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Yes my bad

\frac{3\sin x}{2 + \cos x} < \frac{3x}{2+\cos x} \\ \Rightarrow \ x - \frac{3x }{2 + \cos x} = \frac{x(1-\cos x)}{2 + \cos x}> 0 \ \ $as$ \ 1 \geq \cos x \\ \therefore \ \frac{3x}{2+\cos x} < x \\ \therefore \ \frac{3\sin x}{2+ \cos x} < x

Re: HSC 2014 4U Marathon 1. His brain 2. Well below 50s lol

Re: HSC 2014 4U Marathon - Advanced Level oops

Re: HSC 2014 4U Marathon - Advanced Level Nope that's fine well done (though you might want to show that the sequence in n is increasing/decreasing to show it is not periodic) My method involved getting an upper and lower bound for \sum_{k=1}^n \ln k using upper/lower rectangles on y=ln(x)...

Re: HSC 2014 4U Marathon - Advanced Level \\ $Using squeeze theorem, find the real function$ \ f \ $so that$ \ \lim_{n \to \infty} (n!)^{f(n)} \ $converges and is positive$

Re: HSC 2014 4U Marathon First place got ~67 iirc and came first in the state, and second place got one mark below and came 7th

Please don't do that, Anna Wintour is a homosexual atheist Also there is no need to respond to this, if you actually check some of the references then you will find what the verse actually says. But it is not actually a rebuttal to say "here look at the Bible". But of course non-Muslims who...

Responded to in my post How would it be applicable if by your own statement this would not be a true reflection of the religion? e.g. The Ahmadiyya community consider themselves Muslims, but they aren't.

ITT: People who have no idea how Islamic law is formulated

What do you mean what does "religion" not explicitly prohibit? If by religion you mean your layman willy-nilly flick through the Bible/Qur'an, then that is a very weak religion. But if by religion you mean an entire Orthodoxy built upon the Qur'an and Sunnah, then you will find a source for...

It is well established in the Shari'a that consummation of marriage cannot take place until the wife/husband is sexually mature. There is a difference in definition of a child in the Western world and the Shari'a. In the Western World, someone is considered an adult once they pass an arbitrary...

Re: HSC 2014 4U Marathon arg(xy) = arg(x) + arg(y) \ $not$ \ arg(x) \cdot arg(y)

Re: HSC 2014 4U Marathon - Advanced Level I have not given it enough thought but it looks like a formal explanation using inequalities is the way to go.

Re: HSC 2014 4U Marathon - Advanced Level On your case where the numbers add up to 1 You say: "my next guess would be a=3, b=4, c=5 and d=6 which gives sum 0.95, while any other combination should give an even smaller number as we must pick larger numbers." THis is nto true since I can pick...

Re: HSC 2014 4U Marathon (leave this marathon for the 2014ers please) \\ $i) Expand the complex number and find it in mod-arg form$ \\ (1+i)(1+2i)(1+3i) \\ $ii) Hence show that by finding each individual product in mod-arg form$ \\ \tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1}3 = \pi

Re: HSC 2014 4U Marathon - Advanced Level If we take the trivial case: a = N b = 2N c = 3N d = 6N then: the sum is 2/N, and this is an integer for N= 1 and N=2, so there must be more than one solution

Re: HSC 2014 4U Marathon - Advanced Level \\ $To find the number of natural pairs$ \ (a,b) \ $such that$ \ a,b \ $are factors of$ \ 6^6 \ $and$ \ a \ $is a factor of$ \ b \\ $Consider the prime factorisation of$ \ 6^6 = 3^6 \cdot 2^6 \\ $Some factor of$ \ 6^6 \ $can be represented as some...

Re: HSC 2014 4U Marathon - Advanced Level \\ 6^6 = 3^6 \cdot 2^6 To find the number of solutions (a,b) that satisfies the above conditions, then we look at the number of ways we can pick from 6 3s and 6 2s, such that when we pick our second number we are picking from the remaining numbers to...

Re: HSC 2014 4U Marathon - Advanced Level For the polynomials one what I did was: \\ f(x) + f(1-x) = 1 \\ f'(x) = f'(1-x) \\ \therefore \ f'(x) = \sum_{k=0}^n c_k x^{a_k} (1-x)^{b_k} \\ $Then$ \ f(x) = \int \sum_{k=0}^n c_k x^{a_k}(1-x)^{b_k} \ dx + c \\ \\ $where$ \ \ c = \frac{1}{2} -...