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Re: HSC 2015 4U Marathon don't you have to answer the previous question before posting a new one? anyways i) $ (x+y+z)((x+y+z)^2 -3(xy + xz + zy)) $ \\ $ ii) Just expand $ \\ $ iii) let \alpha = (b-c)(x-a), \beta = (c-a)(x-b), \gamma = (a-b)(x-c). $ $Then, f(x) = (\alpha + \beta +...

Re: MX2 2015 Integration Marathon $ \ \int_0^a f(x) \ dx = - \int_a^0 f(a-u) \ du = \int_0^a f(a-u) \ du $ $which the same as what's desired by change of variables (use substituation x=a-u) $ \\ $ \ \int_0^{\pi/4} \ln (1 + \tan x) \ dx = \int_0^{\pi/4} ln2 \ dx - \ \int_0^{\pi/4} \ln (1+tanx)...

Re: HSC 2015 4U Marathon I think I saw a good question earlier on this forum like 3 days ago or something, but for some reason I think it's gone? And I don't remember who the original poster was. But this was the question and it required some thinking! \text{Given that} x^{5x} = y^{y}...

Re: HSC 2015 4U Marathon $ Okie. When you differentiate the function, you get $ P'(x)=5x^4 - a $ Suppose alpha is a root to this function, then you find that $ alpha = (\frac{a}{5})^{1/4} $. Since there's only two turning points (at x= plus minus alpha), for there to be three real roots there...

Re: MX2 2015 Integration Marathon advanced 4u I think. They have some very interesting problems

Re: MX2 2015 Integration Marathon Lol why are people only active on this marathon? Can someone plz post a problem in the other one (advanced)?

Re: MX2 2015 Integration Marathon Yeah sorry I might've sounded rude/aggressive cause you know it's the internet but I didn't mean to be rude lol. But yeah I see ur point sir

Re: MX2 2015 Integration Marathon Actually, contrary to what you think, my method does work. Of course what you stated is probably a great alternative, but ur post sounds as if my method was invaild, when it clearly works

Re: MX2 2015 Integration Marathon After a few manipulations with t form, the expression becomes 338t/((1+t^2)((5t+12)) dt(unless I made a silly somewhere...but u get the idea) which u can then integrate by using partial fractions.

Re: MX2 2015 Integration Marathon change the tanx tan2xtan3x into tan3x -tan2x - tanx by considering the expansion of tan (2x+x). then pretty straightforward from there

Re: HSC 2015 4U Marathon - Advanced Level whoops hahaha I see my mistake in case 1 lol mustve been too tired. but nice solution sir!

Re: MX2 2015 Integration Marathon Just a hint for those, divide top and bottom by b^2 cos^2x and you get a constant multiplied by a standard tan invere integral

Re: HSC 2015 4U Marathon - Advanced Level Okay, so the expression just becomes xy(x^2-2y)+1=(x-1)(y-1) Case 1: Suppose both x,y are positive. If x^2>2y the LHS grows faster than the RHS (xy>(x-1)(y-1) for positive x,y) so equality cannot be achieved. However if x^2<2y, then the second factor...

Re: HSC 2015 4U Marathon - Advanced Level Lol I remember this question from my exam 3 weeks ago lol. What I did during the exam was apply AM-GM (they made us prove the general form in the previous parts) for each factor on the LHS, i.e. (1+a3)^3>= 3^3 /2^2 a3 then I just multiplied all the...