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# C4 in January? Post here watch

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1. I thought we could find out which tsrians are doing c4 and then create a revision thread soon.
2. i did it last june.
3. (Original post by Undry1)
i did it last june.
Thanks for that.
4. I am doing it in June which is kind of a waste since I've finished sylabuss and am at my peak now ,6months down line probably would have forgotton everything. Dont mind revising with you though name a topic

5. --------------

(Original post by Malik)
I am doing it in June which is kind of a waste since I've finished sylabuss and am at my peak now ,6months down line probably would have forgotton everything. Dont mind revising with you though name a topic
vectors.
6. (Original post by Undry1)
vectors.
A vector is a quantity that has both magnitude and direction
7. full marks!!
8. I saw this in a P3 paper, or something similar.

Split into partial fractions.

4x+5/(1+x²)(1-x)(2+4x)

Never seen one like this before. Is it just the same method?
9. 4x+5/(1+x²)(1-x)(2+4x) = A/(1-x) + B/(2+4x) + C/(1+x²)
4x + 5 = 2A(1+2x)(1+x²) + B(1-x)(1+x²) + 2C(1-x)(1+2x)

now let x=1,

9 = 2A(3)(2)
A = 9/12 = 3/4

let x=-1/2

3 = B(3/2)(5/4)
B = 8/5

let x=0

5 = 2A(1)(1) + B(1)(1) + 2C(1)(1)
5 = 2A + B + 2C
C = (5 - 2A - B)/2
C = 19/20

So 4x+5/(1+x²)(1-x)(2+4x) = 3/ 4(1-x) + 4/ 5(1+2x) + 19/ 20(1+x²)

I hope..
10. (Original post by john!!)
4x+5/(1+x²)(1-x)(2+4x) = A/(1-x) + B/(2+4x) + C/(1+x²)
4x + 5 = 2A(1+2x)(1+x²) + B(1-x)(1+x²) + 2C(1-x)(1+2x)

now let x=1,

9 = 2A(3)(2)
A = 9/12 = 3/4

let x=-1/2

3 = B(3/2)(5/4)
B = 8/5

let x=0

5 = 2A(1)(1) + B(1)(1) + 2C(1)(1)
5 = 2A + B + 2C
C = (5 - 2A - B)/2
C = 19/20

So 4x+5/(1+x²)(1-x)(2+4x) = 3/ 4(1-x) + 4/ 5(1+2x) + 19/ 20(1+x²)

I hope..
Why 2A and 2C?
11. (Original post by john!!)
4x+5/(1+x²)(1-x)(2+4x) = A/(1-x) + B/(2+4x) + C/(1+x²)
4x + 5 = 2A(1+2x)(1+x²) + B(1-x)(1+x²) + 2C(1-x)(1+2x)

now let x=1,

9 = 2A(3)(2)
A = 9/12 = 3/4

let x=-1/2

3 = B(3/2)(5/4)
B = 8/5

let x=0

5 = 2A(1)(1) + B(1)(1) + 2C(1)(1)
5 = 2A + B + 2C
C = (5 - 2A - B)/2
C = 19/20

So 4x+5/(1+x²)(1-x)(2+4x) = 3/ 4(1-x) + 4/ 5(1+2x) + 19/ 20(1+x²)

I hope..
Ah okay. my teacher said there is a completely different method for these ones.
12. maybe there is, because what I just wrote down is completely wrong.
13. (Original post by Widowmaker)
I saw this in a P3 paper, or something similar.

Split into partial fractions.

4x+5/(1+x²)(1-x)(2+4x)

Never seen one like this before. Is it just the same method?
(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 = i gives:

4i + 5 = A(1 - i)(2 + 4i)

4i + 5 = A(6 + 2i)

A = (4i + 5)/(2i + 6)

Therefore:

(4x + 5)/[(1 + x²)(1 - x)(2 + 4x)] = (4i + 5)/[(2i + 6)(1 + x²)] + 3/[4(1 - x)] + 8/[10(1 + 2x)]

I'm not quite sure how the imaginary bits got in there ! Not done any of these questions before, thought I'd give it a stab for the hell of it !
14. (Original post by Widowmaker)
I saw this in a P3 paper, or something similar.

Split into partial fractions.

4x+5/(1+x²)(1-x)(2+4x)

Never seen one like this before. Is it just the same method?
(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 !
15. Isn't the general form Ax + B for the numerator of a quadratic factor?

i.e, (4x + 5)/[(1 + x²)(1 - x)(2 + 4x)] ≡ (Ax+B)/(1 + x²) + C/(1 - x) + D/(2 + 4x)

...and then proceed as usual.
16. (Original post by mockel)
Isn't the general form Ax + B for the numerator of a quadratic factor?

i.e, (4x + 5)/[(1 + x²)(1 - x)(2 + 4x)] ≡ (Ax+B)/(1 + x²) + C/(1 - x) + D/(2 + 4x)

...and then proceed as usual.
I thought as much. I think you are right
17. (Original post by mockel)
Isn't the general form Ax + B for the numerator of a quadratic factor?

i.e, (4x + 5)/[(1 + x²)(1 - x)(2 + 4x)] ≡ (Ax+B)/(1 + x²) + C/(1 - x) + D/(2 + 4x)

...and then proceed as usual.
It could well be ... I'll leave it for someone else to have fun with ...
18. Partial fractions are NOT my friend. Whenever I'm doing any kind of integration question, I try to avoid partial fractions at all cost. They're just uncool.
So I'll leave it for someone else to do as well...
19. im taking c1 to c4 in jan
20. (Original post by john!!)
So 4x+5/(1+x²)(1-x)(2+4x) = 3/ 4(1-x) + 4/ 5(1+2x) + 19/ 20(1+x²)

I hope..
Incorrect.

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

(Original post by Simba)
(4x + 5)/[(1 + x²)(1 - x)(2 + 4x)] = 33/[20(1 + x²)] + 3/[4(1 - x)] + 8/[5(2 + 4x)]
Correct.

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