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# C4 in January? Post here watch

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1. (Original post by V.P. Keys)
Correct.
If mine is correct, why does Mockel say we need 'Ax + B', not just A? Is this a special case? Or do we just need A?
2. (Original post by Simba)
If mine is correct, why does Mockel say we need 'Ax + B', not just A? Is this a special case? Or do we just need A?
(4x + 5)/[(1 + x²)(1 - x)(2 + 4x)]

can be written in the form (Ax+B)/(1 + x²) + C/(1 - x) + D/(2 + 4x)

or ... A/(1 + x²) + B/(1 - x) + C/(2 + 4x)

Mockel's is also correct because the term
(1 + x²) in the denominator can be written as (1 - x)(1 + x)

Mockel's way would give a 'nicer' result though.
3. (Original post by V.P. Keys)
(1 + x²) in the denominator can be written as (1 - x)(1 + x)
Wouldn't that give (1 - x²)?
4. (Original post by Simba)
Wouldn't that give (1 - x²)?
haha, yeah sos mistake!

Had it been an (1-x)(1+x) it would have worked for Mockel's way.

5. (Original post by V.P. Keys)
haha, yeah sos mistake!

Had it been an (1-x)(1+x) it would have worked for Mockel's way.

Ok hehe, good to know ^_^ ! Thanks for the confidence boost ! I got all worried the other day lol, someone asked an integration question and said it was for C3... in my C3 textbook there is no integration, so I started to worry hehe... but, it turns out that the integration they were talking about actually is in C4 ... Some of the integration in C4 looks a little nasty, I suppose it's just a matter of learning the basic ideas though, and it all falling into place from there ...
6. (Original post by V.P. Keys)
Incorrect.

--------------

Correct.
wonderful insight... but I already said mine was wrong, and simba's is wrong too (counterexample, let x = pi). plus (1+x^2) is not (1+x)(1-x)

I forgot that the quadratic factor needs (Ax+B) on the top. mockel was right.
7. (Original post by Simba)
(4x + 5)/[(1 + x²)(1 - x)(2 + 4x)] ≡ A/(1 + x²) + B/(1 - x) + C/(2 + 4x)

4x + 5 ≡ A(1 - x)(2 + 4x) + B(1 + x²)(2 + 4x) + C(1 + x²)(1 - x)

x = 1 gives:

9 = B(2)(6)

12B = 9

B = 3/4

x = -1/2 gives:

3 = C(5/4)(3/2)

15C/8 = 3

C = 8/5

x = 2 gives:

13 = A(-1)(10) + 0.75(5)(10) + 1.6(5)(-1)

13 = -10A + 37.5 - 8

10A = 16.5

A = 1.65 = 33/20

Therefore:

(4x + 5)/[(1 + x²)(1 - x)(2 + 4x)] = 33/[20(1 + x²)] + 3/[4(1 - x)] + 8/[5(2 + 4x)]

Hehe, I hope this is right ! Not done any questions like this before, so might be wrong sorry ...

Doing my C4 exam in June, sorry I can't join you in January !
sorry simba. I think it's wrong.
Try x = 0
5/2 ≠ 33/20 + 3/4 + 8/10
5/2 ≠16/5
8. To lazy to attempt it, but the answer your looking for is:

(7x+19)/20(x²+1) - 3/4(x-1) + 4/5(2x+1)
9. Hehe, ok, wasn't really expecting to be right to be honest ^_^;;; ... thanks for the answer, Malik !
10. C4 in January for me - w00t....might even be fun..
11. (Original post by Malik)
To lazy to attempt it, but the answer your looking for is:

(7x+19)/20(x²+1) - 3/4(x-1) + 4/5(2x+1)
(4x + 5)/[(1 + x²)(1 - x)(2 + 4x)] ≡ A/(1 + ix) + B/(1 - ix) + C/(2 + 4x)+D/(x-1)
using the cover up method
C=(4x + 5)/[(1 + x²)(1 - x)] when x=-1/2
=3/(5/4)(3/2)=24/15=8/5

D=(4x + 5)/[(1 + x²)(2 + 4x)] when x=1
=9/(2)(6)=3/4

A=(4x + 5)/[(1 -ix)(1 - x)(2 + 4x)] when x=i
=5+4i/(2)(1-i)(2+4i)

B=(4x + 5)/[(1 + ix)(1 - x)(2 + 4x)] when x=-i
=5-4i/(2)(1+i)(2-4i)

combine A/(1 + ix) + B/(1 - ix) to get

½(1+x²)[(5+4i)(1-ix)/2[3+i]+(5-4i)(1+ix)/2[3-i]
¼(1+x²)[(5+4i+4x-5ix)(3-i)+(5-4i-4x+5ix)(3+i)/10]
= 1/40(1+x²)[38+14x]
=1/40(1+x²)[19+7x]
12. (Original post by distortedgav)
C4 in January for me - w00t....might even be fun..
Ah good luck. You're with me!
Damn slow school. I would have done this last year.
Some people doing C1-C4 and FP1-FP2 as one a-level have only sat C1 and C2 so far and got U's!
Only 5 have sat C3 including me.
A bit pathetic really.
13. (Original post by Widowmaker)
Ah good luck. You're with me!
Damn slow school. I would have done this last year.
Some people doing C1-C4 and FP1-FP2 as one a-level have only sat C1 and C2 so far and got U's!
Only 5 have sat C3 including me.
A bit pathetic really.
Good luck to you too. We were supposed to do C4 last year, but there wasn't really enough time to realstcally do it all, there was a phase where our teacher missed a lot of lessons, so yeah it didn't really happen. She didn't quite finish C2 with the single maths students, which they weren't too happy about. Ah well, it happens lol. Not that it matters though cos the college I'm at now weren't doing C4 until this year anyhoo lol. Good luck with C4 anyway
14. I'm doing C4 in January as well. Last year I did A2 Maths privately so I could join the Further Maths class at school. I've already done C3 and I'll be finishing A2 Maths in January (C4 and S1). Then I have all my Further Maths units in June
15. I'm retaking C3 and C4 in Jan.
16. (Original post by ArVi)
Then I have all my Further Maths units in June
Bloody hell! 6 maths units in june, fun fun fun

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