Search results

Re: HSC 2013 3U Marathon Thread 29+12\sqrt5=29+2\sqrt{180}=(\sqrt{20})^2+(\sqrt9)^2+2\sqrt{20}\times\sqrt9=(\sqrt{20}+\sqrt9)^2 \therefore \sqrt{29+12\sqrt5}=\sqrt9+\sqrt{20}=3+2\sqrt5 $ and $ 11+6(3+2\sqrt{5})=29+12\sqrt5 $ again$ \therefore \sqrt{11+6\sqrt{29+12\sqrt5}}=3+2\sqrt5...

Re: HSC 2013 3U Marathon Thread Let the point be P. From similar triangles, P divides internally the interval AB in the ratio 3:2.

Re: HSC 2014 4U Marathon I would also do complementary. But do you mean two R's are different while 2C's, 2S's, 2T's are each pair identical? lol

Re: HSC 2014 4U Marathon - Advanced Level 5\times{8 \choose 2}^2\times{6 \choose 2}^2\times{4 \choose 2}^2=31752000

Re: MX2 Integration Marathon J_{n+1}+J_n=\int_0^{\frac{\pi}{4}} \tan^{2n}x\sec^2xdx=\int_0^{\frac{\pi}{4}} \tan^{2n}xd\tan x=\frac{1}{2n+1}$(1)$ $ so $ \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}=(J_1+J_0)-(J_2+J_1)+(J_3+J_2)-\cdots=J_0\pm J_\infty=\frac{\pi}{4} $ because $...

Re: HSC 2013 3U Marathon Thread Nice one. $ Consider $ b_n=\frac{1}{a_n}. $ flip the recursive relationship, it is clear that $ b_n $ is an AP. $

Re: MX2 Integration Marathon $ Let $ I_n=\int_0^{\frac{\pi}{2}} \frac{\sin ((2n+1)x)}{\sin x}dx $ Then $ I_n-I_{n-1}=\int_0^{\frac{\pi}{2}} \frac{\sin ((2n+1)x)-\sin ((2n-1)x)}{\sin x}dx =\int_0^{\frac{\pi}{2}}\frac{2\cos((2n)x)\sin x}{\sin x}dx...

Re: HSC 2014 2U Marathon a few typros: part (i) =b^3 $ at the end should be $ +b^3 part (iii) 1010 should be 5050 part (v) the result should be 338350

Re: HSC 2014 4U Marathon - Advanced Level Nice proof by contradiction. and the conclusion can be a tiny bit stronger: the inequality sign can be strictly greater than. lol

Re: HSC 2014 4U Marathon for part (ii): $Consider the geometric series of complex numbers$ \sum_{n=0}^\infty[(\cos\theta+i\sin\theta)\cos\theta]^n Because the modulus of common ratio is less than 1, the limiting sum exists and S_\infty=\frac{1}{1-\cos\theta(\cos\theta+i\sin\theta)}...

Re: HSC 2014 4U Marathon The easier way for part (i): \cos n\theta+\cos (n-2)\theta=2\cos\theta\cos (n-1)\theta

Re: MX2 Integration Marathon yeah this way there is no singularity at x=0

Re: MX2 Integration Marathon $ At $ x=0, \ln x $ has no definition, so the lower limit $ x=0 $ is understood in the sense of $ x\rightarrow0^+ $ Due to the boundedness of sine and cosine function, we have $ \lim_{x\rightarrow0^+}x\sin\ln x=\lim_{x\rightarrow0^+}x\cos\ln x=0 Now...

It seems to me what you listed all belong to pure mathematics, although part of which have found wide applications in science, engineering and econimics etc. In my opinion, applied and computational mathematics are also interesting branches. Applications can be found in HSC maths such as...

Re: MX2 Integration Marathon $ Let $ u=\sqrt{\tan x}, $ then $ x=\tan^{-1}u^2 $ and $ dx=\frac{2u}{1+u^4}du $ Therefore $ I=\int\sqrt{\tan x}dx=\int\frac{2u^2}{1+u^4}du the integral is converted into the integral of a rational function, which can be always solved. However the solution is...

Re: MX2 Integration Marathon which? \int\sqrt{\tan x}dx ?

Re: MX2 Integration Marathon $ Find $ \int\frac{dx}{\sqrt[3]{(x-1)(x+1)^2}}

Re: HSC 2014 4U Marathon - Advanced Level well, it seems to me it is a theorem in Number Theory. would love to see a solution using HSC method

Try to sumarize the formulas, ideas, and typicial questions, by yourself. Once you have solved one question, ask yourself what is the key step to the solution, what is useful to solve other similar questions? To think more, to compare more, and to analyse more, is the key to learn maths.

Dr. Xie Maths Coaching