Timing in the MX2 exam (1 Viewer)

[sikeveo]

Imo, 80 would not give a band 6(if the rank correlation was correct). I doubt we can have an increase of 15 band 6's in maths regardless of what ppl say about our grade.

U sure danza? Cos even Vishan got around 80's.

 

[haboozin]

All these talk about the SBHS test...

i wanna do it :'(

 

[ishq]

Thanks people! Much appreciated!

Your talkin about the Sydney High 4u trial right? I personally thought Kourty went really soft, all of Q7 andQ 8, cept for hte projectile was free marks.

I liked that projectile question. Took me a while, but I got it.

Thats the trouble - if you have time to think about it, you can probably do it. If not, then take a reasonably good stab at it. But, in exam conditions, last part of last question with next to no time left, I would just go 'ahh fuck it' and start doing other things. Like wondering how my parents would take my failure.

 

[Antwan23q]

...I would just go 'ahh fuck it' and start doing other things. Like wondering how my parents would take my failure.

lmao. stay positive people!

 

[KFunk]

because from the u minus the first from the second.

so you get
arg (z+2) - arg(z-2) = pi/2

I got that far too. But how do you find the complex number z from there?

If you do what antwan2bu suggested you'll get the equation of the circle x² + y² = 4 (or half of it at least). Since you have two equations containing relations of z you actually get a specific value (though this value does all on the circle). If you were just given arg (z+2) - arg(z-2) = pi/2 then arg(z+2) and arg(z-2) can take on a number of different values as long as they satisfy that equation.

You can solves this using sumultaneous equations. If arg(z+2) = π/6 then z lies on a ray passing through the point (-2, 0) with a gradient of tan(π/6) giving the line:

y = (1/√3)(x + 2)

If arg(z-2) = 2π/3 then z lies on a ray passing through (2, 0) with a gradient of -tan(π/3) giving the line:

y = -√3(x - 2)

Solving these simultaneously you find that x + iy = 1 + √3i. You can see that this sits on the circle x² + y² = 4.


EDIT: Also, an alternative method would be to forget about the conceptual stuff and use the fact that arg(z+2) = tan^-1[y/(x+2)]=π/6 and arg(z-2) = tan^-1[y/(x-2)]=2π/3. Take tan of both sides and you get the same equations without the need to explain anything.

 

[Antwan23q]

wow, that like cuts down all the working!

 

[~ ReNcH ~]

Can someone post up this SBHS paper? Now I'm getting curious...

 

[haboozin]

Can someone post up this SBHS paper? Now I'm getting curious...

yea so am i

so

<b> can someone post the SBHS paper</b>

 

[Vampire]

It's already up isnt it? tywebb put it up...

 

[sikeveo]

If someone doesn't do it soon, ill put it up tomorrow or the day after.

 

[haboozin]

http://www.boredofstudies.org/cours..._Mathematics_Extension_2_Trial_Paper_SBHS.pdf

 

[haboozin]

I might be wrong in this but using what they give you:

&int; ln(tanx) dx = &int; ln(tan{&pi;/2 -x}) dx (both integrals between 0 and &pi;/2)

&int; ln(tanx) dx = &int; ln(1/tanx) dx ... (since tan(&pi;/2 - x) = 1/tanx)

&int; ln(tanx) dx = - &int; ln(tanx) dx ... ( 'cause ln(1/tanx) = -ln(tanx)

so 2 &int; ln(tanx) dx = 0

&there4; &int; ln(tanx) dx = 0

how about this??

ln(tan{&pi;/2 -x} = ln((tan&pi;/2 - tanx)/1-tan&pi;/2tanx))

lim as @ -->&pi;/2
tan@ --> infinity

so ln(infinity/infinity)

= ln 1
= 0

I = 0 dx
=0 ????

is this ok?

 

[~ ReNcH ~]

http://www.boredofstudies.org/cours..._Mathematics_Extension_2_Trial_Paper_SBHS.pdf

Thanks haboozin

 

[~ ReNcH ~]

With Q1(e) of the SBHS paper, even though it says "evaluate", are you justified in sketching 2-|x| and finding the sum of the areas between the curve and the x-axis between -3 and 3? In general, I'm just wondering whether you're allowed to use any available method to answer a question or whether "evaluate" implies an alegbraic method must be used.

 

[KFunk]

how about this??

ln(tan{&pi;/2 -x} = ln((tan&pi;/2 - tanx)/1-tan&pi;/2tanx))

lim as @ -->&pi;/2
tan@ --> infinity

so ln(infinity/infinity)

= ln 1
= 0

I = 0 dx
=0 ????

is this ok?

I'd be careful with that because if you have F(x)/G(x) where F(x) --> &infin; and G(x) --> &infin; when x-->k then that doesn't necesarily mean that F(k)/G(k) = 1.

e.g. as x-->0 , [1/x²]/[1/(2x²+x)] &ne; 1

You could bring something outside of the syllubus like l'hopitals rule into it or perhaps rearrange it so that it works but I'd hesitate before claiming that infinity/infinity was equal to one.

 

[haboozin]

I'd be careful with that because if you have F(x)/G(x) where F(x) --> &infin; and G(x) --> &infin; when x-->k then that doesn't necesarily mean that F(k)/G(k) = 1.

e.g. as x-->0 , [1/x²]/[1/(2x²+x)] &ne; 1

You could bring something outside of the syllubus like l'hopitals rule into it or perhaps rearrange it so that it works but I'd hesitate before claiming that infinity/infinity was equal to one.

i was thinking of eg. when finding horisontal assymptode kinda thing ?

 

[haboozin]

With Q1(e) of the SBHS paper, even though it says "evaluate", are you justified in sketching 2-|x| and finding the sum of the areas between the curve and the x-axis between -3 and 3? In general, I'm just wondering whether you're allowed to use any available method to answer a question or whether "evaluate" implies an alegbraic method must be used.

anything?

yea sketching will do

thats how i did it, easy as that way someone else did it by integrating in this thread.

 

[KFunk]

ln[tan(&pi;/2 -x)] = ln(cotx)

as x--> &pi;/2 , cotx ---> 0 &there4; ln[tan(&pi;/2 -x)] ---> -&infin;

I think you need to use the integral properties for this one because the question hinges around the symmetry (I think) the graphs have around x=&pi;/4. You can't really say "ln[tan(&pi;/2 -x)] = 0 hence &int; ln(tanx) dx = 0" if you know what I mean.

 

[Stefano]

L'Hospitale's rocks.

 

[jake2.0]

L'Hospitale's rocks.

what is his rule?