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Thread: HSC 2018-2019 MX2 Marathon

  1. #151
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by CapitalSwine View Post
    Cant seem to figure out how to do this question:
    https://imgur.com/a/jRaWrmq
    BSQ = PQB (alternate angle theorem)
    PQB = 180-RPQ (co-interior angles in a parallelogram are supplementary)
    BSQ + RPQ = 180

    Hence opposite angles are supplementary so RSQP is a cyclic quad

    Hope this reasoning is legit, forgot maths

    EDIT: Nvm too slow
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  2. #152
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    Re: HSC 2018 MX2 Marathon

    It is easy to prove RSQP is cyclic; so that this can be easily shown in many ways.

    e.g.

    Let angle AQS = @; .: angle QBS = @; .: angle PRS = @ (PR // QB)

    .: angle AQS = angle PRS

    .: external angle AQS of quadrilateral RSQP = int opp angle PRS

    .: RSQP must be cyclic

    QED
    Last edited by Drongoski; 29 May 2018 at 9:29 AM.
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  3. #153
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    Re: HSC 2018 MX2 Marathon











    HSC 2018

    English Adv - MX1 - MX2 - Physics - Economics

  4. #154
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    Re: HSC 2018 MX2 Marathon

    a)


    b)



    c)

    For , .

    If , then clearly .

    Therefore, .

    From b) . Since ,



    Replacing with , we get with and so



    Giving, as required,



    d)

    From b),

    .







    From c),

    altSwift likes this.
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    Re: HSC 2018 MX2 Marathon





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  6. #156
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    Re: HSC 2018 MX2 Marathon

    yoyoyo
    there was a nice question i found a while back and thought i would post it here:


    given that e^x can be written as a sum of an odd and even function, find the two functions.
    Last edited by mrbunton; 20 Nov 2018 at 11:20 AM.

  7. #157
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by mrbunton View Post
    yoyoyo
    there was a nice question i found a while back and thought i would post it here:


    given that e^x can be written as a sum of an odd and even function, find the two functions.
    Is the intended answer the Taylor series for ?

    Edit: nevermind, I found some others
    Last edited by fan96; 20 Nov 2018 at 12:02 PM.
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by fan96 View Post
    Is the intended answer the Taylor series for ?

    Edit: nevermind, I found some others


    its Hyperbolic function of sin and cos
    so e^x = sinhx+coshx
    Its legit like deriving cosx and sinx in complex field


    The taylor expansion is just an approximate for e^x for values less than 1 or close to 0 but can get quite close. You would just factorise the even and odd parts of the function. However, that does not mean you cant use it, I tried it with Maclaurin series and it seems to look solid.
    Last edited by HeroWise; 20 Nov 2018 at 2:05 PM.

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    Re: HSC 2018 MX2 Marathon


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    Re: HSC 2018 MX2 Marathon

    Yeah it clicked in when i was thinking of expressing these interms of Complex version of cos and sin and basically simultaneous

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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by InteGrand View Post
    ya. You can generate that through utilising the definition of even and odd functions which was the intent of the question.

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    Re: HSC 2018 MX2 Marathon

    More qtns

  13. #163
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    Re: HSC 2018 MX2 Marathon



    fan96 likes this.

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    Re: HSC 2018 MX2 Marathon

    fan96 likes this.

  15. #165
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by InteGrand View Post
    tsk tsk trying to introduce multivariable optimisation in a sidestepping manner
    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.

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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by Paradoxica View Post
    tsk tsk trying to introduce multivariable optimisation in a sidestepping manner
    Haha, that wasn't the motivation at all.

  17. #167
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by Sy123 View Post


    The equation of the tangent at the hyperbola is given by



    Because the shortest distance between a point and a line is the perpendicular distance, is the intersection between the tangent line and the line perpendicular to the tangent which also passes through the origin.

    That line is



    Solving these two equations gives



    Now,





    Hence the locus of is given by the equation .







    And similarly,



    So,









    Last edited by fan96; 1 Jan 2019 at 3:35 PM.
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    Re: HSC 2018 MX2 Marathon

    Last edited by stupid_girl; 17 Feb 2019 at 12:16 AM.

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    Re: HSC 2018 MX2 Marathon


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