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    (Original post by Tiwa)
    Am I the only one who got an area of 36 when measuring the area of the trapezium?
    =½(30+6)*3 = 54

    (Original post by tanyapotter)
    for the last question, because i couldn't factorise and didn't get the correct critical values, can i still get one mark if i made a sign diagram of what i did get and then do p < x < q? or have i lost all marks completely?
    You would probably get a few marks, maybe 1 or 2.


    (Original post by iAstro)
    For question 6, I differentiated it wrong I got completely put off by the signs.

    Will there be any carry over marks for the other parts, even though I differentiated it wrong I got a maximum value (by doing d^y/dx^2) and found a value when dy/dx=0.
    There should be.
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    (Original post by student0042)
    =½(30+6)*3 = 54



    You would probably get a few marks, maybe 1 or 2.




    There should be.

    Oh ok hopefully there is, does this mean that I would able to get full marks from those questions or just some?
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    (Original post by brokenlevel23)
    Sign diagram is a mark.
    really? that's great! thank you.

    also, what do you think the grade boundary for an A will be? judging by what ive read about the correct solutions, i think i've got 67/75 - would this secure an A?
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    (Original post by iAstro)
    Oh ok hopefully there is, does this mean that I would able to get full marks from those questions or just some?
    You definitely won't be able to get full marks, as one mark is for the correct answer usually. (as accuracy mark). You will be able to gain marks for method though, so you won't lose too many I don't think.
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    (Original post by brokenlevel23)
    FINAL MARKSCHEME:

    Question 1
    (a) -3/5 [2]
    (b)(i)
    perp. gradient = 5/3
    y-?=5/3(x-?)
    5x-3y+1=0 [3]
    (b)(ii) A(9,-4) [3]

    Question 2
    7+√15
    n=7 [5]

    Question 3
    (a) y=-10x-4 [4]
    (b)(i) 108/5 (oe, e.g. 21.6) [5]
    (b)(ii). 162/5 (oe, e.g. 32.4) [3]

    Question 4
    (a) (x+1)^2 + (y-3)^2 = 50 [2]
    (b)(i). C(-1,3) [1]
    (b)(ii) 5√2 [2]
    (c)
    k^2-6x-16
    (k-8)(k+2)
    k=8, k=-2 [2]
    (d) min. distance = 7 [2]

    Question 5
    (a) p = 3/2, q = -¼ [3]
    (b)(i) (-3/2, -¼) [2]
    (b)(ii) line of symmetry is x=-3/2 [1]
    (c) y=x^2-x+4 [3]

    Question 6
    **IT WAS AN OPEN TOP CYLINDER DAMNIT**
    (a)(i) h = 24/r -r/2 [3]
    (a)(ii) Sub πr^2 into part (i) to get V = 24πr -π/2 r^3 [3]
    (b)(i) 24π -3π/2 r^2 [2]
    (b)(ii)
    r=4 when dy/dx=0
    r=4 => d^2y/dx^2(=-12π)<0, therefore maximum [4]

    Question 7
    (a) http://i.imgur.com/rVJ169A.png [3]
    (b)(i) R = 36 [2]
    (b)(ii) R= 0 therefore root [2]
    (b)(iii)(x-2)(x^2-5x+10) [2]
    (b)(iv) x=2 [3]

    Question 8
    (a) (Show) x^2 + 3(k-2)x -13-k=0 [1]
    (b) 9k^2-32k-16<0 [3]
    (c) -4/9< k<4 [4]
    I am sure that question 7biv was x=-2
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    The cylinder question was very hard
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    (Original post by JacobC1998)
    I am sure that question 7biv was x=-2
    The equation was x^2(x-3) + 20 which the root is (x-2) hence x = 2.
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    (Original post by iAstro)
    Oh ok hopefully there is, does this mean that I would able to get full marks from those questions or just some?
    Because it was worth 4 marks total, and judging from past years, I'd say that you only get an M1 mark for the sign table
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    (Original post by JacobC1998)
    I am sure that question 7biv was x=-2
    (Original post by student0042)
    The equation was x^2(x-3) + 20 which the root is (x-2) hence x = 2.
    I remember it being -2 as it was (x+2)...
    Might have remembered incorrectly though...
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    (Original post by l1lvink)
    I remember it being -2 as it was (x+2)...
    Might have remembered incorrectly though...
    The question said that the factor was (x+2) i am certain of it
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    (Original post by 1423al-plleshi)
    did anyone know how many marks question 6 was?
    cylinder one was 11
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    (Original post by brokenlevel23)
    FINAL MARKSCHEME:

    Question 1
    (a) -3/5 [2]
    (b)(i)
    perp. gradient = 5/3
    y-?=5/3(x-?)
    5x-3y+1=0 [3]
    (b)(ii) A(9,-4) [3]

