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# MEI P5 - January 2005 watch

1. Just wondering if anyone has the answers for this paper, or alternatively if anyone could work some of them out? Question 3(b) parts (ii)-(iv) would be useful in particular. Paper attached. Thanks.
Attached Images
2. P5 Jan 05.pdf (39.2 KB, 96 views)
3. bi) zn + 1/zn = 2cos nθ zn - 1/zn = 2jsin nθ

bii)
[(z-1/z)(z+1/z)]³ = [4jsinθcosθ]³ = -64jsin³θcos³θ (1)

[(z-1/z)(z+1/z)]³ = [z²-1/z²]³ = z6 -3z³ + 3/z³ -1/z6
= z6 -1/z6 - 3(z³-1/z³)
=2jsin6θ - 6jsin2θ (2)

(1) = (2)
-64jsin³θcos³θ = 2jsin6θ - 6jsin2θ
sin³θcos³θ = 3/32 sin2θ - 1/32 sin6θ
4. (Original post by geekypoo)
bi) zn + 1/zn = 2cos nθ zn - 1/zn = 2jsin nθ

bii)
[(z-1/z)(z+1/z)]³ = [4jsinθcosθ]³ = -64jsin³θcos³θ (1)

[(z-1/z)(z+1/z)]³ = [z²-1/z²]³ = z6 -3z³ + 3/z³ -1/z6
= z6 -1/z6 - 3(z³-1/z³)
=2jsin6θ - 6jsin2θ (2)

(1) = (2)
-64jsin³θcos³θ = 2jsin6θ - 6jsin2θ
sin³θcos³θ = 3/32 sin2θ - 1/32 sin6θ
Thanks.
5. b iii)

sin³θcos³θ = 3/32 (2θ - 4θ³/3 + 4θ5/15-...) - 1/32(6θ-36θ³+324θ5/5 -..)
= 3θ/16θ - 3θ/16 - θ³/8 + 9θ³/8 + θ5/40 - 81θ5 / 40
= θ³ - 2θ5

b iv)
θ³cos2θ = θ³[1-2θ² + 2θ4/3 + ...] = θ³ - 2θ5
which approx = sin³θcos³θ
6. (Original post by geekypoo)
b iii)

sin³θcos³θ = 3/32 (2θ - 4θ³/3 + 4θ5/15-...) - 1/32(6θ-36θ³+324θ5/5 -..)
= 3θ/16θ - 3θ/16 - θ³/8 + 9θ³/8 + θ5/40 - 81θ5 / 40
= θ³ - 2θ5

b iv)
θ³cos2θ = θ³[1-2θ² + 2θ4/3 + ...] = θ³ - 2θ5
which approx = sin³θcos³θ
Yay! I got the same. Thanks very much for your help.
7. S'ok, i had that paper as a mock, did quite well in it 54/60 It's quite an easy paper isnt it
8. (Original post by geekypoo)
S'ok, i had that paper as a mock, did quite well in it 54/60 It's quite an easy paper isnt it
Do you know how to do Q4b)i)?
9. No i dont sorry, as i havent done conics for p5,

I've just found out the answers i have though so i could give u answers they got;

4b

2x/a² - 2y/b².(dy/dx) = 0
dy/dx= b²x/a²y
At point (xi, yi)

dy/dx = b²xi/a²yi
Eqn of tangent is

y-yi=b²xi/a²yi(x-xi)
=> a²yiy - a²yi² = b²xix-b²xi²

b²xix - a²yiy = b²xi² - a²yi² (*)

i) As (xi,yi) lies on the hyperbola:
xi²/a² - yi²/b² = 1
b²xi² - a²yi² = a²b² (**)

put (**) into (*)

b²xix - a²yiy = b²xi² - a²yi²
b²xix - a²yiy = a²b²
xix/a² - yiy/b² = 1
10. thank you!! much appreciated
11. i really hate complex numbers!!!!!! so aint going to do any q of complex nos in this june exam. but i will also try this paper. shame ther is no mark schme innit?
12. Well any questions that people have trouble with i can post the answers which the maths department of my college did.
13. Well any questions that people have trouble with i can post the answers which the maths department of my college did.
thanks. will defo try the paper tommorrow. aint liking quartics through. but u know what? i got a strong feelin there wont be any quartics question this june. the quartics seem to be asked only in the jan papers (notice carefully from jan 2001-jan 2005), and quardratics and cubics are asked in june. i hope its true cos i cant stand quartic roots!!!!
14. But of course quartics/remainder theorm will be asked, then there will be a question on inverse trig functions, a question on hyperbolic functions and a question on conics.

geekypoo pls check if they are right

q 1-
(a) y^4+8y³+12y²+6y+1 = 0
(b) k = 3 and m = -10
(ii) 3x+6
(iii) yeah toodles
(iv) -30x-52

doing q 2 now......
16. arghhhhh q 2 not working
geeky poo pls post the ans to q 2.
17. and 4 (ii) and 4 (iii)
18. (Original post by chubby)
arghhhhh q 2 not working
geeky poo pls post the ans to q 2.
here's a solution.
Attached Images

19. Solutions for Q4.
Attached Images

20. thanks a lot fermat!!!! i learnt something new today.
21. Also fermat in 4(i) the curve is between 0 and pi as mentioned in the question so i am guessing all the four loops lie above the xaxis (two between 0 and pi/2 and 2 between pi/2 and pi). the second and fourth loop starting from the 0 angled line are dotted lines. am i right?

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