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March 29, 2024, 09:54:22 pm

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164631 times)  Share 

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TrueTears

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Re: Specialist 3/4 Question Thread!
« Reply #15 on: November 27, 2011, 10:26:07 pm »
+3
ill send u the ebook? pm me ur email brah :D
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Hutchoo

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Re: Specialist 3/4 Question Thread!
« Reply #16 on: November 27, 2011, 10:32:46 pm »
0
I'm doing specialist next year, and I'm looking at these questions having no idea what's going on hahaha. How do you guys know this stuff already, was it in the year 11 course?? feelsbadman.jpeg
heh, no need to worry, just do some self study and this stuff will be easy ^^
Good times will come, young padawan.

monkeywantsabanana

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Re: Specialist 3/4 Question Thread!
« Reply #17 on: December 05, 2011, 07:08:09 pm »
+1
Illustrate this on an Argand diagram:



Can someone explain why there's only HALF a hyperbola?

I've done this so far:



let



















Asymptotes:

Answers only show right half of the graph.

Bachelor of Commerce (Economics & Finance)

TrueTears

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Re: Specialist 3/4 Question Thread!
« Reply #18 on: December 05, 2011, 07:26:08 pm »
+2
i'll give you a hint, it lies in this step





now think about the restrictions on x ;)
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HERculina

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Re: Specialist 3/4 Question Thread!
« Reply #19 on: December 05, 2011, 07:38:38 pm »
0
ill send u the ebook? pm me ur email brah :D
do you have ebook of heinemann spesh book? ;)
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TrueTears

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Re: Specialist 3/4 Question Thread!
« Reply #20 on: December 05, 2011, 07:47:01 pm »
0
ill send u the ebook? pm me ur email brah :D
do you have ebook of heinemann spesh book? ;)
nope :(
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Bhootnike

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Re: Specialist 3/4 Question Thread!
« Reply #21 on: December 05, 2011, 07:56:35 pm »
0
Find x such that:

4 tan x = 6 tan 2x, 0 ≤ x ≤ 2π

----------------------------------------

I am stuck on this!

4 tan x= 6tan 2x
        =6( 2tan x/1-tan^2 x)
        = 12 tan x/1-tan^2 x
4 tan x(1-tan^2 x) = 12 tan x
1-tan^2 x = 3
3+tan^2 x = 1
root3 +tan x = 1

ehhh
i probs made some huge and obvious and rather silly mistake:p

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TrueTears

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Re: Specialist 3/4 Question Thread!
« Reply #22 on: December 05, 2011, 08:02:03 pm »
+1
3+tan^2 x = 1
root3 +tan x = 1

lol u cant do that
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Bhootnike

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Re: Specialist 3/4 Question Thread!
« Reply #23 on: December 05, 2011, 08:04:51 pm »
0
yeahh thats right haha,
hence 'im stuck' :p

I know i shouldnt have even gone there..., so do you think it was something to do with not converting tanx to sin/cos ?
or just some arithmetic?
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b^3

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Re: Specialist 3/4 Question Thread!
« Reply #24 on: December 05, 2011, 08:05:32 pm »
0
Find x such that:

4 tan x = 6 tan 2x, 0 ≤ x ≤ 2π

----------------------------------------

I am stuck on this!

4 tan x= 6tan 2x
        =6( 2tan x/1-tan^2 x)
        = 12 tan x/1-tan^2 x
4 tan x(1-tan^2 x) = 12 tan x
1-tan^2 x = 3
3+tan^2 x = 1
root3 +tan x = 1

ehhh
i probs made some huge and obvious and rather silly mistake:p


Use the tan double angle formula then see where you can get to.
tan(2x)=(2tan(x))/(1-tan^2(x))


EDIT: wait didnt READ!
« Last Edit: December 05, 2011, 08:13:36 pm by b^3 »
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Bhootnike

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Re: Specialist 3/4 Question Thread!
« Reply #25 on: December 05, 2011, 08:07:39 pm »
0
Find x such that:

4 tan x = 6 tan 2x, 0 ≤ x ≤ 2π

----------------------------------------

I am stuck on this!

4 tan x= 6tan 2x
        =6( 2tan x/1-tan^2 x)
        = 12 tan x/1-tan^2 x
4 tan x(1-tan^2 x) = 12 tan x
1-tan^2 x = 3
3+tan^2 x = 1
root3 +tan x = 1


Use the tan double angle formula then see where you can get to.
tan(2x)=(2tan(x))/(1-tan^2(x))

:\ i suck haha
2011: Biol - 42
2012: Spesh |Methods |Chemistry |English Language| Physics
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khuda ne jab tujhe banaya hoga, ek suroor uske dil mein aaya hoga, socha hoga kya doonga tohfe mein tujhe.... tab ja ke usne mujhe banaya hoga

b^3

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Re: Specialist 3/4 Question Thread!
« Reply #26 on: December 05, 2011, 08:13:11 pm »
+1
move everytihng to the one side then used null factor law.

4tan(x)-4tan3(x)-12tan(x)=0
take tan(x) as a factor
tan(x)(-8-4tan2(x))=0
tan(x)=0 or -8-4tan2(x)=0
x=0 or tan2(x)=2
x=0, or   or no solution

So x=0, ,

EDIT: forgot sols
« Last Edit: December 05, 2011, 08:16:57 pm by b^3 »
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pi

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Re: Specialist 3/4 Question Thread!
« Reply #27 on: December 05, 2011, 08:14:35 pm »
+1
x also equals pi and 2pi from b^3's working

b^3

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Re: Specialist 3/4 Question Thread!
« Reply #28 on: December 05, 2011, 08:16:16 pm »
+2
x also equals pi and 2pi from b^3's working
Yeh I'm a bit rusty, I clicked post, then realised and my com froze.

Moderator action: removed real name, sorry for the inconvenience
« Last Edit: January 02, 2017, 09:48:32 pm by pi »
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TrueTears

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Re: Specialist 3/4 Question Thread!
« Reply #29 on: December 05, 2011, 08:18:27 pm »
+2
yeah other mistake was

4 tan x(1-tan^2 x) = 12 tan x

cancelling the tan killed off solns
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