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    I got 72/75
    96% WOOT!!
    Lost marks because of silly mistakes
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    thanks a lot! i wish u luck to get what you want for all ur exams
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    (Original post by john1234567)
    meh i guess. i just recognised that \frac {1}{1+x^2} is the derivative of arctanx, which is what the question is getting us to find, so i was just lucky i suppose although i don't think it would've confused me even if i hadn't known that lol
    Well i got there in the end (last 5 mins or so) when going over it
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    (Original post by kosy91)
    2 + 2
    Ok, was cot2x=0 definitely an answer then as some people seemed to have debated it. Another question I messed up was the one with Rcos(O+a). Got alpha wrong and carried it into the following parts, how many marks will I be docked for that? Thanks.
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    (Original post by NAJC)
    Ok, was cot2x=0 definitely an answer then as some people seemed to have debated it. Another question I messed up was the one with Rcos(O+a). Got alpha wrong and carried it into the following parts, how many marks will I be docked for that? Thanks.
    yes, cot2x definately had answers!!
    it depemds what went wrong with your alpha. there are 2 marks for it: M1A1. you may lose both or just 1
    your answer of apha is carried forward so you are not penalised again for it
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    (Original post by kosy91)
    yes, cot2x definitely had answers!!
    it depemds what went wrong with your alpha. there are 2 marks for it: M1A1. you may lose both or just 1
    your answer of apha is carried forward so you are not penalised again for it
    Somehow I think I got to: Rsina=3
    Rcosa=5
    so tana=5/3

    Then carried it from there.
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    (Original post by Arsey)
    that is a very kind post; however, I have absolutely no idea what all this rep business is about. Just trying to help and I also benefit from learning from your experiences of the paper(s).
    Arsey,
    for 3 b) I had the right info and wrote the equation, showed x+alpha = cos^-1 (4/root 34)
    and got the 0.815
    how many marks is this because I think i got the solutions after wrong.

    also
    for 7 b) I forgot to put 3sec3xtan3x and used 3secxtanx, but continued and factorised it correctly and used this in part c).
    I made dy/dx = 0 to find x also then subbed this value back in. How many marks will i have lost on both of these parts?

    Finally for 8 I showed cot2x = 0 but managed to get 0,180 () is that only a answer mark gone?

    Thanks if you can answer these
    without all these marks i get 49/75 so hoping i get some method marks from them
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    Got everything right according to the unofficial mark scheme.
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    I probably got mid 50s to high 50s ;_;

    EDIT: Besides, we could simply say that 1/cosx = (cosx)^-1.
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    (Original post by munaaf)
    Arsey,
    for 3 b) I had the right info and wrote the equation, showed x+alpha = cos^-1 (4/root 34)
    and got the 0.815
    how many marks is this because I think i got the solutions after wrong.

    also
    for 7 b) I forgot to put 3sec3xtan3x and used 3secxtanx, but continued and factorised it correctly and used this in part c).
    I made dy/dx = 0 to find x also then subbed this value back in. How many marks will i have lost on both of these parts?

    Finally for 8 I showed cot2x = 0 but managed to get 0,180 () is that only a answer mark gone?

    Thanks if you can answer these
    without all these marks i get 49/75 so hoping i get some method marks from them
    You might get lucky and have 49/75 as the grade boundary for an A. Like they did for M1 last year in January.

    Method marks are usually awarded if your method is correct but your value is wrong, so I think you'd get some method marks.
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    (Original post by NAJC)
    Somehow I think I got to: Rsina=3
    Rcosa=5
    so tana=5/3

    Then carried it from there.
    what you found is cota and not tana
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    (Original post by kosy91)
    what you found is cota and not tana
    I can't remember exactly what I wrote in the exam, I may have put Rsina=5 and Rcosa=3, but I definitely remember getting tana=5/3.
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    (Original post by john1234567)
    sorry, how? it's clearly two separate parts of the question

    The notation of the i and ii are often used to continue the same situation of the question
    in my experience, it usually goes
    8) a) i)
    ii)
    etc.
    It's fairly easy to see how they considered that i and ii were linked
    and it is fairly often seen in the past papers that answers from part i, or part a, are used in other parts
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    did anyone else run out of time?
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    (Original post by tiggerplease)
    The notation of the i and ii are often used to continue the same situation of the question
    in my experience, it usually goes
    8) a) i)
    ii)
    etc.
    It's fairly easy to see how they considered that i and ii were linked
    and it is fairly often seen in the past papers that answers from part i, or part a, are used in other parts
    yes. i completely agree with you there
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    do invigilators give you extra time in UK???
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    (Original post by Elitist1991)
    I see your point, but let me ask this question.

    if fg(x) = (x-1)^2 + 3, where whatever g(x) is the input, and that g(x) can equal 1, such that fg(x) = 3.

    And surely if the range of g(x) which includes 1, becomes the domain of f(x) in the composite function, as you claim, then fg(x) = 3 is a plausible argument. Basically, that just validates my point.
    no it doesn't.

    a composite function works in the following way.

    fg(x) an input value x is applied to the function g (the input value x would have to belong to the domain of g,hence originally x>1)

    the output value g(x) then becomes the input value of fg(x) the composite function. Now since the range of g(x) is any number, any number can enter the function f.

    the function f is e^2x + 3 - the range of this function is >3 not >=3 for the reasons I said earlier.

    Thus whatever value x you put it, the output value fg(x) has to be >3 not >=3


    On to your second point,

    fg(x) = (x-1)^2 + 3

    This does indeed have a minimum point at (1,3). However, as I said earlier the original value of x is first applied to the function g, so has to belong to its domain. Hence x>1; hence fg(x)>3.

    This was the quick answer as to why the range wasn't >=3 in the first place. The domain of g does matter in the composite function.

    Make sense?
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    (Original post by kosy91)
    i got only 2 solutions for Q8(22.5, 112.5). my range for Q9(last part was wrong)
    by mistake i differentiated the ln question as 1/x^2 x 2x instead of 1/x^2+1 x 2x. hence my final answer is wrong.
    so i am guesssing about 5 marks off. how much UMS could that be?
    My guess! would be very close to 100
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    last question, last part, fg(x) >=3 right??? arsey's answer is wrong, or mine is??
    and I thought that cot (2x) =0 has no value, what a dumbass am I
    first question, I forgot to simplify it more
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    (Original post by sb1234)
    Even though it specifically asked for it in the form 1/cos x?
    that is fine - it is the same thing.
 
 
 
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