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    (Original post by SeanFM)
    :hmmmm2: do you have an example of what you mean?
     \frac{x}{2-x} > \frac{x+3}{x}
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    (Original post by kkboyk)
     \frac{x}{2-x} > \frac{x+3}{x}
    You never know, so you have to multiply both sides by (2-x)^2 and x^2 unless the domain is restricted to 0<x<2.
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    (Original post by SeanFM)
    You never know, so you have to multiply both sides by (2-x)^2 and x^2 unless the domain is restricted to 0<x<2.
    Sometimes the marks scheme does it normally by multiplying the denominators to both sides, which reay confuses me :s
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    (Original post by kkboyk)
    Sometimes the marks scheme does it normally by multiplying the denominators to both sides, which reay confuses me :s
    :hmmmm2: have you got an example or mark scheme that does this?
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    (Original post by SeanFM)
    :hmmmm2: have you got an example or mark scheme that does this?
    June 2011

     \frac{3}{x+3} > \frac{x-4}{x}
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    (Original post by Zacken)
    Ah, your hint gave it away.

    It's not hard to prove that \tan \frac{x}{2} = \frac{\sin x}{1+\cos x} from which you argue that \cot \frac{x}{2} = \csc x - \cot x and hence \csc x = \cot \frac{x}{2} - \cot x.

    So your sum is: \cot 1 - \cot 2 + \cot 2 - \cot 4 + \cdots  + \cot 2^{n-1} - \cot 2^n which telescopes down to \cot 1 - \cot 2^n.
    III 2006 reminds me of those identities. Thank god.


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    Please could anyone explain how to do question 17b? I don't know if I'm missing something ovbious but wouldn't multiplying by theta make it very difficult to integrate? Thanks Name:  image.jpg
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    (Original post by economicss)
    Please could anyone explain how to do question 17b? I don't know if I'm missing something ovbious but wouldn't multiplying by theta make it very difficult to integrate? Thanks Name:  image.jpg
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    You use by parts from C4
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    Why does (cos theta)= cos(-theta )
    and -sin(theta) = sin(-theta )

    and how do I remember this and not get confused
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    (Original post by Rkai01)
    Attachment 543309
    How did you get it down to the second line please?
    I've written it as if it were an actual FP2 question, if that gives any hints.

    (a) Show that

    \ln |\sec 2x| = \ln 2 + \ln |\sin 2x| - \ln |\sin 4x|

    (b) Use the method of differences to show that

    \displaystyle \sum_{n=1}^{m}\ln |\sec 2^{r}\theta| = m\ln 2 + \ln |\sin 2\theta| - \ln |\sin 2^{m+1}\theta|

    (c) Hence, or otherwise, show that

    \displaystyle \sum_{n=1}^{m}\tan \left(2^{r}\theta\right)  = 2 \cot 2\theta - 2^{m+1}\cot\left( 2^{m+1} \theta \right)
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    Name:  image.jpg
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    (Original post by Patrick2810)
    do we have to know how to do these? they seem a lot harder than any of the exam qs

    can i ask how you would approach 9a?
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    (Original post by Cpj16)
    Why does (cos theta)= cos(-theta )
    and -sin(theta) = sin(-theta )

    and how do I remember this and not get confused
    Draw the graphs. The cosx graph is reflected in the y axis so any negative value for x could be positive and give the same result. The sinx graph when x<0 is just the negative of the sinx graph when x>0.
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    (Original post by Rkai01)
    You use by parts from C4
    Thank you
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    (Original post by Craig1998)
    Draw the graphs. The cosx graph is reflected in the y axis so any negative value for x could be positive and give the same result. The sinx graph when x<0 is just the negative of the sinx graph when x>0.
    thank you. you are a genius
    I just went on desmos to sketch them
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    (Original post by kkboyk)
    June 2011

     \frac{3}{x+3} &gt; \frac{x-4}{x}
    :hmmmm2: I do not know, sorry!

    :bump: - why can they multiply both sides of this by the denominators of each side, rather than the square? There is nothing else to the question.
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    Please could anyone do me a massive favour and post a worked solution to question 16b as I always seem to just get stuck at the same stage in these types, thanks Name:  image.jpg
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    (Original post by SeanFM)
    :hmmmm2: I do not know, sorry!

    :bump: - why can they multiply both sides of this by the denominators of each side, rather than the square? There is nothing else to the question.
    Thanks anyway

    I realised that squaring the denominator would work although its much longer.
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    (Original post by kkboyk)
    Thanks anyway

    I realised that squaring the denominator would work although its much longer.
    Stick to multiplying by squares, you cannot be sure of the sign of what your multiplying if you do not.
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    is anyone else finding the latest GCE and IAL papers (~2013 onwards) relatively more difficult than previous ones?
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    (Original post by waqasabbasi)
    is anyone else finding the latest GCE and IAL papers (~2013 onwards) relatively more difficult than previous ones?
    I think that's been a trend for most modules, but in my opinion FP2 has stayed relatively the same, (hence why we are seeing GBs of 71 for A*).
 
 
 
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