Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    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
  1. File Type: pdf P5 Jan 05.pdf (39.2 KB, 96 views)
    Offline

    0
    ReputationRep:
    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
    • Thread Starter
    Offline

    0
    ReputationRep:
    (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.
    Offline

    0
    ReputationRep:
    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
    • Thread Starter
    Offline

    0
    ReputationRep:
    (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.
    Offline

    0
    ReputationRep:
    S'ok, i had that paper as a mock, did quite well in it 54/60 It's quite an easy paper isnt it
    Offline

    0
    ReputationRep:
    (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)?
    Offline

    0
    ReputationRep:
    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
    Offline

    0
    ReputationRep:
    thank you!! much appreciated
    Offline

    0
    ReputationRep:
    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?
    Offline

    0
    ReputationRep:
    Well any questions that people have trouble with i can post the answers which the maths department of my college did.
    Offline

    0
    ReputationRep:
    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!!!!
    Offline

    0
    ReputationRep:
    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.
    Offline

    0
    ReputationRep:
    answers so far:

    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......
    Offline

    0
    ReputationRep:
    arghhhhh q 2 not working
    geeky poo pls post the ans to q 2.
    Offline

    0
    ReputationRep:
    and 4 (ii) and 4 (iii)
    Offline

    8
    ReputationRep:
    (Original post by chubby)
    arghhhhh q 2 not working
    geeky poo pls post the ans to q 2.
    here's a solution.
    Attached Images
      
    Offline

    8
    ReputationRep:
    Solutions for Q4.
    Attached Images
       
    Offline

    0
    ReputationRep:
    thanks a lot fermat!!!! i learnt something new today.
    Offline

    0
    ReputationRep:
    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?
 
 
 
Turn on thread page Beta
Updated: June 27, 2005
The home of Results and Clearing

1,723

people online now

1,567,000

students helped last year

University open days

  1. SAE Institute
    Animation, Audio, Film, Games, Music, Business, Web Further education
    Thu, 16 Aug '18
  2. Bournemouth University
    Clearing Open Day Undergraduate
    Fri, 17 Aug '18
  3. University of Bolton
    Undergraduate Open Day Undergraduate
    Fri, 17 Aug '18
Poll
Do you want your parents to be with you when you collect your A-level results?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.