Edexcel A2 C4 Mathematics June 2015 - Official Thread

Original post by imedico10

is it compulsory to divide it by 2

No at all

Reply 1682

math42

20

Original post by imedico10

is it compulsory to divide it by 2

nah it's just more aesthetic really; you could technically put any multiple of that vector there and it'd be correct

edit: well not zero lol

(edited 8 years ago)

Reply 1683

Dingdongdidly

Original post by tawaz1997

Can someone please help me with Q 4b) June 2013 Replacement, I don't understand how they got 0.1 for x. Here's the paper and mark scheme...

https://a086a5a2f39bda93734c56a63fab1d7be0a9ba38.googledrive.com/host/0B1ZiqBksUHNYQXE5T2xiNDBRd2s/June%202013%20(R)%20QP%20-%20C4%20Edexcel.pdf

https://a086a5a2f39bda93734c56a63fab1d7be0a9ba38.googledrive.com/host/0B1ZiqBksUHNYQXE5T2xiNDBRd2s/June%202013%20(R)%20MS%20-%20C4%20Edexcel.pdf

Thanks in advance!

In the question it says x<8/9 so you need to adjust the cuberoot(7100) by taking out cuberoot(1000) which gives 10cuberoot(7.1). I think there is an error in the mark scheme here as it is printed as 71 not 7.1. Then you set 8-9x = 7.1 which gives x=0.1 then sub into your expansion to get the answer :smile:

Reply 1684

anyone know how to do this question? Solomon L, question 3a and 3b!
mark scheme is confusing

intergration.PNG3.5KB

Reply 1685

johnh545

7

Original post by Anonymous696061

(dy/dt) / (dx/dt)

(So differentiate x & y separately, with respect to t; then divide the derivative of y over the derivative of x)

Hope this helps!

Thats differentiation.
To integrate a parametric you integrate y*(dx/dt).dt
Kind of like integrating with a substitution.

Reply 1686

Kevin De Bruyne

21

Original post by ember8

anyone know how to do this question? Solomon L, question 3a and 3b!
mark scheme is confusing

3b - factor formulae may help.

3a - Find du/dx, substitute that in and the u stuff in.

Reply 1687

anonwinner

19

Original post by Gilo98

Yep. just stick with it and be methodical - then it really isn't that bad

I managed to get it all right except for part b which was only 1 mark lol

Reply 1688

FirdousH

Jan 15 IAL Paper link please thanks

Reply 1689

Original post by SeanFM

3b - factor formulae may help.

3a - Find du/dx, substitute that in and the u stuff in.

yeah done that, but i got -1/4ln2u as oppose to their -1/2lnu for 3a but no idea where i went wrong

Reply 1690

tawaz1997

7

Original post by Dingdongdidly

In the question it says x<8/9 so you need to adjust the cuberoot(7100) by taking out cuberoot(1000) which gives 10cuberoot(7.1). I think there is an error in the mark scheme here as it is printed as 71 not 7.1. Then you set 8-9x = 7.1 which gives x=0.1 then sub into your expansion to get the answer :smile:

So it's that simple, haha.Thank you!

Reply 1691

Maham88

Original post by ember8

anyone know how to do this question? Solomon L, question 3a and 3b!
mark scheme is confusing

for part a when you get ∫ x/u -1/2x du both of the x cancel out leaving you with
∫ 1/u (-1/2) du then take the -1/2 out and then integrate

Reply 1692

Egg1

Done all the main papers, whats best next- gold standard or the reserve, as apparently theyre quite hard.

Reply 1693

Kevin De Bruyne

21

Original post by ember8

yeah done that, but i got -1/4ln2u as oppose to their -1/2lnu for 3a but no idea where i went wrong

What did you get as dx? Did you substitute u in after substituting du in? what did you integrate overall?

Reply 1694

Anon-

6

Original post by physicsmaths

Yes a very wordy problem i thinkz

Posted from TSR Mobile

What do you mean by a wordy question?

Reply 1695

Anonymous696061

Original post by johnh545

Thats differentiation.
To integrate a parametric you integrate y*(dx/dt).dt
Kind of like integrating with a substitution.

Oh crap sorry I misread that! -Thanks for correcting!!

Reply 1696

Dingdongdidly

Original post by ember8

anyone know how to do this question? Solomon L, question 3a and 3b!
mark scheme is confusing

Hopefully you can read my writing! I integrated by parts twice then replaced the last part with 9 times the original integral and then took it over to the other side. :smile:

Reply 1697

Original post by SeanFM

What did you get as dx? Did you substitute u in after substituting du in? what did you integrate overall?

just got it! when i multiplied it by my dx/du which was -1/2(2-u)^1/2 i accidently multiplied it and then took the two out, and then wrote the 2 in again for the denominator in later working! silly mistake hahah

thanks for your help!

Reply 1698