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Re: HSC 2016 Complex Numbers Marathon $\noindent By the way, we don't just want the solutions for $\theta$ above (in fact we don't need $\theta$ itself), we really want the solutions for $z$, which is $\cos \theta + i\sin \theta$. So we should find that $\cos \theta=\frac{1}{2}$ is a double...

Note that p4'(x) = p3(x) for all x. Also, p4"(x) = p2(x) for all x. We've shown that p2(x) is a positive definite quadratic, so that means the second derivative of p4 is always positive. This means the minimum of p4 will be a global min. Now, to find this min., set p4'(x) = p3(x) = 0. (There is...

If f' is differentiable and monotone on an interval I, we must have (f')' = f" stay the same sign (possible 0 some places) over that interval. Let's just suppose it's always non-negative. Since f is twice differentiable, the sign of f" dictates the concavity of f over the interval. Since f" is...

Re: HSC 2016 3U Marathon That's equivalent to arcsin(3/5) + pi/2 – arctan(24/7) = pi/2 – arcsin(3/5) <==> 2arcsin(3/5) = arctan(24/7). The above can be proved by taking the sines of both sides (you can draw a triangle to assist with finding sin(RHS) and use double-angle formula and...

Re: HSC 2016 3U Marathon Yeah, in fact there's equality if and only if the right-triangle is isosceles.

Well essentially you'd first find those x-values that satisfy the inequation. Whether you do this algebraically or graphically is up to you. Once you've found these values, the region would be found by drawing vertical lines at each of these x-values.

Basically solving the inequation for x is the first step. Have you done that yet? Once you've done that, it's fairly easy.

If we're talking about sharing that as a region in the x-y plane, since y doesn't appear in that inequation, y can be anything at all. So you just need to find the set of all x-values satisfying that equation, and then essentially draw vertical lines at each of these places (which will end up...

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread Ah OK, that would explain the difference. Though if davidgoes4wce's version of the Q. is correct, then yes, the answers had a mistake, they did the probability all days were fine. This mistake could be added to the textbook...

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread In leehuan's version of the question (the one he typed up some posts ago), part a) just asked for the probability that the first two days are wet, which is different to your part a), which also requires the remaining days to be...

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread I found how they got their answer to part a): they did (12/30)^5 (but this should be the probability that all days are fine, not the probability that the first two days are fine).

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread For the first two days chosen to be fine, the probability is: (prob. a day is fine)^2 = (12/30)^2 This Q. is like a coin toss Q. basically. (We are assuming for this Q. that the weather of the days are independent of each other...

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread n!/(r!.(n-r)!) n will be 5 because we have 5 days total.

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread Use standard binomial probability formulas with p = 18/30 if you want to define 'success' as a rainy day, or p = 1 – 18/30 = 12/30, if you want to define 'success' as a fine day.

For b), try defining f(t) = -ln(1-t) for 0 <= t<= x, where x is some arbitrary fixed number in (0,1).

Re: HSC 2016 MX2 Combinatorics Marathon Ah yeah Monty Small is another. Knew there was another variant rhyming with 'Hall' that I'd forgotten.

Basically yeah. But even if we define it, we wouldn't say that 1/x is 1 at 0. We'd say f is 1 at 0, where f(x) is only 1/x away from x = 0. 1/x itself would always be undefined at x = 0 because we can't divide by 0.

It's undefined. If we consider the function f: R -> R, defined as f(x) = 1/x for x =/= 0 and f(0) = 1 (say), then f will have an essential discontinuity at 0, since the limit from a side (in fact both sides here) is infinite (where infinite means limit being +/- infinity).

Re: HSC 2016 MX2 Combinatorics Marathon Yeah, and (unrelated) he also happens to be a really good chess player I think. Also there's a lot of Monty Hall variants out there, with amusing names like 'Monty Fall' and 'Monty Crawl'.

Sorry what I said about 'not necessarily undefined' was actually in reference to 'jump discontinuity', got my terms mixed up. To distinguish between a removable discontinuity and another discontinuity (jump or essential), the former type is one where the function can be defined at the point in...