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FP1 AQA 15th June 2016 UNOFFICIAL MARK SCHEME (what you have been looking for) Watch

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    (Original post by Sameechoudhury;[url="tel:65832083")
    65832083[/url]]Question 8d is wrong, it's meant to be 2root3 2
    It asked for the original coordinates before the transformation I got ((3^0.5)/2, 2)
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    (Original post by JackLloyd98)
    5b is n(2n+1)(2n-3)(n-1)
    Do you know where i went wrong?
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    (Original post by henryhong)
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    Do you know where i went wrong?
    You forgot to to take out the n with the 3n
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    (Original post by henryhong;[url="tel:65832655")
    65832655[/url]]Name:  IMG_3824.jpg
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    Do you know where i went wrong?
    Those are my workings
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    (Original post by JackLloyd98)
    It asked for the original coordinates before the transformation I got ((3^0.5)/2, 2)
    I got (-root3/2 , -2)
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    (Original post by moodyloo1998)
    I got (-root3/2 , -2)
    Maybe I read it wrong, I thought the question was stretch followed by reflection.
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    (Original post by JackLloyd98)
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    Those are my workings
    Thank you man, ill update the Unofficial MS
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    (Original post by JackLloyd98)
    Maybe I read it wrong, I thought the question was stretch followed by reflection.
    You are correct about the question, it did say stretch followed by reflection so
    reflection matrix in front and stretch matrix behind. I think the problem might be caused by the simultaneous equation.
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    (Original post by moodyloo1998)
    You are correct about the question, it did say stretch followed by reflection so
    reflection matrix in front and stretch matrix behind. I think the problem might be caused by the simultaneous equation.
    Name:  image.jpg
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    (Original post by JackLloyd98)
    Name:  image.jpg
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    I'm pretty sure the question told us that the point transformed to (0, 4), not (0, -4). Well actually, the answer I told you is the one I just worked out by memory now, my answer might be the same as yours during the exam if it is (0,-4), I just can't really remember anything right now. All the calculation seems correct though.
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    (Original post by henryhong)
    My answers for this morning's exam, please let me know if you agree/disagree on answers!

    Q1 Roots of Polynomials

    (a) alpha+beta=6 [2marks]
    alpha*beta=14

    (b) 7X^2-4X+7 [5 marks]

    Q2 Gradients using h

    (a) -h-3 [3 marks]

    (b) as h-->0 gradient-->-3 [2 marks]
    when h=0 gradient= -3

    Q3 Log graph
    (a) show linear relationship log(Y)= Xlog(B) + Log(A) [2 marks]

    (b)gradient was -0.4 [1marks]

    (c) a= 316 [4 marks]
    b=0.398

    Q4 General Solutions

    (a) k=6 [1 marks]

    (b) x= πn+ π/2 [4marks]
    x= πn+ π/3

    (c) root 3 [2 marks]

    Q5 summations

    (a) show it equal to 3n(4n^2-1) [5 marks]

    (b) n(2n+1)(2n-3)(n-1) [4marks]

    Q6 parobola

    (a) show a=2 [3 marks]

    (b) k= 1/2 [6 marks]
    k= -2
    k=0

    Q7 imaginary

    (a) -2+4i [3 marks]
    -2-4i

    (bi) i+qi=0 therefore q must beequal to -1 [2 marks]

    (bii) p=4 q= -12-i [5marks]
    p=-4 q= 4-i

    Q8 matrix

    (a) matrix 4 0 [1 marks]
    0 1
    (b) stretch in x-directionSF4 [1 marks]

    (c)matrix 0.5 -root3/2 [2marks]
    -root3/2 -0.5

    (d) matrix p= -2root3 [6marks]
    -2

    Q9 graph

    (a) Asymptotes: x=0.5 x=2 y=0 [2 marks]

    (b) Intersections: x=5/2 x=1 x=0 [3 marks]

    (c) a sketch [3 marks]

    (d) x<equal to 0 [3marks]
    0.5< to x<equal to1
    2<x<equal to 2.5
    Q1 - part a seems about right. But part b, i'm pretty sure it asked for an equation, so in that case, you need to put 7x^2-4x+7=0, the =0 matters you see. Stop being so sloppy.

    Q2, Q3, Q4 & Q5 - looks about right

    Q6 - (a) seems familiar. Pretty sure (b) asked for a range, or did it not?

    Q7 - the answers seem right. But the question title "imaginary", none of these answers were imaginary, -2+4i is complex, -2-4i is complex, -12-i is complex, 4-i is complex, so where the hell did you get "imaginary" from?!?

    Q8 - parts (a) through to (c) seem familiar, but I think part d, the COORDINATES, not matrix, P were (2rt3, 2). Also, the question should probably be titled "matrices" since we were asked about two of them, not just one.

    Q9 - all good.
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    Can someone post the question for the general solution? I can't remember my answer so I want to try it again to see if I got it right
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    How did you do Q7bii? I spent ages on it in the exam and couldn't work out how to do it at all
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    The only thing I got wrong by the looks of it was Q4 part c, I thought it was 1 anyone mind clearing up with it was root 3?
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    (Original post by rektangle)
    Q1 - part a seems about right. But part b, i'm pretty sure it asked for an equation, so in that case, you need to put 7x^2-4x+7=0, the =0 matters you see. Stop being so sloppy.

    Q2, Q3, Q4 & Q5 - looks about right

    Q6 - (a) seems familiar. Pretty sure (b) asked for a range, or did it not?

    Q7 - the answers seem right. But the question title "imaginary", none of these answers were imaginary, -2+4i is complex, -2-4i is complex, -12-i is complex, 4-i is complex, so where the hell did you get "imaginary" from?!?

    Q8 - parts (a) through to (c) seem familiar, but I think part d, the COORDINATES, not matrix, P were (2rt3, 2). Also, the question should probably be titled "matrices" since we were asked about two of them, not just one.

    Q9 - all good.
    are you my teacher? lol I just got a *******ing. but yeah ill corect them now
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    (Original post by henryhong)
    are you my teacher? lol I just got a *******ing. but yeah ill corect them now
    b o ll o c king was what I wrote initially but student room didn't allow it
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    (Original post by Chickenslayer69)
    Can someone post the question for the general solution? I can't remember my answer so I want to try it again to see if I got it right
    help pls thx
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    I did product or roots and sum of roots for complex numbers and got p=+-4 (from POR= (p+2i)(p-2i)= p^2-4i^2 = 20 so p^2=16....) then SOR ( -(4+i+iq)= p+2i + p-2i) so -4-i-qi = 2p, sub in p= 4 and -4, so qi+i= 4 so I did q = (4-i)/i to get 4/i -1 and the other q= -12/i -1) anyone get this or do this?😂😶


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    (Original post by Olsmarto)
    I did product or roots and sum of roots for complex numbers and got p=+-4 (from POR= (p+2i)(p-2i)= p^2-4i^2 = 20 so p^2=16....) then SOR ( -(4+i+iq)= p+2i + p-2i) so -4-i-qi = 2p, sub in p= 4 and -4, so qi+i= 4 so I did q = (4-i)/i to get 4/i -1 and the other q= -12/i -1) anyone get this or do this?😂😶


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    I didn't do that (I was completely stuck) but seems like a really good way of doing it haha
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    Wht are the steps for q)7b)ii)

    I'm really confused
 
 
 
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