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Edexcel Core 3 - 21st June 2016 AM Watch

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    (Original post by Bigboss27)
    How did people do q1😭
    substitute the function g into f, didn't it ask for fg(x)?
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    (Original post by Whizbox)
    Nah I used algebraic long division
    did you remainder end up like so

    4x+12

    4(x+3)/(x+3)(x-2)

    so 4/x-2
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    (Original post by timothykizito123)
    Arcsin(x+1)=pi/3
    X+1=root3/2
    X=root3/2-1
    Arcsin = -pi/3.
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    (Original post by js1953)
    what was Q 1a?
    What did all of question 1 involve i cant remember😂
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    Hoping the grade boundary to be lower. Last question was horrible so lost 5 or 6 marks, so I'm hoping I just managed to get an A*
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    My sister just sat this exam and she realised afterwards that she completely missed out the last question because she didn't see it.

    Will this be catastrophic if she wants a high A?
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    where the **** is arsey im having a meltdown
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    (Original post by mangoli)
    I used long division + factor theorum, although now i've seen it partial fractions would've been easier
    (Original post by Yua)
    did you remainder end up like so

    4x+12

    4(x+3)/(x+3)(x-2)

    so 4/x-2
    ????
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    (Original post by Iamvery1/cos(c))
    My sister just sat this exam and she realised afterwards that she completely missed out the last question because she didn't see it.

    Will this be catastrophic if she wants a high A?
    not if everything else was right
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    (Original post by jwanjwan)
    where the **** is arsey im having a meltdown
    He doesn't do model answers anymore
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    (Original post by Yua)
    am i the only one that didn't use partial fractions?
    Nope I just did it the core 3 way, divided, got a remainder, put that over the original fraction, factorised top and bottom and cancelled.
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    (Original post by blahblah21)
    Q1a, x=6, x=-5
    b. a=6
    2a. dy/dx = (20-4x^2)/(x^2+5)^2
    b. x>rt(5) x<-rt(5)
    3a. rt(5)cos(theta +26.57)
    b. theta = 33.0, 273.9
    c. theta = 86.1
    4a. y=21, x=ln(2.5)
    y=25 (k=25)
    b. Proof
    c. x1= 1.4368
    x2= 1.4373
    d. Show the change in sign. (gradient function)
    5i. Turning points: 0.9463
    ii. 0.5cosec(4y)
    6a. A=3 B=4
    b. x+2y-35=0 (or equivalent)
    7a. arcsin(x) graph
    b. x= -(2+rt(3))/2
    8a. Proof
    b. x= 0.294, -1.277, 1.865 and 1 more I think
    9a. 6.740
    b. Proof
    c. T = 5ln(2+(2/e))
    5 ii was 2cosec4y I think
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    (Original post by mangoli)
    I used long division + factor theorum, although now i've seen it partial fractions would've been easier
    To be fair long division is remarkably quick for this question in particular.

    The remainder was nice and cancelled down into the form required beautifully.
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    (Original post by Yua)
    am i the only one that didn't use partial fractions?
    nope, i used polynomial division
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    how many marks was Q9?
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    x4+x3−3x2+7x−6x2+x−6=x2+x 6


    Anyone got the Algebraic Fraction method for me please?
    I have never been taught long division, and I know I got the answers wrong, just looking to see if I got some method marks.
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    Can anybody remember how much was question 1 a) worth?


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    (Original post by Anon-)
    how did you show that it was the right root to 3 decimal places using an appropriate interval?
    g(x) - (2(x)+43) with the appropriate x values will get you a change of sign
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    (Original post by AhnafR)
    Guys which equation did you substitute in the values to get the sign change to prove x=1.473?
    It was g(x) - (what g(x) was = to in that part)
    So the one with the es minus the one that was ax+b...
    That was probably useless sorry, but you might know what I mean? x
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    b + b/e = [7.5/(15e^-1 + 15)]^-1

    (be+b)/e = [7.5/(15e^-1 + 15)]^-1

    be+b = [7.5/(15e^-1 + 15)]^-1 * e

    b(e+1) = [7.5/(15e^-1 + 15)]^-1 * e

    b = [7.5/(15e^-1 + 15)]^-1 * e / e + 1

    b = 2

    (This looks horrible but LaTeX doesn't seem to be working, sorry)
 
 
 
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