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Edexcel A2 C4 Mathematics June 2016 - Official Thread Watch

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    (Original post by Zacken)
    I got full UMS and I barely did 6 or 7 of Edexcel's papers, let alone other exam boards.
    Nice
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    (Original post by ULTRALIGHT BEAM)
    Only for student aiming for 75/75

    Get rich or die tryin

    

y= f(x)

y = \left( x^{2} - 1\right) ^ {-\dfrac{3}{2}}

R = \int^{2}_{\sqrt {2}}ydx
    R rotated through 2\pi about the X axis
    Show the solid formed is
    

\dfrac {\pi}{8}\left[ 3\ln \left( \dfrac {1+a}{b}\right) + \dfrac {7}{288} - \dfrac {a}{8}\right]
    For 20 marks
    a and b are square roots
    Which solomon paper is that from? Mind posting the answer?
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    (Original post by Zacken)
    I got full UMS and I barely did 6 or 7 of Edexcel's papers, let alone other exam boards.
    But you are already working at a higher level of maths. You have no need to do more than a few C3/4 papers, as it's fairly routine for you.

    However many students are only working at that C3/4 level, and haven't gone above that level. So for them doing more past papers and practice is necessary, I would say.


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    (Original post by kingaaran)
    But you are already working at a higher level of maths. You have no need to do more than a few C3/4 papers, as it's fairly routine for you.

    However many students are only working at that C3/4 level, and haven't gone above that level. So for them doing more past papers and practice is necessary, I would say.


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    I agree.


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    (Original post by kingaaran)
    But you are already working at a higher level of maths. You have no need to do more than a few C3/4 papers, as it's fairly routine for you.

    However many students are only working at that C3/4 level, and haven't gone above that level. So for them doing more past papers and practice is necessary, I would say.


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    You've just summed it up perfectly
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    (Original post by kingaaran)
    But you are already working at a higher level of maths. You have no need to do more than a few C3/4 papers, as it's fairly routine for you.

    However many students are only working at that C3/4 level, and haven't gone above that level. So for them doing more past papers and practice is necessary, I would say.

    I was talking about physics.
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    (Original post by Zacken)
    I was talking about physics.
    Any hard integration questions boss
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    (Original post by BBeyond)
    Any hard integration questions boss
    (Original post by Zacke)
    Use the substitution y = ux where u is a function of x to solve the DE:

    \displaystyle

\begin{equation*}\frac{\mathrm{d  }y}{\mathrm{d}x} = \frac{x}{y} + \frac{y}{x} \quad \quad (x>0, y>0)\end{equation*}

    that satisfies y=2 when x=1 in the form y = f(x) \quad (x > e^{-2}).
    It's a 'teaching' question - i.e: something to ease you into the problem and teach you a technique that I'll then test with harder ones after you're done with this.
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    (Original post by Zacken)
    It's a 'teaching' question - i.e: something to ease you into the problem and teach you a technique that I'll then test with harder ones after you're done with this.
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    (Original post by BBeyond)
    ...
    Ta! Well done. Speedy as well. (you can drop the modulo signs since x > e^{-2} > 0).

    Now use a substitution to find the solution of the differential equation

    \displaystyle 

\begin{equation*}\frac{\mathrm{d  }y}{\mathrm{d}x} = \frac{x}{y} + \frac{2y}{x} \quad \quad \left(x>0, y>0\right)\end{equation*}

    that satisfies y=2 when x=1.
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    (Original post by Zacken)
    Ta! Well done. Speedy as well. (you can drop the modulo signs since x > e^{-2} > 0).

    Now use a substitution to find the solution of the differential equation

    \displaystyle 

\begin{equation*}\frac{\mathrm{d  }y}{\mathrm{d}x} = \frac{x}{y} + \frac{2y}{x} \quad \quad \left(x>0, y>0\right)\end{equation*}

    that satisfies y=2 when x=1.
    Regrettably, I am horrible at spotting substitutions although I can generally monkey through the working when given them.
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    (Original post by Zacken)
    Ta! Well done. Speedy as well. (you can drop the modulo signs since x > e^{-2} > 0).

    Now use a substitution to find the solution of the differential equation

    \displaystyle 

\begin{equation*}\frac{\mathrm{d  }y}{\mathrm{d}x} = \frac{x}{y} + \frac{2y}{x} \quad \quad \left(x>0, y>0\right)\end{equation*}

    that satisfies y=2 when x=1.
    y=u^{2}x ?

    Not in a position to be able to try it I'm afraid. Though it would cancel 2u from both sides and leave you with 2ux(du/dx) = 1/u

    I think

    EDIT: Nevermind I'm waffling :laugh: it would be 2ux(du/dx) + u^2 , not + 2u
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    (Original post by Kvothe the arcane)
    Regrettably, I am horrible at spotting substitutions although I can generally monkey through the working when given them.
    The first question gave you a substitution and you're meant to understand why that substitution worked in that case and adapt it for it to work here.
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    (Original post by Zacken)
    The first question gave you a substitution and you're meant to understand why that substitution worked in that case and adapt it for it to work here.
    I know.
    But it would be nice to know what sub to use at sight.
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    (Original post by edothero)
    y=u^{2}x ?
    There are two substitutions that are both useful in their own way here, but yours isn't one of them. It's not usually the case that your substitution is imposing some extra function on the u (if that makes sense?). So if you find yourself doing that, you should make sure that that's really the sub you want to use.

    Edit: to clarify, a substitution of that form is more often y = uf(x) instead of y = xf(u).
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    (Original post by Kvothe the arcane)
    I know.
    But it would be nice to know what sub to use at sight.
    Practice. :-)
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    (Original post by edothero)
    y=u^{2}x ?

    Not in a position to be able to try it I'm afraid. Though it would cancel 2u from both sides and leave you with 2ux(du/dx) = 1/u

    I think
    Nah that won't work
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    (Original post by Zacken)
    There are two substitutions that are both useful in their own way here, but yours isn't one of them. It's not usually the case that your substitution is imposing some extra function on the u (if that makes sense?). So if you find yourself doing that, you should make sure that that's really the sub you want to use.

    Edit: to clarify, a substitution of that form is more often y = uf(x) instead of y = xf(u).
    It's weird how much more difficult a question gets when you add a 2 into it
    Only other thing I can think of is y=ux^2
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    (Original post by Zacken)
    Ta! Well done. Speedy as well. (you can drop the modulo signs since x > e^{-2} > 0).

    Now use a substitution to find the solution of the differential equation

    \displaystyle 

\begin{equation*}\frac{\mathrm{d  }y}{\mathrm{d}x} = \frac{x}{y} + \frac{2y}{x} \quad \quad \left(x>0, y>0\right)\end{equation*}

    that satisfies y=2 when x=1.
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Size:  112.1 KB

    Not so sure about this one

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    (Original post by Zacken)
    There are two substitutions that are both useful in their own way here, but yours isn't one of them. It's not usually the case that your substitution is imposing some extra function on the u (if that makes sense?). So if you find yourself doing that, you should make sure that that's really the sub you want to use.

    Edit: to clarify, a substitution of that form is more often y = uf(x) instead of y = xf(u).
    looks like y=(root2)ux or ux^2


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