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# P6 June 2003 watch

1. could sum1 explain how 2 do Q 4 (ii), 5 (b) and 7 please i've looked at the mark scheme but it don't make ney senes 2 me!! i'm soooo confused the june 2002 ppr was way easier than this ppr!!
2. This is for Edexcel , yes ?
3. yeah....sorry ^_^ i 4got 2 say
4. 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
================
5. 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

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!!
7. thanx fermat! ^_^ cud u show me how 2 do Q 4 the 2nd part please
8. 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|
===========
9. (Original post by Fermat)

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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