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Edexcel A2 C4 Mathematics June 2016 - Official Thread Watch

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    When doing an integration question that asks for the region R for example. If you area is negative, is it ok to leave it as negative??
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    (Original post by boyyo)
    When doing an integration question that asks for the region R for example. If you area is negative, is it ok to leave it as negative??
    If its asks for the area of the region, then no - you need to give it as a positive answer.
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    (Original post by edothero)
    First you deal with the x

    So you would have

    0 < 2x < 720

    Then if you're adding/subtracting anything

    -60 < 2x-60 < 660

    Then once you have your values you do it in the opposite order

    so \dfrac{(x+60)}{2}


    Exactly as edothoro says. When working out your values using your CAST diagram or graph, you'll be finding all the values for 2x-60, so to get x you can always just set your list of answers equal to 2x-60, and just rearrange to get x.
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    (Original post by Zacken)
    If its asks for the area of the region, then no - you need to give it as a positive answer.
    Oh ok, so when should you leave it as negative?
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    (Original post by boyyo)
    Oh ok, so when should you leave it as negative?
    When it asks you to integrate something without context. "Evaluate this integral" or "do this integral" or w/e.
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    (Original post by Zacken)
    When it asks you to integrate something without context. "Evaluate this integral" or "do this integral" or w/e.
    ahh ok cool thanks
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    Hi, I think the question attached is improper, if so why don't you carry out long division before finding the partial fractions?
    Thanks
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    (Original post by ryandaniels2015)
    Hi, I think the question attached is improper, if so why don't you carry out long division before finding the partial fractions?
    Thanks
    It is improper - however, it is accounted for by the 'A' term.
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    (Original post by SeanFM)
    It is improper - however, it is accounted for by the 'A' term.
    so do you do long division to find the A term?
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    (Original post by ryandaniels2015)
    so do you do long division to find the A term?
    Proceed as normal - multiply both sides by (x-1)(x+2). Work out B and C first and then you should be able to work out what A is after that.
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    (Original post by SeanFM)
    Proceed as normal - multiply both sides by (x-1)(x+2). Work out B and C first and then you should be able to work out what A is after that.
    Thanks, very strange because on some questions you have to do long division first, why is it different here?
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    (Original post by ryandaniels2015)
    Thanks, very strange because on some questions you have to do long division first, why is it different here?
    You can use long division if you want. They are two different methods to get the same answer.
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    (Original post by ryandaniels2015)
    Thanks, very strange because on some questions you have to do long division first, why is it different here?
    :holmes: I do not know, sorry

    Another way of looking at it (without long division) is to use a tricks like so:

    (x-1)(x+2) = x^2 + x -2

    \frac{2x^2+5x-10}{x^2 + x -2} =\frac{(2x^2+2x -4) + (3x-6)}{x^2 + x -2} as we wanted to find a multiple of the bottom on top, and then add the rest of the fraction that isn't in the first.

    = 2 + \frac{(3x-6)}{x^2 + x -2} == 2 + \frac{(3x-6)}{(x-1)(x+2)}

    Not useful now but with integration it's useful, as the second fraction can be turned into something that looks like \frac{kf'(x)}{f(x)} where k is some constant, which can be integrated to give klnmod(f(x)) and then the final term is a constant that can be integrated using partial fractions. This is, of course, useless in C4 but the more you know...
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    Do we need to know proof by contradiction? It comes up in the Solomon papers but I don't think I've seen it in a regular paper, easy enough to do just can be a bit awkward
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    Thanks makes sense! Thanks for the help!

    (Original post by SeanFM)
    :holmes: I do not know, sorry

    Another way of looking at it (without long division) is to use a tricks like so:

    (x-1)(x+2) = x^2 + x -2

    \frac{2x^2+5x-10}{x^2 + x -2} =\frac{(2x^2+2x -4) + (3x-6)}{x^2 + x -2} as we wanted to find a multiple of the bottom on top, and then add the rest of the fraction that isn't in the first.

    = 2 + \frac{(3x-6)}{x^2 + x -2} == 2 + \frac{(3x-6)}{(x-1)(x+2)}

    Not useful now but with integration it's useful, as the second fraction can be turned into something that looks like \frac{kf'(x)}{f(x)} where k is some constant, which can be integrated to give klnmod(f(x)) and then the final term is a constant that can be integrated using partial fractions. This is, of course, useless in C4 but the more you know...
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    FML ... I thought equilateral triangle = 45 degrees(was thinking of C3 triangles :shot: ) :facepalm: :rip: @C2

    I worked out that If I don't get A*(90+90) in C3 & C4 + B(70) in M1 I'm not going to get my A* nor will I scrap 480/600 ums for 'A' grade. :argh:

    Pressure is definitely on arguably at the highest it has ever been :lol:
    I can't afford to think about previous exams but whats left and whats required to get the grades ... :sigh:
    90, 90, 70 --> C3, C4, M1 + A* in unit 4, 5 Physics for my A*A* offers.

    Its do or die. :cry2: ----> :argh: ----> back to :work: :lol:

    Thankfully, I've been scoring 92-95% ish in C4 and 90-95% in C3 ... damn this pressure make me :puke:
    Time to finish up all C34 IAL papers before end of May and then grind Madasmaths papers for the entire June month.
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    How does the c4 mark scheme work, for example on the question where they ask to find cartesian equation from parametric equations, if I work it out using a different method to the mark scheme but get the same answer, do I get all the marks?
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    (Original post by ryandaniels2015)
    How does the c4 mark scheme work, for example on the question where they ask to find cartesian equation from parametric equations, if I work it out using a different method to the mark scheme but get the same answer, do I get all the marks?
    Depends on if your method is mathematically sound or not. Though if you end up with the same answer chances are your Maths is correct. You would be fine.
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    (Original post by edothero)
    Depends on if your method is mathematically sound or not. Though if you end up with the same answer chances are your Maths is correct. You would be fine.
    Thanks for the reply, so the method does not have to be the same as the mark scheme?
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    (Original post by XxKingSniprxX)
    I thought equilateral triangle = 45 degrees(was thinking of C3 triangles :shot: )
    Dude, what's a C3 triangle? Is that on the spec? :eek:
 
 
 
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