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Edexcel A2 C4 Mathematics June 2015 - Official Thread

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Original post by Bustamove
After C3, chances of me getting A* is slim to zero... so I'll probably be aiming as high an A as I can get overall... Hopefully university will give me some leeway for having a high ums score overall for maths rather than an A*...


As long as i get a C im god dam happy!!!
Original post by DomRauba
Someone answer this, more interested in the whether you lose marks part, cheers.


you only put the modulus in if there is a chance that the thing you are taking the ln of is negative. in some cases, the modules is therefore not needed, so we omit - i doubt you will drop marks for including it when it is not nesscesary - failing to add it in when you should do will inevitably lose you marks however, as you will fail to take the mod of a value later in the question when you should have. although in this case, it would become apparent you have forgotten it as your calculator will not like you asking it to ln a -ve number. hope that makes sense.
Original post by Bustamove
Ah.. when you find a improper fraction for partial fractions... do you divide by it first? or can you just express the letter "A" on it's own first and work it out from there?
Hope this makes sense LOL


Could you make up an example?
Original post by frozo123
What does everyone need to get an A approximately?
or are you all looking at A*s?


Hoping for an A*
I think core 3 went really well, so just hoping for a decent paper tomorrow
Does anyone know how to integrate 1/(sin2x)^2 ?
Original post by frozo123
What does everyone need to get an A approximately?
or are you all looking at A*s?


Looking at an A for C4 to get an A overall. Highly doubt it lol.
Original post by Poughkeepsie
Does anyone know how to integrate 1/(sin2x)^2 ?


Cosec^2(2x)


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I have tonight to quickly re-learn the whole of C4
fun....
Original post by physicsmaths


how was your STEP?
Original post by physicsmaths

that integrates to -1/2 cot2x right?
(edited 8 years ago)
Original post by Gilo98
I think that is what i shall do - i have a history exam tomorrow AM so after dinner in 5 min I'll casually do a 'difficult' paper, brush over some history notes then get to bed early. I seem to go nocturnal on study leave it's so annoying


ah I hate having exams on consecutive days.... even worse when you have 2 hard exams on the same day....

I made a mistake before my C3 exam, I had economics unit 3 and C3 on consecutive days and I just focused too much on econ rather than C3 (I'm applying for economics and it would be awks to get a bad grade in it)...
At the time, I was really good at C3 but I couldnt do a economics unit 3 paper at the time and I had 2 days before the exam... so I panicked and just focused on that...

I focused too much on econ 3 and got a little out of touch with C3 in the end... I ended up making soo many silly mistakes... May have blown my chances for an A*, I probably just got borderline A which is not good enough.
I hope this doesn't happen to you tomorrow...it may be even harder for you since you have 2 exams in the same day... I hear history is pretty hard O.o
What is the hardest c4 paper that you guys have come across?
How would you implicitly differentiate -2xy a bit confused because of the minus, thanks!
To confirm: C4 tomorrow is definitely in the afternoon?
Original post by TheRaspberry
How would you implicitly differentiate -2xy a bit confused because of the minus, thanks!


Use the product rule.

Let u = -2x and v= y and then differentiate implicitly :smile:


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Original post by TheRaspberry
How would you implicitly differentiate -2xy a bit confused because of the minus, thanks!


Use the product rule with u=-2x and v=y
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mb_12
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Original post by MrBowcat
Yes please, would greatly appreciate that!


Hang on, you can use the rules below for odd and even powers (I just checked in A2 Maths textbook)
The general rules are:
For powers of cosx: integral of (sinx)*(cosnx) dx = (-1/(n+1))*cosn+1x + c
For powers of sinx: integral of (cosx)*(sinnx) dx = (1/(n+1))*sinn+1x + c


So the answer to your question is:
integral of (sinx)*(cos3x) dx = (-1/4)*cos4x + c


This should make the question waaaaaayyy easier.
(I tried to attach a pic but it didn't work)
(edited 8 years ago)
Original post by SeanFM
Could you make up an example?


ummm....https://5a302d29ff432793ce79475dd3dba997851918fc.googledrive.com/host/0B1ZiqBksUHNYNElsMEE2MDU0eWM/13%20Gold%203%20-%20C4%20Edexcel.pdf

question 3...

I can't paste it on here sorry...

I watched exams solutions and his method is to divide 9x^2 +20x -10 by (x+2)(3x-1) and then what ever value he got as a remainder, he used that for partial fractions...

I'm just wondering, instead of doing that, could I do:
step 1) 9x^2 +20x -10/(x+2)(3x-1) = A + B/(x+2) + C/(3x-1) ?
and then
step 2) 9x^2 +20x -10 = A(x+2)(3x-1) + B(3x-1) + C(X+2) ?

Sorry, it looks really messy, I hope this makes sense... I don't know if this method works, but I don't really like the dividing method and I'm just wondering if this works... if it wasn't a improper fraction, usually the letter "A" wouldn't be on it's own like the first step that I've put down.
Original post by randlemcmurphy
Use the product rule with u=-2x and v=y


Thanks, always get a bit confused whether the y should be minus or not, basic, I know! :smile:
Original post by TheRaspberry
How would you implicitly differentiate -2xy a bit confused because of the minus, thanks!


you would use the product rule with u=-2x and v=y, so du/dx = -2, dv/dy=1(dy/dx) so you get -2y + (-2x dy/dx) so if you rearragne, dy/dx = 2y/-2x if that makes sense?

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