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HSC Revision Thread 2006 (1 Viewer)

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Riviet

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A thread for 4u topic revision and questions from past HSCs, past trials etc. Feel to post them up or help others.
 
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pLuvia

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I'm sure some of you have done or have the 1992 HSC paper could someone confirm with me the question 3b (i). I think it's a typo or my copy is just wrong
 

onebytwo

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do you have the compressed version....from the sites most generously provided by buchanan? because there the ones i got and they seem to have a couple errors - or at i think they do!
 

onebytwo

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1992 3b(i)
ADE is a line, not a triangle....LOL
you would think they mean ACE
 
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pLuvia

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onebytwo said:
do you have the compressed version....from the sites most generously provided by buchanan? because there the ones i got and they seem to have a couple errors - or at i think they do!
I got my copy from Buchanan which had this typo
 
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pLuvia said:
I'm sure some of you have done or have the 1992 HSC paper could someone confirm with me the question 3b (i). I think it's a typo or my copy is just wrong
ADE should be ADC
 
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pLuvia

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1992 HSC

I had a bit of trouble on
2d(ii) is there an easier way of finding the answer then just subbing in the given roots?
3b(iii)
Q6 altogether :(
7a(iii)
8a(iii)
 
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Riviet

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1992 Q3 b) (iii)

AD.DE=BD.DC (product of intersecting chords in a circle are equal) [1]

But DE=AE-AD

Substitute this into [1]:

AD(AE-AD)=BD.DC

AD2=AB.AC-BD.DC

.'. AD2=AD.AE-BD.DC (using result in ii)
 
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housemouse

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Got a question
In the 1998 HSC Question 3 (a)(i)

How exactly do you get the parabolic asymptote? Please show working please :)
 
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pLuvia

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housemouse said:
Got a question
In the 1998 HSC Question 3 (a)(i)

How exactly do you get the parabolic asymptote? Please show working please :)
What parabolic asymptote? You are talking about the sketch f(x) one right? The asymptotes are just y=x and x=0
 
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housemouse

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Well this is what i thought, theres asymptote at y=x and y =0 however the asnwers from the MANSW say that it does not have asymptote at y =0 instead at y=-4/x
 

speedie

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q2 d (ii)

here, you go (x-1+i)(x-1-i) is a root
= (x^2-2x+1)

thus using complimenting, P(z)= (x^2-2x+1)(2x+1)
(x=z), sorry
 

Riviet

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housemouse said:
Well this is what i thought, theres asymptote at y=x and y =0 however the asnwers from the MANSW say that it does not have asymptote at y =0 instead at y=-4/x
That dotted y=-4/x is not an asymptote :p it's there to illustrate the addition of y=x and y=-4/x and what it look like when all three are graphed on the same number plane.
 
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pLuvia

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housemouse said:
Well this is what i thought, theres asymptote at y=x and y =0 however the asnwers from the MANSW say that it does not have asymptote at y =0 instead at y=-4/x
y=0 is the x axis, which isn't an aysmptote. And the y=-4/x dotted line, isn't an asymptote it's just an outline of the y=-4/x graph
 

Trebla

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Do you have solutions (particularly Q5-8) for 1995 HSC?
 

Riviet

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Unfortunately not, as the MANSW book only has 96-05 solutions. Post up any specific problems you have.
 
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