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Re: HSC 2013 4U Marathon Sorry?

Re: HSC 2013 2U Marathon What has maths come to when you're expecting to be able to answer new questions straight away, first try playing around with the problem for a bit And by no means is ColdLipstick's questions too hard for 2U

Re: HSC 2013 2U Marathon g'(a) = 0

Re: HSC 2013 2U Marathon Translate the words into mathematics If x=a is where min of g(x) occurs thennnnnnnn?

Re: HSC 2013 4U Marathon Ah ok I'm getting 2\pi^2 r now,

Re: HSC 2013 2U Marathon $Prove that the volume of a sphere with radius$ \ \ r \ \ $is given by$ \frac{4\pi r^3}{3}

Re: HSC 2013 4U Marathon Its just the area of circle with radius l minus circle radius r isn't it? --- $Consider the sequence$ u_{n} = 4u_{n-1} - 4u_{n-2} \ \ u_1=2 \ \ u_2 = 8 $Find a formula for$ \ \ u_n $And then hence without induction prove $ \sum_{r=1}^n u_n = 2 + (n-1)2^{n+1}

Re: HSC 2013 2U Marathon $Let$ \ \ y=uv \ \ $where$ \ \ u,v \ \ $are functions of$ \ \ x $i) Find an expression for$ \ \ y'', y''', y'''' $ii) Expand$ \ \ (1+x)^2, \ (1+x)^3, \ (1+x)^4 $iii) Notice the pattern and hence find$ \frac{d^5}{dx^5}(e^{-x} (1+x^2))

Re: HSC 2013 2U Marathon $Find the area of the region enclosed by the curves$ y=\frac{1}{x} \ \ y=\sqrt{3} x \ \ $and$ \ \ y=\frac{1}{\sqrt{3}}x

Re: HSC 2013 2U Marathon \int_0^{\pi/4} 1 + \sin^2 x + \sin^4 x + \sin^6x + \dots \ dx

bumpppp

Re: MX2 Integration Marathon Summing the geometric series we get \sum_{n=1}^{\infty} \frac{1}{n^2} = \int_0^{\infty} \frac{x}{e^x - 1} \ dx Any ideas on how to squeeze this integral to somehow get pi^2/6?

Re: MX2 Integration Marathon Yep

Re: HSC 2013 2U Marathon Yep =)

Re: HSC 2013 2U Marathon Yep x=/= 0, pi, 2pi, 3pi, 4pi, ...... which means.......

Re: HSC 2013 2U Marathon Well cosx can be less than 1, for instance if x=pi/4 1 + cos(pi/4) + cos^2(pi/4) + ... has a limiting sum

Re: HSC 2013 2U Marathon I'm not sure where you got that from

Re: HSC 2013 2U Marathon Still not there This is a rather deceptive question lol

Re: HSC 2013 4U Marathon Yea something like that, though I'm getting: \frac{1 - x\cos \theta + x\sin \theta}{x^2 - 2x\cos \theta + 1} = 1 Which yields, x=1 - (cos t + sin t)

Re: HSC 2013 2U Marathon Not quite