Search results

Re: HSC 2018 MX2 Integration Marathon If you don't mind me asking, where do you get these integrals from?

Formulae for the intersection of tangents and normals of a parabola. If necessary, remainder theorem and factor theorem.

\cot \theta = x + 1 \implies \cos \theta = (x+1)\sin\theta Note that \cot \theta \in \mathbb{R} \implies x \in \mathbb{R} So {\rm cis\,} \theta = \sin \theta(x + 1) + i \sin \theta = (x+a) \sin \theta And \sin n \theta = \mathrm{Im} ({\rm cis \,}^n \theta) So we have...

Does the same also apply to Physics for Engineering?

Vectors are defined by two things: - a magnitude e.g. |z| - a direction e.g. \arg z So if two vectors have the same magnitude and direction then they are equal.

The green vector and the vector from w to z are the same vector. The position doesn't matter.

Why is that unfortunate?

Also the "identity" \arg\left(\frac z w\right) = \arg z - \arg w isn't actually true because \arg z - \arg w isn't guaranteed to be in the range [\pi, \, -\pi) . If \arg z - \arg w was instead expressed as the principal value for that angle then I believe it would be true. So...

i) \frac{z_1 - z_2}{z_1 + z_2} = ki z_1 - z_2 = ki(z_1 + z_2)\quad ( z_1 + z_2 \neq 0) What does this mean? It means that the vector z_1 - z_2 is perpendicular to z_1 + z_2 . That is basically the condition given to you. z_1 - z_2 and z_1 + z_2 are the diagonals of the parallelogram...

Re: HSC 2018 MX2 Integration Marathon Maybe something a bit easier: \int_0^\pi \frac{1}{3 + 2\cos x}\,dx

It's called tex. Here is a good guide to get started: http://community.boredofstudies.org/10/maths/234259/short-guide-latex.html

something that should be explicitly noted is that \sqrt{1-\sin^2 \theta} = |\cos \theta| (and in general, \sqrt{x^2} = |x|) usually the bounds of the integral will mean that |\cos \theta| = \cos \theta but one should still be careful when dealing with situations like this.

Re: HSC 2018 MX2 Marathon Is the intended answer the Taylor series for e^x ? Edit: nevermind, I found some others

It's easier to solve this equation without expanding it. First note that if two numbers multiply to give zero, then that means at least one of them must be zero. (x+5)(x-4) = 0 \begin{cases}x+5=0 \\ x-4=0 \end{cases} x = -5 \quad {\rm OR} \quad x = 4 You've also missed a...

I'm going to play a bit of devil's advocate here. The only ATAR you need is the one that will get you into the uni course you want. What do you want to do at uni? If your course is a 93 (consider adjustment factors/bonus points as well) then it doesn't matter whether you get 93.00 or 99.95...

Re: HSC 2018 MX2 Integration Marathon \int^{\pi/4}_0 \frac{2 \sin 8x \cos 4x \cos 2x}{(1 + \cos 8x)(1+ \cos 4x)(1 + \cos 2x) \sqrt{2 + \cos 2x}}\, dx Using the identities: 1 + \cos kx = 2 \cos ^2 (kx/2) \sin kx = 2 \sin (kx/2) \cos (kx/2) the integral reduces to: \int^{\pi/4}_0...

Part b) is a standard maximisation question. A will be a quadratic function of r_1 . By inspecting the coefficient of {r_1}^2 we can see this is a concave-up ("smiley face") parabola and hence A has a global minimum, though we will need to determine if this global minimum lies in...

A = \pi ({r_1}^2 + {r_2}^2) and k = r_1 + r_2 \implies r_2 = k - r_1 noting that k is a constant, we can then eliminate r_2 by doing one substitution.

If your first pair were queens then your ^{48}C_3 could choose 2 kings, 2 aces, 2 jacks etc. which would be more than 1 pair.

^{48}C_3 does not exclude the possibility of having another pair. i.e. the question (probably) asks for the probability of exactly one pair, but you calculated the probability of having at least one pair. Doing ^{12}C_3 (and then choosing suits afterwards) guarantees that all of the cards...