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Edexcel FP2 Official 2016 Exam Thread - 8th June 2016 watch

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    Hi guys. Hope everything is going smooth and well. I Have a little problem. I am a self taught student. currently doing past papers and very happy with results. the problem is that the notation i use is different to the markscheme slightly. but i get the same answer as them. will this affect the way im graded?
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    (Original post by ChuckNorriss)
    Hi guys. Hope everything is going smooth and well. I Have a little problem. I am a self taught student. currently doing past papers and very happy with results. the problem is that the notation i use is different to the markscheme slightly. but i get the same answer as them. will this affect the way im graded?
    Most likely not - can you give us a few concrete examples of differences in your notation?
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    Could someone help me understand where I'm going wrong here?

    Question 3: https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf

    I ended up with y = \dfrac{3cosx-cos3x+c}{2sinx}

    Though the MS somehow manages to put a 3 infront of the cos3x

    If you look at the line right on top of that it says

    ysinx= \dfrac{3}{2}\left[- \dfrac{1}{3}cos3x + cosx \right] + C

    I got this, am I missing something blindly obvious?
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    (Original post by edothero)
    I got this, am I missing something blindly obvious?
    I would think so.

    Remember that a(b+c) = ab + ac \neq ab + c.

    So \displaystyle \dfrac{3}{2}\left[- \dfrac{1}{3}\cos3x + cosx \right] = \frac{3}{2} \times \frac{-1}{3} \cos 3x + \frac{3}{2} \times \cos x
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    (Original post by Zacken)
    I would think so.

    Remember that a(b+c) = ab + ac \neq ab + c.

    So \displaystyle \dfrac{3}{2}\left[- \dfrac{1}{3}\cos3x + cosx \right] = \frac{3}{2} \times \frac{-1}{3} \cos 3x + \frac{3}{2} \times \cos x
    so ysinx = -\dfrac{1}{2} cos3x + \dfrac{3}{2} cosx

    \therefore ysinx = \dfrac{3cosx-cos3x}{2}

    \therefore y = \dfrac{3cosx-cos3x}{2sinx}

    That's what I did
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    (Original post by edothero)
    so ysinx = -\dfrac{1}{2} cos3x + \dfrac{3}{2} cosx

    \therefore ysinx = \dfrac{3cosx-cos3x}{2}

    \therefore y = \dfrac{3cosx-cos3x}{2sinx}

    That's what I did
    Ah, sorry - I misread. Yes, that's fine - the markscheme is in error.
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    (Original post by Zacken)
    Ah, sorry - I misread. Yes, that's fine - the markscheme is in error.
    No problem, thank you.

    Quite scary how there's a mistake in a 2015 paper's mark scheme.
    One can only assume, and hope, that it was fixed in time so candidates weren't marked down
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    (Original post by edothero)
    No problem, thank you.

    Quite scary how there's a mistake in a 2015 paper's mark scheme.
    One can only assume, and hope, that it was fixed in time so candidates weren't marked down
    Nah, the examiners get a different copy of the markscheme that's far more detailed than this one which is produced for the singular use of teachers/candidates. Plus they have a meeting before the marking commences and any such errors would be spotted immediately, although failing that (in the most extreme case) - they will notice something is up as soon as candidates seem to be making the same "error" repeatedly.
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    (Original post by Zacken)
    Most likely not - can you give us a few concrete examples of differences in your notation?


    So i know this is an fp2 thread but with m3. instead of writing v dv/dx = .... i just write a= ..... and then below straight to 0.5v^2.
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    (Original post by ChuckNorriss)
    So i know this is an fp2 thread but with m3. instead of writing v dv/dx = .... i just write a= ..... and then below straight to 0.5v^2.
    That's not different notation at all.
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    Please could anyone explain exercise A question 7a https://8fd9eafbb84fdb32c73d8e44d980...lZweHc/CH6.pdf I really can't follow the working! Thanks
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    (Original post by economicss)
    Please could anyone explain exercise A question 7a https://8fd9eafbb84fdb32c73d8e44d980...lZweHc/CH6.pdf I really can't follow the working! Thanks
    What's confusing you?

