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Edexcel A2 C4 Mathematics June 2016 - Official Thread

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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
Original post by ULTRALIGHT BEAM
Only for student aiming for 75/75

Get rich or die tryin

[br]y=f(x)[br]y=(x21)32[br]R=22ydx[br][br]y= f(x)[br]y = \left( x^{2} - 1\right) ^ {-\dfrac{3}{2}}[br]R = \int^{2}_{\sqrt {2}}ydx[br]
R rotated through 2π\pi about the X axis
Show the solid formed is
[br]π8[3ln(1+ab)+7288a8][br]\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?
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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(edited 7 years ago)
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 :tongue:
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. :tongue:
Original post by Zacken
I was talking about physics. :tongue:


Any hard integration questions boss
Original post by BBeyond
Any hard integration questions boss


Zacke
Use the substitution y=uxy = ux where uu is a function of xx to solve the DE:

Unparseable latex formula:

\displaystyle[br]\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=2y=2 when x=1x=1 in the form y=f(x)(x>e2)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.
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.


ImageUploadedByStudent Room1460372696.868991.jpg


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Original post by BBeyond
...


Ta! Well done. Speedy as well. (you can drop the modulo signs since x>e2>0x > e^{-2} > 0).

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

Unparseable latex formula:

\displaystyle [br]\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=2y=2 when x=1x=1.
Original post by Zacken
Ta! Well done. Speedy as well. (you can drop the modulo signs since x>e2>0x > e^{-2} > 0).

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

Unparseable latex formula:

\displaystyle [br]\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=2y=2 when x=1x=1.


Regrettably, I am horrible at spotting substitutions although I can generally monkey through the working when given them.
Original post by Zacken
Ta! Well done. Speedy as well. (you can drop the modulo signs since x>e2>0x > e^{-2} > 0).

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

Unparseable latex formula:

\displaystyle [br]\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=2y=2 when x=1x=1.


y=u2xy=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
(edited 7 years ago)
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.
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. :smile:
But it would be nice to know what sub to use at sight.
Original post by edothero
y=u2xy=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 uu (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)y = uf(x) instead of y=xf(u)y = xf(u).
(edited 7 years ago)
Original post by Kvothe the arcane
I know. :smile:
But it would be nice to know what sub to use at sight.


Practice. :-)
Original post by edothero
y=u2xy=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
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 uu (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)y = uf(x) instead of y=xf(u)y = xf(u).


It's weird how much more difficult a question gets when you add a 2 into it :colondollar:
Only other thing I can think of is y=ux^2
(edited 7 years ago)
Original post by Zacken
Ta! Well done. Speedy as well. (you can drop the modulo signs since x>e2>0x > e^{-2} > 0).

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

Unparseable latex formula:

\displaystyle [br]\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=2y=2 when x=1x=1.


ImageUploadedByStudent Room1460374282.993128.jpg

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 uu (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)y = uf(x) instead of y=xf(u)y = xf(u).


looks like y=(root2)ux or ux^2


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