https://i.imgur.com/DOI8Tanr.jpg
Hopefully im right, correct me if wrong
yup
Last edited by mrbunton; 4 Feb 2018 at 9:50 PM.
polynomial/complex number question:
cot(x)= y+1
a,b are roots of z^2-2z+2=0
prove that ( (y+a)^n - (y+b)^n ) / (a-b) = sin(nx) / sinx^n
Last edited by mrbunton; 12 Feb 2018 at 4:56 PM. Reason: error
This is an interesting question I found in the SGS notes.
If is a complex number, find the locus of if
HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]
HSC 2018
English Adv - MX1 - MX2 - Physics - Economics
As I said above, that's not the correct answer.
The locus is not .
There are certain values of (specifically ) that need to be considered.
(By the way, is not on the unit circle so it's not necessary to explicitly exclude it if the locus was just the unit circle)
I'll post the solution tomorrow if nobody gets the right answer.
Last edited by fan96; 10 Feb 2018 at 5:43 PM.
HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]
For anyone interested, the Patel question I was talking about is:
HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]
It definitely seems to be something that most people forget. For this type of question it tends to not matter (in terms of marks), but being able to spot when a variable's domain has been restricted has saved me a few marks in exams.
(This might help with the locus problem I posted before...)
HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]
Last edited by jathu123; 10 Feb 2018 at 9:07 PM.
2017
4u99 - 3u98 - EngAdv88 - Phys94 - Chem94
Atar : 99.55
Course: AdvSci(Hons)/Engineering(Hons) @ UNSW
A classic one
2017
4u99 - 3u98 - EngAdv88 - Phys94 - Chem94
Atar : 99.55
Course: AdvSci(Hons)/Engineering(Hons) @ UNSW
Yep.
If you used , you ended up with the locus
Which is the correct answer.
It's tempting to divide both sides by , but if you performed the division you were implicitly assuming that , which lost solutions.
Clearly if , any value of is a solution, except for zero since .
That can also be shown by considering
which is zero for all .
To graph , you could consider two separate cases: , and .
This is the solution from the SGS notes:
HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]
Last edited by mrbunton; 12 Feb 2018 at 1:32 AM.
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