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    (Original post by ollycostello)
    I was wondering the same (did same thing)
    (Original post by yelash)
    How did people solve the equation 0.5(X-1)>(X-1)/(2X-1)(X+2) ? I cancelled out the two (X-1) brackets but that left me with two intersections, rather than the three evident from the graph.
    You probably divided by (x-1) rather than factorising. You lose the solution x=1 if you do that.
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    (Original post by Chickenslayer69)
    Yes, that was the wording, I remember.
    I did that first but it says p is translated on to point (0,-4) by.... (Not exact wording )
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    (Original post by B_9710)
    You probably divided by (x-1) rather than factorising. You lose the solution x=1 if you do that.
    Ah yeah. Why can't you divide by X-1 out of interest? I know we've been find not to but unless i know why I'll irritably do the same in the future
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    [QUOTE=yelash;65815503]Ah yeah. Why can't you divide by X-1 out of interest? I know we've been find not to but unless i know why I'll irritably do the same in the future [/QUOTE

    Division by zero. If x=1 then you would be dividing by zero. So unless x is definitely not allowed to be 1 (this would be in the domain probably) you are not allowed to divide both sides by x-1.
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    For the matrices question, for the last part, instead of doing BA^2 can you do B(0,-4) =(x',y') then do A^2(x',y')? which gives (8sqrt3.2) ?
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    boundary for an A??
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    (Original post by yaz1423)
    boundary for an A??
    My guess is 60-62 - it wasn't too bad a paper imo, although I had C3 straight after which probably gave me strong contrast. Core 3 was ****ing awful.
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    You guys are making me feel really bad for thinking q had to be real but was evidently complex, I wish I realised it earlier 😂😂😂😂😂
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    I have been reading this thread and I feel really pleased about how well I did in FP1. It was pretty nice unlike core 3.
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    (Original post by Red_Inferno112)
    My guess is 60-62 - it wasn't too bad a paper imo, although I had C3 straight after which probably gave me strong contrast. Core 3 was ****ing awful.
    Yeah I heard about that. Apparently students who usually get near our at 100 ums thought it was really hard, and everyone else was just crying really (at my school)
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    (Original post by yelash)
    Ah yeah. Why can't you divide by X-1 out of interest? I know we've been find not to but unless i know why I'll irritably do the same in the future

    (Original post by Physics Optimist)
    Division by zero. If x=1 then you would be dividing by zero. So unless x is definitely not allowed to be 1 (this would be in the domain probably) you are not allowed to divide both sides by x-1.
    Ok thanks! So how would you actually solve it then?
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    (Original post by Simsllama)
    I have been reading this thread and I feel really pleased about how well I did in FP1. It was pretty nice unlike core 3.
    Looking at the c3 thread I feel lucky that I did it last year (ours was a real nice paper)
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    (Original post by yelash)
    Ok thanks! So how would you actually solve it then?
    Multiply both sides by denominator, factorise, then simplify to get 3 linear factors.
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    (Original post by -jordan-)
    That's the right answer if the question was finding the image of (-4, 0) under the combined matrix, but people seem to disagree as to what the question was.

    I seem to remember the almost exact wording of the question being Find the image of (-4, 0) under the transformation represented by A^2 followed by a reflection in the line x+root(3)y. But maybe it wasn't.
    Might ask my teacher for the wording haha
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    guys what did you get for the last question on the ranges for which one equation was > than another. how did you do it
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    I thiught the paper went quite well. For the dodgy imaginary numbers bit i used the fact that the sum of the roots would be the negative coefficent of z and the product would be the coefficrnt of the integer
    The 2 roots being given to you
    Hence solving p to be 4 and -4 and substituting into the product equation to obtain q.
    I got q to just be 2 ingegers
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    Was the answer to the last question actually (sqrt3/2 , 2) ? If so I have no idea how I managed to get that one right. Overall I thought it went well but I was resitting it so I will have different views to those doing it for the first time. Low 60's for an A?
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    (Original post by tomdavis.)
    What did everyone get on part b of the matrices question? And the values of q on the complex numbers question?
    you can check your answers here on the Unofficial Mark scheme

    http://www.thestudentroom.co.uk/show....php?t=4168081
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    (Original post by Kedric123)
    Was the answer to the last question actually (sqrt3/2 , 2) ? If so I have no idea how I managed to get that one right. Overall I thought it went well but I was resitting it so I will have different views to those doing it for the first time. Low 60's for an A?
    this is what I got : http://www.thestudentroom.co.uk/show....php?t=4168081
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    (Original post by sufiyan1999)
    I thiught the paper went quite well. For the dodgy imaginary numbers bit i used the fact that the sum of the roots would be the negative coefficent of z and the product would be the coefficrnt of the integer
    The 2 roots being given to you
    Hence solving p to be 4 and -4 and substituting into the product equation to obtain q.
    I got q to just be 2 ingegers
    Q was non real.
 
 
 
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