Search results

1 > cos \theta \ \frac{1}{1} < \frac{1}{\cos \theta} 1 < \sec \theta

\\ $We can re-write$ \ (a+b), (a+c), (b+c) \ $as$ \\ (a+b+c) - c, (a+b+c) - b, (a+b+c) - a \\ $We know from the original polynomial that$ \ \ a+b+c = 4/2 = 2 \\ $Therefore to find a polynomial of roots$ \ \ (a+b), (a+c), (b+c) \ \ $is same as finding polynomial of roots$ \ \ (2-a), (2-b), (2-c)...

Conics is just algebra, become fast at algebra and you become fast at Conics (mostly) To get fast at algebra you just need to increase your computational speed, this can be done through practice. Some things you can do are: Don't write down every step, skip steps here and there and be...

Re: HSC 2014 4U Marathon The first is correct, for the second one, try finding (z+z^2) in modulus argument form by considering the vector geometrically (use the first part)

Re: HSC 2014 4U Marathon (Definitely within syllabus) \\ $In the complex plane let$ \ z = \cos \theta + i \sin \theta \ $and$ \ 0 < \theta < \frac{\pi}{2} \\ $i) Show that a rhombus is formed with the complex numbers$ \ \ 0,z,z^2, (z+z^2) \\ $ii) By finding$ \ (z^2+z) \ $in modulus-argument...

Re: HSC 2014 4U Marathon Oh I was under the impression it was (still a good problem though it was not useless)

Re: HSC 2014 4U Marathon But remember y=/=1 here! its not z=x+i, its z=(x+i)^2 which is completley different just imagine its z=t + i instead of x + i just so its not confusing so z= x+iy = (t+i)^2 = (t^2-1) + 2t i Hence x=(t^2-1) , y= 2t t = y/2, x= (y^2/4-1) Hence y^2 = 4(x+1)...

Re: HSC 2014 4U Marathon How do you let z=x+iy? Think of it this way since z=(x^2-1) + 2x i What is z if x=1? or x=2? or x=3? You get a series of dots, but they follow a certain curve, this curve is the locus. How do you find this locus given z?

Re: HSC 2014 4U Marathon So z^2 = (x+i)^2 = (x^2-1) + i \cdot 2x What kind of parabola would it be? How did you know it was a parabola?

Re: HSC 2014 4U Marathon \\ $i) Find the area underneath the curve$ \ f(x) = \frac{1}{x} \ $for$ \ 1 \leq x \leq n \\ $ii) Find the volume when the area in (i) is rotated about the$ \ x$-axis by 360^{\circ}$ \\ $iii) What happens when$ \ n \to \infty \ $what does this mean?$

Re: HSC 2014 4U Marathon - Advanced Level Attempt #2 \\ $Letting the roots of$ \ P \ $be$ \ \alpha_k \ (k=1,2,\dots ,n) \\ $It is clear that$ \ \ \sum \alpha_k = 5 \ , \ \sum \alpha_i \alpha_j = 12 \\ \Rightarrow \ \sum \alpha_k^2 = 1 \\ $Assume that$ \ \alpha_k > 0 \ \ $then it is clear that$...

Re: HSC 2014 4U Marathon - Advanced Level oops

Re: HSC 2014 4U Marathon Yep the line y=1 (When using x,y we are talking about reala numbers so it doesn't make sense to say y=i) Think of it this way z is some number on the complex plane, where the real part can be anything (x is real), but its imaginary part has to always be 1, this is of...

Re: HSC 2014 4U Marathon - Advanced Level It is the same thing as \sum but for products =) i.e. \\ \sum_{k=1}^n a_k = a_1 + a_2 + \dots + a_n \\ \\ \prod_{k=1}^n a_k = a_1 \cdot a_2 \cdot \dots \cdot a_n

Re: HSC 2014 4U Marathon Hopefully my proof is correct. \\ $Letting the roots of$ \ P(x) \ $be$ \ \alpha_k \ , \ k=1,2,\dots,n \\ $It is clear that$ \\ \sum \alpha_k = 5 \ , \sum \alpha_i \alpha_j = 12 \Rightarrow \ \sum a_k^2 = 1 \\ \\ $Now assume that$ \ P(x) \ $has all real positive roots$...

Re: MX2 Integration Marathon Yep that is what I got.

Re: MX2 Integration Marathon \int_0^{2\pi} \sqrt{1+\cos x} \ dx

Re: HSC 2014 4U Marathon \\ $For some complex$ \ z \ $and real$ \ x \ $let$ \ z=x+i \ \ $describe the locus in the complex plane of$ \\ $i)$ \ z \\ $ii)$ \ z^2

Re: HSC 2014 4U Marathon Yea something like that Ah well, ok

Re: HSC 2014 4U Marathon \\ $Let$ \ \ I_n = \int_0^{\pi/4} \tan^{2n}x \ dx \\ \\ $i) Show that$ \ \ I_n + I_{n+1} = \frac{1}{2n+1} \ \ \ (n \geq 0) \ \ \ \ \fbox{2} \\ \\ $ii) Show that$ \ \ \ \sum_{n=0}^{N} \frac{(-1)^n}{2n+1} = I_0 + (-1)^N I_{N+1} \ \ \ \fbox{1} \\ \\ $iii) Show that$ \...