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    (Original post by cashfield777)
    Do you think you would still get the marks if you left it in a correct but messier way?

    I didn't simplify it as much as you did :o:
    :yes:
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    made stupid mistakes. like the x=tany question. i was puttin in back into part 1. lol. evn the 1st question was kinda hard.
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    (Original post by dancing.barefoot)
    Will I lose marks if I get the right answer but use a different method?

    For the x = tan y question:

    \frac {dx}{dy} = \sec^2 y = \frac {1}{\cos^2 y}

    \frac {dy}{dx} = \cos^2 y

    Then I drew a triangle, put the adjacent side as 1 and the opposite as x, therefore the hypotenuse is \sqrt {1 + x^2}. So, since cos y = A/H,

    \cos^2 y = \frac {1}{1 + x^2}


    For 9b) I did:

    

3^{x} = e^{xln3}

    3^{x} e^{7x+2} = e^{xln3} e^{7x+2}

     = e^{xln3 + 7x + 2} = 15

    x ln 3 + 7x + 2 = ln 15

    x (ln 3 + 7) = ln 15 - 2

    x = \frac {ln 15 - 2}{ln 3 + 7}



    Anybody know if I'll lose marks for that?
    that looks a nice way of doing it.
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    (Original post by wake123)
    I did something really stupid and forgot to label the axis for the graphs.
    I got the write shape and co-ordinates, and in previous mark schemes this is what the marks are given for, but am I likely to lose marks because of forgetting to label the axis?
    no
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    (Original post by Get Cape.Wear Cape.Fly.)
    where is it?
    It's on page 33.
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    (Original post by john1234567)
    :yes:
    :top:
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    (Original post by john1234567)
    3a) Express 5cosx - 3sinx in the form Rcos(x + a), where R > 1/2 and 0 < a < pi/2.



    i worked out R and a correctly but i didn't "express" it in that form - that is, i didn't explicitly say "5cosx - 3sinx = (root34)sin(x + 0.54)", will i lose a mark?
    I would have said no, but looking at previous mark schemes, you will probably lose 1 mark.
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    just under full marks. i think i made a rounding error on Q7b. was a resit though =P
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    (Original post by Remarqable M)
    No you won't lose mark, because examiner is looking for the value of R and alpha, but you expressed it inroot34sin(x-alpha)=4 anyways when you tried to find alpha
    Mathematically, I think you are right.

    I only checked the June 09 MS, on that one accuracy mark was for writing out the equation with R and a put in.
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    (Original post by klgyal)
    how were you able to get the paper and how did you remember all your answers???
    I am not a student
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    Well according to Arsey's solutions I got 68/75 which isn't too bad but I was so stupid by mistaking the lnIxI graph for Iln(x)I

    As long as I get over 90% then I'm happy though :p:
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    (Original post by Milan.)
    number 8
    you solutions include 45 and 135 i plugged that into the top original equation and you get math error ...
    please tell me you dont need those two aswell!!!!!!!
    thanks


    and number 9
    part ii) b) when drawing that graph (x-1)^2 + 3 you can include the value of 3 therefore isn't it fg >= 3

    thanks
    tan 2x = 1/0 which is infinity
    and arctan (infinity) = 90, 270, etc

    i think for bii) it was x2 - 2x + 4. differentiate and its 2x = 2, x = 1 as the minimum. so i had the range as fg(x) >= 1
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    ^ THIS
    as soon as i got out of the exam i was kicking myself
    i threw away an easy few marks =/ i can do those differentiation questions easily

    The first question i completely screwed up
    I did 2 pages of working and the closest i got was
    4(x-1) over 3(x^2-1)(x+1)
    or something
    which is also weird because im usually great at those too =|

    Unless the boundaries are really low i'm pretty sure i can only get a B at best
    and i need an A
    so it also looks like a resit in june for me
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    (Original post by Elitist1991)
    Appreciate the markscheme, however, for the last question:

    so for g(x) = 1 ln(x-1) = 1 therefore, x -1 = e, therefore x = (1 +e)>1

    Therefore fg(e+1) = f(1)

    therefore fg(x)>=3 . The fact that x>1 for g(x) is irrelevent, as g(x) is the input. to f(x)
    nah, you are wrong.

    in fg(x) the range of g(x) becomes the domain for f(x) in the composite function

    f(x) = e^2x + 3 this has an asymptote at y=3 thus can not be equal to 3. Therefore the range of fg(x) > 3 NOT >=
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    It was sooo hard I hate edexcel!
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    (Original post by sb1234)
    Lose all 3 marks?

    By writing sec x as 1/cos x, show that d(sec x)/dy = secxtanx

    How many marks would I lose if I used (cos x)^-1 and differentiated that way.
    no I meant you will only get 3 marks instead of 4

    hahah i did it that way too and used chain rule to full MARK!
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    (Original post by tiggerplease)
    ...
    i love it when people don't recognise the difference of two squares
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    (Original post by Arsey)
    nah, you are wrong.

    in fg(x) the range of g(x) becomes the domain for f(x) in the composite function

    f(x) = e^2x + 3 this has an asymptote at y=3 thus can not be equal to 3. Therefore the range of fg(x) > 3 NOT >=
    i said fg(x)>4 would i still get mark? i wasn't too bothered about finding domain and range question.
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    I entered the exam hall confident and pumped to get my A*.
    I left the exam hall completely raped from all angles by that paper.
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    (Original post by Playboy King)
    I entered the exam hall confident and pumped to get my A*.
    I left the exam hall completely raped from all angles by that paper.
    haha me too, but i felt much better about it when i saw the paper again this morning
 
 
 
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