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    Grrr. Not looking forward to this exam. Can someone explain 2ii. Jan 2010 to me please? I realise that I should try and find a ratio, but the ratios I find don't give the correct answer. Maybe telling me what I should find a ratio between could be useful.

    Thanks.
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    (Original post by m00c0w)
    I did think it was a bit weird having a negative log :P So basically whenever you integrate something that becomes log, just put it inside mod? My teacher had said that you can generally just ignore the mod when I pointed it out to him earlier in the year, and it isn't that he is a bad teacher. In fact, he is the best teacher I have ever had, and most of the people in my class would agree, just strange that he didn't tell us this bit.

    Anyway, thanks for that, hopefully I will remember it in the exam!
    To be honest it is more that it has come up in papers so rarely (that you need the modulus sign) that your teacher probably neglected to mention it. I mean so far that is the only question (to my memory) that i've done in past papers where the modulus sign was actually needed. But yeah it is good to remember to include the modulus sign as the examiners could indeed put it in our paper to try and trick us

    I doubt they would penalise you for putting modulus signs in log integral questions where the limits do not require a modulus sign as we are still saying something that is mathematically correct.


    (Original post by m00c0w)
    Grrr. Not looking forward to this exam. Can someone explain 2ii. Jan 2010 to me please? I realise that I should try and find a ratio, but the ratios I find don't give the correct answer. Maybe telling me what I should find a ratio between could be useful.

    Thanks.
    Find the direction vectors AB and AC first (i.e. B-A and C-A) then if you look at the two results a ratio should seem obvious between those two directions
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    find  \int secxtan^2xdx\

    Spoiler:
    Show
     \int secxtan^2x dx\

= \int secx(sec^2x-1)dx\

= \int sec^3xdx\ -\int secxdx\

= \int sec^2xsecxdx\ -\int secxdx


    and do the first integral by parts



    note: you may assume  \int secxdx\ = ln|secx+tanx| + c
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    (Original post by jassi1)
    find  \int secxtan^2xdx\

    Spoiler:
    Show
     \int secxtan^2x dx\

= \int secx(sec^2x-1)dx\

= \int sec^3xdx\ -\int secxdx\

= \int sec^2xsecxdx\ -\int secxdx


    and do the first integral by parts



    note: you may assume  \int secxdx\ = ln|secx+tanx| + c
    Is this a question you have?
    If so you need to use a recursive integral - naming your integral I may help.
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    (Original post by jamesaung)
    Completely agree, there isn't very much variation between papers at all.
    I also agree apart from the june 2009 q9 threw me off abit.
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    Hmm I'm doing Jan 12 and it's a tricky one
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    I'm finding C4 better than C3 in some respects as the papers seem more straightforward (so far). But the grade boundaries seem pretty high. :/
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    Anyone know why I can't split x^2/(x-1)^2(x-2) into partial fractions like A/(x-1)^2 + B/(x-2). Instead it's A/(x-1)^2 + B/(x-2) + C /(x-1)
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    (Original post by eggfriedrice)
    Anyone know why I can't split x^2/(x-1)^2(x-2) into partial fractions like A/(x-1)^2 + B/(x-2). Instead it's A/(x-1)^2 + B/(x-2) + C /(x-1)
    You can't split that expression like you have because it has a repeated factor. So you treat it differently if it has a repeated factor like so:

     \frac{ax^2+bx+c}{(x-e)^2(x-f)} = \frac{ \alpha }{(x-e)^2} + \frac{ \beta }{(x-e)} + \frac{ \gamma }{(x-f)}   for  e \not= f

    And if you're interested the pattern carries on:

     \frac{p(x)}{(x-e)^3(x-f)} = \frac{ \alpha }{(x-e)^3} + \frac{ \beta }{(x-e)^2} + \frac{ \gamma }{(x-e)} + \frac{ \delta}{(x-f)}  for  e \not= f where  p(x) has degree less than or equal to 3.

    however C4 only has either a factor of multiplicity 2 or no repeated factors.
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    (Original post by joostan)
    Is this a question you have?
    If so you need to use a recursive integral - naming your integral I may help.
    I was just posing this question to others because I found it quite fun and interesting
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    Ahhhhhh parametrics someone help please
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    Hi guys does anyone have the OCR jan 13 paper?


    This was posted from The Student Room's iPhone/iPad App
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    (Original post by SexyAndIKnowIt.)
    Ahhhhhh parametrics someone help please
    Do you have a specific question you are stuck on?
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    Can anyone explain how this equals this please? Thanks Name:  ImageUploadedByStudent Room1371462796.589262.jpg
Views: 125
Size:  133.2 KB


    Posted from TSR Mobile
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    (Original post by ninuzu)
    Do you have a specific question you are stuck on?
    Jow do I know when's to integrate/ differentiate or substitute when dealing with parametrics
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    If you are given eg. Cos 4x how do you convert this into the double angle formulae that we know
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    (Original post by Hello_ImJess)
    Can anyone explain how this equals this please? Thanks Name:  ImageUploadedByStudent Room1371462796.589262.jpg
Views: 125
Size:  133.2 KB


    Posted from TSR Mobile
    you know cos4x = 2cos^2(2x) -1 from C3
    rearrange to cos4x + 1 = 2cos^2(2x)
    then 0.5cos4x + 0.5 = cos^2(2x)
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    (Original post by SexyAndIKnowIt.)
    If you are given eg. Cos 4x how do you convert this into the double angle formulae that we know
    You can treat cos4x very similarly to cos2x when doing the multiple angle formulae.
    So, in the double angle formula for cos: cos2x = cos^2(x) - sin^2(x), for cos 4x, you can simply deduce that: cos4x = cos^2(2x) - sin^2(2x)
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    (Original post by SexyAndIKnowIt.)
    Jow do I know when's to integrate/ differentiate or substitute when dealing with parametrics
    If you are asked to find the equation of the tangent/gradient of a curve, then you have to differentiate the parametric equations, so if you are told x=*some function involving t* and y=*another function involving t*, then to find the gradient: dy/dx = (dy/dt)/(dx/dt).
    You use substitution when you need to find the Cartesian form (i.e. an equation that doesn't involve t).
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    Help with question 4 please

    Name:  ImageUploadedByStudent Room1371465715.798886.jpg
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