Edexcel A2 C4 Mathematics June 2015 - Official Thread

Scroll to see replies

We've just finished the trapezium rule in integration today, not much left to go but we haven't started D1. However we can decide to do M1, what would people advise?

Posted from TSR Mobile

I did D1 last year as an AS Further Maths module, and I'm doing M1 this year as an A2 Maths module.
I really enjoyed doing D1 and got 100 UMS in the exam. It involves a lot of algorithms so if you're good at following methods and like working out values with diagrams rather than lots of formulae then you will probably enjoy it.
I found M1 quite difficult at first because it was unlike normal maths, but I am finding it easier now I have got the hang of using the relevant formulae and drawing the diagrams correctly.
I would highly recommend examsolutions.com for useful videos that explain the difficult topics very clearly.
I would advise that you look through the syllabus for M1 and D1 if you haven't yet made your decision and see which best suits your strengths and interests.

Reply 81

Hudl

I need help with this question, I dont get why when integrating the term [ (3+2x)^7] / 7

it is ( 3 + 2x)^8 / ( 7 x 16)

Where is the 16 from? Shouldn't it be / (7 x 8 )

http://gyazo.com/42b1c7f62c1e5d6b747c20c7917d4353

Also I am stuck on question 3e on the c4 heinmann book. This is my working up to now but it's like the book has gone from DV/dx to v = sec^2x where I have gone to tan^2x. Picture of working attached below

(edited 9 years ago)

Reply 82

Hudl

ImageUploadedByStudent Room1424659830.222636.jpg

Posted from TSR Mobile

EDIT: Ive figured out what I did wrong here, thanks for the help

(edited 9 years ago)

Reply 83

Random1357

Original post by Hudl

To integrate, add one to power, then divide by new power and divide by derivative of thing that was to power (derivative was 2)

Reply 84

Hudl

Original post by Random1357

To integrate, add one to power, then divide by new power and divide by derivative of thing that was to power (derivative was 2)

Not for the 2sec^2xtanx integral it is a site integral and will be done by trying y = tan^2x therefore dy/dx is 2sec^2xtanx and this 2sec^2xtan x integrates to tan^2x

Posted from TSR Mobile

Reply 85

Random1357

Original post by Hudl

Not for the 2sec^2xtanx integral it is a site integral and will be done by trying y = tan^2x therefore dy/dx is 2sec^2xtanx and this 2sec^2xtan x integrates to tan^2x

Posted from TSR Mobile

Or more formally by substitution: I didn't bother looking at the attached hence just explained the former.

Reply 86

Hudl

Original post by Random1357

Or more formally by substitution: I didn't bother looking at the attached hence just explained the former.

It's a by parts question

Posted from TSR Mobile

Reply 87

Random1357

Original post by Hudl

It's a by parts question

Posted from TSR Mobile

No: that's a clumsy method if it even works: inspection for a general Q and substitution for being formal (e.g. Proof)

Oh sorry, just actually read it :tongue:

Yeah use both

(edited 9 years ago)

Reply 88

Random1357

Original post by Hudl

Posted from TSR Mobile

You have 2 problems: 1) your working is very hard to follow: I used to be bad, trust me, learning to be neat is essential.

2) the integral of tan^2(x) is integral of (sec^2-1) is tan(x)-x+C

Reply 89

Mary562

guys if you had this bracket : -2( 2x - 1 )^-1

Can I change it to (-1 + 2x) then expand etc or not?

(binomial question)

Reply 90

Random1357

Original post by Mary562

guys if you had this bracket : -2( 2x - 1 )^-1

Can I change it to (-1 + 2x) then expand etc or not?

(binomial question)

If you mean can I write as -2[(-1) + (2x)]^(-1) then that is a true statement but useless as can't be expanded!

We could do: 2[(1) + (-2x)]^(-1), valid for IxI<1/2 by pulling out a factor of (-1)^(-1)=-1

Reply 91

Mary562

Original post by Random1357

Yep that is what I'm saying - thanks the bottom one is correct

:smile:

Reply 92

Hudl

Can you integrate Ln^2 x using a chain rule method or site integral method

chain rule method

(Ln^3 x)/3 * x

Site integral method

Try y = Ln^3 x
dy/dx = 3Ln^2 x * 1/x

Therefore legit y and thus answer of integral is x/3Ln^3 x

Ive seen the by parts answer but why doesnt the chain rule / site integral method work?

Reply 93

tiny hobbit

Original post by Hudl

If the bit you gain when you differentiate isn't a constant, you can't "undo" it when you want to integrate.

Reply 94

Hudl

Original post by tiny hobbit

If the bit you gain when you differentiate isn't a constant, you can't "undo" it when you want to integrate.

Yeah thanks a lot, whats wrong with the chain rule method...why doesnt that work because you're allowed to differentiate the term in the brackets and even if it gives you a non linear term like x^2 you have to do 1/x^2 as you're integrating.

Reply 95

tiny hobbit

Original post by Hudl

If you differentiate your answer, using the product or quotient rule, you won't get the thing you are trying to integrate.

Reply 96

john122334455

Hey im just starting doing maths paper, can someone give me a estimate on how many papers i should do to fully master the syllabus, and how many times i should re do the papers. Thank you guys

Reply 97

Gilo98

Original post by john122334455

Hey im just starting doing maths paper, can someone give me a estimate on how many papers i should do to fully master the syllabus, and how many times i should re do the papers. Thank you guys

I'm also doing A2 at the moment but am using same process as last year which worked out well (got full 300 UMS). 1 past paper every other day starting in February right up until the exams, do them daily as of May 1st - this includes Solomon Papers and Edexcel ones, swapping between the two in blocks of 2 weeks. Seems a bit excessive when you consider this will total to about 90 past papers but I self teach so have daily free periods in which to do them. Personally, I would never re-do past papers - I don't see the point, surely you'll subconciously remember answer/method? For me at least there is more than enough papers available to not re-do any but what works for you, works for you!! :h: