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# Edexcel A2 C4 Mathematics June 2015 - Official Thread watch

1. (Original post by TeeEm)
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I'm resitting C4 and I need to be able to get 90 UMS on the paper, I've been doing the harder (second half) of the solomon papers. Any resources you would suggest I use to make sure I'm prepared for any hard questions? Wasn't sure if you had your challenging questions for C4 too.
2. (Original post by Genesis2703)
Haven't done C4 in awhile so I hope I'm not talking garbage here. I believe the limits are the other way around to what you'd intuitively believe because at t=0 you are at (5,0) and at t=2pi you are at (0,4). As you know from normal integration to find the area under a curve you need to have the upper limit being the large x-coordinate, otherwise you get a negative value

trivial e.g. if you want to find the area under y=1 between x=1 and 2 you have 2 as the upper limit and 1 as the lower limit, the other way around gives you -1 rather than +1 for the area. So similarly in your question t=0 gives you the larger x-coordinate rather than t=2pi, so t=0 is your upper limit.

Hope I made sense! (and hope that I am remembering C4 correctly also :P )
Thanks - that does make sense actually.

If I came across it in an exam I'd probably be ripping my hair out thinking why am I getting a negative value, ha!
3. (Original post by physicsmaths)
Limits of y with respect to dy x^2

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Would you have to rearrange the equation ? i.e make "x" the subject instead?
4. (Original post by Genesis2703)
Haven't done C4 in awhile so I hope I'm not talking garbage here. I believe the limits are the other way around to what you'd intuitively believe because at t=0 you are at (5,0) and at t=2pi you are at (0,4). As you know from normal integration to find the area under a curve you need to have the upper limit being the large x-coordinate, otherwise you get a negative value

trivial e.g. if you want to find the area under y=1 between x=1 and 2 you have 2 as the upper limit and 1 as the lower limit, the other way around gives you -1 rather than +1 for the area. So similarly in your question t=0 gives you the larger x-coordinate rather than t=2pi, so t=0 is your upper limit.

Hope I made sense! (and hope that I am remembering C4 correctly also :P )
Is this always the case, i.e. making the larger 'x' value the upper limit even if actuality it is the other way around?
5. (Original post by TheAnnabelle)
Is this always the case, i.e. making the larger 'x' value the upper limit even if actuality it is the other way around?
Yes the larger limit is always on top, unless you're integrating using substitution, then you put the largest x value on top, but say when you change the limits to u, the equivalent numbers stay in the original position of the x values, so a smaller u value could end up being on top
6. (Original post by TheAnnabelle)
Is this always the case, i.e. making the larger 'x' value the upper limit even if actuality it is the other way around?
I believe so, when finding an area under a curve in the region a<x<b you integrate with respect to x with lower limit a and upper limit b as you have done for a few units now.

When we have parametric equations, we still want to integrate over that x range, but we have to do it in terms of t, or theta (or whatever parameter is being used) as it makes the integral easier. But SIMILARLY (but not the same) to integration by substitution, you need to change the limits and put them with respect to t.

So if your larger x-value (b, from above) occurs at time t2, and the smaller one (a) occurs at time t1, then your integral is with upper limit t2 and lower limit t1, regardless of whether t2>t1 or not.

Hope I cleared things up!
7. (Original post by Genesis2703)
Haven't done C4 in awhile so I hope I'm not talking garbage here. I believe the limits are the other way around to what you'd intuitively believe because at t=0 you are at (5,0) and at t=2pi you are at (0,4). As you know from normal integration to find the area under a curve you need to have the upper limit being the large x-coordinate, otherwise you get a negative value

trivial e.g. if you want to find the area under y=1 between x=1 and 2 you have 2 as the upper limit and 1 as the lower limit, the other way around gives you -1 rather than +1 for the area. So similarly in your question t=0 gives you the larger x-coordinate rather than t=2pi, so t=0 is your upper limit.

