Official BOS Trial 2014 Thread (2 Viewers)

Carrotsticks · Sep 25, 2014

Here is one of the integration problems I was considering putting in the Extension 2 paper.

$\int \frac{\cos ^2 x}{1+ \tan x}dx$

Carrotsticks · Sep 25, 2014

Here's another problem I was considering adding into the paper. Bonus points if you can find a method to solve this in less than 3 lines.

$\\ $Let $ z $ be a point on the unit circle other than $ z=-1. \\\\ $ Show that the origin, $ \frac{1+z}{1+\bar{z}} $ and $ z $ are collinear.$$

Immortality · Sep 25, 2014

Carrotsticks said:
$\int \frac{\cos ^2 x}{1+ \tan x}dx$

Carrotsticks said:
$\\ $Let $ z $ be a point on the unit circle other than $ z=-1. \\\\ $ Show that the origin, $ \frac{1+z}{1+\bar{z}} $ and $ z $ are collinear.$$

Oooh these questions are pretty slick (Y)

clockworkorange · Sep 25, 2014

Please put me in for both Ext 1 and Ext 2

trapizi · Sep 26, 2014

Carrotsticks said:
Here is one of the integration problems I was considering putting in the Extension 2 paper.

$\int \frac{\cos ^2 x}{1+ \tan x}dx$

Is it too hard or too easy that makes you remove this question from the paper ?

SittingForMathsExams · Sep 26, 2014

I'll sit for the tests

lilytao2014 · Sep 26, 2014

Please put me down for Ext 1 and Ext 2 exams thanks!

megusta · Sep 26, 2014

Carrotsticks said:
Here's another problem I was considering adding into the paper. Bonus points if you can find a method to solve this in less than 3 lines.

$\\ $Let $ z $ be a point on the unit circle other than $ z=-1. \\\\ $ Show that the origin, $ \frac{1+z}{1+\bar{z}} $ and $ z $ are collinear.$$

$\\ $arg$ \left (\frac{1+z}{1+\bar{z}} \right ) = $ arg$(1+z) $ $- $ arg$(\overline{1+z}) = 2$ arg$(1+z) \\\\ $From sketch, $ 2$ arg$(z+1)= $ arg$(z) $ (isosceles triangle, exterior angle of triangle equals sum of opposite interior angles)$ \\\\ $Thus, $ $arg$\left(\frac{1+z}{1+\bar{z}}\right) = $ arg$(z) $ so the origin, $ \frac{1+z}{1+\bar{z}} $ and $ z $ must be collinear.$

3 lines?

dunjaaa · Sep 26, 2014

damn, beat me to posting

, as for the integration question:

Screen shot 2014-09-26 at 1.47.57 AM.png

(I haven't thought of a more cleaner way besides t-formula)

Swaan · Sep 26, 2014

I'm interested in ext 1, is there still space?
And am I still welcome to sit it even though I'm looking at a 43-44 atm?

RealiseNothing · Sep 26, 2014

Carrotsticks said:
Here's another problem I was considering adding into the paper. Bonus points if you can find a method to solve this in less than 3 lines.

$\\ $Let $ z $ be a point on the unit circle other than $ z=-1. \\\\ $ Show that the origin, $ \frac{1+z}{1+\bar{z}} $ and $ z $ are collinear.$$

$|z| + |\frac{1+z}{1+\bar{z}}| = 1+1 = 2$

(should be trivial that the modulus of the fraction is 1 since the bottom is the conjugate of the top)

$|z+\frac{1+z}{1+\bar{z}}| = |\frac{1+z\bar{z}+2z}{1+\bar{z}}| = |2\frac{1+z}{1+\bar{z}}| = 2$ (as $z\bar{z} = |z|^2 = 1$ )

Hence:

$|z| + |\frac{1+z}{1+\bar{z}}| = |z+\frac{1+z}{1+\bar{z}}|$

So the three points are collinear due to the condition that equality holds for the triangle inequality IFF the triangle is degenerate, i.e. the vertices are all collinear.

RealiseNothing · Sep 26, 2014

Carrotsticks said:
Here's another problem I was considering adding into the paper. Bonus points if you can find a method to solve this in less than 3 lines.

$\\ $Let $ z $ be a point on the unit circle other than $ z=-1. \\\\ $ Show that the origin, $ \frac{1+z}{1+\bar{z}} $ and $ z $ are collinear.$$

$\frac{1+z}{1+\bar{z}} \times \frac{z}{z} = \frac{z(1+z)}{z+z\bar{z}} = \frac{z(1+z)}{1+z} = z$

They are in fact the same point, etc.

nomadz · Sep 26, 2014

I would like to attend Ext2, thanks!

FallenDepths · Sep 26, 2014

I'd be interested in both Ext 1 and 2, thanks

megusta · Sep 26, 2014

RealiseNothing said:
$\frac{1+z}{1+\bar{z}} \times \frac{z}{z} = \frac{z(1+z)}{z+z\bar{z}} = \frac{z(1+z)}{1+z} = z$

They are in fact the same point, etc.

damn gg
hats off

mojodojo · Sep 26, 2014

Awww yeee Defs gonna want to try out the Ext 1 and Ext 2 Paperr

aDimitri · Sep 26, 2014

dunjaaa said:
damn, beat me to posting , as for the integration question: View attachment 30951 (I haven't thought of a more cleaner way besides t-formula)

This was my method too except i stopped halfway through the partial fractions because i was disappointed i couldn't find a cleaner way to do it

galligalli · Sep 26, 2014

I would like to attend the 3u and 4u exams

Nukeboy · Sep 26, 2014

Can I please confirm a +1 for the 2unit tests?

Carrotsticks · Sep 26, 2014

RealiseNothing said:
$\frac{1+z}{1+\bar{z}} \times \frac{z}{z} = \frac{z(1+z)}{z+z\bar{z}} = \frac{z(1+z)}{1+z} = z$

They are in fact the same point, etc.

Exactly what I was looking for, well done.

dunjaaa said:
damn, beat me to posting , as for the integration question: View attachment 30951 (I haven't thought of a more cleaner way besides t-formula)

There is a cleaner way.

Hint: Auxiliary Angle Formula.

Official BOS Trial 2014 Thread (2 Viewers)

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