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  1. N

    justifying a trig for inequality

    \frac{\pi}{2}<2<\pi (1) 0<\pi-2<\frac{\pi}{2} so its in the first quadrant or using your logic: \cos(\pi-2)=-\cos(2) but \cos(2)<0 from 1 \cos(\pi-2)>0
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    complex question

    z=w(|z|-1) |z|=|w|||z|-1| |z|^2=|w|^2||z|-1|^2 |z|^2=|w|^2(|z|^2-2|z|+1) |z|^2(1-|w|^2)+2|w|^2|z|-|w|^2=0 then just solve the quadratic for |z| Drogonskis method is better and less algebra but this method requires less thinking
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    MX2 Integration Marathon

    I=\int_{0}^{\frac{\pi}{8}}\frac{\left(\sin x-\cos x\right)\ln\left(\tan2x\right)}{\left(\sin x+\cos x\right)\sqrt{\cos4x}}dx I=\int_{0}^{\frac{\pi}{8}}\frac{\left(\sin x-\cos x\right)^2\ln\left(\tan2x\right)}{\left(\sin x+\cos x\right)\left(\sin x-\cos x\right)\sqrt{\cos4x}}dx...
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    Complex number questions

    Yes I guess I should have clarified that this same argument can be repeated for different orders of 0,z_1,z_2,z_3 on the circle, but I assumed by the symmetry it is quite easy to see how.
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    Complex number questions

    For part 3: Unfortunately circle geometry has been removed from the syllabus but here is a circle geometry proof anyways: We essentially have to prove: arg\left(\frac{1}{z_2}-\frac{1}{z_1}\right)=arg\left(\frac{1}{z_3}-\frac{1}{z_2}\right) As the two vectors must have the same direction...
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    Hard questions

    Fair enough, and I agree nearer to your exam you should definitely do exams in timed conditions. I was mainly talking about when you first start off past papers (atleast for me), if I got the question myself without any outside help I improved much better than looking at the solutions, because I...
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    Hard questions

    Yes and you also have to shift your perspective during an exam. Whilst in practice your sole purpose is to answer all the questions not caring too much about time, in an exam you have to be realistic in how you spend your time (you probably already know this). But many people advise you practice...
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    Hard questions

    Theres a difference between you personally getting a question out yourself and the tutor doing it for you, or even guiding you. Basically try hard questions all by yourself trying every method or thought that comes to mind until you get the answer. Then think to yourself why that path worked...
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    HSC Extension 2 Mathematics Predictions / Thoughts

    Yeah standard textbook question really. Should have been either vectors or proofs.
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    Advance Math HSC

    I would expect atleast some probability as it is one of the topics that changed a lot with the new syllabus. So expect conditional probability, venn diagrams etc.
  11. N

    Hard Proofs Question

    Definitely tricky I think it has something to do with squeeze theorem but not exactly sure.
  12. N

    MX2 Integration Marathon

    Not quite sure where I missed the half, but I might not have checked it properly.
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    MX2 Integration Marathon

    Not sure if this is still active but I'll give this a go for fun (let me know if I've made any mistakes or there is an easier way). The x-ints are x= \pm \frac{1}{2\sqrt{2}} Using the corrected identity (Thanks to vernburn for showing that there is a factor of a half): \int_{-a}^{a}...
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    Can you use bounds in 3u

    Oh yep thats perfectly fine, unless the question specifically states use the substitution u=... in which case you have to use that substitution.
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    Can you use bounds in 3u

    It was perfectly ok (and even encouraged) to use bounds in the old syllabus and i'll assume it is too for the new syllabus . Not sure what you mean for the second question you will have to give an example or clarify.
  16. N

    Very hard mechanics question

    Part ii/iii written out: In part i you would find: \dot{x}=-Ae^{-kt}(n\sin(nt)+k\cos(nt)) To find when particle is rest set \dot{x}=0 which you would do in part ii -Ae^{-kt}(n\sin(nt)+k\cos(nt))=0 -Ae^{-kt} \neq 0 so solve: (n\sin(nt)+k\cos(nt))=0 mucking around with auxiliary angle (or you...
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    Atar estimate

    If you can get 95+ for math and 90+ for Buisness and 80+ for everything else than you will easily get 90+ atar.
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    Help pleaseee

    Hey, I fixed it up, for B_1 there is a payment of 5000 which I missed, which gives your answer.
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    Help pleaseee

    When we say pa (per annum) it is compounded yearly. So: A_1=5000(1+0.06) A_2=5000(1.06)^2+5000(1.06) A_3=5000(1.06)^3+5000(1.06)^2+5000(1.06) and so on for 10 years: A_{10}=5000(1.06)^{10}+5000(1.06)^9+...+5000(1.06) Using the GP sum formula...
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