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Re: HSC 2014 4U Marathon what you can let is a=sin x, b=sin y (because |a|<=1, |b|<=1, you can always do this), and try to show x,y are complementary. then you will have the result

Re: HSC 2014 4U Marathon well done

Re: HSC 2014 4U Marathon This 'let...' is illegal, coz if you already have this, don't you really need to prove a^2+b^2=1? it is just pythagoras

Re: HSC 2014 4U Marathon $ Given that $ a\sqrt{1-b^2}+b\sqrt{1-a^2}=1, $ prove $ a^2+b^2=1 (i) by using a trigonometric substitution (ii) without using trigonometry. (not an HSC-like question lol)

lol there should be one more typo

Re: Geomietry I believe there were some typos: if angle ABC=90 degree, how can AB=AC? and since the point E seems irrelevant, i guess u missed something or should angle BFC be BFE?

vertical asymptote or vertical tangent? two different things

Here is a little thing I find out: f(x+2)-f(x)=\int_0^{\frac{\pi}{2}}\frac{\sin[(x+2)t]-\sin(xt)}{\sin{t}}dt=-\frac{2}{x+1}\sin\left(\frac{x+1}{2}\pi\right) From which some results may be derived?

Sy, that's not the correct way you differentiate an integral with an unknown in it (unless i am mistaken). f^\prime(x)=\int_0^{\frac{\pi}{2}}\frac{d}{dx} \left( \frac{\sin(xt)}{\sin{t}} \right)dt=-\int_0^{\frac{\pi}{2}}\frac{t\cos(xt)}{\sin{t}}dt...

which one? all questions?

Re: MX2 Integration Marathon well done

Re: MX2 Integration Marathon Hey guys we can not rest for so many days. Come warm up with this one: $ Evaluate $ \int_0^1\frac{\tan^{-1}x}{1+x}dx.

$ Drawing a neat diagram, the desired area $ A= $ area of the triangle enclosed by the tangent, the y-axis and the line $ y=\frac{\pi}{3} $ minus the area of the region bounded by $ x=\sin y, $ the the y-axis and the line $ y=\frac{\pi}{3} $ Therefore $...

Re: HSC 2014 4U Marathon - Advanced Level Now that you nominated me, i just have my try, not an inspiring method though. $ First of all i'll base my solution on the facts that $ \cos72^\circ=\frac{\sqrt5-1}{4}, \cos36^\circ=\frac{\sqrt5+1}{4}, $ which (by themselves are good classroom...

Re: HSC 2014 4U Marathon - Advanced Level well done, so smart! Here is my proof: $ let $ (1+m)^{\frac1n}=1+\alpha, (1+n)^{\frac1m}=1+\beta, $ so $ \alpha>0, \beta>0 $ and the desired result is $ \frac1{1+\alpha}+\frac1{1+\beta}>1, $ which is equivalent to $...

Re: HSC 2014 4U Marathon - Advanced Level $ yep the series converges, i can show that it is bounded above by $ 10(1+\frac12+\frac13+\cdots+\frac18)

Re: HSC 2014 4U Marathon - Advanced Level oh yep my bad

Re: HSC 2014 4U Marathon - Advanced Level For part (b) (i): H_{2n}+S_{2n}=\sum_{k=1}^{2n}\left(\frac1k+\frac{(-1)^k}{k}\right)=\sum_{m=1}^n\frac{2}{2m}=H_n (since odd numbers go away, even numbers k=2m get doubled) Part (b)(ii): $ We again need the assumption that $...

Re: HSC 2014 4U Marathon - Advanced Level Part (a) Consider the area under the curve y=1/x between 1 and n. Divide the interval into (n-1) equal parts. If we use upper rectangles to approximate the area, we have 1+\frac12+\frac13+\cdots+\frac{1}{n-1}>\ln{n} While we use lower rectangles to...

yep seems so, or the function is a constant function, i.e., the function value is the same for all x values.