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No worries

$Draw the diagram. \\Let \angle TLP=\alpha \\\angle PGM=\alpha (\because$ angle in the same segment) \\\therefore \angle SPG=\frac{\pi}{2}-\alpha(\because$sum of \angle's$ in a \triangle=\pi$ and \angle PSG=\frac {\pi} {2}) \\ \\$Let \angle TPL=\beta \\\angle MPS=\beta(\because $ vertically...

They are not trial papers.

My study time varies but on average, I study for about 2 hours per day, I guess. I can study less than many other people because I only have 10 units and do ESL lol.. And I go to GRC Oatley Senior Campus (sadly ranked 254 last year)

shouldn't you say that coefficient of x^2>0 to get the mark if you use the discriminant method?

HSC markers will probably not pay you full marks for that though.

\Delta =16(m-1)^2+48(m+1) \\=16m^2-32m+16+48m+48 \\=16m^2+16m+64 \\=16(m^2+m+4) \\=16((m+\frac{1}{2})^2+\frac{15}{4})\geq 0 \\\therefore$ Roots are real for any real value of m.

Depends on how confident you are about your subjects. Do 10 units if and only if you are confident that you know what you want to do for the HSC.

I don't think this works because I remember doing one integration problem in two different methods and one method gave answers like Ctan^(-1) (ktan(pi/2)) where C and k were integral, C,k>1 while the other method gave out an actual real number like pi/3 or something like that. When I tried using...

How did you know this?

if you see the working first part, towards the end, you can see n in the demominator. Thus n cannot be 0.

Oh true that... No wonder people call you the maths version of Einstein in BoS lol

<!-- google_ad_section_start(weight=ignore) -->eldore44How did you handle tan(pi/2) when you used the t method?

U(0) \\=\frac{2}{5}[U(1)+U(-1)] \\=\frac{2}{5}\int_{0}^{\pi}\frac{cosx+cos(-x)}{5-4cosx}dx \\=\frac{1}{5}\int_{0}^{\pi}\frac{4cosx-5+5}{5-4cosx}dx \\=\frac{1}{5}\int_{0}^{\pi}(-1+\frac{5}{5-4cosx})dx \\=\frac{-\pi}{5}+U(0) \\\frac{\pi}{5}=0$ which is not true... LOL This method doesn't work...

well i was solving at the same time i was typing

Whoops, made a mistake solving and writing at the same time in LaTex is quite hard lol... I'll fix it up later

tan\frac{\pi}{2}=undefined \\\therefore 3tan\frac{\pi}{2}=undefined \\tan^{-1}(3tan\frac{\pi}{2}) \\=tan^{-1}(undefined) \\=\frac{\pi}{2}

Yea, that works nicely but the question says "then find U(0)" or something similar so I thought I should use the previous part of the question.

LHS=U(n+1) + U(n-1) - \frac{5}{2}U (n) \\\because $We are given that \int_{0}^{\pi}\frac{cosnx}{5-4cosx}dx, \\$ LHS=\int_{0}^{\pi}\frac{cos(n+1)x+cox(n-1)x- \frac{5}{2}cosnx}{5-4cosx}dx \\=\int_{0}^{\pi}\frac{2coxnxcosx- \frac{5}{2}cosnx}{5-4cosx}dx...

Re: Westinghouse & edison??? It's all because of those people who studied history and philosophy of science (NOT a real science) in uni changing science curricula by getting a job in Boardofstudies.