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Thread: HSC 2017 MX2 Marathon

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    Re: HSC 2017 4U Marathon

    Quote Originally Posted by Paradoxica View Post
    Is x a real or complex number?
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    Re: HSC 2017 4U Marathon

    Quote Originally Posted by frog1944 View Post
    Is x a real or complex number?
    It doesn't matter for the first proof.
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    Re: HSC 2017 4U Marathon

    Quote Originally Posted by mini8658 View Post
    For z^5-1=0, the roots are 1, w^2-w^4, and
    z^5-1=(z-1)(z-cis2pi/5)(z-cis-2pi/5)(z-cis4pi/5)(z-cis-4pi/5) How do you get back to the LHS from expanding the RHS?
    By inspection.
    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 2017 4U Marathon

    Quote Originally Posted by mini8658 View Post
    For z^5-1=0, the roots are 1, w^2-w^4, and
    z^5-1=(z-1)(z-cis2pi/5)(z-cis-2pi/5)(z-cis4pi/5)(z-cis-4pi/5) How do you get back to the LHS from expanding the RHS?
    This is true by definition.

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    Re: HSC 2017 4U Marathon

    http://prntscr.com/db7gfc

    How would you sketch this? It's a HSC question.
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    Re: HSC 2017 4U Marathon

    Quote Originally Posted by pikachu975 View Post
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    How would you sketch this? It's a HSC question.
    The first locus is a Half- plane locus. You sketch the original locus and you add in the inequality.

    The 2nd locus is an Argand locus. It has the ranges of pi/4 and -pi/4.
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    Re: HSC 2017 4U Marathon

    Quote Originally Posted by pikachu975 View Post
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    How would you sketch this? It's a HSC question.
    The first part is a linear line that is a perpendicular bisector from points that subtend from 0 and 2. (make z=x+iy and you will get a linear eqn)
    Second part is just the argument from the origin from -pi/4 to pi on 4

    equate the two inequalities
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    Re: HSC 2017 4U Marathon

    Quote Originally Posted by si2136 View Post
    The first locus is a Half- plane locus. You sketch the original locus and you add in the inequality.

    The 2nd locus is an Argand locus. It has the ranges of pi/4 and -pi/4.
    Quote Originally Posted by Rathin View Post
    The first part is a linear line that is a perpendicular bisector from points that subtend from 0 and 2. (make z=x+iy and you will get a linear eqn)
    Second part is just the argument from the origin from -pi/4 to pi on 4

    equate the two inequalities
    Oh yeah i get it now thanks!
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    Re: HSC 2017 4U Marathon

    Quote Originally Posted by Rathin View Post
    a linear line
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    Re: HSC 2017 4U Marathon

    Quote Originally Posted by seanieg89 View Post
    Had to make sure lol
    In sciences a LINE of best fit can be a linear line of best fit or a curve of best fit
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    Re: HSC 2017 4U Marathon

    Quote Originally Posted by Rathin View Post
    Had to make sure lol
    In sciences a LINE of best fit can be a linear line of best fit or a curve of best fit
    Actually, I don't even know why a line of best fit can be a curve. I would just call a curve of best fit its own seperate thing. (Polynomial of best fit, blah blah)

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    Re: HSC 2017 4U Marathon

    Quote Originally Posted by Rathin View Post
    Had to make sure lol
    In sciences a LINE of best fit can be a linear line of best fit or a curve of best fit
    That's just hazy high school science definitions. There are various precise definitions of lines/curves of best fit (regression analysis, interpolating polynomials etc), and whenever such an object is named a line it is definitely a line in the usual sense of the word (at least as far as I have seen).
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    Re: HSC 2017 4U Marathon

    Is there a way to expand it so the cis2kpi/5's cancel out, leaving only the z^5-1?

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    Re: HSC 2017 4U Marathon

    Quick question: When we are finding the roots of unity (or any complex number) and leaving it in mod-arg form do we have to change the argument to the principle range or just leave it as it as (e.g. for k=5 arg = 15pi/8). The answers in past papers leave it as it is...so not sure what to do.
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    Re: HSC 2017 4U Marathon

    Quote Originally Posted by Rathin View Post
    Quick question: When we are finding the roots of unity (or any complex number) and leaving it in mod-arg form do we have to change the argument to the principle range or just leave it as it as (e.g. for k=5 arg = 15pi/8). The answers in past papers leave it as it is...so not sure what to do.
    I always converted it to the principle range, technically you'd be right, but it's better to be safe.
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    Re: HSC 2017 4U Marathon

    Quote Originally Posted by Rathin View Post
    Quick question: When we are finding the roots of unity (or any complex number) and leaving it in mod-arg form do we have to change the argument to the principle range or just leave it as it as (e.g. for k=5 arg = 15pi/8). The answers in past papers leave it as it is...so not sure what to do.
    I like to use -2, -1, 0, 1, 2 instead of 0, 1, 2, 3, 4 because it'll always give you arguments in the principal argument range.

    Just a side note it's principal argument not principle
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    Re: HSC 2017 4U Marathon






    Last edited by InteGrand; 6 Dec 2016 at 8:27 PM.

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    Re: HSC 2017 4U Marathon


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    Re: HSC 2017 4U Marathon

    Quote Originally Posted by InteGrand View Post
    This is probably a vague explanation but...


    In the case where both roots are real.
    The sum of roots = -b/a, which is negative if a,b are either both positive or both negative. This implies that at least one of the roots is negative.
    Product of roots = c/a, if the roots are both negative, then the product must be positive therefore a,c must be both positive or both negative.

    If the roots are complex, they must be in conjugate pairs due to real coefficients, this means that the real parts must be the same.
    So if the real part is negative, then the sum of roots must be negative, and the product of roots must be positive.
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    Re: HSC 2017 4U Marathon

    here's a nice binomial followed by a harder binomial
    133.PNG
    134.PNG

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    Re: HSC 2017 MX2 Marathon

    This geometry question doesn't involve any advanced knowledge but requires a bit of creativity.

    ABCD is a quadrilateral with three equal sides AB,BC and CD. Show that the mid-point of AD lies on a circle with diameter BC if and only if the area of ABCD is a quarter of the product of its diagonals.

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    Re: HSC 2017 MX2 Marathon

    A question posted in the 2016 MX2 marathon by Paradoxica that wasn't answered:
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    Re: HSC 2017 MX2 Marathon

    Quote Originally Posted by frog1944 View Post
    A question posted in the 2016 MX2 marathon by Paradoxica that wasn't answered:
    It approaches infinity.

    lim x--> 0, sin x / (1-cos x)

    = lim x--> 0 (1+cos x) / sin x

    Apply l'hopital, sin^2 x, sin^4 x, all approach infinity.

    Therefore the limit approaches infinity.

    Could someone check my answer please? Thanks.
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    Re: HSC 2017 MX2 Marathon

    Quote Originally Posted by si2136 View Post
    It approaches infinity.

    lim x--> 0, sin x / (1-cos x)

    = lim x--> 0 (1+cos x) / sin x

    Apply l'hopital, sin^2 x, sin^4 x, all approach infinity.

    Therefore the limit approaches infinity.

    Could someone check my answer please? Thanks.
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    Re: HSC 2017 MX2 Marathon

    Suppose you have n points on a circle such that no three distinct chords coincide at any single point.

    How many regions do the nC2 chords divide the interior of the circle into?

    (You don't have to provide rigorous proof for this question if you can guess the answer correctly).

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