Hope I made sense! (and hope that I am remembering C4 correctly also :P )
t=2π is also (5,0) ?
8. Does anyone have any tough C4 questions?
9. Want this A* so much, just think a*s always seem to be hit and miss, there are papers that suit me well, and papers i just hate, like last years c4 paper. Ive sat that paper 3 times and haven't hit higher then 63/75. Compared to other papers where I've been getting 100ums on.
10. (Original post by joanneeve)
take the minus out i would say then use the bracket so theres no confusion of signs
danke
11. Sir do the iygb papers increase in difficulty from a to z?

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12. (Original post by raypalmer)
Sir do the iygb papers increase in difficulty from a to z?

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On his website it says the difficulty ranges
13. (Original post by Gome44)
t=2π is also (5,0) ?
A typo, I meant pi/2 (and then you multiply by 4 to get the whole region!)

if you want to do it all in 1 integral like in the markscheme, I believe you need to do it from 0 to 2pi due to the direction you travel around the curve as t-varies.

It is the same principle as what I said though. Since integration gives a Net-Signed-Area, to get a positive integral value you need to:
(a) if the curve is ABOVE x-axis, you integrate such that the higher 'x'-value is the upper limit.
(b) if the curve is BELOW x-axis, you integrate s.t. the lower 'x'-value is the upper limit.

using (a) and (b) together shows that having 0 as an upper limit gives a positive value.

If you need a further explanation I can get a drawing or something
14. (Original post by raypalmer)
Sir do the iygb papers increase in difficulty from a to z?

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It explains it on the website but it's something like most of the papers are 'standard', U to Z are 'hard' and S and T are god level.
15. (Original post by Genesis2703)
A typo, I meant pi/2 (and then you multiply by 4 to get the whole region!)

if you want to do it all in 1 integral like in the markscheme, I believe you need to do it from 0 to 2pi due to the direction you travel around the curve as t-varies.

It is the same principle as what I said though. Since integration gives a Net-Signed-Area, to get a positive integral value you need to:
(a) if the curve is ABOVE x-axis, you integrate such that the higher 'x'-value is the upper limit.
(b) if the curve is BELOW x-axis, you integrate s.t. the lower 'x'-value is the upper limit.

using (a) and (b) together shows that having 0 as an upper limit gives a positive value.

If you need a further explanation I can get a drawing or something
Nvm, I went through my working and realised I'd thought cos(2x)=2sin^2(x)-1 :/

16. Hello, can someone help me? Either I'm very confused, or Solomon papers have got the mark scheme wrong. Paper A, Question 5. Trapezium rule. Equation is 4(x^0.5)(e^-x).
My value for x=2 is 0.766 (3sf), but on the mark scheme it says 1.083. Can someone please help me to understand this discrepancy? All my other values are correct.

Question and mark scheme attached.
Attached Images

17. (Original post by Tazmain)
Does anyone have any tough C4 questions?
Haven't seen this on a c4 paper and there is a very slim chance it will come up but if you want a challenge:

∫ cos(x)e^x dx

It's pretty easy once you spot the trick
18. (Original post by Gome44)
Haven't seen this on a c4 paper and there is a very slim chance it will come up but if you want a challenge:

∫ cos(x)e^x dx

It's pretty easy once you spot the trick
When I started reading I thought it'd be this kinda one, thanks for sharing
I don't think its in our spec, but you never know with edexcel..

Is it...?
Spoiler:
Show
1/2(cos(x)e^x + sin(x)e^x)
19. Can someone please help, on Solomon paper F question 7 do you NEED a minus sign infront of the constant for the equation, obviously if you don't use it you get a different answer , but I preferr not to use minus signs and consider it a decrease. Thanks a lot!
20. (Original post by NikolaT)
It explains it on the website but it's something like most of the papers are 'standard', U to Z are 'hard' and S and T are god level.
Thank you

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