    Question 2
    7+√15
    n=7 [5]

    Question 3
    (a) y=-10x-4 [4]
    (b)(i) 108/5 (oe, e.g. 21.6) [5]
    (b)(ii). 162/5 (oe, e.g. 32.4) [3]

    Question 4
    (a) (x+1)^2 + (y-3)^2 = 50 [2]
    (b)(i). C(-1,3) [1]
    (b)(ii) 5√2 [2]
    (c)
    k^2-6x-16
    (k-8)(k+2)
    k=8, k=-2 [2]
    (d) min. distance = 7 [2]

    Question 5
    (a) p = 3/2, q = -¼ [3]
    (b)(i) (-3/2, -¼) [2]
    (b)(ii) line of symmetry is x=-3/2 [1]
    (c) y=x^2-x+4 [3]

    Question 6
    **IT WAS AN OPEN TOP CYLINDER DAMNIT**
    (a)(i) h = 24/r -r/2 [3]
    (a)(ii) Sub πr^2 into part (i) to get V = 24πr -π/2 r^3 [3]
    (b)(i) 24π -3π/2 r^2 [2]
    (b)(ii)
    r=4 when dy/dx=0
    r=4 => d^2y/dx^2(=-12π)<0, therefore maximum [4]

    Question 7
    (a) http://i.imgur.com/rVJ169A.png [3]
    (b)(i) R = 36 [2]
    (b)(ii) R= 0 therefore root [2]
    (b)(iii)(x-2)(x^2-5x+10) [2]
    (b)(iv) x=2 [3]

    Question 8
    (a) (Show) x^2 + 3(k-2)x -13-k=0 [1]
    (b) 9k^2-32k-16<0 [3]
    (c) -4/9< k<4 [4]
    For 8b how do u get to that equation? I know that we should use the discriminant but i still don't get the equation required
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    (Original post by SunDun111)
    Guys what was the integration question,can everyone remember the equation we had to integrate
    I think it was x^4 + 3x^2 + 2
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    (Original post by l1lvink)
    I remember it being -2 as it was (x+2)...
    Might have remembered incorrectly though...
    Actually you may be right because it was p(-2) = 0 there x + 2 =0 and x = -2
    Honestly I can't remember anymore.
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    (Original post by SunDun111)
    For 3a did you get gradient as -7 or -10
    Is this the tangent to the curve question if so it was -10
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    (Original post by snuvisol)
    How many marks will I get for drawing y=-x^2(x-3) instead of y=x^2(x-3) Anyone??
    How many marks was that question?
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    There are quite a few questions I've DEFINITELY got wrong, but how many working marks would I get if:

    For Q1Bii) To find A, I attempted a simultaneous equation with the equation given, and the one calculated, but have the wrong answer?

    Q2) I didn't distinctly state that N=7, however I got 7+rt15

    Q4d) To find the smallest length, I used pythagoras with the radius 5rt2 and minused it from another length I had (which is wrong), but would I get any marks for working??

    Q5c) I identified that the values in the complete square form changed, to (x+0.5)^2 + (y+2.5)^2, however I left 'c' the same and didn't minus the 1/4 from 0.5^2, so my quadratic was incorrect. How many working marks, if any?

    Question 6b) I stupidly differentiated 24pi(r)-pi^4/2, instead of 24pi(r)-pi^3/2 (I differentiated with the power 4 instead of 3). I think I did dy/dx and d2y/dx2 correctly, however I differentiated the wrong expression! Would I get any marks carried forward for this, or would I get 0?

    Question 7a) The cubic graph, I wrote down that x=0 and x=3, however the graph that I drew (despite going through 0 and 3) was wrong. Would I get any of the 3 marks?

    Question 8c) Another ridiculous mistake, I couldn't figure out the factors to factorise with (36 and -4), so put b^2-4ac from the previous question into the quadratic formula, however I didn't solve anything as I ran out of time. Would I get a mark for working for doing this?

    If you read all of this post, I thank you enormously!
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    (Original post by Emad1993)
    For 8b how do u get to that equation? I know that we should use the discriminant but i still don't get the equation required
    b^2 - 4ac < 0
    (3k-6)^2 - 4(13-k) < 0
    9k^2 + 36 - 36k -52 + 4k < 0
    9k^2 -32k -16 < 0 Q.E.D.
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    Did anyone do the AQA core 1 paper today?
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    I can probably post an unofficial mark scheme if someone could remember the questions, because I don't remember them.

    And yeah, for the cylinder question, the top was open. So the equation for the surface area was : 2(pi)rh + (pi)r^2 = 48.
    You get an equation for h in terms of r, and then you substitute that equation into V=2(pi)r^2h.
 
 
 
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