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    (Original post by StephHowarth)
    I got that!
    I also got that
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    (Original post by a10)
    today is getting better xD so far I reckon iv got 100ums if im lucky with the grade boundaries.

    wat did u get for the question that said wat values of x is the curve decreasing?
    X <1/4


    (Original post by StickySteve)
    It was to solve a quadratic, with x^6 + x^3 + (..) or something and a typical method would be to substitute a = x^3 or something, and rewrite it as a^2 + b + c

    I got x = + - 1/2

    and my other x value was cuberoot of -1 ? But i mistook that for a square root though X(
    Ah I remember, I got 1/2 and -1.
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    (Original post by StickySteve)
    It was to solve a quadratic, with x^6 + x^3 + (..) or something and a typical method would be to substitute a = x^3 or something, and rewrite it as a^2 + b + c

    I got x = + - 1/2

    and my other x value was cuberoot of -1 ? But i mistook that for a square root though X(
    Wasnt it just x = -1/2 & x = -1?
    I thought y = -1/8 & y = -1 so you just cube root both values to get x which would be -1/2 & -1
    Or did I miscalculate something. :s
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    K=-5
    A=(-2,-27)
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    last question was -3,-27
    k= -5
    the paper in general was good i thought. The first surd was booooky tho lol
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    (Original post by Jack_King)
    Question 10iii) For the coordinates of A, I set the line equal to the curve, and then equated it to 0, then solved it using the remainder theorem, to obtain a quadratic, solved the quadratic and got X to be -2, subbed X value into the curve and got Y to be -27. So the point = (-2,-27) Would I get 5/5 marks for this method?
    I did the same thing as you but because I had my K as -4, I got completely different values. But that is a completely valid method I believe.
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    (Original post by XiLE)
    Wasnt it just x = -1/2 & x = -1?
    I thought y = -1/8 & y = -1 so you just cube root both values to get x which would be -1/2 & -1
    Or did I miscalculate something. :s
    I got the same, think its correct
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    What did people get for question 8? (the other 7 marker)
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    (Original post by eggfriedrice)
    X <1/4




    Ah I remember, I got 1/2 and -1.
    1/4???? wer did u get that from I put

    x>-3/2
    x < 4
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    what do you think the minimum mark will be for an A?
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    (Original post by Lilhazzaman)
    First you say the dy/dx = -3x^2 ect.
    Then you say dy/dx = 3x^2 ect.

    I dont understand where the Negative has come from. Was the original equation; (x-1)(x^2+4x+k)?
    yeah sorry, typing on phone.

    dy/dx=-3x^2-6x+9
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    fuuuuuuuccckk cube rt -1 is still -1, i was thinking of sqrt -1 giving no solution. well, i was bound to lose one mark
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    (Original post by Newbie95)
    I did the same thing as you but because I had my K as -4, I got completely different values. But that is a completely valid method I believe.
    Same, I got K as -4! Don't know where I went wrong, but used all correct methods so out of the 12 marks for both the tangent and stationary point I hope to have got 8/9-ish

    Posted from TSR Mobile
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    (Original post by mikemoore64)
    douse any one have the paper or is any one working on an unofficial mark scheme i would really like to know what i got ruffly.
    *roughly

    and mr m will put up unofficial MS tonight
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    I hope someone's cookin' up a mark scheme...
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    How many marks can you drop and still get 100ums, or do you need full raw marks for 100ums?
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    (Original post by eggfriedrice)
    Guys, for the inequalities question
    3-8x>4
    Did you get x<-1/8 or x>-1/8
    the first one
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    (Original post by a10)
    1/4???? wer did u get that from I put

    x>-3/2
    x < 4
    This is the quadratic right?
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    Question 8?
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    (Original post by GeorgeBarrett1)
    How many marks can you drop and still get 100ums, or do you need full raw marks for 100ums?
    if the A is 57 it may drop down to 71, and so on. its usually full raw in summer as everyone in yr 13 is resitting, or atleast thats the case in my 6th form
 
 
 
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