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    (Original post by UKBrah)
    Yep I agree with your answer for z2.

    The diagram was really cluttered with tons of surds. I think I had like 3 complex numbers in the positive x quadrant and 1 somewhere else.
    Yeah, in terms of them all I got:

     z_{1} = 3-2j

z_{2} = \frac{5\sqrt{2}}{2} + \frac{5\sqrt{2}}{2}j



z_{1} + z_{2} = \frac{6+5\sqrt{2}}{2} + \frac{-4+5\sqrt{2}}{2}j



z_{1} - z_{2} = \frac{6-5\sqrt{2}}{2} - \frac{4+5\sqrt{2}}{2}j


    z2 and z1+z2 were both in the top right quadrant, and z1 was in the bottom right quadrant, and z1-z2 was in the bottom left quadrant.

    I'm more than likely entirely wrong though..! Need to have a look back through the paper.
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    (Original post by UKBrah)
    -3+j/-3-j
    Thank god lol i thougt i made a mistake
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    (Original post by Political Cake)
    Yeah, in terms of them all I got:

     z_{1} = 3-2j

z_{2} = \frac{5\sqrt{2}}{2} + \frac{5\sqrt{2}}{2}j



z_{1} + z_{2} = \frac{6+5\sqrt{2}}{2} + \frac{-4+5\sqrt{2}}{2}j



z_{1} - z_{2} = \frac{6-5\sqrt{2}}{2} - \frac{4+5\sqrt{2}}{2}j
    I think I made an error with z1+z2/z1-z2 or with one of them. The position of the minus sign must have flopped me.

    all in all lost 3 marks in the paper, was doing FP1 for the first time.. would this be 100ums? 69/72
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    (Original post by abdcefghi)
    What was the original p coordinate can you remember?
    i got P to be (-2,2)
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    The induction question was quite alright this time, I've done it here:


    \sum_{r=1}^{n}[r(r-1)-1] = \frac{1}{3}n(n+2)(n-2)

    The inductive step:

    \frac{1}{3}k(k+2)(k-2) + (k+1)((k+1)-1)-1

    = \frac{1}{3}[k(k+2)(k-2) + 3k(k+1) -3]

    = \frac{1}{3}[k(k^{2}-4) + 3k^{2} + 3k -3]

    = \frac{1}{3}[k^{3}-4k + 3k^{2} + 3k -3]

    = \frac{1}{3}[k^{3} + 3k^{2} - k -3]

    = \frac{1}{3}(k+1)(k+3)(k-1)

    = \frac{1}{3}(k+1)((k+1)+2)((k+1)-2)
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    (Original post by Political Cake)
    The induction question was quite alright this time, I've done it here:


    \sum_{r=1}^{n}[r(r-1)-1] = \frac{1}{3}n(n+2)(n-2)

    The inductive step:

    \frac{1}{3}k(k+2)(k-2) + (k+1)((k+1)-1)-1

    = \frac{1}{3}[k(k+2)(k-2) + 3k(k+1) -3]

    = \frac{1}{3}[k(k^{2}-4) + 3k^{2} + 3k -3]

    = \frac{1}{3}[k^{3}-4k + 3k^{2} + 3k -3]

    = \frac{1}{3}[k^{3} + 3k^{2} - k -3]

    = \frac{1}{3}(k+1)(k+3)(k-1)

