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Edexcel Core 3 - 21st June 2016 AM Watch

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    (Original post by ememoville)
    Consider where you can draw a horizontal line and get two values.

    So between the turning point (y co, ordinate of A and 0) that's the range for k
    Could you draw a quick diagram to explain what you mean pls?
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    (Original post by Ainsleyy)
    A good thing you can do is the gold papers from physicsandmathstutor. They are a compilation of all hard Q's from past papers. Otherwise most of June 2013 was more challenging especially Q8. June 2015 Q 8 is quite hard.
    Ah I forgot about those, thank you very much

    Surprisingly I managed to do q8 on June 2013 but I lost quite a few marks on some other questions in that paper
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    (Original post by philo-jitsu)
    Also when your trying to find the x values with the modulus you use both negative and positive version of the modulis....but when your equating .....you know what its easier if i show the question

    Its question 7 of soloman j.....i got i right but i dont get why its correct
    You're only using the positive version because you're only thinking about where the graphs of f(x) and g(x) cross. You could indeed use the negative form to see if there are any points at which the modulus line of f(x) (i.e. the line y=5-2x) crosses g(x) (and whether these would be negative or positive x values).

    However, here it tells you that x is positive (lies between 3 and 4), and is greater than 5/2 (the point at which the modulus of f(x) touches the x-axis), so you know you're dealing with the part of f(x) which is the 'normal' part.

    If you were looking for solutions to f(x)=k (where k is a constant), you'd need to consider both positive and negative values of x, because negative x values will become positive when put into the function.

    Hope this helps! Always do a quick graph sketch if you're unsure.
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    Guys
    Can someone please list all the proofs we have to know for c3?
    Thank you!


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    Can anyone explain how to do Q3c) from the June 2015 paper? Thanks for any help
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    (Original post by ememoville)
    Consider where you can draw a horizontal line and get two values.

    So between the turning point (y co, ordinate of A and 0) that's the range for k
    Thank you v much x
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    have any past papers ever required us to divide a cubic by a quadratic?
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    (Original post by 13 1 20 8 42)
    when 2x - 5 < 0 we have f(x) = -(2x - 5) i.e. 2x - 5 = -f(x) = -(15 + x)
    When you sketch the graphs in part a), how are you supposed to know that they graphs will cross twice :s?
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    Is g(x) =g^-1(x) the same as saying g(x) =x?

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    What should I do to revise for the exam which papers to do last minute
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    (Original post by Azzer11)
    Is g(x) =g^-1(x) the same as saying g(x) =x?

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    Yes, because the graphs intersect where y = x

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    (Original post by Azzer11)
    Is g(x) =g^-1(x) the same as saying g(x) =x?

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    Yes - I believe that is the only function where that is true.
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    (Original post by Mattematics)
    No chance. Prepare for more p^2 + qr + c PTSD.
    What do you mean?
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    (Original post by Pablo Picasso)
    k > Turning points
    And k<0, as above this there is only one solution for each x value
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    (Original post by pineneedles)
    Yes, because the graphs intersect where y = x

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    (Original post by SeanFM)
    Yes - I believe that is the only function where that is true.
    Hmm interesting, thanks!
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    I hope this makes more sense Don Pedro K.
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    (Original post by Bloom77)
    Guys
    Can someone please list all the proofs we have to know for c3?
    Thank you!


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    The only proofs I've come across on C3 is to prove the formula for tan2A using the double angle formula. They would only ever ask simple proofs though
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    (Original post by kkboyk)
    What do you mean?
    Reference to FP2 and M2 where questions that have always involved numbers previously randomly became that equation :P
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    Does anyone know why the time period for one revolution is found by equalling 30t to 360?

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    (Original post by ememoville)
    I hope this makes more sense Don Pedro K.
    Ahh that makes perfect sense, thank you!
 
 
 
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