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    (Original post by atofu)
    Heya, does anyone have the 2010 paper and mark scheme? It's secure/locked on the edexcel website

    Also, what things will I need to know how to do without a calculator? xD All I can think of so far are:
    -Trig functions for 30, 45, 60, 90 etc.
    - Working with surds
    An easy way to remember those trig values easily without drawing triangles is that:

    30 45 60 90

     \frac{\sqrt{1}}{2} \frac{\sqrt{2}}{2} \frac{\sqrt{3}}{2} \frac{\sqrt{4}}{2} sin

    The reverse is true for cos, and then tan = sin/cos for each value.
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    (Original post by AnonyMatt)
    An easy way to remember those trig values easily without drawing triangles is that:

    30 45 60 90

     \frac{\sqrt{1}}{2} \frac{\sqrt{2}}{2} \frac{\sqrt{3}}{2} \frac{\sqrt{4}}{2} sin

    The reverse is true for cos, and then tan = sin/cos for each value.
    That is genius! :hat2:
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    I'm struggling to understand question 7(d) in the June 2008 AEA paper (paper and mark scheme attached). The question requires that one finds the centre of a circle, that is contained in a triangle and is just touching (tangent to) all the sides of the triangle.

    The triangle is isoceles (proved in 7(c)), so I understand that the centre of the circle lies on the midpoint of BC. But, I'm stuck on how to progress after this. The mark scheme simply states that the centre of the circle is the intersection between the perpendicular bisector of BC and the angle bisector of ABC -- but, I don't understand why the centre of the circle must lie on the angle bisector of ABC? I fear I may be lacking some geometrical knowledge necessary for the understanding of this question. If someone could explain this to me, it would be much appreciated.

    Thanks
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  1. File Type: pdf 9801_01_rms_20080807.pdf (365.3 KB, 103 views)
  2. File Type: pdf 9801_01_que_20080625.pdf (685.7 KB, 115 views)
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    (Original post by ArchimedesSpiral)
    I'm struggling to understand question 7(d) in the June 2008 AEA paper (paper and mark scheme attached). The question requires that one finds the centre of a circle, that is contained in a triangle and is just touching (tangent to) all the sides of the triangle.

    The triangle is isoceles (proved in 7(c)), so I understand that the centre of the circle lies on the midpoint of BC. But, I'm stuck on how to progress after this. The mark scheme simply states that the centre of the circle is the intersection between the perpendicular bisector of BC and the angle bisector of ABC -- but, I don't understand why the centre of the circle must lie on the angle bisector of ABC? I fear I may be lacking some geometrical knowledge necessary for the understanding of this question. If someone could explain this to me, it would be much appreciated.

    Thanks
    The angle bisector of ABC is also the perpendicular bisector of the line segment AC, I believe. So since it is perpendicular to AC it must pass through S. And since both L1 and L pass through S their point of intersection is S. :wink2:
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    (Original post by AnonyMatt)
    An easy way to remember those trig values easily without drawing triangles is that:

    30 45 60 90

     \frac{\sqrt{1}}{2} \frac{\sqrt{2}}{2} \frac{\sqrt{3}}{2} \frac{\sqrt{4}}{2} sin

    The reverse is true for cos, and then tan = sin/cos for each value.
    ahh thanks for the tip!
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    Does anyone have answers/markscheme for last year's paper (2010)? I've spent a lot of time searching for it, with no luck
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    (Original post by sijapu17)
    Does anyone have answers/markscheme for last year's paper (2010)? I've spent a lot of time searching for it, with no luck
    Raimu posted the paper here.

    But no mark scheme :sad:
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    (Original post by sijapu17)
    Does anyone have answers/markscheme for last year's paper (2010)? I've spent a lot of time searching for it, with no luck
    Here's AEA 2010 mark scheme:

    http://www.scribd.com/doc/57071274/1...heme-June-2010
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    Guys can you help me with 6)d from the 08 paper. I looked at the mark scheme but don't get the bit about reflection.

    Paper and mark scheme is here (a couple of posts up)
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    This probably sounds like a ridiculous question and I guess it's just the nervousness before the exam, but could someone tell me what I need to bring on the day of the test?

    I believe we are provided a formula booklet, so is it just a pencil and a biro? Do I need to bring any paper at all?
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    (Original post by w34)
    This probably sounds like a ridiculous question and I guess it's just the nervousness before the exam, but could someone tell me what I need to bring on the day of the test?

    I believe we are provided a formula booklet, so is it just a pencil and a biro? Do I need to bring any paper at all?
    Pen, pencil, ruler, rubber (+ sharpener maybe)
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    (Original post by ElMoro)
    Pen, pencil, ruler, rubber (+ sharpener maybe)
    I for one welcome our vengeful Imperial Overlord.
    Thank you very much!
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    Is it possible that vectors don't come up... I hate them!
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    (Original post by soczek322)
    Is it possible that vectors don't come up... I hate them!
    Sadly, there's pretty much always a epic vector Q at the end. :dontknow:
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    Oh, but vector is quite predictable... It's always going to ask you to find the equation, or prependicular distance first... Then it's the hard stuff, oh well. As long as you find c4 vectors okay, then you should be fine tbh!
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    grrrr can anyone explain to me q6 d) and 7b) from the 2008 paper

    Thanks!
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    (Original post by ElMoro)
    Yeah, I would appreciate that thanks!

    Have you gone through that booklet with old S-level questions, yet? It's good prep and means you can save the papers till your ready to do them in timed conditions. I've attached it for you. :wink2:
    hi could u please let em know where i couls find the solutions to thjese.. particularly stuck on Q 8
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    (Original post by ilovespidergirl)
    hi could u please let em know where i couls find the solutions to thjese.. particularly stuck on Q 8
    The solutions to all qns are found in this thread:
    http://www.thestudentroom.co.uk/showthread.php?t=859837

    (Original post by ArchimedesSpiral)
    The mark scheme simply states that the centre of the circle is the intersection between the perpendicular bisector of BC and the angle bisector of ABC -- but, I don't understand why the centre of the circle must lie on the angle bisector of ABC?
    Referring to the attached diagram Qn 7(d), you should realise triangle BXS' and triangle BS'Y are congruent, where S' is the centre of the circle. This means angle XBS' and angle S'BY are equal and it proves L passes through S'.

    (Original post by ElMoro)
    Guys can you help me with 6)d from the 08 paper. I looked at the mark scheme but don't get the bit about reflection.
    Referring to the attached diagram Qn 6(d) the graph y=f(x-2)+2, the red line is a reflection of the green line on line y=x, both lines have equal gradient, 4, as stated in the question. On the graph y=f(x-2)+2, if the red line has equation y=4x-h, then the green line has equation y=4x+h, as both lines are reflections of each other. This means you have to find the equation of the red line on the graph y=f(x-2)+2 to find the equation of the green line on graph y=f(x-2)+2. Then, you work back to find the equation of the green line on graph y=f(x).

    (Original post by soczek322)
    grrrr can anyone explain to me q6 d) and 7b) from the 2008 paper
    For Qn 7(b), you realise |\vec{AB}| \not=|\vec{BC}| . To give you a hint, try to find a vector that is parallel to \vec{BC} but whose magnitude is the same as |\vec{AB}| by multiplying \vec{BC} with a constant.
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    Anyone know what the last few grade boundaries were? I only saw the ones before 2008 on edexcel website
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    Erm jun 2010 - 81 - disintction, 61 - merit
 
 
 
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