Search results

Re: HSC 2014 4U Marathon Oh great it's not the triangular inequality is it..... so like ar^2 \leq ar+a \Rightarrow 0<r\leq\frac{1+\sqrt{5}}{2}

Re: HSC 2014 4U Marathon value(s) heh

Re: HSC 2014 4U Marathon well it can haha I just realised there are 2 sets of solutions Another case is where a^2 is the longest side instead (where 0<r<1 ) And you'd be solving \\ \\ r^4+r^2-1 \Rightarrow r^2 = \frac{-1+\sqrt{5}}{2}

Re: HSC 2014 4U Marathon Well clearly r is at least positive since you can't have negative sides :P (am I missing something else?) r>1 ?? for part ii let the sides be a,ar,ar^2 since the triangle is right-angled then the Pythagoras theorem applies (assuming r >1, implying ar^2 is the...

Re: MX2 Integration Marathon I haven't seen the Q so yeah I'll assume alpha is positive here otherwise it wouldn't work :P $using integration by parts with$ \ u = \tan^{-1}x \ \ v = 1/x \\ \tan^{-1}(a)ln(a) - \tan^{-1}(1/a)ln(1/a) -...

LHS = \frac{sinA}{cosA+sinA}+ \frac{sinA}{cosA-sinA} = \frac{sinA(cosA-sinA+cosA+sinA)}{\cos^2A-\sin^2A} = \frac{2sinAcosA}{\cos^2A-\sin^2A} = \frac{sin2A}{cos2A} \\ \\ tan2A

yr 7 maths too

As a 4U student, the only 2U content i know none of is superannuation and loan repayments :D

Yeah i just saw the solution and i get what you mean I did it like this lol..... after expanding everything you get n+nx+nx^2+...nx^{n-1} -(x+2x^2+3x^3+...(n-1)x^{n-1}) \\ \\ = n(1+x+x^2+...x^{n-1}) - x(1+2x+3x^2+...(n-1)x^{n-2}) \\ \\ = n(\frac{1-x^n}{1-x}) -...

Nope cause the second expression if you expanded properly is x(1+2x+3x^2+…(N-1)x^(N-2)) Which is not a gp but... You know Sent from my C5303 using Tapatalk

Lol i was wondering how part ii) works then i realised you had to consider \frac{d}{dx} that was pretty cunning of them

Might consider doin Ext 1 :P and what would be the prize for getting over 70 in Ext 2?

Yeah in 3u and 2u last year some people actually got that Sent from my C5303 using Tapatalk

How can you tell.. I know you used product of roots and that the product of the other 2 roots is 5..it could be like 1 and 5 for all we know :P

Re: MX2 Integration Marathon You can't use the limiting sum formula here since it diverges at x = pi/2 (or the indefinite integral which is tan x is undefined at pi/2)

Cause basically there's a lot of students in those school that a capable of doing 3U/4U maths... so much that they don't have enough teachers and classes to cater for all of them...

Re: MX2 Integration Marathon I don't think yiu can take it outside of sigma since the expression is in terms of n Sent from my C5303 using Tapatalk

Re: MX2 Integration Marathon right...................... \int_{0}^{\frac{\pi}{4}} \sum\limits_{k=0}^\infty sin^{2k+1}x dx

Re: MX2 Integration Marathon I_{2n+1} \frac{2n+1}{2^{2n+1}} = $that question$ But after taking the sum likethe question states, I dont know how to simplify it... what am I missing haha

Re: MX2 Integration Marathon $let I_{2n+1} = $ \int_{0}^{\frac{\pi}{2}} sin^{2n+1}x \\ \\ $using integration by parts with u = sin^{2n}x \ \ dv = sinx $ \\ \\ I_{2n+1} = 2n\int_{0}^{\frac{\pi}{2}}sin^{2n-1}x - sin^{2n+1}x \ dx \\ = 2n(I_{2n-1} - I_{2n+1}) \\ \\ \Rightarrow I_{2n+1} =...