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Edexcel Maths FP1 UNOFFICIAL Mark Scheme 19th May 2017 Watch

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    (Original post by Joanna Chan)
    i got a=9/2 but i know i missed one more possible value of a, but i dont know how do you guys get a might also equals to -27/2?
    Image= obj times the modulus of determinant of a so det can equal plus or minus 18
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    (Original post by robinhood111)
    for the matrices question how many marks do i lose if i only got 9/2 ???
    i'm guessin you only lose 2 marks out of the 5
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    (Original post by DystopiaisReal)
    Could someone please explain why for 4b p=20?? If arg(w)=0.25pi, doesn't that mean 2p+12=3p+8?
    it was 3p-8 ^^
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    (Original post by Joanna Chan)
    i got a=9/2 but i know i missed one more possible value of a, but i dont know how do you guys get a might also equals to -27/2?
    Suppose we have a linear transformation

     T(\vector{x}) = A\vector{x}

     A is the 2 x 2 matrix which mapped the triangle T, to the triangle T'.

    It can be shown that iff T is a linear mapping, then  \det(A) \times Area of the original shape  = the area of the new of the shape. However, recall that  \det(A) can be negative and in that case would produce a negative area. Area cannot be negative and thus we include a modulus sign.

    So...

     |\det(A)| \times Area of the original shape  = the area of the new of the shape

    I think in that question we had

     |2p+9|=18

    so p is the solutions to the following equations  2p+ 9 = 18 and  -(2p+9) = 18
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    (Original post by Apachai Hopachai)
    Suppose we have a linear transformation

     T(\vector{x}) = A\vector{x}

     A is the 2 x 2 matrix which mapped the triangle T, to the triangle T'.

    It can be shown that iff T is a linear mapping, then  \det(A) \times Area of the original shape  = the area of the new of the shape. However, recall that  \det(A) can be negative and in that case would produce a negative area. Area cannot be negative and thus we include a modulus sign.

    So...

     |\det(A)| \times Area of the original shape  = the area of the new of the shape

    I think in that question we had

     |2p+9|=18

    so p is the solutions to the following equations  2p+ 9 = 18 and  -(2p+9) = 18
    wrong!! completely false!1!11!!
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    jk, but the matrix was m and the variable was a

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    (Original post by Apachai Hopachai)
    Suppose we have a linear transformation

     T(\vector{x}) = A\vector{x}

     A is the 2 x 2 matrix which mapped the triangle T, to the triangle T'.

    It can be shown that iff T is a linear mapping, then  \det(A) \times Area of the original shape  = the area of the new of the shape. However, recall that  \det(A) can be negative and in that case would produce a negative area. Area cannot be negative and thus we include a modulus sign.

    So...

     |\det(A)| \times Area of the original shape  = the area of the new of the shape

    I think in that question we had

     |2p+9|=18

    so p is the solutions to the following equations  2p+ 9 = 18 and  -(2p+9) = 18
    I just equated it to both 18 and minus 18, quicker than multiplying brackets.
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    (Original post by 04MR17)
    Question 4: Complex Numbers
    i.) a.) ((2p+2)/13) + ((3p-8)/13)i [3]
    b.) p=20 [2]
    ii.) +/- 4sqrt(5) [3]
    sorry i cannot remember what was the question for ii.)?
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    (Original post by k.russell)
    wrong!! completely false!1!11!!
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    jk, but the matrix was m and the variable was a


    Yeah, I've forgotten all the questions already lol.

    This still could be wrong though, I don't know, if that little fact holds for non-linear transformations...
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    (Original post by Wendy_ZHANG)
    sorry i cannot remember what was the question for ii.)?
     |z| = 25

     z = (1-\lambda i)(2+3i) or something like that
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    (Original post by Apachai Hopachai)
     |z| = 25

     z = (1-\lambda i)(2+3i) or something like that
    THANKS SO MUCH!! I was slightly lost!! =.=
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    (Original post by Apachai Hopachai)
     |z| = 25

     z = (1-\lambda i)(2+3i) or something like that
    pretty sure it was 45
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    (Original post by glad-he-ate-her)
    pretty sure it was 45
    Probably
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    (Original post by Apachai Hopachai)
     |z| = 25

     z = (1-\lambda i)(2+3i) or something like that
    are you sure it was equal to 25? not something else???
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    (Original post by Apachai Hopachai)
    nah defo 25
    No it was definitely 45


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    (Original post by goatfarmer)
    No it was definitely 45


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    ok then it was 45
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    (Original post by 04MR17)
    Question 5 Matrices
    i.) a.) ((12-5p) (4p-10))
    ..........((6p-15) (12-5p)) [2]
    b.) p=1.5, k=7.5 [4]
    ii.) a=4.5 [5]
    or a=-13.5
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    (Original post by Apachai Hopachai)
    ok then it was 45
    thanks guys, i thought it was 45,
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    (Original post by diogenesax)
    Can someone explain how k was 2/9?
    You had to use geometric series: C2 style. I didn't spot that and decided to substitute the whole thing for that term. Don't know if I'll get the marks.
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    (Original post by 04MR17)
    Question 4: Complex Numbers
    i.) a.) ((2p+2)/13) + ((3p-8)/13)i [3]
    b.) p=20 [2]
    ii.) +/- 4sqrt(5) [3]
    It wanted simplest form so a should be:
    (2(p+6)/13) + ((3p-8)/13)i
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    (Original post by 04MR17)
    Question 4: Complex Numbers
    i.) a.) ((2p+12)/13) + ((3p-8)/13)i [3]
    b.) p=20 [2]
    ii.) +/- 4sqrt(5) [3]
    lplp
 
 
 
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