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Thread: HSC 2018 MX2 Marathon

  1. #126
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by mrbunton View Post
    i) the line created is bx+ay=0.since you said i) is not the locus of z; and z can be truly anything; then we can observe that bx+ay=0 only yields to purely real values; so it becomes the line y=0 on the imaginary number plane

    ii) (x-p)^2+(y-q)^2= k^2
    sub y=0 and expand and bring to one side
    x^2 -(2p)x +(p^2+ q^2 - k^2) =0
    product of roots: z1*z2= p^2 + q^2 - k^2

    which is the same thing as we are meant to show. normally one would say the product of the roots are just the real parts of z1 and z2; but there is no imaginary part so condition is still true. But in an exam u would of been expected to spend a couple more lines to actually be rigorous but im lazy.
    How can you say that the equation is y=0 on the imaginary plane but then sub into an equation that's on the real number plane? Wouldn't you have to sub in bx + ay = 0?
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  2. #127
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    Re: HSC 2018 MX2 Marathon

    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.

  3. #128
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by altSwift View Post




    Quote Originally Posted by mrbunton View Post
    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.
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  4. #129
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    Re: HSC 2018 MX2 Marathon

    What is the solution of z^2 = i*(conjugate of z)

  5. #130
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    Re: HSC 2018 MX2 Marathon

    x² − y² = y & 2xy = x
    x=0, y = -1,0
    or
    y=½, x = ±½√3



    Sent from my iPhone using Tapatalk
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    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.

  6. #131
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by altSwift View Post




    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.

  7. #132
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    Re: HSC 2018 MX2 Marathon

    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?
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  8. #133
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    Re: HSC 2018 MX2 Marathon

    Yep, it's broken for me as well.

  9. #134
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by altSwift View Post
    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?
    Quote Originally Posted by fan96 View Post
    Yep, it's broken for me as well.
    Thanks for notifying us!
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  10. #135
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by fan96 View Post


    where did you get that part from? nice solution btw
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    Re: HSC 2018 MX2 Marathon

    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.

  12. #137
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by altSwift View Post
    where did you get that part from? nice solution btw


    and satisfies the locus in i), so



    which gives us



    therefore

    Last edited by fan96; 24 Mar 2018 at 1:34 AM.

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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by Paradoxica View Post
    Find separate expressions for the real and imaginary parts of

    Hence determine as a sum of independent real and imaginary parts
    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.

  14. #139
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by fan96 View Post
    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.
    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.

  15. #140
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by Paradoxica View Post
    Woops. Forgot to rationalise the denominator.



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    Re: HSC 2018 MX2 Marathon

    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.
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  17. #142
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by altSwift View Post
    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
    Hint: conjugate root theorem.

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    Re: HSC 2018 MX2 Marathon

    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.

  19. #144
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by CapitalSwine View Post
    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.
    If you know the stationary points you can get a rough sketch of the graph and hence find the k values (which shift the graph up and down) which satisfy the equation having one real root
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  20. #145
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by CapitalSwine View Post
    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.
    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.
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  21. #146
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    Re: HSC 2018 MX2 Marathon

    For which a ∈ R is the sum of the squares of the zeroes of x for
    x^2 − (a − 2)x − a − 1 minimal

  22. #147
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by mrbunton View Post
    For which a ∈ R is the sum of the squares of the zeroes of x for
    x² − (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.

  23. #148
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by Paradoxica View Post
    P(√x)=0 has roots α², β²


    clearly the minimum is 5, for a=1
    correct

  24. #149
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    Re: HSC 2018 MX2 Marathon

    Cant seem to figure out how to do this question:
    https://imgur.com/a/jRaWrmq

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    Re: HSC 2018 MX2 Marathon

    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
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