    = \frac{1}{3}(k+1)((k+1)+2)((k+1)-2)
    Do you think some kind of explicit working was required to go from the polynomial to the factorised form? :/ Seems kinda weak seeing as the answer was given, but it was awkward to do that on top of everything else. I was a bit thrown by this because I couldn't believe they'd give us something which didn't factorise neatly without needing complete expansion first :/ I would have much preferred the so called 'horrible' proofs from previous papers, I quite enjoy the algebra for those.
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    (Original post by UKBrah)
    I got that for part iii or whatever it was before, not for the last part!
    sorry you were right I did the question again, 9)vi) is y=x
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    (Original post by BeatlesFan96)
    Do you think some kind of explicit working was required to go from the polynomial to the factorised form? :/ Seems kinda weak seeing as the answer was given, but it was awkward to do that on top of everything else. I was a bit thrown by this because I couldn't believe they'd give us something which didn't factorise neatly without needing complete expansion first :/ I would have much preferred the so called 'horrible' proofs from previous papers, I quite enjoy the algebra for those.
    That's exactly what I was thinking. This induction question just ruined my day. I got to the cubic and though, there must be a way I can make this work without getting a cubic. spend so much time trying to find a solution that I had to do question 9 in 5 minutes. Luckily I found it Q9 really easy as I remember a very similar question from the 2005/2006 past papers.

    all in all, the exam was fairly straightforward, but all the questions were long and was weird to explain.

    For question 7, When it says determine, could you just give the right answers? It was 4 marks and 3 answers so I guess they wanted working out, but I didn't know what to put up lol.
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    Does anyone have a past paper infront of them. If so could you tell me how many marks Q's: 6, and each part of 9, were worth please, thanks
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    (Original post by Political Cake)
    Yeah, in terms of them all I got:

     z_{1} = 3-2j

z_{2} = \frac{5\sqrt{2}}{2} + \frac{5\sqrt{2}}{2}j



z_{1} + z_{2} = \frac{6+5\sqrt{2}}{2} + \frac{-4+5\sqrt{2}}{2}j



z_{1} - z_{2} = \frac{6-5\sqrt{2}}{2} - \frac{4+5\sqrt{2}}{2}j


    z2 and z1+z2 were both in the top right quadrant, and z1 was in the bottom right quadrant, and z1-z2 was in the bottom left quadrant.

    I'm more than likely entirely wrong though..! Need to have a look back through the paper.
    I think I got exactly that, and put them on the graph how you described.
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    (Original post by dada55)
    That's exactly what I was thinking. This induction question just ruined my day. I got to the cubic and though, there must be a way I can make this work without getting a cubic. spend so much time trying to find a solution that I had to do question 9 in 5 minutes. Luckily I found it Q9 really easy as I remember a very similar question from the 2005/2006 past papers.

    all in all, the exam was fairly straightforward, but all the questions were long and was weird to explain.

    For question 7, When it says determine, could you just give the right answers? It was 4 marks and 3 answers so I guess they wanted working out, but I didn't know what to put up lol.
    I just wrote what the asymptotes were. Then said that the fractiony asymptote must be the one with a coeffecient in front of an x, the one without a fraction must be the other vertical asymptote. And then found the horizontal asymptote which was 3/2, then said that c must be 3 and b must be 2.
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    (Original post by Law-Hopeful)
    I did all the induction (and got it right), apart from the last bit '((k+1)+2)' I tried but ended up doing something like '((k)+3)' how many marks would I lose for this, 0 or 1?
    As long as you got the final k+1 form, then you shouldn't technically lose any, as you only put the "k+1"s where the "k"s are just to show that the k is being replaced. But as long as you got the action form right, then you should be fine. Although if they are being harsh, you might lose 1 at the most, but it shouldn't jeopardise any other part of the question.
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    (Original post by BeatlesFan96)
    Do you think some kind of explicit working was required to go from the polynomial to the factorised form? :/ Seems kinda weak seeing as the answer was given, but it was awkward to do that on top of everything else. I was a bit thrown by this because I couldn't believe they'd give us something which didn't factorise neatly without needing complete expansion first :/ I would have much preferred the so called 'horrible' proofs from previous papers, I quite enjoy the algebra for those.
    I was wondering this myself, I just wrote something weird like (k-1) is a factor and made it look like I did the polynomial division in my head. I doubt that it would limit you to the marks.
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    (Original post by Aaqil)
    Guys what did you get in the matrix questions because I was kind of baffled except the first and the last parts. I think the (ii) I got p'(2,-2), (iii) y=-x, (iv) y =6
    i) Rotation clockwise 90 degrees about the origin
    ii) (2,-2) or (-2,2), I forgot which way around it was
    iii) y=-x
    iv) y=6
    v) Showed that the determinant=0 then said it was singular and baffled on about what it implies for the transformation
    vi) I found R, which was:
    (01)
    (01)
    Then showed that it was y=x
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    (Original post by jonny7bell)
    i) Rotation clockwise 90 degrees about the origin
    ii) (2,-2) or (-2,2), I forgot which way around it was
    iii) y=-x
    iv) y=6
    v) Showed that the determinant=0 then said it was singular and baffled on about what it implies for the transformation
    vi) I found R, which was:
    (01)
    (01)
    Then showed that it was y=x
    Hold tite 11/12 marks, did the EXACT same thing as yourself.

