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2006 HSC Mathematics Solutions (1 Viewer)

T.Goodhew

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Here are my worked solutions to the Mathematics (2Unit) HSC 2006. I apologise to those who downloaded the earlier versions, there were some minor calculation errors that have been fixed in the newer version of the solutions available here.

Fixes:
  • The proof of Q6 a) iii. is now included as is Q7 which were accidentally left out.
  • Small trig evaluation errors made in evaluating sin of 2pi/3 were corrected.
  • The final two questions Q10 b) iv. and v. will be added shortly.
Once again I sincerely apologise if you were worried because you downloaded a earlier solution file. You must excuse me I was working the solutions late between 11-12pm at night and small errors seem to be the consequence of that.

Mathematics (2Unit) HSC Solutions - 2006 (v1.25)
(Question 10 b) iv. v. will be included in v1.3)
 
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2006hsc

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just a few mistakes but..but u did well. But you missed a whole question! 7 i think..
 

T.Goodhew

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Oh :eek: , I did too... i have written the solutions to Q7... they will be included in this file and available to download by 12:30pm Today.

if you could forward the mistakes you found to my email that would be apprecitated!! :D
 
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mlinger

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Question 4a ii and iii are wrong.. use cosine rule instead... answer for bd should have been the square root of 27
 

Bijs

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Isn't the sine rule question in q1 wrong too?
 

Keskimo

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gah how could you be bothered doing the whole paper for the second time? i never want to see maths again - except your solutions to see how i went! oh, and chemistry mole/equation work...
 

Keskimo

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don't worry - thought you were a current hsc student :D
 

da1

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why doesn't it let me look at the solutions stupid comp
 

sando

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yeh its not working for me either.. wats the deal ?
 

lissamaher

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Question 3: c) (ii) you have 13/2, shouldnt it be 14/2?? as n=14?? or am i wrong
 

shinji

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i have a lil discrepancy with 2)c)
i got a far more complicated equation than that.
i got the same gradient.

but instead of doing x=0 and y=1,
i used found out y via x-> y=cos2x
which gave y= 1/2

and then did y-y1=m(x-x1)
which equaled
y- 1/2 = -(sqrt3)[x- pi/6]
2y - 1 = -x[sqrt3] + [[pi](sqrt3)]/6
-> x[sqrt3] + 2y - 1 - (pi[sqrt3])/6 = 0



welll that's just me .. . lol
 

sando

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in 2 (c) ... shouldnt the y-value be ...... y = cos2 (pi/6) = 1/2

im no maths expert.. but correct me if im wrong
 

sando

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yes.. i got same at Shinji.. and he is like boy genius lol
 

T.Goodhew

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shinji said:
i have a lil discrepancy with 2)c)
i got a far more complicated equation than that.
i got the same gradient.

but instead of doing x=0 and y=1,
i used found out y via x-> y=cos2x
which gave y= 1/2

and then did y-y1=m(x-x1)
which equaled
y- 1/2 = -(sqrt3)[x- pi/6]
2y - 1 = -x[sqrt3] + [[pi](sqrt3)]/6
-> x[sqrt3] + 2y - 1 - (pi[sqrt3])/6 = 0



welll that's just me .. . lol
y - 1/2 = -(sqrt3)[x - pi/6]
2y -1 = -2(sqrt3)[x - pi/6]
2y -1 = -2(sqrt3)x +([sqrt3]pi/3)
2(sqrt3)x +2y -1 -(pi[sqrt3]/3) = 0

i think you will find :D ... Changed the solutions accordingly grr.. stupid mistakes i made late at night.
 

shinji

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woops. mybad. lol
a little misclculation error for me as well there. haha
im pretty sure ididn't do it in the exam~ XD

and ur welcome =D

@sando: me? lil genius boy? i wish! haha
 

falbrav

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I've been trying to open these solutions all day but it just wont seem to work and i was wondering whether any1 had any suggestions?
 

rolfey24

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falbrav said:
I've been trying to open these solutions all day but it just wont seem to work and i was wondering whether any1 had any suggestions?

yeh i had probs as well. insted of opening jsut save it to ur pc. worked 4 me.
 

T.Goodhew

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rolfey24 said:
yeh i had probs as well. insted of opening jsut save it to ur pc. worked 4 me.
I've moved the file to a faster server, so you should be able to download the solutions easier now, thanks shinji for your help :D
 

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