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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by mrbunton View Post
    glad someone noticed my question
    make sure to use all the given equations.
    eg: a and b represent their own complex numbers (on the lhs), try finding them;
    its easier approaching it from the lhs to the rhs;without trying to take out any common factors immediately
    Are you sure you wrote the question right? Cause a and b are real numbers (-1+-√2) not complex numbers in the form a+ib (b=/=0) (assuming that's what you intended it to be)
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by mrbunton View Post
    glad someone noticed my question
    make sure to use all the given equations.
    eg: a and b represent their own complex numbers (on the lhs), try finding them;
    its easier approaching it from the lhs to the rhs;without trying to take out any common factors immediately
    I used all the stuff you gave me and went from lhs -> rhs but I end up with crappy trig inside the ()^n that I haven't been able to transform into cosx - isinx to use with de moivre

    ill write it up in latex later this afternoon
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by jathu123 View Post
    Are you sure you wrote the question right? Cause a and b are real numbers (-1+-√2) not complex numbers in the form a+ib (b=/=0) (assuming that's what you intended it to be)
    Quote Originally Posted by altSwift View Post
    I used all the stuff you gave me and went from lhs -> rhs but I end up with crappy trig inside the ()^n that I haven't been able to transform into cosx - isinx to use with de moivre

    ill write it up in latex later this afternoon
    soz dudes i made a typo; z^2-2z+2=0 where a and b are roots
    i apologise for any inconvenience this may have caused
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by mrbunton View Post
    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
    Quote Originally Posted by mrbunton View Post
    soz dudes i made a typo; z^2-2z+2=0 where a and b are roots
    i apologise for any inconvenience this may have caused





















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    Re: HSC 2018 MX2 Marathon

    yup

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    Re: HSC 2018 MX2 Marathon

    Oh whoops... I read that as .
    HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]

    1(3√3) t2 dt cos(3π/9) = log(3√e) | Integral t2 dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.

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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by jathu123 View Post
    A classic one

    Before I write up the latex, is the answer

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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by altSwift View Post
    Before I write up the latex, is the answer

    Correct!

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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by jathu123 View Post
    A classic one

    Quote Originally Posted by InteGrand View Post
    Correct!




































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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by altSwift View Post








































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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by InteGrand View Post




    Question 7 Of Ekman's Compilation
    for the working out of integrands clever solution if anyones interested
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by InteGrand View Post




    Ah I remember seeing telescopic stuff in an old MX2 marathon as a hard Q7 or Q8

    Are there any resources in particular to practice Q7/Q8 level questions apart from past hsc papers?
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by altSwift View Post
    Alternatively, you can use the fact that the Imaginary part of 1 is 0 to do this:

    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

    I'm not sure what you did on this line, I'm guessing its in regard to the fact that Im(1) = 0



    Also, did you mean:

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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by altSwift View Post
    Ah I remember seeing telescopic stuff in an old MX2 marathon as a hard Q7 or Q8

    Are there any resources in particular to practice Q7/Q8 level questions apart from past hsc papers?
    One thing (probably not the only thing) you can try is searching up past years' HSC 4U marathons here on BOS (including the Advanced Level ones, though these are typically harder than HSC style Q's or would be broken into multiple steps if asked in the HSC exam).
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by altSwift View Post
    Also, did you mean:

    That does not follow. You can verify it for yourself by converting the conjugates to reciprocals and clearing them.
    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
    That does not follow. You can verify it for yourself by converting the conjugates to reciprocals and clearing them.
    Sorry if I'm missing something obvious, but just to clarify





    When you take the imaginary part of the sum of GP formula, does that instantly disregard the z^0 (=1) in the original series becuase Im(1) = 0? It seems weird because you use a = 1 for the formula, maybe I'm thinking about this too much haha
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by altSwift View Post
    Sorry if I'm missing something obvious, but just to clarify





    When you take the imaginary part of the sum of GP formula, does that instantly disregard the z^0 (=1) in the original series becuase Im(1) = 0? It seems weird because you use a = 1 for the formula, maybe I'm thinking about this too much haha








    Last edited by InteGrand; 14 Feb 2018 at 6:25 AM.
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by InteGrand View Post








    Yeah that makes more sense now, its pretty creative. Thank you
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    Re: HSC 2018 MX2 Marathon



    HSC 2018

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    Re: HSC 2018 MX2 Marathon

    Is the answer ?

    (I intend to post a writeup later)
    HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]

    1(3√3) t2 dt cos(3π/9) = log(3√e) | Integral t2 dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.

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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by fan96 View Post
    Is the answer ?

    (I intend to post a writeup later)
    yes it's 4
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    Re: HSC 2018 MX2 Marathon



    Converting to polar form,



    Since ,



    The cosine function is even, so the first term on the LHS and RHS are equal. Cancelling,



    Since the sine function is odd, for all .



    And clearly this is only true when .

    So therefore,
    Last edited by fan96; 14 Feb 2018 at 3:24 PM.
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    HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]

    1(3√3) t2 dt cos(3π/9) = log(3√e) | Integral t2 dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.

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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by fan96 View Post


    Converting to polar form,



    Since ,



    The cosine function is even, so the first term on the LHS and RHS are equal. Cancelling,



    Since the sine function is odd, for all .



    And clearly this is only true when .

    So therefore,
    Nice!











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    Re: HSC 2018 MX2 Marathon

    2u integration+circle geo
    i made this question myself in paint so dont mind the bad production quality.
    edit: look for the question which has the circle with one edge on -5 instead of -3. ignore the question where the circle touches -3.
    integral composite graph.png
    Attached Images Attached Images
    Last edited by mrbunton; 17 Feb 2018 at 1:41 PM.

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