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    (Original post by Bronze92)
    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
    This is exactly what i was thinking, I'm sure it is >= 3 because if you dy/dx the minimum point is y=3 at x=1
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    (Original post by Arsey)
    My guess! would be very close to 100
    thanks a lot.....my fingers are crossed!!!
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    (Original post by Anonymous0155)
    Will you be doing the C4 paper on monday?
    should do, I can not remember, is it an AM or PM exam?

    I can imagine many people will be doing C4, predicting grade boundaries is a minefield. The larger the cohort taking it, the easier it is.
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    (Original post by Arsey)
    My guess! would be very close to 100
    I've lost the range mark on the last question but I think I've got the rest right. Do you think 74/75 can get 100 ums?
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    (Original post by Arsey)
    that is fine - it is the same thing.
    cool....because i did cosx^-1 and used the chain rule
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    (Original post by Arsey)
    should do, I can not remember, is it an AM or PM exam?

    I can imagine many people will be doing C4, predicting grade boundaries is a minefield. The larger the cohort taking it, the easier it is.
    Ty; its an AM paper. Im hoponh to improve on that 53% that ruined my A last year.
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    Well I got the ln|x| question wrong =.= forgot to draw the curve for negative x values...Btw in question 9, for range and domain, does it exactly have to be x>3, or would x>/=3 be okay?
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    (Original post by cashfield777)
    I've lost the range mark on the last question but I think I've got the rest right. Do you think 74/75 can get 100 ums?
    ARE YOU MAD???? that is 100 boy....go enjoy
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    (Original post by Noble.)
    How do you think the paper compared to June 2006's paper? Do you think 73/74 would be sufficient for full UMS marks on this paper? (I know it's pretty hard to say, but the 100 UMS cap, for June 2006, was 70).
    I think it is harder personally. I am expecting lower boundaries.

    That said, there were quite a lot of easy question - Q1-Q6 were pretty straight forward.
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    nope....strictly >3
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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
    Luckily I had done a VERY similar question to that of part ii) the day before the exam, so I did it fine Most people at my school were also saying that the paper was quite hard. So maybe the grade boundaries will be lowered.
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    (Original post by kosy91)
    nope....strictly >3
    Damn..so there goes another 2 marks -.-
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    (Original post by NAJC)
    How many marks do you reckon I will drop for the following? Q7c I worked out that the min. point was either -/+0.196 but used the positive solution with its y equivalence. Q8 I left out cotx=0.
    I would guess - 2 marks for Q8. Not sure about 7, you've clearly lost the accuracy mark for b but I am not sure if you will lose one for a.

    If that is all you have got wrong you are looking at 71/72 which I think will be 100UMS
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    (Original post by Doughboy)
    I probably got mid 50s to high 50s ;_;

    EDIT: Besides, we could simply say that 1/cosx = (cosx)^-1.
    this is what i did, then just use chain rule.

    Argh so annoyed, i didn't simplify 4i) enough (left it as 2nd from last in scheme) had no idea wtf to do in 4ii) forgot to divide by 3 at last stages in 7c) only got 2 solutions in 8. Thats my 100 UMS down the sh!tt3r. I'll be lucky to make it past 90 now (
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    (Original post by cOnfusedtOm)
    Damn..so there goes another 2 marks -.-
    no
    range is only 1 mark
    showing that fg(x)=(x-1)^2 + 3 or any other form is the other mark
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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
    3b) I would say 2/5

    I think the error is Q7 will cost you big time; you are looking at 3/8

    8) prob 1 mark


    so you are prob looking at 55/75 which may be enough for an A.
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    Um..in the previous question they asked for the domain of f^-1 (x) and I answered it as x>/=3 when it was meant to be just x>3.. ¬_¬
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    (Original post by Arsey)
    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?
    Yeah, I see. Sneaky question. Thanks. I assume that's one mark lost with the 2 lost for not realising cot 2x = 0 has solutions.

    On another note, do you believe A-Level maths has become easier, and also is yesterday's paper the governments attempt at 'stretch and challenge'?
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    (Original post by Bronze92)
    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
    you're wrong, sorry to be blunt but I have explained why it isn't >=3 in great detail in previous posts.
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    I think I should be saying goodbye to the potential of getting an A* ):
 
 
 
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