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Edexcel A2 C4 Mathematics June 2016 - Official Thread

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Original post by Jackpalmer1996
Don't worry if you haven't and others have, there's still plenty of time until the exam! Loads of cramming to do :smile:
Best of luck!!!



Thank you 😄 you just made me feel better
Reply 861
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bob.123
Good luck guys
Reply 862
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Danllo
2
Any new C3/4 crash maths papers yet ?
Reply 863
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imran_
2
hey I need help with JAN IAL C34 question 4 im stuck on it
Reply 864
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SSD07
3
Is there a thread for C3???
Reply 865
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SSD07
3
and how is everyone revising for maths? just practice? is that the only thing i can do?
Original post by imran_
hey I need help with JAN IAL C34 question 4 im stuck on it


yeah?
Reply 867
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imran_
2
Original post by Zacken
yeah?


sorry, idk how to type it on here but heres the link

https://a086a5a2f39bda93734c56a63fab1d7be0a9ba38.googledrive.com/host/0B1ZiqBksUHNYQXE5T2xiNDBRd2s/January%202015%20(IAL)%20QP%20-%20C34%20Edexcel.pdf

its question 4
Original post by SSD07
Is there a thread for C3???


http://www.thestudentroom.co.uk/showthread.php?t=3847081&p=63691635#post63691635
Original post by SSD07
Is there a thread for C3???


http://www.thestudentroom.co.uk/showthread.php?t=3847081
Original post by imran_


What are you stuck on? the limits becomes 0 0 \, and π3\frac{\pi}{3} your dxdu=cos⁡θ\frac{\mathrm{d}x}{\mathrm{d}u} = \cos \theta so your integral is transformed into:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\int_0^{\pi/3} \frac{\cos \theta}{2\sqrt{1-\sin^2 \theta}} \, \mathrm{d}\theta\end{equation*}



Now just use the fact that 1−sin⁡2θ=cos⁡2θ=⋯\sqrt{1-\sin^2 \theta} = \sqrt{\cos^2 \theta} = \cdots
Reply 871
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imran_
2
Original post by Zacken
What are you stuck on? the limits becomes 0 0 \, and π3\frac{\pi}{3} your dxdu=cos⁡θ\frac{\mathrm{d}x}{\mathrm{d}u} = \cos \theta so your integral is transformed into:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\int_0^{\pi/3} \frac{\cos \theta}{2\sqrt{1-\sin^2 \theta}} \, \mathrm{d}\theta\end{equation*}



Now just use the fact that 1−sin⁡2θ=cos⁡2θ=⋯\sqrt{1-\sin^2 \theta} = \sqrt{\cos^2 \theta} = \cdots


what happens to the power 3/2
Original post by imran_
what happens to the power 3/2


Aye, sorry I mis-read that as 12\frac{1}{2} because I'm blind. We have that (1−sin⁡2θ)3/2=1−sin⁡2θ(1−sin⁡2θ)(1-\sin^2 \theta)^{3/2} = \sqrt{1-\sin^2 \theta}(1- \sin^2 \theta) by indices laws, so your integrand is:

cos⁡θ21−sin⁡2θ(1−sin⁡2θ)=cos⁡θ2cos⁡θcos⁡2θ\displaystyle \frac{\cos \theta}{2\sqrt{1-\sin^2 \theta}(1-\sin^2 \theta)} = \frac{\cos \theta}{2\cos \theta \cos^2 \theta}
Reply 873
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SSD07
3
Original post by Zacken


thanks! what about s2?
Original post by SSD07
and how is everyone revising for maths? just practice? is that the only thing i can do?


Pretty much all I'm doing, just loads and loads of past papers and exam questions, the solomon papers are great for the tougher questions if you're aiming for A/A* !

http://www.physicsandmathstutor.com/a-level-maths-papers/c4-solomon/

Good luck!
Is anyone else finding the Solomon papers for c3 easier than the actual edexcel pprs


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Severely stuck on method for C4 June 2014 (R) Q7 integration, any help?
Original post by Jacob2215
Severely stuck on method for C4 June 2014 (R) Q7 integration, any help?


Link the paper? Cba to go find it rn
Original post by Jacob2215
Severely stuck on method for C4 June 2014 (R) Q7 integration, any help?


Have you seperated the variables?
Put the functions of t on one side(right side preferably) and N on Left side.


Posted from TSR Mobile
Original post by physicsmaths
Have you seperated the variables?
Put the functions of t on one side(right side preferably) and N on Left side.


Posted from TSR Mobile


Yeah I think i see what I'm doing now, cheers cheers :smile:

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