ya rip me. i misinterpreted what z as a variable point meant. i thought it meant z could be anything and that the line was dependent on what z was; although that doesnt make much sense once you think about it.
ya rip me. i misinterpreted what z as a variable point meant. i thought it meant z could be anything and that the line was dependent on what z was; although that doesnt make much sense once you think about it.
Last edited by mrbunton; 28 Feb 2018 at 9:56 PM.
What is the solution of z^2 = i*(conjugate of z)
x² − y² = y & 2xy = x
x=0, y = -1,0
or
y=½, x = ±½√3
Sent from my iPhone using Tapatalk
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
i)
i.e. . But,
Therefore the locus of is the line , which is linear and passes through .
The exception is when , where the locus of is all complex numbers . Going off the question, I will assume is nonzero.
ii)
Using a substitution obtainable from the locus from i), , we have
Simplify and we get the quadratic equation
Let the roots be . Then, using the product of roots formula,
Which I will call equation . Now,
Using equation ,
Therefore,
, for .
__________________________________________________ _
Considering the case where :
So, either of or or both are zero.
If and it can be proved in a similar manner that .
If , which I addressed in part i).
Last edited by fan96; 19 Mar 2018 at 2:26 PM.
I can't load any latex in this thread anymore, I've tried on 3 computers, is anyone else getting this issue or just me?
HSC 2018
English Adv - MX1 - MX2 - Physics - Economics
Yep, it's broken for me as well.
--------------------------------------------------------------------------------
Buy my books/notes cheaply here!
--------------------------------------------------------------------------------
Uni Course: Actuarial Studies and Statistics at MQ -- PM me if you have questions
2017 HSC Subjects: Eng Adv / 3u+4u Maths / Bio /Phys
ATAR: 99.75
--------------------------------------------------------------------------------
Find separate expressions for the real and imaginary parts of
Hence determine as a sum of independent real and imaginary parts
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
Let .
(Note that if then , and we have . Otherwise,)
Expanding, we get .
Equating real and imaginary parts gives us
Squaring,
Adding and using the identity with , we get
Since and are both positive, we can take the positive square roots of both sides to get
and since ,
So cleaning up, we have
and we can get rid of that second sign if desired:
Where is the sign function.
Last edited by fan96; 24 Mar 2018 at 3:13 PM.
No idea how to do ii):
capture.PNG
I know this isn't the best place to put this but i didn't want to dedicate a thread to a 1 mark question that's apparently simple
Last edited by altSwift; 1 Apr 2018 at 8:57 PM.
HSC 2018
English Adv - MX1 - MX2 - Physics - Economics
I dont really understand why you have to find the stationary points (says so in the answers), would appreciate an expanation on how to solve this problem.
--------------------------------------------------------------------------------
Buy my books/notes cheaply here!
--------------------------------------------------------------------------------
Uni Course: Actuarial Studies and Statistics at MQ -- PM me if you have questions
2017 HSC Subjects: Eng Adv / 3u+4u Maths / Bio /Phys
ATAR: 99.75
--------------------------------------------------------------------------------
For simplicity, let .
First, sketch the graph of (the quickest way to do this is to find the turning points of the graph). The graph of can be obtained by shifting the graph of by units up or down.
Finally, a polynomial has as many real roots as its graph has -intercepts. So a polynomial has exactly one real root if and only if it crosses the -axis exactly once.
For which a ∈ R is the sum of the squares of the zeroes of x for
x^2 − (a − 2)x − a − 1 minimal
P(√x)=0 has roots α², β²
Simplify the equation to form a standard quadratic equation.
x - (a+1) = (a-2)√x
x² - 2(a+1)x + (a+1)² = (a-2)²x
x² - 2(a+1)x + (a+1)² = (a²-4a+4)x
x² - (a²-4a+4+2a+2)x + (a+1)² = 0
x² - (a²-2a+6)x + (a+1)² = 0
α²+β² = a²-2a+6 = (a-1)²+5
clearly the minimum is 5, for a=1
Last edited by Paradoxica; 29 Apr 2018 at 3:32 PM.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
Cant seem to figure out how to do this question:
https://imgur.com/a/jRaWrmq
RPQ and PQB are supplementary (parallel lines)
PQB = QSB (angle in alt segment)
hence RPQ + PSB = 180
since opposite angles are supplementary, RSQP is cyclic?
been 4 years hope I am right :P
CHEM + PHYS TRIAL QUESTIONS SORTED BY DOT POINT
http://community.boredofstudies.org/...t-answers.html
EXAM RESPONSES + FREE NOTES (99.90)
http://community.boredofstudies.org/...m-phys%5D.html
There are currently 1 users browsing this thread. (0 members and 1 guests)
Bookmarks