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    (Original post by RDoh)
    For f(x)=|x-1|-2=5x, I think the answer is -3/4.

    For the water flow question, I got the cross-sectional area as 43.86m^2 and the volume in a minute as 5260m^3 to 3sf.
    I think I got 5260 too - I did 43.86 x 120 = 5263.2 = 5260
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    damn there goes like 4 marks.
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    I did it like this

    distance = speed* time
    = 2*120
    =240, and then times this by the 44, gives 10560, maybe wrong though
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    (Original post by spikeyguru)

    ..volume of water and couldnt remember volume of a cone so lift out the second part of the integration question.( think it coulda bin done another way but i wanted to get on to the trig asap)

    wot did u lot get for your answers?
    I think that there was a more simple way using the information given about the tangent, but like youself i was eagar to get on to the next question, therefore just guessed it quickly at the end.

    The binomial: p = -2 , q = 24

    The fg (x) = 1/4 what did people get for this (if i remembered the question correctly)
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    (Original post by sequence123)
    I did it like this

    distance = speed* time
    = 2*120
    =240, and then times this by the 44, gives 10560, maybe wrong though
    I thought they wanted a volume?
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    Surely volume = distance(ie length) * crossesctional area?
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    I did the speed * 60 * answer to part b
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    For the volumes of revolution question I got

    a) π(8/3 - 2ln3) by integration
    b) π(2ln3 - 5/3) by working out the volume of revolution of the triangle* and subtracting it from a)

    * I did this by realising that it would create a cone of radius 2 and height 2. I didn't know the formula for the volume of a cone but I worked it out (I think) by using the formula for the area of a circle, multiplied by height to get volume of cylinder and divide by 2 to get the volume of the cone. This gave (πr2h)/2 and by subbing in r=2 and h=2, I got 4π.
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    (Original post by sequence123)
    I did it like this

    distance = speed* time
    = 2*120
    =240, and then times this by the 44, gives 10560, maybe wrong though
    This would have been correct but it looks like you've copied the value of time wrong - it's 60 seconds, not 120 seconds.
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    (Original post by topdude)
    does any1 remember what the trig question for tan theta actually was......(so i can work out what i done, and if i got the right answer!)
    (dno if uve got this answered already but anyways)

    i think it was...
    given that 2sin(theta + 30) = cos (theta+60)
    find tantheta

    i got -1/(3root3)


    stoopidly, i thought you had to find theta, so i did, but didnt get it as an exact figure so i did the WHOLE thing again and got theta=30degrees. then realised i only had to find tantheta, which id just done twice...grrr! :mad:
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    I think it went really well.
    for the ccross seciton of area i got around 48m 40 soemtihng anyway!
    so for volume i got 48x2 m/s
    x60 to get the amount per minute

    so it is 5760ish

    for the volume under the curve i think i got 2ln2+2 2/3pi then the volume under the triangle2 2/3pi took one away from other and got volume of r just to be 2ln2 (might have been 3ln2 now i come ot think of it something ln2 anyway) but i did it a relly long way, first i foudn the equation of the straight line then i squared it, and integrated that times pi, input values. then take away Volume triangle from volume under whole curve! There must be a simpler way to find the volume cut out by the triangel but i just couldnt figure it out.

    p=-2 q=24 for binomial

    for the trigonemtry qs i manged them all except part ii to find the value of the angle, i relaised afterwards it is a sin^2-sin+c quadratic kinda thing but nevermind

    the function quesiton where you ahve to input values and get a sign change didnt really work out for me until right at the end so i may have got that wrong.
    and the modulus x-1 find the value of x question didnt work out 100% either
    i got -3/4 but when i put it back in it didnt work at all!

    added up afterwards think i got around 55-60/75 which is really good for me

    fgx=1/4 when x=2

    erm my a and b for the graphs were sensible one of them was 2 or -2
    thought the log questiosn were really easy, for the change of base rule u end up with ln8/ln5 iirc put in calculator-done, i know my mates struggled with this. the trapezium rule was pretty straight forward but i made a daft mistake with copying numbers which i noticed in time phew.

    yep, i am very happy
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    (Original post by Star_inda_sky)
    (dno if uve got this answered already but anyways)

    i think it was...
    given that 2sin(theta + 30) = cos (theta+60)
    find tantheta

    i got -1/(3root3)


    stoopidly, i thought you had to find theta, so i did, but didnt get it as an exact figure so i did the WHOLE thing again and got theta=30degrees. then realised i only had to find tantheta, which id just done twice...grrr! :mad:
    I got 1/3root3 no minus sign but its pretty close lol :cool:
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    yeh got the same as most of you guys, coundnt do the first trig q and the last mod 1 though. For fg(x) i got 2, and i got 5000 something for the volume but didnt convert to 3 sfg would i lose any marks. I however got something like pie(2ln3 - 4) for the volume of R. 1.29 and 1/6 for the first question. For the 3rd trig i got the values 30, 270 and 150. a = -2 and b=-1. The binomial was q=-2 and p=24
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    (Original post by sequence123)

    What did people get for the value whereby f(x) = 5x?
    I got -1/3
    Yeh, there was only one value for x and it was -1/6 because the question was:

    |x-1|-2 = 5x
    |x-1| = 5x+2
    x-1 = -(5x+2)
    -6x = 1
    x = -1/6


    I thought the paper was pretty easy overall...
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    Can somebody type up the question numbers with however many marks each one was worth?
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    method marks all the baby
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    what was the log questions answers?

    first one was lg8/lg5 I think

    the integration question wasnt part b just ANS- 2π?
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    (Original post by chats)
    what was the log questions answers?

    first one was lg8/lg5 I think

    the integration question wasnt part b just ANS- 2π?
    yea lg8/lg5 or ln8/ln5 dont think it matters. my mate did ans-2pi because she thought the volume under the traingle was just area x pi but i doubted this anyone know the formula for a cone, that would have prob solved it easy peezy lemon squeezy- i hope that ist something that is in the massive data booklet i didnt think to look :eek:

    bring on s2 and no more maths alevel yay
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    I did this by realising that it would create a cone of radius 2 and height 2. I didn't know the formula for the volume of a cone but I worked it out (I think) by using the formula for the area of a circle, multiplied by height to get volume of cylinder and divide by 2 to get the volume of the cone. This gave (πr2h)/2 and by subbing in r=2 and h=2, I got 4π.
    The volume of a cone is 1/3π(r^2)h i think, this gave a volume of 2πln3 for the region R.
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    (Original post by marisad_uk)
    yea lg8/lg5 or ln8/ln5 dont think it matters. my mate did ans-2pi because she thought the volume under the traingle was just area x pi but i doubted this anyone know the formula for a cone, that would have prob solved it easy peezy lemon squeezy- i hope that ist something that is in the massive data booklet i didnt think to look :eek:

    bring on s2 and no more maths alevel yay
    I thought it was just 2π (area of triangle * π )

    oh well
 
 
 
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