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Thread: HSC 2018-2019 MX2 Marathon

  1. #101
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    Re: HSC 2018 MX2 Marathon

    rip the circle was meant to have radius 2; i changed it now so the circle touches -5 to -1.
    edit: i keep writing questions incorrectly; i check next time btw.
    Last edited by mrbunton; 17 Feb 2018 at 3:15 PM.

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    Re: HSC 2018 MX2 Marathon



    HSC 2018

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    Re: HSC 2018 MX2 Marathon

    where


    __________________________________________________ ___

    If is real, then



    The smallest value of is .
    __________________________________________________ ___

    If is purely imaginary, then



    The smallest value of is .
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by fan96 View Post
    where


    __________________________________________________ ___

    If is real, then



    The smallest value of is .
    __________________________________________________ ___

    If is purely imaginary, then



    The smallest value of is .
    Correct!
    Here is another (I really like these questions)





    Last edited by altSwift; 18 Feb 2018 at 1:08 PM.
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    Re: HSC 2018 MX2 Marathon

    The equation does not seem to hold for .

    https://www.wolframalpha.com/input/?...D2%5E(2(3)%2B1)

    Is there supposed to be an somewhere?
    Last edited by fan96; 18 Feb 2018 at 1:32 AM.
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by fan96 View Post
    The equation does not seem to hold for .

    https://www.wolframalpha.com/input/?...D2%5E(2(3)%2B1)

    Is there supposed to be an somewhere?
    Yes it should be
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    Re: HSC 2018 MX2 Marathon









    Since is an even function and is an odd function,








    __________________________________________________ ___________

    If , then



    Which is the general solution for



    Last edited by fan96; 18 Feb 2018 at 12:27 PM.
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by fan96 View Post
    The equation does not seem to hold for .

    https://www.wolframalpha.com/input/?...D2%5E(2(3)%2B1)

    Is there supposed to be an somewhere?
    Quote Originally Posted by jathu123 View Post
    Yes it should be
    My bad, yeah thats the correct equation
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    Re: HSC 2018 MX2 Marathon

    An interesting roots of unity question from the SGS notes.

    i) Find the fourth roots of in the form .

    ii) Hence write as a product of two quadratic factors with real co-efficients.
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by fan96 View Post
    An interesting roots of unity question from the SGS notes.

    i) Find the fourth roots of in the form .

    ii) Hence write as a product of two quadratic factors with real co-efficients.










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    Re: HSC 2018 MX2 Marathon



    Last edited by altSwift; 19 Feb 2018 at 10:35 PM.
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by altSwift View Post


    Trivially, .

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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by sharky564 View Post
    Trivially, .
    Yeah lol, I was trying to force the weird perfect square thing that cambridge does. But that works too
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by sharky564 View Post
    Trivially, .
    Quote Originally Posted by altSwift View Post
    Yeah lol, I was trying to force the weird perfect square thing that cambridge does. But that works too













    Last edited by altSwift; 20 Feb 2018 at 7:55 AM.
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by sharky564 View Post
    Trivially, .
    What made the polynomial so easy to factor?
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  16. #116
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    Re: HSC 2018 MX2 Marathon

    probability: how many 3 digits numbers exist where the middle digit is the average of the first and last digit.

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    Re: HSC 2018 MX2 Marathon

    Note that only even numbers can be halved to obtain an integer, and and . Finally, the average of any two digits is never greater than 9 or less than 0.

    If the first digit is odd (1, 3, 5, 7, 9), then the last digit must be odd too (1, 3, 5, 7, 9).

    If the first digit is even (2, 4, 6, 8), then the last digit must be even or zero (0, 2, 4, 6, 8).

    Any two digits can only have one possible average, so the middle digit doesn't matter.

    Therefore the number of digits is
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    Re: HSC 2018 MX2 Marathon

    yup

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    Re: HSC 2018 MX2 Marathon

    Show that the polynomial (where is an integer such that ) has exactly one real zero where .
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    Re: HSC 2018 MX2 Marathon

    derive equation to get turning points at -1 and 1
    find y values of these turning points:
    (-1,2+k) and (1,k-2)
    since the poly is odd it will have at least one root.
    when k>2; both turning points are above the x-axis; so no extra roots are possible.

    proving that r<-1 or similar value:
    when letting k=2; that poly is less then 0. the poly can be factored into (x+2)(x-1)^2<0
    condition is only true at x<-2 or r<-2
    -2<-1 so r<-1
    Last edited by mrbunton; 24 Feb 2018 at 10:32 AM.

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    Re: HSC 2018 MX2 Marathon





    Last edited by altSwift; 25 Feb 2018 at 7:24 PM.
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    Re: HSC 2018 MX2 Marathon

    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.
    Last edited by mrbunton; 27 Mar 2018 at 6:57 PM.

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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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    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.

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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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