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    (Original post by Hot&SpicyChicken)
    For wuestion 8) (c) it is impossible for the tangent to meet the curve again, bu the normal does, so i think the question is wrong?
    Yep, it's a typo. Meant to be "normal".
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    (Original post by Hot&SpicyChicken)
    https://7cba9babeb0db0ff9468853e0b2d...%20Edexcel.pdf

    Even the mark scheme uses the normal? Am i correct or just liost?
    There's an error in the paper. It's meant to read 'normal' not 'tangent'
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    (Original post by connorbarr)
    Haven't done the question but it is possible for the tangent to meet and xy graph again i think
    but ye the mark scheme uses normal so must be wrong
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    In the FP1 textbook, pg 82, there's something about linear transformations. What do we need to know about this? I dont really understand it too well
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    (Original post by dididid)
    Hey could someone please link/ send me some hard proofs by induction for each of the proofs mentioned on the spec
    This was a pretty cool question on an old fp2 paper - although you can only do it if you have done c3/c4 first.
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    (Original post by AmarPatel98)
    In the FP1 textbook, pg 82, there's something about linear transformations. What do we need to know about this? I dont really understand it too well
    ye its well **** i had to look online, havent seen it in a past paper before but i think for it to be a linear transformation u can only add x and y values and times by a constant, might be wrong
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    (Original post by iMacJack)
    But I was unable to get the one in the IAL 2015 paper using it.
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    Thats how I done the question, not sure if you'll know this method. Don't use it if you don't.
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    Good luck everyone!!!
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    Good luck for tomorrow everyone! I'm sure it won't be too bad. Well, I hope so anyway...
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    I was gone pull a all nighter but then I was like I'll just wake up early tomorrow to cram some last minute revision in. I hope its not one of them wake ups that you say now just to make yourself feel better.
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    good luck

    r= 0 summations, is it just the constant that gets affected by the n+1 at the end? and not the sum of r^2 + sum of r bit?
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    Some one plz help
    cant understand 8b even after looking at Mark Scheme

    https://edba9d72a6846be4dc688090a802...%20Edexcel.pdf
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    (Original post by dragozox)
    Some one plz help
    cant understand 8b even after looking at Mark Scheme

    https://edba9d72a6846be4dc688090a802...%20Edexcel.pdf
    Multiply the exression by A^-1 to the right
    AAA^-1 = 7AA^-1 + 2IA^-1
    Becomes
    A= 7I +2A^-1
    Then rearrange
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    (Original post by Windowswind123)
    Multiply the exression by A^-1 to the right
    AAA^-1 = 7AA^-1 + 2IA^-1
    Becomes
    A= 7I +2A^-1
    Then rearrange
    I Did do that but shouldnt it be:

    A^-1AA = 7A^-1A+2A^-1I

    which gives A=7 + 2A^-1I

    I dont get how 7 and I get together?? Thank You!!
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    (Original post by dragozox)
    I Did do that but shouldnt it be:

    A^-1AA = 7A^-1+2A^-1I

    which gives A=7 + 2A^-1I

    I dont get how 7 and I get together. Thank You!!
    AA^-1 =I
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    (Original post by anndz3007)
    AA^-1 =I
    ooooh I C!!!

    Thanks For clarifying!!
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    (Original post by dragozox)
    I Did do that but shouldnt it be:

    A^-1AA = 7A^-1A+2A^-1I

    which gives A=7 + 2A^-1I

    I dont get how 7 and I get together?? Thank You!!
    A*A-1 =I
    And AI = A and IA^-1 = A^-1
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    (Original post by Windowswind123)
    A*A-1 =I
    And AI = A and IA^-1 = A^-1
    Thank you Sooo much!!

    This makes so much sense now!!
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    Good luck everyone.
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    (Original post by tazza ma razza)
    good luck

    r= 0 summations, is it just the constant that gets affected by the n+1 at the end? and not the sum of r^2 + sum of r bit?
    the general summations dont change because you're just adding 0, 0^2 or 0^3 to the start, but for the constants, you're multiplying it by the number of time the sum is done, so when r=0, there's one more term than you'd usually expect, so if the constant was '5', instead of making it 5n, it become 5(n+1) to account for when r=0
 
 
 
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