    From that you can work out x and y
    Thats what i thought you did but i couldnt get 3y + 2x =14 ... I got y-7=3/2(x-3) and thought it simplified to 3x-2y+5=0 :s:s How did you get that equation?


    Oh wait you use the equation from 1a! I get it! But is the answer to 1bii) 3x-2y+5=0?
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    (Original post by qwerty54321)
    So sorry about the delay of copying this up.
    Here is the mark scheme that i copied down from class:

    1a) y=-2/3x +14/3
    gradient is -2/3
    b) y=-2/3x+9
    ii) you'll know if you got this one right or not if you got to the right answer.. if you didn't quote me and i'll do it later

    c) y=2, x=4 --> (4,2)

    2a) 14-6root5
    b)-11+5root5


    3ai) likewise you'll know if yuo got this one right.. you must say x+3 is a factor

    ii) long division correct --> (x+3)(x+5)(x-1)

    b)p(2)=35
    ci)p(-1) = -16
    p(0) = -15
    therefore p(-1) is less, meaning the lowest point on the graph is nearer to when y=-1, to y = 0

    cii) draw correctly, going through -5, -3, and 1 on the x axis, and -15 at the bottom
    IMPORTANT: because for part ci we found that the lowest point (the vertex) is around where y=-1, the lowest point on yourgraph CAN'T be where it crosses the y axis. the vertex must be slightly to the left of the y axis. quote me if you need help.

    4a) 1/5x^5 - 4x^2 + 9x (+x)
    42/5
    4aii) y=9 integrate 9
    9x
    sub in 2
    =18
    18-42/5 = 48/5

    b) dy/dx = 4x^3 -8
    where x = 1
    = -4
    ii) where x=1, y=2
    so equation is y=-4x+6

    5a) (x+5)^2 + (y-6)^2 = 5^2
    bi) 25=25
    bii) gradient = -4/3
    y=-4/3x-2/3
    iii) m = -7/2,4
    PM^2 =2^2 + (3/2)^2
    pm = 5/2
    PO = 2root2

    sp p is closer to M


    6ai)you'll know if you got this one right or not
    ii)
    v=6x^2y
    rearrange 12=x^2 + y -> y=12/x -x
    v = 6x^2(12/x-x)
    v = 72x - 6x^3

    bi) dv/dx = 72 - 18x^2.. when x=2 it equals 0
    c) d^2v/dx^2 = -36x
    sub in when x=2 = -72 <0 so a maximum

    7) 2(x-5)^2 +3
    no real roots - when try to solve can't squareroot a negative number

    bi) you'll know if you got this right, ask me if you want me to go through it.
    ii) (7k+1)(k-1) <0 (PLEASE NOTE THERE SHOULD BE A LESS THAN OR EQUALS SIGN, WHERE I HAVE JUST PUT A LESS THAN because i dont know how to do it otherwise!)
    iii) -1/7 < k < 1

    again, there should be a less than or equals sign, not just less than. !!

    Hope this helped everyone, sorry again about the delay, and if you're stuck quote me and i'll try and help
    what do you mean about 3cii? The y intercept is not -15?
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    (Original post by Shazui)
    what do you mean about 3cii? The y intercept is not -15?
    Yes it is?
    Y intercept is where x = 0
    so x^3 + 7x^2 + 7x -15

    so, when x is 0, y=-15

    ?

    Or have i made a really silly mistake? what do you think it is?
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    (Original post by qwerty54321)
    Yes it is?
    Y intercept is where x = 0
    so x^3 + 7x^2 + 7x -15

    so, when x is 0, y=-15

    ?

    Or have i made a really silly mistake? what do you think it is?
    He /sheprobably got confused because of what you stated below about the vertex. =]
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    (Original post by qwerty54321)
    Yes it is?
    Y intercept is where x = 0
    so x^3 + 7x^2 + 7x -15

    so, when x is 0, y=-15

    ?

    Or have i made a really silly mistake? what do you think it is?
    No, I'm pretty sure the -intercept is at -15 but I was asking what you meant by what you stated after. What would the graph look like?
    Im supposing you get a mark each respectively for stating the 3 points at which it crosses the x-axis and the y intercept so I don't see what else there would be to draw?
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    Oh basically, you know how p(-1)<p(0)
    that means, the y value when x=-1 is less than the y value when x=0
    so you know the vertex has to be around when x=-1, therefore to the left of the yaxis. basically the vertex isn't dead on the yaxis. sorry i wasn't too clear and probably made a mistake typing up the answer to that part of the question
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    hmm....
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    does anyone know the link of where i can get the c1 aqa maths paper june 2010 online ?
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    (Original post by shaheen :))
    does anyone know the link of where i can get the c1 aqa maths paper june 2010 online ?
    http://www.fileserve.com/file/ZUT6QEt
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    anyone get 6). a). ii).
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    My Maths teacher did a set of answers for this paper to help us revise. This is what she came up with, I don't quite understand how she's done it, but it may help..

    6)a)ii) Volume = 6X^2 x Y
    12-X^2/X
    V = 6X^2(12-X^2)/X
    V = 6X(12-X^2)
    V = 72X - 6X^3
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    how do you work out the gradientin 1a
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    im only joking
    does1 know where i can get the mark scheme for this paper
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    [/QUOTE]I am looking for core 4 aqa june 2010
    anyone?
    thanks
    [/COLOR][/COLOR]
 
 
 
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