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    I'm struggling to make sense of this diagrammatically. Can anyone help?Name:  Untitled4.png
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    Also, have I gone mad, or is this forgetting the little bit at the corner?Name:  Untitled5.png
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    (Original post by jadoreétudier)
    I'm struggling to make sense of this diagrammatically. Can anyone help?Name:  Untitled4.png
Views: 247
Size:  41.8 KB

    Also, have I gone mad, or is this forgetting the little bit at the corner?Name:  Untitled5.png
Views: 248
Size:  15.3 KB
    The second attachment is just wrong! You're correct, it DOES neglect the corner. As for the first, take an equation Ax+By=C. Think about how it'll look as you vary the signs of A and B between positive and negative. You're looking for the case where there will be a FINITE number of positive integer solutions, that is, a finite portion of the first quadrant will satisfy Ax+By<C. This immediately rules out positive slope of x axis. Negative slope can be achieved by keeping A and B both positive or by keeping them both negative. But in the second case, the region satisfying Ax+By<C is the upper region, so there are infinite solutions. In the first case, it is the lower region, so it has finitely many solutions. Now replace A, B by the suitable constants and set them positive. Does this make sense? (I don't remember the question properly, so I might have got something wrong. Please tell me if I messed up or didn't explain properly.)

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    (Original post by souktik)
    The second attachment is just wrong! You're correct, it DOES neglect the corner. As for the first, take an equation Ax+By=C. Think about how it'll look as you vary the signs of A and B between positive and negative. You're looking for the case where there will be a FINITE number of positive integer solutions, that is, a finite portion of the first quadrant will satisfy Ax+By<C. This immediately rules out positive slope of x axis. Negative slope can be achieved by keeping A and B both positive or by keeping them both negative. But in the second case, the region satisfying Ax+By<C is the upper region, so there are infinite solutions. In the first case, it is the lower region, so it has finitely many solutions. Now replace A, B by the suitable constants and set them positive. Does this make sense? (I don't remember the question properly, so I might have got something wrong. Please tell me if I messed up or didn't explain properly.)

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    Makes perfect sense, thank you so much!
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    (Original post by jadoreétudier)
    Makes perfect sense, thank you so much!
    You're welcome!
    One question, if you don't mind. Err, where is that second attachment from? Is that on any website?

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    (Original post by souktik)
    You're welcome!
    One question, if you don't mind. Err, where is that second attachment from? Is that on any website?

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    Yea, slide 35
    http://www.drfrostmaths.com/resource.php?id=11226
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    (Original post by jadoreétudier)
    Also, have I gone mad, or is this forgetting the little bit at the corner?Name:  Untitled5.png
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Size:  15.3 KB
    you are indeed correct
    I calculated r=3-2root2
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    Does anyone know where/if the mark scheme for the 2006 paper can be found? I can't seem to find it anywhere...
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    (Original post by bluebell_flames)
    Does anyone know where/if the mark scheme for the 2006 paper can be found? I can't seem to find it anywhere...
    there isn't one but we discussed our answers on this thread yesterday

    around here: http://www.thestudentroom.co.uk/show...4#post45034414
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    (Original post by sun_tzu)
    you are indeed correct
    I calculated r=3-2root2
    me too

    i actually got (root2 - 1)/ (root2 +1), which when I put into my calculator gives me what you got.

    yay, i just simplified it myself by multiplying and dividing by (root2 - 1)
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    (Original post by jadoreétudier)
    there isn't one but we discussed our answers on this thread yesterday

    around here: http://www.thestudentroom.co.uk/show...4#post45034414
    Thank you!!
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    Hey guy's can someone help me on section A question H.

    The answers say it's pi / 9 - 1/6 but I keep getting pi / 9 - root 3 / 12.

    Can someone check where the error is in my method?

    What I've done is found the sector, with radius 1 angle 2pi/3 (using inverse trig). Then I took away the triangle from it which was side length 1 and 1 and angle 2pi/3.

    I then doubled this to find the shaded area, giving me 2pi /3 - root3 / 2.
    I then divided by 6, the area of the rectangle.

    Thanks
    Attached Files
  1. File Type: docxRZC-MAT-Shape.docx (38.2 KB, 82 views)
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    I got pi/9 - root3 /12
    same as you so we are wrong together (or right hopefully)
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    (Original post by jadoreétudier)
    me too

    i actually got (root2 - 1)/ (root2 +1), which when I put into my calculator gives me what you got.

    yay, i just simplified it myself by multiplying and dividing by (root2 - 1)

    how did you get that?
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    (Original post by IceKidd)
    how did you get that?

    The square root of 2 (from the centre to the top right corner) is equal to 1 + 2r + the little bit.

    Find the little bit in terms of r and then solve the above equation for r. I'm sure you can do it
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    (Original post by BankOfPigs)
    Hey guy's can someone help me on section A question H.

    The answers say it's pi / 9 - 1/6 but I keep getting pi / 9 - root 3 / 12.

    Can someone check where the error is in my method?

    What I've done is found the sector, with radius 1 angle 2pi/3 (using inverse trig). Then I took away the triangle from it which was side length 1 and 1 and angle 2pi/3.

    I then doubled this to find the shaded area, giving me 2pi /3 - root3 / 2.
    I then divided by 6, the area of the rectangle.

    Thanks
    Where did you get the root3 / 2 from???
    Using basically the same method, I obtain pi/9 - 1/6, and the only difference I can see is that when you "took away the triangle from it which was side length 1 and 1 and angle 2pi/3", I took away the triangle whose base is 2 and its height is 1/2.

    That's why there shouldn't be any root three anywhere... WOW wait a minute... Nope, I'm wrong, but I see why the answer is wrong too. They supposed the length of the triangle was two (when it actually is root 3). So you're both correct!
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    Does it matter if we do the exam in pencil or pen?
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    Yes - you must use pen. With pencil answers could be 'tampered with'.
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    (Original post by jadoreétudier)
    Does it matter if we do the exam in pencil or pen?
    I just realised how little I know about the technical aspects of the test. Can anyone answer this? Does crossed out work get checked if no other working is present?

    Sent from my GT-N7100 using Tapatalk 4
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    (Original post by jadoreétudier)
    The square root of 2 (from the centre to the top right corner) is equal to 1 + 2r + the little bit.

    Find the little bit in terms of r and then solve the above equation for r. I'm sure you can do it
    but im stumped on how you could find the little bit!
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    you need a ruler, rubber, protractor, compass, sharpener, highlighters, at least 10 colouring pencils, 20 is ideal. It's also recommended to bring an umbrella and a spare pair of socks

    kidding just bring a pen
 
 
 
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