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Calculations are the free marks in the HSC because either you know the method or you don't. The short/long responses are where you lose your marks because the marking criteria can be very specific or you may have forgotten bits of content. The chemistry skills you acquire (mole calculations...

\frac{\pi \theta}{12} = 0, \pi, 2\pi... \frac{\theta}{12} = 0, 1, 2, ... \theta = 0, 12, 24, ... \therefore $ Lowest positive integer n = 12$

I won't draw the region for you, but i'll explain it to you. 0 \leq \frac{\pi}{4} + arg(z) \leq \frac{\pi}{2} -\frac{\pi}{4} \leq arg(z) \leq \frac{\pi}{4} The region is in between arg(z) = pi/4 and arg(z) = -pi/4 First draw these two rays emanating from the origin (with an open circle at...

arg((1 + i)z) arg(1 + i) + arg(z) \frac{\pi}{4} + arg(z) $I'm sure you can do the rest yourself.$

We know that if z = cistheta (i.e. |z| = 1), z + 1/z = 2Re(z) If |z| does not equal to 1, you won't get a real result when adding z + 1/z. Let z = cistheta, cistheta + 1/cistheta = 2costheta 2costheta = root2 costheta = (root2)/2 = 1/root2 Therefore theta = pi/4 We also know that z^n +...

1. x(x+2) - \frac{35}{x(x+2)} = 34 $Let u = x(x + 2) because it would look a bit messy and would take me longer to type$ u - \frac{35}{u} = 34 u^2 - 35 = 34u u^2 - 34u - 35 = 0 $You find a quadratic which you can easily solve$ (u - 35)(u + 1) = 0 u = 35, \; u = -1 x(x + 2) = 35, \; x(x +...

Linear function, so it is in the form of g(x) = mx + b g(3) = 3m + b = 2 ... (1) g(4) = 4m + b = 4 ... (2) (2) - (1): m = 2 From (1), 6 + b = 2 b = -4 Therefore g(x) = 2x - 4 Range: g(x) >= -4

log[f(x)] f(x) < 0, log[f(x)] is undefined f(x) = 0, asymptote at this point in log[f(x)] f(x) = 1, log[f(x)] = 0 f(x) = A, where A is the base of the log, log[f(x)] = 1 f(x) = B, where 0 < B < 1, log[f(x)] < 0 f(x) = C, where C > 1, log[f(x)] > 0

Haha i don't know why i said 5..

9) z^5 = -1 $Let $ z = cis\theta cis5\theta = -1 5\theta = \pi, 3\pi, 5\pi, 7\pi, 9\pi \theta = \frac{\pi}{5}, \frac{3\pi}{5}, \pi, \frac{7\pi}{5}, \frac{9\pi}{5} \therefore z = cis\frac{\pi}{5}, cis\frac{3\pi}{5}, cis\pi, cis\frac{7\pi}{5}, cis\frac{9\pi}{5} $To show this on the argand...

Carrot, for question 3, isn't it n = 7?

x = 0 is an x-intercept.

2) ii) (rcis\theta)^n r^ncisn\theta $For it to be real, Im(r^ncisn\theta) = 0 r^nsinn\theta = 0 sinn\theta = 0 n\theta = 0, \pi, 2\pi, ... $Replace the theta with the argument of the complex number to find the lowest integer n$

Re: HSC 2012 Chemistry Marathon If half the heat is lost to the surroundings, you'll need to produce double the amount of heat. That means you'll need c = (210 \times 4.18 \times 2)J of energy n(ethanol) = \frac{m}{46.068} Finding molar heat of combustion of ethanol: H_c = \frac{(210...

Intensity of light is inversely proportional to the distance. If an object experiences 100J/m^2 at 1m distance, it would experience, 25J/m^2 at 2m distance.

Hint: Complete the... Found another question.. Factorise: x^2 - a^2 + 4xy + 2ab - b^2 + 4y^2

Was intended for OP but okay.. lol. Another one, factorise: 4x^4 - 2x^3y - 3xy^3 - 9y^4

lol.. I've changed the question.

$Factorise a^4 + 4b^4$

Wait.. do you do MX2 (your signature)? |z| means the displacement of the complex number, z, from the origin. Similarly, by shifting it upwards 1 unit, |z - i|, means the displacement of the complex number, z, from 1 unit above the origin, i.e. (0, 1). Same thing with |z - 1|, just meaning the...