    Part v, and the argand diagram were my only errors of 3 marks..
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    (Original post by jonny7bell)
    I just wrote what the asymptotes were. Then said that the fractiony asymptote must be the one with a coeffecient in front of an x, the one without a fraction must be the other vertical asymptote. And then found the horizontal asymptote which was 3/2, then said that c must be 3 and b must be 2.
    I just gave the right answers and was planning to come back to it, but because I spend half an hour trying to work out how to solve the cubic in the induction question properly (and I still didn't manage) I didn't have time to show what I did.

    Hopefully I'll lose a maximum of 1 mark, What I'm fearing however is that the mark scheme will be like this:

    A1 mark for a =2
    A1 mark for b = 2
    M1 mark for doing 3x^2/2x^2
    A1 mark for c = 3 dependant on M1 being shown

    Which means I'll lose 2 marks. If they wanted working out for "a" and "b" then I'll be way pissed as that will mean 4 marks lost LOL
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    (Original post by UKBrah)
    Hold tite 11/12 marks, did the EXACT same thing as yourself.

    Part v, and the argand diagram were my only errors of 3 marks..
    The only bit I am worried about is the last question, and maybe the equation with the 1/3x+1 roots in it, as I got a load of big numbers as my coefficients. The Argand diagram I am reasonably happy about, although I only labelled my points as z1, z2 ect because I didn't want to write out a giant surd.
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    (Original post by dada55)
    I just gave the right answers and was planning to come back to it, but because I spend half an hour trying to work out how to solve the cubic in the induction question properly (and I still didn't manage) I didn't have time to show what I did.

    Hopefully I'll lose a maximum of 1 mark, What I'm fearing however is that the mark scheme will be like this:

    A1 mark for a =2
    A1 mark for b = 2
    M1 mark for doing 3x^2/2x^2
    A1 mark for c = 3 dependant on M1 being shown

    Which means I'll lose 2 marks. If they wanted working out for "a" and "b" then I'll be way pissed as that will mean 4 marks lost LOL

    I am not sure how harsh they will be, although I feel a lot of people will have probably jumped to the answers so it shouldn't be too bad. The induction wasn't too bad for me. I messed it up a couple of times by trying to factor everything, but then just ended up expanding most if it out and getting it in the right form at the end. Although you should only lose 2/3 marks as long as you did everything else right.
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    (Original post by jonny7bell)
    The only bit I am worried about is the last question, and maybe the equation with the 1/3x+1 roots in it, as I got a load of big numbers as my coefficients. The Argand diagram I am reasonably happy about, although I only labelled my points as z1, z2 ect because I didn't want to write out a giant surd.
    I screwed up the z1+z2/z1-z2 parts. When I added/subtracted, I miscarried the signs so 2/4 marks on that part gone.

    Everything else is tip top condition, gonna focus on C2/NM now!
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    (Original post by theCreator)
    Does anyone have a past paper infront of them. If so could you tell me how many marks Q's: 6, and each part of 9, were worth please, thanks
    Question 6 was worth 7 marks, and as for 9, each part (there were six parts) were worth 2 marks.
 
 
 
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