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    could sum1 explain how 2 do Q 4 (ii), 5 (b) and 7 please :p: i've looked at the mark scheme but it don't make ney senes 2 me!! :confused: i'm soooo confused the june 2002 ppr was way easier than this ppr!!
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    This is for Edexcel , yes ?
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    yeah....sorry ^_^ i 4got 2 say
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    Q5 b)
    let x = cosθ, => |x| <= 1
    then,
    cos5θ = -1
    5θ = pi, 3pi, 5pi, 7pi, 9pi
    θ = pi/5, 3pi/5, pi, 7pi/5, 9pi/5
    cosθ = 0,809, -0.309, -1, -0.309, 0.809
    x = 0.809, -0.309, -1
    ================
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    Question 7 is troubling me...I dont understand how they have found out that the normal to the plane is as they say :/

    thanks for any help
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    yeh, someone else posted about this q.

    It looks like a mistake in the paper.

    The direction vector of r (-1,2,3) isn't parallel to the normal of pi_1, taken from the eqn of the plane. x - 5y - 3z = -6 => n = (1,-5,-3) so it doesn't look like r can be perpindicular to pi_1!!
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    thanx fermat! ^_^ cud u show me how 2 do Q 4 the 2nd part please
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    Here's the first part.

    4) ii) a)

    w = (z-1)/z
    |w| = |z-1|/|z|
    |z-1| = |w||z|

    but |z-1| = 1

    giving,

    1 = |w||z|
    |w| = 1/|z|
    =========

    from w = (z-1)/z

    w - 1 = (z-1)/z - 1
    w - 1 = (z - 1 - z)/z
    w - 1 = -1/z
    |w - 1| = 1/|z|
    ============

    substituting for 1/|z|, in the earlier eqn,

    |w| = |w - 1|
    ===========
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    (Original post by Fermat)
    yeh, someone else posted about this q.

    It looks like a mistake in the paper.

    The direction vector of r (-1,2,3) isn't parallel to the normal of pi_1, taken from the eqn of the plane. x - 5y - 3z = -6 => n = (1,-5,-3) so it doesn't look like r can be perpindicular to pi_1!!
    aces

    been tryin to do question 7 for ages and couldnt see how it worked, then i thought the marking scheme might be wrong. but due to my own obvious idiocy i decided it was me

    so that means i got 61/75
 
 
 
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