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    (Original post by Steve8052)
    Shudnt this be 2pi(2 2/3-ln3) = 4 2/3pi - 2pi ln 3.

    You got (2pi/3)(4+ln3) = 4 2/3pi - 2/3pi ln 3
    v=pi[x+2lnx-(1/x)] with limits 1,3.
    v=pi[(8/3)+2ln3]
    v=(pi/3)(8+6ln3)
    v=(2pi/3)(4+3ln3)
    There's no subtraction between logarithmic and fraction.
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    I think the equation of the line was y=(x-1)/x tho
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    I thought the paper was really nice (and im by no means a mathmatician - I reckon i got around 73/75)

    1. a) x = lg8/lg5 = 1.29
    b) x = 1/6

    2a) 1 + 12px + 66p²x²
    b) P = -2 q = 24

    3a) x = 4 y = 1.600, x = 12 y = 3.394, x = 16 y = 3.200
    b) 43.86m²
    c) 6263.2m³

    4a) Im not gonna write out my working but it all worked out ok,
    b) f^-1(X) = 2/x + 1
    c) X = +/-2 (the reason i said it could be -2 is that in function g x can be any real number, function g contains x² therefore this will make any negative positive allowing for the condition on f that x > 1 - did anyone else do this???)

    5a) π(8/3 + 2ln3)
    b) i couldnt remember the formula for volume of cone - will hopefully get one mark for subtracting something - i tried to integrate the formula for curved surface area this gave me 1/2πr²h - so close but so far

    6a) 3e^x - 1/2x
    b) it worked with a bit of rearranging
    c) x1 = 0.0613
    x2 = 0.1568
    x3 = 0.1425
    x4 = 0.1445

    d) f'(0.14425) = -0.000685
    f'(0.14435) = 0.002
    change in sign therefore root intbetween

    7.a) see attached picture
    b) see attched picture
    c) a = -2 b = -1
    d) x = -1/6

    8. a) tan θ = -1/3√3
    b) i) use of double angle formula and cos²x + sin²x = 1
    ii) x = 3π/2 π/6 5π/6
    iii) it all cancelled down to
    cos²x + sin²x = 1
    then with a little substitution it showed that 1=1

    I have a copy of the paper if anyone would like specific questions posted.
    Let me know what u think
    Cheers
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    (Original post by Gaz031)
    v=pi[x+2lnx-(1/x)] with limits 1,3.
    v=pi[(8/3)+2ln3]
    v=(pi/3)(8+6ln3)
    v=(2pi/3)(4+3ln3)
    There's no subtraction between logarithmic and fraction.
    i left my answer as the bold bit above
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    (Original post by Steve8052)
    I think the equation of the line was y=(x-1)/x tho
    The equation of the curve was y=(x+1)/x. I remember that distinctly as when flicking back through the paper f^-1(x)=(x+2)/x in the previous question was very similar to the equation of the curve in this question.
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    c) X = +/-2 (the reason i said it could be -2 is that in function g x can be any real number, function g contains x² therefore this will make any negative positive allowing for the condition on f that x > 1 - did anyone else do this???)
    I left it as +-2 as fg(x) implies g is carried out first and g is an even function.
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    (Original post by josklkr)
    I thought the paper was really nice (and im by no means a mathmatician - I reckon i got around 73/75)

    1. a) x = lg8/lg5 = 1.29
    b) x = 1/6

    2a) 1 + 12px + 66p²x²
    b) P = -2 q = 24

    3a) x = 4 y = 1.600, x = 12 y = 3.394, x = 16 y = 3.200
    b) 43.86m²
    c) 6263.2m³

    4a) Im not gonna write out my working but it all worked out ok,
    b) f^-1(X) = 2/x + 1
    c) X = +/-2 (the reason i said it could be -2 is that in function g x can be any real number, function g contains x² therefore this will make any negative positive allowing for the condition on f that x > 1 - did anyone else do this???)

    5a) π(8/3 + 2ln3)
    b) i couldnt remember the formula for volume of cone - will hopefully get one mark for subtracting something - i tried to integrate the formula for curved surface area this gave me 1/2πr²h - so close but so far

    6a) 3e^x - 1/2x
    b) it worked with a bit of rearranging
    c) x1 = 0.0613
    x2 = 0.1568
    x3 = 0.1425
    x4 = 0.1445

    d) f'(0.14425) = -0.000685
    f'(0.14435) = 0.002
    change in sign therefore root intbetween

    7.a) see attached picture
    b) see attched picture
    c) a = -2 b = -1
    d) x = -1/6

    8. a) tan θ = -1/3√3
    b) i) use of double angle formula and cos²x + sin²x = 1
    ii) x = 3π/2 π/6 5π/6
    iii) it all cancelled down to
    cos²x + sin²x = 1
    then with a little substitution it showed that 1=1

    I have a copy of the paper if anyone would like specific questions posted.
    Let me know what u think
    Cheers
    What was the equation to the graph for finding the volume of revolution?
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    (Original post by Steve8052)
    What was the equation to the graph for finding the volume of revolution?
    It was y = (x + 1)/x
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    LOL thanks!! oops my bad memory!! Thought I worked through the question with y=(x-1)/x but remembered now.
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    Btw could you mind posting the marks up for the questions thanks. Trying to work out how many marks I lost.
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    ^^ yeh
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    (Original post by Steve8052)
    LOL thanks!! oops my bad memory!! Thought I worked through the question with y=(x-1)/x but remembered now.
    no problem, what do u think about the +/- 2 thing in question 4??
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    (Original post by Steve8052)
    Btw could you mind posting the marks up for the questions thanks. Trying to work out how many marks I lost.
    Lol same here
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    http://www.thestudentroom.co.uk/t121212.html

    the raw marks needed for an A has never been more than 61, was this paper really so easy????
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    (Original post by josklkr)
    no problem, what do u think about the +/- 2 thing in question 4??
    Sounds decent, I only wrote down the positive answer tho because like you said it said something like x>1?? cant remember, maybe no. of marks will depict what is required.
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    (Original post by tobyclark)
    http://www.thestudentroom.co.uk/t121212.html

    the raw marks needed for an A has never been more than 61, was this paper really so easy????

    June 2003, Raw Mark 65/75 = 80UMS

    Regards
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    well if it is +-2 then i think i will be getting a very high ums...
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    Alright here are the marks for each questoin
    1a) 3
    b) 3

    2a) 2
    b) 4

    3a) 2
    b) 4
    c) 2

    4 a) 4
    b) 3
    c) 3

    5a) 7
    b) 3

    6a) 3
    b) 2
    c) 2
    d) 2

    7. a) 2
    b) 3
    c)2
    d)4

    8a) 5
    b i) 2
    ii) 5
    iii) 3
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    (Original post by devesh254)
    well if it is +-2 then i think i will be getting a very high ums...
    Will be 66-67/75 IMO, although i messed up as usual haha...
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    pi (8/3 + 2ln3) ?

    Couldnt you have left it like that?

    Oh and, i dont think the grade boundries will be low or high. i think 60-62 for an A sounds about right - the general feeling about this paper is people either found it quite easy or quite hard.

    And also, a paper easy to make a mistake in especially the trig question.
 
 
 
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