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    Post all your thoughts on today's exam here. What questions did you struggle with?
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    Wait 24hours
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    Some people haven't finished yet?
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    You have to wait til midnight guys.
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    your gona get warning points mate ..
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    (Original post by DalekLover)
    You have to wait til midnight guys.
    It's 4.30AM for afternoon exam discussion. Thread closed.
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    can we discuss this at 12'o clock. iam staying awake waiting for an unofficial mark scheme to be posted.

    any 1 else doing the same?/
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    4:30 am i think
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    4.30am
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    waiting for the unofficial mark scheme too.
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    4.30 omg why??
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    (Original post by victor_b1992)
    4.30 omg why??
    Waiting for Californian and other islands students to finish c3 exam.
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    hw did u find the exam ??
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    tell i said that and i got a Warnings Level: 10 for that so i advise no to reply to this. lol
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    4.30am for this exam, as you would have seen from the announcement. Locked until then =)
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    just a few questions i remember

    --------------------------------------------------

    3^xe^{7x+2} = 15

    3^xe^{7x}e^2 = 15

    ln(3^xe^{7x}e^2) = ln(15)

    ln(3^x) + ln(e^{7x}) + ln(e^2) = ln(15)

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

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

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


    --------------------------------------------------

    cosec^22x - cot2x = 1

    --------------------------------------------------


    x = tany

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

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

    \frac {dy}{dx} = \frac {1}{1 + tan^2y}

    \frac {dy}{dx} = \frac {1}{1 + x^2}


    --------------------------------------------------


    y = secx = \frac {1}{cosx}

    u = 1

    v = cosx

    \frac {du}{dx} = 0

    \frac {dv}{dx} = -sinx

    \frac {dy}{dx} = \frac {d(secx)}{dx} = \frac {0 - - sinx}{cos^2x} = \frac {sinx}{cos^2x} = \frac {tanx}{cosx} = tanxsecx

    --------------------------------------------------

    this is the one with the minimum (a,b) and you have to find a and b, did anyone get the same as me?

    y = e^2xsec3x

    u = e^{2x}

    v = sec3x

    \frac {du}{dx} = 2e^{2x}

    \frac {dv}{dx} = 3sec3xtan3x

    \frac {dy}{dx} = 2e^{2x}sec3x + 3e^{2x}sec3xtan3x = e^{2x}sec3x(2 + 3tan3x)

    e^{2x}sec3x(2 + 3tan3x) = 0

    3tan3x = -2

    tan3x = \frac {-2}{3}

    x = \frac {arctan(\frac {-2}{3})}{3}

    so a = -0.196

    then put x = -0.196 into the original equation to find b = 0.812


    --------------------------------------------------


    graph of y = ln(|x|) was pretty simple, i checked on wolframalpha, got it right
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    Very hard I thought. Here are my solutions (updated!)
    Attached Images
  1. File Type: pdf C3 Jan 2010 Solution V2.pdf (400.2 KB, 1512 views)
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    Piss easy exam, I made a mistake with the cosec^2x - cot2x question because I did it so many times on the Elmwood papers that I rushed this question, I still got 22.5 as one value so it's probably 2 lost marks.
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    (Original post by Arsey)
    Very hard I thought. Here are my solutions
    for question 8 you could change cot2x=0 to cos2x/sin2x=0 and so you have cos2x=0. i did the same as you though
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    (Original post by Arsey)
    Very hard I thought. Here are my solutions
    also for the last question (the range of fg) i got it's >4 because the minimum is already at the point (1,1) before the +3
 
 
 
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