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    (Original post by john1234567)
    :confused: i'm not sure i agree with that. if cot2x = 0 then you should be able to put x=45 in and get 0
    (Original post by MrBo)
    I got the same first 2 solutions as this, but for the last ones i'm not too sure. Dosn't cotx have asymtotes every 90 plus/minus 180? So if this is true 2x can't equal either 45 or 135
    There's another way then: cot2x = 0

    That's equal to cos2x/sin2x = 0

    Thus you get cos2x = 0

    and now that is definitely x= 45 and x= 135

    Don't worry about it....like hardly any one got it so the grade boundaries would probably take off at least 2 marks to accommodate the loss
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    (Original post by john1234567)
    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
    mate i think you got the 1/cosx wrong you can't use product rule here because it is constant/function and so need to use chain rule by taking the reciprocal of cosx
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    The exam was pretty easy, so easy that I ended up making some really silly mistakes, I think I forgot to put ln on both sides for the last question when taking logs to both sides. and I totally ****** up question one by not looking at it correctly, that's 4 marks gone.
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    (Original post by milliondollarcorpse)
    I don't know how to differentiate arcsinx, lucky I still managed to do it anyway
    actually i meant arctanx. and the question got us to do just that

    (Original post by Remarqable M)
    mate i think you got the 1/cosx wrong you can't use product rule here because it is constant/function and so need to use chain rule by taking the reciprocal of cosx
    both ways work, i've tried it
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    How On Earth Did I Get 1 Wrong!
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    (Original post by Arsey)
    Very hard I thought. Here are my solutions
    haha i got every differentiation alright, but thing is for the x=tany differentiation i laid down two workings using different method to check if i get the same answeer
    will i get mark???
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    In the graph for ln(mod(x)) if I have drawn only one part of the graph how many marks would i lose out of 3
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    That went so bad... i couldn't do most of the paper ... i am now dreading the results day
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    (Original post by alithegreat)
    In the graph for ln(mod(x)) if I have drawn only one part of the graph how many marks would i lose out of 3
    probably 2. coz one mark for the real ln, one mark for the mirrored and i think 1 mark for the asymptote bit
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    How did everyone find this exam.I just want to know if the general opinion was that it was hard or easy. I found the test quite hard...similar to c2 last june.
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    I really hope the grade bondaries are low...I made so many silly mistakes
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    Ye it was a tough exam like C2 ... just hope the boundaries are low ...
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    my mates who did it found it real easy.. but thats only cuz they had a friend from another school who finished the same exams half an hour earlier who gave them the questions!
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    (Original post by hammerhasni5)
    my mates who did it found it real easy.. but thats only cuz they had a friend from another school who finished the same exams half an hour earlier who gave them the questions!
    how unfair:mad: . I made so many silly mistakes in my exam, i'm just hoping the grade boundaries are low-like they were for C2. However some people found the exam easier than C2, so you never know.
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    Pretty easy

    Got 97% worst case scenario
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    we all have different answers for number 8 and Q9 last part
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    (Original post by Arsey)
    of course, so 45 and 135 will also work.
    wait a minute. it wont because if u put 45 or 135 back in the equation

    cosec^2x- cot 2x u dont get one. its maths error!! so it dont work

    u cant have 1/ tan 90 because tan 90 is an asemtope
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    (Original post by crazygirl29)
    wait a minute. it wont because if u put 45 or 135 back in the equation

    cosec^2x- cot 2x u dont get one. its maths error!! so it dont work
    dont use the calculator. try it on paper and treat the math errors as 1/0

    i think it does work
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    why is everyone so quite???
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    That was ****. Everyone in my class found it hard but all you guys found it so easy. Can't believe I got R = 6 for one of the questions....epic fail.
    I think my lowest is going to be 65/75. 1 word...............RETAKE!
 
 
 
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