    \displaystyle \frac{d}{dx}(\ln (x + \sqrt{1+x^2}) = \frac{\frac{d}{d}x(x + \sqrt{1+x^2})}{x + \sqrt{1+x^2}} = \frac{1 + \frac{2x}{2x\sqrt{1+x^2}}}{x + \sqrt{1+x^2}}

    Simplify that down to \frac{1}{\sqrt{1+x^2}} yourself.

    Then f'(x) = \frac{1}{\sqrt{1+x^2}} \iff \sqrt{1+x^2}f'(x) = 1
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    (Original post by Zacken)
    What's confusing you?

    \displaystyle \frac{d}{dx}(\ln (x + \sqrt{1+x^2}) = \frac{\frac{d}{d}x(x + \sqrt{1+x^2})}{x + \sqrt{1+x^2}} = \frac{1 + \frac{2x}{2x\sqrt{1+x^2}}}{x + \sqrt{1+x^2}}

    Simplify that down to \frac{1}{\sqrt{1+x^2}} yourself.

    Then f'(x) = \frac{1}{\sqrt{1+x^2}} \iff \sqrt{1+x^2}f'(x) = 1
    Thanks, I think it's the differential of root (1+x squared) that I'm getting wrong, please could you explain how you do it? I got x(1+x squared)^-0.5, as the differential for it :/ thank you
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    (Original post by economicss)
    Thanks, I think it's the differential of root (1+x squared) that I'm getting wrong, please could you explain how you do it? I got x(1+x squared)^-0.5, as the differential for it :/ thank you
    \displaystyle 

\begin{align*}\frac{\mathrm{d}}{  \mathrm{d}x}(\sqrt{1+x^2}) &= \frac{\mathrm{d}}{\mathrm{d}x}((  1+x^2)^{1/2}) \\&= 2x \cdot \frac{1}{2} \cdot (1+x^2)^{-1/2} \\&= \frac{2x}{2\sqrt{1+x^2}} \\&= \frac{x}{\sqrt{1+x^2}} \end{align*}
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    (Original post by Zacken)
    \displaystyle 

\begin{align*}\frac{\mathrm{d}}{  \mathrm{d}x}(\sqrt{1+x^2}) &= \frac{\mathrm{d}}{\mathrm{d}x}((  1+x^2)^{1/2}) \\&= 2x \cdot \frac{1}{2} \cdot (1+x^2)^{-1/2} \\&= \frac{2x}{2\sqrt{1+x^2}} \\&= \frac{x}{\sqrt{1+x^2}} \end{align*}
    Thanks, understand that now, how does the expression simplify down please as I have 1+ x/root( 1+x^2) divided by x+root(1+x^2) but I can't seem to simplify it. Thanks
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    (Original post by economicss)
    Thanks, understand that now, how does the expression simplify down please as I have 1+ x/root( 1+x^2) divided by x+root(1+x^2) but I can't seem to simplify it. Thanks
    Common denominators on the numerator.
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    (Original post by economicss)
    Thanks, understand that now, how does the expression simplify down please as I have 1+ x/root( 1+x^2) divided by x+root(1+x^2) but I can't seem to simplify it. Thanks
    See if any of this makes sense to you. It can be hard to see what to do, but if the result is given, you can try and 'force it'.
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    Oops, didn't realize Zacken was online...
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    Hi guys, need a little help with a method of differences question if that's okay. The partial fractions gives three terms and I wrote down the expansion (I did this correctly as I checked it against the mark scheme), however I saw no pattern in cancelling out the terms. What is the protocol for going about this when there are three terms instead of 2?

    I have linked the mark scheme's expansion. Name:  fp2 capture.PNG
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    (Original post by Alby1234)
    Hi guys, need a little help with a method of differences question if that's okay. The partial fractions gives three terms and I wrote down the expansion (I did this correctly as I checked it against the mark scheme), however I saw no pattern in cancelling out the terms. What is the protocol for going about this when there are three terms instead of 2?

    I have linked the mark scheme's expansion.
    Popular question, this one.



    Sum along the diagonals.
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    (Original post by EricPiphany)
    See if any of this makes sense to you. It can be hard to see what to do, but if the result is given, you can try and 'force it'.
    Name:  Capture.PNG
Views: 137
Size:  30.2 KBThank you and thanks Zacken
    Oops, didn't realize Zacken was online...
    Thank you, and thanks Zacken
 
 
 
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