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Thread: HSC 2017 MX2 Marathon

  1. #26
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    Re: HSC 2017 4U Marathon

    Are calculators allowed in the 4u test
    MX2 - ADV - ECO - BUS

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    Re: HSC 2017 4U Marathon

    Quote Originally Posted by si2136 View Post
    Are calculators allowed in the 4u test
    Yes.

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    Re: HSC 2017 4U Marathon

    Prove De Moivre's Theorem by Mathematical Induction where |z|=1
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    Re: HSC 2017 4U Marathon

    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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    -insert title here- Paradoxica's Avatar
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    Re: HSC 2017 4U Marathon

    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. #31
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    Re: HSC 2017 4U Marathon

    For the first part about the limit:

    The best I could come up with was that (based on what my teacher said) is that all exponential functions dominate over all polynomial functions which dominate over all log functions. Therefore, based on the second line of the statement, because both g(t) (the function g denotes that left hand expression just before log(t)) and f(t) = t^a both tend to 0, and f(t) dominates over log(t), it must follow that the limit is true because g(t) also tends to 0 and takes precedence over the log function.

    Hopefully that made sense...

    I'm still trying to practice on latex - I got the second part but I don't think it technically uses parts...

  7. #32
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    Re: HSC 2017 4U Marathon

    Quote Originally Posted by TaxiCab1729 View Post
    For the first part about the limit:

    The best I could come up with was that (based on what my teacher said) is that all exponential functions dominate over all polynomial functions which dominate over all log functions. Therefore, based on the second line of the statement, because both g(t) (the function g denotes that left hand expression just before log(t)) and f(t) = t^a both tend to 0, and f(t) dominates over log(t), it must follow that the limit is true because g(t) also tends to 0 and takes precedence over the log function.

    Hopefully that made sense...

    I'm still trying to practice on latex - I got the second part but I don't think it technically uses parts...
    You will need to flesh out your explanation more for it to be valid.

    The idea is to use the inequalities:



    To bound the function behaviour near the origin and squeeze both sides.
    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.

  8. #33
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    Re: HSC 2017 4U Marathon

    Quote Originally Posted by Paradoxica View Post
    The idea is to use the inequalities:



    To bound the function behaviour near the origin and squeeze both sides.
    Like, is anyone even gonna get this?

  9. #34
    Ancient Orator leehuan's Avatar
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    Re: HSC 2017 4U Marathon

    Quote Originally Posted by TaxiCab1729 View Post
    Like, is anyone even gonna get this?
    This is a marathon. They purposely throw in questions that can get overcomplicated. 100% of the time they are doable using just HSC techniques however if it were the HSC exam you'd be guided step by step

  10. #35
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    Re: HSC 2017 4U Marathon

    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.

  11. #36
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    Re: HSC 2017 4U Marathon

    Quote Originally Posted by Paradoxica View Post
    If anyone wants a hint, consider the expansion (1+x)^2n and what is is equal to for a), and for b), note that nCk=nC(n-k), and group together the like terms. Do consider the even and odd values of n separately for ease.
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  12. #37
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    Re: HSC 2017 4U Marathon

    Quote Originally Posted by calamebe View Post
    If anyone wants a hint, consider the expansion (1+x)^2n and what is is equal to for a), and for b), note that nCk=nC(n-k), and group together the like terms. Do consider the even and odd values of n separately for ease.
    There is no need to consider parity cases if you're at that level.
    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.

  13. #38
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    Re: HSC 2017 4U Marathon

    Quote Originally Posted by Paradoxica View Post
    There is no need to consider parity cases if you're at that level.
    Eh, it works.
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    Ancient Orator leehuan's Avatar
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    Re: HSC 2017 4U Marathon





    And I probably did something wrong.

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    Re: HSC 2017 4U Marathon

    Quote Originally Posted by leehuan View Post




    And I probably did something wrong.
    Yeah you left off the squares. Put the squares on, then use the result from a) and you should be fine.

    EDIT: Though I don't see a need to break the terms up as you did in the second line. You can do it easily without breaking it up.
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  16. #41
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    Re: HSC 2017 4U Marathon

    Quote Originally Posted by calamebe View Post
    Yeah you left off the squares. Put the squares on, then use the result from a) and you should be fine.

    EDIT: Though I don't see a need to break the terms up as you did in the second line. You can do it easily without breaking it up.
    Was wondering where the squares vanished to.

    There wasn't a need for it; I did it purely for my visual purposes.
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  17. #42
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    Re: HSC 2017 4U Marathon

    You really don't need to split based upon parity. It only makes things worse.

    You haven't considered



    enough.

    Side note your conclusion is wrong.
    calamebe likes 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.

  18. #43
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    Re: HSC 2017 4U Marathon

    Here's a nice anecdote.


    "One day Gauss' teacher asked his class to add together all the numbers from 1 to 100, assuming that this task would occupy them for quite a while. He was shocked when young Gauss, after a few seconds thought, wrote down the answer 5050. The teacher couldn't understand how his pupil had calculated the sum so quickly in his head, but the eight year old Gauss pointed out that the problem was actually quite simple.

    He had added the numbers in pairs - the first and the last, the second and the second to last and so on, observing that 1+100=101, 2+99=101, 3+98=101, ...so the total would be 50 lots of 101, which is 5050"
    calamebe and frog1944 like 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.

  19. #44
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    Re: HSC 2017 4U Marathon

    Damn it...I was doing that

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    Re: HSC 2017 4U Marathon

    Binomial Question.pdf

    I think I got it, but it looks dodgy...

  21. #46
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    Re: HSC 2017 4U Marathon

    Quote Originally Posted by TaxiCab1729 View Post
    Binomial Question.pdf

    I think I got it, but it looks dodgy...
    For odd n, I don't think your S_2 is an integer. I think you don't need that subtraction by 1.

    Also note that you can evaluate the sums C(n,k)^2 and kC(n,k)^2 combinatorially. (This takes less effort than other methods of computation imo.)

    These sums amount to:

    A: The number of ways of choosing a team of n people out of 2n people, n of whom are in year 12 and n of whom are in year 11.

    This can be done in C(2n,n) ways.

    B: Same as above, but also selecting one of the year 12 students to be team captain.

    This can be done in nC(2n,n)/2 ways. (The number of ways of choosing a team and specifying a captain is clearly nC(2n,n). By symmetry, this captain will be in year 12 exactly half of the time.)
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    Re: HSC 2017 4U Marathon

    Note that the above yields an end answer of something like



    Busy prepping for a meeting right now but paradoxica can confirm if that is what he was looking for.
    calamebe likes this.

  23. #48
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    Re: HSC 2017 4U Marathon

    Quote Originally Posted by seanieg89 View Post
    Note that the above yields an end answer of something like



    Busy prepping for a meeting right now but paradoxica can confirm if that is what he was looking for.
    Yeah that's what I got too.
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  24. #49
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    Re: HSC 2017 4U Marathon

    Yeah...I realized my mistake. My bad...

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    Re: HSC 2017 4U Marathon

    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?

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