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  1. GoldyOrNugget

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon at work, so ill be answering these intermittently i) use sin(2a) = 2sin(a)cos(a) repeatedly, expanding the sin each time. ii) i don't get this question. \ \lim_{n \to \infty} \frac{\sin \theta}{\theta} = \frac{\sin \theta}{\theta}
  2. GoldyOrNugget

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon It's a neat trick that I've seen crop up in a number of places, so it's definitely worth keeping in the back of your mind: $ $\log(x)$ is strictly increasing, so $\log(a) > \log(b) \iff a > b, \quad a,b > 0$ \\ \\ $\log(9999^{10000}) = 10000 \log(9999) \approx...
  3. GoldyOrNugget

    How much money have you got saved in bank?

    ...you can't leave it at that! What book?
  4. GoldyOrNugget

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon Here's a sexy beast of a geometry question, by yours truly. $ A circle on the Cartesian plane with diameter 5 and center $(0; \frac{5}{2})$ completes a $360^\circ$ clockwise revolution around the $z$-axis (as if you pinned a point of its circumference to $(0; 0)$ and...
  5. GoldyOrNugget

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon Slightly offended that Fermat is worth more than Goldbach.
  6. GoldyOrNugget

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon IMO problems are floating around in this thread? o_O *steps back to make room for the pros* $i) Grouping terms as $P(x) = c_0(1 + x^{2n+1}) + c_1(x + x^{2n}) + ..., P(-1) = c_0(1-1) + c_1(-1+1) + ... = 0$ so $x=-1$ is a root. $ $ii) Let $Q(x) = a_0 + a_1x +...
  7. GoldyOrNugget

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon The pressure of doing the exam is a pretty big factor. I did around 10 ext2 HSC past papers and my lowest mark on those would still have scaled far higher than my actual exam mark. A friend of mine is a maths genius and for his study, he started doing past HSC papers...
  8. GoldyOrNugget

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon Hmph. Fine, I'll accept it :P the solution I have in mind uses logs.
  9. GoldyOrNugget

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon Sneaky... I haven't seen that approach before. What if I changed it to 9999^{10001} $ vs $ 10001^{9999}? The method I have in mind can compare any pair of large exponents regardless of the values of the bases and exponents.
  10. GoldyOrNugget

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon $Which is greater: $9999^{10000}$ or $10000^{9999}$? Justify your answer$
  11. GoldyOrNugget

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon Oooh I love this proof! It's so clean and elegant. $Let $y=f(x)$ be a function that satisfies $y=y'$, so $f(x) = f'(x)$, i.e. $f(x) - f'(x) = 0$. \\ Define $g(x) = \frac{f(x)}{e^x}$. By the quotient rule , $g'(x) = \frac{e^xf'(x) - f(x)e^x}{e^{2x}} = \frac{f'(x) -...
  12. GoldyOrNugget

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon The hint helped -- that question was (and still is) making me feel pretty stupid :( hope I didn't make a mistake here. $If $\alpha$ is a stationary point of $E(x)$, then $E'(\alpha) = 0$. But we established that $E'(x) = E(x) - \frac{x^n}{n!}$ so $E(\alpha) =...
  13. GoldyOrNugget

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon $i) Double roots occur when $E(x) = E'(x) = 0$ for some real $x$. Now $ E(x) = 1 + \frac{x}{1!} + \frac{x^2}{2!} + ... + \frac{x^n}{n!} \\ E'(x) = 1 + \frac{x}{1!} + \frac{x^2}{2!} + ... + \frac{x^{n-1}}{(n-1)!} = E(x) - \frac{x^n}{n!}$, so $E(x) = E'(x)$ iff...
  14. GoldyOrNugget

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon Yep. Out of interest, what method did you use?
  15. GoldyOrNugget

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon Hope people don't mind that I'm not a 2013-er -- just trying to keep sharp over the holidays. Solve for n: \ln (\prod_{k=1}^n 13^k) = 11325 \ln 13
  16. GoldyOrNugget

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon \lim_{n \to \infty} \int_{0}^1 \left (\frac{1}{x} \int \frac{1-x^n}{1-x} \ dx \right) \ dx = \lim_{n \to \infty} \int_{0}^1 \left (\frac{1}{x} \int (1 + x + x^2 + ... + x^{n-1}) \ dx \right) \ dx = \lim_{n \to \infty} \int_{0}^1 \left (\frac{1}{x} \int (x +...
  17. GoldyOrNugget

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon Oops, sorry. Wasn't thinking carefully enough Not always true. Picking 9 apples is proportional to picking 3 apples, for example.
  18. GoldyOrNugget

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon Sounds good Hm ok... but what about the arrangement (3 grapes, 3 apples)? This is an arrangement for 6 fruits, and it's not proportional to any arrangement of 3 fruits , but yet it isn't legal... You are almost there :) go deeper!
  19. GoldyOrNugget

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon It looks like you have some good ideas, but it needs refining. For your case 3, for example, the correct answer is 35, but your method gives 30. I think it's because of this: "we can pick Case 2 becasue we just add on one fruit to each permutation from Case 2". But...
  20. GoldyOrNugget

    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon seanie, Correct! :) I used the same dots/slashes reasoning, but phrased it as 'choosing slash positions between dots'. For some fixed number of fruit to purchase 'n', this is given by n+5-1C5-1. Once you get into the rhythm of button-mashing, it only takes a few seconds...
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