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Edexcel Core 3 - 21st June 2016 AM Watch

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    does anyone know where to get the IAL jan 2016 paper?
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    (Original post by lucabrasi98)
    Lads, what's the hardest paper apart from june 13 and all the gold papers+ solomon papers?
    I found june 2015 pretty hard towards the end
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    (Original post by jessicaxxx1)
    does anyone know where to get the IAL jan 2016 paper?
    https://0025309b76bc88a0e4c3444acd03...%20Answers.pdf
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    (Original post by Pablo Picasso)
    None of them are hard.
    great input pablo...
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    (Original post by Questioner1234)
    Please can someone explain solomon paper F question 3b? Thanks
    Well lny < 0 for 0<y<1, so you just need to find a value of x which gives 0 < 3x^2+5x+3 < 1
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    (Original post by imran_)
    do we need to know about the factor theorem ??? Ive never seen it on a past paper question for ededxcel
    Nope, but it doesn't really hurt to know it (as it can make it easy when it comes to simplifying algebraic fractions). All you have to know is that given a function f(x) = ax^2 + bx + c (or a cubic or even a quartic), then (x-a) is a factor if f(a) = 0
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    If I was solving |3x+2| = x+4 for example, I set 3x+2=x+4, then is the other one -3x+2=x+4 or -(3x+2)=x+4??
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    Guys,

    Should I do 4 Solomon papers today, or 2 solomon papers and 2 gold papers?
    (I've done Solomon A-D and the first gold already)
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    do we need to be able to divide quartics/cubics by quadratics and stuff using the remainder/factor theorem?
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    (Original post by abbey1)
    If I was solving |3x+2| = x+4 for example, I set 3x+2=x+4, then is the other one -3x+2=x+4 or -(3x+2)=x+4??
    3x + 2 = x + 4 and -(3x+2) = x + 4. Same rule applies to other modulus functions

    (Original post by alevelstresss)
    do we need to be able to divide quartics/cubics by quadratics and stuff using the remainder/factor theorem?
    Yes, but don't be discouraged as it's exactly the same way as dividing a quadratic by (x-a) (except sometimes you're left with a remainder function).
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    (Original post by abbey1)
    If I was solving |3x+2| = x+4 for example, I set 3x+2=x+4, then is the other one -3x+2=x+4 or -(3x+2)=x+4??
    Yes -(3x+4). Always put it into the calc if unsure
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    Thank you!
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    Name:  image.png
Views: 87
Size:  80.4 KBSomeone help on 4d?
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    [QUOTE=kkboyk;65958248]3x + 2 = x + 4 and -(3x+2) = x + 4. Same rule applies to other modulus functions


    (Original post by Pablo Picasso)
    Yes -(3x+4). Always put it into the calc if unsure
    Thank you!!
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    Could someone explain how you would go from the second step to the third one please?

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    (Original post by blackdiamond97)
    Could someone explain how you would go from the second step to the third one please?

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    Multiply by 2

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    (Original post by blackdiamond97)
    ...
    its because sin^2x + cos^2 x is equal to one

    2sinxcosx= sin2x

    therefore, sinxcosx= 0.5sin2x
    1 divided by 0.5 =2
    thats how you get 2 on the top
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    anyone recommend doing madasmaths papers?
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    (Original post by heyitskim)
    Name:  image.png
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Size:  80.4 KBSomeone help on 4d?
    If you look at the graph, the range of possible values will be inbetween the two red lines I've drawn on.


    This is because at all of these points, there are 2 possible values for what you get when you input a value into the function (if that makes sense). Anything above the top line would only have 1 solution, anything below the bottom line will have either 1 or no solutions. The 1 is important for having the right answer.

    You'll then need to work out where the graph intersects the y axis at x=0, and where the minimum point of the graph is (where the modulus is 0).
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    I keep getting stuck on how to sketch graphs, any tips?
 
 
 
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