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    (Original post by tomiswellcool)
    On question 4)i) Did the solution have to be simplified, because i lieterally put everything into the quotient rule, then just left it. Will i be docked marks for this?
    I did that too, i don't think you should lose marks there because there really wasn't much to simplify
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    (Original post by crazedmonkey09)
    Did anyone solve the Cosec^2 - Cot=1 by
    1/sin^2 - CosSin/Sin^2 =1
    cosSin/Sin^2 = 1
    Cos/Sin=1
    cot=1

    This working is completely wrong because i negated the 1 (i only just realised now ) but i still got 2 solutions, do you think i will still get the method marks?
    thats would get one of the anwers but not the one you got.

    you should've got cos(2x) = 0 from that
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    this paper was bloody hard compared to the one june 09...i did it as mock and got 80% on that with no revision at all....this time i would be happy to get half that :eek: :mad:

    oh well there is always a retake possible...:confused:
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    Looked through the markscheme...and was just wondering about the cosec^2x - cotx = 1

    I got the cotx(cotx - 1) = 0 ...(or whatever that turned out to be)

    ..but surely cotx can't = 0, as rewritten that would be 1/tanx = 0 ... multiply both sides by tanx, and that leaves you with 1 = 0 which is incorrect? And tanx can never be 0, because 1/0 is impossible..

    Have I got that wrong?
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    (Original post by Anonymous0155)
    thats would get one of the anwers but not the one you got.

    you should've got cos(2x) = 0 from that
    hows that? from this crazy working i got the solutions 22.5, 112.5 its a sin squared on the bottom and a single sin on top so i get cot but why =0?
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    (Original post by crazedmonkey09)
    Did anyone solve the Cosec^2 - Cot=1 by
    1/sin^2 - CosSin/Sin^2 =1
    1 - cos/Sin/Sin^2 = sin^2
    1 - Cos/Sin^3 = sin^2

    This working is completely wrong because i negated the 1 (i only just realised now ) but i still got 2 solutions, do you think i will still get the method marks?
    actually, thats just kinda wrong; but you got the right answer somehow.

    the bold is how it should've gone, coz if you multiply you gotta do it both sides.
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    (Original post by Smeh)
    That was my plan too!
    I wish I was born year earlier...None of this A* business.
    I should really be revising chemistry right now but instead i'm moping :emo:
    Well after spending several hours looking on the bright side only to have my retinas burnt by more incorrect answers :mad: the grade boundaries should be quite low and if they are as low as 55 for example we could be looking at 93+.
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    (Original post by crazedmonkey09)
    I did that too, i don't think you should lose marks there because there really wasn't much to simplify

    Yeah, the way i looked at it this was: They asked for dy/dx of f(x), and i gave them dy/dx of f(x). They didn't ask for it simplified so i didn't :p:
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    So for question 8 there was meant to be four solutions?  cot2x = 0

     \frac{cos2x}{sin2x} = 0 multiply through by sin2x

     cos2x = 0 which equals zero at  \frac{\pi}{2} , \frac{3\pi}{2}
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    (Original post by Anonymous0155)
    actually, thats just kinda wrong; but you got the right answer somehow.

    the bold is how it should've gone, coz if you multiply you gotta do it both sides.
    i didn't multiply the whole thing by sin i just wrote the fraction differently

    cos2x/sin2x
    as
    Cos2xSin2x/Sin^2 2x

    those fractions are identical because sin can be canceled. My wrong working was where it should become 1-Cos2xSin2x/Sin^2 but don't include the 1- and i get the right answer
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    (Original post by Robbyv)
    ^ so much this.
    Having 8 exams this month and having sat 4 of them and finding them piss, i thought this would be another cakewalk. First question stumped me for a bit. the X=Tany, i got the the dy/dx= 1/sec^2(y) and got stuck (i'm really annoyed i did now :mad:) The graph of ln|x| was fail, i've only drawn lnx for +ve values (another thing :mad:) and then there was that 9a)ii). I figured it out, literally when they said pens down, so i couldn't write it :mad:. Oh and now theres a debate whether cot2x could give you solutions for x. well June here i come
    When i got to 1/sec^2(y) i changed it to 1/1+tan^2(y) and x=tany
    so 1/1+tan^2(y) = 1/1+x i hope thats right =S
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    But that could also be written as:

    1/tan2x = 0

    ...meaning tan2x couldn't = 0 ?
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    (Original post by crazedmonkey09)
    i didn't multiply the whole thing by sin i just wrote the fraction differently

    cos2x/sin2x
    as
    Cos2xSin2x/Sin^2 2x

    those fractions are identical because sin can be canceled. My wrong working was where it should become 1-Cos2xSin2x/Sin^2 but don't include the 1- and i get the right answer
    Yes you had

    (1-cos2xsin2x)/sin^2(2x) = 1

    but you can't just make it 1 - cos2xsin2x = 1

    1 - cos2xsin2x = sin^2(2x) (you need a common denominator for the whole equation in order to cancel it out; not just one side.
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    (Original post by Riten Patel)
    When i got to 1/sec^2(y) i changed it to 1/1+tan^2(y) and x=tany
    so 1/1+tan^2(y) = 1/1+x i hope thats right =S
    thats correct.
    yo, do u go to a catholic 6th form?
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    (Original post by Anonymous0155)
    Yes you had

    (1-cos2xsin2x)/sin^2(2x) = 1

    but you can't just make it 1 - cos2xsin2x = 1

    1 - cos2xsin2x = sin^2(2x) (you need a common denominator for the whole equation in order to cancel it out; not just one side.
    I never did make it that
    1 stays on the other side of the equals sign i never canceled the 1's i just forgot about the other 1 which should have been on top of the fraction, thats how i managed cot2x=1. call it serendipity if you will but i doubt i'll get the marks because my working is badly wrong

    from your working you multiplied the whole thing by sin^2 but somehow divided it as well, i can't understand what you've done there
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    Cot 2x can equal zero; look at the graph of cot x. It gives 2 extra solutions, 45, 135 which ARE valid in the original equation. I know it is confusing but it has been explained it detail earlier. If you didn't include them you've lost 2 marks.
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    can someone tell me what page i can download the unofficial mark scheme from

    thank you
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    easiest paper every i think i got 70/75
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    (Original post by Catherine91)
    Looked through the markscheme...and was just wondering about the cosec^2x - cotx = 1

    I got the cotx(cotx - 1) = 0 ...(or whatever that turned out to be)

    ..but surely cotx can't = 0, as rewritten that would be 1/tanx = 0 ... multiply both sides by tanx, and that leaves you with 1 = 0 which is incorrect? And tanx can never be 0, because 1/0 is impossible..

    Have I got that wrong?
    you've kind of proved why the solutions work. If the logic is that the solutions are not valid then because 1/0 is impossible tanx can never be zero; but it can, an infinite number of times in fact.
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    (Original post by aswat300)
    can someone tell me what page i can download the unofficial mark scheme from

    thank you
    33 I think
 
 
 
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