-
Finished your HSC? Check out our University Discussion, Non-School forums or help out next year's HSC students! -
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page
Ext2 2024 Live Attempt in Progress (1 Viewer)
- Thread starter epicmaths
- Start date
2024 HSC Ext2 epicmath rough solutions from livestream.pdf
drive.google.com
Nice to see you doing this again 
Your concern about 16 (b) (ii) caught my attention -- I haven't thought too much about it and I may have made a mistake but I think it's really just a case of induction.
If you consider the sequence of unordered sets,cos(2^{k+1}\pi/9),cos(2^{k+2}\pi/9)\} )
, you can show by induction that 
for all integers 
. The base case is clear and the induction step is effectively the exact logic you showed.
In my opinion it's quite unclear how much of that would be required for what can surely only be 2 marks. I suppose we'll find out.
Your concern about 16 (b) (ii) caught my attention -- I haven't thought too much about it and I may have made a mistake but I think it's really just a case of induction.
If you consider the sequence of unordered sets
In my opinion it's quite unclear how much of that would be required for what can surely only be 2 marks. I suppose we'll find out.
Yes induction like you said works out quite quickly...I tried to be special and uploaded a solution involving modulo arithmetic, conceding that induction is probably what they wanted.Nice to see you doing this again
Your concern about 16 (b) (ii) caught my attention -- I haven't thought too much about it and I may have made a mistake but I think it's really just a case of induction.
If you consider the sequence of unordered sets
In my opinion it's quite unclear how much of that would be required for what can surely only be 2 marks. I suppose we'll find out.
Fair