The Student Room Group

Edexcel A2 C4 Mathematics June 2015 - Official Thread

Scroll to see replies

anyone know what the hardest c4 paper so far has been or where I can find it? really need to practice hard questions if C4 is anything like the C3 paper!
Hey guys, really confused right now. I was taught that you must only multiply or divide when separating variables in differential equations. But on this IYGB paper on one question you can add ?



Posted from TSR Mobile
ImageUploadedByStudent Room1434202501.438728.jpg143.7KB
Original post by Medicjug
Hey guys, really confused right now. I was taught that you must only multiply or divide when separating variables in differential equations. But on this IYGB paper on one question you can add ?

Posted from TSR Mobile


Yeah that's fine.
You just want to get it all of the x's are on one side, and the y's are on the other.
Reply 983
Avatar for Rushad
Rushad
When it comes to marking papers, how strict are the examiners? If you have your own method for working out an answer which is correct, and you've shown sufficient working to get there, will they more or less give you the method marks even if they are not on the mark scheme? Or do they need to see some particular method or line in that method before they award you those marks?
Original post by XxKingSniprxX
Time to do 3x C3 past papers today, 3x tomorrow and 2x on Monday.
Hopefully that should be enough practice. (prays Edexcel doesn't screw us over)


Do you not have any other exams till then?
Original post by Moniii16081997
How do you do part aImageUploadedByStudent Room1434195501.704682.jpg


Posted from TSR Mobile


Volume of a cone = (pi.r^2.h)/ 3

If you draw out the shaded triangle bit, the height of the triangle is h. If the biggest angle is 90, and angles in a triangle add to 180 then each of the other angles is 90/2 = 45. As the angles are the same it's an isosceles triangle. If you then split this triangle into two smaller triangles, you will see that h = x (the radius) because it's also an isosceles triangle as the position of the right angle has changed. Then just substitute back into the original volume of a cone formula.

Hope that helps x
Original post by Lolaa123
anyone know what the hardest c4 paper so far has been or where I can find it? really need to practice hard questions if C4 is anything like the C3 paper!


http://www.physicsandmathstutor.com/a-level-maths-papers/c4-edexcel-bronze-silver-gold/ gold papers :]
Original post by dominicwild
It becomes

22cos⁡2y\dfrac{2}{2\cos^2y}

Which you then make:

1cos⁡2y=sec⁡2y\dfrac{1}{\cos^2y} = \sec^2y


Where did you get this from? Is this even mathematically correct?


Posted from TSR Mobile
Original post by kingaaran
Where did you get this from? Is this even mathematically correct?


Posted from TSR Mobile


I can confirm the following is correct:

∫2cos(2y)+1 dy=∫1cos2y dy=∫sec2y dy=tany+c\displaystyle \int \frac{2}{cos(2y)+1} \ dy=\int \frac{1}{cos^2y} \ dy= \int sec^2y \ dy =tany+c
Original post by rayquaza17
I can confirm the following is correct:

∫2cos(2y)+1 dy=∫1cos2y dy=∫sec2y dy=tany+c\displaystyle \int \frac{2}{cos(2y)+1} \ dy=\int \frac{1}{cos^2y} \ dy= \int sec^2y \ dy =tany+c


This must be FP3, right?


Posted from TSR Mobile
I was just wondering if someone could help. I think so far this year I have got :
C1- 67/75
C2- 68/75
C3- 40/75 .
S1 - 58/75
D1 - 65/75
What does this currently put me on with UMS marks roughly this year and what would I need in order to get a B overall with just C4 left ?
Original post by kingaaran
This must be FP3, right?


Posted from TSR Mobile


No that's pretty standard c4 :/
Original post by kingaaran
This must be FP3, right?


Posted from TSR Mobile


C3 identity
Original post by kingaaran
This must be FP3, right?


Posted from TSR Mobile


No this can be answered entirely with C4 knowledge.

You can go from the first integral to the second integral using the cos(2y) identity.

You can go from the second integral to the third integral by knowing the definition of sec(y)

You can go from the third integral to the answer by looking in your formula book and looking at where I've circled
Screen Shot 2015-06-13 at 15.31.12.png
Original post by rayquaza17
No this can be answered entirely with C4 knowledge.

You can go from the first integral to the second integral using the cos(2y) identity.

You can go from the second integral to the third integral by knowing the definition of sec(y)

You can go from the third integral to the answer by looking in your formula book and looking at where I've circled
Screen Shot 2015-06-13 at 15.31.12.png


I understand that - I'm not saying in terms of the integral - but how does this link to the integral of cos^2(y) dy, for example?

The post I quoted seems to have said that, cos^2(y) can be written as cos(y)^2, which magically leads to 2/(cos(2y)+1)...?


Posted from TSR Mobile
Original post by Chardy12
I was just wondering if someone could help. I think so far this year I have got :
C1- 67/75
C2- 68/75
C3- 40/75 .
S1 - 58/75
D1 - 65/75
What does this currently put me on with UMS marks roughly this year and what would I need in order to get a B overall with just C4 left ?


I might roughly guess like 90 + 90 + 50 + 80 + 90 for 400 (but from what I've heard of C1, C2 and D1 those could certainly be better and C3 too)
B is 420, A is probably actually still doable by the way (480) though of course that depends on your C4 preparation
Original post by kingaaran
I understand that - I'm not saying in terms of the integral - but how does this link to the integral of cos^2(y) dy, for example?

The post I quoted seems to have said that, cos^2(y) can be written as cos(y)^2, which magically leads to 2/(cos(2y)+1)...?


Posted from TSR Mobile


I'm confused...
Are you asking why cos2(y)=(cosy)2cos^2(y)=(cosy)^2?

It's hard to see from cos^2(y) goes to 2/(cos(2y)+1), but it's easier to see the other way around.
Original post by rayquaza17
I'm confused...
Are you asking why cos2(y)=(cosy)2cos^2(y)=(cosy)^2?

It's hard to see from cos^2(y) goes to 2/(cos(2y)+1), but it's easier to see the other way around.


No, I'm not asking the first.

2cos(2y)+1=2cos2(y)−sin2(y)+1 \frac{2}{cos(2y)+1} = \frac{2}{cos^2(y)-sin^2(y)+1}

1cos2(y) \frac{1}{cos^2(y)}

But I can't see how that equals cos2(y)cos^2(y), like the person I quoted originally said.

I'm not confused with the identities or how to integrate, I'm sure I'm competent with that, but I'm not sure how cos2(y)=2cos(2y)+1cos^2(y) = \frac{2}{cos(2y)+1}?


Posted from TSR Mobile
(edited 8 years ago)
Thank you so much . I thought I had blown my chances at gettin the B because C3 went awful .
Original post by kingaaran
No, I'm not asking the first.

2cos(2y)+1=2cos2(y)−sin2(y)+1 \frac{2}{cos(2y)+1} = \frac{2}{cos^2(y)-sin^2(y)+1}

1cos2(y) \frac{1}{cos^2(y)}

But I can't see how that equals cos2(y)cos^2(y), like the person I quoted.

I'm not confused with the identities or how to integrate, I'm sure I'm competent with that, but I'm not sure how cos2(y)=2cos(2y)+1cos^2(y) = \frac{2}{cos(2y)+1}?


Posted from TSR Mobile


They didn't say that? They said sec2(y)=2cos(2y)+1\displaystyle sec^2(y) = \frac{2}{cos(2y)+1}

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you